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Contents
Generalized Cayley maps and their Petrie duals
Robert Jajcay, Jozef Širáň, Yan Wang . . . . . . . . . . . . . . . . . . . . 385
On p-gonal fields of definition
Ruben A. Hidalgo . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405
An online bin-packing problem with an underlying ternary structure
Helmut Prodinger . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415
Variants of the domination number for flower snarks
Ryan Burdett, Michael Haythorpe, Alex Newcombe . . . . . . . . . . . . . 427
Optimal strategies in fractional games: vertex cover and domination
Csilla Bujtás, Günter Rote, Zsolt Tuza . . . . . . . . . . . . . . . . . . . . 453
The Sierpiński domination number
Michael A. Henning, Sandi Klavžar, Elżbieta Kleszcz, Monika Pilśniak . . 473
On the minisymposium problem
Peter Danziger, Eric Mendelsohn, Brett Stevens, Tommaso Traetta . . . . . 489
Two kinds of partial Motzkin paths with air pockets
Jean-Luc Baril, Paul Barry . . . . . . . . . . . . . . . . . . . . . . . . . . 509
Independent coalition in graphs: existence and characterization
Mohammad Reza Samadzadeh, Doost Ali Mojdeh . . . . . . . . . . . . . . 535
Spectra of signed graphs and related oriented graphs
Zoran Stanić . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 553
Volume 24, Number 3, Fall/Winter 2024, Pages 385–566
xix

ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 24 (2024) #P3.01 / 385–403
https://doi.org/10.26493/1855-3974.2408.42e
(Also available at http://amc-journal.eu)
Generalized Cayley maps and their Petrie duals*
Robert Jajcay †
Comenius University, Bratislava, Slovakia
Jozef Širáň ‡
Slovak University of Technology, Bratislava, Slovakia
Yan Wang §
Yan Tai University, Yan Tai, China
Received 12 August 2020, accepted 22 October 2023, published online 6 June 2024
Abstract
Cayley maps are embeddings of Cayley graphs in orientable surfaces which possess a
group of orientation preserving automorphisms acting regularly on the vertices. We gen-
eralize the concept of a Cayley map by considering embeddings of Cayley graphs in both
orientable and non-orientable surfaces and by requiring a group of automorphisms acting
regularly on vertices that does not have to consist entirely of orientation preserving auto-
morphisms. This leads to new families of maps in both the orientable and non-orientable
cases. Since the Petrie dual operator preserves the property of being a generalized Cayley
map, throughout the paper we consider the action of this operator on our maps.
Keywords: Cayley map, regular embedding, automorphism group of a map.
Math. Subj. Class. (2020): 05C10, 05C25, 05E18
*We wish to thank our referees for their repeated careful readings of our article and their helpful and knowl-
edgeable suggestions that helped considerably in improving our article and making it more focused.
†Corresponding author. Supported by VEGA Research Grant 1/0437/23 and APVV Research Grant 19-0308.
‡Supported by APVV Research Grants 19-0308 and 22-0005, and VEGA Research Grants 1/0567/22 and
1/0069/23.
§Supported by NSFS No. ZR2020MA044.
E-mail addresses: robert.jajcay@fmph.uniba.sk (Robert Jajcay), jozef.siran@stuba.sk (Jozef Širáň),
wang−yan@pku.org.cn (Yan Wang)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
386 Ars Math. Contemp. 24 (2024) #P3.01 / 385–403
1 Introduction
A map is a cellular embedding of a connected graph on some surface; the map is orientable
if the surface is. By a widely adopted definition (cf. [16]), an orientable map M is a Cayley
map if the group of orientation-preserving automorphisms of M contains a subgroup acting
regularly on the vertex set of M. This way of introducing Cayley maps has been motivated
by Sabidussi’s well known characterization of Cayley graphs in terms of a subgroup of
automorphisms of the graph acting regularly on its vertex set [18]. Apart from the history
of the development of the theory of maps, however, when carrying Sabidussi’s characteri-
zation over to maps there is no reason to restrict to orientation-preserving automorphisms,
or even to orientable surfaces. The latter may have been first realized by Tucker [19], and
in this paper we argue for the return of his way of treating Cayley maps.
In order not to clash with customary terminology we will say that a map M on an
arbitrary surface (orientable or not) is a generalized Cayley map if the automorphism group
of M contains a subgroup H that is a regular permutation group on the vertex set of M.
In our view, Tucker’s definition by means of existence of a regular action of a group of
automorphisms has two major advantages: it is a generalization of the currently used defi-
nition that is simple, and in line with the original definition of a Cayley map. To the best of
our knowledge there have been two other attempts at defining non-orientable Cayley maps:
the one by Abas [1] via strategically placed cross-cups on the carrier surface, and another
one by Kwak and Kwon [12] who developed a full theory of generalized Cayley maps
(equivalent to Tucker’s and our approach) but starting off from introducing generalized
Cayley maps in a way that appears to be hard to use in applications.
The main difference between (original and orientable) Cayley maps and generalized
Cayley maps is that the carrier surfaces of the latter may be non-orientable, and even in
the orientable case the subgroup required to act regularly on vertices does not have to be
orientation-preserving. Maps admitting such subgroups appear frequently in the context of
regular maps with automorphism groups acting quasi-primitively on the vertices [6], and
our study of such maps was the main source of motivation for the present paper.
At several places, we employ the Petrie dual operator (e.g., [22]) which proves par-
ticularly useful since it preserves the class of generalized Cayley maps. For example, we
consider orientable generalized Cayley maps whose Petrie dual is again an orientable gen-
eralized Cayley map, as well as non-orientable generalized Cayley maps whose Petrie dual
is orientable.
The outline of the paper is as follows. In order to emphasize parallels between the
original Cayley maps on orientable surfaces and the generalized Cayley maps, the three
sections following this Introduction are devoted to reviewing the background and context
of highly symmetric maps, Cayley maps, and the Petrie dual operator; with Sections 5 and 6
containing the new results. We begin in Section 2 by reviewing some of the most important
properties of regular maps, followed in Section 3 by reviewing algebraic machinery for
dealing with orientable Cayley maps in their original setting. Section 4 discusses Petrie
duals of maps in some detail. We discuss generalized Cayley maps on orientable surfaces
in Section 5 and generalized Cayley maps on non-orientable surfaces in Section 6.
We can summarize the results obtained in our paper as follows. A generalized Cayley
map may be both orientable and non-orientable. The orientable generalized Cayley maps
come in two kinds. First, there are the original orientable Cayley maps in which the group
acting regularly on the vertices of the map consists entirely of orientation preserving auto-
morphisms. Then there are orientable generalized Cayley maps with bipartite underlying
R. Jajcay et al.: Generalized Cayley maps and their Petrie duals 387
graphs in which the group acting regularly on the vertices of the map contains a subgroup
of index 2 consisting of orientation preserving automorphisms while the other half of the
regularly acting group consists of orientation reversing automorphisms (Theorems 5.1 and
5.2). There are infinitely many examples of orientable generalized Cayley maps which
are not the original Cayley maps (Lemma 5.3 and Example 5.5), and infinitely many ori-
entable generalized Cayley maps which are self-Petrie-dual (Remark 5.6). In Remark 5.7
we present orientable generalized Cayley maps that simultaneously admit an orientation
preserving group acting regularly on their vertices and a vertex-regular group half of which
consists of orientation reversing automorphisms (hence, these maps are classical Cayley
maps and orientable maps of the second kind at the same time).
In Remark 6.2 we present an infinite family of non-orientable generalized Cayley maps.
Taking advantage of the well-known fact that the Petrie dual of an orientable map M
is orientable if and only if the underlying graph of M is bipartite (Theorem 4.2 in our
paper), we argue the existence of infinitely many examples of the original orientable Cayley
maps whose Petrie duals are orientable (Example 6.6) and infinitely many examples of
the original orientable Cayley maps whose Petrie duals are non-orientable (any orientable
Cayley map with a non-bipartite underlying graph). The Petrie dual of a non-orientable
generalized Cayley map may be both orientable and non-orientable. We classify the non-
orientable generalized Cayley maps whose Petrie dual is orientable (Corollary 6.5) and
exhibit infinitely many examples of both non-orientable generalized Cayley maps whose
Petrie dual is orientable and non-orientable generalized Cayley maps whose Petrie dual is
non-orientable (Remark 6.2).
2 Regular, orientably-regular, reflexible and chiral maps
For a map M, regions (faces) of its barycentric subdivision are the flags of the map. Every
flag is a triangular region; informally, its three ‘corners’ are a vertex, the ‘midpoint’ of an
edge incident with the vertex, and the ‘centre’ of a face incident with both the vertex and the
edge. As long as no face of the map contains two occurrences of an edge on its boundary
(no maps with such a degeneracy will be considered here), a flag can be identified with a
triple (v, e, F ), where v is a vertex, e an edge incident with v, and F is a face incident with
both v and e. A pair of distinct flags (v, e, F ) and (v′, e′, F ′) of M are incident if they
share a segment of the skeleton of the barycentric subdivision; in case when no face of the
map contains two occurrences of an edge on its boundary, two flags are incident if precisely
two of the three equalities v = v′, e = e′, F = F ′ hold.
Let r0, r1 and r2 be involutory fixed-point-free permutations of the flag set of a map
M formed by two-cycles consisting of incident flags, with r0 swapping faces that share
a face-center-to-edge-midpoint segment, r1 swapping faces that share the vertex-to-face-
center segment, and r2 swapping faces that share the vertex-to-edge-midpoint segment;
again, if no face of the map contains two occurrences of an edge on its boundary, the flag
(v, e, F ) is mapped by r0 to (v′, e, f), v ̸= v′, by r1 to (v, e′, F ), e′ ̸= e, and by r2 to
(v, e, F ′), F ′ ̸= F . The (flag-transitive) permutation group generated by r0, r1 and r2 is
the monodromy group Mon(M) of the map. The three generators of the monodromy group
can be interpreted as ‘gluing instructions’ to assemble the map from its set of flags; in other
words, knowledge of Mon(M) in terms of the action of the generators r0, r1 and r2 on the
set its flags is equivalent to knowing the map M. Note that r0 and r2 commute; if the map
is finite, then the orders of r0r1 and r1r2 are equal to the least common multiples of face
388 Ars Math. Contemp. 24 (2024) #P3.01 / 385–403
lengths and vertex valencies, respectively.
It is well known that the carrier surface of M is orientable if and only if ⟨r1r2, r0r2⟩ =
⟨r1r2, r0r1⟩ is a subgroup of index 2 in Mon(M). In such a case we speak about an
orientable map; letting ρ = r1r2 and λ = r0r2 one may consider the ‘orientable part’
Mon+(M) = ⟨ρ, λ⟩, the index-2-subgroup of Mon(M). Here, Mon+(M) can be re-
garded as a permutation group acting on the dart set D of the map (i.e., on the set of
directed edges of the map). Then ρ is a permutation that cyclically permutes, at each vertex
v, the darts emanating from v in accord with a chosen orientation of the carrier surface of
the map, and λ is an an involution interchanging the two darts belonging to the same edge.
An automorphism of a map M with monodromy group Mon(M) = ⟨r0, r1, r2⟩ is
any permutation of the flag set F of M that preserves incidence of flags. Equivalently, a
permutation of F is an automorphism of M if and only if it commutes with all of r0, r1
and r2; hence the automorphism group Aut(M) of the map is simply the centralizer of the
monodromy group Mon(M) in the symmetric group SF on the set F . Since Mon(M) is
transitive on F , it follows that Aut(M) acts freely on the set set F .
For an orientable analogue, let now M be an orientable map with dart set D and with
Mon+(M) = ⟨ρ, λ⟩ being a subgroup of Mon(M) of index 2. An orientation-preserving
automorphism of M is any permutation of D commuting with ρ and λ. It follows that the
group Aut+(M) of all orientation-preserving automorphisms of an orientable map is the
centralizer of Mon+(M) in the symmetric group SD on the set D. Again, transitivity of
Mon+(M) on D implies that the group Aut+(M) acts freely on the set D. If such an
orientable map M admits an automorphism commuting with λ but inverting ρ, then the
automorphism is orientation-reversing.
Finally, let us discuss the highest ‘level of symmetry’of maps. A map M with mon-
odromy group Mon(M) is called regular if the automorphism group M acts regularly
on the set F of flags. In this case, the groups Aut(M) and Mon(M) are abstractly iso-
morphic, so that in the case of a finite map the order of both groups is equal to |F|. If
M is a regular orientable map, then its automorphism group Aut(M) contains the group
Aut+(M) of orientation-preserving automorphisms as a subgroup of index 2, its other
coset in Aut(M) being the collection of all orientation-reversing automorphisms of the
map. The orientation-reversing automorphisms are sometimes referred to as reflections,
giving such maps the name reflexible. An orientable map M is orientably-regular if the
group Aut+(M) of orientation preserving automorphisms of M acts regularly on its set
of darts. An orientably regular map that is not reflexible is chiral, and in such a case
Aut(M) = Aut+(M).
3 Cayley maps
A Cayley map is an orientable map M that admits a group of orientation preserving auto-
morphisms G acting regularly on its set of vertices. To make this definition more precise,
one needs to consider the induced action of map automorphisms of M on the vertex set
of the underlying graph associating the map automorphism ψ with the vertex permutation
ψ mapping v to v′ whenever ψ(v, e, F ) = (v′, e′, F ′). It is easy to see that ψ is a well
defined graph automorphism of the underlying graph of M defined via its action on the
vertices (of the map or the graph). The induced action of Aut(M) on the vertices is almost
universally faithful; with the very few exceptions listed in [14, Proposition 9]. The auto-
morphisms contained in the automorphism group of a Cayley map which acts regularly on
R. Jajcay et al.: Generalized Cayley maps and their Petrie duals 389
the vertices of the map correspond in this way to automorphisms of its underlying graph
forming a group acting regularly on the graph’s vertices, making it into a Cayley graph [18].
A Cayley graph C(G,X) is a graph whose vertex set can be identified with the elements
of a group G generated by a set X closed under taking inverses and not containing the
identity 1G, with the pairs of adjacent vertices consisting of all pairs g, gx with g ∈ G and
x ∈ X . A graph Γ is isomorphic to a Cayley graph C(G,X) if and only if AutΓ contains
a subgroup G acting regularly on the vertices of Γ [18], which justifies the remark we have
made about the underlying graphs of Cayley maps being Cayley. Hence, the automorphism
group of a Cayley map M containing a group G of orientation preserving automorphisms
acting regularly on the vertices of M is necessarily contained in the automorphism group
of its underlying Cayley graph C(G,X), G ≤ AutC(G,X) (after applying the necessary
restriction to vertices). The action of G on C(G,X) is defined for each g ∈ G via left
multiplication: Ag(h) = gh for all g, h ∈ H , and we shall denote the group {Ag | g ∈ G}
byGL; it is always a subgroup of AutC(G,X). The set of darts of the Cayley map M thus
takes the form {(h, hx) | h ∈ G, x ∈ X} and each Ag ∈ GL ‘extends’ to a permutation
of the darts of M mapping the dart (h, hx) to (gh, ghx). If each of the extensions of Ag ,
g ∈ G, is to preserve the orientation of M, there must exist a cyclic permutation p of X
having the property that the dart (g, gp(x)) is always the dart lying immediately next to the
dart (g, gx) in the local surface neighborhood of g (sharing the face with (g, gx)), for all
g ∈ G and x ∈ X [16]. Thus, we can talk about the local permutation of the darts (g, gx),
x ∈ X , emanating from g, determined by the cyclic permutation p of X . This gives rise
to an equivalent definition of a Cayley map as an orientable embedding of a Cayley graph
C(G,X) in an orientable surface determined by the local rotation scheme with the property
that the local surface ordering corresponding to the vertex g ∈ G, ρg , acting cyclically on
the darts (g, x), x ∈ X , does not depend on g, and is determined by a fixed cyclic permu-
tation p of X . Such Cayley map is denoted by CM(G,X, p), and the equivalence of this
‘local rotation definition’ and the definition via the existence of an orientation preserving
automorphism group acting regularly on its vertices is a well established fact in the theory
of Cayley maps [16].
As argued above, the group GL = {Ag | g ∈ G} (isomorphic to G) is always a
subgroup of the group Aut+CM(G,X, p) of orientation preserving automorphisms, and
CM(G,X, p) is orientably regular if and only if there exists an orientation preserving
automorphism A ∈ Aut+CM(G,X, p) which fixes the identity 1G and maps the darts
emanating from 1G in the same order as p: A((1G, x)) = (1G, p(x)), for all x ∈ X . This
has been shown to be equivalent to the existence a special identity-fixing permutation φ
called skew-morphism in [7], where it was first defined. Given a group G, a permutation
φ : G → G of the elements of G that fixes the identity of G, φ(1G) = 1G, is said to be a
skew-morphism of G with associated power function π : G→ N if the equation
φ(gh) = φ(g)φπ(g)(h) (3.1)
is satisfied for all g, h ∈ G. A Cayley map CM(G,X, p) is orientably regular if and
only if there exists a skew-morphism φ of G with the property φ(x) = p(x), for all
x ∈ X . Furthermore, a group G admits the existence of an orientably regular Cayley
map CM(G,X, p) if and only if G admits the existence of a skew-morphism φ with an
orbit X that generates G and is closed under taking inverses (in which case, the desired
orientably regular map is the map CM(G,X,φ|X)) [7]. The classification of finite groups
G that admit the existence of an orientably regular Cayley map CM(G,X, p) is a hard
390 Ars Math. Contemp. 24 (2024) #P3.01 / 385–403
problem and the topic of many articles. The orientation preserving automorphism group of
an orientably regular Cayley map with skew-morphism φ and power function π takes the
form Aut+(CM(G,X, p)) ∼= GL ⟨φ⟩ with the product multiplication defined by the rule
aφ = φπ(a)φ(a),
for all a ∈ G. A Cayley map CM(G,X, p) with the property p(x−1) = (p(x))−1 satisfied
by all x ∈ X is called a balanced Cayley map. Since a map CM(G,X, p) is balanced
if and only GL is normal in Aut+(CM(G,X, p)), we will break with the long line of
tradition, and we will call such map a normal Cayley map. This is in line with the name
given to Cayley graphs C(G,X) with GL normal in AutC(G,X).
Let us complete this section with a necessary condition for a Cayley map to be regular.
Even though it can be deduced from [5] and [7], we are not aware of this condition being
stated explicitly before, and so we also provide a short proof.
Theorem 3.1. If a Cayley map M = CM(G,X, p) is regular, there exists a pair of skew-
morphisms φ,ψ of G that preserve X and the restriction of φ to X is equal to p, while the
restriction of ψ to X is equal to p−1.
Proof. If M = CM(G,X, p) is regular, it is also orientably regular, hence there exists
a skew-morphism φ of G that preserves X and whose restriction to X is equal to p [7].
Moreover, if M is regular, it is isomorphic to its mirror reflection CM(G,X, p−1), which
is therefore also regular, hence orientably regular, and there exists a skew-morphism ψ of
G that preserves X and whose restriction to X is equal to p−1.
It is interesting to point out that the above necessary condition is not sufficient. For
example, the skew-morphism whose existence guarantees the orientable regularity of a
normal Cayley map CM(G,X, p) is well-known to be a group automorphism of G [21].
The inverse of a group automorphism is always a group automorphism, and hence a skew-
morphism. Thus, every orientably regular normal Cayley map admits a skew-morphism
whose restriction to X is p and whose inverse is a skew-morphism (and its restriction to X
is p−1). However, not every normal orientably regular Cayley map is regular. To mention
just one famous example, all orientably regular embeddings of complete graphs have been
shown to be normal Cayley maps, however, the only orientably regular embeddings of
complete graphs which are also regular are the orientable embeddings of complete graphs
of prime power order not exceeding 4 [8].
4 Petrie dual
The well-known duality operation switching the roles of vertices and faces of a map M
preserves some of the most important topological characteristics of M, such as the ori-
entability and the genus, but generally changes the underlying graph of M. The less well-
known Petrie duality has the advantage of preserving both the embedded graph and the
action of Aut(M) on it. This makes the Petrie dual operation more useful when dealing
with maps with automorphism groups acting regularly on vertices of the underlying graph.
The Petrie dual of a map M is the map P (M) with the same vertices and edges as M
(and thus, the same underlying graph), however, the faces of P (M) are determined by the
Petrie (or zig-zag) walks of M which visit vertices of M along the edges while switching
sides (i.e., faces) in the middle of every next edge situated along the boundary of the face
R. Jajcay et al.: Generalized Cayley maps and their Petrie duals 391
of M. The key point of our interest in the Petrie dual lies in the fact that the two maps
M and P (M) have the same vertex set and the same automorphism group. To put this
claim on a more precise footing, we turn again to the monodromy groups. If ⟨r0, r1, r2⟩ is
the monodromy group of a map M, the monodromy group of the Petrie dual P (M) is the
group ⟨r0r2, r1, r2⟩ (which can be easily seen from the fact that the two flags associated
with the opposite dart get swapped in the Petrie dual construction). Thus, P (P (M)) = M,
and, since ⟨r0, r1, r2⟩ = ⟨r0r2, r1, r2⟩, we obtain
Aut(M) = CSF (⟨r0, r1, r2⟩) = CSF (⟨r0r2, r1, r2⟩) = AutP (M).
Therefore, the two automorphism groups act on the same set of flags in exactly the same
way, and any properties of the automorphism group of M with regard to the vertices of
M are shared by the automorphism group of P (M). In particular, if Aut(M) contains a
subgroup acting regularly on the set of vertices of M, so does the group AutP (M). Since
we make repeated use of this observation throughout our paper, we state it in the form of a
theorem.
Theorem 4.1. A map M is a generalized Cayley map if and only if its Petrie dual P (M)
is also a generalized Cayley map.
Even though the Petrie dual pair shares with the original map the same underlying graph
and the same automorphism group, the two maps have usually quite distinct topological
properties. It is, for example, possible (even common) that M is orientable while P (M)
need not be. For an example, consider M to be the tetrahedron as an embedding of K4 on
the sphere. Then P (M) is an embedding of K4 in the projective plane as a map of type
{4, 3}.
Since we will have to be able to distinguish between various situations with regard to
the orientablity vs. non-orientability, we recall the following well-known result (e.g., [15,
Remark 7]).
Theorem 4.2. If M is orientable, then P (M) is orientable if and only if the underlying
graph of M is bipartite.
Next, recall the concept of a ribbon graph which is constructed from a map M by
cutting out an open neighborhood of each face-center and keeping small bands around the
edges of M and small circles around the vertices. The ribbon graph of the Petrie dual of
an orientable embedding of a bipartite graph is the ribbon graph of the original embedding
with all bands twisted (cut off at one of the end vertices and glued back after being rotated
by 180 degrees) [4].
Remark 4.3. There is a large number of examples of Cayley maps throughout the litera-
ture. For example, four of the five Platonic solids (all but the dodecahedron) are Cayley
maps. All orientably regular embeddings of complete graphs are also Cayley maps [8].
Since none of the complete graphs butK2 are bipartite, their Petrie duals are non-orientable
generalized Cayley maps.
On the other hand, the underlying graph of the dodecahedron is not a Cayley graph,
and hence the dodecahedron and its Petrie dual are neither Cayley maps nor generalized
Cayley maps.
