ARS MATHEMATICA CONTEMPORANEA Volume 14, Number 2, Spring/Summer 2018, Pages 209-454 Covered by: Mathematical Reviews Zentralblatt MATH COBISS SCOPUS Science Citation Index-Expanded (SCIE) Web of Science ISI Alerting Service Current Contents/Physical, Chemical & Earth Sciences (CC/PC & ES) The University of Primorska The Society of Mathematicians, Physicists and Astronomers of Slovenia The Institute of Mathematics, Physics and Mechanics The Slovenian Discrete and Applied Mathematics Society The publication is partially supported by the Slovenian Research Agency from the Call for co-financing of scientific periodical publications. ARS MATHEMATICA CONTEMPORANEA The 2018 Petra Šparl Award Dr Petra Šparl was a talented woman mathematician with a promising future who worked in graph theory and combinatorics, but died mid-career in 2016 after a battle with cancer. In her memory, the Petra Šparl Award was established recently to recognise in each even-numbered year the best paper published in the previous five years by a young woman mathematician in one of the two journals Ars Mathematica Contemporanea (AMC) and The Art of Discrete and Applied Mathematics (ADAM). Nominations for the inaugural award were invited in AMC in 2017, and cases were considered by a committee (consisting of the three of us) appointed by Dragan Marušic and Tomaž Pisanski as editors of AMC and ADAM. As judges we were impressed by the large number of papers in AMC over the five years 2013-2017 having a woman author or co-author: almost 60 in total, with well over half of those being women in the early stages of their career. With helpful commentaries from co-authors (in some cases) we drew up a long list of candidates for the 2018 award, sought reports from referees on those, and also considered the papers themselves, before making a decision, which was unanimous. The winner of the Petra Šparl Award for 2018 is Dr Monika Pilsniak (Department of Discrete Mathematics, AGH University, Kraków, Poland), for her paper 'Improving upper bounds for the distinguishing index', in Ars Mathematica Contemporanea 13 (2017), 259-274. Monika Pilsniak published four papers in AMC in 2016 and 2017, but one stands out: a single-author paper in 2017 on the distinguishing index of a graph. This is the smallest number of classes in a partition of the edge-set such that the only class-preserving automorphism of the graph is the identity automorphism. Monika helped introduce this concept in 2015, and in her 2017 paper in AMC, she classified all graphs with distinguishing index being at least equal to the maximum vertex degree. The main theorem is impressive, and difficult to prove, and it improves on the analogous theorem from 2005 on the distinguishing number (for partitions of vertices). Dr Monika Pilsniak ARS MATHEMATICA CONTEMPORANEA In summary, and quoting a referee: "Monika richly deserves the Petra Sparl Award: four papers in ACM pioneering a new concept, and one solo paper with an outstanding theorem, worthy of an award by itself." We would also like to make special mention of other high quality papers, by Sophie Decelle, María del Río Francos, Klavdija Kutnar, Klara Stokes and Aleksandra Tepeh. Monika Pilsniak will be awarded a certificate and invited to give a lecture in the Mathematics Colloquium at the University of Primorska, and to give lectures at the University of Maribor and the University of Ljubljana. Finally, we encourage nominations for the next Petra Sparl Award in 2020, and submissions of high quality new papers that will be worthy of consideration for future awards. Marston Conder, Asia Ivic Weiss and Aleksander Malnic Members of the 2018 Petra Sparl Award Committee ARS MATHEMATICA CONTEMPORANEA Contents Groups of Ree type in characteristic 3 acting on polytopes Dimitri Leemans, Egon Schulte, Hendrik Van Maldeghem .........209 On hypohamiltonian snarks and a theorem of Fiorini Jan Goedgebeur, Carol T. Zamfirescu.....................227 Inherited unitals in Moulton planes Gábor Korchmáros, Angelo Sonnino, Tamás Szonyi.............251 Trilateral matroids induced by n3-configurations Michael W. Raney...............................267 Growth of face-homogeneous tessellations Stephen J. Graves, Mark E. Watkins.....................285 The 4-girth-thickness of the complete graph Christian Rubio-Montiel...........................319 A note on the thickness of some complete bipartite graphs Siwei Hu, Yichao Chen............................329 Alphabet-almost-simple 2-neighbour-transitive codes Neil I. Gillespie, Daniel R. Hawtin......................345 Congruent triangles in arrangements of lines Carol T. Zamfirescu .............................. 359 A note on the directed genus of Kn,n,n and Kn Rong-Xia Hao.................................375 On domination-type invariants of Fibonacci cubes and hypercubes Jernej Azarija, Sandi Klavzar, Yoomi Rho, Seungbo Sim..........387 On t-fold covers of coherent configurations Alyssa D. Sankey...............................397 On the number of additive permutations and Skolem-type sequences Diane M. Donovan, Michael J. Grannell ................... 415 Classification of regular balanced Cayley maps of minimal non-abelian metacyclic groups Kai Yuan, Yan Wang, Haipeng Qu......................433 A note on extremal results on directed acyclic graphs Álvaro Martínez-Pérez, Luis Montejano, Deborah Oliveros.........445 Volume 14, Number 2, Spring/Summer 2018, Pages 209-454 ARS MATHEMATICA CONTEMPORANEA ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 14 (2018) 209-226 https://doi.org/10.26493/1855-3974.1193.0fa (Also available at http://amc-journal.eu) Groups of Ree type in characteristic 3 acting on polytopes* Dimitri Leemans Université Libre de Bruxelles, Département de Mathématique, C.P.216 - Algebre et Combinatoire, Boulevard du Triomphe, 1050 Brussels, Belgium Egon Schulte Northeastern University, Department of Mathematics, 360 Huntington Avenue, Boston, MA 02115, USA Hendrik Van Maldeghem Vakgroep Wiskunde, Universiteit Gent, Krijgslaan 281, S22, 9000 Gent, Belgium Received 8 September 2016, accepted 14 March 2017, published online 4 September 2017 Every Ree group R(q), with q = 3 an odd power of 3, is the automorphism group of an abstract regular polytope, and any such polytope is necessarily a regular polyhedron (a map on a surface). However, an almost simple group G with R(q) < G < Aut(R(q)) is not a C-group and therefore not the automorphism group of an abstract regular polytope of any rank. Keywords: Abstract regular polytopes, string C-groups, small Ree groups, permutation groups. Math. Subj. Class.: 52B11, 20D05 1 Introduction Abstract polytopes are certain ranked partially ordered sets. A polytope is called "regular" if its automorphism group acts (simply) transitively on (maximal) flags. It is a natural question to try to classify all pairs (P, G), where P is a regular polytope and G is an automorphism group acting transitively on the flags of P. An interesting subclass is constituted *This research was sponsored by a Marsden Grant (UOA1218) of the Royal Society of New Zealand. The authors also thank an anonymous referee for useful comments on a preliminary version of this paper. E-mail addresses: dleemans@ulb.ac.be (Dimitri Leemans), schulte@neu.edu (Egon Schulte), hendrik.vanmaldeghem@ugent.be (Hendrik Van Maldeghem) Abstract ©® This work is licensed under http://creativecommons.org/licenses/by/3.0/ by the pairs (P, G) with G almost simple, as then a lot of information is available about the maximal subgroups, centralizers of involutions, etc., of these groups, making classification possible for some of these families of groups. Potentially this could also lead to new presentations for these groups, as well as a better understanding of some families of such groups using geometry. The study of polytopes arising from families of almost simple groups has received a lot of attention in recent years and has been very successful. Mazurov [23] and Nuzhin [25,26, 27, 28] established that most finite simple groups are generated by three involutions, two of which commute. These are precisely the groups that are automorphism groups of rank three regular polytopes. The exceptions are PSL3(q), PSU3(q), PSL4(2n), PSU4(2n), A6, A7, M11, M22, M23, McL, PSU4(3), PSU5(2). The two latter, although mentioned by Nuzhin as being generated by three involutions, two of which commute, have been found to be exceptions recently by Martin Macaj and Gareth Jones (personal communication). We refer to [18] for almost simple groups of Suzuki type (see also [16]); [4, 20, 21] for groups PSL2(q) < G < PrL2(q); [2] for groups PSL3(q) and PGL3(q); [1] for groups PSL(4, q); [9] for symmetric groups; [3, 10, 11] for alternating groups; and [13, 19, 22] for the sporadic groups up to, and including, the third Conway group Co3, but not the O'Nan group. Recently, Connor and Leemans have studied the rank 3 polytopes of the O'Nan group using character theory [5], and Connor, Leemans and Mixer have classified all polytopes of rank at least 4 of the O'Nan group [6]. Several attractive results were obtained in this vein, including, for instance, the proof that Coxeter's 57-cell and Griinbaum's 11-cell are the only regular rank 4 polytopes with a full automorphism group isomorphic to a group PSL2 (q) (see [20]). Another striking result is the discovery of the universal locally projective 4-polytope of type {{5, 3}5, {3,5}10}, whose full automorphism group is J1 x PSL2 (19) (see [14]); this is based on the classification of all regular polytopes with an automorphism group given by the first Janko group J1. The existing results seem to suggest that polytopes of arbitrary high rank are difficult to obtain from a family of almost simple groups. Only the alternating and symmetric groups are currently known to act on abstract regular polytopes of arbitrary rank. For the sporadic groups the highest known rank is 5. The Ree groups R(q), with q = 32e+1 and e > 0, were discovered by Rimhak Ree [29] in 1960. In the literature they are also denoted by 2G2(q). These groups have a subgroup structure quite similar to that of the Suzuki simple groups Sz(q), with q = 22e+1 and e > 0. Suzuki and Ree groups play a somewhat special role in the theory of finite simple groups, since they exist because of a Frobenius twist, and hence have no counterpart in characteristic zero. Also, as groups of Lie-type, they have rank 1, which means that they act doubly transitively on sets of points without further apparent structure. However, the rank 2 groups which are used to define them, do impose some structure on these sets. For instance, the Suzuki groups act on "inversive planes". For the Ree groups, one can define a geometry known as a "unital". However, these unitals, called Ree unitals, have a very complicated and little accessible geometric structure (for instance, there is no geometric proof of the fact that the automorphism group of a Ree unital is an almost simple group of Ree type; one needs the classification of doubly transitive groups to prove this). Also, Ree groups seem to be misfits in a lot of general theories about Chevalley groups and their twisted analogues. For instance, there are no applications yet of the Curtis-Tits-Phan theory for Ree groups; all finite quasisimple groups of Lie type are known to be presented by two elements and 51 relations, except the Ree groups in characteristic 3 [12]. Hence it may be clear that the Ree groups R(q), with q a power of 3, deserve a separate treatment when investigating group actions on polytopes. Now, the regular polytopes associated with Suzuki groups are quite well understood (see [16, 18]). But the techniques used for the Suzuki groups are not sufficient for the Ree groups. In the present paper, we carry out the analysis for the groups R(q). In particular, we ask for the possible ranks of regular polytopes whose automorphism group is such a group, and we prove the following theorem. Theorem 1.1. Among the almost simple groups G with R(q) < G < Aut(R(q)) and q = 32e+1 = 3, only the Ree group R(q) itself is a C-group. In particular, R(q) admits a representation as a string C-group of rank 3, but not of higher rank. Moreover, the non-simple Ree group R(3) is not a C-group. In other words, the groups R(q) behave just like the Suzuki groups: they allow representations as string C-groups, but only of rank 3. Although Nuzhin proved in [27] that these groups allow representations as string C-groups of rank 3 for every q, we will describe a string C-group representation for R(q), q = 3, for each value of q to make the paper self-contained. Also, almost simple groups R(q) < G < Aut(R(q)) can never be C-groups (in characteristic 3). Rephrased in terms of polytopes, Theorem 1.1 says that among the almost simple groups R(q) < G < Aut(R(q)), only the groups G := R(q) are automorphism groups of regular polytopes, and that these polytopes must necessarily have rank 3. Ree groups can also be the automorphism groups of abstract chiral polytopes. In fact, Sah [30] showed that every Ree group R(32e+1), with 2e + 1 an odd prime, is a Hurwitz group; and Jones [15] later extended this result to arbitrary simple Ree groups R(q), proving in particular that the corresponding presentations give chiral maps on surfaces. Hence the groups R(q) are also automorphism groups of abstract chiral polyhedra. It is an interesting open problem to explore whether or not almost simple groups of Ree type also occur as automorphism groups of chiral polytopes of higher rank. Note that the Ree groups in characteristic 2 are also very special: they are the only (finite) groups of Lie type arising from a Frobenius twist and having rank at least 2. This makes them special, in a way rather different from the way the Ree groups in characteristic 3 are special. We think that in characteristic 2, quite different geometric methods will have to be used in the study of polytopes related to Ree groups. 2 Basic notions 2.1 Abstract polytopes and string C-groups For general background on (abstract) regular polytopes and C-groups we refer to McMullen & Schulte [24, Chapter 2]. A polytope P is a ranked partially ordered set whose elements are called faces. A polytope P of rank n has faces of ranks -1,0,..., n; the faces of ranks 0, 1 or n - 1 are also called vertices, edges or facets, respectively. In particular, P has a smallest and a largest face, of ranks -1 and n, respectively. Each flag of P contains n + 2 faces, one for each rank. In addition to being locally and globally connected (in a well-defined sense), P is thin; that is, for every flag and every j = 0,..., n - 1, there is precisely one other (j-adjacent) flag with the same faces except the j-face. A polytope of rank 3 is a polyhedron. A polytope P is regular if its (automorphism) group r(P) is transitive on the flags. If r(P) has exactly two orbits on the flags such that adjacent flags are in distinct orbits, then P is said to be chiral. The groups of regular polytopes are string C-groups, and vice versa. A C-group of rank n is a group G generated by pairwise distinct involutions p0,..., pn-1 satisfying the following intersection property: 0, is a group of order q3(q -1)(q3 + 1). It has a faithful permutation representation on a Steiner system S := (Q, B) = S(2, q + 1, q3 + 1) consisting of a set Q of q3 + 1 elements, the points, and a family of (q + 1)-subsets B of Q, the blocks, such that any two points of Q lie in exactly one block. This Steiner system is also called a Ree unital. In particular, G acts 2-transitively on the points and transitively on the incident pairs of points and blocks of S. The group G has a unique conjugacy class of involutions (see [29]). Every involution P of G has a block B of S as its set of fixed points, and B is invariant under the centralizer CG(p) of p in G. Moreover, Cg(p) = C2 x PSL2(q), where C2 = 1 be integers such that m | n. The normalizer ND2n (D2m) of any subgroup D2m of D2n coincides with D2m if n/m is odd, or is isomorphic to a subgroup D4m of D2n if n/m is even. 3 Proof of Theorem 1.1 The proof of Theorem 1.1 is based on a sequence of lemmas. We begin in Lemma 3.1 by showing that if R(q) < G < Aut(R(q)) then G can not be a C-group (with any underlying Coxeter diagram). Thus only the Ree groups R(q) themselves need further consideration. Then we prove in Lemma 3.3 that R(q) does not admit a representation as a string C-group of rank at least 5. In the subsequent Lemmas 3.9, 3.11 and 3.12 we then extend this to rank 4 and show that R(q) can also not be represented as a string C-group of rank 4. Finally, in Lemma 3.15 we construct each group R(q) as a rank 3 string C-group. All information that we use about the groups R(q) can found in [17]. We repeatedly make use of the following simple observation. If A : B is a semi-direct product of finite groups A, B such that B has odd order, then each involution in A : B must lie in A. In fact, if p = ap is an involution, with a G A, p G B, then 1 = p2 = a(pap-1)p2, where a(pap-1) G A and p2 G B; hence p2 = 1, so p =1 and p = a G A. 3.1 Reduction to simple groups R(q) We begin by eliminating the almost simple groups of Ree type that are not simple. Lemma 3.1. Let R(q) < G < Aut(R(q)), where q = 32e+1. Then G is not a C-group. Proof. Since Aut(R(q)) = R(q): C2e+1 and 2e +1 is odd, every involution in Aut(R(q)) lies in R(q) (by the previous observation), and hence any subgroup of Aut(R(q)) generated by involutions must be a subgroup of R(q). Thus no subgroup G of Aut(R(q)) strictly above R(q) can be a C-group. (When e = 0 we have Aut(R(3)) = R(3), so the statement holds trivially.) □ 3.2 String C-groups of rank at least five By Lemma 3.1 we may restrict ourselves to Ree groups G = R(q). We first rule out the possibility that the rank is 5 or larger. Lemma 3.2. Let G be a simple group. Suppose G has a generating set S := {p0,..., pn-1} of n involutions such that (G,S) is a string C-group. Then |pjpi+1| > 3 for all i = 0,...,n — 2. Proof. This is due to the fact that, as G is simple, G is not directly decomposable, that is, G cannot be written as the direct product of two nontrivial normal subgroups of G. □ Lemma 3.3. Let G = R(q), where q = 32e+1 = 3. Suppose G has a generating set S of n involutions such that (G, S) is a string C-group. Then n < 4. Proof. Let S = {p0,..., pn-1}, so in particular, G = {p0,..., pn-1). Then p0 commutes with p2,..., pn-1, since the underlying Coxeter diagram is a string. However, by Lemma 3.2, p0 does not commute with p1 and pn-1 does not commute with pn-2. Now suppose n > 5 and consider the subgroup H := {p0, p1, pn-2, pn-1) of G. Then H must be isomorphic to D2c x D2d for some integers c,d > 3. Inspection of the list of maximal subgroups of R(q) described above shows that direct products of (non-abelian) dihedral groups never occur as subgroups in G. So n is at most 4. □ 3.3 String C-groups of rank four Next we eliminate the possibility that the rank is 4. We begin with a general lemma about string C-groups that are simple. Lemma 3.4. Let (G, S) be a string C-group of rank n, and let G be simple. Then Ng(Go1)\Ng(Go) must contain an involution (namely p0). Proof. The involution p0 centralizes G01 and hence must lie in NG(G01). On the other hand, p0 cannot also lie in NG (G0) for otherwise G0 would have to be a nontrivial normal subgroup in the simple group G. □ The next two lemmas will be applied to dihedral subgroups in subgroups of type PSL2(q) or C2 x PSL2(q) of R(q), respectively. Lemma 3.5. Let q = 32e+1 and e > 0. Then the order 2d of a non-abelian dihedral subgroup of PSL2 (q) must divide q — 1 or q + 1. Moreover, d ^ 0 mod 4, and d is even only if 2d divides q +1. Proof. Suppose D2d is a non-abelian dihedral subgroup of PSL2(q), so d > 3. We claim that 2d must divide q +1 or q — 1. Recall that under the assumptions on q, the order 2d must either be 6 or must divide q — 1 or q +1. It remains to eliminate 6 as a possible order. In fact, since q is an odd power of 3, the only maximal subgroups of PSL2 (q) with an order divisible by 6 are subgroups PSL2(q0) with q0 a smaller odd power of 3 (see [8] for a list of the subgroups of PSL(2, q)). If we apply this argument over and over again with smaller odd powers of 3, we eventually are left with a subgroup PSL2(3). However, PSL2(3) = A4 and hence cannot have a subgroup of order 6. Thus 2d must divide q + 1 or q — 1. This proves the first statement of the lemma. The second statement follows from the fact that q = 3 mod 8. □ Lemma 3.6. Let q = 32e+1 and e > 0, let 2d divide q — 1 or q +1, and let D2d be a non-abelian dihedral subgroup of a group C := C2 x PSL2 (q). (a) Then there exists a dihedral subgroup D in PSL2 (q) such that D2d is a subgroup of C2 x D of index 1 or 2, and NC (D2d) = NC (C2 x D) = C2 x NPSL2 (q)(D). Here the normalizer NPSL2(q)(D) must lie in a maximal subgroup Dq+1 or Dq-1 of PSL2(q), and coincide with NDq+1 (D) or NDq-1 (D), according as 2d divides q — 1 or q + 1. (b) Let D2d = D (that is, the index is 2). If 2d | (q — 1) or if 2d | (q + 1) and (q +1)/2d is odd, then NPSL2(q)(D) = D and NC (D2d) = C2 x D2d. If 2d | (q + 1) and (q + 1)/2d is even, then NpsL2(q)(D) = D4d and Nc(D2d) = C2 x D4d. (c) If D2d = C2 x D (that is, d is even, d/2 is odd, D = Dd, and the index is 1), then NpsL2(q)(D) = D2d and Nc (D2d) = C2 x D2d (regardless of whether 2d | (q — 1) or 2d | (q +1)). (d) The structure of NC(D2d) only depends on d and q, not on the way in which D2d is embedded in C. Proof. For the first part, suppose C2 = (p) and D2d = (a0, 01) where a0, a1 are standard involutory generators for D2d. Write a0 = (p1, a'0) and a1 = (pj, o'x) for some i, j = 0,1 and involutions a0 in PSL2(p). Then D := (a0 ) is a dihedral subgroup of PSL2(p), and D2d lies in C2 x D. Since the period of a0divides that of a0a1, the order of D is at most 2d and D2d has index 1 or 2 in C2 x D. If this index is 1 then D2d = C2 x D (and d is even and D = Dd). If the index of D2d in C2 x D is 2, then D = D2d and D2d n {1} x D must have index 1 or 2 in {1} x D. If the index of D2d n {1} x D in {1} x D is 1 then clearly D2d = {1} x D and D2d can be viewed as a subgroup of PSL2 (q). If the index of D2d n {1} x D is 2, then D2d n {1} x D is of the form {1} x E where E is either the cyclic subgroup Cd of D, or d is even and E is one of the two dihedral subgroups of D of order d. (Note here that D2d cannot itself be a direct product in which one factor is generated by p, since p cannot lie in D2d.) Next we investigate normalizers. First note that the normalizer of a direct subproduct in a direct product of groups is the direct product of the normalizers of the component groups. Thus Nc (C2 x D) = C2 x NpsL2(q)(D). We now show that the normalizers in C of the subgroups D2d and C2 x D coincide. There is nothing to prove if D2d = C2 x D or D2d = {1} x D. Now suppose that D2d has index 2 in C2 x D and E is as above. Then it is convenient to write D2d in the form D2d = ({1} x E) u ({p} x (D\E)). (3.1) If (a, fi) € C then (a, fi)D2d(a, fi)-1 = ({1} x fiEfi-1) U ({p} x fi(D\E)fi-1). (3.2) Now if (a, fi) € NC (D2d) then the group on the left in (3.2) is just D2d itself and therefore fiEfi-1 = E and fi(D\E)fi-1 = D\E. It follows that fi normalizes both E and D, so in particular (a, fi) € Nc(C2 x D). Hence Nc(D2d) < Nc(C2 x D). Now suppose that (a, fi) € NC (C2 x D). Then fi normalizes D. But fiEfi-1 must be a subgroup of D of index 2 isomorphic to E, and hence fiEfi-1 and E are either both cyclic or both are dihedral. Clearly, if both subgroups are cyclic then PEP-1 = E. However, the case when both subgroups are dihedral is more complicated. First recall that then d must be even. Now the normalizer NPSL2(q) (D) of the dihedral subgroup D of PSL2 (q) in PSL2(q) either coincides with D (that is, D is self-normalized), or is a dihedral subgroup containing D as a subgroup of index 2. We claim that under the assumptions on q, the second possibility cannot occur. In fact, in this case the normalizer would have to be a group of order 4d, and since d is even, its order would have to be divisible by 8; however, the order of PSL2 (q) is not divisible by 8 when q is an odd power of 3, so PSL2 (q) certainly cannot contain a subgroup with an order divisible by 8. Thus NPSL2(q)(D) = D. But P belongs to the normalizer of D in PSL2(q), so then p must lie in D. In particular, PEP-1 = E since E is normal in D. Thus, in either case we have PEP-1 = E, and since PDP-1 = D, also P(D\E)p-1 = D\E. Hence, (3.2) shows that (a,P) G Nc(D2d). Hence also NC(C2 x D) < NC(D2d). To complete the proof of the first part, note that D must lie in a maximal subgroup Dq±1 of PSL2(q) and Nps^(q)(D) = ND,±1 (D). The second and third part of the lemma follow from Lemma 3.2 applied to the dihedral subgroup D of Dq± 1. In particular, D is self-normalized in Dq± 1 if (q± 1)/1D| is odd, and NPSL2(q)(D) is a dihedral subgroup of Dq±1 of order 2|D| if (q ± 1)/|D| is even. Bear in mind that (q - 1)/2 is odd, and (q + 1)/2 is even but not divisible by 4. To establish the last part of the lemma, note that NC(D2d) = C2 x D2d, except when D = D2d, 2d | (q +1) and (q + 1)/2d is even. However, since q = 3 mod 8, if 2d | (q +1) and (q + 1)/2d is even then d must be odd. In other words, the situation described in the third part of the lemma cannot occur as this would require d to be even. Thus, if 2d | (q +1) and (q + 1)/2d is even, then we are necessarily in the situation described in second part of the lemma, and so necessarily NC (D2d) = C2 x D4d. □ Our next lemma investigates possible C-subgroups of G = R(q) of rank 3. The vertex-figure of a putative regular 4-polytope with automorphism group G would have to be a regular polyhedron with a group of this kind. Lemma 3.7. The only proper subgroups of R(q) that could have the structure of a C-group of rank 3 are Ree subgroups R(qo) with qo = 3 or subgroups of the form PSL2(qo), C2 x PSL2(qo), or R(3)' = PSL2(8). Proof. It is straightforward (sometimes by applying Lemma 3.2) to verify that only subgroups of maximal subgroups of R(q) of the second and third type can have the structure of a rank 3 C-group. Therefore we are left with Ree subgroups R(q0) and subgroups of groups C2 x PSL2 (q0), with q0 an odd power of 3 dividing q, as well as subgroups of type R(3)' = PSL2(8) inside a subgroup R(3). A forward appeal to Lemma 3.15 shows that Ree groups R(q0) with q0 = 3 do in fact act flag-transitively on polyhedra, and by [31], so does R(3)' = PSL2(8). The complete list of subgroups of PSL2(q0) is available, for instance, in [20]. As q0 is an odd power of 3, the group PSL2(q0) does not have subgroups isomorphic to A5, S4, or PGL2(q1) for some q1. Hence, none of the subgroups of PSL2 (q0), except for those isomorphic to a group PSL2(q1), with q1 an odd power of 3 dividing q0 (and hence q), admits flag-transitive actions on polyhedra. Now the maximal subgroups of C2 x PSL2(q0) consist of the factor PSL2(q0), as well as all subgroups of the form C2 x H where H is a maximal subgroup of PSL2 (q0) from the following list: Eq0 : Cqo-1 , Dq0-1, Dq0 + 1, PSL2(q1). A subgroup of C2 x PSL2(q0) of the form C2 x Dq0_i is isomorphic to D2(q0-1) (since q0 = 3 mod 8), so none of its subgroups (including the full subgroup itself) can act regularly on a non-degenerate polyhedron (that is a polyhedron with no 2 in the Schlafli symbol). Similarly, a subgroup of C2 x PSL2(q0) of the form C2 x Dq0+1 is isomorphic to C2 x C2 x D(q0+1)/2, so again none of its subgroups (including the full subgroup itself) can act regularly on a non-degenerate polyhedron. Finally, a subgroup of C2 x PSL2(q0) of the forms C2 x (Eq0 : C(q0-1)/2) has an order not divisible by 4. Hence, as in the two other cases, none of its subgroups (including the full subgroup itself) can act regularly on a non-degenerate polyhedron. In summary, the only possible candidates for rank 3 subgroups of R(q) are of the form R(qo), PSL2(qo), C2 x PSL2(qo), and R(3)' = PSL2(8). We can further rule out a subgroup of type R(3), since R(3) = PrL2 (8) is not generated by involutions. □ For a subgroup B of A we define (B) := (a | a e NA(B),a2 = 1). If B is generated by involutions then B < NO(B) < NA(B). We first state a lemma that will be useful in several places. Lemma 3.8. Let H := R(3) = PrL2(8), and let D := D2d be a dihedral subgroup of H of order at least 6. Then d =3, 7 or 9, and in all cases NH (D) = D. Proof. Straightforward. □ The following lemma considerably limits the ways in which Ree groups R(q) might be representable as C-groups of rank 4. Lemma 3.9. If the group G := R(q) can be represented as a string C-group of rank 4, then N° (G01 ) = N°o(p0)(Go1). (3.3) Proof. Suppose that G admits a representation as a string C-group of rank 4. Thus G = (P0,P1,P2,P3). Since R(3) is not generated by involutions, we must have q = 3. The subgroup G01 = (p2,p3) is a dihedral subgroup D2d (say) of the centralizer Cg(po) of p0, and CG(p0) = (p0) x PSL2(q). Here d > 3, by arguments similar to those used in the proof of Lemma 3.3. Thus D2d = (P2,P3) = G01 < G1 = (P0,P2,P3) < Cg(po) = C2 x PSL2(q). (3.4) By Lemma 3.6 applied to G01 and CG(p0), there exists a dihedral subgroup D in the PSL2(q)-factor of CG(p0) such that G01 is a subgroup of (p0) x D = C2 x D of index at most 2 and NCG(P0)(G01) = NCa (P0)(C2 x D) = C2 x Npsl2(,)(D). In fact, the proof of Lemma 3.6 shows that this subgroup C2 x D is just given by G1. But p0 e G01, so G01 has index 2 in C2 x D = G1, and D = G01 = D2d. Then Lemma 3.5, applied to D, shows that 2d must divide either q +1 or q - 1. The structure of the normalizer Ncg(p0)(G01) can be obtained from Lemma 3.6. In fact, Ncg(p0)(G01) = C2 x D2d, unless 2d | (q + 1) and (q + 1)/2d is even; in the latter case Ncg(po)(G01) = C2 x D4d. In particular, Ncg(po)(G01) is generated by involutions and its order is divisible by 4. We will show that the normalizer of G01 in CG(p0) captures all the information about the full normalizer NG (G01) of G01 in G that is relevant for us. A key step in the proof is the invariance of the structure of the normalizer of G01 in arbitrary subgroups of G of type C2 x PSL2 (q); more precisely, the structure only depends on d and q, not on the way in which G01 is embedded in a subgroup C2 x PSL2 (q) (see Lemma 3.6). The full normalizer NG(G01) of G01 in G must certainly contain Ncg(po)(G01) and also have an order divisible by 8. We claim that all involutions of the full normalizer NG(G01) must already lie in CG(p0) and hence in Ncg(po)(G01). First note that NG(G01) must certainly lie in a maximal subgroup M of G and then coincide with NM(G01). (Since G is simple, the normalizer of a proper subgroup of G cannot coincide with G.) Inspection of the list of maximal subgroups of G shows that only maximal subgroups M of type R(q0), C2 x PSL2(q) or NG(A1) have an order divisible by 4. Only those maximal subgroups could perhaps contain Ngg(po)(G01) and hence NG(G01). We investigate the three possibilities for M separately. Suppose M is a group of type C2 x PSL2(q). Then the invariance of the structure of the normalizer of G01 shows that NM(G01) = Ncg(po)(G01). However, Ncg(po)(G01) < Ng(Go1) and Ng(G01) = Nm(G01), so this gives Ng(Go1) = Ncg(p0)(G?01). But Ncg(po)(Go1) is generated by involutions, so Ngg(po)(Go1) = N£g(po)(Go1) and (3.3) must hold as well. Let M be a group of type NG(A1) = (C2 x D(q+1)/2) : C3 where A1 is a group C(q+1)/4 (recall that (q + 1)/4 is odd). Then all involutions of M must lie in its subgroup K := C2 x D(q+1)/2 = C2 x Dq+1. In particular, all involutions of NM(G01) must lie in K and hence in NK(G01); that is, NM(G01) < NK(G01). Also, G01 itself must lie in K and its order 2d must divide q +1. The subgroup K lies in the centralizer C of the involution generating the C2-factor in the direct product factorization C2 x Dq+1 for K, and NK(G01) < NC(G01). This subgroup C is of type C2 x PSL2(q), and so again the invariance of the structure of the normalizers implies that NC (G01) = Ncg(po)(G01). But Ng(G01) = Nm(G01) and therefore Ncg(po)(G01) = N^(po)(Go1) < NG(G01) = NM(G01) < Nk(G01) < NG(G01). Thus NG(G01) = N°g(po)(Go1), as required. Now let M be a Ree group R(q0) where (q0)p = q and p is a prime. We first cover the case when M is a Ree group R(3) = PSL2(8) : C3, that is, q = 3p where p is a prime. In that case, by Lemma 3.8, NG(G01) = NM(G01) = G01. Hence, since also G01 < N0G(po)(Go1) < NG(G01), we must have NG(Go1) < N°g(po)(Go1). Now suppose q0 = 3, so in particular M is simple. Then 2d must divide q0 ± 1, since Ngg(p0)(Go1) lies in M and therefore po € M, giving G01 < Ngm(po)(Go1) = C2 x PSL2(q0). Since the subgroup NG(G01) of M must have an order divisible by 4, it must lie in a maximal subgroup M' of M of type R(q1), C2 x PSL2(q1), or NR(q0)(A1) with A1 — C(q0+1)/4. The maximal subgroups M' of M = R(q0) of types C2 x PSL2(q0) and NR(q0)(A1), respectively, lie in maximal subgroups of G of type C2 xPSL2(q) or NG (A1), so they are subsumed under the previous discussion. (Alternatively we could dispose of these cases for M' directly, using arguments very similar to those in the two previous cases for M.) Then this leaves the possibility that M' is of type R(q1), in which case we are back at a Ree group. Now continuing in this fashion to smaller and smaller Ree subgroups that could perhaps contain NG(G01), we eventually arrive at either a Ree subgroup M(k) (say) whose parameter q(k) ± 1 (say) is no longer divisible by 2d, or a Ree group R(3). In the first case, R(q0) does not contribute anything new to NG(G01), and the normalizer Ng(G01) must already lie in one of the maximal subgroups of type C2 x PSL2(q) or Ng(Ai) discussed earlier; in particular, Nq(G0i) = NCG(po)(G01), as required. In the second case, the normalizer Ng(G01) lies in a Ree subgroup R(3) = PSL2(8) : C3, and its involutory part N0(G01) must lie in the PSL2(8) subgroup. Again, by Lemma 3.8, we have N0(G01) = NM(G01) = G01, hence (3.3) must also hold in this case. □ Lemma 3.10. If the group G := R(q) can be represented as a string C-group of rank 4, then q = 3 and both the facet stabilizer G3 and vertex stabilizer G0 have to be isomorphic to PSL2(8) = R(3)' (i.e. the commutator subgroup of R(3)) or a simple Ree group R(q0) with q = qm for some odd integer m. Proof. We consider the possible choices for G0 in the given C-group representation of G of rank 4. Our goal is to use Lemma 3.4 to limit the choices for G0 to just R(3)' or R(q0). First recall from Lemma 3.7 that the only possible candidates for G0 are either Ree subgroups R(q0) with q0 = 3 or subgroups of the form PSL2(q0), C2 x PSL2(q0), or R(3)' = PSL2 (8). To complete the proof we must eliminate the second and third types of candidates. This is accomplished by means of Lemmas 3.4 and 3.6, proving in each case that Ng(G01)\Ng(G0) cannot contain an involution, or equivalently N^ (G01) < Ng(G0). Bear in mind that G01 < G0. First observe that all subgroups of G of the form C2 x PSL2(q0) are self-normalized in G; and the normalizer of a subgroup of G of the form PSL2 (q0) is isomorphic to C2 x PSL2(q0). In other words, Ng(G0) = G0 if G0 is of type C2 x PSL2(q0), and Ng(G0) = C2 x G0 if G0 is of type PSL2^). We show that N° (G01) < Ng(G0) for each of these two choices of G0. Suppose that G0 = C2 x PSL2(q0). We first claim that then 2d | q0 ± 1 (where 2d = |G01|). To see this, note that the intersection of G01 with the PSL2(q0)-factor of G0 is a subgroup of index 1 or 2 in G01. If the index is 1, the statement is clear by Lemma 3.5, since then G01 lies in the PSL2(q0)-factor; and if the index is 2 and the intersection is a cyclic group Cd, the statement follows by inspection of the possible orders of cyclic subgroups of PSL2 (q0). Now if the index is 2 and the intersection is a dihedral group Dd, then Lemma 3.6 shows that d must be even, 2d | q +1, and d/2 must be odd; moreover, d | q0 + 1 since Dd lies in PSL2(q0), and hence 2d | q0 + 1 since q0 + 1 is divisible by 4. Thus 2d | q0 ± 1, as claimed. Now, since G0 = C2 x PSL2(q0), the normalizer Ngg(po)(G01) coincides with the normalizer NH(G01) of G01 taken in a suitable subgroup H of CG(p0) of type C2 x PSL2 (q0). In fact, from Lemma 3.6 we know that Ngg(po)(G0i) < C2 x Dq± 1 < Cg(P0) = C2 x PSL2(q). But 2d | q0 ± 1, so we must have Ncg(po)(G01) < C2 x Dqo±1. However, C2 x Dqo±1 lies in a subgroup H of CG(p0) isomorphic to C2 x PSL2 (q0). To complete the argument (for any given type of group G0) we show that N0 (G01) must lie in NG 0 (G01) and therefore also in G0 and NG(G0). When G0 is a group of type C2 x PSL2 (q0), the normalizer Ngo (G01) can be determined using Lemma 3.6 (with q replaced by q0). In fact, by the invariance of the normalizers of G01 we know that Ngo (G01) and NH(G01) are isomorphic and that both subgroups are generated by involutions. However, then by Lemma 3.9, Ngo (Goi) = NO o (G01) < NG (Goi) = NO G(po) (G01) = K (G01) = Nh (Goi), so clearly Noo (Goi) = Nh (Goi). Thus N° (G01) = Noo (Goi) < Go < NG (Go). Now let G0 beoftype PSL2(q0). Then C := No(Go) is a group of type C2 xPSL2 (q0) containing G0, so we can replace G0 by C and argue as before. In fact, using the same subgroup H, we see that the normalizers NO (G01) and NH(G01) are isomorphic subgroups generated by involutions. In particular, No(G01) = NO(G01) < NO(G01) = NO^)^) = NH(G01) = Nh(G01), and therefore No(G01) = Nh(G01). Hence NO(G01) = No(G01) < C = No(Go). □ Let us now show that Go ^ R(3)'. Lemma 3.11. If R(q) has a representation as a string C-group of rank 4 with G0 = R(3)', then q = 27. Proof. Suppose G := R(q) is represented as a string C-group of rank 4 with generators p0,..., p3. Then we know that G01 < G1 < CG(p0) = C2 x PSL2(q). The abstract regular polyhedra with automorphism group R(3)' = PSL2(8) are all known and are available, for instance, in [22]. There are seven examples, up to isomorphism, but not all can occur in the present context. In fact, the dihedral subgroup G01 of G0 must also lie CG(p0) = C2 x PSL2(q) and hence cannot be a subgroup D18. It follows that the polyhedron associated with G0 (that is, the vertex-figure of the polytope for G) must have Schlafli symbol {3,7}, {7,3}, or {7, 7}. We can further rule out the possibility that G01 = D6 by Lemmas 3.5 and 3.6, giving that C2 x PSL2 (q) has no dihedral subgroup of order 6. Hence G01 = D14. The fixed point set of every involution in G is a block of the corresponding Steiner system S(2, q + 1, q3 + 1), and vice versa, every block is the fixed point set of a unique involution. Hence, two involutions with two common fixed points must coincide, since their blocks of fixed points must coincide. Suppose B0 denotes the block of fixed points of p0. As p2 and p3 centralize p0, they stabilize B0 globally but not pointwise. However, p2 cannot have a fixed point among the q +1 points in B0, since otherwise two points of B0 would have to be fixed by p2 since q + 1 is even. Thus p2, and similarly p3, does not fix any point in B0. Moreover, in order for G01 = D14 to lie in a subgroup of G of type C2 x PSL2(q), we must have 7 | q +1 or 7 | q - 1. Using q = 32e+1 and working modulo 7 the latter possibility is easily seen to be impossible. On the other hand, the former possibility occurs precisely when e = 1 mod 3, and then 3 | 2e +1. Hence G must have subgroups isomorphic to R(27) = R(33). We claim that G itself is isomorphic to R(27), that is, q = 27. Now the subgroup G0 = R(3)' lies in a unique subgroup K = R(3) of G, namely its normalizer No(Go). Indeed, Figure 1 tells us that G0 = R(3)' is in a unique subgroup isomorphic to R(27) (because of the lower 1's on the edges joining the boxes). This subgroup K, in turn, lies in a unique subgroup H = R(27) of G. All Ree subgroups of G are self-normalized in G, so in particular K and H are self-normalized. Relative to the Ree subgroup H, the normalizer NH (C7) in H of the cyclic subgroup C7 of G01 is a maximal subgroup of type NH (Ai) = (Cf x D14) : C3 in H, which also contains G01 (see Section 2.2 or [7, p. 123]). Note here that this subgroup C7 is a 7-Sylow subgroup of both K and H, and is normalized by G01. Thus, NH(C7) = (Cf x D14) : C3. We claim that NH(G01) = NH(C7). Clearly, Nh(G01) < Nh(C7). For the opposite inclusion observe that (Cf x D14) : C3 has four subgroups isomorphic to D14, including G01. The subgroup G01 is normalized by the C3-factor, and the three others are permuted under conjugation by C3. This is due to the fact that if it were otherwise, the number of subgroups R(3)' containing G01 would not be an integer but 4/3. Hence, among these four subgroups only G01 is normal and can be thought of as the subgroup D14 occurring in the factorization of the semi-direct product. It follows that the subgroups Cf and C3 normalize G01. Thus Nh(G01) = Nh(C7) = (Cf x D14) : C3. Figure 1 shows the sublattice of the subgroup lattice of G that is relevant to the current situation. Each box contains two pieces of information: a group that describes the abstract structure of the groups in the conjugacy class of subgroups of G depicted by the box, and a number in the lower left corner that gives the number of subgroups in the conjugacy class. This number is the order of G divided by the order of the normalizer in G of a representative subgroup of the conjugacy class. Two boxes are joined by an edge provided that the subgroups represented by the lower box are subgroups of some subgroups represented by the upper box. There are also two numbers on each edge. The number at the top gives the number of subgroups in the conjugacy class for the lower box that are contained in a given subgroup in the conjugacy class for the upper box. The number at the bottom similarly is the number of subgroups in the conjugacy class for the upper box that contain a given subgroup in the conjugacy class for the lower box. If we know the lengths of the conjugacy classes for the upper box and lower box, then knowing one of these two numbers on the connecting edge gives us the other. For instance, in Figure 1, if we know that there are 36 (conjugate) subgroups D14 in a given subgroup R(3)', then there are |G| |G| |R(3)| . / |22 . 3 .14| (conjugate) subgroups R(3)' containing a given subgroup D14. Returning to our line of argument, as already pointed out above, Figure 1 tells us that G0 = R(3)' is in a unique subgroup isomorphic to R(27), namely H (because of the lower 1's on the edges joining the boxes). It also shows that G01 is contained in a unique subgroup (Cf x D14) : C3, which, in turn, is contained in a unique R(27), namely H. As we saw above, this subgroup (Cf x D14) : C3 is necessarily the normalizer NH(G01) of G01 in H. Moreover, p0 has to lie in this unique subgroup (Cf x D14) : C3, which itself is a subgroup of H, and therefore (p0, G0) < H. This holds because Ng(G01) = NH(G01). That these normalizers coincide can be seen as follows. Clearly, NH(G01) < Ng(G01). Now for the opposite inclusion observe that for g G Ng(G01) we have G01 = gG01g-1 < gNH(G01)g-1 and (trivially) G01 < NH(G01). But then Figure 1 shows that a subgroup D14 of H must lie in a unique conjugate of (Cf x D14) : C3 = NH(G01), so necessarily gNH(G01)g-1 = Nh(G01). Similarly, since2 Nh(G01) < H and hence Nh(G01) = gNH (G01)g-1 < gHg-1, Figure 1 (at box R(27)) gives gHg-1 = H, so g G H since H is self-normalized. Thus G = (p0, G0) = H = R(27). □ Figure 1: A sublattice of the subgroup lattice of R(q). Lemma 3.12. The group R(27) cannot be represented as a string C-group of rank 4. Proof. Let G = R(27). By the previous lemmas we may assume that G0 = G3 = PSL2(8). In all other cases we know that G cannot be represented as a rank 4 string C-group. Moreover, from the proof of the previous lemma we already know that G01 = D14 and Ng(G01) = (Cf x D14) : C3. As there is a unique conjugacy class of subgroups R(3)' in R(27), and there is also a unique conjugacy class of subgroups D14 in R(3)', the choice of p2, p3 is therefore unique up to conjugacy in R(27). Once p2, p3 have been chosen, there are three candidates for p0, namely the elements of the subgroup Cf that centralizes D14, and these are equivalent under conjugacy by C3. Hence there is a unique choice for {p0, p2, p3} up to conjugacy. By similar arguments we also know that G3 = R(3)' and G23 = D14, and that the pair (G23, G3) is related to (G01, G0) by conjugacy in R(27). Hence there must exist an element g G R(27) such that • Po = P3 pf = p^ P3 = P0, or g g g • p0 = P3, p2 = po, p3 = P1. The second case can be reduced to the first, as the centraliser of p0 contains an element that swaps p2 and p3 (any two involutions in D14 are conjugate). Hence, we may assume without loss of generality that g swaps p0 and p3. In particular, (p0, p3) is an elementary abelian group of order 4 normalized by g. All such subgroups are known to be conjugate and have as normalizer a group (Cf x D14) : C3. In this group, there is no element that will swap p0 and p3 under conjugation. All elements that will conjugate p0 to p3 will necessarily conjugate p3 to p0p3. Hence we have a contradiction. □ We therefore know that if a string C-group representation of rank 4 exists for R(q), both G0 and G3 must be subgroups of Ree type. Thus from now on we can assume G0 = R(q0) with q0 > 3. In a Ree group, the dihedral subgroups D2n are such that n must divide one of 9, q - 1, q + 1, aq := q + 1 - 3e+1, Pq := q + 1 + 3e+1. Note that aqPq = q2 - q + 1, so in particular if H is a Ree subgroup R(q0) of G then similarly aq0 Pq0 = q2 - q0 + 1. Lemma 3.13. Let G = R(q) with q = 32e+1 and (p0, p^ p2, p3) be a string C-group representation of rank 4 of G. Then (1) G01 is a dihedral subgroup D2d with d a divisor of q +1 and of either aq0 or Pq0 for some q0 such that q = q™ with m odd (where q0 is determined by G0 = R(q0)); (2) m = 3, and G0 and G3 are conjugate Ree subgroups R(q0) with q = q^; (3) G03 is a dihedral subgroup D2t with t a divisor of aq0 or Pq0. Proof. (1) By Lemmas 3.10, 3.11 and 3.12, we may assume that G0 is a simple Ree subgroup of G. Let G0 be a Ree subgroup R(q0), with q0 = 3 such that q™ = q with m a positive odd integer and let G01 = D2d. As G01 < CG(p0) we have that 2d | q ± 1 by Lemma 3.5. In order to have involutions in Ng(G01)\Ng(G0), the only possibility is that NG (G01) (of order divisible by 4) lies in a maximal subgroup of type NG(A1) but not in a maximal subgroup NG0 (Cq0+i) of G0; for otherwise, the same techniques as in 4 Lemma 3.10 show that there is no involution in Ng(G01)\Ng(G0). Hence 2d divides q + 1. Observe that Ng(A ) = (C22 x D q+i) : C3 has exactly four subgroups D q+i because of the subgroup C|. These four subgroups are not all normalised by the C3 because of the semi-direct product. Hence the C3 must conjugate three of them and normalise the fourth one. Similarly, in R(q0) there are four subgroups Dq0+1 in each NR(q0)(A1) and it is obvious that NR(q0)(D2d) = NR(q) (D2d) for every divisor d of q0 + 1. Hence, in order to find some involutions in NG(G01)\Ng(G0), we need to have that 2d does not divide q0 + 1. Moreover, since q0 - 1 divides q - 1, we have also that (q0 - 1, q + 1) = 2. That forces d not to be a divisor of q0 - 1 as d > 2. Hence, looking at the list of maximal subgroups of R(q0) we can conclude that d is a divisor of either aq0 or Pq0 in order for D2d to be a dihedral subgroup of R(q0). (2) Observe that q3 + 1 = (q0 + 1)aq0Pq0 divides q3 + 1. Let us first show that 2e +1 must be divisible by 3 in order for d to satisfy (1). Suppose (3,2e +1) = 1. Then q0 = 32f +1 with 2e + 1 = m(2f + 1) and (3, m) = 1. Let p be an odd prime dividing (aq0 Pq0, q + 1) but not dividing q0 + 1. Then p divides (q3 + 1, q + 1) and hence p divides (q6 -1, q2m -1) = q2(3,m) -1 = q02 -1 = (q0 + 1)(q0 -1) and hence also q0-1. Asp divides q+1,and q0-1 divides q-1, and since (q-1, q+1) = 2, we have that p | 2, a contradiction. Hence m must be divisible by 3 and so does 2e +1. Suppose m = 3. Then m = 3m' and given a Ree subgroup R(q0) of R(q) with q™ = q, there exists a Ree subgroup R(q3) such that R(q0) < R(q3) < R(q). Using similar arguments as in the proof of Lemma 3.11, it is easy to show that, since aq0Pq0 divides q3 + 1, we must have (p0, p1, p2, p3) = R(q3) and therefore m = 3. Indeed, as we stated in (1), )(D2d) = Nfl(q)(D2d) for every divisor d of q3 + 1. Hence po € R(q3). This implies that m = 3 and G0 = R(q0) with q3 = q. Dually, G3 = R(q0). As all subgroups R(q0) are conjugate in R(q), we have that G0 and G3 are conjugate. (3) is due to the fact that G0 n G3 = G03 and that, by (2), G0 and G3 are conjugate in G. Hence, NG (G03) \ G0 has to be nonempty and G03 must not be contained in a subgroup H of G0 such that NG(H) > NG(G03), for if such a subgroup H exists, then G0 n G3 > H. If t divides 9 or one of q0 ± 1, this does not happen. Hence t divides one of aqo or . □ Lemma 3.14. The small Ree groups have no string C-group representation of rank 4. Proof. Suppose G is a Ree group that has a string C-group representation of rank 4. By Lemma 3.10 and part (2) of Lemma 3.13 we may assume that G := R(q) where q = q3 with q0 = 3m for an odd integer m. Moreover, G0 and G3 are conjugate simple Ree subgroups isomorphic to R(q0). By part (3) of Lemma 3.13, if G03 = D2t then t must be a divisor of either aqo or ftqo, and since q = q3, we also have q +1 = (qo + 1)(q^ - qo + 1) = (qo + 1)aqoftqo. Thus t is also a divisor of q +1. We claim that then G0 n G3 > G03, which gives a contradiction to the intersection property. Indeed, since G03 lies in a subgroup H := Ct : C6 of G0, and the normaliser of G03 is not contained in G0 (for otherwise, D2t would have to lie in a unique subgroup R(q0), whereas already G0 and G3 give two examples of such subgroups, by the previous lemma), we have NG(G03) = (C2 x C2 x D2t) : C3. This group contains H = Ct : C6 = D2t : C3 as a normal subgroup, and G03 is normal in H. We also have that NG(H) = NG(G03). But then, as G03 is normal in H, any subgroup R(q0) containing G03 must contain H. In particular this applies to G3. Thus G0 n G3 > H > G03, and the intersection property fails. □ 3.4 String C-groups of rank 3 It remains to investigate the possibility of representing R(q) as a string C-group of rank 3. Nuzhin already showed in [27] that there exist triples of involutions, two of which commute, that generate R(q) for every q. This completes the proof of Theorem 1.1. However, we decided to give here another way to construct an example of a rank three regular poly-tope for R(q) for the paper to be self-contained. Lemma 3.15. Let G = R(q), with q = 3 an odd power of 3. Then there exists a triple of involutions S := {p0, p1, p2} in G such that (G, S) is a string C-group. Proof. Recall that the fixed point set of an involution in G is a block of the Steiner system S := S(2, q + 1, q3 + 1).Pick two involutions p0, p 1 from a maximal subgroup M of G of type Ng(A3) such that p0p1 has order q + 1 + 3e+1, and let B0, B1, respectively, denote their blocks of fixed points. Obviously, B0 n B1 = 0, for otherwise (p0, p1) would lie in the stabilizer of a point in B0 n B1, which is not possible because of the order of p0p1. Recall here that the point stabilizers are maximal subgroups of the form NG(A) = A : Cq-1, where A is a 3-Sylow subgroup of G. Now choose an involution p2 in CG(p0) distinct from p0 such that its block of fixed points B2 meets B1 in a point. Such a p2 exists as all involutions of CG(p0) have pairwise disjoint blocks of size q + 1 and therefore they cover all of the q3 + 1 points. Then B1 n B2 must consist of a single point p (say), and B0 n B2 = 0 since the stabilizer of a point does not contain Klein 4-groups. Then (pi, p2) lies in the point stabilizer of p, and hence must a dihedral group D2n, with n a power of 3. As (p0, pi) is a subgroup of index 3 in M, and p0 does not belong to M, we see that (po, pi, p2) = G. Moreover, since the orders of p0pi and pip2 arecoprime, the intersection property must hold as well. Thus (G, S), with S := {p0, pi, p2}, is a string C-group of rank 3. □ We have not attempted to enumerate or classify all representations of R(q) as a string C-group of rank 3. References [1] P. A. Brooksbank and D. Leemans, Polytopes of large rank for PSL(4, Fq), J. Algebra 452 (2016), 390-400, doi:10.1016/j.jalgebra.2015.11.051. [2] P. A. Brooksbank and D. A. 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Math Soc. 66 (1960), 508-510, doi:10.1090/s0002-9904-1960-10523-x. [30] C.-H. Sah, Groups related to compact Riemann surfaces, Acta Math. 123 (1969), 13-42, doi: 10.1007/bf02392383. [31] D. Sjerve and M. Cherkassoff, On groups generated by three involutions, two of which commute, in: The Hilton Symposium 1993 (Montreal, PQ), Amer. Math. Soc., Providence, Rhode Island, volume 6 of CRM Proc. Lecture Notes, pp. 169-185, 1994. [32] H. Van Maldeghem, Generalized Polygons, volume 93 of Monographs in Mathematics, Birkhauser, Basel, 1998, doi:10.1007/978-3-0348-0271-0. ARS MATHEMATICA CONTEMPORANEA ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 14 (2018) 227-249 https://doi.org/10.26493/1855-3974.1176.9b7 (Also available at http://amc-journal.eu) On hypohamiltonian snarks and a theorem of Fiorini* * Jan Goedgebeur, Carol T. Zamfirescu Department of Applied Mathematics, Computer Science & Statistics, Ghent University, Krijgslaan 281-S9, 9000 Ghent, Belgium In loving memory of Ella. Received 10 August 2016, accepted 1 May 2017, published online 4 September 2017 Abstract In 2003, Cavicchioli et al. corrected an omission in the statement and proof of Fiorini's theorem from 1983 on hypohamiltonian snarks. However, their version of this theorem contains an unattainable condition for certain cases. We discuss and extend the results of Fiorini and Cavicchioli et al. and present a version of this theorem which is more general in several ways. Using Fiorini's erroneous result, Steffen had shown that hypohamiltonian snarks exist for some orders n > 10 and each even n > 92. We rectify Steffen's proof by providing a correct demonstration of a technical lemma on flower snarks, which might be of separate interest. We then strengthen Steffen's theorem to the strongest possible form by determining all orders for which hypohamiltonian snarks exist. This also strengthens a result of Macajova and Skoviera. Finally, we verify a conjecture of Steffen on hypohamiltonian snarks up to 36 vertices. Keywords: Hypohamiltonian, snark, irreducible snark, dot product. Math. Subj. Class.: 05C10, 05C38, 05C45, 05C85 1 Introduction A graph G is hypohamiltonian if G itself is non-hamiltonian, but for every vertex v in G, the graph G - v is hamiltonian. A snark shall be a cubic cyclically 4-edge-connected graph * We thank Nico Van Cleemput for providing us with a script which greatly enhanced the quality of our figures. We would also like to thank Martin ¡Skoviera for informing us about the equivalence of irreducible and vertex-critical graphs. Finally, we also wish to thank Eckhard Steffen for useful suggestions. Both authors are supported by a Postdoctoral Fellowship of the Research Foundation Flanders (FWO). E-mail addresses: jan.goedgebeur@ugent.be (Jan Goedgebeur), czamfirescu@gmail.com (Carol T. Zamfirescu) ©® This work is licensed under http://creativecommons.org/licenses/by/3.0/ with chromatic index 4 (i.e. four colours are required in any proper edge-colouring) and girth at least 5. We refer for notions not defined here to [22] and [7]. Motivated by similarities between the family of all snarks and the family of all cubic hy-pohamiltonian graphs regarding the orders for which such graphs exist, Fiorini [8] studied the hypohamiltonian properties surrounding Isaacs' so-called "flower snarks" [13] (defined rigorously below). The a priori surprising interplay between snarks and hypohamiltonian graphs has been investigated extensively—we now give an overview. Early contributions include Fiorini's aforementioned paper [8], in which he claims to show that there exist infinitely many hypohamiltonian snarks. (In fact, according to Macajova and Skoviera [18], it was later discovered that a family of hypohamiltonian graphs constructed by Gutt [12] includes Isaacs' snarks, thus including Fiorini's result.) Skupien showed that there exist exponentially many hypohamiltonian snarks [20], and Steffen [22] proved that there exist hypohamiltonian snarks of order n for every even n > 92 (and certain n < 92)—we will come back to this result in Section 3. For more references and connections to other problems, see e.g. [3, 18, 20, 23]. Hypohamiltonian snarks have also been studied in connection with the famous Cycle Double Cover Conjecture [3] and Sabidussi's Compatibility Conjecture [9]. The smallest snark, as well as the smallest hypohamiltonian graph, is the famous Petersen graph. Steffen [21] showed that every cubic hypohamiltonian graph with chromatic index 4 is bicritical, i.e. the graph itself is not 3-edge-colourable but deleting any two distinct vertices yields a 3-edge-colourable graph. Nedela and Skoviera [19] proved that every cubic bicritical graph is cyclically 4-edge-connected and has girth at least 5. Therefore, every cubic hypohamiltonian graph with chromatic index 4 must be a snark. This article is organised as follows. In Section 2 we discuss the omission in Fiorini's theorem on hypohamiltonian snarks [8]—first observed by Cavicchioli et al. [5]—and its consequences and state a more general version of this theorem. In Section 3 we first rectify a proof of Steffen on the orders for which hypohamiltonian snarks exist which relied on Fiorini's theorem—this erratum is based on giving a correct proof of a technical lemma concerning flower snarks, which may be of separate interest. We then prove a strengthening of Steffen's theorem, which is best possible, as all orders for which hypohamiltonian snarks exist are determined. Our result is stronger than a theorem of Macajova and Skoviera [17] in the sense that our result implies theirs, while the converse does not hold. Finally, in Section 4 we comment upon and verify a conjecture of Steffen on hypohamiltonian snarks [23] for small orders. 