392 Ars Math. Contemp. 24 (2024) #P3.01 / 385–403
5 Orientable generalized Cayley maps and their Petrie duals
Let us begin by pointing out that the underlying graph of any generalized Cayley map
(orientable or not) is a Cayley graph. This observation follows from the same line of
arguments as that included in our section on Cayley maps, namely, it follows from the fact
that the group of automorphisms of the underlying graph induced by the group G of map
automorphisms acting regularly on the vertices of the map acts regularly on the vertices
of the graph. Thus, as is well-known, the set of vertices of the underlying graph of a
generalized Cayley map can be identified with the elements of G, and the action of these
automorphisms on the elements of G is that of left-multiplication in G.
It follows that generalized Cayley maps M are precisely embeddings of Cayley graphs
C(G,X) into (orientable or non-orientable) surfaces satisfying the property that all left
multiplications in G viewed as permutations of the elements of G ‘extend’ into automor-
phisms of the embedding, and that the order of the automorphism group of M acting
regularly on the vertices of M must be equal to its number of vertices, i.e., must be equal
to |G|.
Thus, orientable generalized Cayley maps come in two kinds. First, there are the em-
beddings M of Cayley graphs C(G,X) in orientable surfaces having the property that all
left multiplications in G extend into orientation preserving automorphisms of M, which
are precisely the classical Cayley maps M = CM(G,X, p) [16]. An orientable general-
ized Cayley map M that is not of this kind, i.e., that is not a Cayley map, must then be an
embedding of a Cayley graph C(G,X) in an orientable surface having the property that at
least one of the left multiplications in G extends to an orientation-reversing automorphism
of M. Recall that any automorphism group of an orientable map that contains at least
one orientation reversing automorphism must contain an index 2 subgroup of orientation
preserving automorphisms. This means that orientable generalized Cayley maps which are
not Cayley maps are embeddings of Cayley graphs C(G,X) in which a subgroup HL of
GL of index 2 extends into orientation preserving automorphisms and the other coset of
this subgroup in GL extends into orientation reversing automorphisms.
Recall that an orientable embedding of C(G,X) is determined by choosing a cyclic
permutation ρg of the elements in X for each g ∈ G. In the case of Cayley maps
CM(G,X, p), for every g ∈ G, the permutation ρg of the elements of X must be equal
to p. This, of course, cannot be the case for orientable generalized Cayley maps which are
not Cayley maps, i.e., there must exist elements f, g ∈ G such that ρf ̸= ρg . However, the
distribution of local rotations in orientable generalized Cayley maps which are not Cayley
maps is only slightly more complicated than that of Cayley maps. Namely, all such maps
are still determined by a single cyclic permutation p of X which becomes the local rotation
for the vertices contained in a subgroup H of index 2 in G and whose inverse p−1 becomes
the local rotation for the rest of them. We shall denote these maps by GCM(G,H,X, p)
and note that each such map is an orientable embedding of a Cayley graph C(G,X) having
the properties that H is a subgroup of index 2 in G, p is a cyclic permutation of X , and the
embedding of C(G,X) is determined by the local rotations ρh = p, for all h ∈ H , and
ρk = p
−1, for all k ∈ G \H .
Using the notation introduced above, we classify in the following theorem orientable
generalized Cayley maps in terms of their local orientations.
Theorem 5.1. Let M be an orientable generalized Cayley map. Then M is either a Cayley
map, M = CM(G,X, p), or a map M = GCM(G,H,X, p) defined above.
R. Jajcay et al.: Generalized Cayley maps and their Petrie duals 393
Conversely, everyCM(G,H,X, p), i.e., every orientable embedding of a Cayley graph
C(G,X) having the property that ρh = p, for all h ∈ H , whereH is a subgroup of index 2
in G and p is a cyclic permutation of X , and ρk = p−1, for all k ∈ G \H , is an orientable
generalized Cayley map.
Finally, the maps GCM(G,H,X, p) and GCM(G,H,X, p−1) are isomorphic.
Proof. Let M be an orientable generalized Cayley map that is not a Cayley map, let
G ≤ Aut(M) act regularly on the vertices of M , and identify the vertices of M with
the elements of G. Let H be the subgroup of index 2 in G of automorphisms preserv-
ing the orientation in M whose existence has been argued prior to the statement of this
theorem.
First, let h be an element of H . This means that the associated map automorphism Ah
mapping the darts (g, x), g ∈ G, x ∈ X , to the darts (hg, x) preserves the orientation
in M. It follows that Ahρ(g, x) = ρAh(g, x), for all g ∈ G and x ∈ X , and thus
(hg, ρg(x)) = (hg, ρhg(x)), for all g ∈ G and x ∈ X . Since multiplication by h preserves
H and its coset, the above identities yield that ρh = ρh′ , for all h, h′ ∈ H , and ρk = ρk′ ,
for all k, k′ ∈ G \H . Taking next an element k ∈ G \H , its corresponding automorphism
Ak reverses the orientation in M, and hence satisfies the identity Akρ = ρ−1Ak. Thus,
(kg, ρg(x)) = (kg, ρ
−1
kg (x)), for all g ∈ G and x ∈ X . It follows that ρg = ρ
−1
kg , for all
g ∈ G. If we denote the cyclic permutation of X assigned to the elements of H by p, using
the fact that the multiplication by k swaps H and its coset yields that the elements in G\H
are assigned the local permutation p−1. This completes the proof of the first statement of
our theorem.
The proof of the second part of the theorem is essentially the reverse of the above. If
one chooses the local orientations as described in the theorem, it is easy to see that left
multiplication by the elements of H preserves and left multiplication by the elements from
G \H reverses the orientation of the obtained map.
The veracity of the final statement of our theorem can be verified by showing that
the mapping φk defined on the set of darts of GCM(G,H,X, p) by mapping the dart
(g, gx) to the dart (kg, kgx) in GCM(G,H,X, p−1), for any fixed k ∈ G \ H and all
g ∈ G and x ∈ X , is a map isomorphism between the maps GCM(G,H,X, p) and
GCM(G,H,X, p−1). This relatively simple task is left to the reader.
To obtain examples of orientable generalized Cayley maps that are not Cayley maps,
one needs to consider embeddings of Cayley graphs of even order. To obtain an ‘easy’
example, one could consider the generalized Cayley mapGCM(G,H,X, p) for the groups
G = Z2n, n ≥ 1, H = ⟨2⟩, X = G \ {0}, and any cyclic permutation p of X . All of these
maps are embeddings of the complete graphs K2n in orientable surfaces. However, it is
easy to see that the cases n = 1 and n = 2 result in Cayley maps, and even for n ≥ 3, it is
hard to show that the resulting maps are not Cayley. Thus, to construct an infinite family of
orientable generalized Cayley maps which are provably not Cayley maps, one might rely
on the Petrie dual operator again. Specifically, in what follows, we shall consider orientable
generalized Cayley maps whose Petrie dual is orientable again. Because of Theorem 4.2,
each such map must be bipartite (which is also a sufficient condition). Thus, we obtain the
following:
Theorem 5.2. Let M be an orientable generalized Cayley map. The Petrie dual of M is
orientable if and only if M = CM(G,X, p) is a Cayley map for a group G that contains
394 Ars Math. Contemp. 24 (2024) #P3.01 / 385–403
a subgroup H of index 2 for which X ⊆ G \H or if M = GCM(G,H,X, p) where H is
a subgroup of G of index 2 for which X ⊆ G \H .
Furthermore, if the Petrie dual of an orientable generalized Cayley map M is ori-
entable, the two maps CM(G,X, p) and GCM(G,H,X, p) are mutual Petrie duals.
Proof. In order for a Cayley graph C(G,X) to be bipartite, it is easy to see (and well
known) that it must be a Cayley graph of an even order group G that possesses a subgroup
H of index 2, and X must be a subset of the non-trivial coset of H in G. The rest of the
first part of the theorem follows from Theorem 5.1.
Suppose now that M = CM(G,X, p) with X satisfying the required condition with
respect to a subgroup H of G. It is again not too hard to see that the Petrie dual to M
reverses the order of elements fromX for each element ofH (or for each element ofG\H;
the two maps GCM(G,H,X, p) and GCM(G,H,X, p−1) are isomorphic as argued in
Theorem 5.1).
Clearly, an orientable generalized Cayley map M = GCM(G,H,X, p) is not simulta-
neously a Cayley map if and only if the group Aut(M) contains no orientation preserving
subgroup acting regularly on the vertices of M. One way to make sure no such group
exists, is to find a pair of vertices u, v of M for which there is no orientation preserving
automorphism mapping u to v. This is the approach we take to construct an infinite family
of orientable generalized Cayley maps which are not Cayley maps.
Lemma 5.3. Let G be a finite group with a subgroup H of index 2 and a generating set
X ⊆ G \H . If CM(G,X, p) is chiral, its Petrie dual GCM(G,H,X, p) is a generalized
Cayley map that is not a Cayley map.
Proof. Recall that an orientably regular map is chiral if it admits no orientation reversing
automorphisms. Note also, that since CM(G,X, p) is bipartite and connected, any auto-
morphism φ ∈ AutCM(G,X, p) that maps a vertex u ∈ H to a vertex v ∈ G \H maps
all elements of H onto the elements of G \ H . This means that all the automorphisms
in AutGCM(G,H,X, p) mapping elements of H to elements in G \ H are orientation
reversing, and no orientation preserving automorphisms of GCM(G,H,X, p) map the el-
ements of H to the elements in G \H . It follows that GCM(G,H,X, p) is not a Cayley
map.
Remark 5.4. There are many examples of chiral bipartite Cayley maps. One good source
of such maps is the paper [10] the authors of which construct infinitely many orientably
regular embeddings of Kn,n, with n = pe where p is an odd prime and e ≥ 1, which are
Cayley maps for cyclic and dihedral groups and are chiral. Lemma 5.3 yields that all their
Petrie duals are generalized Cayley maps that are not Cayley maps. A simpler family of
chiral bipartite Cayley maps, brought to our attention by one of our referees, is also formed
by the torus maps {4, 4}b,c, for b, c both non-zero, b + c even, and b ̸= c to guarantee
chirality, and the torus maps {6, 3}b,c. The chiral bipartite {4, 4}b,c maps have orders b2 +
c2 and arise by identifying opposite sides of squares with corners (0, 0), (b, c), (b−c, b+c)
and (−c, b) in a unit rectangular grid.
Example 5.5. Consider the dihedral groups
Dn =
〈
a, b | an = b2 = 1, bab−1 = a−1
〉
,
R. Jajcay et al.: Generalized Cayley maps and their Petrie duals 395
n ≥ 3. Taking G = Dn, H = ⟨a⟩ and X = {b, ba, . . . , ban−1} yields a pair of mutu-
ally Petrie dual generalized Cayley maps CM(G,X, p) and GCM(G,H,X, p), for every
cyclic permutation p of X . All such maps are orientable embeddings of the complete bi-
partite graph Kn,n. Taking the specific permutation p = (b, ba, . . . , ban−1) yields two
(possibly isomorphic) embeddings of Kn,n. First, the Cayley map
CM(Dn, {b, ba, . . . , ban−1}, (b, ba, . . . , ban−1))
is an orientable embedding of Kn,n with n faces of length 2n and of genus
(n−1)(n−2)
2 ,
while its Petrie dual
GCM(Dn, ⟨a⟩ , {b, ba, . . . , ban−1}, (b, ba, . . . , ban−1))
is an orientable embedding of Kn,n with the same number of faces of the same length and
of the same genus.
The excluded case n = 2 results in an embedding of K2,2 in the sphere of the form
CM(Z22, {(0, 1), (1, 1)}, ((0, 1), (1, 1)))
which is its own Petrie dual because the inverse of the permutation ((0, 1), (1, 1)) is
((0, 1), (1, 1)) again.
The toroidal maps for case n = 3 are pictured below.
b
a2
ba2 a
ba
1
b
a
ba2 a2
ba
1
Figure 1: Maps CM(D3, {b, ba, ba2}, (b, ba, ba2)) and GCM(D3, ⟨a⟩ , {b, ba, ba2},
(b, ba, ba2)).
Clearly, the two maps pictured in Figure 1 are isomorphic. Hence, the smallest maps
CM(D3, {b, ba, ba2}, (b, ba, ba2)) and CM(Z22, {(0, 1), (1, 1)}, ((0, 1), (1, 1)))) are self-
Petrie-dual. This might come as a surprise, as the same permutations of vertices (com-
ing from left multiplications) cannot simultaneously extend into orientation preserving
and orientation reversing automorphisms. It is, however, easy to notice that relabeling
the vertices of CM(D3, {b, ba, ba2}, (b, ba, ba2)) changes the permutation actions of left-
multiplications, and hence left multiplications with respect to one labeling may extend to
orientation preserving while left multiplications with respect to another labeling may ex-
tend to orientation reversing automorphisms. Since in non-bipartite cases the Petrie dual of
an orientable map is non-orientable, the case of bipartite orientable maps is the only case
where one may encounter orientable self-Petrie-dual generalized Cayley maps.
396 Ars Math. Contemp. 24 (2024) #P3.01 / 385–403
We devote the rest of this section to the investigation of such possibility, i.e., to the study
of orientable self-Petrie-dual generalized Cayley maps. Namely, suppose an orientable
generalized Cayley map M is self-Petrie-dual. Then, M is bipartite and Theorem 5.1 im-
plies that the automorphism group of M contains two subgroups acting regularly on the
vertices, a subgroup of orientation preserving automorphisms as well as a subgroup of au-
tomorphisms half of which is orientation reversing. This yields a necessary condition for
M being self-Petrie-dual, namely, M must be a Cayley map admitting at least one orienta-
tion reversing automorphism. Thus, chiral bipartite Cayley maps are never self-Petrie-dual
(in Example 5.4, we have already encountered infinitely many such maps constructed in
[10]). In fact, it is well-known that no chiral maps are self-Petrie-dual.
These observations appear to suggest that in order to construct bipartite Cayley maps
that are self-Petrie-dual, one should consider maps with many orientation preserving and
many orientation reversing automorphisms. In fact, searching through the literature for
orientable self-Petrie-dual maps resulted only in regular orientable self-Petrie-dual maps
[2, 11, 17], with the methods used in [11] or [17] relying on regularity and producing large
maps of large genera. An infinite family of regular examples also arises as follows.
Remark 5.6. For every positive integer n ≥ 2 that is relatively prime to φ(n) (i.e., there
exists a unique group of order n), there exists exactly one orientably regular embedding
of Kn,n [9]. That means that such embedding is necessarily isomorphic to its mirror re-
flection, hence regular, as well as self-Petrie dual. Kwak and Kwon in [13] classified the
orientable regular self-Petrie embeddings of Kn,n.
Nevertheless, even in the extreme case when M is regular, it might happen that any
orientation preserving subgroup K ≤ Aut(M) of order half the number of vertices of
M paired with any orientation reversing automorphism ψ ∈ Aut(M) generate a group of
order larger than the number of vertices of M. In that case, Aut(M) does not contain a
subgroup of automorphisms containing an orientation reversing automorphism and acting
regularly on the vertices. Hence, M cannot be isomorphic to any GCM(G,H,X, p), and
hence cannot be self-Petrie-dual. To illustrate this possibility, in the following example we
present an infinite family of bipartite regular Cayley maps none of which are self-Petrie-
dual. We are thankful to Gareth Jones who pointed out this example to us.
Remark 5.7. The authors of [3] considered regular embeddings of Kn,n having the prop-
erty that the orientation preserving automorphism group of the embedding that does not
move the bipartite sets is not metacyclic; in which case n is necessarily a power of 2.
For each such n ≥ 8, they construct four non-isomorphic regular orientable embeddings
of Kn,n, denoted N (n; k, l), k, l ∈ {0, 1}. They prove that the maps N (n; 0, 0) and
N (n; 1, 0) are self-Petrie-dual, while N (n; 0, 1) and N (n; 1, 1) are Petrie duals of each
other. They also show that all four maps are Cayley maps. This means, in particular, that
each of the maps N (n; 0, 1) and N (n; 1, 1) is both a Cayley map and an orientable gener-
alized Cayley map, and as such, they both admit both an orientation preserving as well as
an orientation reversing automorphism group acting regularly on the vertices of the maps.
Hence, not even the existence of both types of vertex-regular groups is sufficient for
making a generalized Cayley map self-Petrie. It is hard to say whether these non-metacyclic
maps are in any way typical, and whether a regular bipartite Cayley map is more likely to
be self-Petrie-dual or not. The following highly specialized result appears to be of some
relevance toward answering this question:
R. Jajcay et al.: Generalized Cayley maps and their Petrie duals 397
Theorem 5.8 ([17]). There are no regular, self-dual, self-Petrie-dual, normal Cayley maps
with odd vertex degree.
6 Non-orientable generalized Cayley maps and their Petrie duals
Even though constructing non-orientable generalized Cayley maps is relatively easy – it is
enough to consider the Petrie dual of any non-bipartite orientable generalized Cayley map
– a more explicit construction of such maps is needed. When considering only the Petrie
duals of orientable generalized Cayley maps, one would never encounter a non-orientable
generalized Cayley map whose Petrie dual is not orientable. To see that such maps might
exist, one just needs to realize that the Petrie dual of a non-orientable generalized Cayley
map whose underlying graph is bipartite (we will construct such maps) is necessarily non-
orientable. The present section differs from the previous section which was concerned with
maps whose groups are transitive on arcs (regular and orientably regular). In this section,
we abandon the focus on such highly symmetric generalized Cayley maps and seek only to
provide examples of various non-orientable generalized Cayley maps.
To specify an embedding of a graph in a non-orientable surface, one needs to spec-
ify the local orientation of outgoing darts around every vertex as well as to specify for
each edge whether the two local orientations associated with its end-vertices are consistent
or not. More precisely, for any given edge e incident with vertices u and v, let U be an
open neighborhood of e which includes both u and v and all of e, but no other vertices
and no other complete edges. Then U is homeomorphic to an open disk, and is therefore
an orientable topological object. By calling the two local orientations associated with the
end-vertices of e consistent, we mean that the local orientations around u and v within U
are both clockwise or both anti-clockwise. We call them inconsistent otherwise. Refer-
ring to the ribbon graph associated with the map, the edges with inconsistent end-vertex
orientations are sometimes also called twisted, as the strip containing the edge e and the
vertices u and v is twisted before being attached to the rest of the ribbon graph through the
vertices u and v. To indicate whether the end-vertex orientations of an edge are consistent
or not, one usually assigns 1 or −1 to the edge, respectively (or, in a visualization of the
graph embedding using local rotations, an edge with inconsistent end-vertex orientations is
marked by an ‘x’). Perhaps the biggest disadvantage of this description of an embedding of
a graph is the fact that one does not have a guarantee that the resulting embedding is indeed
non-orientable. For example, the bipartite orientable maps GCM(G,H,X, p) defined in
the previous section can also be described using this ‘non-orientable’ description by saying
that GCM(G,H,X, p) is the map in which the local rotation at each vertex is equal to p
and all edges have inconsistent end-vertex orientations.
As explained in the previous sections, a generalized Cayley map must be an embedding
of a Cayley graph C(G,X) with the property that each left multiplication by the elements
of G extends into a map automorphism of the embedding. One of the finer points to be
dealt with in the previous section was the fact that the extensions might be either orientation
preserving or orientation reversing. We do not have two kinds of map automorphisms in
non-orientable maps which makes the situation a bit easier.
Let us start by describing general embeddings of Cayley graphs. Let C(G,X) be a
Cayley graph, let ρg denote the local rotation of the elements of X around g, and let
ι{g,gx} ∈ {1,−1} be the label assigned to the edge {g, gx}. The three defining involu-
398 Ars Math. Contemp. 24 (2024) #P3.01 / 385–403
tions r0, r1, r2 are then defined as follows:
(g, c{g,gx}, cF )
r0 =
{
(gx, c{g,gx}, cF ), if ι{g,gx} = 1,
(gx, c{g,gx}, cF ′), if ι{g,gx} = −1
,
(g, c{g,gx}, cF )
r1 = (g, c{g,gρι{g,gx}g (x)}
, cF ), and (g, c{g,gx}, cF )r2 = (g, c{g,gx}, cF ′),
where F ′ is the face adjacent to F across the edge {g, gx}.
The following characterization of generalized Cayley maps can be already found in
both [19] and [12]. As stated above, it covers both the orientable and the non-orientable
generalized Cayley maps.
Theorem 6.1 ([12, 19]). A (orientable or non-orientable) map M is a generalized Cayley
map if and only if it is an embedding of a Cayley graph C(G,X) with all local cyclic per-
mutations ρg , g ∈ G, equal to a fixed cyclic permutation p of X , and the twist distribution
ι satisfying the property ι{g,gx} = ι{g′,g′x}, for all g, g′ ∈ G and x ∈ X .
The simplest interpretation of the last condition is that all edges labelled by the same
generator (or its inverse) must be either all simultaneously un-twisted or all twisted, and
hence, from now on, we will assume that ι acts on the set X , ι : X → {−1, 1}, mapping
x ∈ X to ι{g,gx} (thus, in particular, ι(x) = ι(x−1)), for all x ∈ X . As each generalized
Cayley map is determined by the four-tuple (G,X, p, ι), we will denote these maps by
GCM(G,X, p, ι).
The authors of [12] generalized this simplification even further and introduced the fol-
lowing convenient notation.
Let M = GCM(G,X, p, ι) be a generalized Cayley map. Without loss of generality,
we may assume that X = {x0, x1, . . . , xd−1}, while p(xi) = xi+1, for all i ∈ [d] =
{0, 1, 2, . . . d−1} (with the addition performed modulo d). If we denote the distribution of
inverses in X by the function κ : [d] → [d], satisfying the property (xi)−1 = xκ(i), and use
the fact that ι assigns the same value to all edges arising from right mutiplication by xi to
define (with just a hint of abuse of the notation) ι : [d] → {−1, 1}, ι(i) = ι{1g,xi} (note that
ι(κ(i)) = ι(i), for all i ∈ [d]), we may associate the flags of M with the ordered triples
from G × [d] × {−1, 1}, with the neighboring flags (g, c(g,gxi), cF ) and (g, c(g,gxi), cF ′)
corresponding to the triples (g, i, 1) and (g, i,−1). The defining involutions take then the
particularly simple form:
(g, i, j)r0 = (gxi, κ(i), ι(i)j), (g, i, j)
r1 = (g, i+ j, j), and (g, i, j)r2 = (g, i,−j).
Remark 6.2. It is now easy to construct infinitely many non-orientable generalized Cayley
maps. As is well-known (see e.g., [4]), an embedding of a graph determined by local ro-
tations and a twisting function is non-orientable if and only if it contains at least one cycle
with an odd number of twisted edges. Hence, for example, it is easy to construct infinitely
many non-orientable embeddings of the complete bipartite graphs Kn,n. To get such an
embedding, one can take a bipartite C(G,X) with G containing a subgroup H of index 2,
X = G \H = {x0, x1, . . . , x|G|/2−1}, satisfying the only additional condition that x20 ̸=
x−12 x
−1
1 , choose any cyclic permutation p of X , twist the edges labeled by x0, and observe
that the 4-cycle consisting of the vertices 1G, x0, x0x1, x0x1x2, x0x1x2(x0x1x2)−1 = 1G
obtained by successive multiplications by x0, x1, x2 and (x0x1x2)−1 (which, being an odd
product of elements from X , necessarily belongs to X) contains the twisted edge labeled
R. Jajcay et al.: Generalized Cayley maps and their Petrie duals 399
x0 exactly once (if (x0x1x2)−1 were equal to x0, we would get the identity x−12 x
−1
1 = x
2
0
which we explicitely prohibited). Hence, all such generalized Cayley maps GCM(G,X,
p, ι) are unorientable embeddings of complete bipartite graphs, and there are clearly in-
finitely many of them. Note also that all such maps, being embeddings of bipartite graphs,
have the property that their Petrie dual is non-orientable.
Next, we bring in the Petrie dual operator again and fully describe its action on gener-
alized Cayley maps. The mapping −ι used in the statement of the following theorem is the
twisting function −ι(x) = (−1) · ι(x), for all x ∈ X .
Theorem 6.3. If M = GCM(G,X, p, ι), then P (M) = GCM(G,X, p,−ι).
Proof. It is easy to see that −ι is a correctly defined twisting function of a generalized Cay-
ley map, i.e., −ι(x) = −ι(x−1), for all x ∈ X . Since the two maps GCM(G,X, p,−ι)
and P (GCM(G,X, p, ι)) have the same underlying graphs, one way to prove our the-
orem is to show that the Petrie polygons of GCM(G,X, p, ι) and the faces of the map
GCM(G,X, p,−ι) are identical. This can be shown by performing a careful calculation
of the boundaries of the two oriented Petrie polygons of GCM(G,X, p, ι) that start at
(g, i) (which might turn out to be the same polygon visiting (g, i) twice, once followed
along p and once followed along p−1, depending on whether we first turn ‘right’ or ‘left’)
and the boundaries of the corresponding faces in GCM(G,X, p,−ι). As pointed out by
one of our referees, it can also be deduced from the fact that if r0, r1, r2 are the generators
of the monodromy group of GCM(G,X, p, ι), then r0r2, r1, and r2 are the generators for
the monodromy group of P (GCM(G,X, p, ι)) (Section 4) while they can also be easily
seen to be the generators of the monodromy group of GCM(G,X, p,−ι).
Based on the above theorem, obtaining the Petrie dual for a Cayley map is very simple.
Namely, the Petrie dual of a Cayley map CM(G,X, p) is the generalized Cayley map
GCM(G,X, p, ι), where ι twists all the edges of the map, i.e., ι(x) = −1, for every
x ∈ X . Since all the edges are twisted, the Petrie dual of a Cayley map M is non-
orientable if and only if M contains at least one odd cycle (i.e., the underlying graph is
non-bipartite). Hence, Theorem 5.2 is a corollary of Theorem 6.3.
Example 6.4. To obtain a specific example of the use of Theorem 6.3, we go back to a
classical example of a pair of mutually Petrie dual maps, namely the regular embedding of
the tetrahedron on the sphere and its Petrie dual embedded in the projective plane. Since
tetrahedron is the Cayley map CM(Z2 × Z2, {(1, 0), (0, 1), (1, 1)}, ((1, 0), (0, 1), (1, 1)))
with 4 triangular faces, its Petrie dual is the generalized Cayley map
GCM(Z2 × Z2, {(1, 0), (0, 1), (1, 1)}, ((1, 0), (0, 1), (1, 1)), ι),
with ι(x) = −1 for all x ∈ {(1, 0), (0, 1), (1, 1)}. Since the tetrahedron contains odd
cycles (triangles), the dual map is non-orientable, well-known to be an embedding in the
projective plane with 3 faces of length 4 (which can also be easily verified via direct calcu-
lations in the generalized Cayley map).
This example can be viewed as the first member of an infinite family of examples. For
every even n ≥ 4, let Dn = ⟨a, b | an = b2 = 1, bab = a−1⟩ be the dihedral group of
order 2n, X = {b, an2 , ba} be a set of three involutions, and p = (b, an2 , ba). Since n is
assumed even, the Petrie dual of the trivalent Cayley map CM(Dn, X, p) is non-orientable
(as the underlying graph of both maps contains a cycle of lenght n + 1). For even n ≥ 4,
400 Ars Math. Contemp. 24 (2024) #P3.01 / 385–403
the Petrie dual of CM(G,X, p) looks ‘like’ the Petrie dual of the tetrahedron: a 2n-cycle
with inside chords at every other vertex of the cycle, and an outside edge for the rest. All
these Petrie duals are embeddings in the projective plane with one face of length 2n and all
the remaining faces of length 4.
1 ba
b
a
a3
ba2
a2
ba3
Figure 2: Representation of the Petrie dual of the map CM(D4, {b, a2, ba}, (b, a2, ba))
indicating local rotations and marking the edges with inconsistent end-vertex rotations of
its underlying graph.
Next, we characterize non-orientable generalized Cayley maps whose Petrie dual is
orientable. As should be expected, those are the ‘obvious’ Petrie duals of non-bipartite
orientable Cayley maps, i.e., the non-bipartite non-orientable generalized Cayley maps sat-
isfying ι(x) = −1, for all x ∈ X , as well as other maps with a rather special twisting
function. The Petrie duals of all other non-orientable generalized Cayley maps are non-
orientable.
Corollary 6.5. The Petrie dual of a non-orientable generalized Cayley map GCM(G,X,
p, ι) is orientable if and only if the underlying Cayley graph C(G,X) contains an odd
cycle and every cycle of C(G,X) contains an even number of untwisted edges.
Proof. Suppose that M = GCM(G,X, p, ι) is non-orientable while its Petrie dual
P (M) = GCM(G,X, p,−ι) is orientable. This means that M cannot be bipartite as
that would make the orientable P (M) bipartite and hence by Theorem 5.2 its Petrie dual
P (P (M)) = M orientable; which it is not. Thus M contains an odd cycle. If any cycle in
M contained an odd number of untwisted edges, that very same cycle would contain an odd
number of twisted edges in P (M); making it non-orientable. This proves one implication.
Suppose now that M is non-orientable, contains an odd length cycle and every cycle
of M contains an even number of untwisted edges. Then M is not bipartite and its Petrie
dual contains no cycles with an odd number of twisted edges.
As stated prior to Theorem 6.3, non-orientable non-bipartite generalized Cayley maps
with all edges twisted trivially satisfy the property that their Petrie dual is also non-
orientable. Next, we provide an infinite family of examples that satisfy the conditions
of Theorem 6.3 but contain both twisted and untwisted edges.
R. Jajcay et al.: Generalized Cayley maps and their Petrie duals 401
Example 6.6. Let Γ be a bipartite d-regular graph, d-odd, let M be any orientable em-
bedding of Γ, and ρ and λ be the generators for the monodromy group G of M. Con-
sider the Cayley graph C(G, {ρ, ρ−1, λ}). There is an easy way to visualize this graph.
It is the underlying graph of the truncation T (M) of M, i.e., the map obtained from
M by locally removing its vertices and replacing them by d-cycles attached to the dan-
gling edges in the order determined by the local rotation. We claim that every cycle of
C(G, {ρ, ρ−1, λ}) contains an even number of edges labeled λ. To see this, take any cycle
C of C(G, {ρ, ρ−1, λ}). The edges labeled λ trace the ‘preimage’ of this cycle in Γ, which
is necessarily a union of edge disjoint even cycles. Since every edge labeled λ in C corre-
sponds to exactly one edge of the preimage, the number of edges labeled λ in C is even.
Thus, any choice of p together with the twisting function ι(ρ) = ι(ρ−1) = −1, ι(λ) = 1
makes the non-orientable GCM(G,X, p, ι) satisfy the conditions of Theorem 6.3.
The visualization of this example is in fact quite simple. We start with a ‘bipartite’
map whose ‘vertices’ are the d-cycles formed by edges labelled ρ, all of which are twisted,
with the edges labelled λ and connecting the two sets of d-cycles untwisted. This map is
non-orientable, since d is required to be odd. The orientable dual untwists the d-cycles and
twists the connecting edges.
To conclude this section, we consider one more classical concept from topological
graph theory – the orientable double covering Mo of a non-orientable map M given by its
underlying graph, rotation system and twisting function. The underlying graph of the dou-
ble covering is the Z2-lift of the underlying graph of the original map with the untwisted
edges given the voltage 0 ∈ Z2, and the twisted edges receiving the voltage 1 ∈ Z2. Infor-
mally, it is the double cover of the underlying graph of the original map with the untwisted
edges each lifted into two edges connecting vertices in the same layer, and the two lifts of
the twisted edges crossing from one layer to the other (for more details consult [20]). The-
orem 2.4 in [20] asserts that Aut(M) lifts into the orientation preserving automorphism
group Aut+Mo, and the full group Aut(M)o is the direct product of Aut+Mo with some
orientation reversing involutory automorphism ψ of order 2 (that swaps the two layers of
the double covering). It follows that the direct product of the lift of a subgroup of Aut(M)
acting regularly on the vertices of M with the group ⟨ψ⟩ acts regularly on the vertices of
Mo, and we obtain:
Theorem 6.7. The orientable double covering Mo of any non-orientable generalized Cay-
ley map M is an orientable generalized Cayley map admitting a group of automorphisms
containing an orientation reversing automorphism and acting regularly on its vertices.
7 Regular generalized Cayley maps
In conclusion of our paper, we would like to direct the reader toward [12] where the authors
of that paper developed a complete theory of regular generalized Cayley maps including
necessary and sufficient conditions based on the existence of special permutations of the
elements of the underlying group. Their language, however, is different from the one used
here, and they do not consider the orientable generalized Cayley maps GCM(G,H,X, p)
as a special case.
ORCID iDs
Robert Jajcay https://orcid.org/0000-0002-2166-2092
402 Ars Math. Contemp. 24 (2024) #P3.01 / 385–403
Jozef Širáň https://orcid.org/0000-0002-5901-7646
Yan Wang https://orcid.org/0000-0002-0148-2932
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ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 24 (2024) #P3.02 / 405–414
https://doi.org/10.26493/1855-3974.2570.5e8
(Also available at http://amc-journal.eu)
On p-gonal fields of definition*
Ruben A. Hidalgo †
Departamento de Matemática y Estadı́stica, Universidad de La Frontera, Temuco, Chile
Received 24 February 2021, accepted 28 November 2023, published online 11 June 2024
Abstract
Let S be a closed Riemann surface of genus g ≥ 2 and φ be a conformal automorphism
of S of prime order p such that S/⟨φ⟩ has genus zero. Let K ≤ C be a field of definition of
S. We prove the existence of a field extension F of K, of degree at most 2(p−1), for which
S is definable by a curve of the form yp = F (x) ∈ F[x], in which case φ corresponds to
(x, y) 7→ (x, e2πi/py). If, moreover, φ is also definable over K, then F can be chosen to
be at most a quadratic extension of K. For p = 2, that is when S is hyperelliptic and φ
is its hyperelliptic involution, this fact is due to Mestre (for even genus) and Huggins and
Lercier-Ritzenthaler-Sijslingit in the case that Aut(S)/⟨φ⟩ is non-trivial.
Keywords: Riemann surfaces, p-gonal curves, automorphisms.
Math. Subj. Class. (2020): 30F10, 30F20, 14H37, 14H55
1 Introduction
In [23], H. A. Schwarz proved that the group Aut(S) of conformal automorphisms of a
closed Riemann surface S of genus g ≥ 2 is finite. Later, in [17], A. Hurwitz obtained the
upper bound |Aut(S)| ≤ 84(g − 1) (this is known as the Hurwitz’s bound).
Let p ≥ 2 be a prime integer. We say that a closed Riemann surface S is cyclic p-
gonal if there exists some φ ∈ Aut(S) of order p such that the quotient orbifold S/⟨φ⟩
has genus zero. In this case, φ is called a p-gonal automorphism and the cyclic group
⟨φ⟩ a p-gonal group of S. The case p = 2 corresponds to S being hyperelliptic and φ
its (unique) hyperelliptic involution. The case p = 3 was studied by R. D. M. Accola in
[1]. In [10], G. González-Diez proved that p-gonal groups are unique up to conjugation in
Aut(S). In [13], it was observed that, if p ≥ 5n − 7, where n ≥ 3 is the number of fixed
*The author would like to express their gratitude to both referees for their valuable feedback, suggestions, and
corrections that have significantly improved the paper.
†Partially supported by projects Fondecyt 1230001 and 1220261.
E-mail address: ruben.hidalgo@ufrontera.cl (Ruben A. Hidalgo)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
406 Ars Math. Contemp. 24 (2024) #P3.02 / 405–414
points of φ, then ⟨φ⟩ is the unique p-group in Aut(S). Results concerning automorphisms
of p-gonal Riemann surfaces can be found, for instance, in [2, 3, 4, 5, 11, 27].
As a consequence of the Riemann-Roch theorem, a closed Riemann surface S can be
described by an (either affine or projective) irreducible complex algebraic curve, i.e., after
desingularization (if it is non-smooth) and filling at some punctures in the affine case, it
carries a Riemann surface structure which is biholomorphic to that of S (see Remark 1.1
for the case of cyclic p-gonal surfaces). A subfield K of the field C of complex numbers is
called a field of definition of S (or that S is definable over K) if there is an irreducible al-
gebraic curve representing S, which is defined as the common zeroes of some polynomials
with coefficients in K. The intersection of all the fields of definition of S is called the field
of moduli of S. In general, it is not a field of definition (see Section 3).
If we are given a (finite) group G < Aut(S) and the geometrical structure of the
quotient orbifold S/G, then it is not a simple task to find an algebraic curve for S reflecting
the action of G. A family of surfaces for which algebraic models are well known is the
case of cyclic p-gonal Riemann surfaces, which we proceed to recall below.
Let S be a p-gonal Riemann surface, φ ∈ Aut(S) be a p-gonal automorphism and
π : S → Ĉ be a regular branched cover with ⟨φ⟩ its deck group. Let {a1, . . . , am} ⊂ Ĉ =
C ∪ {∞} be the set of branch values of π. If aj ̸= ∞, for every j = 1, . . . ,m, then there
exist integers n1, ..., nm ∈ {1, ..., p−1}, n1+ · · ·+nm ≡ 0 mod p, such that S is defined
by the affine, irreducible and smooth p-gonal curve with equation
E : yp = F (x) =
m∏
j=1
(x− aj)nj ∈ C[x]. (1.1)
If one of the branch values is equal to ∞, say am = ∞, then in (1.1) we delete the
corresponding factor (x − am)nm and assume n1 + · · · + nm−1 ̸≡ 0 mod p. In this
affine algebraic model, π(x, y) = x and φ(x, y) = (x, ωpy), where ωp = e2πi/p. In the
hyperelliptic case, i.e., p = 2, in the above one has m ∈ {2g + 1, 2g + 2} and nj = 1.
Remark 1.1. The affine curve (1.1) is smooth at those points (x, y), where y ̸= 0. At
a point (aj , 0), the curve is smooth exactly when nj = 1 (anyway, if nj > 1, it has a
neighborhood homeomorphic to a disc). An irreducible projective algebraic curve defining
S is obtained from the above affine one as
Ê : ypzn1+···+nm−p =
m∏
j=1
(x− ajz)nj . (1.2)
As in the affine model, the projective curve Ê is smooth at the points [x : y : 1], where
y ̸= 0. At the points [aj : 0 : 1] it is smooth if and only if nj = 1 (again, in the other cases
there is a neighborhood homeomorphic to a disc). The curve is also non-smooth at the point
[0 : 1 : 0]. After normalization of the curve, one obtains a closed Riemann surface which
is biholomorphic to S. In this case, π([x : y : z]) = x/z and φ([x : y : z]) = [x : ωpy : z].
If F is a subfield of C such that in (1.1) we have F (x) ∈ F[x], then we say that F is
a p-gonal field of definition of S (and that S is cyclically p-gonally defined over F). Note
that there are infinitely many different p-gonal fields of definition for S (for instance, if T
is a Möbius transformation, then we may replace the values aj by T (aj)).
Given a field of definition of a p-gonal Riemann surface S, it is not clear at first sight if
it is a p-gonal field of definition. Also, it might be that a minimal p-gonal field of definition
R. A. Hidalgo: On p-gonal fields of definition 407
is not a minimal field of definition (see the exceptional case (m, p) = (4, 3) in Section 4.1).
This paper aims to provide an argument to show that, given any field of definition K of S,
there is a p-gonal field of definition F which is an extension of degree at most 2(p−1) over
K.
If φ is an automorphism of S, then we say that S and φ are simultaneously defined
over K if there is an algebraic curve model of S, defined over K, such that φ is given by a
rational map on it with coefficients in K.
Theorem 1.2. Let S be a cyclic p-gonal Riemann surface of genus g ≥ 2, with a p-gonal
automorphism φ, and let K be a field of definition of S. Then
(1) There is p-gonal field of definition of S, this being an extension of degree at most
2(p− 1) of K (which is also a field of definition of φ).
(2) If both S and φ are simultaneously defined over K, then there is a p-gonal field of
definition of S, this being an extension of degree at most two of K.
(3) If in Equation (1.1) n1 = · · · = nm, then there is a p-gonal field of definition of S,
this being an extension of degree at most two of K.
Remark 1.3. Theorem 1.2 is still valid if we change C to any algebraically closed field,
where in positive characteristic we need to assume that p is different from the characteristic.
Remark 1.4. For each integer n ≥ 2, not necessarily prime, the definition of cyclic n-
gonal Riemann surface S, n-gonal automorphism φ and n-gonal group ⟨φ⟩ is the same as
for the prime situation. In the particular case that every fixed point of a non-trivial power
φk is also a fixed point of φ, the definition of an n-gonal curve is the same as in (1.1),
but replacing p by n and assuming each the exponent nj to be relatively prime to n. In
this case, under the assumption that S has a unique n-gonal group ⟨φ⟩ (this is the situation
for generalized superelliptic Riemann surfaces [15]), then the arguments of the proof of
Theorem 1.2 allows us to obtain that: if K is a field of definition of S, then there is an n-
gonal field of definition of S, this being an extension of degree at most 2ϕ(n) of K, where
ϕ(n) is the ϕ-Euler function.
2 An application to hyperelliptic Riemann surfaces
Let S be a hyperelliptic Riemann surface (i.e., p = 2) with hyperelliptic involution φ and let
K be a field of definition of S. As φ is unique, one may consider the group Autred(S) :=
Aut(S)/⟨φ⟩, called the reduced group of automorphisms of S.
For even genus, in [22], J-F. Mestre proved that S is also hyperelliptically definable
over K. If the genus is odd, then the previous fact is in general false; as can be seen from
examples in [8, 9, 20, 21]. In [16], B. Huggins proved that if Autred(S) is neither trivial nor
cyclic, then S is also hyperelliptically definable over K. In [21], R. Lercier, C. Ritzenthaler
and J. Sijslingit proved that S can be hyperelliptically defined over a quadratic extension
of K if the reduced group is a non-trivial cyclic group. Our theorem asserts that this fact is
still valid even if the reduced group is trivial.
Corollary 2.1. If K is a field of definition of a hyperelliptic Riemann surface, then it is
hyperelliptically definable over an extension of degree at most two of K.
408 Ars Math. Contemp. 24 (2024) #P3.02 / 405–414
3 An application to fields of moduli
Let S be a closed Riemann surface and let C be an irreducible algebraic curve representing
it. The field of moduli MS of S is the fixed field of the group ΓC = {σ ∈ Aut(C/Q) :
Cσ ∼= C}; this field does not depend on the choice of the algebraic model C. In [18],
S. Koizumi proved that MS coincides with the intersection of all fields of definition of S,
but in general it might not be a field of definition [6, 7, 12, 16, 19]. If Aut(S) is trivial (the
generic situation for g ≥ 3), then Weil’s descent theorem [25] asserts that MS is a field
of definition of S. In [26], J. Wolfart proved that if S/Aut(S) is the Riemann sphere with
exactly 3 cone points (i.e., S is quasiplatonic), then MS is also a field of definition of S.
In a more general setting, if S/Aut(S) has genus zero, then it is known that S is definable
over an extension of degree at most two of MS (see [14] for a more general statement).
Now, let S be a p-gonal Riemann surface of genus g ≥ 2 and let G = ⟨φ⟩ < Aut(S)
be a p-gonal group. As previously noted, S is either definable over MS or over a suitable
quadratic extension of it (but it might not be cyclically p-gonally definable over such a
minimal field of definition). In the case that G is not a unique p-gonal subgroup, in [28],
A. Wootton noted that S can be cyclically p-gonally defined over an extension of degree at
most 2 of its field of moduli. In the case that G is the unique p-gonal subgroup, the quotient
group Aut(S)/G is called the reduced group of S. In [19], A. Kontogeorgis proved that if
the reduced group is neither trivial nor a cyclic group, then S can always be defined over
its field of moduli. So, a direct consequence of Theorem 1.2 is the following.
Corollary 3.1. Let S be a cyclic p-gonal Riemann surface with a p-gonal group G = ⟨φ⟩.
(1) If G is not a normal subgroup of Aut(S), then S is cyclically p-gonally definable
over an extension of degree at most two of MS .
(2) If G is a normal subgroup of Aut(S) and Aut(S)/G is different from the trivial
group or a cyclic group, then S is cyclically p-gonally definable over an extension of
degree at most 2(p − 1) of MS . Moreover, if φ also is defined over MS , then the
extension can be chosen to be of degree at most two.
(3) If G = Aut(S), then S is cyclically p-gonally definable over an extension of degree
at most 4(p− 1) of its field of moduli. Moreover, if φ also is defined over MS , then
the extension can be chosen of degree at most 4.
As every hyperelliptic Riemann surface is definable over an extension of degree at most
two of its field of moduli, Corollary 2.1 asserts the following.
Corollary 3.2. Every hyperelliptic Riemann surface is hyperelliptically definable over an
extension of degree at most 4 of its field of moduli. Moreover, if either (i) the genus is
even or (ii) the genus is odd and the reduced group is not trivial, then the hyperelliptic
Riemann surface is hyperelliptically defined over an extension of degree at most 2 of its
field of moduli.
Examples of hyperelliptic Riemann surfaces with a trivial reduced group that cannot be
defined over their field of moduli were provided by C. J. Earle [6, 7] and G. Shimura [24].
The same type of examples, but with a non-trivial cyclic reduced group, were provided by
B. Huggins [16].
R. A. Hidalgo: On p-gonal fields of definition 409
4 Proof of Theorem 1.2
We assume the p-gonal Riemann surface S to be provided by an irreducible curve C, de-
fined over a subfield K of C. If K is the algebraic closure of K inside C, then (in this
algebraic model) the p-gonal automorphism φ is given by a rational map defined over K.
We divide the arguments depending on the uniqueness of the cyclic group ⟨φ⟩.
4.1 The case when ⟨φ⟩ is not unique
The following result, due to A. Wootton, describes those cases where the uniqueness fails.
Theorem 4.1 ([28, A. Wootton]). Let S be a cyclic p-gonal Riemann surface of genus
g ≥ 2 and let m = 2(g+p−1)/(p−1). If (m, p) is different from any the following tuples
(i) (3, 7), (ii) (4, 3), (iii) (4, 5), (iv) (5, 3), (v) (p, p), p ≥ 5, (vi) (2p, p), p ≥ 3,
then S has a unique p-gonal group.
In the same paper, Wootton describes the exceptional cyclic p-gonal Riemann surfaces,
ie., where the p-gonal group is non-unique.