2 Fiorini's theorem revisited We call two edges independent if they have no common vertices. Let G and H be disjoint connected graphs on at least 6 vertices. Consider G' = G - {ab, cd}, where ab and cd are independent edges in G, put H' = H — {x, y}, where x and y are adjacent cubic vertices in H, and let a', b' and c', d' be the other neighbours of x and y in H, respectively. Then the dot product G ■ H is defined as the graph (V(G) U V(H'), E(G') U E(H') U {aa', bb', cc', dd'}). Two remarks are in order. (1) Under the above conditions, the dot product may be disconnected. (2) In fact, there are eight ways to form the dot product for a specific ab, cd, xy. For the computational results in Section 4 we indeed applied the dot product in all eight possible ways, but for the theoretical proofs in this paper we will perform the dot product in one way, namely as follows. We always construct the dot product by adding the edges aa', bb', cc', dd' and we will abbreviate this as "a, b, c, d are joined by edges to the neighbours of x and y, respectively". The dot product was introduced by Adelson-Velsky and Titov [1], and later and independently by Isaacs [13]. Its original purpose was to obtain new snarks by combining known snarks. Fiorini then proved that the dot product can also be used to combine two hypohamiltonian snarks into a new one. Unfortunately, Fiorini's argument is incorrect. We shall discuss this omission within this section, and correct the proof of a lemma of Steffen [22] which depended on Fiorini's result in Section 3. In a graph G, a pair (v, w) of vertices is good in G if there exists a hamiltonian path in G with end-vertices v and w. Two pairs of vertices ((v, w), (x, y)) are good in G if there exist two disjoint paths which together span G, and which have end-vertices v and w, and x and y, respectively. Claim 2.1 (Fiorini, Theorem 3 in [8]). Let G be a hypohamiltonian snark having two independent edges ab and cd for which (i) each of (a, c), (a, d), (b, c), (b, d), ((a, b), (c, d)) is good in G; (ii) for each vertex v, exactly one of (a, b), (c, d) is good in G — v. If H is a hypohamiltonian snark with adjacent vertices x and y, then the dot product G ■ H is also a hypohamiltonian snark, where ab and cd are deleted from G, x and y are deleted from H, and vertices a, b, c, d are joined by edges to the neighbours of x and y, respectively. Cavicchioli et al. [5] point out the omissions in Claim 2.1: in order for the proof to work, the given vertex pairs need to be good in G — {ab, cd} rather than in G. They give a corrected statement of the theorem envisioned by Fiorini and give a new proof. Claim 2.2 (Cavicchioli et al., Theorem 3.2 in [5]). Let G be a hypohamiltonian snark having two independent edges ab and cd for which (i) each of (a, c), (a, d), (b, c), (b, d), ((a, b), (c, d)) is good in G — {ab, cd}; (ii) for each vertex v, each of (a, b), (c, d) is good in G — {v, ab, cd}. If H is a hypohamiltonian snark with adjacent vertices x and y, then the dot product G ■ H is also a hypohamiltonian snark, where ab and cd are deleted from G, x and y are deleted from H, and vertices a, b, c, d are joined by edges to the neighbours of x and y, respectively. In above statements, the fact that the dot product of snarks is itself a snark had already been shown [1, 13], so indeed only the hypohamiltonicity was to be proven. We point out that the hypotheses in Claim 2.2 are unattainable for v e {a,b,c,d}, since (a, b) and (c, d) cannot both be good in G — {v, ab, cd} if v e {a, b, c, d}. This is tied to the fact that the requirements in (ii) are stronger than what is needed to prove the statement. In [11, Theorem 1], we gave the following (second) restatement of Claim 2.1 which we used to solve a problem of McKay. Note that in [11] the graphs are required to be cubic and below we do not state this requirement—we do however need the two vertices which are removed to be cubic. This allows us to use exactly the same proof as in [11, Theorem 1]. Nevertheless, we now give a sketch of the proof: first, we assume that G ■ H does contain a hamiltonian cycle. This however implies that at least one of the factors is hamiltonian, contradicting their hypohamiltonicity. Second, we prove that every vertex-deleted subgraph of G • H is indeed hamiltonian. This is done with a careful case analysis (depending on where the removed vertex lies) using the goodness of various pairs (and pairs of pairs) of vertices in G — {ab, cd} and G — {v, ab, cd}. Theorem 2.3. Let G be a non-hamiltonian graph having two independent edges ab and cd for which (i) each of (a, c), (a, d), (b, c), (b, d), ((a, b), (c, d)) is good in G — {ab, cd}; (ii) for each vertex v, at least one of (a, b) and (c, d) is good in G — {v, ab, cd}. If H is a hypohamiltonian graph with cubic adjacent vertices x and y, then the dot product G • H is also a hypohamiltonian graph, where ab and cd are deleted from G, x and y are deleted from H, and vertices a, b, c, d are joined by edges to the neighbours of x and y, respectively. If G and H are planar, and ab and cd lie on the same facial cycle, then the dot product can be applied such that G • H is planar, as well. If g and h are the girth of G and H, respectively, then the girth of G • H is at least min{g, h}. If G and H are cubic, then so is G • H. Note that the fact that G is non-hamiltonian together with condition (ii) implies that G must be hypohamiltonian. In the following, we will call the pair of edges ab, cd from the statement of Theorem 2.3 suitable. The Petersen graph is the smallest snark, and the two Blanusa snarks on 18 vertices are the second-smallest snarks. All three graphs are also hypohamiltonian. Due to the huge automorphism group of the Petersen graph, it can be verified by hand that it does not contain a pair of suitable edges. Although both Blanusa snarks are dot products of two Petersen graphs, the Petersen graph does not contain a pair of suitable edges. Thus, in a certain sense, Theorem 2.3 is not "if and only if", i.e. there exist dot products whose factors do not contain suitable edges. Let us end this section with a remark which may prove to be useful in other applications. Throughout its statement and proof, we use the notation from Theorem 2.3. Observation 2.4. We have that G • H + ab, G • H + cd, and G • H + ab + cd are hypo-hamiltonian, as well. Proof. Put N(x) = {a', b', y} and N(y) = {c', d!, x} such that the unique neighbour of a' (b', c', d') in G is a (b, c, d). Assume G • H + ab + cd contains a hamiltonian cycle h. Thus, at least one of ab and cd lies in h, say ab. We treat H — {x, y} as a subgraph of G • H. If aa', bb' e E(h), then h n H U a'xb' U c'yd' gives a hamiltonian cycle in H, a contradiction. If aa', bb' <£ E(h), then the cycle h n G + cd yields a contradiction. So w.l.o.g. aa' e E(h) and bb' <£ E(h). This implies the existence of a hamiltonian path in H — {x, y} with end-vertices a' and u e {c', d'}. But this path together with uyxa' is a hamiltonian cycle in H, a contradiction. It follows that G • H + ab and G • H + cd are non-hamiltonian, as well. □ 3 On a theorem of Steffen on hypohamiltonian snarks 3.1 Rectifying Steffen's proof A snark is irreducible if the removal of every edge-cut which is not the set of all edges incident with a vertex yields a 3-edge-colourable graph. Steffen's article [22] is motivated J. Goedgebeur and C. T. Zamfirescu: On hypohamiltonian snarks and a theorem ofFiorini 231 by the following problem. Problem 3.1 (Nedela and Skoviera [19]). For which even number n > 10 does there exist an irreducible snark of order n? In particular, does there exist an irreducible snark of each sufficiently large order? Steffen settled the second question of Problem 3.1 by giving the following main result from [22]. Theorem 3.2 (Steffen, Theorem 2.5 in [22]). There is a hypohamiltonian snark of order n (1) for each n G {m : m > 64 and m = 0 mod 8}, (2) for each n G {10,18} U {m : m > 98 andm = 2 mod 8}, (3) for each n G {m : m > 20 and m = 4 mod 8}, (4) for each n G {30} U {m : m > 54 and m = 6 mod 8}, and (5) for each even n > 92. Isaacs' flower snark J2k+u see [13], is the graph ({ah bi, Ci, di}fto, {biai, biCi, bidi, ajai+i, cdi+1, diCi+1}fi0) , where addition in the indices is performed modulo 2k +1. However, the proof of [22, Lemma 2.3], which is essential for the proof of the theorem, is erroneous, since it uses Fiorini's erroneous Claim 2.1 (and it does not work with Theorem 2.3). We here give a correct proof of that lemma. d0 b° co Figure 1: The flower snark J9. The suitable edges b0c0 and b4c4 are marked in bold red. Lemma 3.3 (Steffen, Lemma 2.3 in [22]). The flower snarks J9, J11, and J13 satisfy the conditions of Theorem 2.3. Proof. In [22], in each of the graphs J9, J11, and J13, the suitable edges were chosen to be 60 co and 64 c4. However, in [22], for various vertices v, the hamiltonian paths did not satisfy condition (ii) from Theorem 2.3, as the paths used one of the edges 60c0 or 64c4. This was for instance the case for v G {a0, a8, d0} in J9, for v G {a0, a10, d0} in J11, and for v G {a0, c1, c12, d0} in J13, see Claims 6, 7, and 8 in the Appendix of [22]. We will now prove that 60c0 and 64c4 are indeed suitable edges for Theorem 2.3 for J9, J11, and J13. For J9 the proof is given below (and partially in the Appendix), while the technical details of the proofs for J11 and J13 can be found in the Appendix. The mapping between the a.j, 6j, cj, dj (used by Steffen) and the vertex numbers used in the proof is shown in Figures 1-3. We use numbers as labels in the proof to make it easier to read these graphs using a computer for verifying the results. Proof that b0c0 and b4c4 are suitable edges for J9. Figure 1 shows the flower snark J9. In J9, the edges 60c0 and 64c4 correspond to the edges (0,26) and (11,12), respectively. The pairs (0,11), (0,12), (26,11) and (26,12) are good in Jg - {(0,26), (11,12)} due to the following hamiltonian paths, respectively: • 11, 10, 5, 6, 7, 8, 9, 30, 29, 28, 27, 35, 24, 23, 22, 17, 16, 15, 32, 31, 12, 13, 14, 19, 18, 33, 34, 21, 20, 25, 26, 4, 3, 2, 1, 0 • 12, 13, 14, 15, 16, 11, 10, 5,4, 26, 25, 20, 19, 18, 17, 22, 21, 34, 33, 32, 31, 30, 9, 8, 7, 6, 29, 28, 3, 2, 1, 23, 24, 35, 27, 0 • 11, 10, 5,4, 3, 2, 1, 0, 27, 28, 29, 6, 7, 8, 9, 30, 31, 12, 13, 14, 19, 18, 17, 16, 15, 32, 33, 34, 35, 24, 23, 22, 21, 20, 25, 26 • 12, 13, 8, 7, 2, 3,4, 5, 6, 29, 28, 27, 0, 1, 23, 22, 17, 18, 33, 32, 31, 30, 9, 10, 11, 16, 15, 14, 19, 20, 21, 34, 35, 24, 25, 26 Note that ((0, 26), (11,12)) is good in J9 - {(0, 26), (11,12)} due to the following two disjoint paths with end-vertices 0 and 26, and 11 and 12, respectively, which together span J9: • 26, 25, 20, 19, 14, 13, 8, 7, 2, 1, 0 • 12, 31, 32, 15, 16, 17, 18, 33, 34, 21, 22, 23, 24, 35, 27, 28, 3, 4, 5, 6, 29, 30, 9, 10, 11 We showed by computer that at least one of (0, 26) or (11,12) is good in J9-{v, (0,26), (11,12)} for every v G V(J9). In each case we verified that the path found by the computer is indeed a valid hamiltonian path in the graph. Below we explicitly show this for v = 0. The hamiltonian paths for the other choices of v can be found in the Appendix. • v = 0: 12, 13, 14, 15, 32, 31, 30, 29, 6, 5, 10, 9, 8, 7, 2, 1, 23, 24, 25, 26, 4, 3, 28, 27, 35, 34, 33, 18, 19, 20, 21, 22, 17, 16, 11 □ Since Steffen's statement of Lemma 3.3 remains intact, the proof and statement of his main result, reproduced above as Theorem 3.2, are correct as given in [22]. Even though we prove a stronger version of Steffen's theorem in the next section, we think it is important to fix the proof of Lemma 3.3 as there may be others who rely on this lemma, or might want to rely on it in the future. Figure 2: The flower snark Ju. The suitable edges 60 c0 and 64c4 are marked in bold red. Figure 3: The flower snark J13. The suitable edges 60c0 and 64c4 are marked in bold red. 3.2 Orders of hypohamiltonian snarks We shall now prove a strengthening of Steffen's theorem, which in a sense is strongest possible since we will determine all orders for which hypohamiltonian snarks exist. We emphasise that our proof's mechanism contains significantly fewer "moving parts" than Macajova and Skoviera's [17], and, as mentioned in the introduction, our theorem also strengthens their result. We do need the following two easily verifiable lemmas. Lemma 3.4. The second Blanusa snark B2 shown in Figure 4 has a pair of suitable edges. Proof. Figure 4 shows the second Blanusa snark B2. By computer we determined that B2 has exactly three pairs of suitable edges: ((6,8), (10,16)), ((3, 9), (12,17)) and ((4, 7), (13,15)). We will now prove by hand that ((6,8), (10,16)) is a suitable edge pair. The pairs (6,10), (6,16), (8,10) and (8,16) are good in B2 - {(6, 8), (10,16)} due to the following hamiltonian paths, respectively: • 10, 11, 12, 17, 16, 15, 13, 14, 0, 1, 5,4, 7, 8, 9, 3, 2, 6 • 16, 15, 9, 8, 7, 17, 12, 13, 14, 10, 11, 1, 0, 2, 3,4, 5, 6 • 10, 11, 1, 0, 14, 13, 12, 17, 16, 15, 9, 3, 2, 6, 5,4, 7, 8 • 16, 15, 9, 3,4, 5, 6, 2, 0, 1, 11, 10, 14, 13, 12, 17, 7, 8 Note that ((6, 8), (10,16)) is good in B2 - {(6,8), (10,16)} due to the following two disjoint paths with end-vertices 6 and 8, and 10 and 16, respectively, which together span B2: • 8,7,4,5,6 • 16, 17, 12, 11, 1, 0, 2, 3, 9, 15, 13, 14, 10 We now prove that at least one of (6,8) or (10,16) is good in B2 - {v, (6,8), (10,16)} for every v G V(B2). By symmetry we only need to prove this for v = 0, 2,4,6,7,8. • v = 0: 8, 7, 4, 5, 1, 11, 10, 14, 13, 12, 17, 16, 15, 9, 3, 2, 6 • v = 2: 8, 9, 3,4, 7, 17, 16, 15, 13, 12, 11, 10, 14, 0, 1, 5, 6 • v = 4: 16, 15, 13, 14, 0, 1, 5, 6, 2, 3, 9, 8, 7, 17, 12, 11, 10 • v = 6: 16, 15, 9, 8, 7, 17, 12, 13, 14, 0, 2, 3, 4, 5, 1, 11, 10 • v = 7: 8, 9, 15, 16, 17, 12, 13, 14, 10, 11, 1, 0, 2, 3, 4, 5, 6 • v = 8: 16, 15, 9, 3, 2, 6, 5, 4, 7, 17, 12, 13, 14, 0, 1, 11, 10 □ Lemma 3.5. The first Loupekine snark L1 shown in Figure 5 has a pair of suitable edges. Proof. Figure 5 shows the first Loupekine snark Li. By computer we determined that Li has exactly six pairs of suitable edges: ((0,1), (17, 20)), ((0, 2), (8,17)), ((1,5), (14, 20)), ((2,3), (8,10)), ((3,4), (10,12)) and ((4,5), (12,14)). We will now prove by hand that ((3,4), (10,12)) is a suitable edge pair. The pairs (3,10), (3,12), (4,10) and (4,12) are good in Li - {(3,4), (10,12)} due to the following hamiltonian paths, respectively: • 10, 7, 9, 6, 8, 17, 19, 21, 16, 13, 11, 0, 1, 5, 4, 18, 20, 14, 12, 15, 2, 3 Figure 4: The second Blanusa snark. It has 18 vertices. The suitable edges (6, 8) and (10,16) are marked in bold red. • 12, 14, 20, 18, 4, 5, 7, 10, 8, 17, 19, 21, 16, 9, 6, 1, 0, 11, 13, 15, 2, 3 • 10, 7, 5, 1, 0, 11, 13, 16, 9, 6, 8, 17, 20, 14, 12, 15, 2, 3, 19, 21, 18,4 • 12, 14, 11, 0, 1, 6, 9, 16, 13, 15, 2, 3, 19, 21, 18, 20, 17, 8, 10, 7, 5,4 Note that ((3,4), (10,12)) is good in L1 - {(3,4), (10,12)} due to the following two disjoint paths with end-vertices 3 and 4, and 10 and 12, respectively, which together span Li: • 4, 5, 1, 6, 8, 17, 20, 18, 21, 19, 3 • 12, 14, 11,0,2, 15, 13, 16,9,7, 10 We now prove that at least one of (3,4) or (10,12) is good in Li - {v, (3,4), (10,12)} for every v G V(L1). By symmetry we only need to prove this for v = 1,4, 5, 6,7,8, 9, 10,16,17,18, 21. • v = 1: 12, 15, 13, 16, 9, 6, 8, 17, 20, 14, 11, 0, 2, 3, 19, 21, 18, 4, 5, 7, 10 • v = 4: 12, 14, 20, 18, 21, 16, 9, 6, 8, 17, 19, 3, 2, 15, 13, 11, 0, 1, 5, 7, 10 • v = 5: 4, 18, 20, 17, 8, 10, 7, 9, 6, 1, 0, 2, 15, 12, 14, 11, 13, 16, 21, 19, 3 • v = 6: 4, 5, 1, 0, 2, 15, 12, 14, 11, 13, 16, 9, 7, 10, 8, 17, 20, 18, 21, 19, 3 • v = 7: 12, 14, 20, 17, 19, 3, 2, 15, 13, 11, 0, 1, 5,4, 18, 21, 16, 9, 6, 8, 10 • v = 8: 12, 14, 20, 17, 19, 3, 2, 15, 13, 11, 0, 1, 6, 9, 16, 21, 18, 4, 5, 7, 10 • v = 9: 4, 5, 7, 10, 8, 6, 1, 0, 2, 15, 12, 14, 11, 13, 16, 21, 18, 20, 17, 19, 3 • v = 10: 4, 5, 7, 9, 16, 13, 11, 0, 1, 6, 8, 17, 19, 21, 18, 20, 14, 12, 15, 2, 3 • v = 16: 4, 5, 1, 6, 9, 7, 10, 8, 17, 19, 21, 18, 20, 14, 12, 15, 13, 11, 0, 2, 3 • v = 17: 4, 5, 1, 6, 8, 10, 7, 9, 16, 13, 11, 0, 2, 15, 12, 14, 20, 18, 21, 19, 3 • v = 18: 4, 5, 1, 6, 8, 10, 7, 9, 16, 21, 19, 17, 20, 14, 12, 15, 13, 11, 0, 2, 3 • v = 21: 4, 18, 20, 14, 12, 15, 2, 0, 11, 13, 16, 9, 6, 1, 5, 7, 10, 8, 17, 19, 3 □ Figure 5: The first Loupekine snark L1. It has 22 vertices. The suitable edges (3,4) and (10,12) are marked in bold red. The following generalisation of Steffen's Theorem 3.2 is strongest possible. Theorem 3.6. A hypohamiltonian snark of order n exists if and only if n e {10,18, 20, 22} or n is even and n > 26. Proof. For n = 10, it is well-known that the Petersen graph is hypohamiltonian and it is also well-known that no snarks exist of order 12, 14 or 16. In Lemma 3.4 we showed that the second Blanusa snark B2 (which has order 18) contains a pair of suitable edges. In [3] it was proven that hypohamiltonian snarks exist for all even orders from 18 to 36 with the exception of 24 (see Table 1). Let Sn denote a hypohamiltonian snark of order n. Using Theorem 2.3, we form the dot product B2 • Sn for n e {18,20,22,26,28, 30, 32} and obtain hypohamiltonian snarks of all even orders between 34 and 48 with the exception of 40 (recall that the dot product of two snarks is a snark). By Lemma 3.5 we know that the first Loupekine snark L1 (which has order 22) contains a pair of suitable edges. Applying Theorem 2.3 to this snark and a 22-vertex hypohamilto-nian snark, we obtain a hypohamiltonian snark of order 40. We form the dot product B2 • Sn for all even n e {34,..., 48} and obtain hypohamiltonian snarks of all even orders from 50 to 64. This may now be iterated ad infinitum, and the proof is complete. □ 3.3 Hypohamiltonian and irreducible snarks In [17] Macajova and Skoviera proved the following theorem (which fully settles Problem 3.1). Theorem 3.7 (Macajova and Skoviera, Theorems A and B in [17]). There exists an irreducible snark of order n if and only if n e {10,18, 20, 22} or n is even and n > 26. Nedela and Skoviera [19] proved that a snark is irreducible if and only if it is bicritical, and Steffen [21] showed that every hypohamiltonian snark is bicritical—while the converse is not true, as will be shown in Table 1. A graph G without k-flow is k-vertex-critical if, for every pair of vertices (u,v) of G, identifying u and v yields a graph that has a k-flow; see [4] for more details. In [4] Carneiro, da Silva, and McKay determined all 4-vertex-critical snarks up to 36 vertices and Skoviera [24] showed that a snark is 4-vertex-critical if and only if it is irreducible. Cavicchioli et al. [5] determined all hypohamiltonian and irreducible snarks on n < 28 vertices. Later, Brinkmann et al. [3] determined all hypohamiltonian snarks on n < 36 vertices. These counts can be found in Table 1 together with the number of irreducible snarks from [4]. These graphs can also be downloaded from the House of Graphs [2] at http://hog.grinvin.org/Snarks. The number of hypohamiltonian cubic graphs on n < 32 vertices can be found in [10]. As can be seen from Table 1, there is a significant number of irreducible snarks which are not hypohamiltonian. The smallest such snarks have order 26. So Theorem 3.6 implies Theorem 3.7, while the converse is not true. Table 1: Number of irreducible and hypohamiltonian snarks (see [4, Table 1] and [3, Table 2]). Ac stands for cyclic edge-connectivity. The counts of cases indicated with a '>' are possibly incomplete; all other cases are complete. Order irreducible hypo. hypo. and Ac = 4 hypo. and Ac > 5 10 1 1 0 1 18 2 2 2 0 20 1 1 0 1 22 2 2 0 2 24 0 0 0 0 26 111 95 87 8 28 33 31 30 1 30 115 104 93 11 32 13 13 0 13 34 40 328 31 198 29 701 1497 36 13 720 10 838 10 374 464 38 ? ? > 51431 ? 40 ? ? > 8 820 ? 42 ? ? > 20 575 458 ? 44 ? ? > 8 242 146 ? The hypohamiltonian snarks on n > 34 vertices constructed by the dot product in the proof of Theorem 3.6 clearly all have cyclic edge-connectivity 4. By combining this with Table 1 we obtain: Corollary 3.8. Hypohamiltonian snarks of order n and cyclic edge-connectivity 4 exist if and only if n G {18, 26, 28, 30} or n is even and n > 34. As already mentioned, every hypohamiltonian snark is irreducible, thus Corollary 3.8 implies [17, Theorem A]. For higher cyclic edge-connectivity, the following is known. Theorem 3.9 (Macajova and Skoviera [18]). There exists a hypohamiltonian snark of order n and cyclic connectivity 5 for each even n > 140, and there exists a hypohamiltonian snark of order n and cyclic connectivity 6 for each even n > 166. If we relax the requirements from hypohamiltonicity to irreducibility, more is known: Theorem 3.10 (Macajova and Skoviera [17]). There exists an irreducible snark of order n and cyclic connectivity 5 if and only if n G {10, 20, 22, 26} or n is even and n > 30, and there exists an irreducible snark of order n and cyclic connectivity 6 for each n = 4 (mod 8) with n > 28, and for each even n > 210. Note that as every hypohamiltonian snark is irreducible, Theorem 3.9 also implies that n > 210 can be improved to n > 166 in Theorem 3.10. The smallest hypohamiltonian snark of cyclic edge-connectivity 5 has order 10 and is the Petersen graph, and the second-smallest such graph has order 20. The flower snark J7 of order 28 is the smallest cyclically 6-edge-connected hypohamiltonian snark. We conclude this section with the following two problems motivated by Theorem 3.10 and results of Kochol [14, 16]. Problem 3.11 (Macajova and Skoviera [17]). Construct a cyclically 6-edge-connected snark (irreducible or not) of order smaller than 118 and different from any of Isaacs' snarks. Problem 3.12. Determine all orders for which cyclically 6-edge-connected snarks exist. 4 On a conjecture of Steffen on hypohamiltonian snarks Consider a cubic graph G. We denote with ^k(G) the minimum number of edges not contained in the union of k 1-factors of G, for every possible combination of k 1-factors. If (G) = 0, then G is 3-edge-colourable. In [23], Steffen made the following conjecture on hypohamiltonian snarks. Conjecture 4.1 (Steffen, Conjecture 4.1 in [23]). If G is a hypohamiltonian snark, then M3 (G) = 3. If true, this conjecture would have significant consequences, e.g. by Theorem 2.14 from [23], it would imply that every hypohamiltonian snark has a Berge-cover (a bridgeless cubic graph G has a Berge-cover if ^s(G) = 0). We wrote a computer program for computing ^(G) and tested Conjecture 4.1 on the complete lists of hypohamiltonian snarks up to 36 vertices. This leads to the following observation. Observation 4.2. There are no counterexamples to Conjecture 4.1 among the hypohamil-tonian snarks with at most 36 vertices. The authors of [3] noted a huge increase (from 13 to 31198) in the number of hypohamiltonian snarks from order 32 to 34, see Table 1. Using a computer, we were able to determine that 29 365 out of the 29 701 hypohamiltonian snarks with cyclic edge-connectivity 4 on 34 vertices can be obtained by performing a dot product on a hypohamil-tonian snark on 26 vertices and the Petersen graph. We also determined that the remaining hypohamiltonian snarks with cyclic edge-connectivity 4 on 34 vertices are obtained by performing a dot product on the Blanusa snarks. Intriguingly, our computations show that some hypohamiltonian snarks can be obtained by performing a dot product on a hypohamiltonian snark on 26 vertices and the Petersen graph, as well as by performing a dot product on the Blanusa snarks. There is also a (slightly less dramatic) increase in the cyclically 5-edge-connected case—these are obviously not dot products—and we believe it to be due to more general graph products, for instance "superposition" introduced by Kochol [15]. It would be interesting to further explore these transformations in order to fully understand these sudden increases and decreases in numbers. Using a computer, we determined that all hypohamiltonian snarks with cyclic edge-connectivity 4 up to 36 vertices can be obtained by performing a dot product on two hypohamiltonian snarks. This leads us to pose the following question. Problem 4.3. Is every hypohamiltonian snark with cyclic edge-connectivity 4 a dot product of two hypohamiltonian snarks? In [6] Chladny and Skoviera proved that every bicritical snark with cyclic edge-connectivity 4 is a dot product of two bicritical snarks. Since every hypohamiltonian snark is bicritical, this implies that every hypohamiltonian snark with cyclic edge-connectivity 4 is a dot product of two bicritical snarks. Recall that in Theorem 2.3 the graphs G and H are hypohamiltonian, but the theorem is not "if and only if", since although the Petersen graph does not contain a pair of suitable edges, the Blanusa snarks (which are dot products of two Petersen graphs) are hypohamiltonian. Despite the previous paragraph, we believe the answer to Problem 4.3 to be "no" due to the following observation. In order to cover all cases, we would need to add to condition (ii) of Theorem 2.3 the possibility of ((a, b), (c, d)) being good in G - {v, ab, cd}. However, we would then need to require from H that it contains a 2-factor containing exactly two (necessarily odd) cycles. Although we were unable to find a counterexample, we believe that there exist hypohamiltonian snarks which do not possess such a 2-factor. We also determined all hypohamiltonian snarks up to 44 vertices which can be obtained by performing a dot product on two hypohamiltonian snarks. The counts of these snarks can be found in the fourth column of Table 1. We also verified Conjecture 4.1 on these snarks. Observation 4.4. There are no counterexamples to Conjecture 4.1 among the hypohamil-tonian snarks with at most 44 vertices which are a dot product of two hypohamiltonian snarks. References [1] G. M. Adelson-Velsky and V. K. Titov, On edge 4-chromatic cubic graphs (in Russian), in: Proc. Seminar of 1971 at Moscow Univ., Voprosy Kibernetiki, 1973 pp. 5-14. [2] G. Brinkmann, K. Coolsaet, J. Goedgebeur and H. Melot, House of Graphs: a database of interesting graphs, Discrete Appl. Math. 161 (2013), 311-314, doi:10.1016/j.dam.2012.07.018, available at http://hog.grinvin.org/. [3] G. Brinkmann, J. Goedgebeur, J. Hagglund and K. Markstrom, Generation and properties of snarks, J. Combin. Theory Ser. B 103 (2013), 468-488, doi:10.1016/j.jctb.2013.05.001. [4] A. B. Carneiro, C. N. da Silva and B. D. McKay, A faster test for 4-flow-criticality in snarks, Electron. 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[24] M. Skoviera, Flow-critical and critical snarks coincide, 2016, in preparation. Appendix Below we give the technical details which were left out in the proof of Lemma 3.3. Proof that b0c0 and b4c4 are suitable edges for J9 (continued) We will now prove that at least one of (0, 26) or (11,12) is good in J9 - {v, (0, 26), (11,12)} for every v G V (J9) except for v = 0, which we have already shown above in the proof of Lemma 3.3. • v = 1: 26, 4, 5, 6, 7, 2, 3, 28, 29, 30, 31, 12, 13, 8, 9, 10, 11, 16, 17, 18, 33, 32, 15, 14, 19, 20, 25, 24, 23, 22, 21, 34, 35, 27, 0 • v = 2: 12, 31, 32, 15, 14, 13, 8, 7, 6, 5, 10, 9, 30, 29, 28, 3, 4, 26, 25, 24, 23, 1, 0, 27, 35, 34, 33, 18, 19, 20, 21, 22, 17, 16, 11 • v = 3: 26, 4, 5, 6, 7, 2, 1, 23, 22, 21, 20, 25, 24, 35, 34, 33, 32, 15, 14, 19, 18, 17, 16, 11, 10, 9, 8, 13, 12, 31, 30, 29, 28, 27, 0 • v = 4: 26, 25, 20, 19, 18, 17, 22, 21, 34, 33, 32, 31, 12, 13, 14, 15, 16, 11, 10, 5, 6, 7, 8, 9, 30, 29, 28, 3, 2, 1, 23, 24, 35, 27, 0 • v = 5: 26, 4, 3, 2, 1, 23, 22, 21, 20, 25, 24, 35, 34, 33, 32, 15, 14, 19, 18, 17, 16, 11, 10, 9, 30, 31, 12, 13, 8, 7, 6, 29, 28, 27, 0 • v = 6: 26, 25, 20, 19, 18, 17, 22, 21, 34, 33, 32, 31, 12, 13, 14, 15, 16, 11, 10, 5, 4, 3, 2, 7, 8, 9, 30, 29, 28, 27, 35, 24, 23, 1, 0 • v = 7: 26, 4, 5, 6, 29, 30, 31, 12, 13, 8, 9, 10, 11, 16, 17, 18, 19, 14, 15, 32, 33, 34, 35, 24, 25, 20, 21, 22, 23, 1, 2, 3, 28, 27, 0 • v = 8: 26, 25, 20, 19, 18, 17, 22, 21, 34, 33, 32, 31, 12, 13, 14, 15, 16, 11, 10, 9, 30, 29, 28, 3, 4, 5, 6, 7, 2, 1, 23, 24, 35, 27, 0 • v = 9: 26, 4, 3, 2, 1, 23, 22, 21, 20, 25, 24, 35, 34, 33, 32, 15, 14, 19, 18, 17, 16, 11, 10, 5, 6, 7, 8, 13, 12, 31, 30, 29, 28, 27, 0 • v = 10: 12, 13, 14, 15, 32, 31, 30, 9, 8, 7, 2, 3, 28, 29, 6, 5, 4, 26, 25, 24, 23, 1, 0, 27, 35, 34, 33, 18, 19, 20, 21, 22, 17, 16, 11 • v = 11: 26, 4, 3, 2, 1, 23, 22, 21, 34, 35, 24, 25, 20, 19, 14, 15, 16, 17, 18, 33, 32, 31, 12, 13, 8, 7, 6, 5, 10, 9, 30, 29, 28, 27, 0 • v = 12: 26, 25, 20, 19, 18, 17, 22, 21, 34, 33, 32, 31, 30, 9, 8, 13, 14, 15, 16, 11, 10, 5, 4, 3, 2, 7, 6, 29, 28, 27, 35, 24, 23, 1, 0 • v = 13: 12, 31, 30, 29, 6, 5, 10, 9, 8, 7, 2, 1, 0, 27, 28, 3, 4, 26, 25, 20, 21, 22, 23, 24, 35, 34, 33, 32, 15, 14, 19, 18, 17, 16, 11 • v = 14: 26, 25, 20, 19, 18, 17, 22, 21, 34, 33, 32, 15, 16, 11, 10, 9, 8, 13, 12, 31, 30, 29, 28, 3, 4, 5, 6, 7, 2, 1, 23, 24, 35, 27, 0 • v = 15: 26, 4, 3, 2, 1, 23, 22, 21, 20, 25, 24, 35, 34, 33, 32, 31, 12, 13, 14, 19, 18, 17, 16, 11, 10, 5, 6, 7, 8, 9, 30, 29, 28, 27, 0 • v = 16: 12, 13, 8, 7, 2, 3, 28, 29, 6, 5, 4, 26, 25, 24, 23, 1, 0, 27, 35, 34, 33, 18, 17, 22, 21, 20, 19, 14, 15, 32, 31, 30, 9, 10, 11 • v = 17: 26, 4, 3, 2, 1, 23, 22, 21, 20, 25, 24, 35, 34, 33, 18, 19, 14, 13, 12, 31, 32, 15, 16, 11, 10, 5, 6, 7, 8, 9, 30, 29, 28, 27, 0 • v = 18: 26, 25, 20, 19, 14, 15, 32, 33, 34, 21, 22, 17, 16, 11, 10, 9, 8, 13, 12, 31, 30, 29, 28, 3, 4, 5, 6, 7, 2, 1, 23, 24, 35, 27, 0 • v = 19: 26, 4, 3, 28, 27, 35, 34, 21, 20, 25, 24, 23, 22, 17, 18, 33, 32, 31, 12, 13, 14, 15, 16, 11, 10, 5, 6, 29, 30, 9, 8, 7, 2, 1, 0 • v = 20: 26, 25, 24, 35, 27, 28, 29, 6, 5, 4, 3, 2, 7, 8, 13, 12, 31, 30, 9, 10, 11, 16, 17, 18, 19, 14, 15, 32, 33, 34, 21, 22, 23, 1, 0 • v = 21: 26, 4, 3, 2, 1, 23, 22, 17, 18, 19, 20, 25, 24, 35, 34, 33, 32, 31, 12, 13, 14, 15, 16, 11, 10, 5, 6, 7, 8, 9, 30, 29, 28, 27, 0 • v = 22: 26, 25, 20, 21, 34, 33, 32, 15, 14, 19, 18, 17, 16, 11, 10, 9, 8, 13, 12, 31, 30, 29, 28, 3, 4, 5, 6, 7, 2, 1, 23, 24, 35, 27, 0 • v = 23: 26, 4, 3, 28, 27, 35, 24, 25, 20, 19, 18, 17, 22, 21, 34, 33, 32, 31, 12, 13, 14, 15, 16, 11, 10, 5, 6, 29, 30, 9, 8, 7, 2, 1, 0 • v = 24: 26, 25, 20, 19, 14, 15, 32, 33, 18, 17, 16, 11, 10, 9, 8, 13, 12, 31, 30, 29, 28, 3, 4, 5, 6, 7, 2, 1, 23, 22, 21, 34, 35, 27, 0 • v = 25: 26, 4, 3, 2, 1, 23, 24, 35, 34, 33, 18, 17, 22, 21, 20, 19, 14, 13, 12, 31, 32, 15, 16, 11, 10, 5, 6, 7, 8, 9, 30, 29, 28, 27, 0 • v = 26: 12, 13, 8, 7, 2, 3, 4, 5, 6, 29, 28, 27, 0, 1, 23, 22, 21, 34, 35, 24, 25, 20, 19, 14, 15, 16, 17, 18, 33, 32, 31, 30, 9, 10, 11 • v = 27: 26, 4, 5, 6, 7, 2, 3, 28, 29, 30, 31, 12, 13, 8, 9, 10, 11, 16, 17, 18, 19, 14, 15, 32, 33, 34, 35, 24, 25, 20, 21, 22, 23, 1, 0 • v = 28: 12, 13, 14, 15, 32, 31, 30, 29, 6, 5, 10, 9, 8, 7, 2, 3, 4, 26, 25, 24, 23, 1, 0, 27, 35, 34, 33, 18, 19, 20, 21, 22, 17, 16, 11 • v = 29: 26, 4, 5, 6, 7, 8, 13, 12, 31, 30, 9, 10, 11, 16, 17, 18, 19, 14, 15, 32, 33, 34, 35, 24, 25, 20, 21, 22, 23, 1, 2, 3, 28, 27, 0 • v = 30: 26, 25, 20, 19, 18, 17, 22, 21, 34, 33, 32, 31, 12, 13, 14, 15, 16, 11, 10, 9, 8, 7, 2, 3, 4, 5, 6, 29, 28, 27, 35, 24, 23, 1, 0 • v = 31: 12, 13, 8, 7, 6, 5, 10, 9, 30, 29, 28, 27, 0, 1, 2, 3, 4, 26, 25, 20, 21, 22, 23, 24, 35, 34, 33, 32, 15, 14, 19, 18, 17, 16, 11 • v = 32: 26, 25, 24, 23, 1, 2, 7, 6, 5, 4, 3, 28, 29, 30, 31, 12, 13, 8, 9, 10, 11, 16, 15, 14, 19, 20, 21, 22, 17, 18, 33, 34, 35, 27, 0 • v = 33: 26, 4, 3, 28, 27, 35, 34, 21, 20, 25, 24, 23, 22, 17, 18, 19, 14, 13, 12, 31, 32, 15, 16, 11, 10, 5, 6, 29, 30, 9, 8, 7, 2, 1, 0 • v = 34: 26, 25, 24, 35, 27, 28, 29, 6, 5, 4, 3, 2, 7, 8, 13, 12, 31, 30, 9, 10, 11, 16, 17, 18, 33, 32, 15, 14, 19, 20, 21, 22, 23, 1, 0 • v = 35: 26, 4, 3, 2, 1, 23, 24, 25, 20, 19, 18, 17, 22, 21, 34, 33, 32, 31, 12, 13, 14, 15, 16, 11, 10, 5, 6, 7, 8, 9, 30, 29, 28, 27, 0 Proof that b0c0 and b4c4 are suitable edges for Jn Figure 2 shows the flower snark Jn and here 60c0 and 64c4 correspond to the edges (0,32) and (11,12), respectively. The pairs (0,11), (0,12), (32,11), and (32,12) are good in Jn - {(0,32), (11,12)} due to the following hamiltonian paths, respectively: • 11, 10, 5, 6, 7, 8, 9, 36, 35, 34, 33, 43, 30, 29, 28, 23, 22, 17, 16, 15, 38, 37, 12, 13, 14, 19, 18, 39, 40, 21, 20, 25, 24, 41, 42, 27, 26, 31, 32, 4, 3, 2, 1, 0 • 12, 13, 14, 15, 16, 11, 10, 5, 4, 32, 31, 26, 25, 24, 41, 42, 27, 28, 23, 22, 17, 18, 19, 20, 21, 40, 39, 38, 37, 36, 9, 8, 7, 6, 35, 34, 3, 2, 1, 29, 30, 43, 33, 0 • 11, 10, 5,4, 3, 2, 1, 0, 33, 34, 35, 6, 7, 8, 9, 36, 37, 12, 13, 14, 19, 18, 17, 16, 15, 38, 39, 40, 41, 24, 25, 20, 21, 22, 23, 28, 29, 30, 43, 42, 27, 26, 31, 32 • 12, 13, 8, 7, 2, 3, 4, 5, 6, 35, 34, 33, 0, 1, 29, 28, 23, 22, 17, 18, 39, 38, 37, 36, 9, 10, 11, 16, 15, 14, 19, 20, 21, 40, 41, 24, 25, 26, 27, 42, 43, 30, 31, 32 We have that ((0,32), (11,12)) is good in J11 — {(0, 32), (11,12)} due to the following two disjoint paths with end-vertices 0 and 32, and 11 and 12, respectively, which together span Jn. • 32, 31, 26, 25, 20, 19, 14, 13, 8, 7, 2, 1, 0 • 12, 37, 38, 15, 16, 17, 18, 39, 40, 21, 22, 23, 24, 41, 42, 27, 28, 29, 30, 43, 33, 34, 3, 4, 5, 6, 35, 36, 9, 10, 11 The following hamiltonian paths show that at least one of (0, 32) or (11,12) is good in J11 — {v, (0, 32), (11,12)} for every v e V(Jn). • v = 0: 12, 13, 14, 15, 38, 37, 36, 35, 6, 5, 10, 9, 8, 7, 2, 1, 29, 30, 31, 32, 4, 3, 34, 33, 43, 42, 41, 24, 23, 28, 27, 26, 25, 20, 19, 18, 39, 40, 21, 22, 17, 16, 11 • v = 1: 32, 4, 5, 6, 7, 2, 3, 34, 35, 36, 37, 12, 13, 8, 9, 10, 11, 16, 17, 18, 19, 14, 15, 38, 39, 40, 41, 24, 23, 22, 21, 20, 25, 26, 31, 30, 29, 28, 27, 42, 43, 33, 0 • v = 2: 12, 37, 38, 15, 14, 13, 8, 7, 6, 5, 10, 9, 36, 35, 34, 3, 4, 32, 31, 30, 29, 1, 0, 33, 43, 42, 41, 24, 23, 28, 27, 26, 25, 20, 19, 18, 39, 40, 21, 22, 17, 16, 11 • v = 3: 32, 4, 5, 6, 7, 2, 1, 29, 28, 27, 26, 31, 30, 43, 42, 41, 40, 21, 22, 23, 24, 25, 20, 19, 14, 15, 38, 39, 18, 17, 16, 11, 10, 9, 8, 13, 12, 37, 36, 35, 34, 33, 0 • v = 4: 32, 31, 26, 25, 24, 23, 28, 27, 42, 41, 40, 39, 18, 17, 22, 21, 20, 19, 14, 13, 12, 37, 38, 15, 16, 11, 10, 5, 6, 7, 8, 9, 36, 35, 34, 3, 2, 1, 29, 30, 43, 33, 0 • v = 5: 32, 4, 3, 2, 1, 29, 28, 27, 26, 31, 30, 43, 42, 41, 40, 21, 22, 23, 24, 25, 20, 19, 14, 15, 38, 39, 18, 17, 16, 11, 10, 9, 36, 37, 12, 13, 8, 7, 6, 35, 34, 33, 0 • v = 6: 32, 31, 26, 25, 24, 23, 28, 27, 42, 41, 40, 39, 18, 17, 22, 21, 20, 19, 14, 13, 12, 37, 38, 15, 16, 11, 10, 5, 4, 3, 2, 7, 8, 9, 36, 35, 34, 33, 43, 30, 29, 1, 0 • v = 7: 32, 4, 5, 6, 35, 36, 37, 12, 13, 8, 9, 10, 11, 16, 17, 18, 19, 14, 15, 38, 39, 40, 41, 24, 23, 22, 21, 20, 25, 26, 31, 30, 29, 28, 27, 42, 43, 33, 34, 3, 2, 1, 0 • v = 8: 32, 31, 26, 25, 24, 23, 28, 27, 42, 41, 40, 39, 18, 17, 22, 21, 20, 19, 14, 13, 12, 37, 38, 15, 16, 11, 10, 9, 36, 35, 34, 3,4, 5, 6, 7, 2, 1, 29, 30, 43, 33, 0 • v = 9: 32, 4, 3, 2, 1, 29, 28, 27, 26, 31, 30, 43, 42, 41, 40, 21, 22, 23, 24, 25, 20, 19, 14, 15, 38, 39, 18, 17, 16, 11, 10, 5, 6, 7, 8, 13, 12, 37, 36, 35, 34, 33, 0 • v = 10: 12, 13, 14, 15, 38, 37, 36, 9, 8, 7, 2, 3, 34, 35, 6, 5, 4, 32, 31, 30, 29, 1, 0, 33, 43, 42, 41, 24, 23, 28, 27, 26, 25, 20, 19, 18, 39, 40, 21, 22, 17, 16, 11 • v = 11: 32, 4, 3, 2, 1, 29, 28, 27, 26, 31, 30, 43, 42, 41, 40, 21, 20, 25, 24, 23, 22, 17, 16, 15, 14, 19, 18, 39, 38, 37, 12, 13, 8, 7, 6, 5, 10, 9, 36, 35, 34, 33, 0 • v = 12: 32, 31, 26, 25, 24, 23, 28, 27, 42, 41, 40, 39, 18, 17, 22, 21, 20, 19, 14, 13, 8, 9, 10, 11, 16, 15, 38, 37, 36, 35, 34, 3, 4, 5, 6, 7, 2, 1, 29, 30,43, 33, 0 • v = 13: 12, 37, 36, 35, 6, 5, 10, 9, 8, 7, 2, 1, 0, 33, 34, 3, 4, 32, 31, 26, 27, 28, 29, 30, 43, 42, 41, 40, 21, 22, 23, 24, 25, 20, 19, 14, 15, 38, 39, 18, 17, 16, 11 • v = 14: 32, 31, 26, 25, 24, 41, 42, 27, 28, 23, 22, 17, 18, 19, 20, 21, 40, 39, 38, 15, 16, 11, 10, 9, 8, 13, 12, 37, 36, 35, 34, 3, 4, 5, 6, 7, 2, 1, 29, 30,43, 33, 0 • v = 1B: 32, 4, 3, 2, 1, 29, 28, 27, 26, 31, 30, 43, 42, 41, 40, 21, 22, 23, 24, 25, 20, 19, 14, 13, 12, 37, 38, 39, 18, 17, 16, 11, 10, 5, 6, 7, 8, 9, 36, 35, 34, 33, 0 • v = 16: 12, 13, 8, 7, 2, 3, 34, 35, 6, 5, 4, 32, 31, 30, 29, 1, 0, 33, 43, 42, 41, 24, 25, 26, 27, 28, 23, 22, 17, 18, 39, 40, 21, 20, 19, 14, 15, 38, 37, 36, 9, 10, 11 • v = 1T: 32, 4, 3, 2, 1, 29, 28, 27, 26, 31, 30, 43, 42, 41, 40, 21, 22, 23, 24, 25, 20, 19, 18, 39, 38, 37, 12, 13, 14, 15, 16, 11, 10, 5, 6, 7, 8, 9, 36, 35, 34, 33, 0 • v = 18: 32, 31, 26, 25, 24, 23, 28, 27, 42, 41, 40, 39, 38, 15, 14, 19, 20, 21, 22, 17, 16, 11, 10, 9, 8, 13, 12, 37, 36, 35, 34, 3, 4, 5, 6, 7, 2, 1, 29, 30,43, 33, 0 • v = 19: 32, 4, 3, 2, 1, 29, 28, 27, 26, 31, 30, 43, 42, 41, 40, 21, 20, 25, 24, 23, 22, 17, 18, 39, 38, 37, 12, 13, 14, 15, 16, 11, 10, 5, 6, 7, 8, 9, 36, 35, 34, 33, 0 • v = 20: 32, 31, 26, 25, 24, 41, 42, 27, 28, 23, 22, 21, 40, 39, 38, 15, 14, 19, 18, 17, 16, 11, 10, 9, 8, 13, 12, 37, 36, 35, 34, 3, 4, 5, 6, 7, 2, 1, 29, 30,43, 33, 0 • v = 21: 32, 4, 3, 2, 1, 29, 28, 27, 26, 31, 30, 43, 42, 41, 40, 39, 18, 17, 22, 23, 24, 25, 20, 19, 14, 13, 12, 37, 38, 15, 16, 11, 10, 5, 6, 7, 8, 9, 36, 35, 34, 33, 0 • v = 22: 32, 31, 26, 25, 24, 23, 28, 27, 42, 41, 40, 21, 20, 19, 14, 15, 38, 39, 18, 17, 16, 11, 10, 9, 8, 13, 12, 37, 36, 35, 34, 3, 4, 5, 6, 7, 2, 1, 29, 30,43, 33, 0 • v = 23: 32, 4, 3, 2, 1, 29, 28, 27, 26, 31, 30, 43, 42, 41, 24, 25, 20, 19, 18, 17, 22, 21, 40, 39, 38, 37, 12, 13, 14, 15, 16, 11, 10, 5, 6, 7, 8, 9, 36, 35, 34, 33, 0 • v = 24: 32, 31, 26, 25, 20, 21, 22, 23, 28, 27, 42, 41, 40, 39, 38, 15, 14, 19, 18, 17, 16, 11, 10, 9, 8, 13, 12, 37, 36, 35, 34, 3, 4, 5, 6, 7, 2, 1, 29, 30,43, 33, 0 • v = 2B: 32, 4, 3, 2, 1, 29, 28, 27, 26, 31, 30, 43, 42, 41, 24, 23, 22, 17, 18, 19, 20, 21, 40, 39, 38, 37, 12, 13, 14, 15, 16, 11, 10, 5, 6, 7, 8, 9, 36, 35, 34, 33, 0 • v = 26: 32, 31, 30, 29, 1, 2, 7, 6, 5, 4, 3, 34, 35, 36, 37, 12, 13, 8, 9, 10, 11, 16, 17, 18, 19, 14, 15, 38, 39, 40, 41, 24, 25, 20, 21, 22, 23, 28, 27, 42, 43, 33, 0 • v = 2T: 32, 4, 3, 2, 1, 29, 28, 23, 24, 25, 26, 31, 30, 43, 42, 41, 40, 39, 18, 17, 22, 21, 20, 19, 14, 13, 12, 37, 38, 15, 16, 11, 10, 5, 6, 7, 8, 9, 36, 35, 34, 33, 0 • v = 28: 32, 31, 26, 27, 42, 41, 40, 21, 22, 23, 24, 25, 20, 19, 14, 15, 38, 39, 18, 17, 16, 11, 10, 9, 8, 13, 12, 37, 36, 35, 34, 3, 4, 5, 6, 7, 2, 1, 29, 30,43, 33, 0 • v = 29: 32, 4, 3, 34, 33, 43, 30, 31, 26, 25, 24, 23, 28, 27,42,41,40, 39, 18, 17, 22, 21, 20, 19, 14, 13, 12, 37, 38, 15, 16, 11, 10, 5, 6, 35, 36, 9, 8, 7, 2, 1, 0 • v = 30: 32, 31, 26, 25, 20, 21, 22, 23, 24, 41, 40, 39, 38, 15, 14, 19, 18, 17, 16, 11, 10, 9, 8, 13, 12, 37, 36, 35, 34, 3, 4, 5, 6, 7, 2, 1, 29, 28, 27, 42,43, 33, 0 • v = 31: 32, 4, 3, 2, 1, 29, 30, 43, 42, 41, 24, 23, 28, 27, 26, 25, 20, 19, 18, 17, 22, 21, 40, 39, 38, 37, 12, 13, 14, 15, 16, 11, 10, 5, 6, 7, 8, 9, 36, 35, 34, 33, 0 • v = 32: 12, 13, 8, 7, 2, 3, 4, 5, 6, 35, 34, 33, 0, 1, 29, 28, 27, 26, 31, 30, 43, 42, 41, 40, 21, 20, 25, 24, 23, 22, 17, 16, 15, 14, 19, 18, 39, 38, 37, 36, 9, 10, 11 • v = 33: 32, 4, 5, 6, 7, 2, 3, 34, 35, 36, 37, 12, 13, 8, 9, 10, 11, 16, 17, 18, 19, 14, 15, 38, 39, 40, 41, 24, 23, 22, 21, 20, 25, 26, 31, 30, 43, 42, 27, 28, 29, 1, 0 • v = 34: 12, 13, 14, 15, 38, 37, 36, 35, 6, 5, 10, 9, 8, 7, 2, 3, 4, 32, 31, 30, 29, 1, 0, 33, 43, 42, 41, 24, 23, 28, 27, 26, 25, 20, 19, 18, 39, 40, 21, 22, 17, 16, 11 • v = 35: 32, 4, 5, 6, 7, 8, 13, 12, 37, 36, 9, 10, 11, 16, 17, 18, 19, 14, 15, 38, 39, 40, 41, 24, 23, 22, 21, 20, 25, 26, 31, 30, 29, 28, 27, 42, 43, 33, 34, 3, 2, 1, 0 • v = 36: 32, 31, 26, 25, 24, 23, 28, 27, 42, 41, 40, 39, 18, 17, 22, 21, 20, 19, 14, 13, 12, 37, 38, 15, 16, 11, 10, 9, 8, 7, 2, 3, 4, 5, 6, 35, 34, 33,43, 30, 29, 1, 0 • v = 37: 12, 13, 8, 7, 6, 5, 10, 9, 36, 35, 34, 33, 0, 1, 2, 3, 4, 32, 31, 26, 27, 28, 29, 30, 43, 42, 41, 40, 21, 22, 23, 24, 25, 20, 19, 14, 15, 38, 39, 18, 17, 16, 11 • v = 38: 32, 31, 26, 25, 24, 23, 28, 27, 42, 41, 40, 39, 18, 17, 22, 21, 20, 19, 14, 15, 16, 11, 10, 9, 8, 13, 12, 37, 36, 35, 34, 3, 4, 5, 6, 7, 2, 1, 29, 30, 43, 33, 0 • v = 39: 32, 4, 3, 2, 1, 29, 28, 27, 26, 31, 30, 43, 42, 41, 40, 21, 20, 25, 24, 23, 22, 17, 18, 19, 14, 13, 12, 37, 38, 15, 16, 11, 10, 5, 6, 7, 8, 9, 36, 35, 34, 33, 0 • v = 40: 32, 31, 26, 25, 24, 41, 42, 27, 28, 23, 22, 21, 20, 19, 14, 15, 38, 39, 18, 17, 16, 11, 10, 9, 8, 13, 12, 37, 36, 35, 34, 3, 4, 5, 6, 7, 2, 1, 29, 30, 43, 33, 0 • v = 41: 32, 4, 3, 2, 1, 29, 28, 27, 42, 43, 30, 31, 26, 25, 24, 23, 22, 17, 18, 19, 20, 21, 40, 39, 38, 37, 12, 13, 14, 15, 16, 11, 10, 5, 6, 7, 8, 9, 36, 35, 34, 33, 0 • v = 42: 32, 31, 26, 27, 28, 23, 22, 21, 20, 25, 24, 41, 40, 39, 38, 15, 14, 19, 18, 17, 16, 11, 10, 9, 8, 13, 12, 37, 36, 35, 34, 3, 4, 5, 6, 7, 2, 1, 29, 30, 43, 33, 0 • v = 43: 32, 4, 3, 2, 1, 29, 30, 31, 26, 25, 24, 23, 28, 27, 42, 41, 40, 39, 18, 17, 22, 21, 20, 19, 14, 13, 12, 37, 38, 15, 16, 11, 10, 5, 6, 7, 8, 9, 36, 35, 34, 33, 0 Proof that b0c0 and b4c4 are suitable edges for J13 Figure 3 shows the flower snark J13 and here b0c0 and b4c4 correspond to the edges (0,38) and (11,12), respectively. The pairs (0,11), (0,12), (38,11), and (38,12) are good in J13 - {(0,38), (11,12)} due to the following hamiltonian paths, respectively: • 11, 10, 5, 6, 7, 8, 9, 42, 41, 40, 39, 51, 36, 35, 34, 29, 28, 23, 22, 17, 16, 15, 44, 43, 12, 13, 14, 19, 18, 45, 46, 21, 20, 25, 24, 47, 48, 27, 26, 31, 30, 49, 50, 33, 32, 37, 38, 4, 3, 2, 1, 0 • 12, 13, 14, 15, 16, 11, 10, 5, 4, 38, 37, 32, 31, 30, 29, 34, 33, 50, 49, 48, 47, 24, 23, 28, 27, 26, 25, 20, 19, 18, 17, 22, 21, 46, 45, 44, 43, 42, 9, 8, 7, 6, 41, 40, 3, 2, 1, 35, 36, 51, 39,0 • 11, 10, 5,4, 3, 2, 1, 0, 39, 40, 41, 6, 7, 8, 9,42,43, 12, 13, 14, 19, 18, 17, 16, 15, 44, 45, 46, 47, 24, 23, 22, 21, 20, 25, 26, 31, 30, 29, 28, 27, 48, 49, 50, 51, 36, 35, 34, 33, 32, 37, 38 • 12, 13, 8, 7, 2, 3, 4, 5, 6, 41, 40, 39, 0, 1, 35, 34, 29, 28, 23, 22, 17, 18, 45, 44, 43, 42, 9, 10, 11, 16, 15, 14, 19, 20, 21,46,47, 24, 25, 26, 27, 48, 49, 30, 31, 32, 33, 50, 51, 36, 37, 38 The pair of pairs ((0, 38), (11,12)) is good in J13 - {(0,38), (11,12)} due to the following two disjoint paths with end-vertices 0 and 38, and 11 and 12, respectively, which together span J13. • 38, 37, 32, 31, 26, 25, 20, 19, 14, 13, 8, 7, 2, 1, 0 • 12, 43, 44, 15, 16, 17, 18, 45, 46, 21, 22, 23, 24, 47, 48, 27, 28, 29, 30, 49, 50, 33, 34, 35, 36, 51, 39, 40, 3, 4, 5, 6,41, 42, 9, 10, 11 The following hamiltonian paths show that at least one of (0,38) or (11,12) is good in J13 - {v, (0,38), (11,12)} for every v e V(J13). • v = 0: 12, 13, 14, 15, 44, 43, 42, 41, 6, 5, 10, 9, 8, 7, 2, 1, 35, 36, 37, 38, 4, 3, 40, 39, 51, 50, 49, 30, 29, 34, 33, 32, 31, 26, 25, 24, 23, 28, 27, 48, 47, 46, 45, 18, 19, 20,21,22, 17, 16, 11 • v = 1: 38, 4, 5, 6, 7, 2, 3, 40, 41, 42, 43, 12, 13, 8, 9, 10, 11, 16, 17, 18, 19, 14, 15, 44, 45, 46, 47, 24, 23, 22, 21, 20, 25, 26, 31, 30, 49, 48, 27, 28, 29, 34, 35, 36, 37, 32, 33, 50, 51, 39, 0 • v = 2: 12, 43, 44, 15, 14, 13, 8, 7, 6, 5, 10, 9, 42, 41, 40, 3, 4, 38, 37, 36, 35, 1, 0, 39, 51, 50, 49, 30, 29, 34, 33, 32, 31, 26, 25, 24, 23, 28, 27, 48, 47, 46, 45, 18, 19, 20,21,22, 17, 16, 11 • v = 3: 38, 4, 5, 6, 7, 2, 1, 35, 34, 33, 32, 37, 36, 51, 50, 49, 48, 27, 26, 31, 30, 29, 28, 23, 22, 21, 20, 25, 24, 47, 46, 45, 44, 15, 14, 19, 18, 17, 16, 11, 10, 9, 8, 13, 12, 43, 42, 41, 40, 39, 0 • v = 4: 38, 37, 32, 31, 30, 29, 34, 33, 50, 49, 48, 47, 24, 23, 28, 27, 26, 25, 20, 19, 18, 17, 22, 21, 46, 45, 44, 43, 12, 13, 14, 15, 16, 11, 10, 5, 6, 7, 8, 9, 42, 41, 40, 3, 2, 1,35, 36,51, 39,0 • v = 5: 38, 4, 3, 2, 1, 35, 34, 33, 32, 37, 36, 51, 50, 49, 48, 27, 26, 31, 30, 29, 28, 23, 22, 21, 20, 25, 24, 47, 46, 45, 44, 15, 14, 19, 18, 17, 16, 11, 10, 9, 42, 43, 12, 13, 8, 7, 6, 41, 40, 39, 0 • v = 6: 38, 37, 32, 31, 30, 29, 34, 33, 50, 49, 48, 47, 24, 23, 28, 27, 26, 25, 20, 19, 18, 17, 22, 21, 46, 45, 44, 43, 12, 13, 14, 15, 16, 11, 10, 5, 4, 3, 2, 7, 8, 9, 42, 41, 40, 39,51,36, 35, 1,0 • v = 7: 38, 4, 5, 6, 41, 42, 43, 12, 13, 8, 9, 10, 11, 16, 17, 18, 19, 14, 15, 44, 45, 46, 47, 24, 23, 22, 21, 20, 25, 26, 31, 30, 29, 28, 27, 48, 49, 50, 51, 36, 37, 32, 33, 34, 35, 1, 2, 3, 40, 39, 0 • v = 8: 38, 37, 32, 31, 30, 29, 34, 33, 50, 49, 48, 47, 24, 23, 28, 27, 26, 25, 20, 19, 18, 17, 22, 21, 46, 45, 44, 43, 12, 13, 14, 15, 16, 11, 10, 9, 42, 41, 40, 3, 4, 5, 6, 7, 2, 1,35, 36,51, 39,0 • v = 9: 38, 4, 3, 2, 1, 35, 34, 33, 32, 37, 36, 51, 50, 49, 48, 27, 26, 31, 30, 29, 28, 23, 22, 21, 20, 25, 24, 47, 46, 45, 44, 15, 14, 19, 18, 17, 16, 11, 10, 5, 6, 7, 8, 13, 12, 43, 42, 41, 40, 39, 0 • v = 10: 12, 13, 14, 15, 44, 43, 42, 9, 8, 7, 2, 3, 40, 41, 6, 5, 4, 38, 37, 36, 35, 1, 0, 39, 51, 50, 49, 30, 29, 34, 33, 32, 31, 26, 25, 24, 23, 28, 27, 48, 47, 46, 45, 18, 19, 20,21,22, 17, 16, 11 • v = 11: 38, 4, 3, 2, 1, 35, 34, 33, 32, 37, 36, 51, 50, 49, 48, 27, 26, 31, 30, 29, 28, 23, 22, 21, 46, 47, 24, 25, 20, 19, 14, 15, 16, 17, 18, 45, 44, 43, 12, 13, 8, 7, 6, 5, 10, 9, 42,41, 40, 39, 0 • v = 12: 38, 37, 32, 31, 30, 29, 34, 33, 50, 49, 48, 47, 24, 23, 28, 27, 26, 25, 20, 19, 18, 17, 22, 21, 46, 45, 44, 43, 42, 9, 8, 13, 14, 15, 16, 11, 10, 5, 4, 3, 2, 7, 6, 41, 40, 39, 51, 36, 35, 1, 0 • v = 13: 12, 43, 42, 41, 6, 5, 10, 9, 8, 7, 2, 1, 0, 39, 40, 3, 4, 38, 37, 32, 33, 34, 35, 36, 51, 50, 49, 48, 27, 26, 31, 30, 29, 28, 23, 22, 21, 20, 25, 24, 47, 46, 45, 44, 15, 14, 19, 18, 17, 16, 11 • v = 14: 38, 37, 32, 31, 30, 29, 34, 33, 50, 49, 48, 47, 24, 23, 28, 27, 26, 25, 20, 19, 18, 17, 22, 21, 46, 45, 44, 15, 16, 11, 10, 9, 8, 13, 12, 43, 42, 41, 40, 3, 4, 5, 6, 7, 2, 1, 35, 36, 51, 39, 0 • v = 15: 38, 4, 3, 2, 1, 35, 34, 33, 32, 37, 36, 51, 50, 49, 48, 27, 26, 31, 30, 29, 28, 23, 22, 21, 20, 25, 24, 47, 46, 45, 44, 43, 12, 13, 14, 19, 18, 17, 16, 11, 10, 5, 6, 7, 8, 9, 42,41, 40, 39, 0 • v = 16: 12, 13, 8, 7, 2, 3, 40, 41, 6, 5, 4, 38, 37, 36, 35, 1, 0, 39, 51, 50, 49, 30, 29, 34, 33, 32, 31, 26, 25, 24, 23, 28, 27, 48, 47, 46, 45, 18, 17, 22, 21, 20, 19, 14, 15, 44, 43, 42, 9, 10, 11 • v = 17: 38, 4, 3, 2, 1, 35, 34, 33, 32, 37, 36, 51, 50, 49, 48, 27, 26, 31, 30, 29, 28, 23, 22, 21, 20, 25, 24, 47, 46, 45, 18, 19, 14, 13, 12, 43, 44, 15, 16, 11, 10, 5, 6, 7, 8, 9, 42,41, 40, 39, 0 • v = 18: 38, 37, 32, 31, 30, 29, 34, 33, 50, 49, 48, 47, 24, 23, 28, 27, 26, 25, 20, 19, 14, 15, 44, 45, 46, 21, 22, 17, 16, 11, 10, 9, 8, 13, 12, 43, 42, 41, 40, 3, 4, 5, 6, 7, 2, 1, 35, 36, 51, 39, 0 • v = 19: 38, 4, 3, 2, 1, 35, 34, 33, 32, 37, 36, 51, 50, 49, 48, 27, 28, 29, 30, 31, 26, 25, 20, 21, 46, 47, 24, 23, 22, 17, 18, 45, 44, 43, 12, 13, 14, 15, 16, 11, 10, 5, 6, 7, 8, 9, 42,41, 40, 39, 0 • v = 20: 38, 37, 32, 31, 30, 29, 34, 33, 50, 49, 48, 47, 24, 25, 26, 27, 28, 23, 22, 21, 46, 45, 44, 15, 14, 19, 18, 17, 16, 11, 10, 9, 8, 13, 12, 43, 42, 41, 40, 3, 4, 5, 6, 7, 2, 1, 35, 36, 51, 39, 0 • v = 21: 38, 4, 3, 2, 1, 35, 34, 33, 32, 37, 36, 51, 50, 49, 48, 27, 26, 31, 30, 29, 28, 23, 22, 17, 18, 19, 20, 25, 24, 47, 46, 45, 44, 43, 12, 13, 14, 15, 16, 11, 10, 5, 6, 7, 8, 9, 42,41, 40, 39, 0 • v = 22: 38, 37, 32, 31, 30, 29, 34, 33, 50, 49, 48, 47, 24, 23, 28, 27, 26, 25, 20, 21, 46, 45, 44, 15, 14, 19, 18, 17, 16, 11, 10, 9, 8, 13, 12, 43, 42, 41, 40, 3, 4, 5, 6, 7, 2, 1, 35, 36, 51, 39, 0 • v = 23: 38, 4, 3, 2, 1, 35, 34, 33, 32, 37, 36, 51, 50, 49, 48, 27, 28, 29, 30, 31, 26, 25, 24, 47, 46, 45, 18, 17, 22, 21, 20, 19, 14, 13, 12, 43, 44, 15, 16, 11, 10, 5, 6, 7, 8, 9, 42,41, 40, 39, 0 • v = 24: 38, 37, 32, 31, 30, 29, 34, 33, 50, 49, 48, 47, 46, 21, 22, 23, 28, 27, 26, 25, 20, 19, 14, 15, 44, 45, 18, 17, 16, 11, 10, 9, 8, 13, 12, 43, 42, 41, 40, 3, 4, 5, 6, 7, 2, 1, 35, 36, 51, 39, 0 • v = 25: 38, 4, 3, 2, 1, 35, 34, 33, 32, 37, 36, 51, 50, 49, 48, 27, 26, 31, 30, 29, 28, 23, 24, 47, 46, 45, 18, 17, 22, 21, 20, 19, 14, 13, 12, 43, 44, 15, 16, 11, 10, 5, 6, 7, 8, 9,42, 41, 40, 39, 0 • v = 26: 38, 37, 32, 31, 30, 29, 34, 33, 50, 49, 48, 27, 28, 23, 22, 21, 20, 25, 24, 47, 46, 45, 44, 15, 14, 19, 18, 17, 16, 11, 10, 9, 8, 13, 12, 43, 42, 41, 40, 3, 4, 5, 6, 7, 2, 1,35, 36,51, 39,0 • v = 27: 38, 4, 3, 2, 1, 35, 34, 33, 32, 37, 36, 51, 50, 49, 48, 47, 24, 23, 28, 29, 30, 31, 26, 25, 20, 19, 18, 17, 22, 21,46,45,44,43, 12, 13, 14, 15, 16, 11, 10, 5, 6, 7, 8, 9,42, 41, 40, 39, 0 • v = 28: 38, 37, 32, 31, 30, 29, 34, 33, 50, 49, 48, 27, 26, 25, 20, 21, 22, 23, 24, 47, 46, 45, 44, 15, 14, 19, 18, 17, 16, 11, 10, 9, 8, 13, 12, 43, 42, 41, 40, 3, 4, 5, 6, 7, 2, 1,35, 36,51, 39,0 • v = 29: 38, 4, 3, 2, 1, 35, 34, 33, 32, 37, 36, 51, 50, 49, 30, 31, 26, 25, 24, 23, 28, 27, 48, 47, 46, 45, 18, 17, 22, 21, 20, 19, 14, 13, 12, 43, 44, 15, 16, 11, 10, 5, 6, 7, 8, 9,42, 41, 40, 39, 0 • v = 30: 38, 37, 32, 31, 26, 27, 28, 29, 34, 33, 50, 49, 48, 47, 46, 21, 22, 23, 24, 25, 20, 19, 14, 15, 44, 45, 18, 17, 16, 11, 10, 9, 8, 13, 12, 43, 42, 41, 40, 3, 4, 5, 6, 7, 2, 1,35, 36,51, 39,0 • v = 31: 38, 4, 3, 2, 1, 35, 34, 33, 32, 37, 36, 51, 50, 49, 30, 29, 28, 23, 24, 25, 26, 27, 48, 47, 46, 45, 18, 17, 22, 21, 20, 19, 14, 13, 12, 43, 44, 15, 16, 11, 10, 5, 6, 7, 8, 9,42, 41, 40, 39, 0 • v = 32: 38, 37, 36, 35, 1, 2, 7, 6, 5, 4, 3, 40, 41, 42, 43, 12, 13, 8, 9, 10, 11, 16, 17, 18, 19, 14, 15, 44, 45, 46, 47, 24, 23, 22, 21, 20, 25, 26, 31, 30, 49, 48, 27, 28, 29, 34, 33, 50, 51, 39, 0 • v = 33: 38, 4, 3, 2, 1, 35, 34, 29, 30, 31, 32, 37, 36, 51, 50, 49, 48, 47, 24, 23, 28, 27, 26, 25, 20, 19, 18, 17, 22, 21, 46, 45, 44, 43, 12, 13, 14, 15, 16, 11, 10, 5, 6, 7, 8, 9,42, 41, 40, 39, 0 • v = 34: 38, 37, 32, 33, 50, 49, 48, 27, 26, 31, 30, 29, 28, 23, 22, 21, 20, 25, 24, 47, 46, 45, 44, 15, 14, 19, 18, 17, 16, 11, 10, 9, 8, 13, 12, 43, 42, 41, 40, 3, 4, 5, 6, 7, 2, 1,35, 36,51, 39,0 • v = 35: 38, 4, 3, 40, 39, 51, 36, 37, 32, 31, 30, 29, 34, 33, 50,49,48, 47, 24, 23, 28, 27, 26, 25, 20, 19, 18, 17, 22, 21, 46, 45, 44, 43, 12, 13, 14, 15, 16, 11, 10, 5, 6, 41, 42, 9, 8, 7, 2, 1, 0 • v = 36: 38, 37, 32, 31, 26, 27, 28, 29, 30, 49, 48, 47, 46, 21, 22, 23, 24, 25, 20, 19, 14, 15, 44, 45, 18, 17, 16, 11, 10, 9, 8, 13, 12, 43, 42, 41, 40, 3, 4, 5, 6, 7, 2, 1, 35, 34, 33, 50, 51, 39, 0 • v = 37: 38, 4, 3, 2, 1, 35, 36, 51, 50, 49, 30, 29, 34, 33, 32, 31, 26, 25, 24, 23, 28, 27, 48, 47, 46, 45, 18, 17, 22, 21, 20, 19, 14, 13, 12, 43, 44, 15, 16, 11, 10, 5, 6, 7, 8, 9,42, 41, 40, 39, 0 • v = 38: 12, 13, 8, 7, 2, 3, 4, 5, 6, 41, 40, 39, 0, 1, 35, 34, 33, 32, 37, 36, 51, 50, 49, 48, 27, 26, 31, 30, 29, 28, 23, 22, 21, 46, 47, 24, 25, 20, 19, 14, 15, 16, 17, 18, 45, 44, 43,42, 9, 10, 11 • v = 39: 38, 4, 5, 6, 7, 2, 3, 40,41,42, 43, 12, 13, 8, 9, 10, 11, 16, 17, 18, 19, 14, 15, 44, 45, 46, 47, 24, 23, 22, 21, 20, 25, 26, 31, 30, 29, 28, 27, 48, 49, 50, 51, 36, 37, 32, 33, 34, 35, 1, 0 • v = 40: 12, 13, 14, 15, 44, 43, 42, 41, 6, 5, 10, 9, 8, 7, 2, 3, 4, 38, 37, 36, 35, 1, 0, 39, 51, 50, 49, 30, 29, 34, 33, 32, 31, 26, 25, 24, 23, 28, 27, 48, 47, 46, 45, 18, 19, 20,21,22, 17, 16, 11 • v = 41: 38, 4, 5, 6, 7, 8, 13, 12, 43, 42, 9, 10, 11, 16, 17, 18, 19, 14, 15, 44, 45, 46, 47, 24, 23, 22, 21, 20, 25, 26, 31, 30, 29, 28, 27, 48, 49, 50, 51, 36, 37, 32, 33, 34, 35, 1, 2, 3,40, 39, 0 • v = 42: 38, 37, 32, 31, 30, 29, 34, 33, 50, 49, 48, 47, 24, 23, 28, 27, 26, 25, 20, 19, 18, 17, 22, 21, 46, 45, 44, 43, 12, 13, 14, 15, 16, 11, 10, 9, 8, 7, 2, 3, 4, 5, 6, 41, 40, 39, 51, 36, 35, 1, 0 • v = 43: 12, 13, 8, 7, 6, 5, 10, 9, 42, 41, 40, 39, 0, 1, 2, 3, 4, 38, 37, 32, 33, 34, 35, 36, 51, 50, 49, 48, 27, 26, 31, 30, 29, 28, 23, 22, 21, 20, 25, 24, 47, 46, 45, 44, 15, 14, 19, 18, 17, 16, 11 • v = 44: 38, 37, 32, 31, 30, 29, 34, 33, 50, 49, 48, 47, 24, 25, 26, 27, 28, 23, 22, 17, 18, 45, 46, 21, 20, 19, 14, 15, 16, 11, 10, 9, 8, 13, 12, 43, 42, 41, 40, 3, 4, 5, 6, 7, 2, 1, 35, 36, 51, 39, 0 • v = 45: 38, 4, 3, 2, 1, 35, 34, 33, 32, 37, 36, 51, 50, 49, 48, 27, 28, 29, 30, 31, 26, 25, 20, 21, 46, 47, 24, 23, 22, 17, 18, 19, 14, 13, 12, 43, 44, 15, 16, 11, 10, 5, 6, 7, 8, 9, 42,41, 40, 39, 0 • v = 46: 38, 37, 32, 31, 30, 29, 34, 33, 50, 49, 48, 47, 24, 25, 26, 27, 28, 23, 22, 21, 20, 19, 14, 15, 44, 45, 18, 17, 16, 11, 10, 9, 8, 13, 12, 43, 42, 41, 40, 3, 4, 5, 6, 7, 2, 1, 35, 36, 51, 39, 0 • v = 47: 38, 4, 3, 2, 1, 35, 34, 33, 32, 37, 36, 51, 50, 49, 48, 27, 26, 31, 30, 29, 28, 23, 24, 25, 20, 19, 18, 17, 22, 21, 46, 45, 44, 43, 12, 13, 14, 15, 16, 11, 10, 5, 6, 7, 8, 9, 42,41, 40, 39, 0 • v = 48: 38, 37, 32, 31, 30, 49, 50, 33, 34, 29, 28, 27, 26, 25, 20, 21, 22, 23, 24, 47, 46, 45, 44, 15, 14, 19, 18, 17, 16, 11, 10, 9, 8, 13, 12, 43, 42, 41, 40, 3, 4, 5, 6, 7, 2, 1, 35, 36, 51, 39, 0 • v = 49: 38, 4, 3, 2, 1, 35, 34, 33, 50, 51, 36, 37, 32, 31, 30, 29, 28, 23, 24, 25, 26, 27, 48, 47, 46, 45, 18, 17, 22, 21, 20, 19, 14, 13, 12, 43, 44, 15, 16, 11, 10, 5, 6, 7, 8, 9, 42,41, 40, 39, 0 • v = 50: 38, 37, 32, 33, 34, 29, 28, 27, 26, 31, 30, 49, 48, 47, 46, 21, 22, 23, 24, 25, 20, 19, 14, 15, 44, 45, 18, 17, 16, 11, 10, 9, 8, 13, 12, 43, 42, 41, 40, 3, 4, 5, 6, 7, 2, 1, 35, 36, 51, 39, 0 • v = 51: 38, 4, 3, 2, 1, 35, 36, 37, 32, 31, 30, 29, 34, 33, 50, 49, 48, 47, 24, 23, 28, 27, 26, 25, 20, 19, 18, 17, 22, 21, 46, 45, 44, 43, 12, 13, 14, 15, 16, 11, 10, 5, 6, 7, 8, 9, 42,41, 40, 39, 0 ARS MATHEMATICA CONTEMPORANEA ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 14 (2018) 251-265 https://doi.org/10.26493/1855-3974.1285.f3c (Also available at http://amc-journal.eu) Inherited unitals in Moulton planes * Gábor Korchmáros , Angelo Sonnino Dipartimento di Matematica, Informática ed Economia Universita degli Studi della Basilicata Viale dell'Ateneo Lucano 10, 85100 Potenza, Italy Tamás Szonyi ELTE Eotvos Loránd University, Institute of Mathematics and MTA-ELTE Geometric and Algebraic Combinatorics Research Group H-1117 Budapest, Pázmany P. s. 1/c, Hungary Received 11 January 2017, accepted 24 July 2017, published online 4 September 2017 We prove that every Moulton plane of odd order—by duality every generalised Andre plane—contains a unital. We conjecture that such unitals are non-classical, that is, they are not isomorphic, as designs, to the Hermitian unital. We prove our conjecture for Moulton planes which differ from PG(2, q2) by a relatively small number of point-line incidences. Up to duality, our results extend previous analogous results—due to Barwick and Gruning—concerning inherited unitals in Hall planes. Keywords: Unitals, Moulton planes. Math. Subj. Class.: 51E20, 05B25 1 Introduction A unital is a set of q3+1 points together with a family of subsets, each of size q+1, such that every pair of distinct points are contained in exactly one subset of the family. Such subsets are usually called blocks so that unitals are block-designs 2-(q3+1, q +1, 1). The classical example of a unital arises from the unitary polarity in the Desarguesian projective plane PG(2, q2) where the points are the absolute points, and the blocks are the non-absolute lines of the unitary polarity. The name of "Hermitian unital" is commonly used for the * This research was carried out within the activities of the GNSAGA of the Italian INdAM. E-mail address: gabor.korchmaros@unibas.it (Gabor Korchmaros), angelo.sonnino@unibas.it (Angelo Sonnino), szonyi@cs.elte.hu (Tamas Szonyi) Abstract ©® This work is licensed under http://creativecommons.org/licenses/by/3.0/ classical example since the absolute points of the unitary polarity are the points of the Hermitian curve defined over GF(q2). A unital U is embedded in a projective plane n of order q2, if its points are points of n and its blocks are intersections with lines. As usual, we adopt the term "chord" to indicate a block of U. Aline I of n is either a tangent or a (q +1)-secantto U according as nU| = 1 or nU| = q + 1, and in the latter case I n U is a chord. Examples of unitals embedded in PG(2, q2) other than the Hermitian ones are known to exist. A unital is classical if it is isomorphic, as a block-design, to a Hermitian unital. Classical unitals contain no O'Nan configurations, and it has been conjectured that any non-classical unital embedded in PG(2, q2) must contain a O'Nan configuration. In several families of non-desarguesian planes, the problem of constructing and characterizing unitals has also been investigated; see [1, 2, 5, 6, 8, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 22, 23, 24, 27, 28]. Apart from the examples of unitals arising from a unitary polarity in a commutative semifield plane, the known examples are inherited unitals from the Hermitian unital. In a non-desarguesian plane n of order q2 arising from PG(2, q2) by altering some of the point-line incidences, the adjective "inherited" is used for those pointsets of PG(2, q2) which keep their intersection properties with lines when moving from PG(2, q2) to n. In this paper we construct inherited unitals in Moulton planes of odd order q2, and, by duality, in generalised Andre planes of the same order; see Theorem 3.1. We also investigate the problem whether these unitals are classical; see Theorems 3.5 and 3.6. We show that if such a plane differs from PG(2, q2) by a relatively small number of incidences only, then the inherited unital is non-classical. Also, we exhibit non-classical inherited unitals in case of many point-line incidence alterations. Such unitals appear to be of interest in coding theory; see [25]. What emerges from our work leads us to conjecture that the inherited unitals constructed in our paper are all non-classical. It should be noticed that our results extend previous analogous results due to Barwick and Griming concerning inherited unitals in Hall planes which are very special Andre planes; see [8, 16] and Remark 3.4. The methods used in [8] are mostly geometric and involve Baer subplanes and blocking sets. In this paper, we adopt a more algebraic approach that allows us to exploit results on the number of solutions of systems of polynomial equations over a finite field. 