(i) Case (m, p) = (3, 7) corresponds to Klein’s quartic (a non-hyperelliptic Riemann
surface of genus 3) x3y+y3z+z3x = 0, whose group of automorphisms is PGL2(7)
(of order 168). This surface is cyclically 7-gonally defined as y7 = x2(x− z)z4.
(ii) Case (m, p) = (4, 3) corresponds to the genus 2 Riemann surface defined hyperel-
liptically by y2z3 = x(x4−z4), whose group of automorphisms is GL2(3) (of order
48). This surface is cyclically 3-gonally defined as y3z3 = (x2 − z2)(x2 − (15
√
3−
26)z2)2.
(iii) Case (m, p) = (4, 5) corresponds to the genus 4 non-hyperelliptic Riemann surface,
called Bring’s curve, which is the complete intersection of the quadric x1x4+x2x3 =
0 and the cubic x21x3 + x
2
2x1 + x
2
3x4 + x
2
4x2 = 0 in the 3-dimensional complex
projective space. Its group of automorphisms is S5, the symmetric group in five
letters S5. This surface is cyclically 5-gonally defined as y5z5 = (x2−z2)(x2+z2)4.
(iv) Case (m, p) = (5, 3) corresponds to the genus 3 non-hyperelliptic closed Riemann
surface x4+y4+ z4+2i
√
3z2y2 = 0, whose group of automorphisms has order 48.
The quotient of that surface by its group of automorphisms has signature (0; 2, 3, 12).
This surface is cyclically 3-gonally defined as y3z3 = x2(x4 − z4).
(v) Case (m, p) = (p, p), where p ≥ 5, corresponds to the Fermat curve xp+yp+zp = 0,
whose group of automorphisms is Z2p ⋊ S3. This is already in a p-gonal form as
yp = −zp − xp.
(vi) Case (m, p) = (2p, p), where p ≥ 3. There is a 1-dimensional family with group
of automorphisms Z2p ⋊ Z22 (the quotient by that group has signature (0; 2, 2, 2, p)).
Also, there is a surface with group of automorphisms Z2p ⋊D4 (the quotient by that
group has signature (0; 2, 4, 2p). These surfaces are cyclically p-gonally defined as
ypzp = (xp − apzp)(xp − zp/ap) = x2p − (ap + 1/ap)xpzp + z2p.
410 Ars Math. Contemp. 24 (2024) #P3.02 / 405–414
Note that, in all the above exceptional cases, the surface S is cyclically p-gonally de-
fined over an extension of degree at most 2 over the field of moduli. In fact, with only the
exception of case (ii), S is cyclically p-gonally defined over its field of moduli. So, we are
done in this situation.
4.2 The case when ⟨φ⟩ is unique
We now assume that ⟨φ⟩ is unique. Set Γ = Gal(K/K). Let us consider a rational map
π : C → P1K, defined over K, which is a regular branched covering with ⟨φ⟩ as its deck
group and whose branch values are a1, ..., am ∈ C (in fact, these values belong to K). Let
the integers n1, ..., nm ∈ {1, ..., p − 1}, n1 + · · · + nm ≡ 0 mod p, be such that C is
isomorphic to a p-gonal curve E with Equation (1.1).
4.2.1 Proof of Part (1)
Let us recall that φ is already defined over K. In the next, we note that φ is defined over an
extension of K of degree at most p− 1.
Claim 4.2. The rational map φ is defined over an extension K1 of K of degree at most
p− 1.
Proof. If σ ∈ Γ, then φσ is an automorphism of order p of Cσ = C. As we are assuming
the uniqueness of ⟨φ⟩, we must have that φσ ∈ Ω := {φ,φ2, . . . , φp−1}. In particular,
the subgroup A of Γ consisting of those σ such that φσ = φ must have index at most the
cardinality of the set Ω, which is p − 1. This asserts that φ is defined over the fixed field
K1 of A, which is an extension of degree at most p− 1 of K.
Set Γ1 = Gal(K/K1). If τ ∈ Γ1, then (as the identity I : C → C = Cτ conjugates
⟨φ⟩ = ⟨φ⟩τ = ⟨φτ ⟩ to itself), there is a (unique) automorphism gτ of P1K such that π
τ =
πτ ◦ I = gτ ◦ π (see the following diagram).
C
I−−−−→ C = Cτ
π
y πτy
P1K
gτ−−−−→ P1K
As the group of automorphisms of P1K is given by Möbius transformations (i.e., ele-
ments of PGL2(K)), we must have gτ ∈ PGL2(K).
We may apply each σ ∈ Γ1 to the above diagram to obtain the following one
Cσ = C
I−−−−→ C = Cστ
πσ
y πστy
P1K
gστ−−−−→ P1K
The above permits us to obtain the following diagram
R. A. Hidalgo: On p-gonal fields of definition 411
C
I−−−−→ C = Cσ I=I
σ
−−−−→ C = Cστ
π
y πσy πστy
P1K
gσ−−−−→ P1K
gστ−−−−→ P1K
As the transformation gρ is uniquely determined by ρ ∈ Γ1, the collection {gρ}ρ∈Γ1
satisfies the co-cycle relation
gστ = g
σ
τ ◦ gσ, σ, τ ∈ Γ1.
Weil’s descent theorem [25] ensures the existence of a genus zero irreducible and non-
singular algebraic curve B, defined over K1, and an isomorphism R : P1K → B, defined
over K, so that
gσ ◦Rσ = R, σ ∈ Γ1.
Also, for σ ∈ Γ1, we have {σ(a1), ..., σ(am)} = {gσ(a1), ..., gσ(am)}, so it follows
that {R(a1), ..., R(am)} is Γ1-invariant.
Let us denote by A(nj) the set of those ak’s for which nk = nj .
Claim 4.3. Each set R(A(nj)) is Γ1-invariant.
Proof. If σ ∈ Γ1, then (as πσ = gσ ◦ π) the set gσ(A(nj)) corresponds to the set of those
σ(ak) having the same nl (for some l), that is, gσ(A(nj)) = σ(A(nl)). As φσ = φ, we
must have nl = nj , that is, gσ(A(nj)) = σ(A(nj)). This last equality implies the desired
claim.
Claim 4.4. There is an effective K1-rational divisor U ≥ 0 of degree at most two in B.
Proof. We follow similar techniques as used by Huggins in her thesis [16] (and other au-
thors). Let us consider any K1-rational meromorphic 1-form ω in B. Since B has genus
zero, the canonical divisor K = (ω) is a K1-rational of degree −2. In this way, there is a
positive integer d such that the divisor D = R(a1) + · · ·+R(am) + dK is K1-rational of
degree 1 or 2. If D ≥ 0, then we set U := D.
Let us assume D is not effective. Let us consider the Riemann-Roch space L(D),
consisting of those non-constant rational maps ϕ : B → P1K whose divisors satisfy (ϕ) +
D ≥ 0 together with the constant ones. As the divisor D is K1-rational, for every σ ∈ Γ1
and every ϕ ∈ L(D), it follows that ϕσ ∈ L(D). This, in particular, permits us to observe
that we can find a basis of L(D) consisting of rational maps defined over K1. One of
the elements of such a basis must be a non-zero constant map. As, by Riemann-Roch’s
theorem, L(D) has dimension 2 (if D has degree one) or 3 (if D has degree two), we may
find a non-constant f ∈ L(D) belonging to such a basis (defined over K1). In this case, we
may take U = (f) +D ≥ 0.
By Claim 4.4, there is an effective K1-rational divisor U of degree 1 or 2 and U ≥ 0.
We have three possibilities:
(1) U = s, where s ∈ B is K1-rational; or
(2) U = 2t, where t ∈ B is K1-rational; or
412 Ars Math. Contemp. 24 (2024) #P3.02 / 405–414
(3) U = r + q, where r, q ∈ B, r ̸= q, and {r, q} is Γ1-invariant.
In cases (1) and (2) we have the existence of a K1-rational point in B. In this case, we
set K2 = K1. In case (3) we have a point (say r) in B which is rational over a quadratic
extension K2 of K1.
Let b ∈ B be a K2-rational point (whose existence is provided above). By Riemann-
Roch’s theorem, the Riemann-Roch space L(b) (where b is thought of as a divisor of degree
one) has dimension 2. Similarly as above, we may choose a basis {1, L} of L(b), with each
element defined over K2. In this case, L : B → Ĉ turns out to be an isomorphism defined
over K2.
We have that Q = L ◦R ◦ π : C → Ĉ is a Galois (branched) covering with deck group
⟨φ⟩ and whose branch values are {L(R(a1)), ..., L(R(am))}. It follows that S is p-gonally
defined by
yp = F (x) =
m∏
j=1
(x− L(R(aj)))nj .
As the sets {L(R(a1)), ..., L(R(am))} and L(R(A(nj))) are Gal(K/K2)-invariant
(by Claim 4.3 and the fact that K1 is a subfield of K2), it follows that F (x) =∏m
j=1 (x− L(R(aj)))
nj ∈ K2[x]. As K2 is an extension of degree at most two of K1
and the last one is an extension of degree at most p− 1 of K, we are done.
4.2.2 Proof of Parts (2) and (3)
If φ is already defined over K then we assume K1 = K (i.e., we set Γ1 = Γ) in the above
arguments. Similarly, if in Equation (1.1) we have that n1 = · · · = nm = n, then there
will be only one set A(n). In this case, in the previous arguments, we do not need to use
Claim 4.3 (where it was needed for the choice of K1) and we may work as in the proof of
Part (1) with K instead of K1.
ORCID iDs
Ruben A. Hidalgo https://orcid.org/0000-0003-4070-2819
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ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 24 (2024) #P3.03 / 415–425
https://doi.org/10.26493/1855-3974.2701.b7d
(Also available at http://amc-journal.eu)
An online bin-packing problem with an
underlying ternary structure
Helmut Prodinger
Mathematics Department, Stellenbosch University, 7602 Stellenbosch, South Africa and
NITheCS (National Institute for Theoretical and Computational Sciences), South Africa
Received 19 September 2021, accepted 30 June 2023, published online 12 June 2024
Abstract
Following an orginal idea by Knödel, an online bin-packing problem is considered
where the large items arrive in double-packs. The dual problem where the small items
arrive in double-packs is also considered. The enumerations have a ternary random walk
flavour, and for the enumeration, the kernel method is employed.
Keywords: Knödel walks, third-order recursion, kernel method, coefficient extraction, state diagram.
Math. Subj. Class. (2020): 05A15, 68R05
1 Introduction
Walter Knödel introduced the following online bin-packing problem [3]: There are bins of
size 1, and random items of size 23 (large items) and of size
1
3 (small items) appear and are
put into the boxes. A typical scenario is that a number j of partially filled boxes exist, and
the number j becomes j + 1 resp. j − 1, depending on whether a the new item is of large
resp. small type. “At random” means that both types appear with the same probability 12 .
In my collection of examples [4], I showed how to deal with the Knödel problem us-
ing the kernel method. I was, however, not the only author who was intrigued by such
questions; a notable paper is by Michael Drmota [1], which is of a more probabilistic type,
whereas I tried to emphasize the combinatorial point of view.
The present paper has a certain ‘ternary’ flavour: the next section deals with the instance
of large items appearing in double-packs. The handler breaks off the double-packs, and
then treats the items as Knödel would have done. Typically, the number of partially filled
boxes increases by 2 or decreases by 1. In order to keep the system balanced, we assume
that the small items appear twice as often as the double-packs.
E-mail address: hproding@sun.ac.za (Helmut Prodinger)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
416 Ars Math. Contemp. 24 (2024) #P3.03 / 415–425
The last section deals with the dual problem, where the small items appear in double-
packs and the large items as single units.
The kernel method is used to obtain all the relevant enumerations. The recent paper [5]
served as an inspiration, but deals with a different issue. It must be said that, when [4] was
prepared, such ternary questions would have been outside of my reach. Luckily, now, they
are not.
We confine ourselves here just to enumerations, deriving explicit generating functions
in one or two variables. Questions of a more probabilistic nature are not treated.
2 The first model
The following items arrive at random: a double-pack of items, each of size 23 , and an item
of size 13 . We could equip the set-up with general probabilities p and q = 1 − p, but we
restrict ourselves to the ‘balanced’ case where the single items are twice as likely as the
double-packs, so we set p = 13 and q =
2
3 .
The following state diagram (we show only a finite part of it) describes the situation.
There are states representing ‘i boxes filled to 23 ’; a double-pack pushes the i to i+ 2, and
a single item reduces it to i − 1. There is an exceptional state, called β, standing for one
box, filled to 13 . The red edges represent an arrival of a double-pack, and will be labelled
by pz; the black edges represent an arrival of a single item, and will be labelled by qz.
0 1 2 3 4 5 6 7 8
β
From the state diagram, we set off an infinite set of generating functions in the variable
z, where the coefficient of zn is the probability that n random steps lead to state i, for i ≥ 0
or i = β. Mostly, we just write fi instead of fi(z). The following system of recursions can
be read off immediately:
f0 = 1 + qzf1, fβ = qzf0,
f1 = zfβ + qzf2 = qz
2f0 + qzf2,
fi = pzfi−2 + qzfi+1, i ≥ 2.
Our method to solve this system is the kernel method. For that, we introduce a bivariate
H. Prodinger: An online bin-packing problem with an underlying ternary structure 417
generating function F (u, z), but we mostly write just F (u):
F (u) =
∑
i≥0
uifi(z)
= 1 + qzf1 + qz
2uf0 + qzuf2 +
∑
i≥2
ui
[
pzfi−2 + qzfi+1
]
= 1 + qzf1 + qz
2uf0 + pzu
2F (u) +
∑
i≥1
uiqzfi+1
= 1 + qzf1 + qz
2uf0 + pzu
2F (u) +
qz
u
(
F (u)− f0 − uf1
)
= 1 + qz2uf0 + pzu
2F (u) +
qz
u
(
F (u)− f0
)
Note that f0 = F (0). It is beneficial to introduce the new variable u = zU ; doing this,
powers of z that appear are multiples of 3. Later, it will be convenient to set x = z3. As can
be seen, the numbers of steps leading to a state i belong to just one residue class modulo 3.
We compute
F (u) =
−3U − 2z3U2f0 + 2f0
z3U3 − 3U + 2
=
−3U − 2xU2f0 + 2f0
xU3 − 3U + 2
.
As it is common using the kernel method, setting U = 0 leads to a void equation. How-
ever, factorizing the denominator is the method of choice. There is ‘bad’ factor in the
denominator, which must also appear in the numerator, which allows us to compute f0
and consequently the whole bivariate generating function. In order to deal with the ternary
equation successfully, we further set x = z3 = 274 t(1− t)
2 and we find the 3 roots
U1 =
2
3(1− t)
, U2 =
1
σ
, U3 =
1
τ
,
with
σ =
3
4
(t−
√
4t− 3t2 ), τ = 3
4
(t+
√
4t− 3t2).
A motivation for this substitution is the Lagrange inversion formula and/or the enumeration
of ternary trees; see also [6]. Plugging U = 23(1−t) into the numerator (this is the bad factor,
as explained a little bit later), leads to
f0 =
1
(1− t)(1− 3t)
and furthermore to the simplified numerator
−3U − 2xU2f0 + 2f0
U − 23(1−t)
=
1
1− 3t
(
− 3 + 27
2
t(t− 1)U
)
.
418 Ars Math. Contemp. 24 (2024) #P3.03 / 415–425
The variable x is given in terms of t. The inverse relation is of interest. It can be obtained
by the Lagrange inversion formula or, as here, by contour integration:
[xk]t =
1
2πi
∮
dx
xk+1
t =
1
2πi
27
4
( 4
27
)k+1 ∮ dt(1− t)(1− 3t)
tk+1(1− t)2k+2
t
=
1
2πi
( 4
27
)k ∮ dt(1− 3t)
tk(1− t)2k+1
=
( 4
27
)k
[tk−1]
1− 3t
(1− t)2k+1
=
( 4
27
)k[(3k − 1
k − 1
)
− 3
(
3k − 2
k − 2
)]
,
which, after simplification, gives us
t =
∑
k≥1
1
k
(
3k − 2
k − 1
)
22k
33k
xk.
A similar computation leads to
1
1− t
=
∑
k≥0
1
2k + 1
(
3k
k
)
22k+1
33k+1
xk.
From this we infer that for z ∼ 0, U ∼ 23 , or u ∼
2
3z, explaining why we are talking about
the bad factor. We continue the computation:
F (u) =
1
1− 3t
(
− 3 + 27
2
t(t− 1)U
) 1
x(U − 1σ )(U −
1
τ )
=
1
1− 3t
(
− 3 + 27
2
t(t− 1)U
) 9
4 t(t− 1)
x(1− σU)(1− τU)
=
1
(1− 3t)(1− t)
(
1− 9
2
t(t− 1)U
) 1
(1− σU)(1− τU)
.
Partial fraction decomposition leads to (we use the abbreviation W =
√
4t− 3t2 )
1
(1− σU)(1− τU)
=
1
2
(
1− t
W
) 1
1− σU
+
1
2
(
1 +
t
W
) 1
1− τU
=
1
2
[ 1
1− σU
+
1
1− τU
]
+
t
2W
[ 1
1− τU
− 1
1− σU
]
=
1
2
[ 1
1− σU
+
1
1− τU
]
+
3t
4(τ − σ)
[ 1
1− τU
− 1
1− σU
]
.
For the further simplification we will resort to two identities going by the name of Girard-
Waring formula, see e. g. [2]:
Xm + Y m =
∑
0≤k≤m/2
(−1)k
(
m− k
k
)
m
m− k
(XY )k(X + Y )m−2k;
Xm − Y m
X − Y
=
∑
0≤k≤(m−1)/2
(−1)k
(
m− 1− k
k
)
(XY )k(X + Y )m−1−2k.
H. Prodinger: An online bin-packing problem with an underlying ternary structure 419
Of course, we will apply them with X = τ and Y = σ. Then
[Um]
1
2
[ 1
1− σU
+
1
1− τU
]
=
1
2
∑
0≤k≤m/2
(−1)k
(
m− k
k
)
m
m− k
(9
4
t(t− 1)
)k(3
2
t
)m−2k
=
1
2
(3
2
)m ∑
0≤k≤m/2
(−1)k
(
m− k
k
)
m
m− k
tk(t− 1)ktm−2k
=
1
2
(3
2
)m ∑
0≤k≤m/2
(−1)k
(
m− k
k
)
m
m− k
(t− 1)ktm−k
and
[Um]
3t
4(τ − σ)
[ 1
1− τU
− 1
1− σU
]
=
3t
4
∑
0≤k≤(m−1)/2
(−1)k
(
m− 1− k
k
)(9
4
t(t− 1)
)k(3
2
t
)m−1−2k
=
(3
2
)m−1 3t
4
∑
0≤k≤(m−1)/2
(−1)k
(
m− 1− k
k
)
(t− 1)ktm−1−k
=
1
2
(3
2
)m ∑
0≤k≤(m−1)/2
(−1)k
(
m− 1− k
k
)
(t− 1)ktm−k.
Combining the two leads to a pleasant simplification:
[Um]
1
2
[ 1
1− σU
+
1
1− τU
]
+ [Um]
3t
4(τ − σ)
[ 1
1− τU
− 1
1− σU
]
=
(3
2
)m ∑
0≤k≤m/2
(−1)k
(
m− k
k
)
(t− 1)ktm−k,
or simpler
[Um]
1
(1− σU)(1− τU)
=
(3
2
)m ∑
0≤k≤m/2
(−1)k
(
m− k
k
)
(t− 1)ktm−k.
We need a second similar term:
[Um]
(
− 9
2
t(t− 1)U
) 1
(1− σU)(1− τU)
= −9
2
t(t− 1)[Um−1] 1
(1− σU)(1− τU)
= −9
2
t(t− 1)
(3
2
)m−1 ∑
0≤k≤m/2
(−1)k
(
m− 1− k
k
)
(t− 1)ktm−1−k
= −3
(3
2
)m ∑
0≤k≤m/2
(−1)k
(
m− 1− k
k
)
(t− 1)k+1tm−k.
420 Ars Math. Contemp. 24 (2024) #P3.03 / 415–425
Putting everything together we found
F (u) =
1
(1− 3t)(1− t)
∑
m≥0
um
zm
(3
2
)m ∑
0≤k≤m/2
(−1)k
(
m− k
k
)
(t− 1)ktm−k
− 3
(1− 3t)(1− t)
∑
m≥0
um
zm
(3
2
)m ∑
0≤k≤m/2
(−1)k
(
m− 1− k
k
)
(t− 1)k+1tm−k.
Reading off the coefficients of zj for j ≥ 1 as well is now done with Cauchy’s integral for-
mula; the contours are always small circles (or equivalent) around the origin. The starting
point is
[znuj ]F (u) =
(3
2
)j
[zn+j ]
1
(1− 3t)(1− t)
∑
0≤k≤j/2
(−1)k
(
j − k
k
)
(t− 1)ktj−k
−
(3
2
)j
[zn+j ]
3
(1− 3t)(1− t)
∑
0≤k≤j/2
(−1)k
(
j − 1− k
k
)
(t− 1)k+1tj−k
and we will treat the two sums separately. There is only a contribution if n+ j ≡ 0 mod 3.
(This is also clear from the combinatorial context.) Assume this and set N := n+j3 .
Step 1:
[xN ]
1
(1− 3t)(1− t)
∑
0≤k≤j/2
(−1)k
(
j − k
k
)
(t− 1)ktj−k
=
1
2πi
∮
dx
xN+1
1
(1− 3t)(1− t)
∑
0≤k≤j/2
(−1)k
(
j − k
k
)
(t− 1)ktj−k
=
1
2πi
∮
27
4
dt(
27
4 t(1− t)2
)N+1 ∑
0≤k≤j/2
(−1)k
(
j − k
k
)
(t− 1)ktj−k
=
1
2πi
∮ ( 4
27
)N
dt
∑
0≤k≤j/2
(−1)k
(
j − k
k
)
(t− 1)k−2N−2tj−k−N−1
=
( 4
27
)N
[tN−j+k]
∑
0≤k≤j/2
(−1)k
(
j − k
k
)
(t− 1)k−2N−2
=
( 4
27
)N
(−1)N−j
∑
0≤k≤j/2
(
j − k
k
)(
k − 2N − 2
N − j + k
)
.
Step 2:
[xN ]
3
(1− 3t)(1− t)
∑
0≤k≤j/2
(−1)k
(
j − 1− k
k
)
(t− 1)k+1tj−k
= 3
( 4
27
)N
[tN−j+k]
∑
0≤k≤j/2
(−1)k
(
j − 1− k
k
)
(t− 1)k−2N−1
= 3
( 4
27
)N
(−1)N−j
∑
0≤k≤j/2
(
j − 1− k
k
)(
k − 2N − 1
N − j + k
)
.
H. Prodinger: An online bin-packing problem with an underlying ternary structure 421
We put all the results of this section together in a theorem.
Theorem 2.1. The generating function F (u) = F (u, z) has the following explicit form:
F (u) =
1
(1− 3t)(1− t)
(
1− 9
2
t(t− 1)U
) 1
(1− σU)(1− τU)
Here, u = zU , z3 = x = 274 t(1− t)
2, and
σ =
3
4
(t−
√
4t− 3t2 ), τ = 3
4
(t+
√
4t− 3t2 ).