2 Two new results on the Hermitian unital We establish and prove two theorems on Hermitian unitals that will play a role in our study on unitals in Moulton planes. Up to a change of the homogeneous coordinate system (X1, X2, X3) in PG(2, q2), the points of the classical unital U are those satisfying the equation + X2q+1 + X3q+1 = 0. (2.1) In the affine plane AG(2, q2) arising from PG(2, q2) with respect to the line X3 = 0, we use the coordinates (X, Y) where X = X1/X3 and Y = X2/X3). Then the points of U in AG(2, q2) are the solutions of the equation X 9+1 + Y9+1 + 1=0. (2.2) Since GF(q2) is the quadratic extension of GF(q) by adjunction of a root i of the polynomial X2 — s with a non-square element s of GF(q), every element u of GF(q2) can uniquely be written as u = u + iu2 with u^u2 € GF(q). Then uq = u — iu2 and ||u|| = uq+1 = ui — su2. Therefore, the points P(x, y) € U lying in AG(2, q2) are those satisfying the equation x2 — sx2 + y2 — sy2 + 1=0. (2.3) For a subset T C GF(q) \ {0}, let St denote the set of points { (x, y) | ||x|| = t € T }. Hence the pointset St n U comprises all points P(x, y) such that both x2 — sx2 = t and (2.3) hold. Therefore, a point P (x, y) € AG(2,q2) is in St nU if and only if P1(x1,x2) € AG(2, q) lies on the non-degenerate conic C1 : X2 — sY2 — t = 0 while P2(y1 ,y2) € AG(2, q) does lie on the conic C2 : X2 — sY2 + 1 + t = 0. This shows that St n U has size (q + 1)2 apart from the case t = —1 when it consists of the q + 1 points of U lying on the X -axis. Lemma 2.1. Let i be a non-vertical line in AG(2, q2). Then |i nUnSt| € {0,1,2, q +1} for every t € T .If q +1 occurs then i is either a horizontal line, or it passes through the origin. Proof. The points P (x, 0) with ||x|| = t form a Baer subline. As U is classical, i nU is a Baer subline of i, and hence the projection of inU on the X-axis from is a Baer-subline, as well. Since two distinct Baer sublines have at most two common points, the first assertion follows. To prove the second one, we need some computation. If i has equation Y = Xm+ b, we have to count the roots x of the polynomial f (X) = Xq+1 + (Xm + b)q+1 + 1 whose norm ||x|| is equal to t. If ||x|| = t, then f (x) = bmqxq + bqmx +1(1 + mq+1) + bq+1 + 1 and hence xf (x) = bq mx2 + (t(1 + mq+1) + bq+1 + 1)x + bmq t. If we have at least three such roots x then either m = 0 and t + 1 = —bq+1, or b = 0 and t(1 + mq+1) = —1. □ Take any two distinct non-tangent lines i1 and i2 of U. We are interested in the intersection of the projection of i1 n U from P on i2 with i2 n U. For any point P outside both i1 and i2, the projection of i1 to i2 from P takes the chord i1 n U to a Baer subline of i2. Since two Baer sublines of i2 intersect in 0,1, 2 or q +1 points, one may want to determine the size of the sets (i = 0,1, 2, q +1) consisting of all points P for which this intersection number is equal to i. The points in are called elliptic, parabolic, hyperbolic, or full with respect to the pair (i1, i2), according as i = 0, i = 1, i = 2, or i = q + 1, respectively; see [21]. We go on to compute the size of n U. Since the linear collineation group G = PGU(3, q) of PG(2, q2) preserving U acts transitively on the points outside U, we may assume that = i1 n i2. The stabiliser of in G acts on the pencil with center in as the general projective group PGL(2, q) on the projective line PG(1, q2). Therefore, it has two orbits, one consisting of all tangents the other of all chords to U through YTO. This allows us to assume without loss of generality that i1 is the line at infinity. Since i2 is not a tangent to U, its equation is of the form X = c with cq+1 + 1=0. Therefore, cq+1 + 1 is either a non-zero square or a non-square element of GF(q). These two cases occur depending upon whether a linear collineation 7 € PGL(2, q) taking i1 to i2 is in the subgroup isomorphic to the special projective group PSL(2, q) or not. Accordingly, {i1 , i2 } is called a special pair or a general pair. Further, since P is a point outside i1 and i2, it is an affine point P = (a, b) with a = c. ¿2 \ Y = 0 X = 0 X = c Figure 1: The initial configuration. Let P = (a, b) denote a point of U, that is, aq+i + bq+i + 1 = 0. (2.4) Takealine r of equation Y = m(X - a) + b through P = (a, b). A necessary and sufficient condition for r to meet both ^ and in U is the existence of a solution t g GF(q2) of the system consisting of (2.4) together with Cq+1 + t q+l + 1 = mq+1 + 1 = 0. (2.5) (2.6) In fact, Q(c, t) with t = m(c — a) + b is the point of r on ¿2. Then (2.5) holds if and only if Q G U. Furthermore, (2.6) is the necessary and sufficient condition for the infinite point of r to be in U; see Figure 1. The above discussion also shows how to count lines through P meeting both n U and n U. Essentially, one has to find the number of solutions in the indeterminate t of the system consisting of the equations (2.4), (2.5), and (2.6). Observe that (2.4), (2.5), (2.6) are equivalent to a2 - sa2 + &2 - sb2 + 1 = 0, c2 - sc2 + rf - sr22 + 1 = 0, biTi - S&2T2 + aici - sa2C2 + 1 = 0. (2.7) (2.8) (2.9) ¿1 — ¿CO 0 From this the following result is obtained. Proposition 2.2. The number of lines through P meeting both ¿1 nU and ¿2 nU equals the number of solutions (r1, r2), with r1, r2 G GF(q), of the system consisting of (2.7), (2.8), (2.9). In investigating the above system, two cases are distinguished according as (61,62) is (0,0) or not. In the former case, Equations (2.7) and (2.9) read a2 - sa2 + 1 = 0 and a1c1 - sa2c2 + 1 = 0. Geometrically in AG(2, q), the point U = (a1, a2) is the intersection of the ellipse E, with equation X2 — sY2 + 1 = 0, and the line v with equation c1X — sc2Y + 1 = 0. Since cq+1 + 1 = c2 — sc2 + 1 is a non-zero element of GF(q), v must be either a secant, or an external line to E and this occurs according as c2 — sd^ + 1 is a non-zero square or non-square element in GF(q). In fact, from (2.7) and (2.9), sc2&2 — 1 —sc2 ± ic1\Jc2 — sc2 + 1 a1 = -, a2 = -T2-^-. c1 s ( c12 — sc22 ) Therefore, if P is on the X-axis, then P is elliptic in general, apart from the case where cq+1 +1 = c2 — sc2 + 1 is a non-square element in GF(q) and P is one of the two common points of C and v, namely P = P(a, 0) where . —1 ± V1+ c«+1 a = a1 + «a2 = -. cq Further, in the exceptional case, P is a full point as for any c1, c2 G GF(q) with c1 — sc2 + 1 = 0, Equation (2.8) always has q +1 solutions (t1, t2) with t1, t2 g GF(q). In the latter case, either 61 or 62 is not zero. If 61 = 0, retrieving t1 from (2.9) and putting it in (2.8) gives a quadratic equation in the indeterminate t2, namely (s2 62 — s62)t22 — 2s62(a1c1 — sa2c2 + 1)t2 + (a1c1 — sa2c2 + 1)2 + 62(1 + c2 — sc2) = 0, (2.10) whose discriminant is A1 = s62 A with A = (62 — s62)(1 + c2 — sc2) + (a1c1 — sa2c2 + 1)2 which can also be written by (2.7) as A = —(1 + c2 — sc2)(a2 — sa2 + 1) + (a^ — sa2c2 + 1)2. For 62 = 0, retrieving t2 from (2.9) and putting it in (2.8) gives the following quadratic equation in the indeterminate t1: (—61 + 62)tx2 + 2a161c1T1 — a1 — s2a2c2 — 62 c1 + s62c2 + 2sa2c2 — 62 — 1 = 0 (2.11) with discriminant A2 = s362A. Since A1 and A2 are simultaneously zero, or a square, or anon-square in GF(q), each of the equations (2.10) and (2.11) has 2,1 or zero solutions in GF(q), depending upon whether A is a square element, zero, or a non-square element of GF(q), respectively. This leads to the study of the zeroes of the polynomial F (X,Y,Z) = —(1 + c1 — sc2)(X2 — sY2 + 1) + (c1X — sc2Y + 1)2 — Z2. (2.12) Geometrically, F(X, Y, Z) = 0 is the equation of a quadric Q in AG(3, q). Actually, Q Table 1: Elliptic, parabolic, hyperbolic and full points. P(a, 0) P(a,b), b = 0 1± l|c|| G □ 1± l|c||G □ 1± ||c|| G □ 1± |c|G □ N£ i ±1 i — 1 —3 — 9q ± q2 ± q3 —3 — 5q — q2 ± q3 2 2 NP 0 0 2q — 1 0 NH 0 0 (i —21)2 (q ±1) (q ±21)2 (q— 1) Nf 0 2 0 0 is a cone. In fact, the system FX = FY = FZ = 0 has a (unique) solution (c1,c2,0) and hence the point V(c1,c2,0) is the vertex of Q. In particular, the intersection of Q with the plane Z = 0 splits into two lines over GF(q) or its quadratic extension GF(q2), and this occurs according as the infinite points of the conic with equation -(1 + c2 - sc2)(X2 - sY2) + (c1X - sc2Y)2 = 0 lie in PG(2, q) or in PG(2, q2) \ PG(2, q). By a direct computation, this condition only depends on cq+1, namely whether 1 + cq+1 is a square or a non-square element of GF(q). Therefore, Q contains either 2q-1 or 1 points in the plane Z = 0, and this occurs according as the pair [£1, l2} is special or general. Also, in the former case there are exactly 2q - 1 parabolic points P but in the latter case no point P is parabolic. Therefore, the following result holds. Theorem 2.3. Let i1, t2 be any two distinct non-tangent lines of the classical unital U in PG(2, q2) whose common point is off U. The number NE,Np,N%,NS, of elliptic, parabolic, hyperbolic and full points of U with respect to the pair {£1,£2} is given in Table 1. We state a corollary of Theorem 2.3 that will be used in Section 3. For i = 1,2 let Aj be a subset of ^ n U such that | A11 = | A21 = A. Theorem 2.4. If x>yj (i±M (2.13) there exists a non-degenerate quadrangle AiBiA2B2 with vertices Ai, Bi G Aj for i = 1,2 such that its diagonal point AiB2 n BiA2 lies in U. Proof. We prove the existence of a hyperbolic point D in U such that the projection of Ai from D on l2 share two points with A2. From Theorem 2.3, we have at least 2 (q— 1)2 (q±1) hyperbolic points in U. We omit those hyperbolic points projecting Ai = (£i n U) \ Ai to a pointset of l2 meeting l2 n U nontrivially. The number of such hyperbolic points is A(q - 1)(q + 1) with A = q + 1 - A. Similarly we omit all hyperbolic points projecting A2 = (¿2 nM) \ A2 to apointset of ¿1 meeting ¿4 n U nontrivially. Therefore, the total number of omitted hyperbolic points is 2A(q2-1)-A2(q-1) = (q -1)A(2q+2-A(q-1)). From Theorem 2.3, this number is smaller than the total number of hyperbolic points as far as (2.13) holds. □ To state the other new result on the classical unital a couple of ad hoc notation in AG(2, q2) will be useful: For a non-vertical line ¿ with equation Y = Xm + b, A denotes the non-vertical line with equation Y = Xmq + b. Given a point P(a, b) outside U, two lines ¿1 and ¿2 are said to be a good line-pair whenever the lines ¿1 and ¿2 meet in a point of U. Our goal is to show that if a = 0 then there exist many good pairs. For i = 1, 2, write the equations of ¿ in the form Y = (X - a)m¿ + b. Then ¿i has equation Y = Xm? - am¿ + b. Hence P(x, y) = ¿A n ¿2 where a(mi - m.2) and hence a(mi - m2) q -q-q— mi — ami + b. Note that lldl = xq+1 = aq+1 1 (mi — m2)q 1 q+i (mi — m-2 )q' 2- = llaH = 0. The condition for P(x, y) to lie in W is xq+i + yq+i + 1 = aq+i + a' q+i + yq+i + 1 = aq+i + aq+^ i^l_mi V (mi — m2)q q+1 m? — mi +— ) +1=0. 1a Let Then the last equation reads e = — aq+i + 1 aq+i e GF(q). (mi — m2) q --^-mi (mi — m2)q b\ q+i mi — mi +— = e. 1a (2.14) Henceforth we assume that With Ml = —1. mi = a + iß, m2 = 7 + ¿5, — = u + iv, a (2.14) reads (a — 7)+ i(ß — 5) (a — y) — i(ß — 5) (a — iß) — (a + iß) + u + iv q+1 e, qq mm 1 2 y 1 2 a whence (ua — «y — svfi + svJ)2 — s(2fiY — 2aJ — u(fi — J) + v(a — y))2 - C((a — y)2 — s(fi — J)2) = 0, that is, (u(a — y ) — sv(fi — J))2 — s(2fiY — 2aJ — u(fi — J) + v(a — y))2 — ^((a — Y)2 — s(fi — J)2)=0. (2.15) With _ Y = a — Y, J = fi — J, Equation (2.15) becomes («Y — svJ)2 — s(—2fiY — 2aJ — «J + vY )2 — C(Y2 — sJ2) = 0, (2.16) which can be viewed as a quadratic form in y and J: F (y, J) = («2 — v2s + 4vfis — 4fi2s — £) y2 + 2(—2ufis + 2vas — 4afis) YJ + (—u2 s — 4«as + v2 s2 — 4a2 s + s£) J2 (2.17) with discriminant A = —u4s + 2u2v2s2 + 2u2s£ — v4s3 — 2v2s2£ — s£2 + (—4u3s + 4«v2s2 + 4us£) a + (— 4«2vs2 + 4v3s3 + 4vs2£) fi — 8uvs2 afi + (—4u2s + 4s£) a2 + (—4v2s3 — 4s2£) fi2. Note that P(x, y) € U if and only if A = A2 for some A G GF(q). This leads us to consider the quadric Q in AG(3, q) of equation aoo + aoi X + a^Y + a^XY + an! 2 + a22Y2 — Z2 =0, a00 = —u4s + 2u2v2s2 + 2u2s£ — v4s3 — 2v2s2£ — s£2, aoi = —4«3s + 4uv2s2 + 4us£, a02 = —4u2vs2 + 4v3 s3 + 4vs2£, ai2 = —8uvs2, aii = —4u2s + 4s£, a22 = —4v2s3 — 4s2£. The above coefficients are related by the following equations: (i) aoo — 2(èaoi« — ao2v) = s£(u2 — sv2 — £); (ii) 2aoi — è(aii« — èai2v) = 0; where (iii) 2a02 - 1 (2ai2U - 022«) = 0. Therefore, the determinant D of the 4 x 4 matrix associated with Q is equal to -sC(u2 -sv2 - C) multiplied by the determinant of the cofactor of a00. The latter determinant ana22 - 4a22 is equal to Do = s3C(u2 - sv2 - C) = s3(aq+1 + bq+1 + 1)(aq+1 + 1). (2.18) It turns out that D = -(s2C(u2 - sv2 - C))2. Observe that C = 0 if and only if aq+1 = -1, while 2 ^ bq+1 aq+1 + 1 aq+1 + bq+1 + 1 u - sv - C = —tt +--n— =-n- vanishes only for P(a, b) G U. Therefore, Q is non-degenerate. More precisely, the quadric Q is either elliptic or hyperbolic according as q = -1 (mod 4) or q = 1 (mod 4). The plane at infinity cuts out from Q a conic C with homogeneous equation a11X2 + a12XY + a22 Y2 - Z2 =0. Observe that C is non-degenerate by D0 = 0. Thus, the number of points of Q in AG(3, q) is q2 ± q with q = ±1 (mod 4). Furthermore, the point at infinity on the Z-axis does not lie on Q, and it is an external point or an internal point to C according as -D0 is a non-zero square or a non-square in GF(q). Therefore, the number of tangents to Q through in AG(3, q) is equal to q - 1 or q +1 according as -D0 is a (non-zero) square or a non-square in GF(q). From the above discussion, the numbers Ns and Nt of secants and tangents to Q through are those given in the following lemma: Lemma 2.5. For q = -1 (mod 4), either Nt = q + 1, Ns = 1 (q - 1)2, or Nt = q -1, Ns = 1 (q2 - 2q -1), according as D0 is a (non-zero) square or a non-square in GF(q). For q = 1 (mod 4), either Nt = q - 1, Ns = 1 (q2 + 1), or Nt = q +1, Ns = 1 (q2 - 1), according as D0 is a (non-zero) square or a non-square in GF(q). Going back to the discriminant A, we see that A vanishes for Ns + Nt ordered pairs (a, that is, Ns + Nt is the number of lines ¿4 through P(a, b) for which there exists a line ¿2 such that (¿1, ¿2) is a good line-pair. For each counted in Nt (resp. Ns), we have q - 1 (resp. 2(q - 1)) such lines ¿2, since if (2.17) has a non-trivial solution (7, J) in GF(q) x GF(q) then it has exactly q - 1 solutions, the multiples of (7,7) by the non-zero elements of GF(q). If we do not count the q +1 tangents to U through P(a, b), each of the lines through P(a, b) counted in Ns is in at least 2(q - 1) - (q + 1) = q - 3 good line-pairs. Therefore Lemma 2.5 has the following corollary. Theorem 2.6. Let P(a, b) be a point of AG(2, q2) outside U. If a = 0, ||a|| = -1 and q > 3, then there exist at least two non-tangent lines , ¿2 of U through P, such that the non-tangent lines ¿1 and ¿2 meet in a point of U. Further, if q > 5 then and ¿2 may be chosen among the lines through P(a, b) other than the horizontal lines and those passing through the origin. 3 Unitals in Moulton planes Let T be a non-empty subset of the multiplicative group of GF(q). The (affine) Moulton plane MT (q2) which is considered in our paper is the affine plane coordinatized by the left quasifield GF(q2)(+, o) where Jxy if ||x||G T, x o y = < [xyq if ||x|| G T, with ||x|| = xq+1 being the norm of x G GF(q2) over GF(q). Geometrically, MT(q2) is constructed on AG(2, q2) by replacing the non-vertical lines with the graphs of the functions Y = X o m + b. (3.1) This also shows that to the non-vertical line £ of equation Y = Xm + b there corresponds the line of equation £ of equation Y = X o m + b in MT (q2), and viceversa. It is useful to look at the partition of the points outside the Y-axis into q - 1 subsets Si, called stripes, where P(x, y) G Si if and only if ||x|| = wi with w a fixed primitive element of GF(q). Such stripes were already defined in Section 2; here we just abbreviate the subscript wi by i. In fact, moving to MT(q2) the point-line incidences P G £ in AG(2, q2) do not alter as long as P G Si with wi G T. The projective Moulton plane is the projective closure of Mt(q2) and it has the same points at infinity as AG(2, q2). For a similar description of Moulton planes see also [3, 4, 26]. The dual of the Moulton plane is the Andre plane AT (q2 ) coordinatized by the right quasifield GF(q2)(+, *) where x * y xy if | x| G T, xqy if ||x|| G T. In this duality, the correspondence occurs between the point (u, v) of MT(q2) and the line of equation Y = u * X - v, as well as between the line of equation Y = X o m + b and the point (m, -b) of AT(q2). The correspondence between points at infinity and lines through Yto, and viceversa, is the same as the canonical duality between PG(2, q2) and its dual plane PG* (2, q2). If T consists of just one element, then the arising Andre planes are pairwise isomorphic and they are also known as Hall planes. Let U be the classical unital in PG(2, q2) given in its canonical form (2.1). We prove that U is an inherited unital in the Moulton plane, that is, the point-set of U is a unital in Mt(q2) as well. Theorem 3.1. Let U be the classical unital in PG(2, q2) given in its canonical form (2.1). Then, for any T, U is a unital in the projective Moulton plane MT (q2) as well. Proof. In the very special case T = {-1}, the proof is straightforward. It is enough to show that if a non-vertical line I of equation Y = Xm + b meets U in a point P(x, y) with ||x|| = —1 then y = 0 and x = —b/m with (—b/m)q+1 = 1. In fact, the corresponding line ¿ in Mt(q2) has the same property: if P(x, y) G í n U then y = 0 and x = (—b/mq)q+1. Since (-b/m)q+1 = (-b/mq)q+1, the assertion follows for T = {-1}. In the general case, it suffices to exhibit a bijective map from I n U to í n U for every line í of AG(2, q2). We may limit ourselves to non-vertical lines with non zero slopes. Let Y = Xm + b be the equation of such a line t and take any point P (x, y) lying in t nU. Then m = 0 and x = (y — b)m-1. Define the map ^: t ^ t by Obviously, ) = P whenever ||x|| G T. Since ^ is bijective, it suffices to show that P gM yields y(P) G U, and the converse also holds. P(x, y) = ((y - 6)m-1, y) G U if and only if ((y - 6)m-1)q+1 + yq+1 - 1 = (y - 6)q+1(m-1)q+1 + yq+1 - 1 = 0. By (mq)q+1 = mq+1, the latter equation is equivalent to Theorem 3.1 and its proof also show that if i is a tangent to U in AG(2, q2) then the corresponding line i is a tangent to U in the projective Moulton plane, and the converse also holds. In particular, the tangent to U at a point outside the X-axis is the line i of equation Y = X(-cd-1)q - d-q with tangency point P(c, d). Therefore, the corresponding line i of equation Y = X o (-cd-1)q - d-q is a tangent to U at the point y(P) = P(c, d) with c = c or c = c(cd-1)q-1 according as ||c|| G T or ||c|| G T. Since ||c|| = ||c||, the tangency points of i and i lie in the same stripe. The tangents of U with tangency point at infinity contain the origin and each of them has equation Y = Xm with mq+1 + 1 = 0. By the proof of Theorem 3.1, the corresponding lines Y = X o m are the tangents of U in the projective Moulton plane. Now look at dual plane of the projective Moulton plane MT (q2) which is the projective Andre plane AT (q2). In this duality, the tangent line i of U with equation Y = X o (-cd-1)q - d-q corresponds to the point P*(u*,v*) G (q2) where u* = -(-cd-1)q and v* = d-q. Since ((-cd-1)q)q+1 + (d-q)q+1 + 1 = 0, we have u*q+1 + v*q+1 + 1 = 0. Similarly, the tangent line i of U with equation Y = X o m, mq+1 + 1=0, corresponds to the point P* (u* ,v*) G AT(q2) where u* = u and v* = 0. Therefore u*q+1 + v*q+1 + 1 = 0. In terms of PG* (2, q2), the Desarguesian plane which gives rise to the projective Andre plane AT (q2), the points P*(u*, v*) lie on the classical unital U* given in its canonical form. This shows that U* can be viewed as an inherited unital in the projective Andre plane AT (q2). Remark 3.2. If T = {-1} then the unique stripe where incidence are altered meets U in q +1 points lying on the X-axis. The unital U* in the Hall plane is the Gruning unital [16] while for T = {¿} with wJ = -1, U* in the Hall plane is the Barwick unital [7]. A O'Nan configuration of a unital consists of four blocks b1, b2, 63 and 64 intersecting in six points P1, P2, P3, P4, P5 and P6 as in Figure 2. As mentioned in the introduction, the Hermitian unital contains no O'Nan configuration. This fundamental result due to O'Nan dates back to 1972, see [22] and [9, Section 4.2]. ¥>(P) P((y — b)m-1,y) for ||x|| £ T, P((y — b)m-q, y) for ||x|| G T. ((y — b)q+1(m-q )q+1 + yq+1 — 1 = ((y — b)m-q )q+1 + yq+1 — 1 = 0, whence the claim follows. □ Lemma 3.3. If T = { — 1} then the unitalU of (q2) is non-classical. Proof. We show that the unital U in (q2) with T = {-1} contains a O'Nan configuration. Take a e GF(q2) such that ||a|| = -1. The line of equation Y = X - a meets U in Q(a, 0) and q more points. Take m e GF(q2) such that mq-1 = -1. The line i2 of equation Y = Xm + am meets U in R (-a, 0) and q more points. Further, the common point of l1 and l2 is S = -a(m + 1) -2am (m - 1) (m - 1) Since -a(m + 1) (m - 1) ^q+1 (m +1)q+1 = (m - 1)q+1 mq+1 + mq + m +1 mq+1 - mq - m +1 - m2 - m + m + 1 - m2 + m - m + 1 1, the point S is outside U. Further, in the Moulton plane (q2) with T = {-1}, the corresponding lines and ¿2 meet in Q(a, 0) which is a point of U. To show that U is not a classical unital in our Moulton plane (q2), it suffices to exhibit a O'Nan configuration {P0, P1, P2, P3, P4, P5} lying in U. The idea is to start off with P0 = Q(a, 0), and to find four more affine points P1, P2 e ¿4 and P3, P4 e each lying in U, so that U also contains one of the two diagonal points P5 of the quadrangle P1P2P3P4 that are different from P0. First we show that P1 e ¿1. Let P1 = P1 (x1, y1). Then, ||x1| = -1. In fact, otherwise, we would have yq+1 = 0 and hence y1 = 0, contradicting P0 = P1. Similarly, P2 e and P3,P4 e ¿2. Now we use a counting argument in PG(2, q2) to show that the quadrangle P1P2PsP4 can be chosen in such a way that P5 e U. Since S = n ¿2 is outside U, the lines of U joining a point of ^ with a point of cover (q + 1)2(q - 1) points of U other than those lying in ^ U From (q +1)2 (q - 1) > q3 + 1 - 2q, there exists a quadrangle P^1P2P3P4 in PG(2, q2) such that P1, P2 e n U, P3, P4 e ¿2 n U, P5 = P1P3 n P2P4 e U. Since (q +1)2 (q -1) > q3 +1 - 2q + (q +1) we may also assume that either P5 g nU, or P5 = (x5,y5) with ||x5|| = -1. In particular, P5 is not on the X-axis. If Pi, P2 = Q and P3, P4 = R then P5 remains a diagonal point of the quadrangle PiP2P3P4 in (q2), and we are done. Otherwise, take the cyclic subgroup G of PGU(3, q) of order q +1 fixing the point S and preserving each line through S. Since |G| > 4, G contains an element g such that Q G {g(Pi), g(P2)} and R G {g(P3), g(Pi)}. Then g takes the quadrangle P1P2P3P4 to another one, whose vertices are different from both Q and R. The image g(P5) is on the line r through S and P5. Since r n U has at most one point on the X-axis, there exists at most one g G G such that g(P5) lies on the X-axis. Therefore, if |G| > 5, some g G G also takes P5 either to a point of infinity or a point (x5, y5) with ||x51| = -1. In the Moulton plane (q2), the O'Nan configuration P0, g(P1), g(P2), g(P3), g(P4), g(P5) arising from the quadrangle g(P^1)g(P2)g(P3)g(P4) lying in U has also two diagonal points, namely P0 and g(P5), belonging to U. □ Remark 3.4. Lemma 3.3 can also be obtained from Griining's work. In fact, if T = {-1} then U is isomorphic to its dual, see [16, Theorem 4.2], and the dual of U contains some O'Nan configuration, see [16, Lemma 5.4c]. We conjecture that Lemma 3.3 holds true for any T. Theorem 3.5 proves this as long as T is small enough. On the other end, Theorem 3.6 provides Moulton planes with large T for which the conjecture holds. Theorem 3.5. If q > 5 and then U in the Moulton plane MT (q2) is a non-classical unital. Proof. As in the proof of Lemma 3.3, we show the existence of a O'Nan-configuration {Po, Pi, P2, P3, P4, P5} lying in U. For a point P(a, b) G AG(2, q2) with a = 0 and ||«y G T \ { —1}, Theorem 2.6 ensures the existence of two non-vertical lines ¿4 and ¿2 through P such that (i) neither nor ¿2 is horizontal or passes through the origin, (ii) P0 = n 4 g U. From Lemma 2.1, there exist at least q +1 - 2|T| points P(x, y) lying on n U such that ||x|| G T, and the same holds for ¿2 n U. Therefore, Theorem 2.4 applies with A = q +1 - 2|T| showing that if (3.2) is assumed, then the unital U in (q2) contains a O'Nan configuration. □ Theorem 3.6. If q > 5, then there exists a T with |T | > q — 4 such that U is a non-classical unital in (q2). Proof. From the proof of Theorem 3.5, some Moulton plane MT (q2 ) contains O'Nan configurations lying in U .If {P0, P^ P2, P3, P4, P5} one of them, add each non-zero element s G GF(q) to T which satisfies the condition s = ||xj|| for Pj = Pj(xj,yj) with 1 < i < 5. Then T expands and its size becomes at least q—4. 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Sonnino, Non-classical unitals may be code words, submitted. [26] A. Sonnino, Existence of canonically inherited arcs in Moulton planes of odd order, Finite Fields Appl. 33 (2015), 187-197, doi:10.1016/j.ffa.2014.11.011. [27] S. D. Stoichev and V. D. Tonchev, Unital designs in planes of order 16, Discrete Appl. Math. 102 (2000), 151-158, doi:10.1016/s0166-218x(99)00236-x. [28] K. L. Wantz, Unitals in the regular nearfield planes, J. Geom. 88 (2008), 169-177, doi:10.1007/ s00022-007-2021-9. ARS MATHEMATICA CONTEMPORANEA ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 14 (2018) 267-284 https://doi.org/10.26493/1855-3974.953.3e1 (Also available at http://amc-journal.eu) Trilateral matroids induced by n3-configurations Michael W. Raney * Georgetown University Washington, DC, U.S.A. Received 9 October 2015, accepted 9 June 2017, published online 4 September 2017 We define a new class of a rank-3 matroid called a trilateral matroid. When defined, the ground set of such a matroid consists of the points of an n3-configuration, and its bases are the point triples corresponding to non-trilaterals within the configuration. We characterize which n3-configurations induce trilateral matroids and provide several examples. Keywords: Configurations, trilaterals, matroids. Math. Subj. Class.: 05B30, 51E30, 05C38, 05B35 1 Introduction A (combinatorial) n3-configuration C is an incidence structure consisting of n distinct points and n distinct blocks for which each point is incident with three blocks, each block is incident with three points, and any two points are incident with at most one common block. If C may be depicted in the real projective plane using points and having (straight) lines as its blocks, then it is said to be geometric. As observed in [6] (pg. 17-18), it is evident that every geometric n3-configuration is combinatorial, but the converse of this statement does not hold. A trilateral in a configuration is a cyclically ordered set {p0, h0,p1, h1,p2, h2} of pair-wise distinct points pi and pairwise distinct blocks hi such that pi is incident with hi- 1 and hi for each i e Z3 [2]. We may without ambiguity shorten this notation by listing only the points of the trilateral as {p0,p1,p2}, or more simply as p0p1p2. A configuration is trilateral-free if no trilateral exists within the configuration. Unless stated otherwise, the n3-configurations we shall examine are point-line configurations, so that the blocks are lines. But we shall investigate an example of a point-plane configuration in Section 3. Following the terminology of [7], we define a matroid M to be an ordered pair (E, B) consisting of a finite ground set E and a nonempty collection B of subsets of E called bases which satisfy the basis exchange property: *The author wishes to acknowledge the anonymous referee for the suggestion to consider point-plane n3-configurations as potential sources for trilateral matroids. E-mail address: mwr23@georgetown.edu (Michael W. Raney) Abstract ©® This work is licensed under http://creativecommons.org/licenses/by/3.0/ Definition 1.1. If Bl, B2 G B and x G Bl — B2, then there exists y G B2 — Bl such that Bl — x U y G B. It is a consequence of this definition that any two bases of M share the same cardinality; this common cardinality is called the rank of the matroid. See [7], pg. 16-18 for the details. It is a standard result that any n3-configuration C defines a rank-3 linear matroid, or vector matroid, M (C) = (E, B) whose ground set E consists of the points {pl ,p2,...,pn} of C and whose set of bases B consists of the point triples {pa ,pb,pc} which are not collinear in C. Hence the cardinality of B is Q) — n for the linear matroid M (C) induced by C. In this work we pose the following associated question: under what conditions do the trilaterals of an n3-configuration C induce a rank-3 matroid Mtri(C) = (E, B) whose ground set E again consists of the points of C, but now whose bases are the point triples corresponding to non-trilaterals? This question, to our knowledge, has not previously been considered in the literature on configurations and matroids. Definition 1.2. A trilateral matroid Mtri (C) = (E, B), when it exists, is a matroid defined on the set E of points of an n3-configuration C whose set of bases B consists of all of the non-trilaterals of C. When Mtri(C) exists, we say that C induces Mtri(C). We shall see that, in contrast to the linear matroid setting, seldom is it the case that an n3-configuration C induces a trilateral matroid Mtri(C). But thankfully such matroids do exist; for instance, any trilateral-free configuration induces a trilateral matroid, since in this setting every point triple forms a base of the matroid. In other words, if C is a trilateral-free n3-configuration, then Mtri(C) exists and furthermore Mtri(C) = U3,n, the uniform matroid of rank 3 on n points. Thus our initial motivation to define this new class of matroids stems from the desire to enlarge the class of trilateral-free configurations. For purposes of instruction, we regard an example of a 153-configuration which induces a trilateral matroid on its points. Here is a combinatorial description of this configuration. li 12 I3 I4 I5 h I7 1g I9 1l0 1ll 1l2 1l3 114 115 1112233455 7 7 9 10 13 246468 11 68 9 8 9 11 11 14 3 5 7 14 10 12 13 12 10 13 14 15 12 15 15 This configuration has 10 trilaterals: tl ¿2 t3 t4 t5 ie ¿7 tg t9 tl0 1 1 1 2 3 7 9 9 9 11 2 2 4 4 11 14 11 11 13 13 4 6 6 6 12 15 13 15 15 15 In Figure 1 we see both a diagram of this 153-configuration and a geometric representation of its trilateral matroid. In the geometric representation, each trilateral (that is, each nonbasis element) is collinear. Note that the configuration contains two complete quadrangles. The first complete quadrangle is determined by the point set {1, 2,4, 6}, and the second by {9,11,13,15}. This means, for example, that no three points in {1, 2,4,6} are collinear, and each pair of points is incident to a line of the configuration. So all four point triples present within {1, 2,4,6} give trilaterals, and hence are not bases of the matroid. Thus every 2-element subset of {1, 2,4,6} is independent, but no 3-element subset of {1, 2,4, 6} is. Therefore 11 9 13 15 Figure 1: A 153-configuration with 10 trilaterals, and a geometric representation of the matroid induced by these trilaterals. the four-point line that represents the level of dependency of {1,2,4, 6} in the geometric representation is appropriate. This minor is isomorphic to U2,4, which is the unique excluded minor for the class of binary matroids ([7], pg. 501). We must note that there is a fundamental difference between trilateral matroids and linear matroids. Admittedly a finite set of points and lines in the plane gives a (linear) matroid if and only if any pair of lines meet in at most one point. For suppose there exist two points a and b which are met by two lines, so that points a, b, c are collinear, points a, b, d are collinear, but a, b, c, d are not all on one line. Pick a new point e so that c, d, and e are not collinear, and so that a, b, and e are not collinear. Let B\ = abe and B2 = cde G B; both are bases of the linear matroid. We have B1 — B2 = ab and B2 — B1 = cd. Let x = e G B\ — B2, so B\ — x = cd. But if y G B2 — B\ = ab, then B\ — x U y equals either abc or abd, neither of which is a base. Hence a linear matroid cannot have two points common to more than one line. But a trilateral matroid can; if both abc and abd are trilaterals, then the configuration has a chance to induce a trilateral matroid if trilaterals acd and bcd are also present, meaning that points c and d are incident to a particular line of the configuration. In other words, points {a, b, c, d} form a complete quadrangle within the configuration. We shall explore this necessity further in Theorem 1.7. Any point of an n3-configuration is incident to three lines; these three lines are then incident to six points which are distinct from the original point and from each other. Consequently, the maximum number of trilaterals incident to a given point is (3) — 3 = 12, since lines are not trilaterals. This maximum is achieved by every point of the Fano 73-configuration (the smallest n3 -configuration) given in Figure 2. Proposition 1.3. Suppose an n3-configuration C induces a trilateral matroid Mtri(C) = (E, B). Then each point of the configuration is incident to at most six trilaterals. Proof. Let a be a point in C, and let abc, ade, and afg be the lines in C incident to a. Each of these lines belongs to B, and hence there are at most (3) — 3 = 12 trilaterals incident to a, namely abd, abe, abf, abg, acd, ace, acf, acg, adf, adg, aef, and aeg. Figure 2: The Fano 73-configuration. Since Bi = abc and B2 = ade are bases of Mtri(C), the basis exchange property applies to them. This means that if x G Bi — B2 = bc, there must exist some y G B2 — Bi = de such that Bi — x U y G B. Consequently, letting x = b, we find at least one of acd and ace must be a base, hence not a trilateral. Likewise, letting x = c, it follows that at least one of abd and abe is not a trilateral. Applying a similar analysis to the pair of bases Bi = abc, B2 = afg, we find that at least one of acf and acg is not a trilateral, and at least one of abf and abg is not a trilateral. Finally, given Bi = ade, B2 = afg, we find that at least one of aef and aeg is not a trilateral, and at least one of adf and adg is not a trilateral. Hence at least six of the 12 possible non-collinear triples are not trilaterals, so at most six are trilaterals. □ Corollary 1.4. Suppose an n3-configuration C induces a trilateral matroid Mtri(C) = (E, B). Then C contains at most 2n trilaterals. Although Corollary 1.4 admittedly serves as a crude necessary condition for an n3-configuration to induce a trilateral matroid, it does permit us to eliminate some of the smallest n3-configurations from consideration, such as the Fano 73-configuration (which contains 28 trilaterals) and also the Mobius-Kantor 83-configuration (which contains 24 trilaterals). Additionally, two of the three non-isomorphic 93-configurations may be dismissed from consideration by this criterion, although the Pappus 93-configuration, which contains 18 trilaterals, is still a possibility. We shall soon see, though, that the Pappus configuration does not induce a trilateral matroid on its points. The upper bound indicated by Proposition 1.3 is sharp, for it turns out that the Desargues 103-configuration induces a trilateral matroid. Each of the points of the Desargues configuration is incident to six trilaterals. We now establish our main result. This will require the introduction of two types of geometric obstructions (near-complete quadrangles and near-pencils) that, when present within an n3-configuration C, individually preclude the existence of Mtri (C). Definition 1.5. A near-complete quadrangle [ab : cd] consists of four points a, b, c, and d of the configuration, no three of which are collinear, for which five of the six possible lines connecting each pair of points exist within the configuration, except for the pair cd. c X c d Figure 4: Near-complete quadrangle [ab : cd]. For example, we note the presence of the near-complete quadrangle [ab : cd] in the Pappus configuration in Figure 5. Figure 5: The Pappus 93-configuration. It is important to note that, by our conventions, a complete quadrangle determined by points {a, b, c, d} does not contain anear-complete quadrangle [ab: cd], since there exists a line in the configuration incident to both c and d. So the Desargues configuration, for example, possesses five complete quadrangles but no near-complete quadrangle. As we shall witness in greater detail, n3-configurations which induce trilateral matroids may contain complete quadrangles. Indeed, in a linear matroid, given any two points, at most one line passes between them. But, two trilaterals (call them abc and abd) may share the points a, b provided that acd and bcd are also trilaterals, that is, that line cd is also present within the configuration. Definition 1.6. A near-pencil [a : bcd] consists of four points a, b, c, and d of the configuration, with a incident to each of b, c, and d, and with bcd a line of the configuration. We regard the near-pencil [a : bcd] in the Mobius-Kantor 83-configuration given in Figure 7. The notations [ab : cd] and [a : bcd] for a near-complete quadrangle and a near-pencil, respectively, are similar in that the points incident to three of the lines which determine the object appear to the left of the colon, and those points incident to two lines appear to the right of the colon. a Figure 6: Near-pencil [a : bcd]. Figure 7: The Möbius-Kantor 83-configuration. Theorem 1.7. Let C be an n3-configuration, and let B be the set of the non-trilaterals of C. Then C induces a trilateral matroid Mtri(C) if and only if no four points of C determine either a near-complete quadrangle or a near-pencil. Proof. ( w) First suppose that C contains a near-complete quadrangle [ab: cd]. Let e be the third point on line ace. Case 1: bde is a line in C. Then the following subfiguration is present inside C. Let B\ = ace and B2 = bde; both Bi,B2 G B. Then B\— B2 = ac and B2 — B\ = bd. Let x = c G B\ — B2; then B\ — x = ae. But both abe and ade are trilaterals, so B1 — x U y G B for all y G B2 — B1. Hence B cannot be the set of bases of a matroid. Case 2: bde is not a line in C. Then inside of C we have b d ace Note that edge be cannot be present, for if so point b would have four lines incident to it, but every point in an n3-configuration is incident to three lines. Let B\ = abe,B2 = acd G B. Take e G B\ — B2; we have B\ — e = ab. But B2 — B\ = cd, and both abc and abd are trilaterals. Hence B cannot be the set of bases of a matroid. Now suppose C contains a near-pencil [a: bcd] as indicated in the diagram. Let e be the third point on line ace. a em We have Bi = ace, B2 = bcd G B. Choose e G Bi — B2. Then Bi — e = ac. But B2 — B1 = bd, and both abc and acd are trilaterals. Hence B cannot be the set of bases of a matroid. Suppose that C does not induce a trilateral matroid Mtri (C). Since B cannot be the set of bases of a matroid, there must exist a pair Bi ,B2 in B for which the basis exchange property is violated. So there must exist x G Bi — B2 such that for all y G B2 — Bi, Bi — x U y is a trilateral. There are several cases to consider, some of which are vacuous. Case 1: Bi = B2. Then Bi — B2 = 0, so a violation of the basis exchange property cannot occur in this circumstance. Case 2: Bi = abc, B2 = abd (distinct letters label distinct points in C.) Then Bi — B2 = c and B2 — Bi = d. For a violation to occur, we require that Bi — cU d be a trilateral. But Bi — c U d = B2 G B. Hence no violation can occur in this case as well. Case 3: Bi = abc, B2 = ade. Then Bi — B2 = bc and B2 — Bi = de. Without loss of generality we assume that x = b. For a violation of the basis exchange property to occur, both acd and ace must be trilaterals. Subcase 3.1: ade is a non-collinear non-trilateral. Then [ac : de] is a near-complete quadrangle. a c P 13, the cyclic configuration C3(n, 1,4) induces a trilateral matroid on n trilaterals which equals the linear matroid on C3(n, 3,4). In other words, Mtri(C3(n, 1, 4)) = M(C3(n, 3,4)). Moreover, C3(n, 1,4) = C3(n, 3, 4). Proof. In order to determine the trilaterals of C3 (n, 1, 4), it suffices to ascertain the trilaterals which involve 0, and then extend from this via a cyclic pattern. The trilaterals involving 0 are: • 0 3 4 (using the lines {0,1,4}, {3,4, 7}, and {n — 1,0,3}) • n — 4 n — 30 (using the lines {n — 4, n — 3, 0}, {n — 1,0, 3}, and {n — 5, n — 4, n — 1}) • n — 3 0 1 (using the lines {n — 4, n — 3,0}, {n — 3, n — 2,1}, and {0,1, 4}) Since n > 13, no extra trilateral involving 0 is formed (for example, if n = 12, then 0 4 8 would be a trilateral.) Hence we see, after extending cyclically, that the trilaterals of C3(n, 1,4) form their own configuration, namely C3(n, 3,4), and thus Mtri(C3(n, 1,4)) is the linear matroid corresponding to C3(n, 3,4). Finally we may recognize that C3(n, 1,4) is isomorphic to C3(n, 3,4) either by utilizing Theorem 2.1 or by applying the correspondence t ^ (4 — t) mod n. □ It turns out that C3(16,1,4) and C3(16,1,7) are the smallest examples of non-isomorphic cyclic C3(n, k, m) configurations having n trilaterals each, and hence their corresponding trilateral matroids (which are isomorphic to the linear matroids associated with the respective original configurations) are non-isomorphic to each other as well. It is possible, however, for a non-cyclic n3-configuration to induce a trilateral matroid on its n trilaterals, with the trilaterals capable of determining an n3-configuration in their own right, without the original configuration needing to be cyclic. We have already seen Figure 13: A non-cyclic 163-configuration whose trilateral matroid is isomorphic to the linear matroid associated with the configuration. one example of this with the Coxeter 123-configuration given in Figure 9. Another example is the 163-configuration provided in Figure 13 whose automorphism group has order 32. It is additionally possible for an n3-configuration possessing n trilaterals to induce a trilateral matroid that is not isomorphic to the linear matroid associated with the original configuration. Figure 14 gives a diagram of such a configuration, a 203-configuration containing 20 trilaterals. It contains two points which are involved in six trilaterals and four points involved in four trilaterals. A geometric representation of the matroid is also provided. Figure 14: A 203-configuration with 20 trilaterals whose trilateral matroid is not isomor-phic to the linear matroid of the configuration, and a geometric representation of its trilateral matroid. We next offer an example of of an 183 -configuration possessing 20 trilaterals which induces a trilateral matroid. In Figure 15 we provide a picture of this configuration (with several pseudolines) and the accompanying geometric representation of its trilateral matroid. This example presents another instance, in addition to the Desargues 103-configuration, of an n3-configuration containing more than n trilaterals which induces a trilateral matroid. Figure 15: An 183-configuration with 20 trilaterals, and a geometric representation of its trilateral matroid. Note that this configuration contains four complete quadrangles. We now return to the enumeration of the smallest n3-configurations which induce trilateral matroids. There are four 143-configurations which do so. We label these configurations as 14A, 14B, 14C and 14D, and provide combinatorial depictions of them. 11 12 h 14 15 16 17 18 19 110 1ii 1i2 1i3 114 1 1 1 2 2 3 3 4 5 5 6 7 8 9 2 4 6 4 8 7 8 11 6 12 9 10 13 11 3 5 7 9 10 12 11 12 13 14 10 14 14 13 li 12 13 14 15 16 17 18 19 1i0 1ii 1i2 1i3 114 1 1 1 2 2 3 3 4 5 5 6 6 7 8 2 4 6 4 9 7 10 11 10 12 8 9 9 11 3 5 7 8 12 11 12 13 14 13 10 13 14 14 li 12 13 14 15 16 17 18 19 1i0 1ii 1i2 1i3 114 1 1 1 2 2 3 3 4 5 5 6 7 7 10 2 4 6 4 8 6 13 11 8 12 8 9 10 11 3 5 7 9 10 11 14 12 13 14 9 14 12 13 li 12 13 14 15 16 17 18 19 1i0 1ii 1i2 1i3 114 1 1 1 2 2 3 3 4 5 5 6 6 7 7 2 4 6 4 10 8 12 11 8 10 8 10 9 11 3 5 7 9 13 11 14 12 13 14 9 12 14 13 These configurations contain 14, 10, 10, and 6 trilaterals, respectively. Also, their automorphism groups have orders 14, 1, 4, and 8, respectively. Figure 16 gives a realization of Configuration 14A, which is isomorphic to the cyclic configuration 63(14,1,4). Hence we know its trilateral matroid is isomorphic to its linear matroid by Proposition 2.2. Configurations 14B and 14C both contain 10 trilaterals, so it is conceivable that their associated trilateral matroids could be isomorphic. But they are not, for 14B has three points which are each incident to three trilaterals and one point which is incident to only one trilateral, whereas Configuration 14C has two points each incident to three trilaterals 11 Figure 16: Configuration 14A. and no point incident to only one trilateral. Figure 17 gives geometric representations of the trilateral matroids associated with Configurations 14B and 14C, respectively. Figure 17: Geometric representations for trilateral matroids for Configurations 14B and 14C. Figure 18 is a rendering for Configuration 14D with several pseudolines, along with a geometric representation of its associated trilateral matroid. Proceeding to the n = 15 setting, we encounter a substantial increase, to 220, of the number of 153-configurations which induce trilateral matroids. One such example is the Cremona-Richmond configuration provided in Figure 19. It is the smallest example of a trilateral-free n3-configuration. As it is trilateral-free, the trilateral matroid it induces is the uniform matroid on 15 points U3,15. Another example is the cyclic configuration C3(15,1,4), whose induced trilateral matroid (with 15 trilaterals) is isomorphic to the linear matroid on C3(15,1,4) by Proposition 2.2. Its automorphism group has order 30. Each of the other 153-configurations which induces a trilateral matroid contains k trilaterals, where k G {4,6,7,8, 9,10,11,12,13,14}. It is clearly not the case that for all n, there exists a one-to-one correspondence between the trilateral matroids themselves and the n3-configurations which induce them. We know this because there are four non-isomorphic trilateral-free 183-configurations [3], so each consequently must induce the same uniform matroid on 18 points. Figure 18: Configuration 14D and its trilateral matroid. Figure 19: The Cremona-Richmond 153-configuration. It is of interest to contemplate whether smaller non-isomorphic n3-configurations exist that induce isomorphic trilateral matroids, and indeed this turns out to be true. In fact, this property is satisfied by the following pair of non-isomorphic 153-configurations given in Figure 20. Each contains 8 trilaterals and has a symmetry group of order 48. The Figure 20: Non-isomorphic 153-configurations which induce the same trilateral matroid on 15 points. set of points for both configurations consists of the eight vertices of a cube, the centers of the six faces of the cube, and the center of the cube itself. In the former configura- tion the diagonally-opposing points in each face of the cube are incident via a line which passes through the center of the same face, whereas in the latter configuration one pair of diagonally-opposing points in each face are incident via a "line" which passes through the center of the opposite face. The eight trilaterals involved in these respective configurations are identical, and thus their corresponding trilateral matroids are the same. Figure 21 gives this matroid, which is isomorphic to U2,4 © U2,4 © U3,7. Hence the number of Figure 21: The common trilateral matroid. trilateral matroids that are induced from 153-configurations is smaller than the number of 153-configurations which induce trilateral matroids. Our calculations indicate that there are 214 non-isomorphic trilateral matroids that may be found from the 220 153-configurations which induce trilateral matroids. We conclude this section with a table which summarizes the current state of affairs. Here #c(n) denotes the number of non-isomorphic n3-configurations, #tri(n) denotes the number of these configurations which induce trilateral matroids, and #mat(n) denotes the number of non-isomorphic trilateral matroids which arise from these configurations. n #c(n) #tri(n) #mat(n) 7 1 0 0 8 1 0 0 9 3 0 0 10 10 1 1 11 31 0 0 12 229 1 1 13 2036 1 1 14 21399 4 4 15 245342 220 214 3 A point-plane configuration A point-plane n3-configuration is an incidence structure consisting of n distinct points and n distinct planes for which each point is incident with three planes, each plane is incident with three points, and any two points are incident with at most one common plane. In such a configuration, we deem a trilateral to be a cyclically ordered set {p0, n0,p1,n1 ,p2,n2} of pairwise distinct points pi and pairwise distinct planes ni such that pi is incident with ni_1 and ni for each i e Z3. Once more we may without ambiguity shorten this notation by listing only the points of the trilateral as {po, p 1, p2}, or more simply as pop 1p2. In Figure 22 we offer an example of a point-plane 123-configuration which induces a trilateral matroid on its points. The 12 points are selected from the 20 vertices of the regular dodecahedron so that each of the twelve pentagonal faces contains three points; note that each of the 12 points is the intersection of three faces, so a point-plane 123-configuration is achieved. We observe that each of the eight unlabeled red points in the 4 Figure 22: A 123 point-plane configuration which induces a trilateral matroid. diagram corresponds to a trilateral, and that this trilateral may be specified uniquely by cycling through the configuration points that are immediately adjacent to the red point. For example, the triple {1,3, 5} defines a trilateral. We start at 1, then pass through the plane containing both 1 and 3 to 3. We then pass through the plane containing both 3 and 5 to 5, and then finally pass through the plane containing both 5 and 1 back to 1 to complete the cycle. Here are the eight trilaterals. ¿1 ¿2 ¿3 ¿4 ¿5 ¿6 ¿7 ¿g 1 1 2 3 4 4 6 8 2 3 7 6 5 9 11 10 9 5 8 7 11 10 12 12 Figure 23 gives a geometric representation of the trilateral matroid. Figure 23: The trilateral matroid of the 123 point-plane configuration. After identifying each trilateral with its corresponding red point in Figure 22, we recognize that the trilateral matroid may also be represented as a point-plane configuration, namely an (83,122)-configuration. This means the configuration has eight points, with three planes incident to each point, and twelve planes, with two points incident to each plane. References [1] A. Al-Azemi and D. Betten, The configurations 123 revisited, J. Geom. 105 (2014), 391-417, doi:10.1007/s00022-014-0228-0. [2] M. Boben, B. Grunbaum and T. Pisanski, Multilaterals in configurations, Beitr. Algebra Geom. 54 (2013), 263-275, doi:10.1007/s13366-011-0065-3. [3] M. Boben, B. Grunbaum, T. Pisanski and A. Zitnik, Small triangle-free configurations of points and lines, Discrete Comput. Geom. 35 (2006), 405-427, doi:10.1007/s00454-005-1224-9. [4] R. Daublebsky von Sterneck, Die Configurationen 123, Monatsh. Math. Phys. 6 (1895), 223255, doi:10.1007/bf01696586. [5] H. Gropp, On the existence and nonexistence of configurations nk, J. Comb. Inf. Syst. Sci. 15 (1990), 34-48. [6] B. Grunbaum, Configurations of Points and Lines, volume 103 of Graduate Studies in Mathematics, American Mathematical Society, Providence, Rhode Island, 2009, doi:10.1090/gsm/103. [7] J. G. Oxley, Matroid Theory, volume 3 of Oxford Graduate Texts in Mathematics, Oxford University Press, New York, 1992. ARS MATHEMATICA CONTEMPORANEA ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 14 (2018) 285-318 https://doi.org/10.26493/1855-3974.862.bb5 (Also available at http://amc-journal.eu) Growth of face-homogeneous tessellations Stephen J. Graves * The University of Texas at Tyler, Tyler, TX, USA Mark E. Watkins f Syracuse University, Syracuse, NY, USA Received 2 June 2015, accepted 23 July 2017, published online 13 September 2017 A tessellation of the plane is face-homogeneous if for some integer k > 3 there exists a cyclic sequence a = [po,Pi,... ,pk-1] of integers > 3 such that, for every face f of the tessellation, the valences of the vertices incident with f are given by the terms of a in either clockwise or counter-clockwise order. When a given cyclic sequence a is realizable in this way, it may determine a unique tessellation (up to isomorphism), in which case a is called monomorphic, or it may be the valence sequence of two or more non-isomorphic tessellations (polymorphic). A tessellation whose faces are uniformly bounded in the hyperbolic plane but not uniformly bounded in the Euclidean plane is called a hyperbolic tessellation. Such tessellations are well-known to have exponential growth. We seek the face-homogeneous hyperbolic tessellation(s) of slowest growth rate and show that the least growth rate of such monomorphic tessellations is the "golden mean," 7 = (1 + %/5)/2, attained by the sequences [4, 6,14] and [3,4,7,4]. A polymorphic sequence may yield non-isomorphic tessellations with different growth rates. However, all such tessellations found thus far grow at rates greater than 7. Keywords: Face-homogeneous, tessellation, growth rate, valence sequence, exponential growth, transition matrix, Bilinski diagram, hyperbolic plane. Math. Subj. Class.: 05B45, 05C63, 05C10, 05C12 *Much of the material in this work comes from the doctoral dissertation [4] of the first author written under the supervision of the second author. ÎThe second author was partially supported by a grant from the Simons Foundation (#209803 to Mark E. Watkins). E-mail addresses: sgraves@uttyler.edu (Stephen J. Graves), mewatkin@syr.edu (MarkE. Watkins) ©® This work is licensed under http://creativecommons.org/licenses/by/3.0/ Abstract 1 Introduction It has long been known that there are finitely many homogeneous tessellations of the Euclidean plane; they all have quadratic growth rate. However, in the hyperbolic plane, for various definitions of "homogeneity," infinitely many homogeneous tessellations are realizable, and their growth rate, if not quadratic, is always exponential. Presently we will give a rigorous definition of growth rate, but for now one should think of this parameter intuitively as the asymptotic rate at which additional tiles (or faces) accrue with respect to some chosen center of a tessellation. In this schema, all Euclidean tessellations have growth rate equal to 1, and hyperbolic tessellations have growth rate strictly greater than 1. The first author has shown by construction in [5] that, given any e > 0, there exists a hyperbolic tessellation with growth rate exactly 1 + e. In general, these latter tessellations have few if any combinatorial or geometric symmetries. The question then becomes one of determining the growth rates of hyperbolic tessellations when some sort of homogeneity is imposed. In particular, subject to a homogeneity constraint, how small can the gap be between quadratic and exponential growth? In a seminal work [8], Griinbaum and Shephard defined a graph to be edge-homogeneous with edge-symbol (p, q; k, £) if every edge is incident with vertices of valences p and q and faces of covalences k and I. They proved that the parameters of an edge-symbol uniquely determine an edge-homogeneous tessellation up to isomorphism. The notion of homogeneity was extended by Moran [10]. She defined a tessellation to be face-homogeneous with valence sequence [p0,..., pk-1] if every face is a k-gon incident with vertices of valences p0,... ,pk-1 in either clockwise or counter-clockwise consecutive order. Unfortunately, no uniqueness property analogous to the Grunbaum-Shephard result holds in general for face-homogeneous tessellations. Moran's work on growth rates of face-homogeneous tessellations led the authors (together with T. Pisanski) to return to edge-homogeneous tessellations and conclusively determine their growth rates. In [6] they determined the growth rate of any edge-homogeneous tessellation as a function of its edge-symbol and proved that the least growth rate for an exponentially-growing, edge-homogeneous tessellation is 2 (3 + \/5) « 2.618. The goal of this article is to obtain an analogous result for face-homogeneous tessellations. Our main result is that if a face-homogeneous tessellation with exponential growth rate is determined up to isomorphism by its valence sequence, then its growth rate is at least 2 (1 + %/5), namely the "golden mean." Moreover, we determine exactly the valence sequences for which this golden mean is realized. A significant by-product of our investigation is an abundance of machinery for computing the growth rates of many classes of face-homogeneous planar tessellations. Section 2 consists of six subsections. Following some general definitions concerning infinite graphs in the plane, we present (Subsection 2.2) a system for labeling sets of vertices and sets of faces of a tessellation; such a labeling is called a "Bilinski diagram." Subsection 2.3 presents the notion of face-homogeneity and associated notation. Polynomial and exponential growth, defined on the one hand with respect to the standard graph-theoretic metric, and on the other hand with respect to the notion of angle excess, appear in Subsection 2.4. Subsection 2.5 presents a rigorous theoretical treatment of growth rate with respect to regional distance in a Bilinski diagram. Subsection 2.6 concludes the Preliminaries with a review of the completely resolved case of edge-homogeneous tessellations, summarizing results from [8] and [6]. In Subsection 3.1 we lay out our method for filling in the formulas obtained in Sub- section 2.5 while introducing the notion of a transition matrix. Analogous to a Markov process, this matrix encodes for given n > 1 how many faces of each possible configuration are "begotten" at regional distance n +1 from the root of a Bilinski diagram by a face at regional distance n from the root. The maximum modulus of the eigenvalues of the transition matrix are key to the growth rate of T. Subsection 3.2 applies the machinery of Subsection 3.1 to the significant class of valence sequences that are monomorphic, i.e., that are uniquely realizable as a face-homogeneous tessellation and whose Bilinski diagrams are in a certain sense well-behaved, called uniformly concentric. It is shown in Theorem 3.7 that for such valence sequences, the partial order defined in Subsection 2.3 is preserved by their growth rates. The six classes of monomorphic sequences of lengths 3, 4, and 5 whose Bilinski diagrams are not uniformly concentric are identified in Subsection 3.3, where it is proved that they are indeed monomorphic. The exhaustive proof that this list is complete is contained in the Appendix [7]. Finally, we present in Subsection 3.4 the main result of the paper, that the least growth rate of a face-homogeneous tessellation with monomorphic valence sequence is the golden mean 2 (1 + V5). Those valence sequences (described as polymorphic) which admit multiple non-isomor-phic tessellations are alive and well in Subsection 4.1. A general sufficient condition for polymorphism is given. The difficulties posed by polymorphism are illustrated by an example; the polymorphic sequence [4,4,6,8] is considered in some depth in Subsection 4.2. In particular, we see by this example that two different tessellations having the same (polymorphic) valence sequence may well have different growth rates. We conclude the chapter with some conjectures in Subsection 4.3. The appendix [7] alluded to above is to be found with this article on the arXiv, at arXiv:1707.03443. All references therein are to results in the present paper. Due to the considerable length (and tedium!) of the appendix, it will not appear in Ars Mathemat-ica Contemporanea with this article. 2 Preliminaries 2.1 Tessellations For a graph r, the symbol V(r) denotes the vertex set of r. If M is a planar embedding of r, we call M a plane map and denote by F (M) the set of faces of M. A graph r is infinite if its vertex set V (r) is infinite; r is locally finite if every vertex has finite valence. A graph is 3-connected if there is no set of fewer than three vertices whose removal disconnects the graph. It is well known that if the underlying graph r of a plane map M is 3-connected (as is generally the case in this work), then every automorphism of r induces a permutation of F (M) that preserves face-vertex incidence and can be extended to a homeomorphism of the plane. Thus we tend to abuse language and speak of "the faces of r." When a plane map is 3-connected, every edge is incident with exactly two distinct faces. In this case, the number of edges (and hence of vertices) incident with a given face is its covalence. A map is locally cofinite if the covalence of every face is finite. An accumulation point of an infinite plane map M is a point x in the plane such that every open disk of positive radius (in either the Euclidean or hyperbolic metric) containing x intersects infinitely many map objects, be they faces, edges, or vertices. A map is 1-ended when the deletion of any finite submap leaves exactly one infinite component. Definition 2.1. A tessellation is an infinite plane map that is 3-connected, locally finite, locally cofinite, 1-ended, and also admits no accumulation point. In the terminology of Griinbaum and Shepherd's exhaustive work [9] on tilings of the plane, a tessellation T is normal if there is an embedding of T in the plane and radii 0 < r < R under a specific metric such that the boundary of each face lies within some annulus with inner radius r and outer radius R. A Euclidean tessellation is a tessellation that is normal with respect to the Euclidean metric, and a hyperbolic tessellation is one that is normal with respect to the hyperbolic metric but not with respect the Euclidean metric. 2.2 Bilinski diagrams A very useful tool for computing "growth rate" is what we have called a Bilinski diagram, because these diagrams were first used by Stanko Bilinski in his dissertation [1, 2]. Definition 2.2. Let M be a map that is rooted at some vertex x. Define a sequence of sets {Un : n > 0} of vertices and a sequence of sets {Fn : n > 0} of faces of M inductively as follows. • Let U0 = {x} and let F0 = 0. • For n > 1, let Fn denote the set of faces of M not in Fn-1 that are incident with some vertex in Un-1. • For n > 1, let Un denote the set of vertices of M not in Un-1 that are incident with some face in Fn. The stratification of M determined by the set-sequences {Un} and {Fn} is called the Bilinski diagram of M rooted at x. In a similar way one can define a Bilinski diagram of M rooted at a face f. In this case U0 = 0 and F0 = {f}. Given a Bilinski diagram of T, the induced submap (Fn) of T is its nth corona. A Bilinski diagram is concentric if each subgraph (Un) induced by Un (n > 1) is a cycle; otherwise the Bilinski diagram is non-concentric. If a plane map yields a concentric Bilinski diagram regardless of which vertex or face is designated as its root, then the map is uniformly concentric; analogously a map which for every designated root yields a non-concentric Bilinski diagram is uniformly non-concentric. To answer the question as to which tessellations are uniformly concentric we state a sufficient condition and a necessary condition. Let Ga,b denote the class of tessellations all of whose vertices have valence at least a and all of whose faces have covalence at least b. Let Ga+bh be the subclass of Ga,b of tessellations with no adjacent a-valent vertices. Proposition 2.3 ([3, Corollary 4.2], [11, Theorem 3.2]). Every tessellation T e G^e U G3+j5 U G4,4 is uniformly concentric, and in every Bilinski diagram of T, for all n > 1, every face in Fn is incident with at most two edges in (Un-1). Proposition 2.4 ([3, Theorem 5.1]). If an infinite planar map admits any of the following configurations, then the map is not uniformly concentric: 1. a 3-valent vertex incident with a 3-covalent face; 2. a 4-valent vertex incident with two nonadjacent 3-covalent faces; 3. a 4-covalent face incident with two nonadjacent 3-valent vertices; 4. an edge incident with two 3-valent vertices and two 4-covalent faces; 5. an edge incident with two 4-valent vertices and two 3-covalent faces. 2.3 Face-homogeneity and readability Let k > 3 be an integer and let an equivalence relation be defined on the set of ordered k-tuples (po,Pi,... ,pk-i) of positive integers whereby • (po,pi,... ,Pk-i) = (pi,p2,... ,Pk-i,po), and • (po,Pi, . . . ,Pfc-i) = (Pfc-i,Pfc-2, ... ,po). The equivalence class of which (po,... ,pk-i) is a member is the cyclic sequence [po,... ,pk-i], and k is its length. There is a natural partial order < on the set of cyclic sequences: [po,... ,Pk-i] < [qo,... if and only if k < I and there exists a cyclic subsequence qio, q¿1,..., qik-1 occurring in either order in [qo, qi,..., qi-i] such that pj < qij for each j e {0,..., k - 1}. We write ai < a2 if ai < a2 but ai = a2, where ai and a2 are cyclic sequences. Example 2.5. Let ai = [4,6, 8,10], a2 = [6, 8,12,4], and a3 = [10,8,12,6,4]. Then ai < a2 and ai < a3, but a2 and a3 are not comparable. Definition 2.6. Let a = [po,pi,... ,pk-i] be a cyclic sequence of integers > 3. Then a is the valence sequence of a k-covalent face f of a tessellation T if the valences of vertices incident with f in clockwise or counter-clockwise order arepo,pi,... ,pk-i. If every face of T has the same valence sequence a, then T is face-homogeneous and a is the valence sequence of T. Thus, to say briefly that a tessellation T has valence sequence a implies that T is face-homogeneous. Definition 2.7. Let the cyclic sequnce a be realizable as the valence sequence of a tessellation. If every tessellation having valence sequence a is uniformly concentric, then we say that a is uniformly concentric. Otherwise a is non-concentric. If every tessellation having valence sequence a is non-concentric, then a is uniformly non-concentric. Notation. By convention, when distinct letters are used to represent terms in a cyclic sequence (e.g. [p,p, q, r, q]), the values corresponding to distinct letters are all presumed to be distinct; that is, p = q = r = p. Moreover, if some term in the cyclic sequence is given as an integer (usually 3 or 4), then the terms given by letters are presumed to be greater than that integer. For example, if a = [4, p, q], then we understand that p, q > 4 and p = q. When using subscripts in the general form [po,..., pk-i], we do not make this assumption. Remark 2.8. Not all cyclic sequences are realizable as vertex sequences of face-homogeneous tessellations of the plane. For instance, the map with valence sequence [3, 3, 3] (the tetrahedron) is a tessellation of the sphere but not of the plane. More importantly, there are many cyclic sequences for which no face-homogeneous map exists at all. For instance, the valence sequence [4,5, 6,p] for any p > 3 is not realizable, because in any such map the valences of the neighbors of a 5-valent vertex in cyclic order would have to alternate between 4 and 6. However, this does not generalize to all cyclic sequences containing a subsequence [p, q, r] where q is odd and p = r; for instance, [5,4, 5,6,5, 8] is realizable. Conjecture 2.9. Suppose a is the valence sequence of a face-homogeneous tessellation and that a contains [p, q, r] as a subsequence, with q odd and p = r. Then a must contain at least three terms equal to q. 2.4 Polynomial versus exponential growth Let x be a vertex of a connected graph r. For each nonnegative integer n, the ball of radius n about x is the set of vertices of r at distance < n from x, written Bn(x) = {v e V(r) : d(x, v) < n}, (2.1) where d(—, —) is the standard graph-theoretic metric, that is, d(u, v) is the length of a shortest path with terminal vertices u and v. Definition 2.10. An infinite, locally finite, connected graph r has exponential growth if for some vertex x e V(r) there exist real numbers a > 1 and C > 0 such that, for all n > 0, one has |Bn(x)| > Can; otherwise r has subexponential growth. We say that r has polynomial growth of degree d e N if there exist positive constants C1 and C2 such that Cind < |Bn(x) | < C2nd for all but finitely many n. For example, the graph underlying the square lattice in the plane has quadratic growth (d = 2). If x is any vertex, then |Bn(x) | = 2n2 + 2n + 1 for all n > 1, and one can set Ci = 2 and C2 = 3. Continuing the notation of Equation (2.1) and Definition 2.10, we consider the generating function to &(z) = ^ |B„(x)| zn (2.2) n=0 We denote the radius of convergence of by RB and define the ball-growth rate of V about x to be the reciprocal of RB. If r has exponential growth, then we have TO c & (z) >Y\ Canzn = --, (2.3) 1 — az n=0 where a > 1 is the supremum of values for which the series of Equation (2.2) converges. The convergence is absolute if and only if |z| < 1/a < 1. If r has polynomial growth of degree d, then TO TO TO Ci ^ ndzn < ^ |Bn(x)| zn < C2 ^ ndzn. Jn V 0 By the "ratio test," the first and third series converge if and only if |z| < 1. These computations yield the following. Proposition 2.11. Let RB denote the radius of convergence of the generating function of Equation (2.2). Then RB < 1 if and only if r has exponential growth, and RB = 1 if and only if r has polynomial growth. Moreover, RB is independent of the vertex x about which |Bn(x) | is determined. It will be seen in the next subsection (see Theorem 2.16) that the value of RB is independent of the choice of the root vertex x. It is well known (for example, see [9]) that there exist exactly eleven face-homogeneous Euclidean tessellations, namely the Laves nets. Their valence sequences [p0,... ,pk_1] correspond to integer solutions of the equation k-1 _ k - 2 ^pi = 2 ' i=0 ^ : A necessary condition for the existence of a face-homogeneous hyperbolic tessellation with valence sequence [p0,... ,pk-1] is that the inequality k-1 1 k - 2 ^ pi < — (2.4) i=o P: 2 hold. This condition is not sufficient, because as we have seen, not every such integer solution of the inequality (2.4) is realizable as a valence sequence. Definition 2.12. The angle excess of a cyclic sequence a = [p0,... ,pk-1] is given by (s ^)- 2' Motivation for this definition comes from Descartes' notion of angular defect in the Euclidean plane. When n(a) > 0, there are too many faces incident at a vertex for the faces to be regular k-gons in the Euclidean plane. Proposition 2.13. For a cyclic sequence a = [p0,... ,pk-1], inequality (2.4) is equivalent to n(a) > 0 (2.5) and is a necessary condition for a to be a valence sequence of a face-homogeneous hyperbolic tessellation. Angle excess provides a quick gauge of the growth behavior of a tessellation with valence sequence a. If n(a) < 0, the tessellation is finite. If n(a) = 0, the tessellation is one of the Laves nets and has polynomial growth of degree 2. If n(a) > 0, the tessellation has exponential growth. Additionally, we have the following comparison result. Proposition 2.14. Let a1 and a2 be cyclic sequences that are comparable in the partial order Then a1 < a2 if and only if n(a1) < n(a2). Proof. Suppose that a1 < a2, where a1 = [p0,... ,pk-1] and a2 = [q0,.qe-1]. By definition there exist qio,..., qik_1 with pj < q:. for all j = 0,.. .k - 1. So k 1 ¿-v k 1 c\ & 1 n(a1) = s — < S — < S — = n(«2). (2.6) j=0 pj j=0 q:j :=0 q: If k = I, then pj < q:j for some j and the first inequality in (2.6) is strict. If k < I, the second inequality in (2.6) is strict. Since a1 = a2, at least one such strict inequality must hold. □ 2.5 Growth formulas In Definition 2.10, the standard graph-theoretical metric was used to define polynomial and exponential growth of a connected graph. However, to measure growth rates of tessellations, it is more convenient to use the notion of "regional distance;" we will count the number of graph objects in the nth corona of a Bilinski diagram centered at a given vertex, and our working definition of "growth rate" will be the following. Definition 2.15. Let T be a tessellation labeled as a Bilinski diagram rooted at a vertex x. Let R be the radius of convergence of the power series w 3. Hence for any n > 1 and any vertex v G Un+i(x) there exists a vertex u g Un such that d(u, v) < |_fj. By induction on n, we obtain d(x, v) < (n + 1) |_U, yielding n U Ui (x) C BnLfc/2j(x) (2.8) i=0 and similarly, n Bn(x) C U Ui(x). (2.9) i=0 In addition to the power series yx(z) of Definition 2.15 with radius of convergence RF, we require the power series ux(z) = ^ |Un(x)| zn with radius of convergence Ru. Writing / ^ w I n \ w Yx(z) = = £ E|Ui(x)M zn = £ U Ui(x) we have that the radius of convergence of Tx(z) equals min {Ru, 1} < RB by Equation (2.8) (where RB is as in Proposition 2.11). But similarly by Equation (2.9) we have that Rb < min {Ru, 1}. Hence the radii of convergence of Tx(z) and ^x(z) are equal, for any choice of root vertex x. zn If p is the maximum valence of the vertices in T, each vertex is also incident with at most p faces, while each face is incident with k vertices, giving |Un(x)| < k |Fn+i(x)| < pk |Un+i(x)| for each n > 0, or equivalently, 1 |Un (x)| < |F„+i(x)| < p |U„+i(x)| . Hence the radii of convergence of ux (z) and px (z) are equal, and more importantly, RF = Rb ; that is, the rate of ball-growth equals the rate of growth when the Bilinski diagram is labeled from a vertex x. Finally, it follows from Proposition 2.11 that ball-growth rates computed about distinct vertices are asymptotically equal in locally finite, connected, infinite graphs. Hence the radii of convergence of px(z), Px(z), Py (z), and py (z) are equal for all x, y G V. That is to say, the growth rate of the graph is independent of the choice of root vertex. □ Notation. The subscript on the symbol p of Definition 2.15 has now been shown to be superfluous and will henceforth be suppressed. Consider the function t : N0 ^ N0, (where N0 = {0,1,2,...}) given by t(n) = ^ |Fi|. \ - i=i The quantity lim (2.10) u^to T (n) was the definition of the growth rate of a face-homogeneous tessellation used by Moran [10] provided that this limit exists, in which case she called the tessellation balanced. Moran's limit fails to converge only when there exist subsequences of the sequence j ^Tu+y } with distinct limits. The following proposition shows that the parameters of a face-homogeneous tessellation determine an upper bound for the limit in Equation (2.10). Theorem 2.17. Let T be a face-homogeneous tessellation with valence sequence [p0,... ,pk-i], labeled as a Bilinski diagram. Then t (n +1) k— lim sup-——— < 1 + > pi — 2k < to . u^to T (n) ^ i=0 Proof. By hypothesis, each face of the tessellation shares an incident vertex with exactly k-i k-i Yjp>i — 2) = ^ pi — 2k i=0 i=0 other faces. So for n > 0, |Fn+i| < |Fn| ^pi — 2k^ , which in turn gives that for all n > 0, liiiil < 1+ lEnL (£ p, - 2k) t w E,.O vso ) k-i < 1 + P, - 2k < ro, ,=0 since T is locally finite. □ By the "ratio test" of elementary calculus, the above proof implies that in the case of a "balanced" tessellation, Moran's definition of growth rate concurs with Definition 2.15, and 1 t (n +1) t (n +1) — = lim sup-——— = lim -———. R n^TO T (n) n^TO t (n) The definition of growth rate in terms of the radius of convergence of a power series also allows us to prove the following result, which is essential in many comparisons of growth rates of various tessellations. Lemma 2.18 (Comparison Lemma). Let T1 and T2 be tessellations, and for i =1, 2 let \F,,n\ be the number of faces in the nth corona of a Bilinski diagram of T,. Suppose that for some N e N, we have \Fi,n\ < \F2,n\for all n > N. Then y(ti) < 7^2). Proof. Let TO TO ¿1 (z) = £ \Fi,n\zn, ¿2 (z) = £ \F2,n\zn, n=0 n=0 and for i e {1, 2}, let R, be the radius of convergence of ¿,(z) about 0. Then since |,n| < \F2,n\ for sufficiently large n, and lim sup = Y(Ti), n^TO R, we have 7(Ti) < 7^2). □ 2.6 The edge-homogeneous case We conclude our presentation of preliminary material with a quick review of what is known about growth rates of edge-homogeneous tessellations, as this case has been completely resolved and its consequences turn out to be useful here and there in attacking the present problem. The point of departure here is the following classification theorem of Griinbaum and Shephard. (Edge-symbols were defined in Section 1.) Proposition 2.19 ([8, Theorem 1]). Let p,q,k,£ > 3 be integers. There exists an edge-homogeneous, 3-connected, finite or 1-ended map with edge-symbol (p, q; k, t) if and only if exactly one of the following holds: 1. all of p, q, k, t are even; 2. k = t is even and at least one of p, q is odd; 3. p = q is even and at least one of k, t is odd; 4. p = q, k = £, and all are odd. Such a tessellation is edge-transitive, and the parameters p, q, k, £ determine the tessellation uniquely up to homeomorphism of the plane. If p = q, then the tessellation is vertex-transitive. If k = £, then it is face-transitive. Following up on the Griinbaum-Shephard result, the authors together with T. Pisanski completely determined the growth rates of all edge-homogeneous tessellations. Their main result is the following. Proposition 2.20 ([6, Theorem 4.1]). Let the function g : {t e N : t > 4} ^ [1, œ) be given by 1 2 Let T be an edge-homogeneous tessellation with edge-symbol (p, q; k, t), and let g(t)=2 (t - 2+ V(t - 2)2 - ^ . (2.11) t =(P+q - A - 2) (2.12) Then exactly one of the following holds: 1. the growth rate of T is y(T) = g(t); or 2. the edge-symbol of T or its planar dual is (3, p; 4,4) with p > 6, and the growth rate of T is y(T) = g(t - 1). Observe that each value of t > 4 corresponds to only finitely many edge-homogeneous tessellations and that pairs of planar duals correspond to the same value of t. As the growth rates of edge-homogeneous tessellations are determined by an increasing function in one variable, the following is immediate. Corollary 2.21. The least growth rate of an edge-homogeneous hyperbolic tessellation is (3 + a/5)/2. This value is attained only by the tessellations with edge-symbols (3,3; 7, 7), (4,4; 4, 5), (3, 7; 4,4), and their planar duals. Remark 2.22. It is evident from Proposition 2.19 that if a tessellation is both edge- and face-homogeneous, then its edge-symbol and valence sequence have, respectively, either the forms (p, p; k, k) and [p, p,..., p] or the forms (p, q; k, k) and [p, q,..., p, q], the latter pair being possible only when k is even. We mention that, by an argument similar to the proof of Theorem 2.17, one easily obtains the following upper bound for the growth rate of an edge-homogeneous tessellation. Proposition 2.23. Let T be an edge-homogeneous tessellation with edge-symbol (p, q; k, t). Then for any labeling ofT as a Bilinski diagram, one has t (n +1) lim -——— < 1 + max{pk, qk,pt, qt}. n^TO T (n) 3 Accretion and monomorphic valence sequences 3.1 Accretion Given an arbitrary face-homogeneous tessellation T with valence sequence a = [p0,pi,... ,pk-1], we wish to apply Definition 2.15 to determine its growth rate. Letting T be labeled as a Bilinski diagram, we require a means to evaluate | Fn | for all n g N. This is done inductively; each face f g Fn "begets" a certain number of facial "offspring" in Fn+1, and that number is determined by the configuration of f within (Fn), that is, what the valences are of the vertices incident with f (in the rotational order of a) that belong, respectively, to Un-1 and more importantly to Un. A class of identically configured faces (in any corona) is a face type, and is denoted by f for some range of i = 1,..., r. The benefit of using face types is that we can define an r-dimensional column vector vn, called the nth distribution vector, which lists the frequency with which each face type occurs in the nth corona. Thus, if j is the r-dimensional vector of Is, then |Fn| = j vn via the standard dot product. Figure 1 depicts a face f g Fn of some tessellation and the faces in Fn+1 which are determined by the face type of f. These faces are called the offspring of f, and the figure is accordingly called the offspring diagram for f. As the vertex labeled pj is incident with Un U n+1 f' Pj^ f -—*-""""-—1 -—*-""""-—1 f" A B Figure 1: A face f in Fn of a tessellation T, along with the offspring of f in Fn+1. both faces f and f' g Fn, one-half of those faces in Fn+1 labeled as A in the figure count as offspring of f, and one-half are counted as offspring of f'. Similarly, half of the faces labeled by B count as offspring of f and half as offspring of f". All those faces between labels A and B in Figure 1 are wholly offspring of f. Those faces which are offspring of f, or offspring of offspring of f, and so on, are called collectively descendants of f. Definition 3.1. With respect to the labeling of a Bilinski diagram, each vertex incident with a face f g Fn lies in Un-1 or Un. The pattern of valences of vertices in Un-1 and in Un determines the face type of f. The three face types occurring most routinely are called wedges, bricks, and notched bricks. A face f in Fn is a wedge if it is incident with exactly one vertex in Un-1. The face f is a brick if it incident with exactly two adjacent vertices in Un-1 and at least two vertices in Un. Finally, f is a notched brick if it is incident with three consecutive vertices of Un-1, of which the middle vertex is 3-valent, and f is incident with two or more vertices in Un. For a given labeling of a tessellation T as a Bilinski diagram, the face types of T are indexed i\,..., fr for some r e N; we explain the method by which indices are assigned after the statement of Theorem 3.7. An algorithm by which one can describe the faces, corona by corona, of a tessellation labeled as a Bilinski diagram is called an accretion rule. Often some homogeneous system of recurrence relations determines such an accretion rule. In this case, the nth distribution vector vn defined above has the property that the j,th component of vn is the number of faces of type fj in the nth corona. We then encode the system of recurrences into a transition matrix M such that vn+1 = Mvn holds for all n > 1. When M = [m^-] is such a matrix, the entry m^- is the number of faces of type f that are offspring of a face of type fj. We require the following result from [6]. Proposition 3.2 ([6, Theorem 3.1]). Let T be a tessellation labeled as a Bilinski diagram with accretion rule specified by the transition matrix M and first distribution vector v1. Then the ordinary generating function for the sequence j|Fn|}^=1 is ^(z) = |Fo| + z(p ■ (I - zM)-1vi) , (3.1) where I is the identity matrix and j is the vector of 1s. By using Definition 2.15, we can prove the following more directly than we did in Theorem 3.4 of [6]. Theorem 3.3. If M is the transition matrix of a tessellation T and A is the maximum modulus of an eigenvalue of M, then y (T) = A. Proof. We can write the generating function y>(z) of Proposition 3.2 as a rational function u(z)/v(z), with v(z) determined entirely by (I - zM)-1. Specifically, using Cramer's rule where r denotes the order of M, we have (I - zM)-1 = ,Tl—— adj(I - zM) v y det(I - zM) JV y 1 wvadj(I - zM) (3.2) (-z)r det(M - 11) 1z adj(I - zM) (-z)r X( 1) where x( 1) is the characteristic polynomial (in 1) of M. Entries of the adjoint adj(I-zM) are polynomials in z of degree at most r - 1, and so v(z) = (-z)rx( 1). As x( 1) is a polynomial in 1 of degree exactly r, v(z) has a nonzero constant term and the roots of v occur precisely at the roots of x( 1). These are precisely the reciprocals of the eigenvalues of M. Thus the minimum modulus of a pole of <^(z) is 1/A. As this is the definition of the radius of convergence of a power series expanded about 0, we have y(T) = A. □ 3.2 Monomorphic, uniformly concentric sequences As we have already remarked, valence sequences of face-homogeneous tessellations are unlike edge-symbols of edge-homogeneous tessellations in two significant ways: (i) the requirements for realizability of an edge-symbol are simpler and less stringent than the realizability criteria for a cyclic sequence, and (ii) two or more non-isomorphic face-homogeneous tessellations may share a common valence sequence. This latter property motivates the following definition. Definition 3.4. Let a be a cyclic sequence. If there exists (up to isomorphism) a unique face-homogeneous tessellation with valence sequence a, then we say that a is monomor-phic. If there exist at least two (non-isomorphic) tessellations with valence sequence a, then a is polymorphic. Proposition 3.5 (Moran [10]). All realizable cyclic sequences of length 3 are monomor-phic. A second property of interest is whether a given valence sequence is uniformly concentric. These two properties thus yield four classes of valence sequences. Not surprisingly, the class most amenable to an elegant and simple accretion rule consists of those that are both monomorphic and uniformly concentric. One can find in [13] a complete classification of cyclic sequences of length k for 3 < k < 5 in terms of Definition 3.4 which will help us to narrow our investigation. (It is actually the equivalent dual problem that is treated in [13], and the term "covalence sequence" is used. In the present work we have opted to follow Moran [10], speaking rather in terms of "valence sequences.") We now turn to considering the relative growth rates of tessellations with monomor-phic valence sequences. The ideal condition would be to have that the partial order on cycic sequences is mirrored by the natural order on growth rates: that is, if T1 and T2 are tessellations with valence sequences a1 < a2, then y(T1) < y(T2). For monomorphic, uniformly concentric valence sequences, this is precisely the case, as stated below in Theorem 3.7. In order to prove the theorem, we now demonstrate the necessary machinery via the following example, which can be readily generalized. Example 3.6. Consider T1 and T2 to be face-homogeneous tessellations with monomorphic valence sequences a1 = [4, 5,4,5] and a2 = [4,6,6,4, 5], respectively, both labeled as face-rooted Bilinski diagrams. Note that a1 < a2. We continue the convention that Fi,n denotes the set of faces of the nth corona of Ti for i = 1,2. (The reader may follow Figures 2 through 7.) Starting with T1, we construct by induction a sequence {TQ : j e N} of tessellations such that: 1. TQ = T1 as a base for the induction, 2. if we denote by Fj,n the set of faces in the nth corona of TQ, then for each j e N, the unions of the first n coronas of TQ satisfy (U ^ ^ (U ^ as induced subgraphs, and 3. |F1,n| < |Fj n| for all n e N. To construct T[ from Tq , the valence sequence of the root face of Tq must change from a1 to a2. To do so, we augment the valence of a 5-valent vertex v e U1 to 6-valent and then Figure 2: The first three coronas of T1. Figure 3: The first three coronas of T\'. The dark gray region is a subgraph inserted by augmentation of the valence of a vertex from 5 to 6; the light gray region is a subgraph inserted while interpolating a 6-valent vertex along an incident edge. These insertions continue throughout all coronas of T1. subdivide an edge of (Ui) incident with v by inserting a 6-valent vertex. Augmentation and interpolation are both performed via the insertion of an infinite "cone" as follows. We choose a sequence of edges e2, e3, e4,..., with ei e (U), such that e2 and v are incident with a common face in F1, and for each i > 2, ei and ei+1 are incident with a common face in Fj. On each of these edges we interpolate vertices, and we insert edges connecting vertices between Ui and Ui+1 ensuring that every face so created has covalence 5. Furthermore, if a created face is incident only with interpolated vertices, then its valence sequence is 6 Figure 7: In the diagram to the left, the (4, 6)-edge at the bottom must have a 6-valent vertex interpolated, along with the attendant subgraph. However, we wish to avoid non-concentricity; hence the single 4-valent vertex x is expanded to a (4,6,4)-path as in the diagram on the right. 5 x 5 We continue by induction; suppose a tessellation Tj has been created by this process. Then in the jth corona, there are finitely many faces which require a finite number of vertices to have their valences increased and a finite number of edges along which we must interpolate a vertex. This creates Tj+1 such that |F1,n| < |Fj+1,n| = lF2,n| for n < j + 1, as the first j coronas are comprised only of faces with valence sequence a2. Furthermore, |F1,n| < |Fj + 1,n| for all n G N. In this manner we can construct an infinite sequence of tessellations, namely {Tj : j G N}, with the properties that |F1,n| < |Fj,n| for any j, n G N0, and |Fj,n| = |F2,n| whenever j > n. In the previous example, we constructed the sequence in the process of transforming T1 with valence sequence [4,5,4, 5] into T2 with valence sequence [4,6,6,4, 5]; however, the process of creating {Tj : j g N0} is identical in any case where T\ and T2 are face-homogeneous and uniformly concentric with monomorphic valence sequences a1 and a2, respectively, where a1 < a2. Thus by Lemma 2.18, we obtain the following result. Theorem 3.7 (Growth Comparison Theorem). Let a1 and a2 be monomorphic valence sequences realized by tessellations T1,T2 G G4 4 U G3+ 5 U G3 6, with a1 < a2. Then Y(T1) < y(T2). Our convention is to index the face types (f1,..., fr for some r) in the following order: first wedges, then bricks, then notched bricks, and finally, other face types if any. A wedge in Fn with face type fj is incident with a pi-1-valent vertex in Un-1, for i = 1,..., k. Similarly, the indices of face types of bricks begin with a brick in Fn incident with a p0-valent vertex and a pk-1-valent vertex in Un-1. A new index fj is not introduced if there is some fj for i < j with the same configuration of vertices in Un-1 and Un, up to orientation. For example, the valence sequence [4,6,8,8, 6,4] yields seven face types f1,..., f7, of which f1, f2, and f3 are wedge types and f4 through f7 are brick types. When a monomorphic sequence [p0,... ,pk-1] is realized by a tessellation in G4j4 U G3+,5 U G3,6, then every face, with respect to any Bilinski diagram, can be only a wedge, a brick, or a notched brick. The indexing of face types when pj = pj for i = j allows a stricter labeling which we can use in several other cases. A face f in Fn is a wedge of type wj when the vertex incident with f in Un-1 corresponds to valence p4-1 in a. If instead f is a brick with incident vertices in Un-1 corresponding to valences pj-1 and pj-2 (indices here taken modulo k), then f has face type bj. Finally, if f a notched brick whose incident vertices in Un-1 have valences pj, pj-1 = 3, and pj-2, then f has face type nj. It is important to note that if pj-1 = 3, then faces of type nj never occur as offspring. This stricter labeling is used explicitly only for the few theorems which follow, by which we determine the number of offspring of each instance of these general face types. Furthermore, we demonstrate a first application of the accretion rules and half-counting of faces that were introduced in Section 3.1. Notation. Let T be a face-homogeneous tessellation with valence sequence a, labeled as a Bilinski diagram. We denote by Q(f) the number of faces in Fn+1 that are counted as offspring of a single face of face type f in Fn, for any n > 0. For T G G4,4 U G3+,5 U G3,6 we let 0(wj), 0(bj), and 0(nj) denote the number of offspring of a single wedge, brick, or notched brick of, respectively, of the given type. Lemma 3.8. For a face-homogeneous tessellation in G4,4 UG3+,5 UG3,6 with monomorphic valence sequence a = [p0,... ,pk-1], one has for i G {1,..., k}, A(wj) = Pi-22+ Pi - 2k + 3+ £pj, and (3.3) j/h "(bi) = pj-32+ pj - 2k + 5+ £ pj, (3.4) j/l2 where I1 = {i — 2, i — 1, i} and I2 = {i — 3, i — 2, i — 1, i}. Also, when pj-1 = 3, fi(nj) = pj-3 + pj+1 — 2k + 7 + £ pj (3.5) j/l3 with I3 = {i — 3, i — 2, i — 1, i, i + 1}. Proof. The reader is referred to the three offspring diagrams shown in Figure 8. Letting i G {1,..., k}, the first diagram applies when pj-1 > 4. If also pj-2,pj > 4 as in the diagram, then we have n(wi) = + 4 + k — 2 + £ (pj — 3) j/h pj-2 + pj — 2k + 3+ J2 pj. 2 j/h If instead pj-2 = 3, then the number of wedge offspring of wj is pj — 4 + — 3), 2j j/h Un-1 Un Un +1 un-1 Un Un + 1 Pi-1 Wi+1 bi+2 Wi+2 Wi-2 bi-1 } Wi-1 pi-1 Pi-2 H bi N ► Wi+1 ■ bi+2 ► Wi+2 ► Wi-3 bi-2 Wi-2 Un-1 Un Un + 1 Pi Pi-1 Pi-2 Wi+2 bi+3 Wi+3 Wi-3 b i-2 } Wi-2 Figure 8: Offspring diagrams for the three general face types (respectively wedges, bricks, and notched bricks) of a tessellation with monomorphic, uniformly concentric valence sequence [po,... ,pk-1]. the number of brick offspring is k — 3, and the number of notched brick offspring is 1. Thus when pi-2 = 3, "(wi) = 1 + ^ + k - 3+ £ (pj - 3) j/h = — 2 + Pi—4 + k — 2+ £ (Pj — 3) j/h Pi-2 — 4 Pi — 4 ^ = —2— + + k — 2 + ^(Pj — 3) j/ii as before; likewise when Pi = 3. Analogous arguments hold for the offspring of bricks and notched bricks. □ The process of establishing an accretion rule and accompanying transition matrices is considerably simplified for tessellations in G4 4 by virtue of the absence of notched bricks. By applying the following lemma and Theorem 3.3, one can then compute the growth rate explicitly of any monomorphic valence sequence realizable in G4 4. Recall that by Proposition 2.3, all such valence sequences are uniformly concentric. Lemma 3.9. Let [p0, ... ,Pk-1] be the monomorphic valence sequence for a tessellation T € G4,4. Then T has an accretion rule which admits the block transition matrix M = A B C D with A = (ai,j), B = (bi,j), C = (ci,j), and D = (di,j) given by 0 j-i=0 ai,j = ( 1 (-¿-I - 4) j - i e {1, k - 1} (mod k) [Pi-1 - 3 otherwise, (3.6) 0 I pi-1 - 3 j - i G {0,1} (mod k) j - i G {2 otherwise, 1 (pi-1 - 4) j - i G {2, k - 1} (mod k) j - i G {0,1} (mod k) otherwise, j - i G {0,1, k - 1} (mod k) otherwise, (3.7) (3.8) (3.9) for i, j G {1, ... ,k}. Proof. Since all general face types are wedges or bricks, we need demonstrate only that the entries a^ and correspond to numbers of offspring of the k face types in wedge configurations and that the entries 6i,j- and d^- correspond to numbers of offspring of the k face types in brick configurations. U„-1 U„ U n+1 pi-1 pi wi pi-2 2 (pi - 4) faces of type wi+1 } pi+1 - 3 faces of type wi+2 } pi-3 - 3 faces of type Wj-2 2 (pi-2 - 4) faces of type Wj-1 Figure 9: Offspring of a wj face in a tessellation T e G4,4, where i e {1,..., k}. The offspring of wedges of type wj are shown in Figure 9, and the offspring of a brick of type bj is shown in Figure 10. The ordering of face types is w1, w2,..., wk, b1,..., bk. U„_ 1 U„ U n+1 pi-1 pi-2 pi bi .......^ pi-3 2 (pi - 4) faces of type wi+1 } pi+1 - 3 faces of type Wi+2 } pi-4 - 3 faces of type wi-3 2 (pi-3 - 4) faces of type wi-2 b d Figure 10: Offspring of a bj face in a tessellation T e G4,4, where i e {1,..., k}. Recalling that the (i, j)-entry of a transition matrix M is the number of faces of the ith indexed type which are produced in Fn+1 as offspring of a face of the jth indexed type in Fn, it is straightforward to verify from these two offspring diagrams that the entries of M Remark 3.10. We emphasize the breadth of this class of monomorphic, uniformly concentric valence sequences. In addition to the many monomorphic face-homogeneous tessellations in G3,6 U G3+j5 U G4,4, there are many with covalence 3 (cf. Proposition 3.5). By Proposition 2.19, all edge-transitive tessellations of constant covalence are included, except for those of the with valence sequence [3,p, 3,p] (edge-symbol (3,p; 4,4}), as they are not uniformly concentric. By Proposition 2.3, a k-covalent tessellation T is uniformly concentric whenever k > 6. If k > 7 and if a is monomorphic, then a > [3,3,3, 3, 3,3,3]. In that case, Theorem 3.7 and Proposition 2.20 tell us that a has growth rate at least Y([3, 3, 3, 3, 3, 3, 3]) = 2(3 + V5). 3.3 Monomorphic non-concentric sequences The purpose of this section is to characterize the six forms of monomorphic, non-concentric valence sequences with positive angle excess. These sequences give rise to face types other than wedges, bricks, and notched bricks, and so the foregoing methods cannot be applied to compute their growth rates. An interesting situation arises when a tessellation is not uniformly concentric but nonetheless, by prudent selection of the root, admits some Bilinski diagram that is concentric. To illustrate this point, we examine sequences of the form [4,p, q]. Example 3.11. Consider the valence sequence a = [4,p, q] with 4 < p < q, where 1 + 1 < 4, and let T be a face-homogeneous tessellation with valence sequence a. For a to be realizable, clearly p and q must be even. Note as well that the inequality (2.4) is satisfied. While a is monomorphic and admits a concentric Bilinski diagram, a is not uniformly concentric (cf. the second case of Proposition 2.4). When a Bilinski diagram of T admits a 4-valent vertex v0 G Un (for some n) adjacent to the vertices u1, G Un-1 and v1, v2 G Un, then the diagram is not concentric; the vertices v1 and v2 must also be adjacent, as T is 3-covalent. Hence ({v0, v1, v2}} is a cycle within (Un}, causing the Bilinski diagram to be non-concentric. However, it is possible to avoid this configuration by choosing the root of the Bilinski diagram to be either a p-valent or a q-valent vertex. When so labeled, only four face types occur, as demonstrated by the offspring diagrams in Figure 11. Un-1 U„ U„+1 U„_1 U„ U„+1 are correct. □ • : 4-valent ► f1 o: p-valent □: q-valent Figure 11: Offspring diagrams for a concentric tessellation with valence sequence [4,p, q]. One sees here that if the root is taken to be a p-valent vertex, the first corona consists entirely of faces of type fi, which produce in turn only offspring of types f2 and f3. Similarly, given a q-valent root, the first corona consists entirely of faces of type f2, which produce in turn only offspring of types f1 and f4. The non-concentric configuration described above can never be produced among the descendants of faces of types f1 or f2. Inspection of Figure 11 gives the first and second columns of the transition matrix M; the third and fourth columns, corresponding to f3 and f4, merit further explanation. A face of type f3 in Fn+1 has a p-valent vertex in Un+1; this vertex is incident with p - 5 faces of type f1 in Fn+2. So the behavior of a face of type f3 is effectively to collapse one of the faces in Un+2 of type f1 begotten by the adjacent face of type f2. Faces of type f4 behave similarly, collapsing a face of type f2 . These considerations give us M = 0 1 (p - 4) -1 0' 1 (q - 4) 0 0 -1 1 0 0 0 0 10 0 as the transition matrix M for this accretion rule for T. As the characteristic equation for M is of degree 4, it can be solved to determine that the maximum modulus of an eigenvalue of M is A = W[2(p - 4)(q - 4) - 16] + 2^(p - 4)2(q - 4)2 - 16(p - 4)(q - 4). By Theorem 3.3 and Theorem 2.16, A is the growth rate of T. This quantity can be minimized by minimizing pq subject to the initial conditions 1 + 1 < 4 and that p and q be even. We shall see in Section 3.4 the role played by this example. Growth rate formulas for each of the other five classes are derived in the Appendix. Theorem 3.12. Let a be a valence sequence such that n(a) > 0. Then a is both monomor-phic and non-concentric if and only if a is of one of the following six forms: (i) [3,p,p], with p > 14 and even; (ii) [4,p, q], with p and q both even, 4 < p < q, and p + 1 < |; (iii) [3,p, 3,p], with p > 7; (iv) [3, p, 4, p], with p > 5 and even; (v) [3, 3,p, 3, p], with p > 5; or (vi) [3,3,p, 3, q], with p, q > 4 and p + 1 < 2. Proof. The parity conditions and the inequalities bounding the parameters in each case are minimal such that a be indeed realizable as a tessellation with n(a) > 0. As noted in Remark 3.10, all valence sequences of length at least 6 are uniformly concentric. Furthermore, by Proposition 3.5, all valence sequences of length 3 are monomor-phic. Valence sequences [3,p,p], [4,p, q], and [3,p, 3,p] give rise to tessellations exemplifying cases 1, 2, and 4 respectively of Proposition 2.4, and hence cannot be uniformly concentric. As a face-homogeneous tessellation with valence sequence [3,p, 3,p] is also edge-transitive, the sequence must be monomorphic. The proof that the sequence [3, p, 4, p] must be monomorphic and uniformly non-concentric is given in the Appendix, where the growth rate of a corresponding tessellation is determined. We now prove that [3,3,p, 3,p] is monomorphic for all p > 5. As a 3-valent vertex is incident with a common face with any two of its neighbors, every 3-valent vertex must be adjacent to at least two p-valent vertices; otherwise some face would be incident with a (3,3, 3)-path. Consider a p-valent vertex vi. By face-homogeneity, v1 is adjacent to some 3-valent vertex u1, with u1 adjacent in turn to a 3-valent vertex u2 which is not adjacent to v0. But then the other vertex adjacent to u1 must be a p-valent vertex v2. This forces the pattern of valences at regional distance 1 from v1 to be (3,3,p,..., 3, 3,p); as v1 was arbitrary, this must be the pattern of valences at regional distance 1 from any p-valent vertex. As every vertex is at regional distance 1 from some p-valent vertex, [3, 3,p, 3,p] must be monomorphic; the first two coronas of a tessellation with this valence sequence rooted at a p-valent vertex is depicted in Figure 12. Furthermore, this local configuration to a p-valent vertex forces the local behaviors to a (3,3)-edge and a 3-valent vertex shown in Figure 13. Hence when a 3-valent vertex v0 is taken as the root of the Bilinski diagram of a tessellation with valence sequence [3, 3,p, 3, p], a pendant vertex occurs in (U3). This is shown in Figure 14. So [3, 3,p, 3,p] is monomorphic but not uniformly concentric; the argument for [3, 3,p, 3, q] is analogous. We have shown these six forms to be both monomorphic and non-concentric; that these are the only such valence sequences is proved via the exhaustive examination of cases in the Appendix. □ Figure 12: The first two coronas of a tessellation with valence sequence [3, 3, p, 3, p] rooted at a p-valent vertex. Each shaded region indicates p - 3 faces in F2 all having the same face type. 3.4 The main result The following theorem establishes the so-called "golden mean" as the least rate of exponential growth for face-homogeneous tessellations with monomorphic valence sequences. Theorem 3.13 (Least Exponential Growth Rate of Monomorphic Valence Sequences). The least growth rate of a face-homogeneous tessellation with monomorphic valence sequence a such that n(a) > 0 is 2 (1 + V5) and is attained by exactly the tessellations with valence sequences [4,6,14] and [3,4, 7,4]. • : 3-valent o : p-valent • : 3-valent o: p-valent (A) (B) Figure 13: (A) Local configuration along an edge with edge-symbol (3, 3; 5,5} in a face-homogeneous tessellation with valence sequence [3, 3,p, 3,p]. (B) Local configuration in the same tessellation when rooted at a 3-valent vertex. Table 1: Table of the least exponential growth rate within each monomorphic class of valence sequences. All rates of growth have been truncated at four decimal places rather than being rounded. Class a Y(T 4. The following theorem gives a simple sufficient condition under which a realizable valence sequence is polymorphic. U„ Un+1 Figure 15: A configuration of faces demonstrating polymorphicity. Proposition 4.1. Let a = [po,... ,pk-1] be the valence sequence of a face-homogeneous tessellation T G G4,4 U G3+,5. If there exist distinct i, j G {0,..., k — 1} such that Pi,Pi+1 > 4 and either 1. Pi = Pj, Pi+i = Pj+i, andpi+2 = Pj+2, or 2. Pi = Pj, Pi+i = Pj-i, andPi+2 = Pj-2, then a is polymorphic. Proof. As the only two forms of valence sequences of length k = 4 that satisfy the hypothesis, namely [p,p,p, q] and [p,p, q, r], are polymorphic (see Appendix), we assume that k > 5. Also, since condition (2) is identical to (1) save for orientation within the cyclic sequence, it suffices to assume that there are distinct i, j such that (1) holds. Furthermore, we may assume i = 0 due to the rotational equivalence of valence sequences. Since k > 5, there exists for some n a face in Fn incident with three consecutive vertices u0, ui, u2 G Un with valence p(um) = pm for m = 0,1,2. Let b be the brick in Fn+i incident with the edge u0ui, and let b' be the brick (or perhaps notched brick if p2 = 3) in Fn+i incident with the edge uiu2. Let vi,..., vr be the vertices in Un+i incident with ui in consecutive order, so that vi is incident with b and vr is incident with b'. Thus r = pi - 2 > 2. If a contains a subsequence [q,pi,p2] with q = p0, then p(vr) may equal either p0 or q, resulting in a choice of face types for b', and we're done. Otherwise we must have p(vr) = p0, which forces the vertex vr-2 and subsequent alternate neighbors of ui in Un+i also to be p0-valent. If pi is even, then p(vi) may equal either p2 or pj+2 in which case the wedge w G Fn+i incident with vertices vi, ui, v2 may be of either type w2 or type Wj+2, and T is polymorphic. (See Figure 15.) If pi is odd, then working backward as in the even case forces p(vi) = p0, which implies that either p0 = p2 or p0 = Pj+2, and without loss of generality, we assume the former. Now we may assign p(v2) to be either p0 or Pj+2, and the argument proceeds as in the even case. □ The existence of polymorphic valence sequences considerably complicates the computation of growth rates of face-homogeneous tessellations. The above proof suggests that, unlike in the monomorphic case, polymorphic valence sequences may admit many different accretion rules, as we illustrate in the next section. 4.2 Two non-isomorphic tessellations with the same valence sequence The minimal polymorphic valence sequence under the partial order on cyclic sequences, namely [4,4,4, 5], is unfortunately not amenable to study via our methods. In fact, there is no well-defined transition matrix between coronas, and this problem is shared by all valence sequences of the form [4,4,4, q] for q > 4. However, [4,4,6,8] provides us with the opportunity to investigate two distinct (but related) accretion rules. The valence sequence [4,4,6, 8] is representative of form [p,p, q, r] discussed in the Appendix. As every face is incident with a pair of adjacent 4-valent vertices, every realization of this valence sequence contains a countable infinity of pairwise-disjoint double rays, each induced exclusively by 4-valent vertices. Figure 16 (A) shows a strip-like patch bordering a double ray of 4-valent vertices. To obtain Figure 16 (B) from this (or vice versa), one can fix pointwise the half-plane on one side of the double ray while translating the half-plane on the other side along one edge of the double ray. To construct still other such (non-isomorphic) realizations, one can choose to "translate" along any one of these double rays by leaving fixed the half-plane on one side of the double ray but translating the half-plane on the other side by one edge. Since there exists ]- -c p- -c ]- -c -c -c :- - - ]- • : 4-valent o: 6-valent □ : 8-valent (A) (B) Figure 16: Two non-isomorphic patches of a tessellation with valence sequence [4,4,6,8], showing possible neighborhoods of double rays of 4-valent vertices. a countable infinity of double rays along which one may choose to translate one or the other or neither of the adjacent half-planes, there exists an uncountable class of pairwise non-isomorphic tessellations that all have the same valence sequence [4,4, 6,8]. While one might expect that all tessellations having the same valence sequence always have the same growth rate, we show that this is not so. We begin by observing that every 4-valent vertex in a face-homogeneous tessellation with valence sequence [4,4,6,8] is adjacent to two other 4-valent vertices and two vertices with valences 6 or 8; thus any given 4-valent vertex either has exactly one 6-valent and one 8-valent neighbor, has two 6-valent neighbors, or has two 8-valent neighbors. Furthermore, every 4-valent vertex lies on a double ray (two-way infinite path) of 4-valent vertices; if one vertex along this path has a 6-valent neighbor and an 8-valent neighbor, then so does every other vertex along the double ray. This is the behavior demonstrated in Figure 16 (A). If the local configuration specified in Figure 16 (A) is enforced along every double ray of 4-valent vertices, then the tessellation obtained is unique; let this tessellation be T1. We can then construct offspring diagrams for T1 as given in Figure 17. It is interesting to note that T1 is the dual of the Cayley graph of the group with presentation G1 = (a, b,c | a2 = b2 = c2 = (bc)3 = (caba)4 = l) . Encoding the offspring diagrams into a matrix, we obtain the transition matrix M1 of T1 given below. The four entries underlined in the matrix are the only entries which change between this example and the next example, T2, that we construct. "0 0 0 1 0 0 0 0" 00100000 3 10 1110 0 M = 25200220 M1 = 0 1 1 0 0 0 1 0 0 0 1 1 0 0 0 1 1 0 0 1 1 0 0 0 11000100 The characteristic polynomial of M1 is f1(z) = (z - 1)(z + 1) (z2 + 3z + 1) (z4 - 3z3 - 4z2 - 3z +1) , which in turn gives that the eigenvalue of maximum modulus of M1 is A1 = 4 [3 + 733 + 2^1+ « 4.13016. Considering again the double-rays of 4-valent vertices, it is trivial to note that if a vertex on such a double ray has two 6-valent neighbors in the tessellation, then both vertices adjacent to it in the double-ray have two 8-valent neighbors. This local behavior is shown in Figure 16 (B). If this pattern is extended to all such double rays we obtain the tessellation T2, which is also the dual of a Cayley graph. The underlying group of this Cayley graph is G2 = (a, b,c,d | a2 = b2 = c2 = d2 = (ab)2 = (ad)2 = (cd)3 = (bc)4) . The growth behavior of T2 differs from that of T1 only in the offspring of faces of types f3 and f4, as shown in the offspring diagrams in Figure 18. Un-1 Un Un + 1 -1 Un Un + 1 f4 Un — 1 Un Un + 1 Un-1 Un Un+1 Un — 1 Un Un + 1 Kb f8 A f6 £4 Un 1 Un Un n+1 f7 Un — 1 Un Un+1 f8 f6 1 • : 4-valent 6-valent □ : 8-valent Figure 18: Offspring diagrams for T2. f 3 f 3 f 4 1 3 f 2 f 3 f 4 The effect of the change of offspring of types f2 and f3 in the transition matrix of T2 lies only in the underlined 2 x 2 submatrix of Mi, while the remainder of the matrix M2 remains identical to M1. Hence we have M2 = 00100000 00010000 3 10 1110 0 25200220 01100010 00110001 1 0 0 1 1 0 0 0 11000100 The characteristic polynomial of M2 is f2(z) = (z - 1)2 (z6 + 2z5 - 15z4 - 40z3 - 15z2 + 2z + 1) . As polynomials of degree 6 are unfortunately not solvable by radicals, we obtain by approximation that the root of maximum modulus is A2 « 4.14659. As these growth rates are nearly the same, there is only a small difference in corona sizes in the first several coronas. However, the size of the coronas and distribution of face types differs greatly farther from the root. To demonstrate this, Table 2 gives corona sizes in Bilinski diagrams of T and of T2, both rooted at 4-valent vertices. Note that the sizes of the coronas of T2 dominate those of T1 only after the 13th corona. 4.3 Some conjectures Ideally, all tessellations realizing the same polymorphic valence sequence would have the same growth rate. The example of valence sequence [4,4,6,8] illustrates that this is not so. We propose the following definitions. Definition 4.2. Let a be some polymorphic valence sequence, and define Ta to be the set of isomorphism classes of face-homogeneous tessellations with valence sequence a. Let Aff =inf{y(T): T e }, (4.1) Aff =sup{Y(T): T e }, (4.2) L = {T : T e and y(T) = Aff}, and (4.3) H = {T : T e and y(T) = Aff}. (4.4) We conjecture that the lower and upper bounds Aa and ACT for any given valence sequence a are realized. Conjecture 4.3. Let a be a polymorphic valence sequence. Then L and Ha are nonempty. Bearing in mind the polymorphic valence sequence [4,4,6,8] analyzed in Section 4.2, we propose as a conjecture the following sharper version of Theorem 3.7. Conjecture 4.4. Let a1 and a2 be valence sequences such that a1 < a2. Then Affl < Aff2. (4.5) In the spirit of the famous quote of the late George Polya [12] ("If you can't solve a problem, then there is an easier problem you can solve: find it."), we offer the following (perhaps) easier conjecture. Table 2: Corona sizes in Ti and T2; emphasis on the 14th corona beyond which the coronas of T2 appear to exceed in size those of Ti. n |Fl,n| |F2,n| n |F1 n| |F2 n 1 4 4 29 1.20050 x 1018 1.27748 x 1018 2 30 28 30 4.95826 x 1018 5.29701 x 1018 3 110 108 31 2.04784 x 1019 2.19652 x 1019 4 494 468 32 8.45791 x 1019 9.10786 x 1019 5 1938 1900 33 3.49325 x 1020 3.77673 x 1020 6 8272 7956 34 1.44277 x 1021 1.56603 x 1021 7 33464 32868 35 5.95887 x 1021 6.49377 x 1021 8 140046 136380 36 2.46111 x 1022 2.69268 x 1022 9 573610 565956 37 1.01648 x 1023 1.11655 x 1023 10 2.38167 x 106 2.34358 x 106 38 4.19821 x 1023 4.62986 x 1023 11 9.80378 x 106 9.73259 x 106 39 1.73393 x 1024 1.91983 x 1024 12 4.05773 x 107 4.02988 x 107 40 7.16140 x 1024 7.96071 x 1024 13 1.67365 x 108 1.67318 x 108 41 2.95777 x 1025 3.30099 x 1025 14 6.91836 x 108 6.93034 x 108 42 1.22161 x 1026 1.36878 x 1026 15 2.85585 x 109 2.87639 x 109 43 5.04544 x 1026 5.67580 x 1026 16 1.17992 x 1010 1.19181 x 1010 44 2.08385 x 1027 2.35352 x 1027 17 4.87218 x 1010 4.94504 x 1010 45 8.60662 x 1027 9.75910 x 1027 18 2.01257 x 1011 2.04947 x 1011 46 3.55467 x 1028 4.04670 x 1028 19 8.31149 x 1011 8.50179 x 1011 47 1.46814 x 1029 1.67800 x 1029 20 3.43297 x 1012 3.52419 x 1012 48 6.06363 x 1029 6.95799 x 1029 21 1.41782 x 1013 1.46172 x 1013 49 2.50438 x 1030 2.88520 x 1030 22 5.85596 x 1013 6.05990 x 1013 50 1.03435 x 1031 1.19637 x 1031 23 2.41857 x 1014 2.51322 x 1014 60 1.49395 x 1037 1.79797 x 1037 24 9.98918 x 1014 1.04199 x 1015 70 2.15777 x 1043 2.70207 x 1043 25 4.12567 x 1015 4.32117 x 1015 80 3.11654 x 1049 4.06079 x 1049 26 1.70397 x 1016 1.79166 x 1016 90 4.50134 x 1055 6.10274 x 1055 27 7.03766 x 1016 7.42979 x 1016 100 6.50145 x 1061 9.17148 x 1061 28 2.90667 x 1017 3.08066 x 1017 200 2.56861 x 10123 5.38996 x 10123 Conjecture 4.5. Let ai and a2 be valence sequences with ai < a2. Then Affl < Aff2. (4.6) If Conjecture 4.4 holds, then one could delete the condition of monomorphicity from the hypothesis of Theorem 3.7 and therefore from Theorem 3.13 as well. Moreover, the Appendix could be much abbreviated. For example, one could eliminate the exhaustive consideration of the many forms of 6-covalent face-homogeneous tessellations listed and treated there by observing that the least valence sequence a of length 6 with n(a) > 0 is [3, 3, 3,3,3,4]. Thus, if any tessellation with the polymorphic valence sequence [3,3,3, 3, 3,4] has growth rate greater than i(1 + a/5), then so does every tessellation with valence sequence a > [3, 3, 3,3,3,4]. Beyond these conjectures, there are some open questions. Consider the partially ordered set of valence sequences, and in particular, the poset consisting of the polymorphic valence sequences. Question 4.6. As one goes up a chain in the poset, do intervals of the form [act , ACT] become (asymptotically) longer? Question 4.7. Do the intervals in the complement of U { [Act , Act] : a is polymorphic} CT become arbitrarily long? If the answer to Question 4.7 is negative, we pose the following. Question 4.8. If x is a sufficiently large real number, is there always some polymorphic valence sequence a such that ACT < x < ACT ? Or, on the other hand, Question 4.9. Do there exist polymorphic sequences a, t such that [aCT,aCT] n [aT,aT] = 0? References [1] S. Bilinski, Homogene mreze ravnine, Rad Jugoslav. Akad. Znanosti i Umjetnosti 271 (1948), 145-255,http://dizbi.hazu.hr/object/14 96. [2] S. Bilinski, Homogene Netze der Ebene, Bull. Internat. Acad. Yougoslave Cl. Sci. Math. Phys. Tech. (N. S.) 2 (1949), 63-111. [3] J. A. Bruce and M. E. Watkins, Concentric Bilinski diagrams, Australas. J. Combin. 30 (2004), 161-174, https://ajc.maths.uq.edu.au/pdf/30/ajc_v30_p161.pdf. [4] S. J. Graves, Growth of Tessellations, Ph.D. thesis, Syracuse University, May 2009. [5] S. J. Graves, Tessellations with arbitrary growth rates, Discrete Math. 310 (2010), 2435-2439, doi:10.1016/j.disc.2010.04.023. [6] S. J. Graves, T. Pisanski and M. E. Watkins, Growth of edge-homogeneous tessellations, SIAM J. Discrete Math 23 (2009), 1-18, doi:10.1137/070707026. [7] S. J. Graves and M. E. Watkins, Appendix to "Growth of face-homogeneous tessellations", 2017, arXiv:1707.03443 [math.CO]. [8] B. Grunbaum and G. C. Shephard, Edge-transitive planar graphs, J. Graph Theory 11 (1987), 141-155, doi:10.1002/jgt.3190110204. [9] B. Grunbaum and G. C. Shephard, Tilings and Patterns, A Series of Books in the Mathematical Sciences, W. H. Freeman & Company, New York, 1987. [10] J. F. Moran, The growth rate and balance of homogeneous tilings in the hyperbolic plane, Discrete Math. 173 (1997), 151-186, doi:10.1016/s0012-365x(96)00102-1. [11] P. Niemeyer and M. E. Watkins, Geodetic rays and fibers in one-ended planar graphs, J. Combin. Theory Ser. B 69 (1997), 142-163, doi:10.1006/jctb.1996.1733. [12] G. P6lya, Mathematical Discovery: On Understanding, Learning, and Teaching Problem Solving, Volume I, John Wiley & Sons, New York, 1962. [13] J. Siagiova and M. E. Watkins, Covalence sequences of planar vertex-homogeneous maps, Discrete Math. 307 (2007), 599-614, doi:10.1016/j.disc.2006.07.014. /^creative ^commor ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 14 (2018) 319-327 https://doi.org/10.26493/1855-3974.1349.b67 (Also available at http://amc-journal.eu) The 4-girth-thickness of the complete graph Christian Rubio-Montiel UMILAFMIA 3175 CNRS at CINVESTAV-IPN, 07300, Mexico City, Mexico División de Matemáticas e Ingeniería, FES Ac-UNAM, 53150, Naulcalpan, Mexico Received 10 March 2017, accepted 21 June 2017, published online 19 September 2017 Abstract In this paper, we define the 4-girth-thickness 6(4, G) of a graph G as the minimum number of planar subgraphs of girth at least 4 whose union is G. We prove that the 4-girth-thickness of an arbitrary complete graph Kn, 6(4, Kn), is [np] for n = 6,10 and 6(4, K6) = 3. Keywords: Thickness, planar decomposition, girth, complete graph. Math. Subj. Class.: 05C10 1 Introduction A finite graph G is planar if it can be embedded in the plane without any two of its edges crossing. A planar graph of order n and girth g has size at most g—■ (n - 2) (see [6]), and an acyclic graph of order n has size at most n - 1, in this case, we define its girth as to. The thickness 6(G) of a graph G is the minimum number of planar subgraphs whose union is G; i.e. the minimum number of planar subgraphs into which the edges of G can be partitioned. The thickness was introduced by Tutte [11] in 1963. Since then, exact results have been obtained when G is a complete graph [1, 3, 4], a complete multipartite graph [5, 12, 13] or a hypercube [9]. Also, some generalizations of the thickness for the complete graph Kn have been studied such that the outerthickness 6o, defined similarly but with outerplanar instead of planar [8], and the S-thickness 6S, considering the thickness on a surfaces S instead of the plane [2]. See also the survey [10]. We define the g-girth-thickness 6(g, G) of a graph G as the minimum number of planar subgraphs of girth at least g whose union is G. Note that the 3-girth-thickness 6(3, G) is the usual thickness and the TO-girth-thickness 6(to, G) is the arboricity number, i.e. the minimum number of acyclic subgraphs into which E(G) can be partitioned. In this paper, we obtain the 4-girth-thickness of an arbitrary complete graph of order n = 10. E-mail address: christian@cs.cinvestav.mx (Christian Rubio-Montiel) ©® This work is licensed under http://creativecommons.org/licenses/by/3.0/ 2 The exact value of 6(4, K„) for n = 10 Since the complete graph Kn has size and a planar graph of order n and girth at least 4 has size at most 2(n — 2) for n > 3 and n — 1 for n G {1,2} then the 4-girth-thickness of Kn is at least n(n — 1) 2(2n — 4) n + 1 1 n + 2 2n — 4 4 + for n > 3 and also |"n+2"| for n G {1,2}, we have the following theorem. Theorem 2.1. The 4-girth-thickness 0(4, Kn) of Kn equals for n = 6,10 and 0(4, Ke) =3. Proof. Figure 1 displays equality for n < 5. i 2 , 2 4 2 Figure 1: 0(4, Kn) = f^] for n =1,2, 3,4,5. To prove that 0(4, Ke) = 3 > = 2, suppose that 0(4, Ke) = 2. This partition define an edge coloring of Ke with two colors. By Ramsey's Theorem, some part contains a triangle obtaining a contradiction for the girth 4. Figure 2 shows a partition of Ke into tree planar subgraphs of girth at least 4. Figure 2: 0(4, Ke) = 3. For the remainder of this proof, we need to distinguish four cases, namely, when n = 4k — 1, n = 4k, n = 4k + 1 and n = 4k + 2 for k > 2. Note that in each case, the lower bound of the 4-girth thickness require at least k + 1 elements. To prove our theorem, we exhibit a decomposition of K4k into k +1 planar graphs of girth at least 4. The other three cases are based in this decomposition. The case of n = 4k - 1 follows because K4fc_i is a subgraph of K4k. For the case of n = 4k + 2, we add two vertices and some edges to the decomposition obtained in the case of n = 4k. The last case follows because K4fc+1 is a subgraph of K4fc+2. In the proof, all sums are taken modulo 2k. 1. Case n = 4k. It is well-known that a complete graph of even order contains a cyclic factorization of Hamiltonian paths, see [7]. Let G be a subgraph of K4k isomorphic to K2k. Label its vertex set V(G) as {v1; v2,..., v2k}. Let Fi be the Hamiltonian path with edges V1V2, V2V2k, V2kV3, V3V2fc_l, . . . , V2 + fcVi+fc. Let Fi be the Hamiltonian path with edges VjVj+i, Vi+iVj_i, Vj_iVi+2, Vi+2Vi-2, . . . , Vi+fc+lVi+fc, where i e {2,3,..., k}. Such factorization of G is the partition {E(Fi), E(F2),..., E(Fk)}. We remark that the center of Fi has the edge e = vi+ ph-j vi+ p 3k-j, see Figure 3. a) i + 1 M f1 +r t- i-i v,+ r 32k v,+r k i-i +r 21 v,+ r k Vi+k+1 vi+k-1 b) v, ^%i+1 vi+r3í i+1 v,+r k i-1 +r kk i Mt i vi+k-1 vi+k+1 Figure 3: The Hamiltonian path F^ Left a): The dashed edge e for k odd. Right b) The dashed edge e for k even. Now, consider the complete subgraph G' of K4k such that G' = K4k \ V(G). Label its vertex set V(G') as {vl, v2,..., v2k} and consider the factorization, similarly as before, {E(F'), E(F2),..., E(Fk)} where F/ is the Hamiltonian path with edges vivi+l> vi+lvi-l, vi-lvi+2> v'+2v'-2, . . . , vi+fc+lvi+fc, where i € {1,2,..., k}. Next, we construct the planar subgraphs Gl, G2,...,Gk-l and Gk of girth 4, order 4k and size 8k - 4 (observe that 2(4k - 2) = 8k - 4), and also the matching Gk+l, as follows. Let Gj be a spanning subgraph of K4k with edges E(Fj) U E(F/) and vivi+i, ivi+i, vi+ivi_i, vi+ivi_i, vi_ivi+2, vi_ivi+2,...,vi+fc+ivi+fc, vi+fc+ivi+fc v v where i e {1,2,..., k}; and let Gk+1 be a perfect matching with edges vjvj for j e {1, 2,..., 2k}. Figure 4 shows Gj is a planar graph of girth at least 4. b) vi V2 G k+1 V2fc Figure 4: Left a): The graph Gj for any i e {1, 2,..., k}. Right b) The graph Gk+1. v v k + 1 To verify that K4k = |j G^ 1) If the edge v^ vi2 of G belongs to the factor Fj i=1 then vjj vi2 belongs to Gj. If the edge is primed, belongs to Gj. 2) The edge v^ v-2 belongs to Gk+1 if and only if i1 = i2, otherwise it belongs to the same graph Gj as vj1 vj2. Similarly in the case of vvj2 and the result follows. 