Note that (1 − σU)(1 − τU) = 1 − 32 tU +
9
4 t(t − 1)U
2. Written in the new variable U ,
only powers of z that are multiples of 3 appear. Further, we get the representation sorted
by powers of u:
F (u, z) =
1
(1− 3t)(1− t)
∑
m≥0
um
zm
(3
2
)m ∑
0≤k≤m/2
(−1)k
(
m− k
k
)
(t− 1)ktm−k
− 3
(1− 3t)(1− t)
∑
m≥0
um
zm
(3
2
)m ∑
0≤k≤(m−1)/2
(−1)k
(
m− 1− k
k
)
(t− 1)k+1tm−k.
Reading off coefficients of zNuj , where N = n+j3 leads to
[zNuj ]F (u, z) =
( 4
27
)N
(−1)N−j
∑
0≤k≤j/2
(
j − k
k
)(
k − 2N − 2
N − j + k
)
− 3
( 4
27
)N
(−1)N−j
∑
0≤k≤(j−1)/2
(
j − 1− k
k
)(
k − 2N − 1
N − j + k
)
.
For the special state β, the following series representation holds:
fβ(z) =
∑
n≥0
22n+1
33n+1
(
3n+ 1
n
)
z3n+1.
The computation for the special state was not shown yet:
[z3n+1]fβ =
2
3
[xn]
1
(1− t)(1− 3t)
=
2
3
1
2πi
∮
dx
xn+1
1
(1− t)(1− 3t)
=
2
3
27
4
1
2πi
∮
dt
( 274 )
n+1tn+1(1− t)2n+2
=
2
3
( 4
27
)n 1
2πi
∮
dt
tn+1(1− t)2n+2
=
2
3
( 4
27
)n
[tn]
1
(1− t)2n+2
=
22n+1
33n+1
(
3n+ 1
n
)
.
3 The dual model
Now, the red edges mean the arrival of the large objects (size 23 ) and the black edges mean a
double-pack of the small edges (size 13 each). To keep the system balanced, the large objects
should arrive twice as often as the double-packs of small edges. Again, the generating
422 Ars Math. Contemp. 24 (2024) #P3.03 / 415–425
function gi refers to paths of length n leading eventually into state i. After n steps, only
a state i can be reached with n ≡ i mod 3. The state diagram and the recursions are
immediate:
0 1 2 3 4 5 6 7 8
β
We work only with p = 23 , q =
1
3 . Directly from the state diagram,
g0 = 1 + zgβ + qzg2 = 1 + qz
2g1 + qzg2,
gβ = qzg1, g1 = zg0 + qzg3,
gi = pzgi−1 + qzgi+2, i ≥ 2.
Summing the recursions,
G(u) =
∑
i≥0
uigi(z) = g0 + ug1 +
∑
i≥2
ui
(
pzgi−1 + qzgi+2
)
= g0 + uzg0 + qzug3 + pzu
∑
i≥1
uigi +
qz
u2
∑
i≥4
uigi
= g0 + uzg0 + pzuG(u)− pzug0 +
qz
u2
∑
i≥3
uigi
= g0 + uzg0 + pzuG(u)− pzug0 +
qz
u2
(G(u)− g0 − ug1 − u2g2)
= g0 + quzg0 + pzuG(u) +
qz
u2
G(u)− qz
u2
g0 −
qz
u
g1 − qzg2
= g0 + quzg0 + pzuG(u) +
qz
u2
G(u)− qz
u2
g0 −
qz
u
g1 + 1 + qz
2g1 − g0.
Solving, we find with V = uz:
G(u) =
−V 3g0 − 3V 2 − g1V 2z2 + z3g0 + g1V z2
2V 3 − 3V 2 + x
.
Now we factorize the denominator, using the same substitutions x = z3 and x = 274 t(1 −
t)2:
2(V − 32 (1− t))(V − σ)(V − τ) = 2V
3 − 3V 2 + x.
This time, both, (V −σ) and (V − τ) are bad factors. Plugging into the numerator, we find
two equations, and the solutions:
g0 =
4
(1− 3t)(4− 3t)
, g1 =
27t(1− t)
z2(1− 3t)(4− 3t)
.
It can be noted that Ṽ := V −1, with the denominator of the previous section; thus, the
three roots carry over. Dividing out the bad factors, we find
−V 3g0 − 3V 2 − g1V 2z2 + z3g0 + g1V z2
(V − σ)(V − τ)
=
12(t− 1)− 4V
(1− 3t)(4− 3t)
.
H. Prodinger: An online bin-packing problem with an underlying ternary structure 423
Altogether:
G(u) =
6(t− 1)− 2V
(1− 3t)(4− 3t)
1
V − 32 (1− t)
=
6(1− t) + 2V
(1− 3t)(4− 3t) 32 (1− t)
1
1− 23(1−t)V
= 2
2 + 23(1−t)V
(1− 3t)(4− 3t)
1
1− 23(1−t)V
.
Furthermore
[V j ]G(u) =
2
(1− 3t)(4− 3t)
[
2
(2
3
1
1− t
)j
+
(2
3
1
1− t
)j]
=
6
(1− 3t)(4− 3t)
(2
3
1
1− t
)j
, j ≥ 1,
and
[uj ]G(u) = zj
6
(1− 3t)(4− 3t)
(2
3
1
1− t
)j
.
Now let us consider j + 3N steps to reach state j, and then
[zj+3Nuj ]G(u) = [xN ]
6
(1− 3t)(4− 3t)
(2
3
1
1− t
)j
=
1
2πi
∮
dx
xN+1
6
(1− 3t)(4− 3t)
(2
3
1
1− t
)j
=
27
4
( 4
27
)N+1 1
2πi
∮
dt(1− t)(1− 3t)
tN+1(1− t)2N+2
6
(1− 3t)(4− 3t)
(2
3
1
1− t
)j
=
( 4
27
)N(2
3
)j 1
2πi
∮
dt
tN+1(1− t)2N+j+1
6
(4− 3t)
=
( 4
27
)N(2
3
)j−1
[tN ]
1
(1− t)2N+j+1
1
(1− 34 t)
=
( 4
27
)N(2
3
)j−1 N∑
i=0
(3
4
)N−i(2N + j + i
i
)
=
N∑
i=0
22i+j−1
32N+i+j−1
(
2N + j + i
i
)
.
The coefficients of g0 are different:
[z3N ]g0 =
N∑
i=0
22i
32N+i
(
2N + i
i
)
.
Furthermore,
[z3N+1]gβ =
1
3
[z3N ]g1 =
N∑
i=0
22i
32N+i+1
(
2N + 1 + i
i
)
.
Here are the main results of this section:
424 Ars Math. Contemp. 24 (2024) #P3.03 / 415–425
Theorem 3.1. The generating function G(u) = G(u, z) has the following explicit form:
G(u) = 2
2 + 23(1−t)V
(1− 3t)(4− 3t)
1
1− 23(1−t)V
.
Here, u = Vz , z
3 = x = 274 t(1− t)
2. Written in the new variable V , only powers of z that
are multiples of 3 appear. Further, we get the representation sorted by powers of u:
[V j ]G(u) =
6
(1− 3t)(4− 3t)
(2
3
1
1− t
)j
, j ≥ 1,
and
[uj ]G(u) = zj
6
(1− 3t)(4− 3t)
(2
3
1
1− t
)j
.
Reading off coefficients of zj+3Nuj leads to
[zj+3Nuj ]G(u, z) =
N∑
i=0
22i+j−1
32N+i+j−1
(
2N + j + i
i
)
.
For the special cases, the following series representation holds:
[z3N ]g0 =
N∑
i=0
22i
32N+i
(
2N + i
i
)
,
[z3N+1]gβ =
1
3
[z3N ]g1 =
N∑
i=0
22i
32N+i+1
(
2N + 1 + i
i
)
.
ORCID iDs
Helmut Prodinger https://orcid.org/0000-0002-0009-8015
References
[1] M. Drmota, Discrete random walks on one-sided “periodic” graphs, in: Discrete Random
Walks, DRW’03. Proceedings of the conference, Paris, France, September 1–5, 2003, Maison
de l’Informatique et des Mathématiques Discrètes (MIMD), Paris, pp. 83–94, 2003.
[2] H. W. Gould, The Girard-Waring power sum formulas for symmetric functions and Fibonacci
sequences, Fibonacci Q. 37 (1999), 135–140, https://www.fq.math.ca/37-2.html.
[3] W. Knödel, Über das mittlere verhalten von online-packungsalgorithmen, Elektron. Informa-
tionsverarb. Kybernetik 19 (1983), 427–433.
[4] H. Prodinger, The kernel method: a collection of examples, Sémin. Lothar. Comb. 50 (2003),
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to ternary trees, in: Algorithmic Combinatorics: Enumerative Combinatorics, Special Functions
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ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 24 (2024) #P3.04 / 427–452
https://doi.org/10.26493/1855-3974.2710.f3d
(Also available at http://amc-journal.eu)
Variants of the domination number for flower
snarks
Ryan Burdett, Michael Haythorpe * , Alex Newcombe
Flinders University, 1284 South Road, Tonsley Park, SA, Australia
Received 25 October 2021, accepted 14 July 2023, published online 8 July 2024
Abstract
We consider the flower snarks, a widely studied infinite family of 3–regular graphs. For
the Flower snark Jn on 4n vertices, it is trivial to show that the domination number of Jn is
equal to n. However, results are more difficult to determine for variants of domination. The
Roman domination, weakly convex domination, and convex domination numbers have been
determined for flower snarks in previous works. We add to this literature by determining
the independent domination, 2-domination, total domination, connected domination, upper
domination, secure domination and weak Roman domination numbers for flower snarks.
Keywords: Flower, snarks, domination, variants, secure.
Math. Subj. Class. (2020): 05C69
1 Introduction
Consider a graph G containing the vertex set V and edge set E. A subset of the vertices
S ⊂ V is said to be a dominating set if every vertex in V \S is adjacent to at least one vertex
in S. Then, the domination problem is to determine the size of the smallest dominating set
in a given graph G, which is known as the domination number of G and is denoted by
γ(G).
There are obvious real-world applications for the domination problem. For example,
suppose that the vertices of a graph correspond to locations in a secure site, and that each
location needs to remain under observation by guards. If a guard at one location is able
to simultaneously observe another, there is an edge between the corresponding vertices.
Then, by placing guards at the sites corresponding to any dominating set, all locations are
under observation. Clearly, it is desirable to do so with as few guards as possible.
*Corresponding author.
E-mail addresses: ryan.burdett@flinders.edu.au (Ryan Burdett), michael.haythorpe@flinders.edu.au
(Michael Haythorpe), alex.newcombe@flinders.edu.au (Alex Newcombe)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
428 Ars Math. Contemp. 24 (2024) #P3.04 / 427–452
However, in many real-world applications, the definition of domination may be unsuit-
able in some way, and so variants of domination have been described to handle such cases.
Continuing the example in the previous paragraph, suppose that the guards carry out their
observation from the top of large towers. The vantage point from these towers enables
them to observe adjacent locations, but does not permit them to observe their own location.
Then, their location would need to be observed by a guard at an adjacent location. In such a
situation, we are seeking not a dominating set, but a total dominating set, which we define
formally now along with several other variants of dominating sets that we will consider in
this manuscript.
Definition 1.1. Consider a dominating set S ⊆ V . Then:
• S is a (weakly) convex dominating set if S is (weakly) convex in G.
• S is an independent dominating set if S is an independent set in G.
• S is a minimal dominating set if any proper subset of S is not a dominating set.
• S is a 2-dominating set if any vertex v ̸∈ S is adjacent to at least two vertices in S.
• S is a total dominating set if every vertex in V is adjacent to at least one vertex in S.
• S is a connected dominating set if the subgraph of G induced by the vertices in S is
connected.
• S is a secure dominating set if, for every vertex v ∈ V \ S, there exists a vertex
w ∈ S such that vw ∈ E, and (S \ {w}) ∪ {v} is a dominating set.
Analogously to the domination number, we define γwcon(G) to be the weakly convex
domination number, γcon(G) to be the convex domination number, i(G) to be the inde-
pendent domination number, γ2(G) to be the 2-domination number, γt(G) to be the total
domination number, γc(G) to be the connected domination number, and γs(G) to be the
secure domination number. We also define the upper domination number, Γ(G), as follows.
Definition 1.2. The upper domination number, denoted by Γ(G), is equal to the size of the
largest minimal dominating set.
In addition, we consider two more variants of domination. Suppose that for our graph
G, we have a function f : V → {0, 1, 2}. Then, the weight of f , denoted w(f), is equal to∑
v∈V f(v).
Definition 1.3. f is a Roman dominating function on G if it satisfies the condition that
every vertex v for which f(v) = 0 is adjacent to at least one vertex w for which f(w) = 2.
In the following definition, we say that a vertex v is undefended with respect to f if
f(v) = 0 and for every vertex w adjacent to v, f(w) = 0.
Definition 1.4. f is a weak Roman dominating function on G if, for every vertex v such
that f(v) = 0, there is at least one vertex w, adjacent to v and satisfying the following: if
we define a new function g : V → {0, 1, 2} defined by g(v) = 1, g(w) = f(w) − 1, and
g(u) = f(u) for all u ̸= {v, w}, then there are no undefended vertices with respect to g.
R. Burdett et al.: Variants of the domination number for flower snarks 429
Then, the Roman domination number γR(G) is the minimum weight over all Roman
dominating functions, and equivalently for the weak Roman domination number γr(G). It
is worth noting that if we further demand that f(v) ≤ 1 for all v ∈ V then weak Roman
domination is equivalent to secure domination. Hence, it is clear that γr(G) ≤ γs(G).
Domination numbers are known for some infinite families of graphs. Other than trivial
results such as for paths or cycles, perhaps the most famous result is the sprawling effort
over a 27 year period [2, 8, 14, 16, 20, 36] to provide a complete characterisation of domi-
nation numbers for grid graphs G(n,m) of all possible sizes, consisting of 23 special cases
before settling into a standard formula for n,m ≥ 16. Other results for domination include
generalized Petersen graphs [13, 24, 41], Cartesian products involving cycles [1, 9, 28],
King graphs [40], Latin square graphs [29], hypercubes [3], Sierpiński graphs [34], Knödel
graphs [12], and various graphs from chemistry [26, 32], among others.
However, far fewer results are known for variants of domination beyond trivial results
such as for paths, cycles, stars, wheels, or complete graphs. We summarise the most note-
worthy of these results for the variants of domination considered in this paper. For upper
domination, results are known for various graphs based on chessboards [4, 11, 18, 39, 40].
For total domination, results are known for some grid graphs G(n,m) (for n ≤ 6) [15, 21],
Knödel graphs [27], and various graphs from chemistry [26]. For connected domination,
results are known for trees [33], circulant graphs [35], Centipede graphs [38], and various
graphs from chemistry [26]. For weak Roman domination, results are known for various
graphs based on chessboards [30], Cartesian products involving complete graphs [37], and
Helm graphs and Web graphs [23]. For secure domination, results are known for some grid
and torus graphs (for n ≤ 3) [17], various Cartesian products of stars, cycles, paths and
complete graphs [37], and middle graphs [31].
In recent works, the following results were shown for flower snarks (which will be
defined in the next section).
Theorem 1.5 ([25, Maksimovic et al. (2018)], [22, Kratica et al. (2020)]). Consider the
flower snark Jn, for n ≥ 3. Then we have γR(Jn) = γwcon(Jn) = 2n, and γcon(Jn) =
4n.
We now add to the above literature by proving the following additional results for flower
snarks:
Theorem 1.6. Consider the flower snark Jn, for n ≥ 3. Then,
γ(Jn) = i(Jn) = n, γs(Jn) = γr(Jn) =
⌈
3n+ 1
2
⌉
,
γ2(Jn) =
{⌈
5n
3
⌉
, if n ̸= 1 mod 3,
5n+4
3 , if n = 1 mod 3,
γt(Jn) =
{⌈
3n
2
⌉
, if n ̸= 2 mod 4,
3n
2 + 1, if n = 2 mod 4,
γc(Jn) = Γ(Jn) =
{
2n, if n is even,
2n− 1, if n is odd.
In most cases, the proofs will be by induction, and hence it will be necessary to first
prove the results for some number of base cases. Rather than provide these proofs here, we
will simply use mixed-integer linear programming formulations of each variant of domina-
tion from literature to handle the base cases.
430 Ars Math. Contemp. 24 (2024) #P3.04 / 427–452
2 Flower snarks
The chromatic index of a graph is the minimum required number of colours to color the
edges of the graph, such that no two incident edges have the same colour. By Vizing’s
theorem, it is known that all 3–regular graphs have chromatic index 3 or 4. The latter
case is rare, and simple, connected, bridgeless 3–regular graphs with chromatic index 4 are
called snarks. It is common to further add the restriction that the girth should be at least
5, with such graphs known as nontrivial snarks. Flower snarks [19], discovered by Isaacs
in 1975, were the first known infinite family of nontrivial snarks, and are denoted by Jn.
They are defined as follows.
Definition 2.1 (Flower snarks). For n ≥ 3, take the union of n copies of K1,3. Denote the
degree 3 vertex in the i-th copy as ai, and the other three vertices in the i-th copy as bi, ci
and di. Then construct an n-cycle through vertices b1, b2, . . . , bn, and a 2n-cycle through
vertices c1, c2, . . . , cn, d1, d2, . . . , dn.
In order to have the properties of a nontrivial snark, n must be odd and n ≥ 5. However,
for other values of n ≥ 3 a 3–regular graph is nonetheless obtained by this construction. In
this paper we will consider all n ≥ 3, and for convenience we will refer to all of them as
flower snarks.
For the remainder of this document, we will consider various kinds of dominating sets
of flower snarks. As such, it is convenient to go over some brief terms and notation here.
Recall that Jn contains n copies of K1,3. We will refer to the i-th copy as J i, and its four
vertices as ai, bi, ci and di. Note that this notation does not include n, as typically n will be
fixed in our considerations. Also, we will say that a copy J i has weight k in a dominating
set S, if S contains k vertices from J i. We will use the term pattern to describe a sequence
of weights of consecutive copies of Jn in S. For example, if we say that S contains the
pattern 121, it means there is a set of three consecutive copies J i−1, J i, J i+1 which have
weights 1, 2 and 1 respectively in S. Further to this, in a set of consecutive copies of Jn,
we will use the term configuration to describe the specific allocation of its vertices to S.
Finally, we will define wSi to be equal to the number of copies with weight i in S.
Flower snarks can be visualised in various ways. A common method is to distribute the
copies of K1,3 in a circle, using curved edges for the 2n-cycle and straight edges elsewhere.
We display one such drawing of Jn in part (a) of Figure 1, for n = 7. However, for our
purposes it will be convenient to focus only on a small section of a flower snark at a time,
and so we will use the drawing style displayed in part (b) of Figure 1, with each copy
displayed vertically. As indicated in Figure 1 we will assume that in these drawings, the
bottom vertex in copy J i is bi, followed by ai, ci and di. When useful to avoid confusion,
a label will be given above each copy.
It is worth noting that, when viewing only a section of Jn, vertices bi, ci and di are all
essentially equivalent, and this will be useful in simplifying many of the upcoming proofs.
In a global sense this is not the case, as there is a “twist” in the final copy in which cn links
to d1, and dn links to c1. However, due to the symmetry of the flower snark, when viewing
any copy locally we can choose to relabel the vertices so that this twist occurs elsewhere in
the graph. Hence, in the arguments that follow, whenever we are viewing only a portion of
the graph we will always assume that the twist occurs elsewhere in the graph.
Various proofs in this paper will go as follows. We will begin with a set of copies of Jn
for which the pattern is known, as well as possibly knowing in advance that some vertices
are either in S, or not in S. From there, depending on the variant of domination being
R. Burdett et al.: Variants of the domination number for flower snarks 431
(a)
a
b
c
d
J1 J2 J3 J4 J5 J6
(b)
Figure 1: In part (a) a common drawing of the flower snark J7. In part (b), a section of a
larger flower snark consisting of six copies.
considered, and the structure of Jn, we will go on to prove that certain vertices either must
be in S, or must not be in S. For these proofs, figures will be provided, using the following
convention. The pattern known in advance will be indicated by listing the corresponding
weight underneath each copy. Vertices which are known in advance to be in S will be
marked with , and vertices which are known in advance not to be in S will be marked
with . Then, vertices which are subsequently shown (up to equivalence) to be in S will be
marked with , while vertices which are subsequently shown not to be in S will be marked
with . We demonstrate this convention with a simple example.
Example 2.2. Suppose we have four copies J1, J2, J3, J4 which meet the pattern 1111,
and that we know that d2 ∈ S and c1 ̸∈ S. This situation is displayed in Figure 2. Clearly,
since d2 ∈ S and J2 has weight 1, we know that a2 ̸∈ S, b2 ̸∈ S, and c2 ̸∈ S. These
three are marked with in Figure 2 as no argument was needed to establish they are not in
S. Then, the only remaining vertex which can dominate c2 is c3, and hence c3 ∈ S. Since
this was argued in the proof, c3 is marked with a in Figure 2. Also, since J3 has weight
1 the vertices a3, b3, and d3 cannot be in S, and so they are marked with . Then, the
only remaining vertex which can dominate b2 is b1, which we similarly mark in Figure 2.
Finally, the only remaining vertex which can dominate b3 is b4, which we again mark in
Figure 2. Hence, we now know exactly which vertices in J1, J2, J3, J4 are contained in
S.
a
b
c
d
J1 J2 J3 J4
1 1 1 1
Figure 2: The situation described in Example 2.2.
To conclude this section, we note that it is trivial to determine the domination and
independent domination numbers for flower snarks.
Lemma 2.3. For n ≥ 3, we have γ(Jn) = i(Jn) = n.
Proof. The graph Jn contains n copies of K1,3. For each copy J i, the vertex ai is ad-
jacent only to other vertices in J i, and so any dominating set must contain at least one
432 Ars Math. Contemp. 24 (2024) #P3.04 / 427–452
vertex from each copy. Hence, n ≤ γ(Jn) ≤ i(Jn). Then, it suffices to note that the set
{a1, a2, . . . , an} is an independent dominating set, and hence i(Jn) ≤ n, leading to the
result.
Corollary 2.4. For any variant of dominating set considered in this paper, each copy must
have weight at least 1.
3 Upper domination
In this section, we will determine the upper domination number for flower snarks. We
begin by considering what weights are possible for copies in a minimal dominating set.
Lemma 3.1. Consider the graph Jn for n ≥ 3. If S is a minimal dominating set for Jn
then the weight of each copy is either 1, 2 or 3.
Proof. We know from Corollary 2.4 that each copy must have positive weight. Then sup-
pose that a copy J i has weight 4. It can be easily checked that S \{ai} is also a dominating
set, contradicting the assumption that S is minimal.
In the following lemma, we use the term i depends on j to imply that S ∩N [i] = {j}.
Note that since S is minimal, for every vertex in S there must be at least one other vertex
which depends on it.
Lemma 3.2. Consider the graph Jn for n ≥ 4, and a minimal dominating set S. If a copy
J i has weight 3 in S, then both adjacent copies J i−1 and J i+1 have weight 1 in S.
Proof. Suppose it is not the case, that is, J i has weight 3 and at least one of J i−1 and J i+1
has weight greater than 1. Since bi, ci and di are all equivalent in this framing, there are
only two cases to consider; if ai ∈ S and if ai ̸∈ S.