2. Case n = 4k - 1. Since K4k-1 c K4k, we have k +1 < 0(4, K4fc_1) < 0(4, K4fc) < k + 1. 3. Case n = 4k + 2 (for k = 2). Let {G1,..., Gk+1} be the planar decomposition of K4k constructed in the Case 1. We will add the two new vertices x and y to every planar subgraph Gj, when 1 < i < k + 1, and we will add 4 edges to each Gj, when 1 < i < k, and 4k +1 edges to Gk+1 such that the resulting new subgraphs of K4k+2 will be planar. Note that (42kt + 4k + 4k + 1 = (4k+2t. To begin with, we define the graph ffk+1 adding the vertices x and y to the planar subgraph Gk+1 and the 4k +1 edges {xy, xv1,xv2,xv3,xv4,... ,xv2k_1,xv2k, yv'x,yv2,yv3,yv4,..., yv2k_1,yv2k}. The graph Hk+1 has girth 4, see Figure 5. In the following, for 1 < i < k, by adding vertices x and y to Gj and adding 4 edges to Gj, we will get a new planar graph Hj such that {H1,..., Hk+1} is a planar decomposition of K4k+2 such that the girth of every element is 4. To achieve it, the given edges to the graph Hj will be vjx, xvj_1,vjy, yvj_1, for some odd j e {1, 3,..., 2k - 1}. According to the parity of k, we have two cases: Figure 5: The graph Hk+1. • Suppose k odd. For odd i G {1, 2,..., k}, we define the graph Hi adding the vertices x and y to the planar subgraph Gi and the 4 edges {xvi+ [f] -1, xvi+ [32k 1 , yvi+ \32k 1-1' yvi+ [21 } when is even, otherwise {y<+r f 1-1'yVi+[ 32k 1 'XM 32k 1-1'XVI+[ f 1}. Additionally, for even i G {1, 2,..., k}, we define the graph Hi adding the vertices x and y to the planar subgraph Gi and the 4 edges {xvi+r 11-1, xvi+r 11, yvi+ r t1-1, yvi+ r 11} when is even, otherwise {yvi+rk 1-1,yvi+r 11 ,xvi+r 11-1 ,xvi+r 11}. Note that the graph Hi has girth 4 for all i, see Figure 6. • Suppose k even. Similarly that the previous case, for odd i G {1,2,..., k}, we define the graph Hi adding the vertices x and y to the planar subgraph Gi and the 4 edges {xvi+r 32k 1+1, xvi+ r f 1, yvi+ r f 1+1, yv r 32k 1} when is even, otherwise {yvi+ r 32k 1+1, yvi+ r 3^ 1, xvi+r t 1+1, xvi+ r 32k 1}. On the other hand, for even i G {1, 2,..., k}, we define the graph Hi adding the vertices x and y to the planar subgraph Gi and the 4 edges {xvi+r 11, xvi+ r 11 -1, yvi+ r 11, yv*+r k 1 -1} when is even, otherwise {yvi+rk 1, yvI+r 11-1, xvi+r 11, xv r 11 -1}. Note that the graph Hi has girth 4 for all i, see Figure 7. a) Vk+i k+i Pk+i k+i vx Vi+\k 1-1 •Vi+r 11 v,-+r 3k W^Vi + k Vi+k+1 Vi+k-1 vx _«i+1 Vi+r 11-1 •M 11 V,-+r 3k W^Vi + k Vi+k+1 Vi+k-1 V V Figure 6: The graph H when k is odd and its auxiliary graph F*. Above a) When i is odd. Botton b) When i is even. a) Vk + i k + i b) Vk + i k+i vx »M k i+ k -i v,+\ -k l+i »M k i v,+\ * i vi+k-1 Vi+k+1 V, J'i + l »M k i+ k -i Vi+\3k l+i »M k i Vi+\ ¥ l Vi+k+l vi+k-1 V V Figure 7: The graph H when k is even and its auxiliary graph F*. Above a) When i is odd. Botton b) When i is even. In order to verify that each edge of the set [xv'l,xv2,xv'3,XV3,. .. ,xv2fc_i,xv2k,yvi,yv'2,yv3,yv's,. .. ,yv2k-i,yv'^k}. is in exactly one subgraph Hi, for i e {1,..., k}, we obtain the unicyclic graph F* identifying v and vj resulting in vj; identifying x and y resulting in a vertex which is contracted with one of its neighbours. The resulting edge, in dashed, is showed in Figures 6 and 7. The set of those edges are a perfect matching of K2k proving that the added two paths of length 2 in Gi have end vertices vj and vj_1, and the other vj and vj_1. The election of the label of the center vertex is such that one path is vevenxv odd and vevenyvodd and the result follows. 4. Case n = 4k + 1 (for k = 2). Since K4k+1 c K4k+2, we have k +1 < 0(4, K4k+i) < 0(4, K4k+2) < k + 1. For k = 2, Figure 8 displays a decomposition of three planar graphs of girth at least 4 proving that 0(4, Kg) = [^] = 3. By the four cases, the theorem follows. □ About the case of K10, it follows 3 < 0(4, K10) < 4. We conjecture that 0(4, K10) = 4. References [1] V. B. Alekseev and V. S. Goncakov, The thickness of an arbitrary complete graph, Mat. Sb. (N. S.) 101(143) (1976), 212-230, http://mi.mathnet.ru/eng/msb3 8 9 7. [2] L. W. Beineke, Minimal decompositions of complete graphs into subgraphs with embeddability properties, Canad. J. Math. 21 (1969), 992-1000, doi:10.4153/cjm-1969-109-4. [3] L. W. Beineke and F. Harary, On the thickness of the complete graph, Bull. Amer. Math. Soc. 70 (1964), 618-620, doi:10.1090/s0002-9904-1964-11213-1. [4] L. W. Beineke and F. Harary, The thickness of the complete graph, Canad. J. Math. 17 (1965), 850-859, doi:10.4153/cjm-1965-084-2. [5] L. W. Beineke, F. Harary and J. W. Moon, On the thickness of the complete bipartite graph, Math. Proc. Cambridge Philos. Soc. 60 (1964), 1-5, doi:10.1017/s0305004100037385. [6] J. A. Bondy and U. S. R. Murty, Graph Theory, volume 244 of Graduate Texts in Mathematics, Springer, New York, 2008, doi:10.1007/978-1-84628-970-5. [7] G. Chartrand and P. Zhang, Chromatic Graph Theory, Discrete Mathematics and its Applications, CRC Press, Boca Raton, Florida, 2009. [8] R. K. Guy and R. J. Nowakowski, The outerthickness & outercoarseness of graphs, I. The complete graph & the n-cube, in: R. Bodendiek and R. Henn (eds.), Topics in Combinatorics and Graph Theory, Physica-Verlag, Heidelberg, pp. 297-310, 1990, doi:10.1007/ 978-3-642-46908-4_34, essays in honour of Gerhard Ringel, papers from the Graph Theory Meeting held in Oberwolfach, June 3-9, 1990. [9] M. Kleinert, Die Dicke des n-dimensionalen Wurfel-Graphen, J. Comb. Theory 3 (1967), 1015, doi:10.1016/s0021-9800(67)80010-3. [10] P. Mutzel, T. Odenthal and M. Scharbrodt, The thickness of graphs: a survey, Graphs Combin. 14 (1998), 59-73, doi:10.1007/pl00007219. [11] W. T. Tutte, The thickness of a graph, Indag. Math. (Proceedings) 66 (1963), 567-577, doi: 10.1016/s1385-7258(63)50055-9. [12] Y. Yang, A note on the thickness of K;,m,n, Ars Combin. 117 (2014), 349-351. [13] Y. Yang, Remarks on the thickness of Kn>n,n, Ars Math. Contemp. 12 (2017), 135-144, https://amc-journal.eu/index.php/amc/article/view/82 3. ARS MATHEMATICA CONTEMPORANEA ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 14 (2018) 329-344 https://doi.org/10.26493/1855-3974.1236.4d7 (Also available at http://amc-journal.eu) A note on the thickness of some complete bipartite graphs* * Siwei Hu, Yichao Chen t Department of Mathematics, Hunan University, 410082 Changsha, China Received 22 November 2016, accepted 17 July 2017, published online 19 September 2017 The thickness of a graph is the minimum number of planar subgraphs into which the graph can be decomposed. Determining the thickness for the complete bipartite graph is an unsolved problem in graph theory for over fifty years. Using a new planar decomposition for K4k-i,ik (k > 4), we obtain the thickness of the complete bipartite graph Kn,n+4, for n > 1. Keywords: Planar graph, thickness, complete bipartite graph. Math. Subj. Class.: 05C10 1 Introduction In this paper, all graphs are simple. A graph G is denoted by G = (V, E) where V(G) is the vertex set and E(G) is the edge set. A complete graph is a graph in which any two vertices are adjacent. A complete graph on n vertices is denoted by Kn. A complete bipartite graph is a graph whose vertex set can be partitioned into 2 parts, such that every edge has its ends in different parts and every two vertices in different parts are adjacent. We use KPl,P2 to denote a complete bipartite graph in which the ith part contains p vertices, for i = 1, 2. The thickness t(G) of a graph G is the minimum number of planar subgraphs into which G can be decomposed [14]. It is a classical topological parameter of a graph and has many applications, for instance, to graph drawing [12] and VLSI design [1]. Since deciding the thickness of a graph is NP-hard [9], it is very difficult to get the exact number of thickness for arbitrary graphs. Battle, Harary and Kodama [3] in 1962 and Tutte [13] in 1963 independently showed that the thickness of K9 and Kw equals 3. Beineke and *We are grateful to the two anonymous referees for their helpful comments. The second author is supported by the NNSFC under Grant No. 11471106. t Corresponding author. E-mail addresses: husiwei@hnu.edu.cn (Siwei Hu), ycchen@hnu.edu.cn (Yichao Chen) Abstract ©® This work is licensed under http://creativecommons.org/licenses/by/3.0/ Harary [4] determined the thickness of complete graph Kn for n ^ 4 (mod 6) in 1965, the remaining case was solved in 1976, independently by V.B. Alekseev and V.S. Gonchakov [2] and by J.M. Vasak [15]. For complete bipartite graphs, the problem has not been entirely solved yet. By constructing a planar decomposition of Km n when m is even, Beineke, Harary and Moon [5] determined the thickness of Km n for most values of m, n in 1964. Theorem 1.1. [5] For m < n, the thickness of the complete bipartite graph Km,n is t(Km,n) 2(m + n - 2) (1.1) except possibly when m and n are both odd and there exists an integer k satisfying n ' 2fc(m-2) ' (m-2k) We recall that the thickness of Kn n is also obtained in 1968 by Isao and Ozaki [11] independently. The following open problem is adapted from [7] by Gross and Harary. Problem 1.2. [See Problem 4.1 of [7]] Find the thickness of Km n for all m, n. Beineke, Harary and Moon [5] also pointed out that the smallest complete bipartite graph whose thickness is unknown is K17,21. From Euler's Formula, the thickness of K17,21 is at least 5. From Theorem 1.1, we need to determine the thickness of Km n for odd m, n. Since the difference between the two odd numbers is even, we only need to determine the thickness of Kn,n+2k for odd n and k > 0. In this paper, we start to calculate the thickness of Kn n+2k for some small values of k. Indeed, we determine the thickness of Kn n+4. Theorem 1.3. The thickness of Kn,n+4 is t(Knn+4)= j!\3, ifn < 2 ( n,n+4) \ , otherwise. The following corollary follows from Theorem 1.3. Corollary 1.4. The thickness of K17 21 is 5. We may refer the reader to [6, 10, 16] for more background on graph thickness. 2 The thickness of Kn,n+4 To begin with, we define two special graphs called the pattern graph and the kth-order nest graph. Then, we prove a new planar decomposition of K4fc-4 4fc. Finally, we prove the thickness of ^-3,4^1 and Kn,n+4. 2.1 The pattern graph Let U = {u1, u2} and Xn be a set of n vertices. A graph is said to be a pattern graph of order n + 2, denoted by G[u1 Xnu2], if it can be constructed by the following two steps. 1. Arrange the n vertices in a row, and put vertices u1, u2 on the above and below of n vertices, respectively. 2. Join both u1 and u2 to the n vertices using straight lines. From the definition above, the pattern graph is a planar straight-line graph. Figure 1 illustrates the pattern graph G[u1Xnu2]. Remark 2.1. Unless explicitly mentioned, we always join vertices using straight lines in the drawings of the following proofs. Figure 1: The pattern graph G[u1Xnu2]. 2.2 The kth-order nest graph Let Uk = [uil ,ui2 ,...,uik },Vk = [vj1 ,vj2 ,...,vjk} and W2k+2 = [wh ,wh,..., whk+2}, we define a kth-order nest graph G[Uk, Vk, W2k+2] as follows: 1. Arrange 2k + 2 vertices wll, wl2,..., wl2k+2 in a row. 2. For 1 < m < k, place vertices uim and vjm on the above and below of the row, respectively, and join them to wh, whm, w^+i, whm+2. Figure 2 illustrates a third-order nest graph G[U3, V3, W8], where U3 = {u1, u2, u3}, V3 = {vi, v2, V3} and W8 = {wi, w2,..., w8}. V3 Figure 2: The third-order nest graph G[U3, V3, W8]. 2.3 A new planar decomposition of K4k-3,4k+1, for k > 4 In this subsection, we shall construct a planar decomposition for the complete bipartite graph K4k_3j4k+1 with k planar subgraphs G1,G2,... ,Gk. Suppose that the vertex partition of K4k-3,4k+i is (X, Y), where X = {x1,x2,... ,x4k-3}, Y = {yo,y1,y2, ...,V4k }. 2.3.1 The planar decomposition for K4k-4,4k Let the vertex partition of K4fc-4j4fc be (X1, Yi), where X1 = jxi, x2,..., x4k-4}, Y1 = {y0, y1,..., y4k-1}. In this subsection, all subscripts in y- are taken mod 4k. 1. In the graph Gj (1 < i < k), we arrange 4k vertices in a row, and divide the 4k vertices into two subsets L2k and R2k such that each subset contains 2k vertices according to the following steps. 2. In the graph Gj (1 < i < k - 1), we choose four vertices x4j-3, x4j-2, x4j-1, x4j from X1 and construct two pattern graphs G[x4j-3L2kx4j-1] and G[x4j-2R2k x4j]. Then we join both x4j-3 and x4j-1 to the first vertex and the last vertex in R2k. Finally, we label the vertices in ¿2fc and R2k as y1, y3, y5,..., y4k-1 and y2i+6, y2i+8, y2i+10,..., y2i+4k+4 in turn, respectively. 3. In the graph Gk, we label the vertices in L2k and R2k as y1, y3, y5,... y4k-1 and y2,y4,..., y4k-2, y0, respectively. First, we construct a (k - 1)th-order nest graph G[Ufc_1, Vk-1, W2fc], where Ufc-1 = {x2,x6, xu,..., x4fc-6} , V- = {x4,x8, x12,..., x4fc-4, } and W2fc = {y1,y3, y5,... ,y4fc-1}. We join x4j-3 to y2i and y2j+2, for 1 < i < k - 1. Second, we construct a union of paths, if k is even, we join x4j-1 to y2j+2k and y2j+2+2k, for 1 < i < k - 1; otherwise k is odd, we join x4j-1 to y2i+2fc-2 and y2i+2k, for 1 < i < k - 1. 4. In each graph Gj (1 < j < k - 1), we put x4j-2, x4j in the quadrangle x4j-3y4j+1 x4j-1y4j+3, and join them to y4j+1 and y4j+3, for 1 < i < j. We put the vertices x4i-2,x4i in the quadrangle x4j-3y4j-1x4j-1y4j+1, and join both x4j-2 and x4j to y4j-1 and y4j+1, for j < i < k - 1. Next, we put x4j-3 in the quadrangle x4j-2y4j-2i+4x4jy4j-2i+6, and join x4j-3 to y4j-2i+4,y4j-2i+6, for 1 < i < j. We put x4j-3 in the quadrangle x4j-2y4j-2i+4fcx4jy4j-2i+4fc+2, and join x4j-3 to y4j-2i+4fc ,y4j-2i+4fc+2, for j < i < k - 1. For each i (1 < i < k - 1), we define a set Mj = {i + 1, i + 2,..., i + k - 2}. Suppose that m g Mj, if m < k - 1, we let j = m; otherwise, j = m - k + 1. (i) k is even. If i +1 < m < i+ , we put x4j-1 in the quadrangle x4j-2y4m-2j+4 x4jy4m-2i+6, andjoin x4j-1 to y4m-2i+4, y4m-2i+6. If i+ ^ + 1 < m < i + k-2, we put x4j-1 in the quadrangle x4j-2y4m-2j+8x4j y4m-2i+10, andjoin x4j-1 to y4m-2i+8, y4m-2i+10. (ii) k is odd. If i +1 < m < i + ^^^, we put x4j-1 in the quadrangle x4j-2y4m-2j+4 x4jy4m-2i+6, andjoin x4i-1 to y4m-2i+4, y4m-2i+6. If i + ^ + 1 < m < i + k-2, we put x4j-1 in the quadrangle x4j-2y4m-2j+8x4j y4m-2i+10, andjoin x4j-1 to y4m-2i+8, y4m-2i+10. Theorem 2.2. Let G1, G2,..., Gk be the planar subgraphs obtained from steps 1, 2, 3 and 4 above, then {G1, G2,..., Gk} is a planar decomposition of K4fc-4 4fc. Proof. From the constructions above, we have E(Gj) n E(Gj) = 0, for 1 < i = j < k. In order to prove that {G1, G2,..., Gk} is a planar decomposition of K4fc-4j4fc, we need to show that E(G1) u E(G2) u • • • u E(Gfc) = E(K^^^). We denote dGi(v) as the degree of v in Gj, for 1 < i < k. By the construction above, Step 2 contributes to the degrees of v4i_3, v4i_i, v4i-2, and v4i in Gj by terms 2k + 2, 2k + 2,2k +1 and 2k + 1, respectively. In other words, we have do, (v4i-3) = dGi (v4j-1) = 2k + 2 and dGi («44-2) = dGi ) = 2k + 1. For 1 < i < k - 1, Step 3 contributes to dGfc(v4i-3),d0fc(v4i-1),d0fc(v4i-2) and dGfc (v4i) by terms 2,2, 3, and 3, respectively. For 1 < j < k - 1 and i = j, Step 4 contributes to each of dGj(v4i-3), dGj (v4i-1), doj (v4i-2) and do, (^) a term 2. In total, for 1 < i < k - 1 , we have k k k X^do,- (v4i-i) = do,- (v4i-3) = do, (v4i-3) + ^ doj («44-3) + dofc («44-3) j=i j=i i 4. Proof. From Theorem 2.2, Subsection 2.3.2 and Subsection 2.3.3, a planar decomposition of K4k-3,4k+i with k planar subgraphs Gi, G2,..., Gk is obtained. FromEuler's formula, we have "(4k - 3)(4k + 1)" t(K4k-3,4k + i) > k, 2(8k - 4) and so t(K4k-3,4k+i) = k. □ Example 2.4. By using the procedure above, the two planar decompositions of Ki7,2i (k = 5 is odd) and K2i,25 (k = 6 is even) are shown in Appendix A (See Figures 3-7) and Appendix B (See Figures 8-13), respectively. 2.4 Proof of Theorem 1.3 From Theorem 1.1, the proof has two cases: Case 1: n = 4k — 3 (k > 0). When 1 < k < 3, it is routine to check that the theorem is true. For k > 4, 2k(4k-3-2) 4k-3-2k 4k + 1 + 2k-3 4k+1, thus, the thickness of K4fc_3j4fc+i k = can not be determined by Theorem 1.1. By Theorem 2.3, we have t(K4fc_3 4k+1) r ^ i. ' Case 2: n = 4k — 1 (k > 0). Since 4k — 1 and 4k + 3 are both odd and 4k + 3 = 2(fc + 1)(4fc-1-2) 4k-1-2(k+1) (See Lemma 1 of [5] for details), the thickness of K4fc_1j4fc+3 can be determined by Theorem 1.1, thus t(K„,„+4) = t(k4fc-1,4fc+3) (4k - 1)(4k + 3) 2(4k - 1+4k + 3 - 2) k+1 13 k +---- 2 16k n + 3" Summarizing the above, the theorem follows. 4 3 Conclusion In this paper, we determine the thickness for K„,„+4. The proof replies on a planar decomposition of K4fc-3,4fc+1 and the Theorem 1.1 of Beineke, Harary and Moon. We observe that our approach for the construction of a planar decomposition of K„,„+4 is the first step in finding a solution for Problem 1.2. From Theorem 1.1, the next classes of complete bipartite graphs whose thickness is unknown is K4k-1,4k+7, for k > 10. Furthermore, the new smallest complete bipartite graph whose thickness is unknown is K19,29. We hope that the construction here helps establish intuition and structure of the Problem 1.2. Another way of solving the Problem 1.2 is to find a new planar decomposition of Km,„, for odd m, n. Actually, using a new planar decomposition of the complete tripartite graph K1,g,n and a recursive construction, we also [8] obtained the thickness of Ks,t, where s is odd and t > (s-3)3(s-2). Now we split Problem 1.2 into the following two problems. Problem 3.1. Find the thickness of K„,„+4k for odd n and k > 2. Problem 3.2. Find the thickness of K„,„+4k+2 for odd n and k > 0. References [1] A. Aggarwal, M. Klawe and P. Shor, Multilayer grid embeddings for VLSI, Algorithmica 6 (1991), 129-151, doi:10.1007/bf01759038. [2] V. B. Alekseev and V. S. Goncakov, The thickness of an arbitrary complete graph, Mat. Sb. (N. S.) 101(143) (1976), 212-230, http://mi.mathnet.ru/eng/msb3 8 9 7. [3] J. Battle, F. Harary and Y. Kodama, Every planar graph with nine points has a nonplanar complement, Bull. Amer. Math. Soc. 68 (1962), 569-571, doi:10.1090/s0002-9904-1962-10850-7. [4] L. W. Beineke and F. Harary, The thickness of the complete graph, Canad. J. Math. 17 (1965), 850-859, doi:10.4153/cjm-1965-084-2. [5] L. W. Beineke, F. Harary and J. W. Moon, On the thickness of the complete bipartite graph, Proc. Camb. Phil. Soc. 60 (1964), 1-5, doi:10.1017/s0305004100037385. [6] L. W. Beineke and R. J. Wilson (eds.), Topics in Topological Graph Theory, volume 128 of Encyclopedia of Mathematics and its Applications, Cambridge University Press, Cambridge, 2009, doi:10.1017/cbo9781139087223. [7] J. L. Gross and F. Harary, Some problems in topological graph theory, J. Graph Theory 4 (1980), 253-263, doi:10.1002/jgt.3190040302. [8] S. Hu and Y. Chen, The thickness of some complete tripartite graphs, preprint. [9] A. Mansfield, Determining the thickness of graphs is NP-hard, Math. Proc. Camb. Phil. Soc. 93 (1983), 9-23, doi:10.1017/s030500410006028x. [10] P. Mutzel, T. Odenthal and M. Scharbrodt, The thickness of graphs: a survey, Graphs Combin. 14 (1998), 59-73, doi:10.1007/pl00007219. [11] I. Shirakawa, H. Takahashi and H. Ozaki, On the planar decomposition of a complete bipartite graph, SIAMJ. Appl. Math 16 (1968), 408-416, doi:10.1137/0116034. [12] I. G. Tollis, G. Di Battista, P. Eades and R. Tamassia, Graph Drawing: Algorithms for the Visualization of Graphs, Prentice Hall, Upper Saddle River, New Jersey, 1999. [13] W. T. Tutte, The non-biplanar character of the complete 9-graph, Canad. Math. Bull. 6 (1963), 319-330, doi:10.4153/cmb-1963-026-x. [14] W. T. Tutte, The thickness of a graph, Indag. Math. (Proceedings) 66 (1963), 567-577, doi: 10.1016/s1385-7258(63)50055-9. [15] J. M. Vasak, The Thickness of the Complete Graph, Ph.D. thesis, University of Illinois at Urbana-Champaign, ProQuest Dissertations Publishing, 1976, https://search. proquest.com/docview/302820090. [16] Y. Yang, Remarks on the thickness of Kn>n>n, Ars Math. Contemp. 12 (2017), 135-144, https://amc-journal.eu/index.php/amc/article/view/82 3. A A planar decomposition {G1, G2, G3, G4, G5} for K17,2i Figure 4: The graph G2. B A planar decomposition {G1, G2, G3, G4, G5, G6} for K21,25 ARS MATHEMATICA CONTEMPORANEA ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 14 (2018) 345-357 https://doi.org/10.26493/1855-3974.1240.515 (Also available at http://amc-journal.eu) Alphabet-almost-simple 2-neighbour-transitive codes Neil I. Gillespie Heilbronn Institute for Mathematical Research, School of Mathematics, Howard House, University of Bristol, BS81SN, United Kingdom Daniel R. Hawtin * Centre for the Mathematics ofSymmetry and Computation, University of Western Australia, 35 Stirling Highway, Crawley, WA 6009, Australia Received 28 November 2016, accepted 12 June 2017, published online 30 September 2017 Let X be a subgroup of the full automorphism group of the Hamming graph H (m, q), and C a subset of the vertices of the Hamming graph. We say that C is an (X, 2)-neighbour-transitive code if X is transitive on C, as well as Ci and C2, the sets of vertices which are distance 1 and 2 from the code. It has been shown that, given an (X, 2)-neighbour-transitive code C, there exists a subgroup of X with a 2-transitive action on the alphabet; this action is thus almost-simple or affine. This paper completes the classification of (X, 2)-neighbour-transitive codes, with minimum distance at least 5, where the subgroup of X stabilising some entry has an almost-simple action on the alphabet in the stabilised entry. The main result of this paper states that the class of (X, 2) neighbour-transitive codes with an almost-simple action on the alphabet and minimum distance at least 3 consists of one infinite family of well known codes. Keywords: 2-neighbour-transitive, alphabet-almost-simple, automorphism groups, Hamming graph, completely transitive. Math. Subj. Class.: 05E20, 68R05, 20B25 * This author was supported by an Australian Postgraduate Award and a University of Western Australia Safety-Net-Top-Up scholarship while this research was conducted, and would like to acknowledge an accommodation grant from the Heilbronn Institute for Mathematical Research. E-mail addresses: neil.gillespie@bristol.ac.uk (Neil I. Gillespie), dan.hawtin@gmail.com (Daniel R. Hawtin) Abstract ©® This work is licensed under http://creativecommons.org/licenses/by/3.0/ 1 Introduction Ever since Shannon's 1948 paper [18, 19] there has been a great deal of interest around families of error-correcting codes with a high degree of symmetry. The rationale behind this interest is that codes with symmetry should have good error correcting properties. The first families classified were perfect (see [21] or [22]) and nearly-perfect (defined in [12], classified in [15]) codes over prime power alphabets. Note that the classification of nearly-perfect codes follows from the earlier results of [17] on uniformly packed codes, since nearly-perfect codes are uniformly packed codes with maximal packing density. These classifications show that perfect and nearly-perfect codes are rare. In an effort to find further classes of efficient codes, Delsarte [4] introduced completely regular codes, a more general class of codes that posses a high degree of combinatorial symmetry. Much effort has been put into classifying particular classes of completely regular codes (see for instance [1, 2]), and new completely regular codes continue to be found [6]. However, completely regular codes have proven to be hard to classify, and this remains an open problem. Completely transitive codes (first defined in [20], with a generalisation studied in [10]) are a class of codes with a high degree of algebraic symmetry and are a subset of completely regular codes. As such a classification of completely transitive codes would be interesting from the point of view of classifying completely regular codes. This problem also remains open. Here, the conditions of complete transitivity are relaxed and the family of 2-neighbour-transitive codes is studied, a class of codes with a moderate degree of algebraic symmetry. Note that every completely transitive code (see Section 2) is 2-neighbour-transitive. By studying this class of codes we hope to find new codes and gain a better understanding of completely transitive codes. Indeed a classification of 2-neighbour-transitive codes would have as a corollary a classification of completely transitive codes. We also note that codes with 2-transitive actions on the entries of the Hamming graph (which 2-neighbour-transitive codes indeed have), have been of interest lately, where this fact can be used to prove that certain families of codes achieve capacity on erasure channels [14]. The analysis of 2-neighbour-transitive codes is being attacked as three separate problems: entry-faithful (see [7]), alphabet-almost-simple, and alphabet-affine. This paper concerns the alphabetalmost-simple case. The results of this paper do not return any new examples. However, the results here are of interest from the point of view of perfect codes over an alphabet of non-prime-power size, since in this case a code cannot be alphabet-affine (and also not entry-faithful, by [7]), but may be alphabet-almost-simple. The existence of perfect codes over non-prime-power alphabets with covering radius 1 or 2, is still an open question (see [13]). By Theorem 1.1, if such codes exist, then they cannot be 2-neighbour-transitive (unless they are equivalent to the repetition code of length 3). Note that in the prime power case, for each set of parameters for which a perfect code with covering radius p > 2 exists, a 2-neighbour-transitive code with those parameters exists. That is, the repetition and Golay codes are 2-neighbour-transitive. In fact, the repetition, Hamming and Golay codes are completely transitive (by [11, Example 3.1] for the repetition codes, [20, Proposition 7.3] for the Hamming and binary Golay codes, and [10, Example 3.5.6] for the ternary Golay codes). 1.1 Statement of the main results Let X be a subgroup of the full automorphism group S^ X Sm of the Hamming graph r = H(m, q) and let C be a code, that is, a subset of the set of vertices Vr. We say that C is an (X, s)-neighbour-transitive code if X fixes C setwise and acts transitively on C = C0, Ci,..., Cs (where Ci are parts of the distance partition, see Section 2). In joint work with Giudici and Praeger [7], the authors classified all (X, 2)-neighbour-transitive codes for which the group X acts faithfully on the set of entries of the Hamming graph. In this paper, we begin the study of (X, 2)-neighbour-transitive codes such that the action of X on the entries has a non-trivial kernel. Let M be the set of entries of the Hamming graph H(m, q) and Qi be the copy of the alphabet Q in the corresponding entry i e M. Then the vertex set of H(m, q) is: Vr = n Qi. ieM If C is an (X, 2)-neighbour-transitive code with minimum distance S > 3, then the subgroup Xi < X stabilising the entry i e M has a 2-transitive action on the alphabet Qi in that entry (see [7, Proposition 2.7]). Any 2-transitive group G is of affine type (G < AGLd(p) for some integer d and prime p) or almost-simple type (S < G < Aut(S) for some non-abelian simple group S) [5, Theorem 4.1B]. If the action of X on M (see Section 2.1) is transitive with a non-trivial kernel and the action of Xi on the alphabet Qi is almost-simple then we say C is X-alphabet-almost-simple. Our main aim here is to prove the non-existence of codes which are X-alphabet-almost-simple and (X, 2)-neighbour-transitive with minimum distance S > 4. Theorem 1.1. Let C be an X-alphabet-almost-simple and (X, 2)-neighbour-transitive code in H(m, q) with minimum distance S > 3. Then S = 3 and C is equivalent to the repetition code in H(3, q), where q > 5. In Section 2 we define the notation used in the paper. In Section 3 we give some results on the structure of codes that are X-alphabet-almost-simple and (X, 2)-neighbour-transitive, as well as pose some questions about codes for which the action of Xi on the alphabet in the entry i e M is affine. We present some examples of codes with properties of interest in relation to our results in Section 4. Finally, in Section 5, we give a classification of diagonally (X, 2)-neighbour-transitive codes (see Definition 3.1) and prove Theorem 1.1. 2 Preliminaries Throughout this paper we let M = {1,..., m} and Q = {1,..., q}, with m, q > 2, though if q = 2 we will at times use Q = {0,1}. We refer to M as the set of entries and Q as the alphabet. We use Qi to denote the disjoint copy of the alphabet Q in the entry i e M. The vertex set Vr of the Hamming graph r = H(m, q) consists of all m-tuples with entries labeled by the set M, taken from the set Q. An edge exists between two vertices if they differ as m-tuples in exactly one entry. For vertices a, ft of H(m, q) the Hamming distance d(a, ,0) is the number of entries in which a and ft differ, i.e. the usual graph distance in r. For a e Vr, we refer to the element of Q appearing in the i-th entry of a as ai, so that a = (a1,..., am) throughout. A code C is a subset of the vertex set of the Hamming graph. The minimum distance of C is S = min{d(a, p) | a, p G C, a = p}. For a vertex a G H(m, q), define rr (a) = {p G r | d(a, p) = r}, and d(a, C) = min{d(a, p) | p G C}. We then define the covering radius to be p = max{d(a, C) | a G r}. For any r < p, define Cr = {a G r | d(a, C) = r}. Note that Cj is the disjoint union Uaecrj(a) for i < L^J. 2.1 Automorphism groups The automorphism group Aut(r) of the Hamming graph is the semi-direct product B x L, where B = and L = Sm (see [3, Theorem 9.2.1]). We refer to B as the base group, and L as the top group, of Aut(r). Let g = (g^ ..., gm) G B, a G L and a be a vertex in H(m, q). Then g and a act on a as follows: ag = (ag1,..., am") and aCT = (aiff-i ,...,amCT-i). We define the automorphism group of a code C in H(m, q) to be Aut(C) = Aut(r)C, the setwise stabiliser of C in Aut(r). For a subgroup X < Aut(C) we define two other important actions of X which will be useful to us. First, consider the action of X on the set of entries M, which we will write as XM. In particular XM = ^(X), that is, the image of the homomorphism: M : X -^ Sm (hi ,...,hm)a i—> a Note that a here is not necessarily an automorphism of C, that is, a is a permutation of M but may not necessarily fix C setwise, though its pre-image (hi,..., hm)a is an element of Aut(C). We define K to be the kernel of the map m and note that K = X n B. In this paper we are concerned with (X, 2)-neighbour-transitive codes where K = 1. We also consider the action of the stabiliser Xj < X of the entry i G M, on the alphabet Qj in that entry. We denote this action by XQi = ^j(Xj), and it is the image of the homomorphism: : Xj -> Sq (hi, ...,hm )a i—> hj Let C be a code in H(m, q) and let X be a subgroup of Aut(r). Recall that C is (X, s) -neighbour-transitive if each Cj is an X-orbit for i = 0,..., s. Note that this implies X < Aut(C) and C is also (X, r)-neighbour-transitive, for r < s. If s = 1 then C is simply X-neighbour-transitive and if s = p, the covering radius, then C is X-completely transitive. An almost-simple group is a group G where S < G < Aut(S) for some non-abelian simple group S. The socle of a group G, denoted soc(G), is the product of its minimal normal subgroups. The socle of an almost-simple group G is the non-abelian simple group S such that S < G < Aut(S). Recall, if C is a code and X < Aut(C) such that K = 1, XM is transitive on M and the XQi is almost-simple, then we say C is X-alphabet-almost-simple. We may sometimes omit the group X from any of the above terms, if the meaning is clear from the context. We say that two codes, C and C', in H(m, q), are equivalent if there exists x G Aut(r) such that Cx = C'. Since elements of Aut(r) preserve distance, equivalence preserves minimum distance. 2.2 Projections For a subset J = {j1,... , jk} Q M we define the projection of a with respect to J as ■kj (a) = (aj1,..., ajk). For a code C we then define the projection of C with respect to J as nJ(C) = {nJ(a) | a G C}. So nJ maps a vertex or code from H(m, q) into the smaller Hamming graph H(k, q). Let XJ be the setwise stabiliser of a subset J = {j1,... , jk} Q M. For x = (h1,..., hm)a G XJ, we define the projection of x with respect to J as xJ (x) where n j (a)XJ (x) = n j (ax). To be well defined, this requires x g Xj and it follows that XJ(x) = (hj, .. ., hjk) 3 are characterised, are stated below. This is our starting point when looking at codes that are X-alphabet-almost-simple and (X, 2)-neighbour-transitive with S > 3, since we then have that C is indeed X-neighbour-transitive. The following definitions are needed first. For a subgroup T < Sq define Diagm(T) = {(h,..., h) G B | h G T}. Definition 3.1. A code C in H(m, q) is diagonally (X, s)-neighbour-transitive if C is (X, s)-neighbour-transitive and X < Diagm(Sq) x L. Each part of Proposition 3.2 is proved in the relevant citation of [8]. Recall the definitions of: nJ (C) and xj(X) (see Section 2.2), the socle soc(G) and the kernel K = X n B for the action of X on M, where B = Sm is the base group of Aut(r) (see Section 2.1). Note also that G is a sub-direct subgroup of a direct product "=1 T of isomorphic groups T = T, where i G {1,..., n}, if the projection of G in each coordinate is isomorphic to T. Proposition 3.2. Suppose C is an X-neighbour-transitive code in H(m, q) with S > 3. Then the following hold: i) Let J be an X-invariant partition of M and J G J such that nJ(C) is not the complete code. Then nJ(C) is xj(X)-neighbour-transitive [8, Proposition 3.4]. (Note that the assumption that nJ (C) is not the complete code does not appear in [8], but is necessary since the proof assumes that nJ (C )1 is non-empty.) ii) Let J be an X-invariant partition of M and J e J such that nJ(C) is not the complete code. Then nJ(C) has minimum distance at least 2 [8, Corollary 3.7]. iii) If C is also X-alphabet-almost-simple, then soc(K) is a sub-direct subgroup of nieM soc (xq<) [8, Proposition 5.2]. While the next result is not explicitly stated in [8], it is the basis of the characterisation contained within it. Proposition 3.3. Let C be an X-alphabet-almost-simple and X-neighbour-transitive code with S > 3. Then there exists an X-invariant partition J of M such that for all J e J the code nJ(C) is equivalent to a diagonally xJ(X)-neighbour-transitive with minimum distance S(nJ(C)) > 2. Proof. Let T be the non-abelian simple socle of the almost-simple 2-transitive group XQi. By Proposition 3.2-(iii), the group soc(K) is a sub-direct subgroup of ni£M soc (xQi j. Following the discussion after [8, Proposition 5.2], Scott's Lemma [16, p. 328] can be applied to give a partition J of M such that soc(K) = f]JeJ DJ, where each DJ = Diagk(T) acts on -kj(Vr), for all J e J, where k = | J|. Moreover, by [8, Remark 5.5], J is X-invariant. By examining soc(K), it can be shown [8, Section 5] that, up to equivalence, two possibilities occur. Either xJ(X) < Diagk(Sq) x Sk, where k = | J|, for all J e J, or J can be replaced by a more refined X-invariant partition J of M such that Xj(X) < Diagg (Sq) x Sg, where k = | J|, for all J e J. In either case, it follows from Proposition 3.2-(i) and (ii) that, for all J e J or J respectively, xJ(X) acts transitively on nJ(C) and either nJ(C) is the complete code or it is xJ(X)-neighbour-transitive with minimum distance at least 2. Since xJ(X) is a diagonal subgroup, we deduce that nJ(C) is as in the second case, since no diagonal subgroup acts transitively on the complete code. □ Proposition 3.4. Let C be an (X, 2)-neighbour-transitive code with S > 3 in H(m, q), and suppose J is an X-invariant partition of M. Then for all J e J, either; i) nJ(C) is the complete code, S(nJ(C)) = 1, and xJ(X) is transitive on nJ(C); ii) nJ(C) has covering radius 1, S(nJ(C)) = 2 or 3, and is (xJ(X), 1)-neighbour-transitive; or, iii) nJ(C) is (xJ(X), 2)-neighbour-transitive. Proof. Let <5 = nJ(C). The fact that xJ(X) is transitive on C and Ci, if Ci is non-empty, follows from Proposition 3.2-(i). From this we deduce (i) and (ii). In particular, suppose the covering radius of C is at most 1. If the covering radius is 0 then C is the complete code, and if the covering radius is 1 then C is not the complete code and the minimum distance is at most 3 so, by Proposition 3.2-(ii), the minimum distance is at least 2. Therefore, we need only show that when C2 is non-empty xJ (X) is transitive on C2. Suppose C has covering radius at least 2. Let p, v e C2. Then there exists a, ¡3 e C such that d(p, nJ(a)) = d(v, nJ(¡)) = 2. Let k e H(m, q) with ku = vu for u in J and kv = av otherwise. Similarly, let 7 e H(m, q) with /tu = pu for u in J and /tv = 3v otherwise. We claim that v, 7 e C2. We show this for v and note that an identical argument holds for 7. First, note that d(a, k) = 2 and S > 3, so v / C. Suppose v e C1. Then there exists a' e C such that d(v, a') = 1. We then have d(v, nJ(a')) < 1. However, this contradicts v e C2. Hence j, ù e C2. As C is (X, 2)-neighbour-transitive, there exists an x = ha e X mapping l to j. We claim x e XJ. Suppose x e XJ. Then, since J is a system of imprimitivity for the action of X on M, there exists J' e J such that J = J' and J'a = J. Since nJ> (l) = nJ> (a), this implies that nJ(lx) = nJ(ax) e C and hence nJ(lx) = j, which contradicts the fact that vx = j. Thus x e XJ and vXJ (x) = nJ (l)XJ (x) = nJ (lx) = nJ (j) = j. □ Proposition 3.5. Let C be an (X, 2)-neighbour-transitive code in H (m, q) with S > 3, and J be an X-invariant partition of M. Then, for all J e J and i e J, 1. xJ(X)Qi is 2-transitive on Q; and, 2. for a e C, xJ(X)nj(a) is transitive on J. Proof. As C is X-neighbour-transitive with S > 3, we have that XQi is 2-transitive, by [7, Proposition 2.7], and XM is transitive, by [7, Proposition 2.5]. One then deduces that XQi is 2-transitive for all i. Now, because J is an X-invariant partition, it follows that Xi = (XJ)i for all i e J. This in turn implies that xJ(X)i = xJ(Xj). It is now straight forward to show that xJ (Xi)Qi = XQi. Now, since Xa is transitive on M and J is an X-invariant partition of M, it follows that (Xa)J is transitive on J. Thus xJ(Xa) < xJ(X)n(a) is transitive on J. □ The previous two propositions suggest a study of codes that are (X, 2)-neighbour-transitive, have minimum distance S > 2, and where X acts primitively on M. An answer to the following questions would provide us with the building blocks for (X, 2)-neighbour-transitive codes with S > 3. Question 3.6. Can we classify all (X, 2)-neighbour-transitive codes with S > 2 such that XM is primitive and XQi is 2-transitive? Question 3.7. Can we classify all (X, 1)-neighbour-transitive codes with S = 2 or 3 and p =1 such that XM is primitive and XQi is 2-transitive? Let C be a code and X < Aut(C ). If X acts faithfully on M, that is K = X n B = 1, we say C is X-entry-faithful. If K = 1, XM is transitive on M and XQ is affine (X^ < AGLd(p) for some integer d and prime p) we say C is X-alphabet-affine. Questions 3.6 and 3.7 can be further broken down into X-entry-faithful and non-trivial kernel cases, that is, X-alphabet-affine and X-alphabet-almost-simple (see Section 2.1 for the definition of X-alphabet-almost-simple). By the main result of this paper, the outstanding cases of Question 3.6 are X-alphabet-almost-simple and (X, 2)-neighbour-transitive with S = 2, and X-alphabet-affine and (X, 2)-neighbour-transitive, where XM is primitive and XQi is 2-transitive. Given Proposition 3.3, a third question is the following. Question 3.8. Can we construct (X, 2)-neighbour-transitive codes with S > 3 by taking copies of (X, 1)-neighbour-transitive codes with S = 2 or 3 and p =1. 4 Examples We begin this section by considering some examples of codes which have properties relating to the results of the previous section. We first introduce the operators Prod and Rep which allow the construction of new codes from old ones. For an arbitrary code C in H(m, q) we define Prod(C, i) and Rep,(C) in H(mi, q) as Prod(C, i) = {(ai,..., a,) | a e C}, and Rep,(C) = {(a,..., a) | a e C}. The repetition code Rep(m, q) in H(m, q) is the set of all vertices (a,..., a) consisting of a single element a e Q repeated m times. The next two examples present codes which are both X-alphabet-almost-simple and X-completely transitive, though the second example has minimum distance S = 2. Example 4.1. Let C = Rep(3, q), where q > 5, and X = Diag3(Sq) x S3, as in [11, Example 3.1]. Now, Ci = {(a, a, b), (a, b, a), (b, a, a) | a, b e Q; a = b}, and C2 = {(a, b, c) | a, b, c e Q; a = b = c = a}. Since Sq acts 3-transitively on Q and S3 acts transitively on M, it follows that X acts transitively on C, C1 and C2. Thus C is (X, 2)-neighbour-transitive and X-completely transitive, since C has covering radius p = 2. Also, XQi = Sq is almost-simple, since q > 5, and XM = S3 is transitive on M. Hence C is X-alphabet-almost-simple and X -completely transitive. Example 4.2. Let q > 5, i > 2, C = Prod(Rep(2, q), i) and X = (Diag2(Sq))e x U, where Diag2(Sq) is a subgroup of the base group of Aut(H(2, q)) and U = S2 I S, = S| x S, is a subgroup of the top group of Aut(H(2i, q)). Let J = {J1,..., J,}, with Ji = {2i - 1,2i}, be the partition of M preserved by U. Note that S = 2. Let R C {1,..., i} of size s, and v e H(m, q) be such that J (v) = (a, b), where a = b for all i e R, and a = b for all i e R. Any codeword ft is at least distance s from v, since d(nj.(v),nj.(^)) > 1 for each i e R. Also, there exists some codeword a with nj. (a) = (a, a) whenever nj (v) = (a, b) for i e {1,..., i}, and hence d(a, v) = s. So v e Cs. Any vertex v of H(2i, q) can be expressed in this way, for some R, since nj (v) = (a, b) has either a = b or a = b. Thus, for each s, Cs consists of all such vertices v where |R| = s. It also follows from this that p = i. Let v e Cs, with R as above. Let x = (hJl,..., hj)a e X where hJi e Diag2(Sq) such that nJi (v)hj* = (1,2), for i e R, and nJi (v= (1,1), for all i e R. MMore-over, since S, is i-transitive, there exists a e S, < S2 ^ S, mapping {J4l,..., Jis} to {J1,..., Js} (where R = {i1,..., is}), whilst preserving order within each Jj. Then vx = 7 e Cs, where nJi (7) = (1, 2) for all i e {1,..., s} and nJi (7) = (1,1) for all i e {s + 1,..., i}. Since we can map any such v to 7, X is transitive on Cs for each s e {1,..., i}. Hence C is X-completely transitive, and in particular (X, 2)-neighbour-transitive for i > 2. Since XQ = Sq and XM = S2 ? S, is transitive on M, C is X-alphabet-almost-simple. Lemma 4.3. Suppose C is an (X, 2)-neighbour-transitive code in H(m, q), with q > 3, and J is an X-invariant partition of M, such that nJ(C) = Rep(k, q), for all J G J where k = | J|. Then either S = k = 2, or J is a trivial partition. Proof. Let x = (hi,..., hm)a G X and J G J .By the hypothesis it follows that for all a G Q, there exists a G C such that nJ (a) = (a,..., a). Suppose Ja = J' G J. Then nJ / (ax) = (ahii,..., ahik )a = (b,... ,b) for some b G Q, that is, ahis = ahit for all is,it G J. In particular xJ (xa-1) = (h,... ,h) for some h G Sq, and X < Diagk (Sq) IU, where U is the stabiliser of J in the top group. Suppose that the partition J is non-trivial, so that k,l > 2. Since C is a subset of Prod(Rep(k, q), i), which has minimum distance k, it follows that S > k > 2. Suppose S > 3. As C is a subset of Prod(Rep(k, q),i) we can replace C by an equivalent code contained in Prod(Rep(k, q),i) containing a = (1,..., 1) and such that J = {{1, ...,k}, {k + 1,. .., 2k}, • • • , {m - k + 1, ..., m}} . Consider, V = (2,3,1,1,..., 1,1,1,1,..., 1, • • • , 1,..., 1) and v = (2,1,1,1,...,1, 2,1,1,...,1, ••• ,1,...,1). k entries k entries k entries If k = 2, then we claim v G C2. Any vertex ft G Prod(Rep(2, q),£) D C with d(p, ft) = 1 is of the form 7 = (a, a, 1,..., 1), where a = 2 or 3. However, no such 7 is an element of C, since each is distance 2 from a. If k > 3 then v G C2 since d(a, v) = 2 and there is no closer codeword as nJl (p) G nJl (C)2. In both cases v G C2 since d(a, v) = 2 and no codeword is closer, as nJi(v) G nJi(C)1 for i = 1, 2. Let x = (h1,..., hm)a G X such that px = v. We reach a contradiction here, since h1 = h2 = • • • = hk = h cannot, assuming k > 3, map the set {1,2,3} to either of the sets {1,2} or {1}. In the case k = 2, in at least one block we must map the set {1} to {1, 2}, which is not possible. Hence 2 > S > k > 2. □ Suppose J is a system of imprimitivity for the action of X on M and C is an X-neighbour-transitive code, with S > 3. The next example shows that it is possible that the projection of C onto each block of J gives the complete code, though this is not the system of imprimitivity of interest to us in Proposition 3.3. Example 4.4. Let <5 = Prod(C, I) be a code in r = H(m, q), where m = ki and C is an X-neighbour-transitive code in H(k, q) where X n B is transitive on C and S > 3. Let X = ((X n B)e, Diag£(X), Se) preserve the partition J = {{1, ...,k},..., {m - k + 1, ..., m}} = {J1,...,Ji}, of M, where xj((X n B)e) = X n B and xj(Diag£(X)) = X for all J G J, and S£ acts as pure permutations by permuting the blocks of J whilst preserving the order of entries within a given block. It follows that we preserve two X-invariant partitions. These being J and J', where J' is attained by taking the corresponding entries, by order, from each copy of C to form each block: J' = {{1, k + 1,...,m - k + 1}, .. ., {i,k + i,..., m}}. Given any a = (ai,..., af) e C, a, e C, and P = ..., fie) e C, ft, e C there exists an x e (X nB)f mapping a to P since X nB is transitive on C. Hence X is transitive on ¿7. Given any two neighbours v e r1 (a), where v differ from a in the respective blocks J, and Jj, we can map J to J, via some element a e Sf. Then, since Xa. is transitive on ri(aj), there exists an element x e Diagf (X) such that j(vCTX) = j(^). We can then map vto ^ via some element h e (X n B)f, where x j (h) = 1, since each nJt (vCTX) and nJt are elements of C for t = i and X n B is transitive on C. Hence axh maps v to ^ and X is transitive on C1. When we consider the projection nJ(C) for any J e J' we are left with the complete code. To see this, consider that for (a1,..., af) e C, a, e C, we may choose an arbitrary element of C as a, for each i. Since xQ* is 2-transitive on Q,, each element appears in the first entry for some codeword. Thus, as nJ((a1,..., af)) when J = {1, k + 1,..., m -k +1} is the first entry of each a,, we have that nJ(C1) is the complete code. 5 Alphabet-almost-simple (X, 2)-neighbour-transitive codes Before we prove the final results we define the codes used in this section, which first requires the following definition. Definition 5.1. Define the composition of a vertex a e H(m, q) to be the set Q(a) = . . . , (aq,Pq)}, where p, is the number of entries of a which take the value a, e Q. For a e H(m, q) define the set Num(a) = {(p1, S1),..., (pj, sj)}, where (p,, s,) means that s, distinct elements of Q appear precisely p, times in a. Definition 5.2. We define the following codes: 1. Inj(m, q), where m < q, is the set of all vertices a e H(m, q) such that Num(a) = {(1, m)}; 2. for m odd, W([m/2], 2) is the set of vertices in a e H(m, 2) such that Num(a) = {(m + 1)/2,1), (m - 1)/2,1)}; and, 3. All(pq, q), with pq = m, is the set of all vertices a e H(m, q) such that Num(a) = {(p, q)}. More information on these codes is available in [9, Definition 2]. The following lemma is [9, Lemma 4]. Lemma 5.3. Let a be a vertex in H(m, q). Then Num(a) is preserved by Diagm(Sq) x L. The last result, in combination with the classification of diagonally neighbour-transitive codes [9, Theorem 4.3], allows us to prove the next result. Proposition 5.4. Let C be a diagonally (X, 2)-neighbour-transitive code in H(m, q). Then one of the following holds: 1. q = 2 and C = {(a,..., a)}; 2. m = 3 or q = 2, and C = Rep(m, q); 3. C = Inj(3,q); 4. m is odd and C = W([m/2], 2); or, 5. q = 2 or q = m = 3, and there exists some p such that m = pq and C is a subset of All(pq, q). Proof. From [9, Theorem 4.3], we have that a diagonally neighbour-transitive code C is one of: {(a,..., a)} for some a G Q, Rep(m, q), Inj(m, q) with m < q, W([m/2], 2) with m odd, or there exists a p such that m = pq and C is a subset of All(pq, q). Here we consider m > 2, since if m =1 then C2 is empty, so C is not (X, 2)-neighbour-transitive. Also to prove some C is (X, 2)-neighbour-transitive, we need only find some X < Aut(C) such that X < Diagm(Sq) x L and X is transitive on C2, since C is already X-neighbour-transitive, for some X, by [9, Theorem 4.3]. First, if C = Inj(2, q) then C2 is empty. Thus, C is not (X, 2)-neighbour-transitive. Table 1 lists the remaining cases which are not 2-neighbour-transitive. The second and third columns give a pair /, v G C2 such that Num(p) = Num(v). Hence, by Lemma 5.3, X is not transitive on C2. It can be deduced from Num(p), Num(v) that /, v G C2, since this makes it clear that we must change /, v in at least two entries to get a vertex in C. Note that we let a = (1, 2,3,..., q) G H(q, q) and in the second last and last rows we assume a G C and (a,..., a) G C, respectively, and observe for the last row /> = (1,1,1,4,5,..., q), > = (1,1, 3,4,5,..., q) are in r2(a). C Conditions ß e C2 Num(^) v e C2 Num(v) {(a,...,a)} q > 3 (b, b, a,... , a) {(m - 2,1), (2,1)} (b, c, a,... , a) {(m - 2,1), (1, 2)} Rep(m, q) m > q > 3 (2, 2,1,...,1) {(m - 2,1), (2,1)} (2,3,1,...,1) {(m - 2,1), (1, 2)} Inj(m, q) m > 4 (1,1,1, 4, 5,... , m) {(3,1), (1, m - 3)} (1,1, 3, 3, 5, 6,... , m) {(2, 2), (1, m - 4)} C All(q, q) q > 4 (1,1,1, 4, 5,. . . , q) {(3,1),(1,q - 3)} (1,1,3, 3,5, 6,... , q) {(2,2),(1,q - 4)} C All(pq,q) q > p > 2 (/i, a,... , a) {(p - 1, 2), (p,q - 3), (p + 2,1)} (i, i, a,... , a) {(p - 2,1), (p,q - 2), (p + 2,1)} Table 1: Diagonally neighbour-transitive codes C which are not diagonally 2-neighbour-transitive, and elements of C2 which illustrate this. Note: /> = (1,1,1,4,5,..., q), > = (1,1,3,4, 5,..., q) and a = (1,2,3,..., q). Now we prove the result for the cases which are 2-neighbour-transitive. Suppose C = {(a,..., a)} for some a G Q. Let q = 2 and Q = {0,1}. Then L = Sm = Aut(C). Without loss of generality, let a = 0 so that C2 is the set of weight two vertices. Since L is transitive on the sets of weight 2 and weight 1 vertices, it follows C is diagonally (X, 2)-neighbour-transitive. Let C = Rep(m, q). It follows from Example 4.1 that Rep(3, q) is (Diag3(Sq) x S3,2)-neighbour-transitive. If q = 2 then Aut(C) = Diagm(S2) x Sm and C is completely transitive [11, Example 3.1]. Consider C = Inj(m, q) with 3 = m < q and q > 4. If v G C2 then v1 = v2 = v3, since otherwise v G C or Ci. Since Diagm(Sq) < Aut(C), we are transitive on C2. Suppose C = W([m/2], 2) and m is odd. Then by [9, Corollary 3.4] C is Diag(S2) x Sm-completely transitive. Finally, suppose C is a subset of All(pq, q) for some p such that m = pq. Let p > 2, q = 2 and C = All(2p, 2). Then C2 is the set of all weight p ± 2 vertices, which Diag2(S2) x Sm < Aut(C) is transitive on. Let p =1, q = 3 and C = All(3, 3). Then C2 = Rep(3, q) and is Aut(C)-completely transitive by Example 4.1. □ With our classification of diagonally (X, 2)-neighbour-transitive codes from the previous result, Propositions 3.3 and 3.4 mean we are now in a position to prove the main theorem. Proof of Theorem 1.1. Suppose C is an X-alphabet-almost-simple and (X, 2)-neighbour-transitive code with S > 3 such that X n B = 1. By Proposition 3.3, there exists an X-invariant partition J = { Ji,..., J^}, for some i, for the action of X on M. Moreover, j (C) has minimum distance at least 2 and is diagonallyxj (X)-neighbour-transitive. By Proposition 3.4, either j (C) has covering radius p < 1, or j(C) is also (xj(X), 2)-neighbour-transitive. Note p = 0, that is, j (C) is not the complete code, since j (C) has minimum distance at least 2. Suppose j (C) has covering radius p > 2. Since XQi is almost-simple, it follows that q > 5. 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Shannon, A mathematical theory of communication, Bell Syst. Tech. J. 27 (1948), 623656, doi:10.1002/j.1538-7305.1948.tb00917.x. [20] P. Sole, Completely regular codes and completely transitive codes, Discrete Math. 81 (1990), 193-201, doi:10.1016/0012-365x(90)90152-8. [21] A. Tietavainen, On the nonexistence of perfect codes over finite fields, SIAM J. Appl. Math. 24 (1973), 88-96, doi:10.1137/0124010. [22] V. A. Zinoviev and V. K. Leontiev, The nonexistence of perfect codes over Galois fields, Probl. Control Inf. Theory 2 (1973), 123-132, http://real-j.mtak.hu/7 981/. ARS MATHEMATICA CONTEMPORANEA ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 14 (2018) 359-373 https://doi.org/10.26493/1855-3974.982.6d6 (Also available at http://amc-journal.eu) Congruent triangles in arrangements of lines Carol T. Zamfirescu * Department of Applied Mathematics, Computer Science and Statistics, Ghent University, Krijgslaan 281-S9, 9000 Ghent, Belgium Received 27 November 2015, accepted 18 July 2017, published online 30 September 2017 We study the maximum number of congruent triangles in finite arrangements of i lines in the Euclidean plane. Denote this number by f (i). We show that f (5) = 5 and that the construction realizing this maximum is unique, f (6) = 8, and f (7) = 14. We also discuss for which integers c there exist arrangements on i lines with exactly c congruent triangles. In parallel, we treat the case when the triangles are faces of the plane graph associated to the arrangement (i.e. the interior of the triangle has empty intersection with every line in the arrangement). Lastly, we formulate four conjectures. Keywords: Arrangement, congruent triangles. Math. Subj. Class.: 52C10, 52C30 1 Introduction A problem from mathematical folklore asks for bounding four congruent triangles with six matchsticks. This is easily done, and left to the reader. Quite naturally, one can ask whether more congruent triangles may be formed by using the same six matchsticks. It seems that this particular problem has not been treated in the literature. Our main focus lies on constructing planar arrangements in which a fixed number i of lines bound as many congruent triangles as possible. For an excellent overview on arrangements and spreads, see Griinbaum's [11]. The results presented in this article are complementary to work of Erdos and Purdy [4, 5] on sets of n points—see also [6]. In this paper, everything happens in R2. An arrangement (of lines) A shall be a finite family of i lines Li,..., L. In the following, we will ignore the case when there exists a point common to all lines, and thus assume that i > 3. Denote by A the set of all arrangements of i lines. * My research is supported by a Postdoctoral Fellowship of the Research Foundation Flanders (FWO). I thank Benjamin Grothe, Iulia Mihai, and Tudor Zamfirescu for their helpful suggestions. Finally, I am grateful to the anonymous referees, who have dramatically improved the quality of this manuscript with their helpful comments. E-mail address: czamfirescu@gmail.com (Carol T. Zamfirescu) Abstract ©® This work is licensed under http://creativecommons.org/licenses/by/3.0/ We associate to A G A£ a graph T^: the vertices of correspond to the intersection points of lines from A and the edges of correspond to the line-segments between these vertices. is a plane graph. The vertices, edges, and faces of are also said to belong to A. A triangle in A G A£ shall be the convex hull of the set of intersection points of three non-concurrent pairwise non-parallel lines in A. Denote by FA the set of all triangles in A. Whenever Ai, A2 G FA are congruent we write Ai ~ A2. Let Ff4,..., Fp4 be the equivalence classes with respect to ~ such that | F-^ | > ... > |Fp41. Here, |M| denotes the cardinal number of M. We call a triangle A g Fa facial if it is a face of rf, i.e. L n int A = 0 for all L G A. Let GA c FA be the set of all facial triangles in A, and, as before, let Gf,..., Gf be the equivalence classes with respect to ~ such that |GA| > ... > |GA|.Put f(¿) = max |F1A| and o(^) = max |Gf| . v ' AeaJ 1 1 yw AeaJ 1 1 We shall also be considering restrictions relative to a certain arrangement A G A£, namely, for k < fA(k) = max |F1b| and g^(k) = max |Gf| . w BcA, Be aj 11 w BcA, Be aj 11 We call an arrangement A G A£ f -optimal (g-optimal) if |Ff| = f (¿) (|GA | = g(^)). If A is both f -optimal and g-optimal, we simply write optimal. A triangle from FA or GA is said to be good. Note that FA and GA need not be unique. In that case, one makes a choice clearly defining FA and GA. The edges and angles of a good triangle will be called good, too. An arrangement is simple if no three lines are concurrent. The lines of an arrangement are in general position if no two lines are parallel and no three lines are concurrent. Two arrangements A and B are combinatorially equivalent if their associated graphs Ta and rB are isomorphic. (Note that, as mentioned above, we do not consider line arrangements in which all lines meet in a single point.) A G A£ is c-unique if there exists no B G A£ such that A and B are (a) not combinatorially equivalent and (b) |Hf| = |#B |, where H is F or G. In the same vein, we say that two arrangements A G A£ and B g A£ are g-equivalent if A can be obtained from B by translation, rotation, reflection, and scaling. A G Ai is g-unique if there exists no B G A£ such that A and B are (a) not g-equivalent and (b) |Hf| = |Hb |, where H is F or G. A few examples: Three lines in general position yield an arrangement that is c-unique, but not g-unique; any arrangement on four lines that forms exactly two congruent triangles is not c-unique (and thus cannot be g-unique); finally, as we shall see in Theorem 3.5, the f-optimal arrangement from Figure 2 (b) is g-unique (and thus c-unique). F(¿) (G(^)) is defined as the set of all integers u such that there exists an arrangement on I lines having exactly u congruent triangles (congruent facial triangles). We write [s..t] for the set of all integers u with s < u < t, put H for F or G, and h = |f if H = F, g if H = G. Whenever H(¿) = [0..h(^)], we say that H(¿) is complete. In the following, we will tacitly use the fact that G(^) c F (¿). We call an arrangement A G A £ 1-extendable if there exists a line L such that |HfUL| = |Hf| + 1. 2 Preparation We briefly concern ourselves with the following question, since it will shorten later arguments. What if we drop the condition that the triangles need to be congruent? Kobon Fu-jimura asked in 1978 in his book "The Tokyo Puzzle" [8]—see also [10, pp. 170-171 and 178]—what the maximum number K(i) of facial (not necessarily congruent!) triangles realisable by i lines in the plane is. (Griinbaum treated this problem before Fujimura, but he might have only been interested in arrangements in the projective plane [6].) Recently, Bader and Clement [1], improving upon a result of Tamura, showed the following. Lemma 2.1 (Bader and Clement). where I denotes the indicator function. Many arrangements have been constructed in order to find solutions to Fujimura's problem. Fujimura himself gave an example which shows that K(7) > 11, although it was thought for many years that K(7) = 10. In 1996, Grabarchuk and Kabanovitch [13] gave two 10-line, 25-triangle constructions, whereas Lemma 2.1 gives K(10) < 26. Whether K(10) is 25 or 26 is unknown. Other 10-line, 25-triangle arrangements were found independently by Grunbaum [12, p. 400], Wajnberg, and Honma (see [15] for more details). Good overviews of the best (i.e. the greatest number of triangles for a fixed number of lines) known arrangements can be found in [14] and [15]. Bader-Clement bound 1 2 5 7 11 15 21 26 33 39 Best known arrangement 1 2 5 7 11 15 21 25 32 38 Furedi and Palasti [9] construct an arrangement proving K(i) > i(i - 3)/3. See also the article of Forge and Ramirez Alfonsin [7]. We continue with a series of lemmas. Lemma 2.2 is stated without its straightforward proof, but we present the heptagonal case in Figure 1 and its caption. Lemma 2.2. The i lines bounding a regular i-gon determine exactly 2i congruent triangles if i > 7, and therefore 2i e F (i) and f (i) > 2i. With the same construction we obtain for i > 5 that i e G(i) and g(i) > i. Lemma 2.3. Let A e A and h e {f, g}. Then, for 3 < k < i — 1, t(t — 2) K(t) < -3- — :(£ mod 6)e{0,2}}(^)1 Table 1: Bounding K(t) for t < 12. 3 4 5 6 7 8 9 10 11 12 Mt) < t(t — 1)(t — 2) k(k — 1)(k — 2) ■ Mk). Proof. We observe that every subset of k < t lines within the arrangement A cannot have more than hA(k) good triangles (good in A). Counting all together, there are at most Ck) good triangles, each appearing several times. Since each of them lies in (l_33) sub-arrangements of k lines, we obtain hA(0 * ^ U-3/ Thus, for each k we obtain an upper bound for Ha(î). □ Figure 1: Seven lines bounding a regular heptagon. This arrangement contains fourteen congruent triangles: abe, acd, and their symmetric counterparts obtained by rotating around the barycentre of the heptagon by 2nk/7 for k = 1,..., 6. This arrangement proves that f(7) > 14. Lemma 2.4 is a direct consequence of Lemma 2.3. Lemma 2.4. Let h G {f, g}. Then h(£) < min e(f - 1)(/ - 2 • h(k). V ' < 3 h(£) + h(k). Proof. We scale B to B' such that the good triangles of B' are congruent with the good triangles of A. Consider the convex hull Ca (Cb> ) of the intersection points of A (B'). Let PA (Pb> ) be an intersection point of A (B') lying on the boundary of the convex polygon Ca (CB>) and incident with a good angle a a (aB>) of A (B') such that a a = aB>. Denote with ¿A and ¿A (LB and LB ) two of the lines of A (B') intersecting at pa (pB>) and forming the angle a a (aB>). We can now identify ¿A with LB and with LB such that an arrangement C is obtained in which, seeing A and B' as sub-arrangements of C, no good triangle lies in both A and B'. (Note that the number of good triangles in C may be larger than the sum of the number of good triangles in A and B', see e.g. Figure 8, in which the arrangements from Figures 1 and 5 (b) are joined: the original arrangements have 14 and 8 good triangles, respectively, but the new arrangement has 26.) □ We write /(¿) if we consider the values of / (¿) only for arrangements whose good triangles are not right triangles. Proposition 3.2. /(^ +1) < /(¿) + 3^ - 1) and /(^ +1) < /(¿) + 4^ - 1). Proof. Let L be a line being added to A G A^ .If A has the property that |FA| = | F^41, then consider henceforth only the triangles in Fj4, as well as their edges, to be good. We denote the lengths of good edges with a, b, and c. There are at most ¿ - 1 triangles with an edge of length a on L: there are at most ¿/2 lines of one of the two lines needed to make a good triangle with an edge on L, each of these lines is part of at most two triangles with an edge of length a on L, and it is impossible for there to be exactly ¿/2 of them each of which is part of exactly two triangles. Since this can be applied analogously for edges of length b and c, we have / ^ +1) = / (¿) + 3^ -1). For good triangles that are right triangles, we argue in the same manner and obtain that for each of the three types of good edge (i.e. of length a, b or c) there are at most 4^ -1)/3 triangles with a good edge of that type on L. □ Proposition 3.3. /(¿) < ¿^ - 1) and /(¿) < 4¿(¿ - 1)/3. Proof. The idea is the same as the one used in the proof of Proposition 3.2. In the case of non-right triangles, we have established that on each line in A there are at most 3^ - 1) good edges. By multiplying this with ¿, we obtain an upper bound for the number of good edges in A. Now we divide by three (as there are three edges to each triangle) and have the desired bound. The case of right triangles is settled analogously. □ All angles in A G A^ equal to one of the angles of a good triangle which is not a right triangle will be called non-right angles. Proposition 3.4. / (¿) < 2¿(¿ - 2)/3 for simple arrangements. Proof. Consider a simple arrangement A on ¿ lines admitting a good triangle which is not a right triangle. Let V(r^) = V, and write Vk for the set of vertices of degree k. As no three lines are concurrent, in there exist only vertices of degree 2, 3, or 4. Trivially, around a vertex of degree 2 at most one non-right angle resides. Around a vertex of degree 3 likewise (as n/2 is a right angle, and the sum of two non-right angles must be strictly smaller than n), and around a vertex of degree 4 there may be at most two non-right angles. Thus, in A, we have as an upper bound for the maximum number of non-right angles |V2| + |V3| + 2 • |V41 = |V| +1V41. We have |V| < ¿^-1)/2. Also |V4| < |V|-¿, because on every line the first and the last vertex belong to V2 U V3, but any such vertex may appear as first or last vertex on two lines of A. Thus, the bound is ¿2 - 2¿. For odd ¿ > 5, this bound is best possible: for the ¿ lines bounding a regular ¿-gon we have | V21 = ¿, | V31 = 0, and | V41 = ¿^ - 3)/2. One non-right angle cannot lie in more than two triangles which are not right triangles, and every triangle requires three angles, whence, the final bound. (In fact, no good angle can lie in more than two good triangles.) □ 3.2 I < 5 We have f (3) = g(3) = 1 and f (4) = g(4) = 2, and F(3), G(3), F(4), and G(4) are complete. We leave the easy proofs to the reader, but mention that in the 4-line case there exist exactly three combinatorially different solutions with two congruent triangles (these coincide for the g-optimal and the f-optimal case): one with three concurrent lines, one with two parallel lines, and one in general position. We now focus on the first interesting case: I = 5. Figure 2: (a) shows a 5-line arrangement with four congruent triangles constructed as follows. Two lines L1, L2 orthogonal to a third line L3 are considered. Let the intersection points be p1 and p2, resp. A fourth and fifth line are considered such that their intersection point is the midpoint of the line-segment p1p2 and the angle each forms with L3 is n/4. (b) depicts the five lines bounding a regular pentagon. This arrangement contains ten triangles, distributed among two congruence classes of size 5 each. Theorem 3.5. (i) We have f (5) = g(5) = 5 while F(5) and G(5) are complete. Furthermore, the arrangement from Figure 2 (b) is (ii) optimal, and (iii) g-unique among f -optimal and g-optimal 5-line arrangements. Proof. Figure 2 (b) shows that g(5) > 5 (whence, f (5) > 5), with which Lemma 2.1 implies g(5) = 5. f (5) = 5 follows from Lemma 2.4 (with k = 4). We now discuss G(5). We have G(4) = [0..2] c G(5). Consider the four lines bounding a square and add the two lines containing the square's diagonals. By removing one of the four original lines, we have shown that 3 G G(5). Together with the arrangements from Figure 2, we have that G(5) is complete since g(5) = 5. Thus, (i) is proven. (ii) follows directly from (i). We now prove (iii). First, we show that the arrangement from Figure 2 (b) is c-unique. We use the database provided in Christ's Dissertation [3, Chapter 3.2.5]. (A visualisation of Christ's results is available in [2]. Note that this does not coincide with Griinbaum's isomorphism types of arrangements given in [11, p. 5], since Grunbaum discusses the issue in the projective plane, while here we treat the situation in the Euclidean plane.) Among arrangements of five lines in general position, there are exactly six combinatorially different ones, shown in Figure 3. The arrangement in Figure 2 (b) belongs to the combinatorial class (A). Only the arrangements in (A) contain five facial triangles, i.e. triangles which are faces in the associated graph. We leave to the reader the straightforward proof that among arrangements with five lines not in general position (i.e. containing two parallel lines or three concurrent lines), there is none featuring five facial triangles. Note that the occurrence of more triangles is impossible due to Lemma 2.1. We now turn to the case in which triangles are not facial. Let us denote a line-segment between two points x, y with xy and its length with L(xy). We will use the following. Remark. The sum of the measures a and ft of two good angles is n if and only if a = ft = n/2. We write Aijk for the triangle with vertices pi, pj, pk. We will tacitly make use of the fact that if in a given arrangement a triangle A is strictly contained in a triangle A', then A ^ A' and so A and A' cannot lie in the same congruence class. (B) We have A012 C A149 n Am (so A149 ^ A012 ^ A136), A345 c A149 n A248, A569 c A136 n A237 n A578, A578 c A237, A067 c A237, A089 c A248 n A237 n A067. Due to these inclusion relations, only the following set of triangles may form a congruence class of size 5: {A149, A136, A248, A578, A067}. Assume it is indeed a congruence class. Thus, all angles around p6 are right. We apply the Remark to the angles surrounding p6. Combining this with A136 ~ A067 and p0p6 C p1p6, we have L(p6p7) = L(p1p6), L(p0p6) = L(p3p6), and L(p0p7) = L(p^). p^ is the hypothenuse in Am, but as p1p4 is an edge of A149 and A149 ~ A136 we obtain a contradiction, since L(p1p4) > L(p1p3). (C) We have A012 c A134 n A268 n A378, A134 U A239 c A378, A457 U A056 c A158, A049 c A158 n A239 n A378 n A056, A679 c A158 n A268 n A457. There is no congruence class of size 5. (D) We have A012 C A149 c A156, A234 C A039 c A378, A457 U A068 C A258, A679 C A258 n A457 n A068. Once more, all congruence classes have size at most 4. (E) We have A129 C A138 C A145, A237 C A246, A034 C A058 C Aoe7, A789 C A237 n A246 n A569. As above. (F) We have A012 C A139 C A145, A238 U A056 C A246, A678 C A579 C A347, A089 C A238 n A246 n A056. As above. Let us show that in a 5-line arrangement A containing two parallel lines or three concurrent lines, no more than four congruent triangles can be achieved. We first assume that A contains parallel lines L1, L2. If there exists a line L3 parallel to L1, we are done, as in A there are only at most three triples of lines forming triangles. Thus, w.l.o.g. we are in the situation that a third line, L3, intersects L1 and L2. Now assume that a fourth line, L4, is parallel to L3. Note that L1, L2, L3, L4 bound zero triangles. In this situation, a fifth line generates at most four new triangles. Thus, L4 cannot be parallel to L3. We have proven that a 5-line arrangement containing three parallel lines or two parallel pairs of parallel lines cannot have more than four congruent triangles. Denote the open strip bounded by L1 and L2 with S, and the complement of its closure by T. Also, let T and T2 be the connected components of T. In the light of above paragraph, there are three cases (see Figure 4): either (a) L4 is concurrent with L1 and L3 in a point x lying in the closure of T\, (b) L4 intersects L3 in S, or (c) L4 intersects L3 in T2. Denote with L5 the fifth line of A. We know that L5 is not parallel to any of the existing four lines. We write Aijfc for the triangle bounded by the lines Li; Lj, Lk. Figure 4: Cases (a)-(c) occurring in the proof of Theorem 3.5. (a): If x g L5, then the five lines would bound only three triangles, so we can assume x G L5. Case 1: L5 intersects both L3 and L4 in S. We have six triangles, but A345 = A234 n A145 and A245 C A235, so the maximum number of congruent triangles is three. Case 2: L5 intersects L4 in S and L5 intersects L3 in T1U T2. Six triangles appear. Subcase 2.1: L3 and L5 intersect in ï\. But then we have A135 C A345 C A235. Subcase 2.2: L3 and L5 intersect in T2. Here, A235 C A345 C A135, so once more five congruent triangles cannot occur. (If L5 intersects L3 in S and L4 in T2, then we are, combinatorially, in the situation treated in Subcase 2.2.) Case 3: L5 is concurrent with L2 and L3. Subcase 3.1: All intersection points lie in the closure of S. Five triangles appear, but among them one is a subset of another. Subcase 3.2: L4 and L5 intersect in T2. We apply the same argument as before. Subcase 3.3: L4 and L5 intersect in T1 . Once more five triangles occur, but one is contained in another. Case 4: L5 intersects L3 and L4 in points p and p', respectively, which do not lie in S (since this was covered in Cases 1 and 2). Subcase 4.1: If p and p' lie in Ti, six triangles appear, but A345 c A135 c A235. Subcase 4.2: If p and p' lie in T2, again six triangles occur, but A245 c A235 c A135. Subcase 4.3: If p G T1 and p' G T2, six triangles are present in the arrangement, but A245 c A145 c A345, so at most four triangles are congruent. Case 5: L5 is concurrent with L2 and L4. Subcase 5.1: All intersection points lie in the closure of S. Subcase 5.1 coincides with Subcase 3.1. Subcase 5.2: L3 and L5 intersect in T2. But then we are in the same situation as Subcase 3.2. (b) Let L3 and L4 intersect in y. We know that L5 is not parallel to L1. If y G L5, we obtain six triangles. However, either A135 U A145 = A134 and symmetrically A235 u A245 = A234 or A134 U A145 = A135 and A245 U A234 = A235. In either case, the largest congruence class has cardinality at most four. We have treated the cases when L5 is concurrent with L1 and L3, L1 and L4, L2 and L3, or L2 and L4 in (i). We split the remaining cases into four cases according to where the intersection points of L5 with L3 and L4 lie. In each situation, inclusions are given which make the occurrence of a congruence class of cardinality at least five impossible. Case 1: Both intersection points lie in S. However, we then have A135 c A145, A345 = A134 n A235, and A234 c A245. Case 2: The intersection points of L5 with L3 and L4 lie in T1 and T2, respectively. Then A135 U A245 c A345, A135 U A234 c A235, and A134 U A245 c A145. Case 3: The intersection points of L5 with L3 and L4 lie in T\ and S, respectively. We have A135 c A345 c A235, A245 c A234, and A134 c A145. Case 4: Both intersection points lie in Ti. Then A234 c A235, A135 = A235 n A145, and A134 c A145 c A245. As situation (c) uses very similar arguments, we skip it. We have shown that no two lines in A are parallel. Assume now that three lines L1; L2, L3 of A intersect at a point q. If L4 contains q as well, the largest congruence class which may be formed by a fifth line has size 2. So q G L4 and L4 is not parallel to any of L1; L2, L3. W.l.o.g. let the intersection point of L4 with L2 lie between the intersection point of L4 with L1 and the intersection point of L4 with L3. If there are two coincidences (of three lines)—it is easy to see that there cannot be more—we have three combinatorially different cases. W.l.o.g., in each of them L1; L4, and L5 shall be concurrent. We denote this intersection point with a, and the intersection point of L5 with L2 and L3 with b and c, respectively. We differentiate the three cases by the order in which the intersection points occur on L5. Case 1: a - c - b (or equivalently b - c - a): Eight triangles occur. However, we have A124 U A234 = A134 c A345 and A135 U A235 = A125 c A245. Thus, no five triangles can be congruent. Case 2: b - a - c (or equivalently c - a - b): Again, eight triangles appear, but A245 U A124 = A125, A125 U A135 = A235, A234 U A124 = A134, and A134 U A135 = A345. Case 3: a - b - c (or equivalently c - b - a): A124 n A135 = A125, A235 c A234, and A245 c A345. Furthermore, every triangle is contained in A134. We are left with the case that there is exactly one coincidence of three lines (namely in q). Once more, several cases occur. We leave them to the reader—treating them is a straightforward task in exactly the same spirit as above paragraphs. Finally, we prove that the construction from Figure 2 (b) is indeed g-unique. Consider five lines bounding a pentagon P such that we obtain an arrangement A in combinatorial class (A). This implies that no two lines in A are parallel. Combinatorially, there are two types of triangles in P: those sharing exactly two vertices (and thus an edge) with P, and those sharing exactly one vertex with P. Due to straightforward inclusion arguments, all triangles in a congruence class of size 5 are of the same type. Consider the first type, and let A be one of these five congruent facial triangles. Denote the angles of A incident with a vertex of P with a and fi. Applying successively the fact that no two lines in A are parallel, we obtain that a = fi, so A is isosceles. This implies that all angles of P must be equal, and since P is a pentagon, the angles of P measure 3n/5 each. Thus a = 2n/5—in particular, A is not equilateral. Hence, the sides of P must have equal length, so P is a regular pentagon. We treat the second case. We see each triangle of the second type as the union of three faces (of r^): the pentagon P, which lies in all five triangles, and two facial triangles. Since certain pairs of triangles of the second type share a facial triangle, there are at most two congruence classes C1 and C2 of facial triangles. Assume C1 = C2. Thus, there exists a triangle A of second type containing a facial triangle in Ci and a facial triangle in C2. By considering all five congruent triangles of second type, a contradiction is obtained, since necessarily one of these triangles will contain only triangles from either C1 or C2 and thus, it cannot be congruent to A. We have proven that all facial triangles are congruent. Now we may argue as in the preceding paragraph. □ 3.3 I = 6 (a) (b) Figure 5: (a) This arrangement is due to Tudor Zamfirescu and shows that 7 G F(6). To the five lines bounding a regular pentagon a sixth line is added which is parallel to one of the five lines such that seven congruent triangles are present. (b) This arrangement proves that 8 G F(6) and f (6) > 8. In Theorem 3.6 we show that in fact f (6) = 8. The arrangement is obtained by considering six of the seven lines bounding a regular heptagon. Theorem 3.6. We have f (6) = 8, 6 < g(6) < 7, F(6) is complete, and [0..6] c G(6). Proof. The arrangement from Figure 5 (b) proves that f (6) > 8. Theorem 3.5 (iii) states that there is exactly one f-optimal arrangement on five lines, shown in Figure 2 (b). We call this arrangement P. We now show that one cannot produce an arrangement on six lines which has P as a sub-arrangement and features eight (or more) congruent triangles. Assume there exists such an arrangement A. Denote the lines of P by Li;..., L5, and the line added to P in order to obtain A by L. First we prove that the addition of L cannot create a "new" congruence class (i.e. a class the triangles of which are non-congruent to every triangle present in P) of congruent triangles of cardinality at least 8. At least one of the angles n/5, 2n/5, 3n/5, 4n/5 is good in both P and A, since every triangle bounded by L has at least one angle in P. Among all angles in A, each of the aforementioned four angles appears at least ten times in five pairs of opposite angles, since P is a sub-arrangement of A. Thus, L forms at least three copies of the angle a with the lines Li,..., L5, where a G {n/5, 2n/5, 3n/5,4n/5} is fixed. But since the Lj's are pairwise non-parallel, this is only possible if L is parallel to some Lj. But then the addition of L to P yields at most six new triangles—too few. Take the two congruence classes Fp and F2P such that the triangles in Ff are facial in P, and notice that FP = Ff U Ff and |Ff | = |Ff | = 5. Thus, L must add at least three triangles to Fp or Ff. As n/5 belongs to triangles in Fp as well as triangles in F2P, n/5 is a good angle in A, so L makes this angle with a line of P, whence, L is parallel to some Li, say Li. Among all possible positions of L, only three provide new triangles congruent either to a good triangle in Fp or to a good triangle in Ff, see Figure 6. The number of those new triangles is 1, 1,2, respectively. Figure 6: The three essentially different arrangements of five lines bounding a regular pentagon together with a sixth line parallel to one of the five lines forming at least six congruent triangles. We conclude that in an arrangement on six lines which is f -optimal, every sub-arrangement on five lines contains at most four good triangles. With this in mind, by applying Lemma 2.3 (with k = 5), we obtain the desired f (6) = 8. Lemmas 2.1 and 2.2 yield the bounds on g(6). Theorem 3.5 (i) and the Star of David (which proves that 6 G G(6)) imply that [0..6] c G(6). Together with the arrangements from Figure 5, we are done. □ Among arrangements on six lines bounding exactly six congruent facial triangles, we found three combinatorially non-equivalent ones. (It is unknown whether these are all.) In the general case, the solution from Figure 5 (b) seems to be unique; see Conjecture 4.1 (which states that g(6) = 6) in the final section. 3.4 I = 7 Theorem 3.7. We have f (7) = 14, 9 < g(7) < 11, [0..10] U {14} c F(7), and [0..9] c G(7). Proof. By Theorem 3.6, f (6) = 8. Thus, by Lemma 2.2 and Lemma 2.4 (with k = 6), f (7) = 14. For g(7), the lower bound is given by the construction in Figure 7 (a) (by deleting the line marked h), the upper bound by Lemma 2.1. Since the Star of David is 1-extendable, we have 7 g G(7). Removing the line marked h in Figures 7 (a) and (b) shows that 9 g G(7) and 8 g G(7), resp. Thus, [0..9] c G(7). By considering seven of the eight lines bounding a regular octagon, we obtain 10 g F(7). Together with Lemma 2.2, we have [0..10] U {14} c F(7). □ (a) (b) (c) Figure 7: (a) An arrangement proving 12 g G(8). Deleting the line marked h shows that 9 g G(7). (b) An arrangement showing 11 g G(8). Deleting h yields 8 g G(7). (c) This arrangement proves that #(10) > 20. 3.5 I = 8 Theorem 3.8. We have 16 < f (8) < 22, 12 < g(8) < 15, [0..16] \ {13} c F(8) and [0..12] c G(8). Proof. Lemma 2.2 implies the lower bound for f (8), Theorem 3.7 and Lemma 2.4 (with k = 7) the upper bound. For g(8), the lower bound is given by the arrangement from Figure 7 (a), the upper bound by Lemma 2.1. Figures 7 (a) and (b) show that {11,12} c G(8). This, Theorem 3.7 and the fact that the arrangement from Figure 7 (a) minus the line marked h is 1-extendable (which proves that 10 g G(8)) yield [0..12] c G(8). Applying Lemmas 2.2 and 2.4, we obtain [0..16] \ {13} c F(8). □ 3.6 9 < I < 12 As the techniques for proving the following results are very similar to what has been shown, we skip them. A notable exception is the construction from Figure 8. Theorem 3.9. We have 18 < f (9) < 33,15 < g(9) < 21, [0..18] c F(9), [0..15] c G(9), 21 < f (10) < 48, 20 < g(10) < 26, [0..21] c F(10), [0..20] c G(10), 26 < f (11) < 66, 23 < g(11) < 33, [0..26] c F(11), [0..23] c G(11), and 32 < f (12) < 88, 26 < g(12) < 39, [0..28] U {32} c F(12), [0..26] c G(12). Figure 8: An arrangement proving f (11) > 26. It is obtained by joining the two arrangements from Figures 1 and 5 (b) with the technique described in the proof of Proposition 3.1, i.e. such that the two arrangements share a pair of lines (forming the same good angle) in the new arrangement. Deleting the line marked h, one obtains f (10) > 21. By completing the left regular heptagon, we obtain f (12) > 32. 3.7 Summary Consider I < 12 lines in the Euclidean plane, and let f (¿) and g(^) be defined as in the Introduction. Then we have the following bounds. Table 2: Bounding f (¿) and g(^) for t < 12. t 3 4 5 6 7 8 9 10 11 12 f(t) > 1 2 5 8 14 16 18 21 26 32 f(t) < 1 2 5 8 14 22 33 48 66 88 g(t) > 1 2 5 6 9 12 15 20 23 26 g(t) < 1 2 5 7 11 15 21 26 33 39 We were also able to prove that f (13) > 37, f (14) > 44, f (15) > 50, f (16) > 56, and f (17) > 61. 4 Conjectures Conjecture 4.1. g(6) = 6. If Conjecture 4.1 is true, we would have g(7) < 10. Conjecture 4.2. g(7) = 9. If Conjecture 4.2 is true, we would have g(8) < 14. Conjecture 4.3. The f -optimal arrangements on 6 and 7 lines (consider Figures 5 (b) and 1, resp.) are g-unique. Conjecture 4.4. (a) F(7) is not complete, but (b)for every I, G(t) is complete. References [1] J. Bader and G. Clement, Tighter upper bound for the number of Kobon triangles, 2007, draft version, http://www.tik.ee.ethz.ch/sop/publicationListFiles/ cb2007a.pdf (accessed on 16 September 2017). [2] T. Christ, Database of Combinatorially Different Line Arrangements, http://www.inf. ethz.ch/personal/christt/line_arrangements.php (accessed on 15 June 2016). [3] T. Christ, Discrete Descriptions of Geometric Objects, Ph.D. thesis, ETH Zurich, 2011, doi: 10.3929/ethz-a-007018092. [4] P. Erdos and G. Purdy, Some extremal problems in geometry III, in: F. Hoffman (ed.), Proceedings of the 6th Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic University, Boca Raton, Florida), Utilitas Mathematica, 1975 pp. 291-308. [5] P. Erdos and G. Purdy, Some extremal problems in geometry IV, in: F. Hoffman (ed.), Proceedings of the 7th Southeastern Conference on Combinatorics, Graph Theory, and Computing (Louisiana State University, Baton Rouge, Louisiana), Utilitas Mathematica, 1976 pp. 307322. [6] P. Erdos and G. Purdy, Extremal problems in combinatorial geometry, in: R. L. Graham, M. Grotschel and L. Lovasz (eds.), Handbook of Combinatorics, The MIT Press, Cambridge, Massachusetts, volume 1, 1995 pp. 809-874. [7] D. Forge and J. L. Ramirez Alfonsin, Straight line arrangements in the real projective plane, Discrete Comput. Geom. 20 (1998), 155-161, doi:10.1007/pl00009373. [8] K. Fujimura, The Tokyo Puzzles, Charles Scribner's Sons, New York, 1978. [9] Z. Furedi and I. Palasti, Arrangements of lines with a large number of triangles, Proc. Amer. Math. Soc. 92 (1984), 561-566, doi:10.2307/2045427. [10] M. Gardner, Wheels, Life and Other Mathematical Amusements, W. H. Freeman & Co., San Francisco, California, 1983. [11] B. GrUnbaum, Arrangements and Spreads, volume 10 of CBMS Regional Conference Series in Mathematics, American Mathematical Society, Providence, Rhode Island, 1972, http: //bookstore.ams.org/cbms-10. [12] B. Grunbaum, Convex Polytopes, volume 221 of Graduate Texts in Mathematics, Springer, New York, 2nd edition, 2003, doi:10.1007/978-1-4613-0019-9. [13] V. Kabanovitch, Kobon triangle solutions, Sharada (publication of the Russian puzzle club Diogen) 6 (1999), 1-2. [14] N. J. A. Sloane, Sequence A006066 (formerly M1334) in The On-Line Encyclopedia of Integer Sequences, published electronically at https://oeis.org. [15] E. W. Weisstein, Kobon Triangle, from MathWorld—A Wolfram Web Resource, http:// mathworld.wolfram.com/KobonTriangle.html. /^creative ^commor ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 14 (2018) 375-385 https://doi.org/10.26493/1855-3974.911.3b4 (Also available at http://amc-journal.eu) A note on the directed genus of Kn n n and K n Rong-Xia Hao * Department of Mathematics, Beijing Jiaotong University, Beijing 100044, P.R. China Received 7 August 2015, accepted 28 May 2017, published online 24 October 2017 Abstract It is proved that a complete graph Kn can have an orientation whose minimum directed genus is I" 12(n - 3)(n - 4)] if and only if n = 3,7 (mod 12). This answers a question of Bonnington et al. by using a method different from current graphs. It is also proved that a complete symmetric tripartite graph Kn n n has an orientation whose minimum directed genus is 1 (n - 1)(n - 2). Keywords: Digraph, complete tripartite graph, directed genus, surfaces. Math. Subj. Class.: 05C10, 05B05, 05B07 1 Introduction Throughout this paper, all graphs are assumed to be finite, connected and simple. In a directed graph D, the number of in-arcs at a vertex v is called the in-degree of v which is denoted by d- (v); the number of out-arcs at v is called the out-degree of v, denoted by d+(v). The degree of v, denoted by d(v), is the sum of d-(v) and d+(v). A digraph D is Eulerian if it is connected and every vertex has equal in-degree and out-degree. The underlying graph G of a digraph D is a graph obtained from D by suppressing all directions of the arcs in D. The orientable surface of genus h, denoted by Sh, is the sphere with h handles added. A graph is said to be 2-cell embedded in a surface S, if it is embedded in a surface S such that each component, called a region, of S \ D is homeomorphic to an open disk. A 2-cell directed embedding (or 2-cell embedding) of a digraph D on an orientable surface S means that it is a 2-cell embedding of its underlying graph of D in S such that each region is bounded by a directed cycle. In this paper, all embeddings of graphs *The author expresses the sincere thanks to the anonymous referees for their constructive suggestions that greatly improve the quality of this paper. This work was supported by the National Natural Science Foundation of China (Nos. 11371052, 11731002), the Fundamental Research Funds for the Central Universities (Nos. 2016JBM071, 2016JBZ012) and the 111 Project of China (B16002). E-mail address: rxhao@bjtu.edu.cn (Rong-Xia Hao) ©® This work is licensed under http://creativecommons.org/licenses/by/3.0/ and digraphs are assumed to be 2-cell embedded on oriented surfaces. Let the genus of a surface S be denoted by y(S). The directed genus (or simply say genus) of an embeddable digraph D, denoted by y(D), is the smallest of the numbers 7(S) for orientable surfaces S in which D can be directed embedded. Let |X | be the cardinality of a set X. The study of embeddings of a graph began with Euler. By now, there are many results about the genus ([14, 22, 23, 25, 26, 28, 27, 29]), the maximum genus ([24, 30]), and the genus distribution of a graph ([12, 13, 19, 20]). However, a study of the embeddings of a digraph was started in 2002 by Bonnington et al. in [2]. Bonnington, Hartsfield and Sirffl ([3]) gave some obstructions for directed embeddings of digraphs and proved Kuratowski-type theorem for embeddings of digraphs in the plane. This area has remained almost uninvestigated. As we know, genera of only a few kinds of digraphs are known. Hales and Hartsfield calculated the directed genus of the de Bruijn graph in [15]. Hao et al. ([16, 17, 18]) obtained the embedding distributions of some digraphs and maximum embedding properties of digraphs. Chen, Gross and Hu ([4]) derived a splitting theorem for digraph embedding distributions that is analogous to the splitting theorems of [11] and [5] for graph embedding distributions. Let 7(G) denote the genus of a graph G. There are many results on computing genera of undirected graphs. For example, in [25], the genera of the complete graph K„ and the complete tripartite graph Km„,„,„ were given as follows: 7(K„) = [ (n - 3)(n - 4)] and 7(Km„,„,„) = 1 (mn - 2)(n - 1). In [28], 7(K„,„,„-2) = 2 (n - 2)2 for even n > 2 and y(K2„,2„,„) = 2(3n - 2)(n - 1) for n > 1 were derived. In [26], 7(K„,„,„) = 2 (n - 2)(n - 1) was obtained. Up to now, the genera of only a few kinds of digraphs are known. For examples, the directed genus of the de Bruijn graph was derived in [15]. In [2], Bonnington et al. determined the genera of the cartesian product C„ x C„ of two directed cycles, the spoke digraph on n = 2k +1 vertices and the directed antiprism DAk, which are (n2 - 3n + 2)/2, k - 1 and 0, respectively. Let K„ and K„,„,„ be directed graphs gotten from the complete graph K„ and the complete tripartite K„,„,„, respectively, by giving an orientation to each edge. In this paper, we aim to answer the following problem by using a method different from current graphs. Problem 1.1 ([2]). Whichkindsof K„ have y(G ) = [ ± (n - 3)(n - 4)], the genus of K„. A natural question analogue to Problem 1.1 is the following. Problem 1.2. Which kinds of K„,„,„ with n vertices in each parts have directed genus 2(n - 1)(n - 2), the genus of K„,„,„. In this paper, we solve the Problems 1.1 and 1.2. Problem 1.2 is solved by giving the equivalent conditions for the minimum directed genus embedding of a directed graph K„,„,„ and a pair of biembeddable Latin squares with order n in an orientable surface. Furthermore, we prove that there is a one to one correspondence between the set of directed embeddings of a digraph D and the set of face-2-colorable embeddings of the underlying graph of D both on orientable surfaces. The result that there exists an orientation on edges of K„ such that the obtained tournament K„ has the directed genus [ 12 (n - 3)(n - 4)], when n = 3, 7 (mod 12) is gotten which answer the Problem 1.1. 2 Alternating rotations, face-2-colorable embeddings, and Latin squares An alternating rotation at a vertex v of D is a cyclic permutation of the arcs incident at v, such that in-arcs alternate with out-arcs. A list of alternating rotations, one for each vertex, is called an alternating embedding scheme (also called alternating rotation system) for the digraph D. There exists a one to one correspondence between the set of all embeddings (resp. directed embeddings) of a graph G (resp. a digraph D) on orientable surfaces and the set of the embedding schemes (resp. alternating embedding schemes) of G (resp. D). A color class is a set of faces with the same color. A face-2-colorable embedding of a graph G is an embedding which admits a 2-coloring of regions such that no two distinct regions of the same color shares a common edge. Two colors always mean black and white. Regions in an embedding of a graph are also called faces, while regions in a directed embedding of a digraph are partitioned into faces which use the arcs in the forward direction and antifaces which use arcs traversed against the given orientation. An embedding is triangular if all regions are bounded by 3-cycles. Two face-2-colorable embeddings of Kn are said to be isomorphic if there exists a permutation on the n vertices (of the complete graph) such that it maps edges and faces of one embedding to edges and faces of the other one, respectively, see [2]. Equivalently, two face-2-colorable embeddings of Kn are isomorphic if and only if there exists a permutation on the n vertices such that it either preserves the color of the triangles or reverses the color. Let Di and D2 be two digraphs. If D1 is derived from D2 by reversing all arcs of D2, then we say these two digraphs have the opposite orientation. A transversal design TD(3, n) is an ordered triple (V, G, B), where V is a 3n-element set (the points), G is a partition of V into three disjoint sets (the groups) each of which has cardinality n, and B is a set of three-element subsets of V (the triples), such that every unordered pair of elements from V is either contained in precisely one triple or one group, but not both. Example 2.1. An example of a TD(3, n) of n = 3. Let V = {1, 2, 3,..., 9}, G = {{4,5,6}, {7,8,9}, {1,2, 3}}, and B = {(4, 7, 3), (4, 8,1), (4, 9, 2), (5, 7,1), (5, 8, 2), (5, 9, 3), (6, 7, 2), (6, 8, 3), (6, 9,1)}. Then (V, G, B) is a transversal design TD(3,3). A Latin square LS(n) of order n is an n x n array filled with n different entries, each occurring exactly once in each row and exactly once in each column. Example 2.2. A Latin square LS(n) of order n for n = 3. Let M = 3 1 2 1 2 3 2 3 1 Then M is a Latin square LS(3). There are relations among the face-2-colorable triangular embeddings of Kn,n,n on an orientable surface, the transversal design TD(3, n) and the Latin squares as follows. For a given face 2-colourable triangular embeddings of Kn,n,n on an orientable surface, it is proved in [10] that there exists a transversal design which is determined under one of the clockwise and counter-clockwise in each colour class. On the other hand, for a given transversal design TD(3, n) = (V, G, B), there is a Latin square determined by TD(3, n) by assigning the three groups in G as labels for the row, columns and entries of the Latin square. Two color classes A and B of a face-2-colorable triangular embedding of Kn n n on an orientable surface give two Latin squares, corresponding to A and B respectively, which is considered as a biembedding of these two Latin squares with order n. Two Latin squares A and B are biembeddable, denoted by A ixi B, on an orientable surface S if there is a face-2-colorable (black and white) triangular embedding of Kn n n in the orientable surface S such that the white face set is A and the black face set is B. For more details, the readers are referred to [6, 7, 8, 9] and [21]. Example 2.3. Let Vi, V2 and V3 be a partition of V(K3,3,3), where V = {4, 5,6}, V2 = {7, 8,9} and V3 = {1, 2, 3}. For a given embedding p of K3 3 3 on an orientable surface, let pv be the rotation at a vertex v. Let p1 = (7, 5, 9, 6, 8,4); p2 = (7, 6, 9,4, 8, 5); p3 = (7,4, 9, 5, 8, 6); p4 = (7, 3, 9, 2, 8, l); p5 = (7,1, 9, 3, 8, 2); p6 = (8, 3, 7, 2, 9, l); p7 = (l, 5, 2, 6, 3,4); p8 = (2, 5, 3, 6,1, 4); p9 = (2,4, 3, 5,1, 6). Then p = {p4 : i e {1,..., 9}} is a face 2-colourable triangular embedding of K3,3,3 on an orientable surface. In fact, a set of faces with the white color is A1 = {(5, 7,1), (6, 9,1), (4, 8,1), (6, 7, 2), (4, 9, 2), (5, 8, 2), (4, 7, 3), (5, 9, 3), (6, 8, 3)}; while a set of faces with the black color is A2 = {(9, 5,1), (8, 6,1), (7, 4,1), (9, 6, 2), (8,4, 2), (7, 5, 2), (9,4, 3), (8, 5, 3), (7, 6, 3)}. There exists a transversal design TD(3,3), say (V, G, B1), which is determined under the clockwise in white colour class A1. That is, V = {1, 2, 3,..., 9}, G = {{4,5, 6}, {7, 8, 9}, {1, 2, 3}}, and B1 = {(5, 7,1), (6, 9,1), (4, 8,1), (6, 7, 2), (4, 9, 2), (5, 8, 2), (4, 7, 3), (5, 9, 3), (6, 8, 3)}. There exists another transversal design TD(3, 3), say (V, G, B2), which is determined under the counter-clockwise in black colour class A2. That is, V = {1, 2, 3,..., 9}, G = {{4,5, 6}, {7, 8, 9}, {1, 2, 3}}, and B2 = {(5, 9,1), (6, 8,1), (4, 7,1), (6, 9, 2), (4, 8, 2), (5, 7, 2), (4, 9, 3), (5, 8, 3), (6, 7, 3)}. Example 2.4. Let (V, G, B1) be a transversal design given in Example 2.3. Assume that {4, 5,6} labels for the row, {7,8, 9} labels for columns and {1,2,3} labels for entries of the Latin square. Thus B1 = {(5, 7,1), (6, 9,1), (4, 8,1), (6, 7, 2), (4, 9, 2), (5, 8, 2), (4, 7, 3), (5, 9, 3), (6, 8, 3)} determines the matrix A1 as 7 8 9 '3 1 2N 1 2 3 231 (2.1) Thus there is a Latin square A1 determined by (V, G, B1), where Ai 312 123 231 Similarly, for a transversal designs (V, G, B2) given in Example 2.3, there is a Latin square A2 determined by (V, G, B2), where A2 123 231 312 In fact, using V1 = {4,5,6} as labels for the row, V2 = {7, 8,9} as labels for the columns, and V3 = {1,2, 3} as labels for entries of the Latin square, thus B2 = {(5, 9,1), (6, 8,1), (4, 7,1), (6, 9, 2), (4, 8, 2), (5, 7, 2), (4, 9, 3), (5, 8, 3), (6, 7, 3)} determines the matrix A2 as 7 8 9 4/1 2 3\ 5 I 2 3 1 I . (2.2) 6 y3 1 2/ As a result, a face-2-colorable triangular embedding p of K3 3 3 on an orientable surface gives two Latin squares A1 and A2, corresponding to two color classes A1 and A2 respectively. And A1 ix A2 is a biembedding of these two Latin squares with order 3. Because an embedding of an embeddable digraph is an embedding of the underlying graph, the following version of Euler's polyhedral formula holds. Lemma 2.5. Let D = (V, A) be an embedding digraph, then for any alternating embedding scheme p of D, we have |V| — |A| + |R| = 2 - 2g, where |R| is the number of regions in the embeding scheme p and g is the genus of the embedding surface. Lemma 2.6 ([7]). There is a unique regular triangular embedding of a complete tripartite graph Kn,n,n on an orientable surface for n > 2. Lemma 2.7 ([6]). For a triangular embedding of Kn,n,n, it is orientable if and only if it is face-2-colorable embedding. The readers are referred to [1] for any undefined notations. ___ —* 3 The directed genus of Kn,n,n For an embedding a of a given digraph Kn,n,n, the alternating embedding scheme is denoted by pCT, the alternating rotation at a vertex v e V(D) is denoted by pCT(v) (or simply Pv ). Recall that Kn,n,n is a complete tripartite graph. A complete tripartite digraph, denoted by Kn,n,n, obtained from Kn,n,n by giving an orientation for each edge in Kn,n,n. In the following, we find an orientation Kn,n,n of Kn,n,n such that Kn,n,n has the directed genus 1 (n - 1)(n - 2), the same as the genus of Kn,n,n. Theorem 3.1. The following two conditions on an orientation Kn,n,n of the complete tripartite graph Kn,n,n are equivalent: (1) Kn,n,n has a directed embedding on the orientable surface of genus 2 (n — 1)(n — 2), for which we call the sets of faces and antifaces A and B, respectively. (2) The sets A and B of white faces and black faces for a face-2-colorable triangular embedding of Kn,n,n correspond to a pair of biembeddable Latin squares A and B of order n. Proof. We first show that (1) implies (2). Assume Kn,n,n has a directed embedding on an orientable surface of genus 1 (n — 1)(n—2) such that the sets of faces and antifaces A and B, respectively. Let ^: Kn,n,n ^ S be this directed embedding of Kn,n,n and p^ be the alternating embedding scheme of Note that Kn,n,n has 3n vertices, 3n2 arcs and the embedding genus 2 (n — 1)(n — 2). By Euler's formula of Lemma 2.5, the number of regions in p^ is 2n2. This implies that each region is bounded by a directed 3-cycle because there are no ¿-cycles for i = 1,2. Let the embedding scheme p of Kn,n,n be the same as p^ without considering the directions of arcs, then Au B is the facial set of the embedding p of Kn,n,n. We color faces in A with white and antifaces in B with black. By the definition of a directed embedding, each arc appears once in exactly one facial boundary and exactly one antifacial boundary. That is, no two distinct faces in A (resp. B) are incident to the same edge. So p of Kn,n,n is a face-2-colorable triangle embedding with two color classes A and B with |A| = |B| = n2. Note that two color classes A and B of a face-2-colorable triangular embedding of Kn,n,n on an orientable surface give two Latin squares, say A and B, corresponding to A and B respectively, which is a biembedding of these two Latin squares A and B. The result (2) is obtained. Secondly, we show that (2) implies (1). Suppose (2) holds. Note that there exists a face-2-colorable triangular embedding, say of Kn,n,n on an orientable surface with two facial color classes A and B which corresponds a pair of biembeddable Latin squares A and B of order n, respectively. Assume the embedding scheme of the embedding ^ is p^ and the rotation at vertex v in K„,„,„ is denoted by p^(v). Let V(K„,„,„) = Vi U V2 U V3, where {Vl, V2, V3} is a partition of V(Kn,n,n). Suppose Vl = {aL, a2,..., a„|, V2 = {bL, b2,..., bn} and V3 = {ci ,C2 ,...,C„|. Note that A and B determine transversal designs (V, G, A) and (V, G, B) respectively, where V = V(Kn,n,n), G = {Vl, V2, V3| and the faces in each color class form the triples in A and B of the transversal designs. For every edge uv e E(Kn,n,n), without loss of generality, let u = aj e V1, v = bj G V2. By the definition of a transversal design, there is only one triple in A containing aj, bj, say {aj, bj, cx} for some cx G V3. Thus, vertices bj and cx are neighbors of a^ Without loss of generality, let cx be the closest successor of bj in the rotation p^(aj) along the counter-clockwise and the color of the region corresponding to the triple {aj, bj, cx} be white. On the other hand, there is exactly one triple in B containing aj, bj, say {aj, bj ,cy } with cy G V3, so bj is the closest successor of cy in the rotation p^(aj) along the counterclockwise and the color of the region corresponding to the triple {aj, bj, cy } is black which is illustrated in the left one of Figure 1. Figure 1: The rotations at vertices aj and w respectively. Give the orientation of the edge uv = ajbj from u = aj to v = bj, i.e., the color of the left region of the arc uv is white and the color of the right region is black. By the random choice of uv, all edges in Kn,n,n are oriented and the obtained digraph is Kn,n,n. In the following, we only need to show that this orientation makes the in-arcs and out-arcs alternating at p^(v) for any v e V(Kn,n,n). By the contrary, suppose there exists a vertex, say w e V, such that in-arcs and out-arcs at w are not alternative. Without loss of generality, suppose two arcs, say ui W, u2 W, are two neighbor in-arcs of w in p^(w) and P0(w) = (..., u1, u2,...) along counter-clockwise. Let the left face and right face of uiw going from u1 to w be F1 and F2 respectively and the left face and right face of u2 w going along the direction from u2 to w be F3 and F4 respectively. Then F2 = F3. By the principle of the orientation, F2 is colored black because of the direction of arc u 1 w and F3 is colored white because of the direction of arc uw, which is shown in the right graph of Figure 1. It contradicts with face-2-colorable because F2 = F3. As a result, this orientation makes in-arcs and out-arcs alternating at every vertex w e V along the rotation p^(w). As a result, Kn,n,n, obtained from Kn n n by this orientation, has an alternating embedding scheme determined by ^ such that the sets of faces and antifaces of this directed embedding of Kn,n,n are A and B, respectively. Since each region of this directed embedding of Kn n n is a 3-cycle, the number of regions is 2n2. By |V| = 3n, the cardinality of arcs in Kn n n being 3n2 and Lemma 2.5, it follows 3n - 3n2 + 2n2 = 2 - 2g, where g is the genus of this directed embedding. So g = 2 (n - 1)(n - 2). Since neither loop nor 2-cycle is in Kn,n,n, the minimum directed genus of Kn,n,n is 2 (n- 1)(n-2). Thus Kn,n,n has a directed embedding in the orientable surface of genus 1 (n - 1)(n - 2), for which we call the sets of faces and antifaces A and B, respectively. □ Theorem 3.2. Let Kn,n,n be the complete tripartite graph. Then there exists an orientation of Kn,n,n such that the obtained digraph Kn,n,n has the directed genus 2(n — 1 )(n — 2), the same as the genus of Kn,n,n. Proof. Let Kn,n,n be the digraph obtained from Kn,n,n by giving the orientation to each edge in Kn,n,n and g be the directed genus of Kn,n,n. (1) If n =1, then Kn,n,n = Ki,i,i is a triangle. Let Ki,i,i be the digraph obtained by giving an orientation of Ki,i,i such that it is a directed 3-cycle. Hence g = 0. (2) If n > 2, by Lemma 2.6, there is a unique regular triangular embedding of a complete tripartite graph Kn,n,n on an orientable surface. By Lemma 2.7, this regular triangular embedding of a complete tripartite graph Kn,n,n must be a face-2-colorable embedding and two set of color faces are denoted by A and B respectively. By Theorem 3.1 , there is an orientation for Kn,n,n such that the resulting digraph Kn,n,n has a directed embedding in the orientable surface of genus i (n — 1)(n — 2), the set of faces is A and the set of antifaces is B. Thus the result holds. □ 4 The number of different orientations of Kn Theorem 3.1 for a directed triangular embedding of the directed complete tripartite graph can be generalized to Lemma 4.1 for directed embedding of a general digraph. Lemma 4.1. The following two conditions on an orientation G of a graph G are equivalent. (1) G has a directed embedding on an orientable surface of genus g. (2) G has a face-2-colorable embedding on an orientable surface of genus g. Proof. We first show that (1) implies (2). Let G = (V, E) be a graph with n vertices, G = (V, A) be a digraph obtained from G by giving an orientation to each edge. So |V| = n and |E| = |A|. By (1), G has a directed embedding on an orientable surface of genus g. Let p be the alternating embedding scheme and Fi and F2 be the set of faces and antifaces in G, respectively. Note that a directed embedding of G is an embedding of G and Fi U F2 is the set of faces of this embedding of G. We color regions in Fi with white and rigions in F2 with black. From the definition of directed embedding, each arc in GK is incident to exactly one face and exactly one antiface in the directed embedding p of GK, so there is no two distinct regions of the same color sharing a common edge in this embedding of G. It implies that this embedding of G is the face-2-colorable embedding on an orientable surface with genus g. So condition (2) holds. Secondly, we show that (2) implies (1). Suppose that (2) holds. Let p be the embedding scheme of a face-2-colorable embedding of a graph G = (V, E) on an orientable surface S of genus g. And all regions of the embedding p can be colored by white and black. Let Fi and F2 be the set of white and black regions, respectively. For each edge e G E(G), there are exactly two regions sharing the edge e, denoted by Fei and Fe2. By the definition of the face-2-colorable embedding, Fei and Fe2 have different colors. Without loss of generality, suppose that F^ g Fi and F2 G F2. We give the orientation of e such that the left is white region Fei and the right is black region Fe2 (this is known as orientationalprinciple). Since each edge can be oriented, one can obtain a digraph, denoted by G, from the graph G by this orientational principle. Let the alternating embedding scheme of G be the same as p. By the orientational principle and face-2-colorability, the in-arcs and out-arcs alternate at each vertex in p of G. Thus this embedding scheme is an alternating embedding scheme of G as a directed embedding in the same surface S with genus g, so condition (1) holds. □ Theorem 4.2. There is a one to one correspondence between the set of directed embeddings of a digraph D on orientable surfaces and the set of face-2-colorable embeddings of the underlying graph of D on orientable surfaces. Proof. Let D be a digraph and the underlying graph of D be obtained from D by ignoring the direction of arcs. Theorem 4.2 is obtained directly from Lemma 4.1. □ The following Theorem 4.3 give an answer to the problem in [2]. Theorem 4.3. If n = 3,7 (mod 12), then there exists an orientation on edges of Kn such that the obtained tournament Kn has directed genus \^ (n — 3)(n — 4)]. Proof. From Ringel and Youngs' results in [25] and [31], if n = 3,7 (mod 12), there exists a face-2-colorable triangular embedding of Kn on an orientable surface. By Lemma 4.1, there exists an orientation on edges of Kn such that the obtained digraph Kn has a directed triangular embedding on an orientable surface. By Euler's formula, digraph Kn has directed genus \ 12 (n — 3)(n — 4)]. □ 5 Concluding remarks In this paper, we show that there is a one to one correspondence between the set of directed embeddings of a digraph D and the set of face-2-colorable embeddings of the underlying graph of D on orientable surfaces. Furthermore, we show that there exist orientations on Kn,n,n and Kn such that the obtained graph Kn,n,n has the directed genus 2 (n — 1)(n — 2) for n > 1 and Kn has directed genus \ 12 (n — 3)(n — 4)] for n = 3,7 (mod 12) which answers the problem about tournaments given in [2] by using a method different from current graphs which were discussed by the same author et al. References [1] J. A. Bondy and U. S. R. Murty, Graph Theory with Applications, Elsevier, New York, 1976. [2] C. P. Bonnington, M. Conder, M. Morton and P. 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Harris (ed.), Graph Theory and its Applications, Academic Press, New York, pp. 17-50,1970, proceedings of an Advanced Seminar (Mathematics Research Center, University of Wisconsin, Madison, Wisconsin, 1969). ARS MATHEMATICA CONTEMPORANEA ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 14 (2018) 387-395 https://doi.org/10.26493/1855-3974.1172.bae (Also available at http://amc-journal.eu) On domination-type invariants of Fibonacci cubes and hypercubes* Jernej Azarija Institute of Mathematics, Physics and Mechanics, Ljubljana, Slovenia Sandi KlavZar t Faculty of Mathematics and Physics, University of Ljubljana, Slovenia Faculty of Natural Sciences and Mathematics, University ofMaribor, Slovenia Institute of Mathematics, Physics and Mechanics, Ljubljana, Slovenia Yoomi Rho Seungbo Sim Department of Mathematics, Incheon National University, Korea Received 6 August 2016, accepted 11 August 2017, published online 28 October 2017 The Fibonacci cube rn is the subgraph of the n-dimensional cube Qn induced by the vertices that contain no two consecutive 1s. Using integer linear programming, exact values are obtained for Yi(Tn), n < 12. Consequently, 7t(rn) < 2Fn_10 + 21Fn-8 holds for n > 11, where Fn are the Fibonacci numbers. It is proved that if n > 9, then Yt(rn) > |-(Fn+2 - 11)/(n — 3)] - 1. Using integer linear programming exact values for the 2-packing number, connected domination number, paired domination number, and signed domination number of small Fibonacci cubes and hypercubes are obtained. A conjecture on the total domination number of hypercubes asserting that yt(Qn) = 2n-2 holds for n > 6 is also disproved in several ways. Keywords: Total domination number, Fibonacci cube, hypercube, integer linear programming, covering codes. Math. Subj. Class.: 05C69, 68R10, 94B05 *We are grateful to an anonymous referee for a very careful reading on the manuscript. t Author to whom correspondence should be addressed. * On 2 June 2017, Yoomi Rho tragically passed away in the middle of her scientific career. E-mail addresses: jernej.azarija@gmail.com (Jernej Azarija), sandi.klavzar@fmf.uni-lj.si (Sandi KlavZar), tmdqh7507@naver.com (Seungbo Sim) Abstract ©® This work is licensed under http://creativecommons.Org/licenses/by/3.0/ 1 Introduction Fibonacci cubes were introduced by Hsu [19] because of their appealing properties applicable to interconnection networks. Afterwards they have been extensively studied and found additional applications, see the survey [23]. The interest for Fibonacci cubes continues, recent research of them includes asymptotic properties [24], connectivity issues [7], the structure of their disjoint induced hypercubes [14, 30], the (non)-existence of perfect codes [5], and the q-cube enumerator polynomial [31]. From the algorithmic point of view, Ramras [29] investigated congestion-free routing of linear permutations on Fibonacci cubes, while Vesel [34] designed a linear time recognition algorithm for this class of graphs. The domination number of Fibonacci cubes was investigated by now in two papers. Pike and Zou [28, Theorem 3.2] proved that y(T„) > \(Fn+2 - 2)/(n - 2)] for n > 9, where Fn are the Fibonacci numbers: F0 = 0, F1 = 1, Fn = Fn-1 + Fn-2 for n > 2. Exact values of Y(rn) for n < 8 were also obtained in [28]. In the second related paper [9] the domination number of Fibonacci cubes was then compared with the domination number of Lucas cubes. In this note we turn our attention to domination invariants of Fibonacci cubes and of hypercubes with a prime interest on the total domination. We proceed as follows. In the rest of this section we introduce concepts and notation needed. Then, in Section 2, we determine the exact value of the total domination number of rn for n < 12, and obtain an upper bound and a lower bound on jt (rn). In Section 3 we use integer linear programming to either extend or obtain values for several domination-type invariants on Fibonacci cubes and hypercubes. In the final section we consider the total domination of hypercubes with respect to a recent conjecture from [22]. In particular, using known results from coding theory we show that the conjecture does not hold. It is also observed that for any c > 0 there exists n0 G N, such that if n > n0, then 7t(Qn) < 2n-c. The n-dimensional (hyper)cube Qn, n > 1, is the graph with V(Qn) = {0,1}n, two vertices being adjacent if they differ in a single coordinate. For convenience we also set Q0 = K1. The vertices of Qn will be briefly written as binary strings b1... bn. A Fibonacci string of length n is a binary string b1... bn with b • bi+1 = 0 for 1 < i < n. Fibonacci strings are thus binary strings that contain no consecutive 1s. The Fibonacci cube rn, n > 1, is the subgraph of Qn induced by the Fibonacci strings of length n. It is well known that |V (rn) | = Fn+2. If u is a binary string, then the number of its bits equal to 1 is the weight of u. If u and v are binary strings, then uv denotes the usual concatenation of the two strings. If u is a binary string and X a set of binary strings, then uX = {ux : x G X}. Let G be a graph. Then D C V(G) is a dominating set if every vertex from V(G) \ D is adjacent to some vertex from D. The domination number 7(G) is the minimum cardinality of a dominating set of G. D is a total dominating set if every vertex from V(G) is adjacent to some vertex from D. The total domination number Yt(G) is the minimum cardinality of a total dominating set of G. Note that the total domination number is not defined for graphs that contain isolated vertices, hence unless stated otherwise, all graphs in this paper are isolate-free. For more information on the total domination in graphs see the recent book [17] and papers [11, 12]. 