First, consider the case when ai ∈ S. Then there are two other vertices from J i also
in S. Without loss of generality, suppose bi ∈ S and ci ∈ S. This situation is shown
in Figure 3 part (a). Then, since S is minimal, the removal of ai does not result in a
dominating set. This is only possible if di depends on ai. Hence, di−1 ̸∈ S and di+1 ̸∈ S.
Then, consider bi. Since S is minimal, at least one of bi−1 and bi+1 must depend on bi.
Without loss of generality, suppose it is the former. This implies that both bi−1 ̸∈ S and
ai−1 ̸∈ S. Then, ci−1 ∈ S, or else copy Ji−1 has weight 0 which contradicts Lemma 3.1.
Then, since S is minimal, ci+1 must depend on ci, which implies that ci+1 ̸∈ S and
ai+1 ̸∈ S. Hence, from Lemma 3.1 copy J i+1 also has weight 1, contradicting the initial
assumption.
Second, consider the case when ai ̸∈ S. Then, S contains bi, ci, and di. Since S
is minimal and ai does not depend on bi, at least one of bi−1 or bi+1 must depend on
bi. Without loss of generality, suppose it is the former. This implies that both bi−1 ̸∈ S
and ai−1 ̸∈ S. We can make analogous arguments for ci and di. Clearly, the dependent
vertices can not all be from the same copy, or else that copy has weight 0. Hence, two of
the dependent vertices are in one copy and one is in the other; without loss of generality we
will assume that bi−1 is dependent on bi, ci−1 is dependent on ci, and di+1 is dependent on
di. Hence, J i−1 has weight 1, and so by assumption, J i+1 has weight 2, and bi+1 ∈ S and
ci+1 ∈ S. But then, by similar arguments, it must be the case that bi+2 depends on bi+1,
and ci+2 depends on ci+1, implying that J i+2 has weight 1, and di+2 ∈ S. This situation
is displayed in Figure 3 part (b). Finally, it can be seen from this figure that S \ {di} is a
dominating set, contradicting the initial assumption that S is minimal.
R. Burdett et al.: Variants of the domination number for flower snarks 433
a
b
c
d
(a)
a
b
c
d
(b)
Figure 3: The two situations described in the proof of Lemma 3.2.
Recalling that wSi is the number of copies to have weight i in S, Lemma 3.2 implies
the following.
Corollary 3.3. Consider the graph Jn for n ≥ 4. If S is a minimal dominating set for Jn
then wS1 ≥ wS3 , with equality occurring if and only if wS1 = wS3 = n2 .
We are now ready to prove the main result of this section.
Theorem 3.4. Consider a graph Jn for n ≥ 3. Then,
Γ(Jn) =
{
2n, if n is even,
2n− 1, if n is odd.
Proof. We use the first formulation for upper domination from [6] to confirm that Γ(J3) =
5. Then, suppose that S is a minimal dominating set for Jn, for n ≥ 4. From Lemma 3.1
we have wS0 = w
S
4 = 0. Hence, we have w
S
1 +w
S
2 +w
S
3 = n, and w
S
1 +2w
S
2 +3w
S
3 = |S|.
Combining these, we obtain |S| = 2n + wS3 − wS1 . From Corollary 3.3 we know that
wS1 ≥ wS3 , and the inequality is strict if n is odd. Hence, |S| ≤ 2n if n is even, and
|S| ≤ 2n− 1 if n is odd. Then we just need to obtain the corresponding lower bounds.
Suppose that n is even. It is easy to check that if we repeat the configuration displayed
in Figure 4 part (a) n/2 times, what results is a minimal dominating set with weight 2n.
Hence, if n is even, we have Γ(Jn) ≥ 2n. Then suppose that n is odd. Again, it is easy
to check that if we repeat the configuration displayed in Figure 4 part (a) (n− 1)/2 times,
and then use the configuration displayed in Figure 4 part (b) for the final copy, what results
is a minimal dominating set with weight 2n− 1. Hence, if n is odd, Γ(Jn) ≥ 2n− 1.
a
b
c
d
(a)
a
b
c
d
(b)
Figure 4: The configuration for upper domination which gives the desired lower bound for
Γ(Jn) for n ≥ 3. Part (a) can be repeated as many times as necessary. Then, if n is odd,
use part (b) to finish. The result is a minimal dominating set with weight 2n if n is even, or
weight 2n− 1 if n is odd.
434 Ars Math. Contemp. 24 (2024) #P3.04 / 427–452
4 Upper bounds by construction
In the upcoming sections, we will determine lower bounds for the 2-domination, total dom-
ination, connected domination, secure domination, and weak Roman domination numbers
of flower snarks. To obtain equality, we will require corresponding upper bounds, which
we provide here. We leave it as an exercise to the reader to verify that the configurations
given here satisfy the requirements of the various kinds of domination, and that they imply
the appropriate upper bounds for Theorem 1.6.
For 2-domination, the configuration shown in Figure 5 can be repeated as often as
necessary, truncating the final time to get the desired size.
a
b
c
d
Figure 5: The configuration for 2-domination which gives the desired upper bound for
γ2(Jn). The result is a 2-dominating set of weight ⌈ 5n3 ⌉ if n ̸= 1 mod 3, or weight ⌈
5n
3 ⌉+1
if n = 1 mod 3.
For total domination, the configuration shown in Figure 6 part (a) can be repeated as
often as necessary. Then the configurations in parts (b), (c) and (d) can be used to complete
the remaining copies.
a
b
c
d
(a)
a
b
c
d
(b)
a
b
c
d
(c)
a
b
c
d
(d)
Figure 6: The configuration for total domination which gives the desired upper bound for
γt(Jn) for n ≥ 3. Part (a) can be repeated as many times as necessary. If n = 0 mod 4
this is sufficient. If n = 1 mod 4, use (b) to finish. If n = 2 mod 4, use part (c) to finish.
If n = 3 mod 4, use part (d) to finish. The result is a total dominating set of weight ⌈ 3n2 ⌉
if n ̸= 2 mod 4, or weight 3n2 + 1 if n = 2 mod 4.
For connected domination, the configuration shown in Figure 7 can be repeated as often
as necessary, truncating the final time to get the desired size. If n ̸= 1 mod 4 the result is
connected dominating. If n = 1 mod 4, then removing d1 and adding b1 gives the desired
result.
For secure domination, the configuration shown in Figure 8 part (a) can be repeated as
often as necessary. Then the configurations in parts (b), (c), (d), and (e) can be used to com-
plete the remaining copies. Since secure domination is an upper bound for weak Roman
domination, these configurations also give an upper bound for weak Roman domination.
R. Burdett et al.: Variants of the domination number for flower snarks 435
a
b
c
d
Figure 7: The configuration for connected domination which gives the desired upper bound
for γc(Jn) for n ≥ 3. It may be repeated as many times as necessary, truncating the final
time to get the final size. If n = 1 mod 4, then in the first copy, remove d1 and add b1. The
result is a connected dominating set of weight 2n if n is even, or 2n− 1 if n is odd.
a
b
c
d
(a)
a
b
c
d
(c)
a
b
c
d
(d)
a
b
c
d
(b)
a
b
c
d
(e)
Figure 8: The configuration for secure domination which gives the desired upper bound for
γs(Jn) for n ≥ 4. Part (a) can be repeated as many times as necessary. If n = 0 mod 4,
use part (b) to finish. If n = 1 mod 4, use part (c) to finish. If n = 2 mod 4, use part (d) to
finish. If n = 3 mod 4, use part (e) to finish. In all cases, the result is a secure dominating
set of weight ⌈ 3n+12 ⌉.
5 2-domination
In this section, we will determine the 2-domination numbers for flower snarks.
Lemma 5.1. Consider the graph Jn for n ≥ 3, and a 2-dominating set S. If the copy J i
has weight 1 in S, then ai ∈ S.
Proof. Recall that ai is adjacent only to other vertices in J i. From the definition of 2-
domination, if ai ̸∈ S then it must have at least two neighbours in S. Since J i has weight
1, this is impossible, and so ai ∈ S.
Theorem 5.2. Consider the graph Jn for n ≥ 3, and a 2-dominating set S. Then any copy
with weight 1 in S has a neighbouring copy with weight 3 or 4 in S.
Proof. Without loss of generality, suppose that J2 has weight 1. Then, from Lemma 5.1
we have a2 ∈ S. Then, suppose that its neighbours J1 and J3 both have weight less than 3.
At this stage, vertices b2, c2 and d2 have only one neighbour in S, so they each need at least
one more. Without loss of generality, suppose that b1 ∈ S. Then, consider the case when
c1 ∈ S. Since J1 has weight less than 3, this implies that a1 ̸∈ S and d1 ̸∈ S. However,
it is then impossible for d1 to have two neighbours in S. This situation is displayed in part
(a) of Figure 9.
436 Ars Math. Contemp. 24 (2024) #P3.04 / 427–452
a
b
c
d
J1 J2
(a)
a
b
c
d
J1 J2 J3
(b)
Figure 9: The two situations described in the proof of Theorem 5.2. In part (a), d1 cannot
have two neighbours in S. In part (b), copy b3 cannot have two neighbours in S.
Hence, we must have c1 ̸∈ S. Then, c1 must have two neighbours in S, which implies
that a1 ∈ S. Since J1 has weight less than 3, this implies that d1 ̸∈ S. Then, in order for
c2 and d2 to have two neighbours in S, we must have c3 ∈ S and d3 ∈ S, respectively.
Since J3 has weight less than 3, this implies that a3 ̸∈ S and b3 ̸∈ S. However, it is
then impossible for b3 to have two neighbours in S, completing the proof. This situation is
displayed in part (b) of Figure 9.
We are now ready to prove the main result of this section.
Theorem 5.3. Consider the graph Jn for n ≥ 3. Then,
γ2(Jn) =
{⌈
5n
3
⌉
, if n ̸= 1 mod 3,
5n+4
3 , if n = 1 mod 3.
Proof. The upper bound was established in Section 4. Then, from Corollary 2.4 we know
that each copy has weight at least 1. This, combined with Theorem 5.2, implies that any
three set of consecutive copies have weight at least 5. Hence we have γ2(S) ≥ ⌈ 5n3 ⌉.
Hence, Theorem 5.3 is true for n ̸= 1 mod 3.
Suppose that n = 1 mod 3, and we have a 2-dominating set S such that |S| < 5n+43 .
Hence, we must have |S| = 5n+13 . For each copy J
i, denote by w3(i) the combined
weights of J i, J i+1, and J i+2, where the superscripts are taken modulo n. Recall that
w3(i) ≥ 5. Hence there must be a single value k such that w3(k) = 6 and w3(j) = 5 for
all j ̸= k. Consider the copies Jk, Jk+1, Jk+2. We will consider the possible patterns they
could have, and show that each is impossible. Since w3(k) has weight 6, the only possible
patterns (up to symmetry) for copies Jk, Jk+1 and Jk+2 are 123, 132, 222, and 312.
Suppose that Jk, Jk+1 and Jk+2 meet either the pattern 123 or the pattern 132. Since
w3(k + 1) = 5, this implies that Jk+3 has weight 0, which contradicts Corollary 2.4.
Suppose next that Jk, Jk+1 and Jk+2 meet the pattern 222. Since w3(k + 1) =
w3(k+2) = 5, this implies that Jk+3 has weight 1, and Jk+4 has weight 2. However, this
contradicts Theorem 5.2 since Jk+3 has no neighbour with weight 3 or 4. Hence, this is
impossible.
Finally, suppose that Jk, Jk+1 and Jk+2 meet pattern 312. Since w3(k+1) = w3(k+
2) = w3(k+3) = 5, this implies that Jk+3 has weight 2, Jk+4 has weight 1, and Jk+5 has
weight 2. Again, this contradicts Theorem 5.2 since Jk+4 has no neighbour with weight 3
or 4. Hence, all cases are impossible, and so |S| ≥ 5n+43 , completing the proof.
R. Burdett et al.: Variants of the domination number for flower snarks 437
6 Total domination
In this section, we will determine the total domination numbers for flower snarks. We begin
by identifying three patterns which cannot occur in total dominating sets of Jn.
Theorem 6.1. Consider the graph Jn for n ≥ 3, and a total dominating set S. Then S
does not contain the patterns 111, 1121, or 12121.
Proof. From Corollary 2.4, each copy of Jn has weight at least 1 in S. Also, if a copy J i
has weight 1, then ai ̸∈ S because otherwise ai itself is not dominated.
Suppose that S has the pattern 111. Without loss of generality, suppose the three copies
meeting this pattern are J1, J2, J3. As indicated above, a2 ̸∈ S. Regardless of which
vertex from J2 is in S, it does not dominate any of b2, c2, d2. Also, any vertex from J1
dominates at most one vertex from J2, and likewise for any vertex from J3. Hence there
is at least one vertex in J2 which is not dominated, contradicting the assumption that S is
a total dominating set.
Then, suppose that S has the pattern 1121. Without loss of generality, suppose the four
copies meeting this pattern are J1, J2, J3, J4. Using an equivalent argument to above, it
must be the case that b2, c2, d2 are dominated by one vertex from J1 and two vertices from
J3. Hence, a3 ̸∈ S. This means none of the vertices from J3 dominate any of b3, c3, d3.
Also, any vertex from J2 dominates at most one vertex from J3, and likewise for any vertex
from J4. Hence there is at least one vertex in J3 which is not dominated, contradicting the
assumption that S is a total dominating set.
Finally, suppose that S has the pattern 12121. Without loss of generality, suppose the
five copies meeting this pattern are J1, . . . , J5. Since J1 and J3 are both weight 1, they can
collectively dominate at most two vertices from J2. Hence, a2 ∈ S, and by an equivalent
argument, a4 ∈ S. Note that none of the vertices from J3 can dominate any of b3, c3
or d3. Also, the vertices from J2 dominate at most one vertex from J3, and likewise for
the vertices from J4. Hence there is at least one vertex in J3 which is not dominated,
contradicting the assumption that S is a total dominating set.
An immediate observation arising from Corollary 2.4 and Theorem 6.1 is that any four
consecutive copies must collectively have weight at least 6 in S, leading to the following
corollary.
Corollary 6.2. For n ≥ 4, γt(Jn) ≥
⌈
3n
2
⌉
.
We are now ready to prove the main result of this section.
Theorem 6.3. Consider the graph Jn for n ≥ 3. Then,
γt(Jn) =
{⌈
3n
2
⌉
, if n ̸= 2 mod 4,
3n
2 + 1, if n = 2 mod 4.
Proof. The upper bound was established in Section 4. We use the formulation for total
domination from [7] to confirm that γt(J3) = 5. Then, suppose there is some value k ≥
4 such that Theorem 6.3 is false. If k ̸= 2 mod 4 then we have γt(Jk) ≤
⌈
3k
2
⌉
− 1,
contradicting Corollary 6.2. Hence, we must have k = 2 mod 4. Hence, γt(Jk) ≤ 3k2 , and
from Corollary 6.2 this implies that γt(Jk) = 3k2 .
438 Ars Math. Contemp. 24 (2024) #P3.04 / 427–452
Suppose that we have a total dominating set S with weight 3k2 . Note that this implies
that every set of four consecutive copies of Jn has weight 6 in S; call this property 1. It
is clear that property 1 implies that no copy has weight 4 in S. If a copy has weight 3
in S, then in order to satisfy property 1, the next three copies must have weight 1, which
contradicts Theorem 6.1. Hence, every copy has weight 1 or 2 in S; call this property 2.
It can be easily checked that there are only two ways to satisfy properties 1 and 2 without
contradicting Theorem 6.1; either S contains the repeated pattern 1212 . . . 12, or S contains
the repeated pattern 11221122 . . . 1122. In the former case, S contains the pattern 12121
which by Theorem 6.1 is impossible. Hence, we must have the latter case. Also, the latter
case is impossible because k = 2 mod 4. Hence, all cases are impossible, completing the
proof.
7 Connected domination
In this section, we will determine the connected domination numbers for flower snarks.
The following three remarks are obvious, but we list them here to aid the readability of two
upcoming proofs.
Remark 7.1. Suppose that S is a connected dominating set for a graph G. If a new graph
H is created by either (a) adding an edge between two non-adjacent vertices in G, or (b)
deleting a vertex v ̸∈ S from G, then S is also a connected dominating set for H .
Remark 7.2. Suppose that S is a connected dominating set for a graph G, and that G
contains a degree 1 vertex v ∈ S whose neighbour has degree larger than 1. If a new graph
H is created by deleting v from G, then S \ {v} is a connected dominating set for H .
Remark 7.3. Suppose that S is a connected dominating set for a graph G, and that G
contains a triangle uvw such that v is degree 2, and v ∈ S. If a new graph H is created by
deleting v from G, then S \ {v} is a connected dominating set for H .
In the following, we define V S(J i) := S ∩ (ai, bi, ci, di), that is, V S(J i) is the set of
vertices from J i that are contained in S.
Theorem 7.4. Consider the graph Jn for n ≥ 4, and let S be a connected dominating set
for Jn. Suppose that there is a copy J i such that either ai ̸∈ S, or J i has weight 2 in S.
Then S := S \ V S(J i) is a connected dominating set for Jn−1.
Proof. Consider first the situation when ai ̸∈ S, and consider the graph Jn−1. One may
think of Jn−1 as being constructed by starting with Jn and then “smoothing out” copy J i,
in the following sense. First, we add edges connecting bi−1 to bi+1, ci−1 to ci+1 and di−1
to di+1, and then we delete the vertex set V (J i). Suppose that we are midway through this
process, having added all three edges, and having deleted ai. Call this intermediate graph
G, and note from Remark 7.1 that S is a connected dominating set for G. Then, note that in
G, bi is a degree 2 vertex and is part of a triangle bi−1bibi+1. Now, suppose that we delete
bi from G, to obtain a new intermediate graph G2. If bi ̸∈ S then from Remark 7.1 we can
see that S is a connected dominating set for G2. If bi ∈ S then from Remark 7.3 we can
see that S \ {bi} is a connected dominating set for G2. Applying analogous arguments for
ci and di, we obtain the result. The graphs G and G2 are displayed in Figure 10.
We next consider the situation when J i has weight 2 in S. If ai ̸∈ S then the previous
paragraph applies. If ai ∈ S then there must be one more vertex from J i also in S; without
R. Burdett et al.: Variants of the domination number for flower snarks 439
loss of generality, suppose that bi ∈ S. Again, we construct Jn−1 by adding edges to
and deleting vertices from Jn. Suppose that we are midway through this process, having
added all three edges, and having deleted ci and di. Call this intermediate graph G3. From
Remark 7.1 we know that S is a connected dominating set for G3. Note that in G3, ai is a
degree 1 vertex. Remark 7.2 implies that we can obtain a second intermediate graph, G4,
by deleting ai, and that S \ {ai} is a connected dominating set for G4. Finally, note that in
G4, bi is degree 2 vertex and is part of a triangle bi−1bibi+1. Hence, Remark 7.3 implies
the result. The graphs G3 and G4 are displayed in Figure 10.
a
b
c
d
G
a
b
c
d
G2
a
b
c
d
G3
a
b
c
d
G4
Figure 10: Sections of the four intermediate graphs G, G2, G3, G4 from the proof of
Theorem 7.4.
We also make use of the concept of “smoothing out” a copy in the following lemma.
Lemma 7.5. Consider the graph Jn for n ≥ 4, and let S be a connected dominating set
for Jn. Suppose that there is a copy J i with weight 4 in S. Then for any positive integer
k ≤ n−3, we have that S := S \(V S(J i+1)∪ . . .∪V S(J i+k)) is a connected dominating
set for Jn−k.
Proof. Suppose that we delete from Jn all vertices from each of the copies J i+1, . . . , J i+k.
Call this intermediate graph G. Then, we add edges connecting bi to bi+k+1, ci to ci+k+1
and di to di+k+1, and call the resulting graph G2. Since we can always relabel the vertices
so that the twist occurs elsewhere in the graph, it can then be seen that G2 is isomorphic to
Jn−k. Both G and G2 are displayed in Figure 11. Since bi ∈ S, ci ∈ S, and di ∈ S, it
is clear that vertices bi+k+1, ci+k+1 and di+k+1 are dominated in G2. Also, from Corol-
lary 2.4 we know that J i+k+1 has weight at least 1, and so ai+k+1 is dominated. Hence,
S is a dominating set for G2. Next, we need to show that S is a connected dominating
set for G2. Recall the intermediate graph G. It is clear that the subgraph of G induced by
S is either connected, or has exactly two connected components. In the former case, the
subgraph of G2 induced by S is also connected. In the latter case, it is clear that there must
be an edge vw in G2 such that v ∈ J i, w ∈ J i+k+1 and both v, w are contained in S, and
hence the subgraph of G2 induced by S is connected. Either way, we obtain the desired
result.
Lemma 7.6. Suppose that there is an odd value n ≥ 5 such that γc(Jn−1) = 2n− 2. If S
is a connected dominating set for Jn such that |S| = 2n− 1, then wS4 is even.
Proof. Suppose that wS2 > 0. Then there is a copy J
i with weight 2. From Theorem 7.4,
we can obtain a connected dominating set for Jn−1 of cardinality 2n− 3, contradicting the
initial assumption. Hence, wS2 = 0. Then, we have w
S
1 +w
S
3 +w
S
4 = n, and w
S
1 +3w
S
3 +
4wS4 = 2n− 1. Combining these two, we obtain 3wS4 = n− 1− 2wS3 . Since n is odd, this
implies that wS4 is even.
440 Ars Math. Contemp. 24 (2024) #P3.04 / 427–452
a
b
c
d
J i . . . J i+k+1
G
a
b
c
d
J i . . . J i+k+1
G2
Figure 11: Sections of the two intermediate graphs G and G2 from the proof of Lemma 7.5.
Theorem 7.7. Consider the graph Jn for n ≥ 3, and suppose that S is a connected
dominating set. Then S does not contain the patterns 111 or 1312131.
Proof. Any connected dominating set with |S| ≥ 2 is also a total dominating set, and hence
the result for pattern 111 follows immediately from Theorem 6.1.
Next, consider the case when S contains the pattern 1312131. Without loss of gener-
ality, suppose the copies meeting this pattern are J1, . . . , J7. Suppose that a4 ̸∈ S. Then,
without loss of generality, suppose that b4 ∈ S, c4 ∈ S, and d4 ̸∈ S. This situation is
displayed in part (a) of Figure 12. Since S is dominating, it must contain at least one of d3
and d5. Since both J3 and J5 have weight 1, both options are equivalent, so without loss
of generality we will assume that d5 ∈ S. Then a5 ̸∈ S, b5 ̸∈ S, and c5 ̸∈ S. In order for
S to be connected, we must have b3 ∈ S and c3 ∈ S, but this is impossible since J3 has
weight 1. Hence, it must be the case that a4 ∈ S.