2 Total domination in Fibonacci cubes In this section we present exact values of 7t(rn) for n < 12, prove an upper bound on 7t(rn), and a lower bound on 7t(r„). The exact values were obtained by computer and are collected in Table 1, where the order of the cubes is also given so that the complexity of the problem is emphasized. In particular, |V(ri2) | = 377. Table 1: Exact total domination numbers of Fibonacci cubes up to dimension 12. n 1 2 3 4 5 6 7 8 9 10 11 12 |V (r„)| 2 3 5 8 13 21 34 55 89 144 233 377 Yt(rn) 2 2 2 3 5 7 10 13 20 30 44 65 More precisely, the results from Table 1 were obtained using integer linear programming as follows. Suppose we associate to each vertex v e V(T„) a binary variable xv. The problem of determining Yt(rn) can then be expressed as a problem of minimizing the objective function Xv, vev (r„) subject to the condition that for every v e V (T„) we have > i. The value of the objective function is then Yt (T„). We have found out that the most efficient solver for the above problem is Gurobi™ Optimizer [15]. For example, it takes less than 9s to compute Yt(r12) on a standard desktop machine. On the other hand, we were not able to make the computation for 7t(T13) in real time (note that the order of r13 is 610), we could only get the estimates 97 < 7t(ris) < 101. Using the above computations, the following result can be derived. Theorem 2.1. If n > 11, then Yt(r„) < 2F„_io + 21F„_g. Proof. Consider the so-called fundamental decomposition of rn into the subgraphs induced by the vertices that start with 0 and 10, respectively (cf. [23]). These subgraphs are isomorphic to rn-1 and rn-2 respectively, hence we infer that Yt(rn) < Yt(rn-1) + Yt(rn-2). From the above computations we know that Yt(r11) = 44 and Yt(r12) = 65. Define the sequence (an), n > 11, with a11 = 44, a12 = 65, and an = an-1 + an-2 for n > 13. Then one can check by a simple induction argument that an = 2Fn-10 + 21Fn-8 holds for any n > 11. Since Yt(rn) < an the argument is complete. □ Arnautov [3] and independently Payan [27] proved that Y(G) £ 1 (2.1) S +1 j holds for any graph G of minimum degree J. Since J(rn) = |(n + 2)/3j, cf. [25, Corollary 3.5], and because Yt < 27, we get that L nf5 J Yt(rn) < 2iif T j. (2.2) L 3 J j=i j Computing the values of the right-hand side of the bound of Theorem 2.1 and of (2.2) we find out that Theorem 2.1 is better than the bound of (2.2) for n < 33. By using the fact 7t(r13) < 101 that was obtained by our computations, the bound of Theorem 2.1 can be further improved to give Yt(r„) < 601Fn_i - 371F„, n > 12. We continue by establishing a lower bound on Yt (rn). Theorem 2.2. If n > 9, then 7t(r„) > Fn+2 - 11 n3 1. Proof. The proof mimics the proof of [28, Theorem 3.2] which gives a lower bound on the domination number of Fibonacci cubes, hence we will not give all the details. For a graph G and its total dominating set D we introduce the over-total-domination of D in G as ODG(D) = deg(v) - |V(G)|. Consider now rn, n > 9, and let D be a total dominating set of rn. In rn, the vertex 0n is the unique vertex of degree n, vertices 10n-1 and 0n-11 have degree n - 1, and all other vertices of weight 1 have degree n - 2. In addition, the vertices 1010n-3,10n-21, and 0n-3101 are of degree n - 2, while all other vertices of rn have degree at most n - 3, cf. [25]. Let k be the number of vertices of weight 1 from D \ {10n-1,0n-11}. In addition, let t = |D n {1010n-3,10n-21,0n-3101}|. Note that k + i is the number of vertices from D that have degree n - 2. The proof now proceeds by considering the cases that happen based on the membership of the vertices 0n, 10n-1, and 0n-11 in D. Here we consider only the case when {0n, 10n-1, 0n-11} C D. We have: ODg(D) < n + 2(n - 1) + (k + i)(n - 2) + (7t(rn) - 3 - k - i)(n - 3) - F„+2 . Since clearly ODG(D) > 0, from the above inequality we derive that Yt (rn)(n - 3) > Fn+2 - k - i - 7. Because k + i < n +1 we get , . Fn+2 - k - i - 7 ^ F„+2 - (n +1) - 7 7i(in) > -o- > n - 3 n - 3 Fn+2 - 11 - (n - 3) Fn+2 - 11 1 n - 3 n - 3 and the stated inequality holds in this case. All the other cases are treated similarly. □ We conclude the section with Table 2 in which known values and current best bounds on 7i(r„) for n < 33 are collected. The values for n < 12 were computed using the linear program explained above. The bounds for Yt(r13) were established by Gurobi, and we conjecture that in fact rt(r13) = 101. Finally, the remaining bound in Table 2 were obtained by the bounds given in Theorems 2.1 and 2.2. Recall that n = 33 is the last value for which Theorem 2.1 gives a better bound than the bound (2.2). Table 2: Exact values and current best bounds on Yi(Tn), n < 33. n Yt(rn) n Yt(rn) n Yt(rn) 1 2 12 65 23 3749-13276 2 2 13 97-101 24 5779-21481 3 2 14 87-174 25 8926-34757 4 3 15 131-283 26 13816-56238 5 5 16 196-457 27 21424-90995 6 7 17 296-740 28 33280-147233 7 10 18 449-1197 29 51778-238228 8 13 19 682-1937 30 80676-385461 9 20 20 1040-3134 31 125876-623689 10 30 21 1590-5071 32 196649-1009150 11 44 22 2438-8205 33 307580-1632839 3 Additional invariants on small Fibonacci cubes and hypercubes The integer linear programming approach can be used to compute several additional invariants of Fibonacci cubes (and other graphs). This has recently been done by Ilic and Milosevic in [20], where they have computed the domination number, the 2-packing number, and the independent domination number of low dimensional Fibonacci cubes. In particular, they have used integer linear programming to confirm the conjecture from [9] stating that Y(r9) = 17. In addition, an integer linear programming model for the connected domination number has been presented in [13]. In this section we add to the list of integer linear programming models paired domination and signed domination. The concepts mentioned in this paragraph that have not been introduced yet are defined next. A set X C V(G) is a 2-packing if d(x,y) > 3 holds for any x, y G X, x = y. The maximum size of a 2-packing of G is the 2-packing number of G denoted p(G). The independence domination number i(G) of G is the minimum size of a dominating set that induces no edges [26]. The connected domination number yc(G) of G is the order of a smallest dominating set that induces a connected graph [10]. The paired domination number yP(G) is the order of a smallest dominating set S C V(G) such that the graph induced by S contains a perfect matching [2]. Finally, we say that f: V(G) ^ {-1,1} is a signed dominating function if [v] f (u) > 1 holds for every v G V(G), where N[v] is the closed neighborhood of v, that is, N[v] = {v} U {u : vu G E(G)}. The signed domination number ys(G) is the minimum of J2vev(G) f (v) taken over all signed dominating functions f of G, see [18]. We now present the problems to determine the paired domination number of a graph and the signed domination number of a graph as integer linear programs. To model the paired domination problem for a graph G we introduce a binary variable xe indicating whether the edge e G E(G) is present in the graph induced by a paired dominating set of G. Then we can model the problem as follows: Y. xe eEE(G) £ xuv < 1, v e V(G) u^v ££ Xuw > 1, v e V(G) . u^v w^u Similarly, to model the signed domination number we introduce a binary variable xv associated with every vertex v e V(G) indicating whether v is assigned weight 1 or —1, respectively. Then we have the following linear program. minimize (2xv — 1) vev (G) subject to £ (2xu — 1) > 1, v e V(G). ueN [v] Our computational results are collected in Tables 3 and 4. In the rows for Y(rn), p(rn), and i(rn), the results from [20] are in normal font, while the new values are in bold. We have thus extended the results from [20] for one additional dimension. It is interesting to observe that the gap between the independent domination number and the domination in dimension 9 is equal to 2, but then in dimensions 10 and 11 the difference goes down to 1. Table 3: Additional invariants for small Fibonacci cubes and hypercubes. n 1 2 3 4 5 6 7 8 9 10 11 12 Y(r„) 1 1 2 3 4 5 8 12 17 25 39 54-61 p(r„) 1 1 2 2 3 5 6 9 14 20 29 42 i(r„) 1 1 2 3 4 5 8 12 19 26 40 ?-? minimize subject to Table 4: Additional invariants for small Fibonacci cubes and hypercubes. n 1 2 3 4 5 6 7 8 9 10 7c(r„) 1 1 2 3 5 7 10 14 22 Yc(Qn) 1 2 4 6 10 16 28 Yp(r„) 2 2 2 4 6 8 10 14 20 30 Yp ( Qn ) 2 2 4 4 8 14 24 32 Ys(r„) 2 3 3 2 5 9 10 17 25 40 Ys(Qn) 2 2 4 6 12 16 32 4 On total domination in hypercubes It has recently been conjectured in [22, Conjecture 4.6] that Yt(Qn) = 2n-2 holds for n > 6. In [4] Arumugam and Kala first observed that 7t(Qi) = 7t(Q2) = 2 and Yt (Q3 ) = 7i (Q4 ) =4, and then followed by proving that jt (Q5 ) = 8 [4, Theorem 5.1] and 7t(Q6) = 14 [4, Theorem 5.2]. The last result is then a sporadic counterexample to the conjecture. Actually, at this moment the exact value of 7t(Qn) is known for n < 10: 7t(Q7) = 24, 7t(Qs) = 32, 7t(Qg) = 64, and 7t(Q10) = 124, see [33, Appendix B, p. 40]. Hence Q7 and Q10 are additional sporadic counterexamples (and so are Q8 and Qg since Yt(Qs) = 32 = 26 and 7t(Qg) = 64 = 27). Total dominating sets of Qn can be in coding theory equivalently described as covering codes of empty spheres (of length n and covering radius 1). The following result was first proved back in [21], see also [35, Theorem 1(b)]. Let us rephrase the result here in graph-theoretical terms and give a corresponding argument. Proposition 4.1. If n = 2k, k > 0, then 7t(Qn) = 2n-k. Proof. From [32] we know that if n = 2k, then 7(Qn) = 2n-k and from [16] that if n = 2k - 1, then also 7(Qn) = 2n-k. Let n = 2k and consider Qn. Let QL-1 and QR-1 be the subgraphs of Qn induced by the sets of vertices X0 = {0b2... bn : b € {0,1}} and X1 = {1b2... bn : bj € {0,1}}, respectively. Clearly, V(Qn) partitions into X0 and X1, and in Qn every vertex of X0 has a unique neighbor in X1. Moreover, Q^_1 and QR-1 are both isomorphic to Qn-1. Let CL be a perfect code of Q^_1 and let CR be its copy in QR-1. Then CL U CR is a total dominating set of Qn of order 2n-k. Since on the other hand 7t(Qn) > 7(Qn) = 2n-k, the conclusion follows. □ It follows from (2.1) that -,(G) < |V(G)| (i±|+±i!) (4.1) holds for any graph G. Hence, again using the fact that 7t(G) < 27(G), we get for hypercubes that „«,.) < 2n+. (i±M) . Directly from this inequality we infer: Remark 4.2. For any c > 0 there exists n0 € N, such that if n > n0, then Yt(Qn) < 2n-c . Two remarks are in place here. First, (4.1) also follows from a more general result on transversals in hypergraphs due to Alon [1]. Second, the state of the art on the upper bounds on the domination number in terms of the minimum degree and the order of a given graph is given in [8]. It follows from the fact that 7t(Qn) < 27(Qn-1) and from Proposition 4.1 that Yt(Q2fc+1) < 27(Q2k) = 22 -k+1. As proved in [32], the equality actually holds here, that is, 7t(Q2k+1) = 22 -k+1. More generally, 7t(Qn+1) = 27(Qn) holds for any n, a result very recently proved in [6]. Acknowledgment The authors acknowledge the financial support from the Slovenian Research Agency (research code funding No. P1-0297) and from the Basic Science Research Program through the National Research Foundation of Korea funded by the Ministry of Education, Science and Technology grant 2011-0025319. Supported also by the bilateral Korean-Slovenian project BI-KR/13-14-005 and the International Research & Development Program of the National Research Foundation of Korea (NRF) funded by the Ministry of Science, ICT and Future Planning (MSIP) of Korea (Grant number: NRF-2013K1A3A1A15003503). References [1] N. 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ARS MATHEMATICA CONTEMPORANEA 14 (2018) 397-413 https://doi.org/10.26493/1855-3974.938.1ae (Also available at http://amc-journal.eu) On t-fold covers of coherent configurations Alyssa D. Sankey Department of Mathematics & Statistics, University of New Brunswick, P.O. Box4400, Fredericton, N.B., E3B 5A3, Canada Received 15 September 2015, accepted 27 June 2017, published online 29 Qctober2017 We introduce the covering configuration induced by a regular weight defined on a coherent configuration. This construction generalizes the well-known equivalence of regular two-graphs and antipodal double covers of complete graphs. It also recovers, as special cases, the rank 6 association schemes connected with regular 3-graphs, and certain extended Q-bipartite doubles of cometric association schemes. We articulate sufficient conditions on the parameters of a coherent configuration for it to arise as a covering configuration. Keywords: Association scheme, coherent configuration, regular weight, double cover, two-graph, t-graph. Math. Subj. Class.: 05C22, 05C50, 05E30 1 Introduction The Seidel matrix of a graph r may be viewed as a weight on the complete graph: edges of r are weighted (-1) and non-edges (+1). If r is strongly regular with n = 2(2k - A - p), it lies in the switching class of a regular two-graph and we call the weight, analogously, regular on Kn. This condition on r is well known, and dates to 1977, in [25]. The same year, the equivalence of regular two-graphs and antipodal double covers of complete graphs was established in [26]. Martin, Muzychuk and Williford ([18]) defined the extended Q-bipartite double of a cometric association scheme, extending the notion of the bipartite double of a distance regular graph. This construction produces, as special cases, the antipodal double covers of complete graphs from the strongly regular graphs affording regular two-graphs. In recent work, Kalmanovich ([16]) has also generalized the regular two-graph result, working from an unpublished draft of D. G. Higman's ([9]) on regular 3-graphs. As defined in [14], a t-graph weights the edges of Kn with elements of the group of roots of unity of E-mail address: asankey@unb.ca (Alyssa D. Sankey) Abstract ©® This work is licensed under http://creativecommons.org/licenses/by/3.0/ order t, Ut. The regularity condition ensures that the matrix of edge weights has a quadratic minimal polynomial. The work of Kalmanovich-Higman establishes the equivalence of regular 3-graphs with cyclic antipodal 3-fold covers of Kn ([6]). Regular 3-graphs are shown to give rise to certain rank 6 association schemes, and the necessary conditions under which a rank 6 scheme arises in this way are given. In this paper there are two main results. First, working with a regular weight with values in Ut, defined on a coherent configuration (CC), we show that there is always a covering configuration; that is, a CC constructed using a t-fold cover in a natural way, to convert the weight into a CC of higher rank (by a factor of t). As special cases, we recover the equivalence between regular two-graphs and antipodal double covers of complete graphs; some extended Q-bipartite doubles of cometric schemes; the rank 6 schemes associated with regular 3-graphs, and an extension of these to regular t-graphs. A CC with a regular weight has two sets of parameters: the structure constants for the weighted adjacency algebra, }, which lie in C or more specifically in the ring of integers with a primitive tth root of unity adjoined, and the non-negative integers (v)} which count certain triangles with a specified weight. They are related by 4 = E < (v). veUt The weighted adjacency algebra is in general not a coherent algebra, and may in fact have a coherent closure that is much higher in rank than the original CC. In the regular two-graph case, for instance, it is precisely when the (-1) edges form an SRG that we get a minimal closure: a natural fission of the edge set into (+1) and (-1) edges that yields a (rank 3) association scheme. The covering configuration is the realization of a CC whose structure constants are the (v). Some properties, namely homogeneity and commutativity of a CC carry over to the covering configuration. Symmetry is preserved only if t = 2. Metric and cometric properties are not. The second main result of this paper is the articulation of sufficient conditions for a CC to be the covering configuration of a regular weight. In the final section, we describe a family of regular weights on the Hamming Scheme H(n, 2) with values in U4, due to Ada Chan. These weights all fuse to regular 4-graphs, providing an infinite family that may be of interest as complex Hadamard matrices. These regular weights, and their fusions, admit covering configurations of ranks 4(n + 1) and 8 respectively, on 2n+2 points. 2 Preliminaries In this section, we give the definitions that are essential to what follows. Much more can be found in [17] and in the original developments of the area by Weisfeiler and Lehman in [28] and by D. G. Higman in [11, 12], and [14]. 2.1 Coherent configurations Definition 2.1. Let jAi}0 ZC3. □ Observe that E := uj=0Rj is a parabolic in the sense of [10]. Indeed, M0 = Irt implies that R0 is the identity relation of C. Further, E is symmetric, since (x, y) G R for i < t implies that p0, j = 0, so i* is in the same block of Mj as 0. That is, (y, x) G E. Given (x, y) G Rj and (y, z) G Rj with 0 < i, j < t, we see that (x, z) G Rk for some k < t, because k must lie in the same block of Mj as i, since all non-diagonal blocks are zero. Hence, E is a transitive relation. As a parabolic, E induces an equivalence relation on the indices: If there exist x, x', y, y' G X such that (x, x') G E, (y, y') G E, (x,y) G Rj and (x',y') G Rj, then i ~ j. Write [i] for the equivalence class of i. In addition, the parabolic affords a quotient (homogeneous) configuration A := (X, {R[j]}) with an associated partition of the vertex set X into fibres of size t. The fibre containing x is [x] = {y | (x, y) G E}. We will henceforth suppress the bracket notation for fibres, writing x = {xi, x2,... xt}. For j G [0], Lemma 4.4 implies that pkj = 0 for j = 0. But then Rk restricted to x x y has valency at most 1. We conclude that the number of relations occurring between any two fibres is t. We have: For k G I and x G X, |[k]| = |x| = t. Denoting the graph of Rj by rj, we have proved the following: Lemma 4.5. For all j G [0], rj is a t-fold cover of r^. Corollary 4.6. The natural partition of I according to blocks of Mj, for 0 < j < t is the same as that determined by the equivalence classes of the parabolic. That is, [mt] = {mt, mt + 1,. .., mt + t — 1}. Proof. Suppose j G [i] so that there exist xi,x2,yi,y2 G X with (xi,yi) G Rj and (x2,y2) G Rj. Then, by the discussion above, (xi,y3) G Rj for some y3 G y and therefore pjk = 0 for some k < t. But then j = i + k (mod t) by Lemma 4.4. □ Recall that C0jzfc = Rk for k < t, C0jCT has intersection matrix IrZCT, and Cmji = Rmt for 0 < m < r. Fix a fibre a (from here on), and order it so that (aj, aj+i) G , for each i, with addition modulo t. This ensures that the perfect matching induced on a corresponds to the permutation (1,2,..., t) on indices, which in turn corresponds to the permutation of U induced by multiplication by Z. For each x G X, (a, x) G R[mt] for some m. Order x so that (aj,xj) G Cm i. In what follows, we mix the notations regarding indexation of the relations of C. Where two indices are given, we refer to as above; where one index is given we refer to the original numbering of the relations. Lemma 4.7. With notation as above, (x^ xi+i) G for all x G X. Proof. For some a, (x^ xi+i) G C0jCT; (a^ xi+i) G R; for some l, and (a^ x4) G Rmi for some m. Note that l G [m]. Since a4, ai+1, and xi+1 form a triangle of type (0Z, ml, l), we see that p0z m1 = 0. Since C is commutative, R; = by Lemma 4.4. Now observe that aj, xj, and xi+1 form a triangle of type (ml, 0a, mZ), and therefore a = Z. □ Next, following [16] we show that all matchings are cyclic. Lemma 4.8. With notation as above, all matchings between fibres of C are cyclic. Proof. Suppose that (xj, yj) G Rk and (xi+1, y;) G Rk. We must show that l = j + 1. The triangle (xj, xj+1, yj) has type (1, m, k) for some m, indicating that pkm = 0. As in the previous lemma, this implies that k = m +1. On the other hand, the triangle (xj+1, y;-1, y;) has type (b, 1, k) for some 6, hence k = b +1. But then m = b, and by Lemma 4.5, y;-1 = yj as desired. □ Corollary 4.9. For all x G X, (xj, xj+k) G Rk, thus Rk induces on each fibre the perfect matching corresponding to the kth power of the cycle (1, 2,..., t). Proof. The result follows by Lemma 4.7 and induction (on k) applied to the triangles (xj-fc,xj,xj+1) . □ Lemma 4.10. For x G X, (aj,xj+k) G Rmt+fc for 0 < k < t. Proof. The case k = 0 holds by choice of ordering of x. Induction applied to the triangles (aj, xj+k-^ xj+k ) gives the desired result. □ We now define a weight on A by means of Cj1. Let x, y G X and suppose (x, y) G Rj. Then Cj,1 provides a cyclic matching between x and y corresponding to, say, a G U. Set w(x, y) := a. Observe that w(a, x) = 1 for all x. The next lemma shows how to determine the weight of an edge in F^] from any edge in rj. aj l xj+1 aj l xj+1 Figure 4: Triangles (aj, aj+1, xj+1) and (aj,xj,xj+1). Lemma 4.11. If (xi; yj) G Cka, then w(x, y) = aÇj i. Proof. Consider (xi;yj) G CfcjCT. Let l be such that (xi;y;) G Ck1 and note that the triangle (xi; y;, yj) has type (kl, 0Zj—ka). By Proposition 4.6, a = Zj-;. This implies that (xj, y;+m) G Cfcjzm. We conclude that the matching between x and y in CfcjCT is aa, where a = w(x, y). □ We now prove the second main result which is the extension of [16, Prop. 5.4]. Theorem 4.12. Let C = (X, {Ri}) be a commutative CC of rank rt with the first t intersection matrices given by Mj = Ir Zi + M-(g) Z2 where M +, MZ1 and Z2 are defined as in Section 2. It is natural to ask whether C is the centralizer algebra of this permutation representation. In fact, C is properly contained in this centralizer algebra. It affords a CC with valencies 1, 1, 3, 3, 3, 3, 3, 3 which has a fusion to C. 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Tanaka, Distance-regular graphs, 2016, arXiv:1410.6294 [math.CO]. [28] B. Yu. Weisfeiler and A. A. Lehman, Reduction of a graph to a canonical form and an algebra which appears in the process, Nauchno-Tekhnicheskaya Informatsiya Ser. 2 9 (1968), 12-16. ARS MATHEMATICA CONTEMPORANEA ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 14 (2018) 415-432 https://doi.org/10.26493/1855-3974.1098.ca0 (Also available at http://amc-journal.eu) On the number of additive permutations and Skolem-type sequences* * Diane M. Donovan Centre for Discrete Mathematics and Computing, University of Queensland, St. Lucia 4072, Australia Michael J. Grannell * School of Mathematics and Statistics, The Open University, Walton Hall, Milton Keynes MK7 6AA, United Kingdom Received 3 May 2016, accepted 31 May 2017, published online 9 November 2017 Cavenagh and Wanless recently proved that, for sufficiently large odd n, the number of transversals in the Latin square formed from the addition table for integers modulo n is greater than (3.246)n. We adapt their proof to show that for sufficiently large t the number of additive permutations on [—t, t] is greater than (3.246)2i+1 and we go on to derive some much improved lower bounds on the numbers of Skolem-type sequences. For example, it is shown that for sufficiently large t = 0 or 3 (mod 4), the number of split Skolem sequences of order n = 7t+3 is greater than (3.246)6t+3. This compares with the previous best bound of 2 Ln/3J. Keywords: Additive permutation, Skolem sequence, transversal. Math. Subj. Class.: 05B07, 05B10 1 Introduction This paper is concerned with counting additive permutations and Skolem-type sequences. Additive permutations are related to certain kinds of transversals in Latin squares. A Latin square of order n may be envisaged as an n x n array having n distinct entries, each of which appears once in any one row and once in any one column. We adopt a slightly wider * We thank the referees for helpful comments and for drawing to our attention the papers [6, 7, 8]. t Corresponding author. E-mail addresses: dmd@maths.uq.edu.au (Diane M. Donovan), m.j.grannell@open.ac.uk (Michael J. Abstract Grannell) ©® This work is licensed under http://creativecommons.org/licenses/by/3.0/ than usual definition of a transversal in an n x n array (not necessarily a Latin square) as a set of n (row, column, entry) triples that cover every row, every column and n distinct entries from the array. Consequently, a transversal in a Latin square will necessarily cover every entry precisely once. In a recent paper, Cavenagh and Wanless [3] proved that, for all sufficiently large odd n, the number of transversals in the Latin square formed from the addition table for integers modulo n is greater than (3.246)n. This result was a substantial improvement on previous results in this area, although Vardi [10] conjectured that this number exceeds cnn! for some constant c G (0,1). A proof of this conjecture is claimed in the arXiv paper [6]. However, it is important to note that not all transversals are suitable for our purposes. For integers a and b with a < b we use the notation [a, b] to denote the set of integers i such that a < i < b. An additive permutation n on [—t, t] is a permutation of these integers such that {i + n(i) : i G [—t, t]} is also a permutation on the same set of integers. This definition of an additive permutation is the one employed by Abram [1] and others in connection with Skolem sequences, but the reader is cautioned that it differs from that used in [6] where pointwise addition of permutations on [1, n] is taken modulo n. An examination of the proof given in [3] shows that it is possible to adapt the proof to show that the number of additive permutations on [—t, t] is greater than (3.246)2i+1 for all sufficiently large t, and this is done in Theorem 2.1 below. There are strong connections between additive permutations and Skolem-type sequences. We investigate some of these connections and obtain much improved lower bounds on the numbers of some Skolem-type sequences. A pure Skolem sequence, sometimes simply called a Skolem sequence, of order n is a sequence (s1, s2,..., s2n) of 2n integers satisfying the following conditions. (C1) For each k G {1,2,..., n} there are precisely two elements of the sequence, say s¿ and sj, such that s¿ = Sj = k. (C2) If Sj = Sj = k and i < j then j — i = k. For example, (4,1,1, 5,4,2,3, 2, 5, 3) is a pure Skolem sequence of order 5. It is well known that a pure Skolem sequence of order n exists if and only if n = 0 or 1 (mod 4). For this and other existence results mentioned below see, for example, [4, 5]. An extended Skolem sequence of order n is a sequence (s1, s2,..., s2n+1) of 2n +1 integers satisfying (C1) and (C2) above and such that precisely one element of the sequence is zero. An extended Skolem sequence of order n exists for every positive integer n. If the zero element of an extended Skolem sequence of order n appears in the 2n-th position, i.e. s2n = 0, then the sequence is called a hooked Skolem sequence. A hooked Skolem sequence of order n exists if and only if n = 2 or 3 (mod 4). If the zero element of an extended Skolem sequence of order n appears in the (n + 1)-th position, i.e. sn+1 = 0, then the sequence is called a split Skolem sequence or a Rosa sequence. A split Skolem sequence of order n exists if and only if n = 0 or 3 (mod 4). A split-hooked Skolem sequence (also known as a hooked Rosa sequence) of order n is a sequence (s1, s2,..., s2n+2) of 2n + 2 integers satisfying (C1) and (C2) above and such that sn+1 = s2n+1 = 0. A split-hooked Skolem sequence of order n exists if and only if n = 1 or 2 (mod 4) and n =1. The various types of Skolem sequence described above may be used to construct solutions to Heffter's first and second difference problems. These, in turn, may be used to construct cyclic Steiner triple systems. We will refer to the sequences just described, and to some near relatives, somewhat loosely as Skolem-type sequences. It was shown in [1, 2] that for many of these Skolem-type sequences (in particular, pure, hooked, split and split-hooked) the number of them is essentially bounded below by 2 Lf J, where n is the order of the sequence. 2 Additive permutations Define the Latin square An of odd order n = 2t +1 to have its rows and columns indexed by the integers in [—t, t] and the (i, j) entry k G [—t, t] given by k = i + j (mod n). The array An gives the addition table on Zn. A Z-transversal T in An is a transversal in which every (row, column, entry) triple (i, j, k) G T has k = i + j in Z, so that no triples of a Z-transversal have i + j < -t or i + j > t. Not all transversals in An are Z-transversals, for example the transversal formed by the leading diagonal in An is not a Z-transversal. We will only count Z-transversals: in effect the (i, j) cells in An where i + j < -t or i + j > t are ignored. The entries in these cells are therefore irrelevant to our discussions and it will sometimes be helpful to take i + j as the entry, rather than i + j reduced modulo n. This has the advantage that the "ignored" cells are easily identified, particularly when considering subarrays of An - such cells are then precisely those with entries outside the range [—t, t]. Figure 1 shows the array A19 with the "ignored" entries greyed-out, and with a Z-transversal having its entries marked in boxes. For the present, disregard the highlighting of the subarrays. If the (row, column, entry) triples of a Z-transversal in An are listed as a 3 x n array T* with row numbers of An forming the first row of T*, column numbers the second, and entries the third, then each row of this array contains the integers [—t,t], and the entries in the third row are the sums of the corresponding entries in the other two rows. Taking the first row of T* as [—t, t], the second row as a permutation n on [—t, t], and the third row as the vector sum of the first two rows, then n is an additive permutation on [—t, t]. Conversely, if n is an additive permutation on [—t, t] then the (row, column, entry) triples (i, n(i), i + n(i)) for i G [—t, t] form a Z-transversal in An. If the entries in the third row of such a 3 x n array T* are multiplied by ( — 1), then each column of the resulting array sums to zero, and the new array is called a zero-sum array. Thus, there is an equivalence between Z-transversals, additive permutations, and zero-sum arrays. Table 1 (taken from [3]) gives the number of these, here denoted by zn, for n = 2t +1 < 23. These numbers form the sequence A002047 in Sloane's encyclopaedia [9] and have been independently checked by ourselves using our own computer program. The table also gives a rounded down value for (zn)f which will be used subsequently. We remark that z2i+1 is also the number of extremal Langford sequences with defect t + 1 (i.e. starting with t + 1) - see [1, 4] for definitions. Theorem 2.1. Suppose that b and n are odd and that n > 3b > 9. Then zn > (zb)2 L f—J. Proof. Our proof is a re-working of that of Cavenagh and Wanless [3], ensuring that for general b and n the subarrays R and S (defined below) have appropriate (sub-)transversals and that the transversals constructed in An are indeed Z-transversals. We take b = 2a +1 (so that a > 1) and n = 2t + 1. Put k = |_n— J = |_^J, s = t — a — bk, and r = s + b. Then 0 < s < b, b < r < 2b, and one of r, s is odd and the other is even. Next define the subarray M(¿,j),c of An to be the c x c block whose top left entry is in -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 -9 1 2 3 4 5 6 7 8 0 -8 -7 -6 -5 -4 -3 -2 -1 0 -8 2 3 4 5 6 7 8 9 -9 -8 0 -6 -5 -4 -3 -2 -1 0 1 -7 3 4 5 6 7 8 9 -9 H -7 -6 -5 -4 -3 -2 -1 0 1 2 -6 4 5 6 7 8 9 -9 -8 -7 -6 -5 -4 -3 0 -1 0 1 2 3 -5 5 6 7 8 9 -9 -8 -7 -6 -5 -4 0 -2 -1 0 1 2 3 4 -4 6 7 8 9 -9 -8 -7 -6 -5 -4 -3 -2 0 0 1 2 3 4 5 -3 7 8 9 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 0 4 5 6 -2 8 9 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 0 7 -1 9 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 0 5 6 7 8 0 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 0 1 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 0 9 2 -7 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 -9 -8 3 -6 -5 H -3 -2 -1 0 1 2 3 4 5 6 7 8 9 -9 -8 -7 4 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 -9 -8 -7 -6 5 -4 -3 -2 -1 0 □ 2 3 4 5 6 7 8 9 -9 -8 -7 -6 -5 6 -3 -2 -1 1 2 3 4 5 6 7 8 9 -9 -8 -7 -6 -5 -4 7 -2 -1 0 1 H 3 4 5 6 7 8 9 -9 -8 -7 -6 -5 -4 -3 8 -1 0 1 2 3 4 H 6 7 8 9 -9 -8 -7 -6 -5 -4 -3 -2 9 0 1 2 3 4 5 6 0 8 9 -9 -8 -7 -6 -5 -4 -3 -2 -1 Figure 1: The Latin square A19. Table 1: The number of Z-transversals in An 3 5 7 9 11 13 15 17 19 21 23 2 6 28 244 2544 35600 659632 15 106128 425 802 176 14409 526080 577 386122880 (Zn) n > 1.259 1.430 1.609 1.841 2.039 2.239 2.443 2.644 2.845 3.046 3.246 n z n position (i, j) of the array An. For example, if n =19 (see Figure 1) then /-4 -3 -2> M(_6,2),3 = ( -3 -2 -1 V-2 -1 As previously mentioned, it is convenient to take the entries of subarrays unreduced modulo n. Allowing this, M(ij)c has entries from i + j to i + j + 2(c - 1) and, if c is odd, there is a central entry i + j + c - 1. We use the following subarrays: Type 1: A = M{_t+ib_a+ib)ib for i = 0,1,..., (k - 1), Type 2: E = M(-t+r+(k+i)b,-t+ib),b for i = 0,1,..., (k - 1), Type 3: R = M(-t+kb, -a+kb),r and Type 4: S = M(t_s+i,_a-s),s. In Figure 1, n = 19 and b = 3, so that t = 9, a = 1, k = 2, s = 2 and r = 5. The subarrays of types 1 and 2 are lightly shaded and the subarrays R and S are shaded more heavily. Altogether there are 2k + 2 subarrays of the four types. No two of these have a common row and the total number of rows covered is 2bk + r + s = n, so each row of An is covered by precisely one of these subarrays. Similarly, each column of An is covered by precisely one of these subarrays. Consequently we may attempt to construct Z-transversals in An from transversals in the subarrays. The type 1 subarray Di has a central entry 2ib -1 + a. If this value is subtracted from every entry in Di, the resulting array is a copy of Ab. Since Ab has zb Z-transversals, each Di has zb transversals that are symmetric about its central entry, that is to say transversals each covering the entries from 2ib - t to 2ib - t + (b - 1) inclusive. Note that these transversals avoid "ignored" cells of An. Similarly, the type 2 subarray Ei has zb transversals symmetric about its central entry 2ib -1 + 3a + 1, and each of these covers the entries from 2ib -1 + b to 2ib -1 + 2b - 1 inclusive. Collectively the type 1 and type 2 subarrays have (zb)2k transversals covering the entries from -t to t - r - s inclusive. All of these are partial Z-transversals of An. To complete the proof for all b > 3 and n > 3b we must show that the remaining subarrays R and S have appropriate transversals (i.e. avoiding "ignored" cells of An) that cover the entry values from t - r - s + 1 to t inclusive. It will then follow that An has at least (zb)2k = (zb)2LZ-transversals. The subarray R has entries ranging from t - r - s - a + 1 to t + a inclusive, and so R always contains some "ignored" cells of An, namely those with entries exceeding t. The subarray S has entries ranging from t - r - s + 2 + a to t - a - 1 inclusive, so S does not contain any "ignored" cells of An. If we subtract t - s - a from all the entries in R and S we obtain equivalent arrays R' and S', where R' has entries ranging from -(s + 2a) to s + 2a, while S' has entries ranging from - (s -1) to s -1, and we are seeking transversals in these arrays that cover the entry values from -(s + a) to s + a inclusive. Cells in R' that contain entries greater than s + a correspond to the "ignored" cells of R. Our proof that such transversals always exist falls into a number of cases, the details of which are lengthy, and so are postponed until Section 4. □ Corollary 2.2. If n is odd and sufficiently large, then zn > (3.246)n. Proof. Theorem 2.1 gives zn > (zb)2LtitJ > (zb)n_3 for b odd and all sufficiently large n. From Table 1 we have (z23) 23 > 3.246, so taking b = 23 we obtain zn > (3.246)n for all sufficiently large n. □ Putting the corollary into words, the number of additive permutations on [—t, t] is greater than (3.246)2i+1 for all sufficiently large t. 3 Skolem-type sequences A connection between additive permutations and Skolem-type sequences is formed by so-called (m, 3, c)-systems. A set D = {D1, D2,..., Dm}, where each Dj is a triple of positive integers (a^ bj, aj + bj) with aj < b and |Jm=1 Dj = {c, c + 1,..., c + 3m - 1} is called an (m, 3, c)-system. As remarked in [1], such a system exists if and only if (i) m > 2c - 1 , and (ii) m = 0 or 1 (mod 4) if c is odd, or m = 0 or 3 (mod 4) if c is even. Given an (m, 3, c)-system D = {D1, D2,..., Dm}, where Dj = (aj, bj, aj + bj), and putting r = c + 3m - 1, a (Skolem-type) sequence (x_r, x_r+1,..., xr-1, xr) may be constructed by putting x_(ai+bi) = aj, x_bi = aj, x_a = aj + bj, xa = bj, xbi = aj + bj, xai+6i = bj for i = 1, 2,..., m, and xj = 0 for — c < j < c. For example, if c =2 and m = 3, and if D = {D1,D2,D3} where D1 = (2,6, 8), D2 = (3, 7,10), D3 = (4, 5,9) then the constructed sequence is (3,4, 2, 3, 2,4, 9,10, 8, 0, 0,0, 6, 7, 5, 9, 8,10, 6, 5, 7). Observe that in such a sequence, for each k e {c, c + 1,..., r} the two positions occupied by k are precisely k apart. Further observe that, independently for each i e{1,2,..., m}, wemayreplace xj for j e {—aj — bj, —bj, —aj ,aj,bj,aj + bj} by xj where x_(o,) = bj, x'_^ = aj + bj, x_a. = bj, x^. = aj + bj, x^. = aj, x^= aj. Thus we may obtain 2m distinct sequences of length 2r + 1 each of which has the property that for each k e {c, c + 1,..., r}, the two positions occupied by k are precisely k apart. Each such sequence has zeros in the central 2c — 1 positions. If n is an additive permutation on [—t, t] then a (2t + 1,3, t + 1)-system is formed by the set of triples {Dj : i e [—t, t]} where Dj = (i + 2t + 1, n(i) + 4t + 2, i + n(i) + 6t + 3). Note that the first entries in these triples cover the interval [t + 1,3t +1], the second entries cover [3t + 2,5t + 2], and the third entries cover [5t + 3, 7t + 3]. If n1 and n2 are two different additive permutations on [—t, t], then the two resulting (2t + 1, 3, t + 1)-systems contain different triples. Consequently the number of different (2t + 1, 3, t + 1)-systems is bounded below by z2t+1, and hence by (3.246)2i+1 for all sufficiently large t. Combining this with the previous observation that each such sequence gives rise to 22i+1 Skolem-type sequences, we obtain the following result. Theorem 3.1. For all sufficiently large t, there are more than (6.492)2i+1 Skolem-type sequences of length 14t + 7 having the following properties: (a) there are zeros in the central 2t +1 positions, and (b) for k G {t + 1, t + 2,..., 7t + 3}, the two positions occupied by k are precisely k apart. If the central 2t + 1 zero entries in such a Skolem-type sequence are replaced by a split Skolem sequence of order t (which exists for t = 0 or 3 (mod 4)) then a split Skolem sequence of order 7t + 3 is obtained. Hence we have the corollary: Corollary 3.2. For sufficiently large t = 0 or 3 (mod 4), there are more than (6.492)2i+1 split Skolem sequences of order 7t + 3. In fact we can achieve slightly better than this because the number of split Skolem sequences of order t is at least 2L 3 J for all t = 0 or 3 (mod 4) [1, 2], so we have at least that number of choices for replacing the central zeros. The bound (6.492)2i+1 > 25-3973t is (for large t) substantially better than the previous best bound of 2L 73+3j . Given a split Skolem sequence of order n, we can form a pure Skolem sequence of order n +1 by replacing the central zero with n + 1 and placing a further entry n +1 at either the start or the end of the sequence. Hence we obtain: Corollary 3.3. For sufficiently large t = 0 or 3 (mod 4), there are more than (6.492)2i+1 pure Skolem sequences of order 7t + 4. In the next two corollaries, the lower bound of (6.492)2i+1 is extended to hooked and split-hooked Skolem sequences of orders 7t + 4 and 7t + 5 respectively. In each case the basic approach is as follows. For given t, choose a small positive integer c such that t — c = 2 or 11 (mod 12) if c is odd, or t — c = 8 or 11 (mod 12) if c is even. Put m = (t — c +1)/3 and assume that t is large enough to ensure that m > 2c — 1 (i.e. t > 7c — 4). Then there exists an (m, 3, c)-system from which a Skolem-type sequence T may be constructed that has length 2t + 1, has zeros in the central 2c — 1 positions and, for each k G {c, c + 1,..., t}, the two positions occupied by k are precisely k apart. The sequence T is used to replace the central 2t + 1 zeros in each sequence S of length 14t + 7 given by Theorem 3.1. We denote the resulting sequence as T ^ S (T into S), and this sequence has zeros in its central 2c — 1 positions. These are then replaced by a sequence Q of length 2c — 1 to form Q ^ (T ^ S), and a short sequence R of further entries is appended at the right-hand end of this sequence to form a sequence S' = (Q ^ (T ^ S)) A R (where A denotes appending). By choosing c, Q and R appropriately, it is possible to form hooked and split-hooked Skolem sequences S'. To illustrate the procedure, we explain how to convert a Skolem-type sequence S of length 147 of the form described in Theorem 3.1, to a hooked Skolem sequence S' of order 74. Note that for k G {11,12,..., 73}, the two positions in S occupied by k are precisely k apart, and that S has zeros in the central 21 positions. Next take a (3,3,2)-system (for example, the one previously described) and from it form a Skolem-type sequence T of length 21 that has zeros in the central three positions and, for each k G {2,3,..., 10}, the two positions in T occupied by k are precisely k apart. Replace the central 21 zeros of S by the sequence T to form T ^ S. Then T ^ S has length 147, zeros in the central three positions and, for each k G {2,3,..., 73}, the two positions occupied by k are precisely k apart. Finally, replace the central three zeros in T ^ S by the sequence Q = (1,1, 74), and append the sequence R = (0, 74) to the right-hand end of Q ^ (T ^ S). The resulting sequence S' has length 149, has a zero in the penultimate position and, for each k g {1,, 2,..., 74}, the two positions occupied by k are precisely k apart. Hence S' is a hooked Skolem sequence of order 74. Clearly different sequences S give rise to different sequences S'. We now return to the general cases. For given t it is obvious that different sequences S will result in different sequences S' = (Q ^ (T ^ S)) A R. So, to extend the bound, it suffices to specify c (and hence T), Q and R, and to check the parity conditions for t - c. In the next two corollaries to Theorem 3.1, we establish the bound by tabulating appropriate c, Q and R. We leave the reader to check the parity conditions for t - c and that the sequence S' is of the required type. Corollary 3.4. For sufficiently large t = 1 or 2 (mod 4), there are more than (6.492)2i+1 hooked Skolem sequences of order 7t + 4. Proof. Table 2 covers the possible values of t modulo 12. Table 2: Construction of hooked Skolem sequences. t (mod 12) c Q R t > 7c - 4 1,10 2 (1,1, 7t + 4) (0, 7t + 4) t > 10 2, 5 3 (1,1, 2, 7t + 4, 2) (0, 7t + 4) t > 17 6, 9 7 (2,4, 2, 5, 6,4, 3, 7t + 4, 5, 3, 6,1,1) (0, 7t + 4) t > 45 For each Skolem-type sequence S of the form described in Theorem 3.1, the resulting sequence S' = (Q ^ (T ^ S)) A R is a hooked Skolem sequence of order 7t + 4. □ Corollary 3.5. For sufficiently large t = 0 or 3 (mod 4), there are more than (6.492)2i+1 split-hooked Skolem sequences of order 7t + 5. Proof. Table 3 covers the possible values of t modulo 12. Table 3: Construction of split-hooked Skolem sequences. t (mod 12) c Q R t > 7c - 4 0, 3 4 (1,1, 7t + 5, 3, 7t + 4, 0, 3) (7t + 5, 7t + 4, 2,0, 2) t > 24 4, 7 5 (2, 3, 2, 4, 3, 7t + 5, 0, 4, 7t + 4) (1,1, 7t + 5,0, 7t + 4) t > 31 8,11 9 (4, 5, 6, 8, 4, 7, 5, 7t + 5, 6, 7t + 4, 0, 8, 7, 3,1,1, 3) (7t + 5, 7t + 4, 2,0, 2) t > 59 For each Skolem-type sequence S of the form described in Theorem 3.1, the resulting sequence S' = (Q ^ (T ^ S))AR is a split-hooked Skolem sequence of order 7t+5. □ The bound obtained in each of the preceding four corollaries only applies to restricted parts of the appropriate residue classes. We believe that it is possible to extend the bound to all possible residue classes in each case. We do not give a proof of this because our argument breaks into a considerable number of subcases. However, to support our contention, we give one example for split Skolem sequences. Corollary 3.2 gives the bound (6.492)2t+1 when n = 7t + 3 and t = 0 or 3 (mod 4), thereby dealing with n = 3 or 24 (mod 28). The necessary and sufficient conditions on n for the existence of a split Skolem sequence of order n may be written as n = 0, 3,4,7, 8,11,12,15,16,19, 20, 23,24 or 27 (mod 28). For our example, we show how the bound may be extended to n = 0 or 7 (mod 28). Put n = 7t + 7 where t = 0 or 3 (mod 4). Take S to be a Skolem-type sequence as described in Theorem 3.1. Depending on the residue of t modulo 12, take c as in Table 4 and put m = (t - c +1)/3. For t > 7c - 4, use an (m, 3, c)-system to construct a Skolem-type sequence T of length 2t + 1 having zeros in the central 2c - 1 positions and such that for each k G {c, c + 1,..., t}, the two positions occupied by k are precisely k apart. Take Q and R as specified in the table and form S' = (Q ^ (T ^ S)) A R. Then S' is a split Skolem sequence of length 14t + 15 (i.e. of order n = 7t + 7). Hence, for all sufficiently large t, there are more than (6.492)2i+1 split Skolem sequences of order n = 7t + 7 where t = 0 or 3 (mod 4), so that n = 0 or 7 (mod 28). Table 4: Further split Skolem sequences. t (mod 12) c Q R t > 7c - 4 4, 7 5 (4, 7t + 7, 3, 7t + 6,4, 3, 7t + 4, 7t + 5,0) (7t + 7, 7t + 6, 7t + 4, 2, 7t + 5, 2,1,1) t > 31 8,11 9 (1,1, 3, 8,4, 3, 7t + 7, 7,4, 7t + 6, 6, 8,0, 7t + 5, 7, 7t + 4, 6) (5, 7t + 7, 2, 7t + 6, 2, 5, 7t + 5, 7t + 4) t > 59 0, 3 16 (11, 8,4, 7,13, 9, 4,14,10, 8, 7,11,12, 7t + 7, 9,15, 7t + 6,13,10,0, 7t + 5,14, 7t + 4, 6,12, 3,1,1, 3, 6,15) (5, 7t + 7, 2, 7t + 6, 2, 5, 7t + 5, 7t + 4) t > 108 There are methods other than the one described above for generating Skolem-type sequences from additive permutations. For certain orders the construction technique given below can give improved bounds. Suppose that s > i > 0 and that S is a Skolem-type sequence of length 2s + 1 having zeros in the central 2i-1 positions, and such that for k G {i, i+1,..., s} the two positions where k appears in S are precisely k apart. If i =1 then S is a split Skolem sequence of order s (and such a sequence exists if s = 0 or 3 (mod 4)), otherwise the earlier discussion following Corollary 3.3 shows that such a sequence exists when s -i = 2 or 11 (mod 12) if i is odd, or s - i = 8 or 11 (mod 12) if i is even, provided that s > 7i - 4. Let S be indexed by [—s, s]. Construction 3.6. • For j = i, i + 1,..., s, denote by aj , bj (with aj < bj) the positions in S occupied by the entry j, so that bj - aj = j. • For each j = i, i + 1,..., s, let nj be an additive permutation on [-1, t], and denote by n the ordered (s - i + 1)-tuple (n^, ..., ns). • Form a new sequence Sn indexed by [—((2t + 1)s + t), (2t + 1)s + t] by placing the entry (2t + 1)j + nj (i) at positions (2t + 1)aj + i and (2t + 1)bj + i + nj (i) for j = £, £ + 1,..., s and i G [—t, t]. For each j these entries cover the interval [(2t + 1)j — t, (2t + 1)j +1], and so they collectively cover [(2t + 1)£—t, (2t + 1)s+1]. These entries cover the positions [—((2t + 1)s +1), (2t + 1)s + t] \ [—(2t + 1)£ + t +1, (2t + 1)£ — t — 1]. Place zeros in the vacant positions. Then Sn is a Skolem-type sequence of length 2(2st + s + t) + 1 having zeros in the central 2((2t + 1)£ — t) — 1 positions, and such that for k G [(2t + 1)£ — t, (2t + 1)s +1] the two positions where k appears in Sn are precisely k apart. • Now suppose that t = 0 or 3 (mod 4) and let P be a split Skolem sequence of order t, indexed by [—t, t]. Apply the previous three steps to P using additive permutations oj on [—(£ — 1), £ — 1] for j = 1, 2,..., t to form a new sequence Ps, where £ is the ordered t-tuple (o1, o2,..., ot). Then Ps is a Skolem-type sequence of length 2(2t(£ — 1) + t + (£ — 1)) + 1 = 2((2t + 1)£ — t) — 1 having zeros in the central 2((2(£ — 1) + 1) — (£ — 1))) — 1 = 2£ — 1 positions, and such that for k G [£, (2t + 1)£ — t — 1] the two positions where k appears in Ps are precisely k apart. Replace the central zeros of Sn by Ps to form the sequence Ps ^ Sn. Then Ps ^ Sn is a Skolem-type sequence of length 2(2st + s + t) + 1 having zeros in the central 2£ — 1 positions, and such that for k G {£, £ + 1,..., 2st + s + t} the two positions where k appears in Ps ^ Sn are precisely k apart. Given a sequence Ps ^ Sn constructed in this fashion for given £, s and t, the ingredients S, n, P and £ may be recovered by considering entries and positions. Suppose that entry e > 0 occupies positions a and b with a < b. If e > (2t +1)£ — t then e = (2t + 1)j+nj (i) for some i and j, and j = |_(e+t)/(2t +1)_|, while a = (2t +1)aj + i, so aj = |_(a + t)/(2t + 1)J. Similarly, bj = |(b + t)/(2t +1)J, while i,nj(i) G [—t,t] are given by i = a (mod 2t + 1) and nj(i) = e (mod 2t + 1). This process recovers S and n. If e < (2t + 1)£ — t — 1 = (2(£ — 1) + 1)t + (£ — 1), then P and £ may be recovered in the same way from |(e + (£ — 1))/(2£ — 1)J, etc. Hence, varying any of S, n, P and £ will yield different Skolem-type sequences Ps ^ Sn. Disregarding variation due to selection of S, P, and £, the following result is obtained. Theorem 3.7. Suppose that there exists a Skolem-type sequence S of length 2s + 1 having zeros in the central 2£—1 positions, and such that for k G {£, £+1,..., s} the two positions where k appears in S are precisely k apart. Then for t = 0 or 3 (mod 4) there are at least (z2t+1)s—+1 Skolem-type sequences Ps ^ Sn of length 2(2st + s +1) + 1 having zeros in the central 2£ — 1 positions, and such that for k G {£, £ + 1,..., 2st + s + t} the two positions where k appears in Ps ^ Sn are precisely k apart. This result generates lower bounds for the numbers of pure, split, hooked and split-hooked Skolem sequences of various orders. Recall from Corollary 2.2 that for sufficiently large t, z2t+1 > (3.246)2t+1. Corollary 3.8. If s = 0 or 3 (mod 4), then for all sufficiently large t = 0 or 3 (mod 4), the number of split Skolem sequences of order 2st + s +1 is greater than (3.246)2st+s. Proof. Invoke the above construction with i = 1. □ As an example, taking s = 3 gives that for all sufficiently large t = 0 or 3 (mod 4), the number of split Skolem sequences of order n = 7t + 3 is greater than (3.246)6t+3. This is substantially better than the bound given earlier in Corollary 3.2, although this approach does not appear to offer the generalization to n = 0 or 3 (mod 4) mentioned in connection with the previous method. If the central zero in a split Skolem sequence of order n is replaced by n + 1 and an additional entry n + 1 is placed at either the start or the end of the sequence, then a pure Skolem sequence of order n + 1 is formed. Applying this adaptation to the split Skolem sequence constructed in Corollary 3.8 gives the following result. Corollary 3.9. If s = 0 or 3 (mod 4), then for all sufficiently large t = 0 or 3 (mod 4), the number of pure Skolem sequences of order 2st + s +1 + 1 is greater than (3.246)2st+s. To deal with hooked and split-hooked Skolem sequences, we may replace the central 2i - 1 zeros of Vs ^ Sn with an appropriate sequence Q, and append a short sequence R to the right hand end to form (Q ^ (Vs ^ Sn)) A R. Corollary 3.10. If s = 1 or 2 (mod 4) and s > 45, then for all sufficiently large t = 0 or 3 (mod 4), the number of hooked Skolem sequences of order 2st + s +1 +1 is greater than (3.246)(s-6)(2i+1). Proof. Given s and t, put m = 2st + s +1. The sequences Q and R for possible values of s modulo 12 are covered by using Table 2 with t replaced by s, c replaced by i and 7t + 3 replaced by m. Note that in Table 2, c < 7 and so we may assume that i < 7. In each case, the resulting sequence (Q ^ (Vs ^ Sn))A R is a hooked Skolem sequence of order m + 1 = 2st + s +1 +1, and the result follows. □ Corollary 3.11. If s = 0 or 3 (mod 4) and s > 59, then for all sufficiently large t = 0 or 3 (mod 4), the number of split-hooked Skolem sequences of order 2st + s +1 + 2 is greater than (3.246)(s-8)(2i+1). Proof. Given s and t, put m = 2st + s +1. The sequences Q and R for possible values of s modulo 12 are covered by using Table 3 with t replaced by s, c replaced by i and 7t + 3 replaced by m. Note that in Table 3, c < 9 and so we may assume that i < 9. In each case, the resulting sequence (Q ^ (Vs ^ Sn))A R is a split-hooked Skolem sequence of order m + 2 = 2st + s +1 + 2, and the result follows. □ 4 Completing the proof of Theorem 2.1 As previously described, the arithmetic is simplified by subtracting t - s - a from all the entries in R and S to obtain equivalent arrays R' and S', where R' has entries ranging from — (s + 2a) to s + 2a, and S' has entries ranging from — (s - 1) to s - 1. Transversals are sought in these arrays that cover the entry values from -(s + a) to s + a inclusive. We renumber the rows and columns so that for R' the row and column numbers run from 0 to r - 1 = s + 2a and for S' they run from 0 to s - 1. The entry in cell (i, j) of R' is then i + j - (s + 2a), and that in cell (i, j) of S' is i + j - (s - 1). The identification of transversals falls into several cases. Table 5: Case 1 (r even), transversal in S'. Range Row Column Entry (a) j =0,..., 2—1 j 2—1 + j - 2 — 1 +2j (b) f j 2—1 - 1 2—1 + 1+ j j - 2—1 + 1 + 2 j Case 1: r even. Since r is even, s must be odd. Table 5 identifies a suitable transversal in S'. Line (b) of the table, marked with f, is omitted when s = 1. In S', and subject to f, line (a) of the table covers rows 0 to 2—1, and line (b) covers rows 2—1 + 1 to s - 1. For columns, line (b) covers 0 to 2—1 - 1, and line (a) covers 2—1 to s - 1. As regards entries, lines (a) and (b) together cover - 2—1 to 2—1. Subcase 1.1: r = 0 (mod 4). Table 6 identifies a suitable transversal in R'. The line of the table marked with * is omitted when s = b - 2, and the line marked f is omitted when s = 1. Subject to * and f, rows, columns and entries of R' are covered by lines of the Table 6: Subcase 1.1 (r = 0 (mod 4)), transversal in R'. Range Row Column Entry (a) j = 0,..., 4 -1 j a + j -a - s + 2j (b) j = °..., a - r 4 + j a + 4 + s + j 2—1 + 1 + 2 j (c) j = 0,..., 4 -1 a + 1 + j -a - s + 1 + 2j (d) j =0,..., 2—1 a+4+1+j a + 4 + 2!T1 + j a + 1 + 2 j (e) f j =0,..., 2—1 - 1 a + 4 + TT + 2 + j a + 4 + j a + 2 + 2j (f) * j = 0,. .., a - 4 - 1 a + 4 + s + 1+ j 4 + j 2—1 + 2 + 2 j table in the following orders, where notation such as (a&c) means that entries from lines (a) and (c) are interleaved and taken together. Rows 0 to 2a + s by lines (a)(b)(c)(d)(e)(f) in that order. Columns 0 to 2a + s by lines (c)(f)(a)(e)(d)(b) in that order. Entries -a - s to - 2—1 - 1 by lines (a&c), and entries 2—1 + 1 to a + s by lines (b&f)(d&e) in that order; the remaining entries required to complete the transversal values from -(s + a) to (s + a) come from the transversal in S'. Subcase 1.2: r = 2 (mod 4). Table 7 identifies a suitable transversal in R'. The line of the table marked with f is omitted when s = 1. Subject to f, rows, columns and entries of R' are covered by lines of the table in the following orders. Rows 0 to 2a + s by lines (a)(b)(c)(d)(e)(f) in that order. Columns 0 to 2a + s by lines (c)(f)(a)(e)(d)(b) in that order. Entries -a - s to -2—1 - 1 by lines (a&c), and entries 2—1 + 1 to a + s by lines (b&f)(d&e) in that order; the remaining entries required to complete the transversal values from -(s + a) to (s + a) come from the transversal in S'. Case 2: r odd. If r is odd then s must be even. If s = 0 then r = b and R is a copy of Ab Table 7: Subcase 1.2 (r = 2 (mod 4)), transversal in R'. Range Row Column Entry (a) j = 0,...,^ -1 j a + j -a - s + 2j (b) j = 0,. .., a - -+§ + j a + + s + j 1-1 + 2 + 2 j (c) j = 0,..., r-+§ - 2 a + 1 + j j -a - s + 1 + 2j (d) j ^ a + -+§ + j a + + ¥ + j a + 1 + 2j (e) f j =0,..., ^ - 1 a + -+§ + s-1 + 1+j a + + j a + 2 + 2j (f) j = 0,. .., a - -+§ a + -+r§ + s + j - 1+ j 1-1 + 1 + 2 j (with rows and columns appropriately renumbered) and any one of the transversals already identified in Ab provides a suitable transversal in R. So throughout Case 2, we may assume that s > 0, and then Table 8 identifies a suitable transversal in S'. Line (b) of the table, marked with f, is omitted when s = 2. Subject to f, rows, columns and entries of S' are Table 8: Case 2 (r odd), transversal in S'. Range Row Column Entry (a) j =0,..., § - 1 j § - 1+ j - § +2j (b) f j = 0,..., § - 2 § + j j - § + 1 + 2j (c) single cell s - 1 s - 1 s - 1 covered by lines of the table in the following orders. Rows 0 to s - 1 by lines (a)(b)(c) in that order. Columns 0 to s - 1 by lines (b)(a)(c) in that order. Entries -§ to § - 2 by lines (a&b), and entry s - 1 by line (c). To deal with R', we consider four subcases depending on the values of r and s modulo 4. Subcase 2.1: r = 1, s = 0 (mod 4). These conditions imply that b = 1 (mod 4) and we may assume that s > 4. Table 9 identifies a suitable transversal in R'. Lines of the table marked with * are omitted when s = b - 1, and lines marked with f are omitted when s = 4. Subject to * and f, rows, columns and entries of R' are covered by lines of the table in the following orders. Rows 0 to 2a + s by lines (a)(b)(c)(d)(e)(f)(g)(h)(i)(j)(k)(l)(m) in that order. Columns 0 to 2a + s by lines (k)(f)(i)(g)(e)(b)(d)(l)(c)(m)(a)(h)(j) in that order. Entries -a - s to - § - 1 by lines (b&f)(e)(d&g)(i)(c)(a&k) in that order, entries § - 1 to s - 2 by lines (h&l), and entries s to a + s by lines (j&m); the remaining entries required to complete the transversal values from -(s + a) to (s + a) come from the transversal in S'. Subcase 2.2: r = 1, s = 2 (mod 4). These conditions imply that b = 3 (mod 4) and a is odd. Table 10 identifies a suitable transversal in R' when s > 6. The line of the table Table 9: Subcase 2.1 (r = 1, s = 0 (mod 4)), transversal in fi'. Range Row Column Entry (a) * j = o,..., a - 4 -1 j a + s + j -a + 2j (b) j = 0,..., 4 -1 a - s + j 2 4 + j a + 4 + j -a - s + 2j (c) single cell a 2 a + s - 1 -a - 1 (d) t j = 0,..., 4 - 2 a + 1+j a + 22 + j -a - f + 1 + 2j (e) single cell a i s 2 + 4 a + f - 1 2 + 4 1 -a - 2 - 1 (f) t j = 0,..., 4 - 2 a+4+1+j a - 4 + j -a - s + 1 + 2j (g) t j = 0,..., 4 - 2 a + 2 + j a + j -a - 2 + 2j (h) j = 0,..., 4 -1 a + 3s -1+j 3a + 3s + j 2 + 4 +j 2 -1 + 2 j (i) single cell a + s -1 a -1 -a - 2 (j) j = 0,..., a a + s + j 3a + s + j s + 2 j (k) * j = 0,._a - 4 -1 a + s + 1 + j j -a + 1 + 2j (l) j = 0,..., 4 -1 3a + 3f + 1+ j a + 3f - 1+ j 2 + 2 j (m) j = 0,..., a -1 3f + s + 1+ j a + s + j s + 1 + 2 j marked with * is omitted when s = b - 1, and the line of the table marked with f is omitted when s = 6. Subject to * and f, rows, columns and entries of R' are covered by lines of the table in the following orders. Rows 0 to 2a+s by lines (a)(b)(c)(d)(e)(f)(g)(h)(i)(j)(k)(l)(m) in that order. Columns 0 to 2a+s by lines (k)(f)(i)(g)(e)(b)(d)(l)(c)(m)(a)(h)(j) in that order. Entries -a - s to - § - 1 by lines (b&f)(e)(d&g)(c)(i)(a&k) in that order, entries § - 1 to s - 2 by lines (h&l), and entries s to a + s by lines (j&m); the remaining entries required to complete the transversal values from -(s + a) to (s + a) come from the transversal in S'. The case s = 2 may be obtained from the table by omitting lines (b), (d), (e), (f), (g) and (l). Subject to * (i.e. when b = 3), rows, columns and entries of R' are covered by lines of the table in the following orders. Rows 0 to 2a + 2 by lines (a)(c)(h)(i)(j)(k)(m) in that order. Columns 0 to 2a + 2 by lines (k)(i)(c)(m)(a)(h)(j) in that order. Entries -a - 2 to -2 by lines (c)(i)(a&k) in that order, entry 0 by line (h), and entries 2 to a + 2 by lines (j&m); the remaining entries required to complete the transversal values from -(a + 2) to (a + 2) come from the transversal in S'. Subcase 2.3: r = 3, s = 0 (mod 4). These conditions imply that b = 3 (mod 4) and a is odd. Table 11 identifies a suitable transversal in R'. The line of the table marked with * is omitted when s = b - 3, and the lines of the table marked with f are omitted when s = 4. Subject to * and f, rows, columns and entries of R' are covered by lines of the table in the following orders. Rows 0 to 2a + s by lines (a)(b)(c)(d)(e)(f)(g)(h)(i)(j)(k)(l)(m) in that order. Columns 0 to 2a + s by lines (k)(f)(i)(g)(e)(b)(d)(l)(c)(m)(a)(h)(j) in that order. Entries -a - s to - § - 1 by lines (b&f)(e)(d&g)(i)(c)(a&k) in that order, entries § - 1 to s - 2 by lines (h&l), and entries s to a + s by lines (j&m); the remaining entries required to complete the transversal values from -(s + a) to (s + a) come from the transversal in S'. Table 10: Subcase 2.2 (r = 1, s = 2 (mod 4)), transversal in fi'. Range Row Column Entry (a) * j =0,..., ^ -s-2 1 4 1 j a + s + j -a + 2j (b) j = 0,..., ^ - 1 a—1 s-2 i j 2 4 +j a—1 + s —2 + 2 + 4 + 1+ j —a — s + 2j (c) single cell a —1 2 ^ + s - 1 -a — 2 (d) j = 0,..., ^ - 1 s-r + 1+ j ^ + 2 + j —a — f + 2j (e) single cell a—1 + s-2 + 1 2 + 4 + 1 a—1 + s —2 2 + 4 —a — 2 — 1 (f) j = 0,..., ^ - 1 a—1 + s —2 + 2 + 4 + 2 + j a —1 s—2 + j 2 4 +j —a — s + 1 + 2 j (g) t j = 0,..., ^ - 2 a—T + 2 + 1+ j ^ + 1+ j —a — 2 + 1 + 2 j (h) j =0,..., ^ a—1 i 3s —2 i j 2 + 4 +j 3a+1 i 3s —2 i j 2 1 4 +j 2 — 1 + 2j (i) single cell ^ + s a—1 2 —a — 1 (j) j =0,..., ^ a—T + s +1+ j ^ + s + j s + 1 + 2 j (k) * j =0,..., ^ -s-2 i 4 1 a + s + 1 + j j —a + 1 + 2j (l) j = 0,..., ^ - 1 3a+1 i 3s —2 i 2 1 4 + 1+ j a—1 i 3s—2 i j 2 + 4 +j 2 + 2j (m) j = 0,..., ^ 3a+1 + s + j ^ + s + j s + 2j Subcase 2.4: r = 3, s = 2 (mod 4). These conditions imply that b = 1 (mod 4) and a is even. Table 12 identifies a suitable transversal in ñ' when s > 6. The line of the table marked with * is omitted when s = b - 3, and the line of the table marked with f is omitted when s = 6. Subject to * and f, rows, columns and entries of ñ' are covered by lines of the table in the following orders. Rows 0 to 2a+s by lines (a)(b)(c)(d)(e)(f)(g)(h)(i)(j)(k)(l)(m) in that order. Columns 0 to 2a+s by lines (k)(f)(i)(g)(e)(b)(d)(l)(c)(m)(a)(h)(j) in that order. Entries -a - s to -1 - 1 by lines (b&f)(e)(d&g)(i)(c)(a&k) in that order, entries | - 1 to s - 2 by lines (h&l), and entries s to a + s by lines (j&m); the remaining entries required to complete the transversal values from -(s + a) to (s + a) come from the transversal in S'. The case s = 2 may be obtained from the table by omitting lines (b), (d), (e), (f), (g) and (h). Subject to * (i.e. when b = 5), rows, columns and entries of ñ' are covered by lines of the table in the following orders. Rows 0 to 2a + 2 by lines (a)(c)(i)(j)(k)(l)(m) in that order. Columns 0 to 2a + 2 by lines (k)(i)(l)(c)(m)(a)(j) in that order. Entries -a - 2 to -2 by lines (i)(c)(a&k) in that order, entry 0 by line (l), and entries 2 to a + 2 by lines (j&m); the remaining entries required to complete the transversal values from -(a + 2) to (a + 2) come from the transversal in S'. This concludes the proof of Theorem 2.1. Table 11: Subcase 2.3 (r = 3, s = 0 (mod 4)), transversal in R'. Range Row Column Entry (a) j = o,..., ^ - 4 a + s + j -a + 2j (b) j = 0,..., s -1 o±l _ s + j 2 4 + j a-T + 4 + j -a - s + 2j (c) single cell a+1 2 a-T + s - 1 -a - 1 (d) t j = 0,..., s - 2 a+1 + 1+ j a—1 + 22 + j -a - § + 1 + 2j (e) single cell a+1 + s 2 + 4 a — 1 I s i ~ + 4 - 1 -a - 2 - 1 (f) t j = 0,..., s - 2 ^+4+1+j TT - 4 + j -a - s + 1 + 2j (g) t j = 0,..., s - 2 a+1 + 22 + j a-i+j -a - 2 + 2j (h) j = 0,..., s -1 ^ + 3s -1 + j 3a+1 | 3s | j 2 + 4 +j 2 +2j (i) single cell at-1 + s - 1 a — 1 i ~2--1 -a - 2 (j) j =0,..., ^ ^ + s + j ^ + s + j s + 1 + 2j (k) * j 1 -s -1 a + s + 1 + j j' -a + 1 + 2j (l) j = 0,..., s -1 3a+1 , 3s , j 2 + 4 + j a—1 + 3s 2 + 4 1+ j 2 - 1 + 2 j (m) j = 0,..., 1 ^ + s + j a—1 + s + j s + 2j 5 Concluding remarks Our results improve the known lower bounds for the number of additive permutations, zero-sum arrays, some Skolem-type sequences, and some extremal Langford sequences. It seems highly likely that the bounds obtained in this paper apply to all pure, split, hooked and split-hooked Skolem sequences of sufficiently large orders. The recent paper [8] combines such bounds with graph labellings to generate Langford sequences. It seems likely that our new bounds can be combined with these techniques to generate improved estimates for the numbers of Langford sequences. For small orders, the numbers of (pure) Skolem sequences and hooked Skolem sequences (and other related sequences) are tabulated in [4], while Table 8 of [7] gives the numbers of split Skolem (Rosa) sequences of orders n < 12. These numerical results strongly suggest that further improvements to our lower bounds are possible. Since Skolem sequences may be used to construct solutions to Heffter's first and second difference problems, the bounds inform the numbers of these and of resulting cyclic Steiner triple systems. If improved bounds for z2i+1 are obtained in the future, these methods will lead to improved bounds for many related sequences. From Table 1, it will be seen that the ratio z2i+1/z2i-1 appears to increase with t, and to exceed 2t for t > 6, strongly suggesting that z2i+1 > 24t! for all sufficiently large t. This is a weaker bound than might be suggested by Vardi's conjecture, but it is strongly supported by the computational evidence, and one might expect that most transversals are not Z-transversals. Table 12: Subcase 2.4 (r = 3, s = 2 (mod 4)), transversal in R'. Range Row Column Entry (a) 3 = o,..., a - s-2 i a + s + j -a + 2j (b) j = 0,..., ^ - 1 a _ s-2 + j 2 4 +j a + +j -a - s + 2j (c) single cell a 2 a+s -1 -a - 1 (d) j = o,..., -1 a + 1+ j a + 2-1+j -a - § + 2j (e) single cell a 1 s-2 1 1 2 + 4 + 1 a I s — 2 -| 2 + ~ 1 -a - f - 1 (f) j = o,..., -1 a + +2+j a - t2 -1+j -a - s + 1 + 2j (g) t j = 0,..., - 2 a+2+1+j a+j -a - 2 + 1 + 2j (h) j = o,..., ^ - 1 a 1 3s-2 | j 2 + 4 +j 3a | 3s+2 | j 2 + 4 +j 2 +2j (i) single cell a + s - 1 a -1 -a - 2 (j) j = o,..., a a + s + j t + s+j s + 2j (k) * j = o,..., a -s-1 - 2 a + s + 1 + j -a + 1 + 2j (l) j =o,..., ^ 3a , 3s+2 , j 2 + 4 +j a I 3s-6 , j 2 + 4 +j 2 -1 + 2 j (m) j = o,..., a -1 3a + s + 1 + j a+s+j s + 1 + 2 j References [1] J. Abrham, Exponential lower bounds for the number of Skolem and extremal Langford sequences, Ars Combin. 