Again, without loss of generality, suppose that b4 ∈ S, c4 ̸∈ S and d4 ̸∈ S. This
situation is displayed in part (b) of Figure 12. Since S is connected, it must contain at least
one b3 or b5. As in the previous paragraph, without loss of generality we will assume that
b5 ∈ S. Then, in order for S to be dominating, we must have c6 ∈ S and d6 ∈ S. Now,
suppose that b6 ∈ S. Then, since S is connected, it must contain both c7 and d7, but this
is impossible since J7 has weight 1. Hence, it must be the case that b6 ̸∈ S, and hence
a6 ∈ S. Then, since S is connected, we must have b3 ∈ S and b2 ∈ S. In order for S to be
dominating, we must have c2 ∈ S and d2 ∈ S, and since J2 has weight 3 this implies that
a2 ̸∈ S. Finally, to ensure S is connected, it must contain each of b1, c1 and d1, but this is
impossible since J1 has weight 1.
a
b
c
d
J3 J4 J5
1 2 1
(a)
a
b
c
d
J1 J2 J3 J4 J5 J6 J7
1 3 1 2 1 3 1
(b)
Figure 12: The two situations described in the proof of Theorem 7.7. In part (a), J3 has
too many vertices in S. In part (b), J1 has too many vertices in S.
R. Burdett et al.: Variants of the domination number for flower snarks 441
Theorem 7.8. Consider the graph Jn for even n ≥ 6, and a connected dominating set S
for Jn such that |S| = 2n−1. If γc(Jn−1) = 2n−3, and wS4 = 0, then S does not contain
the pattern 112.
Proof. Without loss of generality, suppose that one set of copies meeting the pattern 112
is J2, J3, J4, and consider also J1 and J5. Since J2 and J3 have weight 1, we know that
a2 ̸∈ S and a3 ̸∈ S. Without loss of generality, suppose that b2 ∈ S. Then, suppose that
b3 ∈ S as well. This situation is displayed in part (a) of Figure 13. Since S is dominating,
it must also contain each of c1, d1, c4, and d4. Since J4 has weight 2, we have a4 ̸∈ S
and b4 ̸∈ S. Since S is connected, it implies that b1 ∈ S. Since there are no copies of
weight 4 in S, this implies that a1 ̸∈ S. However, by Theorem 7.4, we can then obtain a
connected dominating set for Jn−1 of cardinality 2n−4, contradicting the assumption that
γc(Jn−1) = 2n− 3. Hence, the assumption that b3 ∈ S must be false.
We instead have b2 ∈ S and b3 ̸∈ S. Without loss of generality, suppose that c3 ∈ S.
This situation is displayed in part (b) of Figure 13. Since S is dominating, it must also
contain d4. Since S is connected, we have c4 ∈ S, and since J4 has weight 2, we have
a4 ̸∈ S and b4 ̸∈ S. Then, since S is dominating, we have b5 ∈ S, and since S is connected
we have c5 ∈ S and d5 ∈ S. Since there are no copies of weight 4 in S, it implies that
a5 ̸∈ S. However, by Theorem 7.4, we can the obtain a connected dominating set for Jn−1
of cardinality 2n − 4, which again contradicts the assumption that γc(Jn−1) = 2n − 3,
completing the proof.
a
b
c
d
J1 J2 J3 J4 J5
1 1 2
(a)
a
b
c
d
J1 J2 J3 J4 J5
1 1 2
(b)
Figure 13: The two situations described in the proof of Theorem 7.8. In both parts, there is
a copy with weight 3, but with the a vertex not contained in S.
In the next theorem, we will require the following concept. Suppose that we have a
connected dominating set S for Jn. Denote by Jn(S) the subgraph of Jn induced by S. By
definition, Jn(S) is connected. Then, consider a set of consecutive copies J i, . . . , J i+k,
and define S = S ∩ (V (J i) ∪ . . . ∪ V (J i+k)). If Jn(S) is a disconnected graph, then we
say that the copies J i, . . . , J i+k are locally disconnected in S. Since S itself is connected,
it is clear that if S contain two sets of locally disconnected consecutive copies, they must
overlap by two or more copies.
Theorem 7.9. Consider the graph Jn for even n ≥ 6, and a connected dominating set S
for Jn such that |S| = 2n − 1. If γc(Jn−1) = 2n − 3, then any set of consecutive copies
meeting the patterns 113 or 3123 are locally disconnected.
Proof. Consider first the pattern 113. Without loss of generality, suppose that one set of
copies meeting this pattern is J1, J2, J3, and that this set of copies is not locally discon-
nected. Suppose that a3 ̸∈ S. Then, from Theorem 7.4 there exists a connected dom-
inating set for Jn−1 with cardinality 2n − 4, which is impossible since by assumption
442 Ars Math. Contemp. 24 (2024) #P3.04 / 427–452
γc(Jn−1) = 2n−3. Hence, a3 ∈ S. Then, without loss of generality, suppose that b3 ∈ S,
c3 ∈ S, and d3 ̸∈ S. This situation is displayed in part (a) of Figure 14. Since J2 has
weight 1, it is clear that a2 ̸∈ S. Then, in order for S to be dominating, it must contain at
least one of d1 and d2. However, if d2 ∈ S, then S does not contain b2 or c2, and hence
J1, J2, J3 are locally disconnected, contradicting the initial assumption. Hence, d2 ̸∈ S,
and so d1 ∈ S. But then, because J1 has weight 1, S does not contain a1, b1, or c1, which
again implies that J1, J2, J3 are locally disconnected, contradicting the initial assumption.
Hence, the initial assumption must be false, and J1, J2, J3 are locally disconnected.
Next, consider the pattern 3123. Without loss of generality, suppose that one set of
copies meeting this pattern is J1, J2, J3, J4, and that this set of copies if not locally dis-
connected. Using an identical argument as from the previous paragraph, we must have
a1 ∈ S and a4 ∈ S. Then, without loss of generality, suppose that b1 ∈ S and c1 ∈ S.
This situation is displayed in part (b) of Figure 14. Since J2 has weight 1, and the set
of copies is not locally disconnected, we have a2 ̸∈ S and d2 ̸∈ S. The remaining two
choices are equivalent; without loss of generality, suppose that b2 ∈ S. Then, since S
is dominating, and the set of copies is not locally disconnected, we must have b3 ∈ S
and d3 ∈ S. Finally, since S is dominating, and the set of copies is not locally discon-
nected, we must have b4 ∈ S, c4 ∈ S and d4 ∈ S. However, this is impossible since J4
has weight 3. Hence, the initial assumption must be false, and J1, J2, J3, J4 are locally
disconnected.
a
b
c
d
J1 J2 J3
1 1 3
(a)
a
b
c
d
J1 J2 J3 J4
3 1 2 3
(b)
Figure 14: The two situations described in the proof of Theorem 7.9. In part (a), the copies
are locally disconnected. In part (b), copy J4 has too many vertices in S.
Theorem 7.7, along with the fact that the patterns 113 and 3123 overlap by at most one
copy, leads to the following corollary.
Corollary 7.10. Consider the graph Jn for even n ≥ 6, and a connected dominating set S
for Jn such that |S| = 2n− 1. If γc(Jn−1) = 2n− 3, then S contains at most one instance
the patterns 113 or 3123.
We are finally ready to prove the main result of this section.
Theorem 7.11. Consider the graph Jn for n ≥ 3. Then,
γc(Jn) =
{
2n, if n is even,
2n− 1, if n is odd.
Proof. The upper bounds are provided in Section 4. We use the MTZ formulation for
connected domination from [10] to confirm that γc(J3) = 5, γc(J4) = 8, γc(J5) = 9,
R. Burdett et al.: Variants of the domination number for flower snarks 443
and γc(J6) = 12. Suppose there is some value k ≥ 7 such that Theorem 7.11 is true for
n = 3, . . . , k − 1, but false for n = k. We will first consider the case when k is odd.
Then there is a connected dominating set S for Jk such that |S| = 2k − 2. Clearly, S
contains a copy, say J i, with weight 1, and ai ̸∈ S. Then from Theorem 7.4, it implies that
there is a connected dominating set for Jk−1 with cardinality 2k− 3, which contradicts the
assumption that Theorem 7.11 is true for Jk−1.
Hence, k must be even. Then, there is a connected dominating set S for Jk such that
|S| = 2k − 1. Suppose that there are two copies of Jk with weight 2 in S. Then, from
Theorem 7.4 we can obtain a connected dominating set for Jk−2 with cardinality 2k − 5,
which contradicts the initial assumption. Hence, there must be at most one copy of Jk with
weight 2 in S. That is, either wS2 = 0 or w
S
2 = 1.
Suppose that wS2 = 0. Then we have w
S
1 +w
S
3 +w
S
4 = k, and w
S
1 +3w
S
3 +4w
S
4 = 2k−1.
From the latter equation, it is clear that wS1 and w
S
3 must have different parity. Hence,
from the former equation, wS4 must be odd. Combining the two equations, we obtain
2wS1 = k + 1 + w
S
4 . Note that this implies that more than half of the copies have weight
1 in S. From Theorem 7.7, S cannot contain the pattern 111. Hence, it can be seen that S
contains the pattern 11 at least wS4 +1 times. From Corollary 7.10 we know there is at most
one instance of the pattern 113 in S. Hence, there is at least wS4 instances of the pattern 11
where both adjacent copies have weight 4, which is impossible.
Hence, we must have wS2 = 1. Then, we have w
S
1 +w
S
3 +w
S
4 = k−1, and wS1 +3wS3 +
4wS4 = 2k − 3. From the latter equation, it is clear that wS1 and wS3 must have different
parity. Hence, from the former equation, wS4 must be even. Combining the two equations,
we obtain 2wS1 = k + w
S
4 . Similar to before, this implies that at least half of the copies
have weight 1 in S. From Theorem 7.7, S cannot contain the pattern 111. Hence, it can
be seen that S contains the pattern 11 at least wS4 times. Now, suppose that S contains the
pattern 4114. From Lemma 7.5, we can then obtain a connected dominating set for Jk−3
with cardinality 2k−7 which has one fewer copy of weight 4 than S, which in turn violates
Lemma 7.6. Hence, every instance of the pattern 11 has a copy of weight 2 or 3 next to it,
and from Corollary 7.10 this means there is at most one instance of the pattern 11. This,
in turn, implies that wS4 ≤ 1, and since wS4 is even, we have wS4 = 0. Hence, wS1 = k2 . If
there are no instances of the pattern 11, then S must contain the pattern 1312131, violating
Theorem 7.7. Hence there is exactly one instance of the pattern 11. This means that there
must be one other instance in S of two consecutive copies having non-unit weight. The
only options are that S contains the pattern 23, or the pattern 33.
Suppose S contains the pattern 23. Without loss of generality, suppose that the copies
J3, J4 meet this pattern. Since this is the only instance of S having two consecutive copies
of weight other than 1, we know that J2 has weight 1. From Theorem 7.8 we know that J1
cannot have weight 1, or else S would contain the pattern 112. The only remaining option
is that J1 has weight 3, and so copies J1, J2, J3, J4 meet the pattern 3123. Then, since
J3 has weight 2 and wS2 = 1, there are no other copies with weight 2. Hence, the one
instance of the pattern 11 which occurs elsewhere in the graph must be followed by a copy
of weight 3. That is, S contains both the patterns 113 and 3123, which from Corollary 7.10
is impossible. Therefore, S must not contain the pattern 23, and instead contains the pattern
33.
Finally, without loss of generality, suppose that the copies J1, J2 meet the pattern 33.
There is one vertex from J1 which is not contained in S. Suppose we define S2 which is
equal to the union of S and this one vertex. Hence, S2 is a connected dominating set for Jk
444 Ars Math. Contemp. 24 (2024) #P3.04 / 427–452
with cardinality 2k. Then, from Lemma 7.5, we can obtain a connected dominating set for
Jk−1 with cardinality 2k−3, which contains exactly one copy with weight 4, contradicting
Lemma 7.6.
Hence, in all cases a contradiction is reached, completing the proof.
8 Weak Roman domination and secure domination
In this section, we determine (simultaneously) the weak Roman domination numbers and
the secure domination numbers for flower snarks. Recall that for the weak Roman domina-
tion number, rather than seeking to construct a set S, we instead look to define a function
f : V → {0, 1, 2}. In order to use language consistent with the rest of this paper, in this
section we define the weight of a copy J i to be equal to f(ai) + f(bi) + f(ci) + f(di).
Note that this is somewhat more ambiguous than in previous sections; for instance, if a
copy has weight 2, it may have two vertices with weight 1, or a single vertex with weight
2. We will deal with these ambiguities when they arise. Similarly to in previous sections,
we will say that f contains a pattern if there is a set of consecutive copies whose weights
meet that pattern.
Although there is a technical definition for weak Roman domination (e.g. see Defini-
tion 1.4), it is useful to provide an intuitive interpretation. Recall that for standard domina-
tion, one interpretation is that the set of vertices needs to be protected. If a guard is placed
at a vertex, then it protects that vertex, and also all adjacent vertices. A configuration of
guards which protects all vertices corresponds to a dominating set. A similar interpreta-
tion applies for weak Roman domination. Suppose that at each vertex we can place up to
two guards. As before, a vertex is protected if there is at least one guard there, or there is
at least one guard among its adjacent vertices. Then we further consider the notion of a
vertex being defended. We say a vertex v is defended if there is at least one guard at v, or
alternatively, if it possible to relocate one guard from an adjacent vertex w to v, in such a
way that all vertices are still protected by the resulting configuration of guards. In the latter
case, we will say that w can defend v. Then a weak Roman dominating function is one in
which every vertex is defended.
Throughout this section, we will often use an argument which goes as follows; suppose
there is exactly one guard at v, and that v defends another adjacent vertex w. In order to
do so, the guard from v would move to w. However, doing so may mean another vertex u,
also adjacent to v, becomes unprotected. In such a case, we will say that v cannot defend
w without leaving u unprotected. Another alternative is as follows; suppose again that v
defends w. In order to do so, the guard from v would move to w. However, doing so may
mean that u will be unprotected unless there is a guard at another vertex, say x. In such a
case, we will say in order for v to defend w, there must be a guard at x to avoid leaving u
unprotected.
Although the following result is simple, we include it here as it will be used many times
in this section.
Lemma 8.1. Consider the graph Jn, for n ≥ 3, and a weak Roman dominating function
f . If a copy J i has weight 1 in f , then none of the vertices from J i can defend any vertices
from J i−1 or J i+1.
Proof. Since J i has weight 1, there is exactly one vertex with positive weight. If f(ai) = 1,
then the result follows immediately since ai is not adjacent to any vertices from J i−1
R. Burdett et al.: Variants of the domination number for flower snarks 445
or J i+1. Then, suppose instead that f(ai) = 0 and one of the other vertices in J i has
weight 1. Then that vertex cannot defend any vertex from J i−1 or J i+1 without leaving ai
unprotected.
Theorem 8.2. Consider the graph Jn, for n ≥ 3, and a weak Roman dominating function
f . Then f does not contain the patterns 111, 1121 or 1212121.
Proof. Suppose that f contains the pattern 111. Without loss of generality, suppose that
copies J1, J2, J3 meet the pattern 111 in f . This situation is displayed in part (a) of
Figure 15. Each vertex in J2 must be defended, but since J1 and J3 have weight 1, from
Lemma 8.1 none of their vertices can defend any vertices from J2. Hence, each vertex in
J2 must be defended solely by vertices in J2. It is clear that this implies that f(a2) = 1.
Then, consider b2. In order for a2 to defend b2, there must be a guard at either c1 or c3 to
avoid leaving c2 unprotected. Likewise, in order for a2 to defend b2, there must be a guard
at either d1 or d3 to avoid leaving d2 unprotected. Since J1 and J3 have weight 1, at most
one vertex from each can have a guard. Without loss of generality, suppose that f(c1) = 1
and f(d3) = 1. Then a2 cannot defend c2 without leaving b2 unprotected. Hence, this is
impossible, and f does not contain the pattern 111.
Next, suppose that f contains the pattern 1121. Without loss of generality, suppose that
copies J1, J2, J3, J4 meet the pattern 1121 in f . Since J2 and J4 have weight 1, from
Lemma 8.1 none of their vertices can defend any vertices from J3. Hence, each vertex
from J3 must be defended solely by vertices in J3. Clearly, this implies that f(a3) ≥ 1.
There are two possibilities, either f(a3) = 2, or f(a3) = 1 and there is a guard at another
vertex in J3. Suppose first that f(a3) = 2. Then there are no vertices from J1 or J3
that can defend any vertices from J2. Hence, J2 is in an equivalent situation to J2 in the
previous paragraph and so this situation is impossible. Instead, suppose that f(a3) = 1
and, without loss of generality, that f(b3) = 1. This situation is displayed in part (b) of
Figure 15. Clearly, vertices c2 and d2 can only be defended by vertices from J2, and hence
we must have f(a2) = 1. Then, in order for a2 to defend c2, there must be a guard at d1 to
avoid leaving d2 unprotected. Similarly, in order for a2 to defend d2, there must be a guard
at c1 to avoid leaving c2 unprotected. This is impossible since J1 has weight 1, and so f
does not contain the pattern 1121.
a
b
c
d
J1 J2 J3
1 1 1
(a)
a
b
c
d
J1 J2 J3 J4
1 1 2 1
(b)
Figure 15: The first two situations described in the proof of Theorem 8.2. In part (a), a2
cannot defend c2 without leaving b2 unprotected. In part (b), there are too many guards in
J1.
Finally, suppose that f contains the pattern 1212121. Without loss of generality, sup-
pose that copies J1, . . . , J7 meet the pattern 1212121 in f . Since J1 and J3 have weight
1, from Lemma 8.1 none of their vertices can defend any vertices from J2. Hence, each
vertex from J2 must be defended solely by vertices in J2, which implies that f(a2) ≥ 1.
446 Ars Math. Contemp. 24 (2024) #P3.04 / 427–452
Analogous arguments imply that f(a4) ≥ 1 and f(a6) ≥ 1. Now, consider J2. There are
two possibilities, either f(a2) = 2, or f(a2) = 1 and there is a guard at another vertex
in J2. Suppose first that f(a2) = 2. Then there are no vertices from J2 that can defend
any vertices from J3, and at most one vertex from J4 can defend a vertex from J3. Hence,
J3 is in an equivalent situation to J2 in the previous paragraph, and so this situation is
impossible.
Instead, suppose that f(a2) = 1 and, without loss of generality, that f(b2) = 1. This
situation is displayed in Figure 16. Then, in order for a2 to defend c2, there must be a
guard at either d1 or d3 to avoid leaving d2 unprotected. Similarly, in order for a2 to defend
d2, there must be a guard at either c1 or c3 to avoid leaving c2 unprotected. Since J1
and J3 have weight 1, at most one vertex from each may have a guard. Without loss of
generality, suppose that f(c1) = 1 and f(d3) = 1. Then, since f is dominating, we must
have f(c4) = 1 in order to protect c3. Now, in order for a4 to defend d4, there must be
a guard at b5 to avoid leaving b4 unprotected. Then, since f is dominating, we must have
f(d6) = 1 in order to protect d5. Finally, consider vertex c5. There is only one adjacent
vertex with a guard that can defend it, c4. However, c4 cannot defend c5 without leaving
c3 unprotected. Thus, c5 is not defended. This is impossible, and hence f does not contain
the pattern 1212121.
a
b
c
d
J1 J2 J3 J4 J5 J6 J7
1 2 1 2 1 2 1
Figure 16: The third situation described in the proof of Theorem 8.2. Here, c4 cannot
defend c5 without leaving c3 unprotected.
Lemma 8.3. Consider the graph Jn, for n ≥ 4, and a weak Roman dominating function f
with weight w(f) ≤ 3n2 . Then any set of four consecutive copies have a combined weight
of 6 in f , every copy has weight either 1 or 2 in f , and w(f) = 3n2 .
Proof. Denote by w4(i) the combined weights of J i, J i+1, J i+2, J i+3 where the super-
scripts are taken modulo n. Now, by Corollary 2.4 we know that no copies have weight
0, and so w4(i) ≥ 4. Suppose w4(i) = 4. This implies that J i, J i+1, J i+2, J i+3 all
have weight 1. However, this means f contains the pattern 111, which by Theorem 8.2 is
impossible. Then, suppose w4(i) = 5. Up to symmetry, there are only two possibilities;
copies J i, J i+1, J i+2, J i+3 either have the pattern 1112 or 1121. Both of these options are
impossible by Theorem 8.2. Hence, we have w4(i) ≥ 6. Then, by assumption, we have∑n
i=1 w
4(i) ≤ 6n. This is only possible if all w4(i) = 6, and also implies that w(f) = 3n2 .
Since each copy has weight at least one, and each set of four consecutive copies has a
combined weight of six, it is clear that any individual copy must have weight at most three.
Suppose there is a copy J i with weight 3 in f , then the next three copies must each have
weight 1. However, by Theorem 8.2 this is impossible. Hence, each copy J i must have
weight either 1 or 2 in f .
R. Burdett et al.: Variants of the domination number for flower snarks 447
Lemma 8.4. Consider the graph Jn, for n ≥ 3, and a weak Roman dominating func-
tion f with weight w(f). Suppose there is an integer k ≥ 3 such that in f , the config-
uration of guards in copies J i+1, . . . , J i+k is identical to the configuration of guards in
J i+k+1, . . . , J i+2k, and suppose that J i+1, . . . , J i+k collectively contain g guards. Then
there is a weak Roman dominating function for Jn−k with weight equal to w(f)− g.
Proof. Suppose that we “smooth out” copies J i+1, . . . , J i+k in the way displayed in Fig-
ure 11. That is, we add edges connecting bi to bi+k+1, ci to ci+k+1, and di to di+k+1,
and then we delete the copies J i+1, . . . , J i+k. Call the resulting graph G2. It is easy to
check that G2 is isomorphic to Jn−k. Then, define a new function f2 such that f2(v) =
f(v) if v ∈ V \ {V (J i+1) ∪ . . . ∪ V (J i+k)}, and f2(v) is undefined otherwise. Since
J i+1, . . . , J i+k collectively contain g guards in f , it is clear that w(f2) = w(f)− g. Then
if we can show that f2 is a weak Roman dominating function for G2, the result follows
immediately.
For a given function, one can check to see if a vertex v is defended by observing the
configuration of guards in vertices no more than distance three away from v. Then, for
any vertex v in G2, consider the configuration of graphs (according to f2) in the subgraph
induced by the vertices at most distance 3 from v. By assumption, this is identical to the
corresponding configuration of guards (according to f ) in the subgraph induced by the
vertices at most distance 3 from v in Jn. Hence, all vertices of G are defended in f2, and
so f2 is a weak Roman dominating function for G, leading to the desired result.
In the proof of the following theorem, whenever there are two guards at a vertex, we
will mark that vertex with a .
Theorem 8.5. Consider the graph Jn, for n ≥ 9, and a weak Roman dominating function
f . If f contains the pattern 21122112, then there exists a weak Roman dominating function
for Jn−4 with weight equal to w(f)− 6.
Proof. Without loss of generality, suppose that copies J1, . . . , J8 meet the pattern
21122112 in f . Consider copy J4, and suppose that f(a4) = 2. This situation is dis-
played in part (a) of Figure 17. Then, no vertices from J4 can defend any vertices from
J3, and by Lemma 8.1 the same is also true for J2. Hence, all four vertices in J3 must be
defended by vertices in J3. This implies that f(a3) = 1. Then, in order for a3 to defend
b3, there must be a guard at c2 to avoid leaving c3 unprotected, and there must also be a
guard at d2 to avoid leaving d3 unprotected. Since J2 has weight 1, this is impossible, and
so f(a4) ̸= 2.