22 (1986), 187-198. [2] G. K. Bennett, M. J. Grannell and T. S. Griggs, Exponential lower bounds for the numbers of Skolem-type sequences, Ars Combin. 73 (2004), 101-106. [3] N. J. Cavenagh and I. M. Wanless, On the number of transversals in Cayley tables of cyclic groups, Discrete Appl. Math. 158 (2010), 136-146, doi:10.1016/j.dam.2009.09.006. [4] C. J. Colbourn and J. H. Dinitz (eds.), Handbook of Combinatorial Designs, Discrete Mathematics and its Applications, Chapman & Hall/CRC, Boca Raton, Florida, 2nd edition, 2006, doi:10.1201/9781420010541, http://www.cems.uvm.edu/~jdinitz/hcd.html. [5] C. J. Colbourn and A. Rosa, Triple Systems, Oxford Mathematical Monographs, Clarendon Press, New York, 1999. [6] S. Eberhard, F. Manners and R. Mrazovic, Additive triples of bijections, or the toroidal semiqueensproblem, 2016, arXiv:1510.05987v3 [math.CO], submitted. [7] M. Gruttmuller, R. Rees and N. Shalaby, Cyclically indecomposable triple systems that are decomposable, J. Combin. Math. Combin. Comput. 63 (2007), 103-122, http://www. combinatorialmath.ca/jcmcc/jcmcc63.html. [8] S. C. Lopez and F. A. Muntaner-Batle, Langford sequences and a product of digraphs, Eur. J. Combin. 53 (2016), 86-95, doi:10.1016/j.ejc.2015.11.004. [9] N. J. A. Sloane (ed.), The On-Line Encyclopedia of Integer Sequences, published electronically at https://oeis.org. [10] I. Vardi, Computational Recreations in Mathematica, Addison-Wesley, Redwood City, California, 1991. ARS MATHEMATICA CONTEMPORANEA ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 14 (2018) 433-443 https://doi.org/10.26493/1855-3974.1043.9d5 (Also available at http://amc-journal.eu) Classification of regular balanced Cayley maps of minimal non-abelian metacyclic groups* Kai Yuan, Yan Wang t School of Mathematics and Information Science, Yantai University, Yan Tai 264005, P.R. China Haipeng Qu School ofMathematics and Computer Science, Shan Xi Normal University, Shan Xi 041000, P.R. China Received 22 February 2016, accepted 31 May 2017, published online 13 November 2017 In this paper, we classify the regular balanced Cayley maps of minimal non-abelian metacyclic groups. Besides the quaternion group Q8, there are two infinite families of such groups which are denoted by Mp,q(m, r) and Mp(n, m), respectively. Firstly, we prove that there are regular balanced Cayley maps of Mp,q (m, r) if and only if q = 2 and we list all of them up to isomorphism. Secondly, we prove that there are regular balanced Cayley maps of Mp(n, m) if and only if p = 2 and n = m or n = m +1 and there is exactly one such map up to isomorphism in either case. Finally, as a corollary, we prove that any metacyclic p-group for odd prime number p does not have regular balanced Cayley maps. Keywords: Regular balanced Cayley map, minimal non-abelian group, metacyclic group. Math. Subj. Class.: 05C25, 05C30 1 Introduction A Cayley graph r = Cay(G, X) is a graph based on a group G and a finite set X = {x\,x2,..., xk} of elements in G which does not contain 1G, contains no repeated elements, is closed under the operation of taking inverses, and generates all of G. In this *The authors want to thank the referees for their valuable comments and suggestions. t Author to whom correspondence should be addressed. Supported by NSFC (No. 11371307, 11671347, 61771019), NSFS (No. ZR2017MA022), J16LI02 and Research Project of Graduate Students (01073). E-mail address: pktide@163.com (Kai Yuan), wang-yan@pku.org.cn (Yan Wang), orcawhale@163.com (Haipeng Qu) ©® This work is licensed under http://creativecommons.org/licenses/by/3.0/ Abstract paper, we call X a Cayley subset of G. The vertices of the Cayley graph r are the elements of G, and two vertices g and h are adjacent if and only if g = hx for some x G X. The ordered pairs (h, hx) for h G G and x G X are called the darts of r. Let p be any cyclic permutation on X. Then the Cayley map M = CM(G, X, p) is the 2-cell embedding of the Cayley graph Cay(G, X) in an orientable surface for which the orientation-induced local ordering of the darts emanating from any vertex g g G is always the same as the ordering of generators in X induced by p; that is, the neighbors of any vertex g are always spread counterclockwise around g in the order (gx, gp(x), gp2(x),..., gpk-1(x)). An (orientation preserving) automorphism of a Cayley map M is a permutation on the dart set of M which preserves the incidence relation of the vertices, edges, faces, and the orientation of the map. The full automorphism group of M, denoted by Aut(M), is the group of all such automorphisms of M under the operation of composition. This group always acts semi-regularly on the set of darts of M, that is, the stabilizer in Aut(M) of each dart of M is trivial. If this action is transitive, then we say that the Cayley map M is a regular Cayley map. As the left regular multiplication action of the underlying group G lifts naturally into the full automorphism group of any Cayley map CM(G, X, p), Cayley maps are proved to be a very good source of regular maps. There are many papers on the topic of regular Cayley maps, we refer the readers to [4, 10] and [11] and the references therein. Furthermore, A Cayley map CM(G, X, p) is called balanced if p(x)-1 = p(x-1) for every x G X .In [11], Skoviera and Siran showed that a Cayley map CM(G, X, p) is regular and balanced if and only if there exists a group automorphism a such that a|X = p, where a|X denotes the restricted action of a on X. Therefore, to determine all the regular balanced Cayley maps of a group is equivalent to determine all the orbits of its automorphisms that can be Cayley subsets. In this paper, a non-abelian group G is called minimal if each of its proper subgroups H (that is H < G but H = G) is abelian. In 1903, Miller and Moreno gave a full classification of minimal non-abelian groups, one may refer to [7] for detailed results. A group G is metacyclic if it has a cyclic normal subgroup N such that the factor group G/N is cyclic. As one can see in [7], there are three classes of minimal non-abelian metacyclic groups: (1) the quaternion group Q8; (2) Mp,q(m,r) = (a, b | ap = 1,bqm = 1,b-1ab = ar}, where p and q are distinct prime numbers, m is a positive integer and r ^ 1 (mod p) but rq = 1 (mod p); (3) Mp(n,m) = (a, b | ap" = bpm = 1,b-1ab = a1+p"-1 ,n > 2, m > 1}. One can also cite [3, Theorem 2.1] for reference or [13, pp. 123] for details. For regular balanced Cayley maps, it has been shown that all odd order abelian groups possess at least one regular balanced Cayley map [4]. Wang and Feng [12] classified all regular balanced Cayley maps for cyclic, dihedral and generalized quaternion groups. In [9], Oh proved the non-existence of regular balanced Cayley maps with semi-dihedral groups. In this paper, we pay our attentions to the regular balanced Cayley maps of minimal non-abelian metacyclic groups. Since the regular balanced Cayley maps of Q8 have been classified in [12] (Q8 has exactly one regular balanced Cayley map up to isomorphism), we only consider the groups Mp,q(m, r) and Mp(n, m). In Section 3, we show that Mp,q(m, r) has regular balanced Cayley maps if and only if q is 2 and we list all of them up to isomorphism. In Section 4, we show that Mp(n, m) has regular balanced Cayley maps if and only if p = 2 and n = m or n = m +1. In either case, it has exactly one regular balanced Cayley map up to isomorphism and the map has valency 4. Moreover, as a corollary any metacyclic p-group for odd prime p doesn't have regular balanced Cayley maps. 2 Preliminaries Lemma 2.1. Take an element b4as G Mp,q(m, r), where t = 0, then the order of b4as is qm if and only if (t, q) = 1. Proof. The group Mp,q(m, r) is the union of one cyclic group of order p and p conjugate cyclic subgroups of order qm. If t = 0, then b4as belongs to one of the cyclic subgroups of order qm. Therefore, the order of b4as is qm if and only if (t, q) = 1. □ Lemma 2.2. The automorphism group of Mp,q(m, r) is Aut(Mp,q(m, r)) = {a | a7 = ab7 = bjak, 1 < i < p-1, 1 < j < qm-1, q | (j-1)}. Proof. Assume a G Aut(Mp q(m, r)). According to Lemma 2.1, a7 = a®, b7 = bjak for some 1 < i < p — 1,1 < j < qm — 1 and (j, q) = 1. If Mp,q(m, r) = (a7, b7}, then we can get the relation q | (j — 1). In fact, since (ar)7 = (b-1ab)7 = (b-1)7a7b7 = b-ja®bj = air' = air, we have air(rj -1) = 1. Moreover, from (ir,p) = 1 and ap = 1, we get (rj-1 — 1) = 0 (mod p), that is rj-1 = 1 (mod p). As rq = 1 (mod p) and q is prime, we have q | (j — 1). □ Lemma 2.3 ([5]). The automorphism group of Mp(n, m) is listed as follows: (i) If n < m, then Aut(Mp(n, m)) = {a | a7 = bja®, b7 = bsar, (i,p) = 1,1 < i < pn, j = kpm-n+1, 0 < k m > 1 or p = 2 and n > m > 1, then Aut(Mp(n, m)) = {a | a7 = bja®, b7 = bsar, (i,p) = 1, 1 < i < pn, 1 < j < pm,r = kpn-m, 0 < k < pm, s = 1 (mod p), 1 < s < pm}. The following Lemma 2.4 is a basic result in group theory and we omit the proof. Lemma 2.4. Let G be a finite group and N be a normal subgroup of G. Take a G Aut(G). If Na = N, then a : Ng ^ Nga is an automorphism of G/N which is called the induced automorphism of a. Lemma 2.5. Let G be a finite group and N be a proper characteristic subgroup of G. Take a G Aut(G) and g G G. If X = is a Cayley subset of G, then X = g = g is a Cayley subset of G = G/N. Moreover, if the order of a is a power of 2 and g is not an involution, then |X | = |X |. Proof. By Lemma 2.4, a is an automorphism of G/N induced by a. Set X = g^, then X = gg^ = g. If X is a Cayley subsetof G, then the relations JX} = G, X = X-1 follow naturally. Since N < G, we have X = and then 1 G X. So, X is a Cayley subset of G. If the order of a is 2s for some positive integer s, then the order of a is 24 for some 2s-1 2s-1 __1 2t _ integer t < s. From ga = g 1, we have ga = g 1. While ga = g, then t > s — 1. So, s = t and |X| = |X|. □ As a direct corollary of Lemmas 2.4 and 2.5, we give the following Corollary 2.6. Corollary 2.6. If a group G has regular balanced Cayley maps, then so does the quotient group G/N for any proper characteristic subgroup N of G. There are many ways to get proper characteristic subgroups. In the following, we give a method to get such subgroups. These results are exercises for students, so we omit the proof. Lemma 2.7. Let G be a finite group, S C G, a e End(G), K be a characteristic subgroup of G and n be a positive integer. Then, (i) (S)- = (S-); (ii) Hi = (xn | x e K) is a characteristic subgroup of G; (iii) H2 = (y | y e G,yn e K) is a characteristic subgroup of G. As for isomorphism of regular maps, one may refer to [10] for the following Lemma 2.8. Lemma 2.8. Assume Mi = CM(G, Xi,pi) and M2 = CM(G, X2, p2) are two regular balanced Cayley maps of the finite group G, where Xi = gl'-1> and X2 = hl'-2> are orbits of two group elements g and h under the action of two automorphisms ai and a2 of G, respectively. Then Mi and M2 are isomorphic if and only if |Xi | = |X21 = k and there is some t e Aut(G) such that h-2 = g-2T, 1 < i < k. As a special case and an application of Lemma 2.8, we have the following Lemma 2.9. Lemma 2.9. Let G be a finite group. Take a e Aut(G) and two elements g,h e G. Assume X = gia> is a Cayley subset of G. If there is some a e Aut(G) such that g- = h, then Y = ha-> is also a Cayley subset of G and Y = X-. Under this situation, the two regular balanced Cayley maps CM(G, X, a|X) and CM(G, Y, a-iaa| Y) are isomorphic. Proof. Because Y = h<--1«-> = g-(--1«-) = g---1 M- = g(«>- = X- and X is a Cayley subset, it follows that Y is also a Cayley subset. The result that CM(G, X, a|X) and CM(G, Y, a-iaa|Y) are isomorphic follows from Lemma 2.8. □ 3 Regular balanced Cayley maps of Mp,q (m, r) As we mentioned in the introduction, to determine all the regular balanced Cayley maps of a group is equivalent to determine all the orbits of its automorphisms that can be Cayley subsets. In this section, we divide our discussion into two parts according to the parity of q. Lemma 3.1. The center Z(Mp,q(m, r)) of Mp,q(m, r) is generated by bq and the quotient group MPq(m, r)/Z(MPtq(m,'r)) = MPq(1, r). Proof. From the defining relation of Mp,q (m, r), we have b-qabq = ar" = a. So, bq e Z(Mp,q(m, r)). Since Mp,q(m, r) is not abelian and generated by a and b, we have a,b g Z(Mpq(m,r)), hence Z(Mpq(m,r)) = (bq). The formula Mp,q(m,r)/Z(Mp,q(m,r)) ^ Mp,q(1, r) follows directly from the definition of Mp,q(m, r). □ Theorem 3.2. If q is odd, then the group Mp,q (1, r) does not have regular balanced Cayley maps. Proof. For brevity, set H = Mp,q (1, r). Suppose there exists a a e Aut(H) and bvau e H such that X = (bvau)l^> is a Cayley subset of H. The derived subgroup of H is H' = (a) which is a characteristic subgroup. Let H = H/H' and a be induced by a. By Lemma 2.2, ba = bak for some integer k and as a result ba = b. So, X = bvau(a) = ¥'a) = jb"}. While X = X-1, o(b) = q and o(F) | o(b), we have b = 1 and so bv e H'. It follows that (X) < H' < H contradicting to H = (X). □ As a corollary of Lemmas 3.1 and 2.5, we have the following Theorem 3.3. Theorem 3.3. If q is odd, then Mp,q (m, r) does not have regular balanced Cayley maps. It is known that = Z2 x Z^n-2 — (-1) x (5), where —1 and 5 denote the class of integers equaling to —1 and 5 modular 2n, respectively. In ap-group G, let ^1(G) = (ap | _2 _ a e G). Then, ) = (5 ) which does not contain —1. Lemma 3.4. For a positive integer n > 2, the equation xk = —1 (mod 2n) holds if and only if k is odd and x = —1 (mod 2n). Proof. It is obviously true when n = 2. So, we may assume n > 3. Let u be a solution of the equation xk = — 1 (mod 2n), then the integer u should be odd, so u e Z2n = (—1) x (5). From the discussion preceding to the lemma, suppose k is even, then —1 = uk = (uk )2 e y1(Z2n), a contradiction. So, k is odd. Let u = ab for some a e (—1) and b e (5) such that uk = —1. Then, uk = akbk = —1. There are two choices of a, that is 1 and —1. But a = 1, for otherwise bk = —1, a contradiction. So, bk = 1 and as a result b =1 and u = —1. □ In a group G, for any element g e G, we use o(g) to denote the order of g. Now we look at the group Mp,2(m, r). In the definition of Mp,2(m, r), one can see that r = —1 (mod p). In particular, if m = 1, then Mp,2(m, r) is a dihedral group of order 2p. One may refer to [12] for the classification of the regular balanced Cayley maps of dihedral groups. For the sake of completeness, We restate the result in the following theorem. Theorem 3.5 ([12, Theorem 3.3]). The dihedral group D2p of order 2p has p — 1 non-isomorphic regular balanced Cayley maps, where p is an odd prime number. When m > 2, we have the following Theorem 3.6. Theorem 3.6. Let G = Mp,2(m, r), where m > 2, p is an odd prime and r = — 1 (mod p). If p — 1 = 2es, where s is odd, then G has s non-isomorphic regular balanced Cayley maps. In particular, ifp is a Fermat prime, then G has exactly one regular balanced Cayley map up to isomorphism. Proof. If the orbit of bvau under the action of a e Aut(G) is a Cayley subset of G, then the integer v must be odd. In fact, both the subgroups (a) and Z(G) = (b2) are characteristic in G, so ((bvau)'CT^) is aproper subgroup of G if (v, 2) = 1. By Lemma 2.2, there is some a e Aut(G) such that (bvau)a = b. According to Lemma 2.9, we only need to consider the orbit of b under the action of a. For brevity, we denote the automorphism a e Aut(G) satisfying aCT = a® and bCT = bjak by aiij-ik and X = by X®¿ik. Let piijjk be the arrangement of the elements in Xi j k which respects the order of the elements in the orbit. Assume Xi j k is a Cayley subset of G for some integer i coprime to p and odd integer j. Note that k ^ 0 (mod p) for otherwise contradicting to the Cayley subset assumption of j . In the quotient group G = G/(a), Xi,j,k = should be a Cayley subset of G. Therefore, there exists some integer t such that b i j'fc = b . Clearly, b i j'fc = _jt __i b = b , so j = —1 (mod 2m). From Lemma 3.4, t is odd and j = —1 (mod 2m). Moreover, as Xf-J'l = Xii_1jfc, we may assume k = 1. Under these conditions, we only need to pay attention to X _11. By direct enumeration one can easily get = bC-i)' a for any positive integer Since X,-1 ^ is a Cayley subset, there exists some positive integer n such that b0*-1,1 = b-1. So, n is odd and in-1 + in-2 + ••• + i +1 = 0 (mod p). If i = 1 (mod p), then b0*,-1,1 = b-1 and X1,-1,1 = {b, b-1a, ba2,. .., bap-1, b-1, (b-1a)-1, ..., (bap-1)-1} is a Cayley subset of G of valency 2p. If 1 < i < p — 1, then in-1 + in-2 + • • • + i + 1 = 0 (mod p) if and only if in = 1 (mod p). Let S = {x | x e Zp, o(x) is odd}, then |S| = s. Since n is odd, any i satisfying in = 1 (mod p) corresponds to i e S. And for any i e S \ {1}, if o( i) = n, then b0*!-1,1 = b-1 and Xi,_i,i = {b, b a, bai+1, b-1 a +i+1,. .., ba* +"+*+1, b_\. .., (ba* + •+i+1)_1} is a Cayley subset of G of valency 2n. From all the above, when i > 1, Xii_1j1 is a Cayley subset of G if and only if i G S and |Xii_1j1| is twice of o( i ). For any two distinct i1 and i2 in S \ {1}, Cayley maps CM(G, Xili_1j1, pi1i_1j1) and CM(G, Xi2i_1j1, pi2i_1j1) are not isomorphic. Otherwise, according to Lemma 2.8, there exists some fi G Aut(G) such that b^ = b and for each I > 1, (b(_1)£ ai1-1+i1-2 + ^+il + 1)^ = b(_1)£ ai22-1+i22-2 + -+i2 + 1. In particular, (b_1a)^ = b_1a and therefore fi is the identical automorphism. Therefore, G has s non-isomorphic regular balanced Cayley maps. When p is a Fermat prime, then p — 1 is a power of 2, so G has exactly one regular balanced Cayley map up to isomorphism. □ 4 Regular balanced Cayley maps of Mp(n, m) For minimal non-abelian p-group, one may refter to [1, 2] or [14] for the following Lemma 4.1. Lemma 4.1 ([14, Theorem 2.3.6]). Let G be a finite p-group, d(G) be the number of elements in a minimal generating subset of G. Then, the followings are equivalent. (i) The group G is a minimal non-abelian group; (ii) d(G) = 2 and |G'| = p; (iii) d(G) = 2 and Z(G) = $(G), where $(G) denotes the Frattini subgroup of G. Lemma 4.2. Assume G is a finite p-group for some prime number p and d(G) = 2. Let P e Aut(G), g e G and X = g<^>. If G = (X), then G = (g, g^). Proof. Because d(G) = 2, it follows that G = G/$(G) = Zp x Zp. Suppose (g, g^) < G, then in the quotient group the subgroup generated by g and g^ has order p, that is |(g, g^)| = p. So, g^ e (g$(G)). As $(G) is a characteristic subgroup of G, for each k > 1 the element g^ e (g^- $(G)). Therefore, X C (g$(G)) and then (X) < (g$(G)) < G, a contradiction. So, G = (g, g^). □ Remark Lemma 4.2 may not be true for a non-p-group. For example, the symmetry group Sn can be generated by two elements (1 2) and (1 2 ... n). Take g = (12) e Sn and P the automorphism of Sn induced from the conjugation by the element (2 3 ... n), then X = = {(1 2), (13),..., (1 n)} is a Cayley subset of S„ and g^ = (1 3). But it is obvious that Sn = ((1 2), (1 3)) when n > 4. Theorem 4.3. Let G = Mp(n, n), where n > 2 and p is an odd prime number. Then, the group G does not have regular balanced Cayley maps. Proof. Let N = (x e G | xp" 1 e G'). According to Lemma 2.7, N is a characteristic subgroup of G. One can see from the defining relations of G that G' = (apn ) = Zp and N = (a,bp). Take a e Aut(G) such that a" = 6fcpai and b" = bsar, where the integers i, s, r satisfy the conditions in Lemma 2.3 and especially s = 1 (mod p). Suppose X = (b"avis a Cayley subset of G. Then b"av e N and therefore (u,p) = 1. In the quotient group G = G/N, X = (b"av)= b" is a Cayley subset of G. So, there exists some integer n such that b-" = bsn". As a result, one can get snu = — u (mod p). While (u,p) = 1, then sn = — 1 (mod p). But this result contradicts to s = 1 (mod p). □ Theorem 4.4. Let G = Mp(n, m), where n > 2, m > 1,m = n and p is an odd prime number. Then, the group G does not have regular balanced Cayley maps. Proof. We firstly assume m > n. Set N = {xp" | x e G}. By Lemma 2.7, N = (bp") is a characteristic subgroup of G. The quotient group G = G/N = (a,b | apn = bpn = 1, ab = a1+p"-1) = Mp(n,n). According to Theorem 4.3 and Lemma 2.5, G does not have regular balanced Cayley maps. When m < n, suppose there exists some a e Aut(G) such that X = (b"av is a Cayley subset of G. Because Z(G) = (ap, bp) is characteristic of G, one can assume u = 0, v = 1 from the results of Lemma 2.3 and Lemma 2.9. That is, X = a1". Assume a" = bja4, o(a) = 2k and t = ak, then aT = a-1, (bja*)T = a-ib-j. Recall that G' = (apn-1) = Zp and [a, bj] e G' < (a), so [a, bj]T = [a, bj]-1. While [a, bj]T = ([a, a®][a, bj][a,bj,a4])T = [a, bja4]T = [aT, (bja4)T] = [a-1,a-ib-j] = [a-1,b-j], and [a-1, b-j ] belongs to the center, the result [a, bj]T = [a-1, b-j] = b-ja-1[a-1, b-j]abj = [a, bj] . _i . 2 follows. Therefore, [a,bj] = [a,bj], that is [a,bj] = 1. But the order of [a,bj] is a power of p which is coprime with 2, we get [a, bj ] = 1. And from Lemma 4.2, one can get G = {a, a17} = {a, ba1}. So G is abelian, a contradiction. Thus in either case, G doesn't have regular balanced Cayley maps. □ Remark 4.5. In the paper of Newman and Xu ([8]), they claimed that for odd primes p every metacyclic p-group is isomorphic to one of the groups G = {a,b | aPr+s+u = 1, Vr+s+t = aPr+s ,b-1ab = a}, (4.1) where r, s, t, u are non-negative integers with r positive and u < r, and these groups are pairwise non-isomorphic. In the following Lemma 4.6, one will see that the metacyclic p-group has an 'intimate' connection with the minimal non-abelian metacyclic p-group. Lemma 4.6. Let G be a metacyclic p-group for some odd prime number p and N < G' be a maximal subgroup of the derived subgroup G'. Then N is a characteristic subgroup of G and the quotient group G = G/N is a minimal non-abelian metacyclic p-group. Proof. Because G' is cyclic and G' is characteristic of G, it follows that N is also characteristic of G. While N is a proper subgroup of G', the quotient group G = G/N is non-abelian and metacyclic, generated by two elements because G is generated by two elements. As G = G' = Zp and so |G | = p. The quotient group G is mininal non-abelian follows from Lemma 4.1. □ From the results of Lemma 2.5 and Theorems 4.3 and 4.4, we get the following Corollary 4.7. Corollary 4.7. For any odd prime number p, the metacyclic p-group does not have regular balanced Cayley maps. Theorem 4.8. Let G = M2(n, m), where m and n are positive integers and m > n > 2. Then G does not have regular balanced Cayley maps. Proof. According to Lemma 2.3, Aut(G) = [a | a7 = bjai, b7 = bsar}, where (is, 2) = 1, 1 < i < 2n, 1 < s < 2m, j = 2m-n+1k, 0 < k < 2n-1, 1 < r < 2n. From the defining relations of G, one can see that both a2 and b2 belong to the center of G. Set N = {a2, b4} = [x G Z(G) | x2m = 1}. By Lemma 2.7, N is a characteristic subgroup of Z(G). Since Z(G) is characteristic in G, N is characteristic in G. Suppose there is some a G Aut(G) and buav G G such that X = (buav)l<7> is a Cayley subset of G. By Lemma 2.9, one may assume u =1 and v = 0, that is, X = bl<7}. Assume a7 = ba1 and b7 = bsar, then 4 | j, (s, 2) = 1 and so s2 = 1 (mod 4). According to Lemma 4.2, G = {b, bsar} = {b, ar} and so (r, 2) = 1. In the quotient group G = G/N, X = b<7 should be a Cayley subset of G. Noticing that 2 | (s + i), 4 | j and G' < N, we have (bsar)7 = (bsar)s(bjai)r = bs2arsbjrair = bs2+jrar(s+{) = bs2. Since o(b) =4 and s2 = 1 (mod 4), we have bs2 = b. So, X = {b, bsar }. But (r, 2) = 1, b G X. Then, X is not a Cayley subset, a contradiction. □ Theorem 4.9. Let G = M2(n, m), where m and n are positive integers, n > m +1 and m > 2. Then G does not have regular balanced Cayley maps. Proof. In this case, Aut(G) = {a | aCT = Va4, bCT = bsar}, where (is, 2) = 1, 1 < i < 2n, 1 < s < 2m, 1 < j < 2m, r = k2n-m, 0 < k < 2m. Let N = (a4, b2) = {x G Z(G) | x2" =1}. According to Lemma 2.7, N is characteristic in Z(G). Since Z(G) is characteristic in G, N is characteristic in G. Similar to the proof of Theorem 4.8, we only need to show that X = al is a Cayley subset of G = G/N. While from 2 | (s + i), 4 | r and G' < N, we have (bja4)CT = (bsar)j (bja4)4 = bsjarjbjiai2 = bj(s+i)ai2+rj = ai2. And from o(a) = 4, i2 = 1 (mod 4), we have a42 = a. So, X = {a, b^a1}. But (j, 2) = 1 implies a -1 G X .So, X is not a Cayley subset, a contradiction. □ In Theorem 4.9, if we allow m =1 and so n > 2, then the group M2(n, 1) belongs to one of the p-groups with a cyclic maximal subgroup which had been considered by D. D. Hou, Y. Wang and H. P. Qu in [6]. We list the result in the following theorem. Theorem 4.10 ([6, Theorem 3.3]). For positive integers n > 2, M2(n, 1) does not have regular balanced Cayley maps. Now, there are still two cases about which we have not said anything, that is M2 (n, n) for n > 2 and M2 (n + 1, n) for n > 1. One may look back at Lemma 2.3 and can easily see that the automorphism groups of both M2 (n, n) and M2 (n + 1, n) are 2-groups. Theorem 4.11. Let G = M2(n, n), n > 2. Then G has exactly one regular balanced Cayley map of valency 4 in the sense of isomorphism. Proof. By Lemma 2.3, Aut(G) = {a | aCT = b2fca4,bCT = bsar}, where (si, 2) = 1, 1 < i, s, r < 2n, 1 < k < 2n-1, and both a2 and b2 belong to Z(G). We firstly show that if for some g G G and a G Aut(G), X = is a Cayley subset of G, then |X| = 4. Set N = {x G G | x2"-2 G G'}. According to Lemma 2.7, N is a characteristic subgroup of G and N = (a2, b4). Without loss of generality, we assume g = b, then in the quotient group G = G/N = Z2 x Z4, the order of b is 4. While there are exactly four order-4 elements in G and X = is a Cayley subset of G, X should contain all these four elements. Because the order of a is a power of 2 and b is not involution, according to the results in Lemma 2.5, we have |X| = |X| = 4. Take a1 G Aut(G) such that aCTl = b2a-1 and bCTl = ba2" -1. By a direct calculation, Xi = b = {b, ba2"-1-1, b-1, (ba2"-1-1)-1} is clearly a Cayley subset of G. For any a2 G Aut(G) such that aCT2 = b2ka4, bCT2 = bsar, where k, i, s, r satisfy the conditions listed in the first paragraph, and X2 = b^2^ = {b, bsar, b-1, (bsar)-1} is a Cayley subset of G, one may take t g Aut(G) such that aT = b1-sa-r(1+2" ) and bT = b. It is easy to check that (ba2" -1)T = bsar. Therefore, by Lemma 2.8, the two regular balanced Cayley maps CM(G, X1, a1 |Xl) and CM(G, X2,a2|X2) are isomorphic. So, G has exactly one regular balanced Cayley map of valency 4 in the sense of isomorphism. □ Theorem 4.12. Let G = M2(n + 1, n), n > 1. Then G has exactly one regular balanced Cayley map up to isomorphism and this map is ofvalency 4. Proof. By Lemma 2.3, Aut(G) = [a | oa = bja4, bCT = bsa2k}, where (si, 2) = 1, 1 < i < 2n+1, 1 < j, s, k < 2n and both a2 and b2 belong to Z(G). We firstly show that if g G G, a G Aut(G) and X = gl = [a, b-1 a, a-1, (b-1a)-1} is a Cayley subset of G. For any a2 G Aut(G) such that aa2 = Wa4, bCT2 = bsa2k, where j, i, s, k satisfy the conditions listed in the first paragraph, and Y2 = oSa2 = [a, Va4, a-1, (bja4)-1} is a Cayley subset of G, one may take t g Aut(G) such that aT = a and bT = b-ja1-i. It is easy to check that (b-1a)T = bja4. Therefore, the two regular balanced Cayley maps CM(G, Y1,a1|Yl) and CM(G, Y2,a2|Y2) are isomorphic and so G has only one regular balanced Cayley map of valency 4 in the sense of isomorphism. □ To be more clear, we list the number of non-isomorphic regular balanced Cayley maps of minimal non-abelian metacyclic groups in Table 1. For brevity, we use |G|, N, RBCM and MNAMG to denote the order of group G, the number of regular balanced Cayley maps up to isomorphism, regular balanced Cayley maps and minimal non-abelian metacyclic groups, respectively. Table 1: Number of RBCM of MNAMG. G |G| N 1 Q8 8 1 2 Mp,2(1,r) = D2p 2p p — 1 3 Mp,2(m, r), m > 2, p - 1 = 2es, (s, 2) = 1 2mp s 4 (m, r), q = 2 pqm 0 5 M2(2, 1) = D8 8 2 6 M2(n, 1), n > 2 2n+i 0 7 M2(n,n), n > 2 22n 1 8 M2(n + 1, n), n > 2 22n+1 1 9 M2 (n, m), m = n and m = n — 1 2n+m 0 10 Mp(n, m), p = 2 pn+m 0 References [1] Y. Berkovich, Groups of Prime Power Order, Volume 1, volume 46 of De Gruyter Expositions in Mathematics, Walter de Gruyter GmbH & Co., Kammergericht, Berlin, 2008, https: //www.degruyter.com/view/product/178 995. [2] Y. Berkovich and Z. Janko, Groups of Prime Power Order, Volume 2, volume 47 of De Gruyter Expositions in Mathematics, Walter de Gruyter GmbH & Co., Kammergericht, Berlin, 2008, https://www.degruyter.com/view/product/3 9455. [3] Z. M. Chen, Inner and OuterS-Groups and Minimal Non-T,-Groups, Southwest University Publishing House, Chongqing, 1988. [4] M. Conder, R. Jajcay and T. Tucker, Regular Cayley maps for finite abelian groups, J. Algebraic Combin. 25 (2007), 259-283, doi:10.1007/s10801-006-0037-0. [5] M. Golasinski and D. L. Goncalves, On automorphisms of split metacyclic groups, Manuscripta Math. 128 (2009), 251-273, doi:10.1007/s00229-008-0233-4. [6] D. D. Hou, Y. Wang and H. P. Qu, Regular balanced Cayley maps of p-groups with a cyclic maximal subgroup, submitted. [7] G. A. Miller and H. C. Moreno, Non-abelian groups in which every subgroup is abelian, Trans. Amer. Math Soc. 4 (1903), 398-404, doi:10.2307/1986409. [8] M. F. Newman and M. Y. Xu, Metacyclic groups of prime-power order, Adv. Math. (China) 17 (1988), 106-107, doi:10.11845/sxjz.1988.17.01.0106. [9] J.-M. Oh, Regular t-balanced Cayley maps on semi-dihedral groups, J. Comb. Theory Ser. B 99 (2009), 480-493, doi:10.1016/j.jctb.2008.09.006. [10] R. B. Richter, J. Siran, R. Jajcay, T. W. Tucker and M. E. Watkins, Cayley maps, J. Comb. Theory Ser. B 95 (2005), 189-245, doi:10.1016/j.jctb.2005.04.007. [11] M. Skoviera and J. Siran, Regular maps from Cayley graphs, Part 1: Balanced Cayley maps, Discrete Math. 109 (1992), 265-276, doi:10.1016/0012-365x(92)90296-r. [12] Y. Wang and R. Q. Feng, Regular balanced Cayley maps for cyclic, dihedral and generalized quaternion groups, Acta Math. Sinica 21 (2005), 773-778, doi:10.1007/s10114-004-0455-7. [13] M. Y. Xu, Introduction to Group Theory I, Science Publishing House, Beijing, 1999. [14] M. Y. Xu and H. P. Qu, Finite p-Group, Peking University Press, Beijing, 2010. ARS MATHEMATICA CONTEMPORANEA ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 14 (2018) 445-454 https://doi.org/10.26493/1855-3974.1107.1c8 (Also available at http://amc-journal.eu) A note on extremal results on directed acyclic graphs Alvaro Martinez-Perez * Departamento de Análisis Económico y Finanzas, Universidad de Castilla-La Mancha, Avda. Real Fabrica de Seda, s/n. 45600 Talavera de la Reina, Toledo, Spain Luis Montejano Deborah Oliveros * Instituto de Matematicas, Universidad Nacional Autonoma de México, Area de la Investigation Científica, Circuito Exterior, C.U., Coyoacan 04510, Mexico D.F., Mexico Received 17 May 2016, accepted 31 May 2017, published online 6 February 2018 This paper studies the maximum number of edges of a Directed Acyclic Graph (DAG) with n vertices in terms of it's longest path £. We prove that in general this number is the Turan number t(n, l +1), the maximum number of edges in a graph with n vertices without a clique of size £ +2. Furthermore, we find the maximum number of edges in a DAG which is either reduced, strongly reduced or extremely reduced and we relate this extremal result with the family of intersection graphs of families of boxes with transverse intersection. Keywords: Directed graphs, Turan numbers, intersection graphs of families of boxes. Math. Subj. Class.: 05C20, 52C99 1 Introduction One of the fundamental results in extremal graph theory is the Theorem of Turan (1941) which states that a graph with n vertices that has more than t(n, k) edges, will always contain a complete subgraph of size k + 1. The Turan graph T(n, k), is a k-partite graph on n vertices whose partite sets are as nearly equal in cardinality, and has the property * Partially supported by MTM 2015-63612P. t Supported by CONACyT 166306. * Partially supported by PAPIIT 104915 and CONACyT 166306. E-mail address: alvaro.martinezperez@uclm.es (Alvaro Martinez-Perez), luis@matem.unam.mx (Luis Montejano), dolivero@matem.unam.mx (Deborah Oliveros) Abstract ©® This work is licensed under http://creativecommons.org/licenses/by/3.0/ that contains the maximum posible number of edges t(n, k) of any graph not containing a clique of size k + 1. It is known that t(n, k) < (1 - 2)\, and equality holds if k divides n. In fact, tnfr = 1 - m. See [1]. Turan numbers for several families of graphs have been studied in the context of extremal graph theory, see for example [3] and [4]. In ([2, 7]) the authors analyze, among other things, the intersection graphs of boxes in Rd proving that, if T(n, k, d) denotes the maximal number of intersection pairs in a family F of n boxes in Rd with the property that no k + 1 boxes in F have a point in common (with n > k > d > 1), then T(n, k, d) = T(n - k + d, d) + T(n, k - d +1,1), with T(n, k, 1) = (n2) - (n-2+1) being the precise bound in dimension 1 for the family of interval graphs. Turan numbers have played and important role for several variants of the Turan Theorem and its relation with the fractional Helly Theorem (see [5, 6]). The purpose of this paper is to study the maximum number of edges in directed acyclic graphs with n vertices with respect to it's longest path. That turns out to be related with the extremal behavior of the family of intersection graphs for a collection of boxes in R2 with transverse intersection. The first result, Theorem 2.10, states that in a directed acyclic graph with n vertices, if the longest path has length then the maximal number of edges is the Turan number t(n, I +1). Theorem 3.19 and its Corollaries state that given a directed acyclic graph G with n vertices such that the longest path has length I, then if G is either reduced, strongly reduced or extremely reduced, G has at most t(n - I + 1,2) + T(n, 1) edges, where again T(n, 1) denotes the maximal number of intersecting pairs in a family F of n intervals in R with the property that no I +1 intervals in F have a point in common. In fact, this bound is best possible. The bound is reached by the intersection graph of a collection of boxes in R2 with transverse intersection (see Proposition 4.6). This graph is extremely reduced (and thus is also strongly reduced and reduced, see Proposition 4.4). 2 Directed acyclic graphs By a directed acyclic graph, DAG, we mean a simple directed graph without directed cycles. A DAG, G = (V, 5), with vertex set V and directed edge set E is transitive if for every x, y, z G V, if {x, y}, {y, z} G E then {x, z} G 5. Definition 2.1. A topological order of a directed graph G is an ordering O of its vertices {v1, v2,..., vn} so that for every edge {vj, Vj} then i < j. The following proposition is a well known result: Proposition 2.2. A directed graph G is a DAG if and only if G has a topological order. Given any set X, by |X | we denote the cardinal of X. Definition 2.3. The indegree, deg- (v), of a vertex v is the number of directed edges {x, v} with x G V. The outdegree, deg+ (v), of a vertex v is the number of directed edges {v, x} with x g V. Notice that each directed edge {v, w} adds one outdegree to the vertex v and one indegree to the vertex w. Therefore, J2veV deg+ (v) = J2veV deg- (v) = | (5) |. The degree of a vertex is deg(v) = deg (v) + deg+(v). Definition 2.4. A vertex v such that deg (v) =0 is called source. A vertex v such that deg+(v) =0 is called sink. It is well known that every DAG G has at least one source and one sink. Definition 2.5. Given a DAG, G = (V, £), a directed path 7 in G is a sequence of vertices {v0,..., vn} such that {vi-1, v4} G £ for every 1 < i < n. Here, 7 has length n, and endpoint vn. Observe that since DAG's are acyclic, all the vertices on a directed path are different. Definition 2.6. Given a DAG, G = (V, £), let r: V ^ N be such that r(v) = k if there exists a directed path 7 in G of length k with endpoint v and there is no directed path 7' with endpoint v and length greater than k. Definition 2.7. Given a DAG, G = (V, £) suppose that i = max{k | r(v) = k for every v G V}. Notice that, since G has no directed cycle, i < |V|. Then, let us define a partition Pr = {V0,..., V£} of V such that V := {v G V | r(v) = i} for every 0 < i < i. Notice that V0 is exactly the set of sources in G and V is contained in the set of sinks in G. Lemma 2.8. V is nonempty for every 0 < i < i. Proof. Let {v0,..., v^} be a directed path of maximal length in G. Clearly, for every 0 < i < i, vj G V if j < i. Suppose vj G Vj with i < j < i. Then, there is a directed path {v0,..., vj = vj} with j > i and {v0,..., vj, vj+1,..., v^} is a directed path with length j + l - i > i which contradicts the hypothesis. □ Lemma 2.9. The induced subgraph with vertices Vj, G[Vj], is independent (has no edges) for every i. Proof. Let vj, vj G Vj and suppose {vj, v'} G £. Let {v0,..., vj} be apath of length i with endpoint v j. Then, {v0,..., v j, v'} defines a directed path of length i +1 which contradicts the fact that v' G Vj. □ Recall that T(n, i) denote the i-partite Turan graph with n vertices and t(n, i) denote the number of edges of T(n, i). Theorem 2.10. Let G = (V, £) be a DAG with n vertices such that the longest directed path has length i. Then, G has at most t(n, i + 1) edges. Proof. Consider the partition Pr = {V0,..., V^} of V. By Lemma 2.9, this defines an (i+1) -partite directed graph. Thus, neglecting the orientation we obtain a complete (i+1) -partite graph with partition sets V0,..., V0 Therefore, the number of edges is at most t(n,i +1). □ Remark 2.11. It is readily seen that the bound in Theorem 2.10 is best possible. Consider the Turan graph T(n, i +1) and any ordering of the i +1 independent sets V0,..., V0 Then, for every edge {vj, vj} in T(n, i) with vj G Vj, vj G Vj- and i < j let us assume the orientation {vj, vj}. It is trivial to check that the resulting graph is a DAG with t(n, i +1) edges. 3 Reduced, strongly reduced and extremely reduced DAGs Let O be a topological ordering in a DAG G. Given any two vertices v, w, and two directed paths in G, 7,7', from v to w, let us define 7 UO 7' as the sequence of vertices defined by the vertices in 7 U 7' in the order given by O. Of course, this need not be, in general, a directed path from v to w. Let r(u, v) be the set of all directed paths from u to v. Let UO{7 | 7 € r(u, v)} represent the sequence of all the vertices from the paths in r(u, v) ordered according to O. Definition 3.1. A finite DAG G is strongly reduced if for any topological ordering O of G, every pair of vertices, v, w, and every pair of directed paths, 7,7', from v to w, then 7 UO 7' defines a directed path from v to w. Remark 3.2. Let G be DAG. Given any two vertices v, w, and two directed paths in G, 7,7', from v to w, let us define 7 < 7' if every vertex in 7 is also in 7'. Clearly, "<" is a partial order. Definition 3.3. A vertex w is reachable from a vertex v if there is a directed path from v to w. Proposition 3.4. Given a finite DAG G = (V, E), the following properties are equivalent: i) For every pair of vertices v, w and every pair of paths, 7,7', from v to w, there exists a directed path from v to w, 7'', such that 7,7' < 7''. ii) For every pair of vertices v, w such that w is reachable from v, there is a directed path from v to w, ym , such that for every directed path, 7, from v to w, 7 < ym . iii) For every topological ordering O of G and any pair of vertices v, w, UO {7 | 7 € r(u, v)} defines a directed path from v to w. Proof. Since the graph is finite and the relation '<' is transitive, i) and ii) are trivially equivalent. If ii) is satisfied, then it is trivial to see that UO{7 | 7 € r(u, v)} = ym and iii) is satisfied. Also, it is readily seen that iii) implies ii) taking ym := UO{7 | 7 € r(u, v)}. □ Definition 3.5. We say that a finite DAG G is reduced if it satisfies any of the properties from Proposition 3.4. Proposition 3.6. If a finite DAG G is strongly reduced, then G is reduced. Proof. Since the graph is finite, it is immediate to see that being strongly reduced implies iii). □ Remark 3.7. The converse is not true. The graph in the left from Figure 1 is clearly reduced. Notice that the directed path ym := {vi, v2, v3, v4, v5} is an upper bound for every directed path from v1 to v5. However, if we consider the directed paths 7 = {v1, v2, v5} and 7' = {v1, v4, v5} with the topological order O = {v1, v2, v3, v4, v5}, then 7 UO 7' = {v1, v2, v4, v5 } which is not a directed path. Figure 1: Being reduced does not imply being strongly reduced and being strongly reduced does not imply being extremely reduced. Definition 3.8. Given a finite DAG G and a vertex v G V we say that w is an ancestor of v if there is a directed path {w = v0,..., vk = v} and w is a descendant of v if there is a directed path {v = v0,..., vk = w}. Definition 3.9. We say that a finite DAG G is extremely reduced if for every pair of non-adjacent vertices x, y, if x, y have a common ancestor, then they do not have a common descendant. Proposition 3.10. If a DAG G = (V, £) is extremely reduced, then it is strongly reduced. Proof. Let 7 = {v, vi,..., vn, w} and 7' = {v, w0,..., wm, w} be two directed paths in G from v yo w. Let O be any topological order in G and consider 7UO 7' = {v, z1,..., zk, w}. First, notice that z1 is either v1 or w1. Therefore, {v, z1} G £. Also, zk is either vn or wm, and {zk, w} G £. Now, for every 1 < i < k, let us see that {zi_1, zj G £. If zi-1,zi G 7 or zi_1, zi G 7', then they are consecutive vertices in a directed path and we are done. Otherwise, since zi-1, zi have a common ancestor v and a common descendant w, then there is a directed edge joining them and, since zi-1, zi are sorted by a topological order, {zi-1, zi} G £. □ Remark 3.11. The converse is not true. The graph in the right from Figure 1 is strongly reduced. However, vertices w2 and w4 are not adjacent and have a common ancestor and a common descendent. Proposition 3.12. If G is transitive, then the following properties are equivalent: • G is extremely reduced, • G is strongly reduced, • G is reduced. Proof. By Proposition 3.10 if G is extremely reduced, then it is strongly reduced. By Proposition 3.6, if G is strongly reduced, then it is reduced. Suppose G is reduced and suppose that two vertices x, y have a common ancestor, v, and a common descendant, w. Then, there are two directed paths 7,7' from v to w such that x G y and y G 7'. By property i) in Proposition 3.4, there exists a path 7'' in G from v to w such that 7,7' < 7''. Inparticular, x, y G 7''. Therefore, either x is reachable from y or y is reachable from x in G. Since G is transitive, this implies that x, y are adjacent. Therefore, G is extremely reduced. □ Definition 3.13. Given a DAG G = (V, ?), the graph with vertex set V and edge set 5' := 5 U {{v, w} | w is reachable from v} is called the transitive closure of G, T[G]. It is immediate to check the following: Proposition 3.14. Given any DAG G, T [G] is transitive. Proposition 3.15. If a DAG G is reduced, then the transitive closure T [G ] is also reduced. Proof. Suppose G satisfies i) in Proposition 3.4 and let 7 = {v = v0,..., vn = w}, 7' = {v = w0,..., wm = w} be any pair of paths from v to w in T[G]. Therefore, vj is reachable from vj_i in G for every 1 < i < n and wj is reachable from wj_i in G for every 1 < i < m. Thus, there exist a path y0 in G such that 7 < y0 and a path 7O in G such that 7' < y0 . By property i), there is a directed path from v to w such that 70,7O < y0'. Therefore, 7,7' < y0' and T[G] satisfies i). □ Then, from Propositions 3.6, 3.10, 3.12, 3.14 and 3.15: Corollary 3.16. If a DAG G is reduced, then the transitive closure T [G] is extremely reduced and strongly reduced. In particular, if G is extremely reduced or strongly reduced, then T [G] is extremely reduced and strongly reduced. Let us recall that t(„,U)=(;;) - (n-2+1)=(n - <+i)(i - D+i^)^ (3.1) As it was proved in [7]: Lemma 3.17. For n > i and d > 1, T(n + d, i, 1) - T(n, i, 1) = d(i - 1). In particular, T(n + 2, i, 1) - T(n, i, 1) = 2(i - 1). Also, from [7]: Lemma 3.18. For 1 < d < n, t(n + d, d) - t(n, d) = (d - 1)n + ^ In particular, t(n + 2, 2) - t(n, 2) = n +1. Theorem 3.19. Let G = ( V, ?) be a DAG with n vertices and such that the longest directed path has length i > 1. If G is extremely reduced, then G has at most t(n - i +1, 2) + T(n, i, 1) edges. Proof. Let us prove the result by induction on n. Suppose that the longest directed path has length £. First, let us see that the result is true for n = £ +1 and n = £ + 2. If n = £ +1 then G has at most = (£-2)2(£-1) + 2(£ - 1) + 1 = T(n, £, 1) + t(n - £ +1,2) edges. The last equation follows immediately from (3.1) and the fact that t(2, 2) = 1. If n = £ +2 then there are £ +1 vertices which define a directed path 7 = {v0,..., v^} and one vertex w such that neither {w, v0} nor {v^, w} is a directed edge. Then, the partition Pr = {V0,..., V^} of G satisfies that vj G Vj for every 0 < i < £. Also, w G Vj for some 0 < j < £ and {w, vj}, {vj, w} are not directed edges. Hence, deg(w) < £. Therefore, G has at most ^f11 + £ = (£-2)2(£-1) + 3(£ - 1) + 2 = T(n, £, 1) + t(n - £ + 1, 2) edges. The last equation follows immediately from (3.1) and the fact that t(3,2) = 2. Suppose the induction hypothesis holds when the graph has n vertices and let #(V) = n + 2. Also, by Proposition 3.15 we may assume that the graph is transitive. Consider the partition Pr = {V0,..., V^} of V. Let #(Vj) = r». Let v G V0 and w be any sink of G. Consider any pair of vertices vj, v- G Vj. Since G is extremely reduced and every two vertices in Vj are non-adjacent, v», vj can not be both descendants of v and ancestors of w simultaneously. Hence, the number of edges joining the sets {v, w} and Vj are at most rj + 1. Therefore, there are at most n + £ - 1 edges joining {v, w} and G \ {v,w} Since G \ {v, w} has n vertices, by hypothesis, it contains at most t(n - £ +1, 2) + T(n, £, 1) edges. Finally, there is at most 1 edge in the subgraph induced by {v, w}. Therefore, by Lemmas 3.17 and 3.18, (G) | < t(n - £ +1, 2) + T(n, £, 1) + n + £ = t(n - £ +3, 2)+ T(n + 2, £, 1). □ By Corollary 3.16 we know that the extremal graph for reduced and strongly reduced graphs is transitive. Thus, from Theorem 3.19 and Proposition 3.12 we obtain the following. Corollary 3.20. Let G = (V, 5) be a DAG with n vertices and such that the longest directed path has length £ > 1. If G is reduced, then G has at most t(n - £ +1, 2) + T(n, £, 1) edges. Corollary 3.21. Let G = (V,5) be a DAG with n vertices and such that the longest directed path has length £ > 1. If G is strongly reduced, then G has at most t(n - £ + 1, 2) + T(n, £, 1) edges. 4 Directed intersection graphs of boxes Definition 4.1. Let R be a collection of boxes with parallel axes in R2. Let G = (V,5) be a directed graph such that V = R and given R, R' gR with R = I x J, R' = I' x J' then {R, R'} G 5 if and only if I c I' and J' C J (i.e. there is an edge if and only if the intersection is transverse and the order is defined by the subset relation in the first coordinate). Let us call G the directed intersection graph of R. Definition 4.2. Let R be a collection of boxes with parallel axes in R2. We say that R is a collection with transverse intersection if for every pair of boxes either they are disjoint or their intersection is transverse. R=|xJ R'=I J' I Figure 2: The transverse intersection above induces a directed edge {R, R'}. J Proposition 4.3. Let R be a collection of boxes with parallel axes in R2 and G be the induced directed intersection graph. If two vertices v, w have both a common ancestor and a common descendant in G, then the corresponding boxes Rv, Rw intersect. Proof. Let a be a common ancestor and Ra = Ia x Ja be the corresponding box. Let b be a common descendant and Rb = Ib x Jb be the corresponding box. Then if Rv = Iv x Jv, Rw = Iw x Jw are the boxes corresponding to v and w respectively, it follows by construction that Ia C Iv ,Iw and Jb C Jv, Jw. Therefore, Ia x Jb c Rv ,Rw and Rv n Rw = 0. □ Proposition 4.4. If R is a collection of boxes with parallel axes in R2 with transverse intersection, then the induced directed intersection graph G is extremely reduced and transitive. Proof. First notice that the transitivity holds simply by the transverse intersection property. Let v, w be two vertices such that there is no edge joining them. This means, by construction, that their corresponding boxes do not have a transverse intersection. Since R has transverse intersection, this implies that these boxes do not intersect. Thus, by Proposition 4.3, if v, w have a common ancestor, then they can not have a common descendant. □ Remark 4.5. Consider the bipartite graph G from Figure 3 with the partition given by {letters, numbers} and assume all directed edges go from letters into numbers. Note that G is extremely reduced, transitive and acyclic. Notice that the induced subgraphs given by the sets Ci := {1, 2, A, B}, C2 := {3,4, C, D} and C3 := {5, 6, E, F} are three cycles of length 4. Furthermore the induced subgraph given by the set of vertices {1, 2,3,4,8, 9, A, B, C, D, H, I} is realizable as boxes in R2 (see Figure 4) note, that contains C1 and C2 and its realization force one of them to be inside the other say C1 inside C2. Similarly the induced subgraphs given by the set of vertices {1, 2, 5,6, A, B, E, F, 7,12, G, L} and the set of vertices {3,4,5,6, C, D, E, F, 10,11, J, K} forces necessarily a system of tree squares one inside the other. However, intervals given by {7,8,9,10,11,12} and {G, H, I, J, K, L} are forced to have more intersections that those given by the graph. In Figure 3: The bipartite, transitive, and extremely reduced DAG, G with partition given by {letters, numbers} and edges directed from letters into numbers. This graph is not realizable as a family of boxes in R2. other words, there is no family of boxes (or intervals) that realizes such a graph or for which it is induced the graph G. Then, the converse of Proposition 4.4 is not true. - 3 — H — 9 1 1 1 D 8 1 C 1 2 1 A I B 4 Figure 4: Realization in R2 of the induced subgraph with vertices {1,2,3,4, 8,9, A, B, C, D, H, I} of the graph shown in Figure 3. Let G[r, l, s] be the graph, G(V, E), such that: • V = {xi, . . . ,Xr ,yi, . . . , yi-1 ,Z1, . . . ,Zs} • {xj, xj} E for any i = j, • {zi,Zj} i Efor any i = j, • {xi,yj} i Efor every i,j, • {yi, yj } i Efor every i < j, • {yi,Zj } i Efor every i,j, • {xi, Zj} i E for every i, j. This is the directed intersection graph from the collection of boxes in Figure 5. C, Cm 1 | 1 | B B Vo V, V,., V, Figure 5: The graph G[r, l, s] corresponds to the directed intersection graph of the collection in the figure where Xj ~ Aiy yj ~ Cj and zk ~ Bk. Notice that the graph is transitive although not every edge is represented in the figure. By Proposition 4.4, G[r, l, s] is a transitive extremely reduced DAG. In particular, G[r,l, s] is strongly reduced and reduced. Now, to prove that the bound obtained in Theorem 3.19 and its corollaries is best possible, it is immediate to check the following: Proposition 4.6. If n -t is even, G[,t, ] has t(n -t +1,2) + T(n, t, 1) edges. If n - t is odd, G[,t, ] has t(n - t+ 1, 2) + T(n, t, 1) edges. References [1] M. Aigner and G. M. Ziegler, Proofs from THE BOOK, Springer-Verlag, Berlin, 1998, doi: 10.1007/978-3-662-22343-7. [2] I. Barany, F. Fodor, A. Martinez-Perez, L. Montejano, D. Oliveros and A. P6r, A fractional Helly theorem for boxes, Comput. Geom. 48 (2015), 221-224, doi:10.1016/j.comgeo.2014.09.007. [3] B. Bollobas, Extremal Graph Theory, Dover Publications, Mineola, New York, 2004. [4] R. Diestel, Graph Theory, volume 173 of Graduate Texts in Mathematics, Springer-Verlag, Berlin, 3rd edition, 2005, http://diestel-graph-theory.com/. [5] G. Kalai, Intersection patterns of convex sets, Israel J. Math. 48 (1984), 161-174, doi:10.1007/ bf02761162. [6] M. Katchalski and A. Liu, A problem of geometry in Rn, Proc. Amer. Math. Soc. 75 (1979), 284-288, doi:10.2307/2042758. [7] A. Martinez-Perez, L. Montejano and D. Oliveros, Extremal results on intersection graphs of boxes in Rd, in: K. Adiprasito, I. Barany and C. 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By mail: Ars Mathematica Contemporanea UP FAMNIT Glagoljaska 8 SI-6000 Koper Slovenia By fax: +386 5 611 75 71 By e-mail: info@famnit.upr.si Special Issue of ADAM on Symmetries of Graphs and Networks - Call for Papers This is a call for submission of papers for a special issue of the journal The Art of Discrete and Applied Mathematics (ADAM), on topics presented or related to talks given at the TSIMF workshop on 'Symmetries of Graphs and Networks' held at Sanya (China) in January 2018. The Sanya workshop added to the series of conferences and workshops on symmetries of graphs and networks initiated at BIRS (Canada) in 2008 and progressed in Slovenia every two years from 2010 to 2016. The Art of Discrete and Applied Mathematics (ADAM) is a modern, dynamic, platinum open access, electronic journal that publishes high-quality articles in contemporary discrete and applied mathematics (including pure and applied graph theory and combinatorics), with no costs to authors or readers. 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