Then, suppose that f(a4) = 1. Then there is another vertex in J4 with a guard. With-
out loss of generality, suppose that f(b4) = 1. This situation is displayed in part (b) of
Figure 17. Using a similar argument as in the previous paragraph, vertices c3 and d3 must
be defended by vertices from J3, and hence f(a3) = 1. Then, in order for a3 to defend c3,
there must be a guard at d2 to avoid leaving d3 unprotected. Likewise, in order for a3 to
defend d3, there must be a guard at c2 to avoid leaving c3 unprotected. Since J2 has weight
1, this is impossible. Hence we can conclude that f(a4) = 0. It is clear that identical
arguments can be made to conclude that f(a1) = f(a5) = f(a8) = 0 as well.
Then, suppose that there is a vertex in J4 with two guards. Without loss of generality,
suppose that f(b4) = 2. This situation is displayed in part (c) of Figure 17. An identical
argument to that in the previous paragraph can be used to show that f(a3) = 1, and this
again implies that f(c2) = f(d2) = 1. Since J2 has weight 1, this is impossible. Hence,
448 Ars Math. Contemp. 24 (2024) #P3.04 / 427–452
exactly two vertices in J4 have weight 1. Again, identical arguments can be made to reach
the same conclusion for each of J1, J5 and J8.
Now, without loss of generality, suppose that f(b4) = f(c4) = 1. Then, d4 must be
defended by either d3 or d5, and by Lemma 8.1 it cannot be d3. Hence, we must have
f(d5) = 1, and there must be one more vertex in J5 with a guard, either b5 or c5. Due to
symmetry, either choice may be made without loss of generality; we will choose f(b5) = 1.
Now, consider vertex d3. From Lemma 8.1 it cannot be defended by d2. Hence, we must
have either f(a3) = 1 or f(d3) = 1.
Suppose that f(a3) = 1. Then, c3 must be defended by at least one of a3 or c4 (from
Lemma 8.1 it cannot be defended by c2). Suppose first that a3 defends c3. This situation is
displayed in part (d) of Figure 17. In order for a3 to defend c3, there must be a guard at d2
to avoid leaving d3 unprotected. Then, in order to defend b2 and c2 we must have f(b1) = 1
and f(c1) = 1 respectively. But then, b1 cannot defend a1 without leaving b2 unprotected,
and c1 cannot defend a1 without leaving c2 unprotected. Hence, a1 is not defended, which
is impossible, and so we conclude that a3 cannot defend c3. Hence, c4 must defend c3.
This situation is displayed in part (e) of Figure 17. In order for c4 to defend c3, there must
be a guard at c6 to avoid leaving c5 unprotected. Also, the only vertex which can defend
d4 is d5, but in order to do so, there must be a guard at d7 to avoid leaving d6 unprotected.
Finally, consider a5, which can only be defended by either b5 or d5. However, b5 cannot
defend a5 without leaving b6 unprotected, and d5 cannot defend a5 without leaving d4
unprotected. Hence, all of these cases are impossible, and so conclude that f(a3) = 0, and
accordingly, f(d3) = 1.
The current situation is displayed in part (f) of Figure 17. Now consider c3. From
Lemma 8.1 it is clear that c2 cannot defend c3, so c4 must defend c3. In order for c4 to
defend c3, there must be a guard at c6 to avoid leaving c5 unprotected. Likewise, from
Lemma 8.1, d3 cannot defend d4, so d5 must defend d4. In order for d5 to defend d4, there
must be a guard at d7 to avoid leaving d6 unprotected. Furthermore, from Lemma 8.1, c6
cannot defend c5, so c4 must defend c5. In order for c4 to defend c5, there must be a guard
at c2 to avoid leaving c3 unprotected. Finally, from Lemma 8.1, the only vertices which
can defend b2 and d2 are b1 and d1 respectively, and so f(b1) = f(d1) = 1. Likewise,
from Lemma 8.1, the only vertices which can defend b7 and c7 are b8 and c8 respectively,
so f(b8) = f(c8) = 1.
At this point, it can be seen that the configuration of guards in J1, J2, J3, J4 is identical
to the configuration of guards in J5, J6, J7, J8. Then, the result follows immediately from
Lemma 8.4.
We are finally ready to prove the main result of this section.
Theorem 8.6. Consider the graph Jn, for n ≥ 3. Then, γs(Jn) = γr(Jn) =
⌈
3n+ 1
2
⌉
.
Proof. The (coincident) upper bounds for both weak Roman domination and secure dom-
ination are provided in Section 4. Note that γr(Jn) ≤ γs(Jn), so a corresponding lower
bound for weak Roman domination will also serve as a lower bound for secure domination.
We use the formulation for weak Roman domination from [7] to confirm that γr(J3) = 5,
γr(J4) = 7, γr(J5) = 8, γr(J6) = 10, γr(J7) = 11, and γr(J8) = 13, and the formu-
lation from [5] to confirm the corresponding results for secure domination. Then, suppose
there is a value k ≥ 9 such that Theorem 8.6 is true for n = 3, . . . , k − 1, but is not
true for n = k. That is, there exists a weak Roman dominating function f for Jk with
R. Burdett et al.: Variants of the domination number for flower snarks 449
a
b
c
d
J2 J3 J4
1 1 2
(a)
a
b
c
d
J1 J2 J3 J4 J5
2 1 1 2 2
(d)
a
b
c
d
J2 J3 J4
1 1 2
(b)
a
b
c
d
J3 J4 J5 J6 J7
1 2 2 1 1
(e)
a
b
c
d
J2 J3 J4
1 1 2
(c)
a
b
c
d
J1 J2 J3 J4 J5 J6 J7 J8
2 1 1 2 2 1 1 2
(f)
Figure 17: The six situations described in the proof of Theorem 8.5. In parts (a), (b) and
(c), there are too many guards in copy J2. In part (d), a1 is undefended. In part (e), a5 is
undefended. In part (f), the configuration of guards in J1, J2, J3, J4 is identical to that in
J5, J6, J7, J8.
weight w(f) <
⌈
3k+1
2
⌉
. This implies that w(f) ≤ 3k2 . Hence, from Lemma 8.3 we have
w(f) = 3k2 , each set of four consecutive copies has weight 6, and each copy has weight 1
or 2.
It is easy to check that there are only two possibilities. Either f has the repeating pattern
121212 . . . 12, or f has the repeating pattern 21122112 . . . 2112. Note that it is impossible
to combine these two patterns without there being a set of four consecutive copies with
weight not equal to 6. Then, since k ≥ 9, by Theorem 8.2 the pattern 121212 . . . 12
is impossible. Hence, f must have the repeating pattern 21122112 . . . 2112. Then, by
Theorem 8.5 there is weak Roman dominating function for Jk−4 with weight equal to
3k
2 − 6 <
⌈
3(k−4)+1
2
⌉
. This contradicts our initial assumption, completing the proof.
ORCID iDs
Michael Haythorpe https://orcid.org/0000-0001-8143-6583
Alex Newcombe https://orcid.org/0000-0002-4401-0951
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ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 24 (2024) #P3.05 / 453–471
https://doi.org/10.26493/1855-3974.2771.4df
(Also available at http://amc-journal.eu)
Optimal strategies in fractional games:
vertex cover and domination
Csilla Bujtás *
Faculty of Mathematics and Physics, University of Ljubljana, Ljubljana, Slovenia and
University of Pannonia, Veszprém, Hungary and
Institute of Mathematics, Physics and Mechanics, Jadranska 19, Ljubljana, Slovenia
Günter Rote
Institut für Informatik, Freie Universität Berlin, Takustr. 9, 14195 Berlin, Germany
Zsolt Tuza †
HUN-REN Alfréd Rényi Institute of Mathematics, Reáltanoda u. 13–15,
1053 Budapest, Hungary and
Department of Computer Science and Systems Technology, University of Pannonia,
Egyetem u. 10, 8200 Veszprém, Hungary
Received 14 December 2021, accepted 27 February 2024, published online 18 July 2024
Abstract
In a hypergraph H = (V, E) with vertex set V and edge set E , a real-valued function
f : V → [0, 1] is a fractional transversal if
∑
v∈E f(v) ≥ 1 holds for every E ∈ E . Its
size is |f | :=
∑
v∈V f(v), and the fractional transversal number τ
∗(H) is the smallest
possible |f |.
We consider a game scenario where two players have opposite goals, one of them trying
to minimize and the other to maximize the size of a fractional transversal constructed in-
crementally. We prove that both players have strategies to achieve their common optimum,
and they can reach their goals using rational weights.
Keywords: Fractional vertex cover, fractional transversal game, fractional domination game.
Math. Subj. Class. (2020): 05C69, 05C65, 05C57
*Corresponding author. The author acknowledges the financial support from the Slovenian Research and
Innovation Agency under the projects N1-0108 and P1-0297.
†The author’s research was supported in part by the National Research, Development and Innovation Office –
NKFIH under the grant SNN 129364.
E-mail addresses: bujtasc@gmail.com (Csilla Bujtás), rote@inf.fu-berlin.de (Günter Rote),
tuza@dcs.uni-pannon.hu (Zsolt Tuza)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
454 Ars Math. Contemp. 24 (2024) #P3.05 / 453–471
1 Introduction
Let H = (V, E) be a finite hypergraph, where V is the finite vertex set and E is the edge
set, a set system over the underlying set V . We assume that every edge contains at least
one vertex; that is, E ⊆ 2V \ {∅}. A hypergraph is k-uniform if |E| = k holds for all
E ∈ E . A set T ⊆ V is a transversal 1 of H if every edge is covered by a vertex of T ,
which formally means that T ∩ E ̸= ∅ holds for all E ∈ E . Its real relaxation, called
fractional transversal, is a function f : V → [0, 1] such that
∑
v∈E f(v) ≥ 1 holds for
every E ∈ E . The size of f is defined as |f | :=
∑
v∈V f(v). The transversal number
τ(H) and the fractional transversal number τ∗(H) of H are the minimum cardinality |T |
of a transversal and minimum value |f | of a fractional transversal, respectively.
The transversal game is a competitive optimization version of hypergraph transversals,
which was introduced in [9] and studied further in [10]. It is played on a hypergraph H by
two players called Edge-hitter and Staller. They take turns choosing a vertex. The game
is over when all edges are covered, and the length of the game is the number of vertices
chosen by the players until the end of the game. Edge-hitter wants to finish the game as
soon as possible, while Staller wants to delay the end. To prevent Staller from making
completely useless moves, we stipulate that the chosen vertex must be contained in at least
one previously uncovered edge.
Assuming that both players play optimally2 and Edge-hitter starts, the length of the
game on H is uniquely determined. It is called the game transversal number of H and is
denoted by τg(H). Analogously, the Staller-start game transversal number of H, denoted
by τ ′g(H), is the length of the game under the same rules when Staller makes the first
move. Among other results, it was proved in [9] that |τg(H) − τ ′g(H)| ≤ 1 always holds.
We further recall that, denoting by n(H) and m(H) the number of vertices and edges in H
respectively, 411 (n(H) +m(H)) is a (sharp) upper bound on τg(H) if H does not contain
one-element edges and it is not isomorphic to the cycle C4.
Below we shall refer to this game as the integer game, as opposed to its fractional
version which we will introduce in the next section.
The important motivation of this approach are the domination game [7] and the to-
tal domination game [17], where in fact the transversal game is played on the ‘closed
neighborhood hypergraph’ and on the ‘open neighborhood hypergraph’ of a graph, re-
spectively.3 Further variants studied so far include the disjoint domination [14], con-
nected domination [2], and fractional domination [15] games on graphs, and the dom-
ination games on hypergraphs [13]. Some of the most recent results can be found in
[3, 4, 8, 11, 12, 16, 18, 19, 20, 21, 22, 23]. For a thorough survey and list of further
1In various areas of discrete mathematics and computer science, a transversal is called vertex cover, or hitting
set, or blocking set. It is also equivalent to the set cover in the dual hypergraph.
2A strategy of a player means that every possible state of the game is associated with a move he/she will play
if that situation arises. From Edge-hitter’s point of view, the value vE(S) of a strategy S is the smallest integer
k such that, if Edge-hitter plays according to S, the transversal game always finishes in at most k moves (no
matter which strategy is applied by Staller). We say that S is an optimal strategy for Edge-hitter, if vE(S) is
the possible smallest value over the family of all strategies. Similarly, from Staller’s point of view, a strategy S
can be associated with the value vS(S) that is the largest integer k such that, if Staller follows strategy S, the
length of the game is always at least k; further S is an optimal strategy for Staller, if vS(S) is the largest value
over the family of all strategies. The reader may find more about optimal strategies and the uniqueness of the
corresponding parameters in [5, Section 1.2].
3Recall from the literature that the closed and open neighborhood hypergraphs of a graph G are defined on the
same vertex set V as G, and the closed (resp. open) neighborhood hypergraph consists of edges corresponding to
the closed (resp. open) neighborhoods of vertices in G.
Cs. Bujtás et al.: Optimal strategies in fractional games: vertex cover and domination 455
references see the book [5].
Our results
In Section 2 we introduce the rules of the game and prove that its value is well-defined. We
present some examples, showing that it makes a difference whether Edge-hitter or Staller
starts. Moreover, edges that are supersets of other edges of the hypergraph may influence
the game value, in contrast to the standard non-game version of the transversal number.
In Section 3 we compare the game transversal number with other related parameters,
and prove a monotonicity property, implying that changing the starting player can affect
the value of the game by at most 1.
The rules of the game allow the players to split their moves into infinitely many sub-
moves. In Section 4 and 5 we give some structural results showing that the full generality
of the moves allowed by our rules is not needed. Namely, any infinite move is equivalent to
a finite move, and Edge-hitter can restrict his strategy to moves in which every permutation
of submoves is equally good.
In Section 6 we prove that the game can be modeled in a way that leads to an opti-
mization problem solvable via the theory of piecewise linear continuous rational functions.
From this, we derive that the game value is rational for every finite hypergraph; moreover
both players can achieve their goals using rational submoves.
Consequences concerning domination games and several conjectures are given in the
concluding Section 7.
2 Fractional transversal game
Let H = (V, E) be a hypergraph. In the context of the fractional transversal game, we will
consider a cover function t : V → [0, 1] that is updated after each move during the game.
We denote by |t| the sum
∑
v∈V t(v). Given a cover function t, the corresponding load
function is ℓ : E → [0, 1] defined by the rule
ℓ(E) = ℓ(E, t) = min
{
1,
∑
v∈E
t(v)
}
for every E ∈ E . If ℓ ≡ 1, we say that H is fully covered. We shall write ti and ℓi for the
cover and load functions after the ith move.
The game begins with t0 ≡ 0 and therefore with ℓ0 ≡ 0. It is finished when the
hypergraph becomes fully covered. Edge-hitter and Staller take turns making moves under
the following rules. As long as ℓ ̸≡ 1, the next player performs a move, which is a sequence
(vi1 , w1), (vi2 , w2), . . . of arbitrary length (possibly infinite). It consists of the submoves
(vik , wk), k = 1, 2, . . . , where vi1 , vi2 , . . . are vertices of H with any number of repetitions
allowed, and the weights w1, w2, . . . are real numbers from [0, 1].
We say that a submove (vik , wk) is legal if it increases the load of some edge by wk. In
a legal move, a player makes a series of legal submoves such that the sum of the weights
equals 1 or the move completes the game, whichever comes first. Formally, the ith move
(vi1 , w1), (vi2 , w2), . . . is legal, if the following conditions hold:
(∗) For every k ≥ 1 there exists an edge E ∈ E such that vik ∈ E and
ℓi−1(E) +
( ∑
vis
∈E
1≤s≤k−1
ws
)
+ wk ≤ 1 .
456 Ars Math. Contemp. 24 (2024) #P3.05 / 453–471
(∗∗) The total weight constraint:
∑
k≥1 wk ≤ 1, and if the move does not end the game,
then
∑
k≥1 wk = 1.
The cover function ti can gradually be reached from ti−1 by adding the weight wk to
t(vik) after each submove; this process converts also the corresponding load function from
ℓi−1 to ℓi.
Suppose that a fractional transversal game G finishes with the qth move. The value |G|
of the game is defined as the value |tq| of the cover function obtained at the end, that is
the sum of the weights that have been spent during the game. The goal of Edge-hitter is to
achieve a value |G| as small as possible, while Staller wants a large |G|.
Assuming that Edge-hitter starts the fractional transversal game on H, we consider the
set of upper bounds,
UH = {a ∈ R : Edge-hitter has a strategy that ensures |G| ≤ a}
and the set of lower bounds,
LH = {b ∈ R : Staller has a strategy that ensures |G| ≥ b}.
Formally the game fractional transversal number τ∗g (H) is defined as
τ∗g (H) = inf(UH).
The Staller-start game fractional transversal number τ∗g
′(H) is defined similarly, under the
condition that the first move is made by Staller.
The following assertion shows that τ∗g (H) is also equal to sup(LH), and the situation is
similar if Staller starts the game. The proof is essentially the same as the one for the game
fractional domination number in [15].
Proposition 2.1. For every hypergraph H we have inf(UH) = sup(LH), and the analo-
gous equality holds for the Staller-start game, too.
Proof. First, assume that inf(UH) < sup(LH) and consequently, there exist two reals x
and y satisfying inf(UH) < x < y < sup(LH). By definition, x ∈ UH and, therefore,
Edge-hitter can ensure that, under every strategy of Staller, the value of the game is at most
x. Similarly, y ∈ LH and Staller has a strategy that ensures |G| ≥ y whatever strategy is
followed by Edge-hitter. This is a contradiction that establishes inf(UH) ≥ sup(LH).
Now, we prove the reverse inequality. By definition, z < inf(UH) implies that Edge-
hitter does not have a strategy to achieve |G| ≤ z. That is, against each strategy of Edge-
hitter there is a strategy of Staller which results in |G| > z. We may infer that z ∈ LH and
therefore z ≤ sup(LH). Since it holds for every z < inf(UH), we conclude inf(UH) ≤
sup(LH). This completes the proof of the proposition.
Later, in Section 6, we will show that inf(UH) = min(UH) and sup(LH) = max(LH).
Therefore, Edge-hitter and Staller have optimal strategies under which, respectively, |G| ≤
τ∗g (H) and |G| ≥ τ∗g (H) are achieved.
Cs. Bujtás et al.: Optimal strategies in fractional games: vertex cover and domination 457
2.1 Examples for the fractional transversal game
(1) Our first example is the 4-cycle C4 = v1v2v3v4v1, which can also be considered
as a 2-uniform hypergraph. It is easy to check that τ∗(C4) = 2, while τg(C4) = 3 and
τ ′g(C4) = 2 were proved for the integer games [9]. Now we prove that τ
∗
g (C4) = 5/2.
• For the upper bound, the following strategy of Edge-hitter ensures that the sum of
the weights spent during the game is at most 5/2. His first move is (v1, 14 ), (v2,
1
4 ),
(v3,
1
4 ), (v4,
1
4 ); it results in ℓ1 ≡
1
2 and
∑
E∈E ℓ1(E) = 2. Then, for the first
1
2
of the weight spent by Staller, each of her submoves necessarily increases the load
of both incident edges; and each of her remaining submoves increases the load of
at least one edge. Therefore, the total load increases by at least 12 × 2 +
1
2 =
3
2 to∑
E∈E ℓ2(E) ≥
7
2 , and Edge-hitter can achieve
∑
E∈E ℓ3(E) = 4 by spending at
most 12 in the final move. This proves τ
∗
g (C4) ≤ 5/2.
• Let us show the reverse inequality. We note that the first move of the game has the
same effect as a sequence (v1, w1), (v2, w2), (v3, w3), (v4, w4) of submoves with∑4
i=1 wi = 1.
4 After this move of Edge-hitter,
∑
E∈E ℓ1(E) = 2 and hence, there
is an edge E with ℓ1(E) ≤ 12 . By symmetry, we may assume that ℓ1(v1v2) ≤
1
2 . Let
Staller play the move (v3, w1+w4), (v4, w2+w3). The move is legal as ℓ1(v2v3) =
w2 + w3 = 1 − (w1 + w4) and ℓ1(v4v1) = w1 + w4 = 1 − (w2 + w3). We then
have ℓ2(v1v2) = ℓ1(v1v2) ≤ 12 and Edge-hitter needs to spend at least
1
2 to finish the
game. This strategy of Staller shows τ∗g (C4) ≥ 5/2.
If Staller starts the fractional transversal game on C4 with the move (v1, w1), (v2, w2),
(v3, w3), (v4, w4), then Edge-hitter can ensure |G| = 2 by playing (v1, w3), (v2, w4),
(v3, w1), (v4, w2). Indeed, ℓ2 assigns w1 +w2 +w3 +w4 = 1 to every edge of the graph.
Therefore, τ∗g
′(C4) ≤ 2. Since τ∗g
′(C4) ≥ τ∗(C4) = 2 also holds5, we get τ∗g
′(C4) = 2.
v4
u4
v1
u1
v2
u2
v3
u3
Figure 1: A hypergraph H with nested edges.
4Theorem 4.1 and Lemma 5.1 will give conditions for this replacement property in general. For the first move
of the game, it is much easier to see as, for each edge E, its load ℓ1(E) equals the sum of the weights assigned to
the vertices of E.
5See the proof of Proposition 3.1(i) for a simple explanation.
458 Ars Math. Contemp. 24 (2024) #P3.05 / 453–471
(2) Now we modify the previous example C4 by adding four new vertices u1, . . . , u4 and
four new edges {v1, v2, u1}, . . . , {v4, v1, u4} to get the hypergraph H shown in Figure 1.
When the fractional (or integer) transversal number is considered, each edge that is a super-
set of another edge can be deleted, which implies τ∗(H) = τ∗(C4) = 2. We show that the
situation is different for the fractional transversal game on H, that is τ∗g (H) = 3 ̸= τ∗g (C4).
Suppose that Edge-hitter starts the fractional transversal game G on H.
• We first show that Edge-hitter can ensure that the value |G| of the game is at most
3. An optimal first move for him is (v1, 14 ), (v2,
1
4 ), (v3,
1
4 ), (v4,
1
4 ). Then, inde-
pendently of Staller’s reply, Edge-hitter plays (v1, w1), (v2, w2), (v3, w3), (v4, w4)
as his second move, where wi equals 14 or, if (vi,
1
4 ) is not a legal submove, wi is
the maximum legal weight for vi. After this move, ℓ3 ≡ 1 and we may infer that
|G| ≤ 3. This proves τ∗g (H) ≤ 3.
• Our second claim is that Staller has a strategy that always results in |G| ≥ 3. As
each vertex of H belongs to at most two 3-element edges, after Edge-hitter’s first
move the sum of the loads of the 3-element edges is at most 2. Thus, Staller can
play a legal move that does not assign weights to v1, v2, v3, v4. After this move, the
sum of the loads of the 2-element edges remains at most 2, and Edge-hitter has to
spend a weight of at least 1 to finish the game. This shows τ∗g (H) ≥ 3. We therefore
conclude τ∗g (H) = 3 > τ∗g (C4).
(3) The removal of the edges which are subsets of other edges in a hypergraph F may
also change the values of the parameters. For instance, let F be the hypergraph obtained
from C4 by adding the 4-element edge E = {v1, v2, v3, v4}. As the load of E equals 1
after the first move in the game, it is easy to see that τ∗g (F) = τ∗g (C4) = 5/2, while the
removal of all 2-element edges results in a one-edge hypergraph F ′ with τ∗g (F ′) = 1.
3 Some basic facts and the Continuation Principle
In this section we first observe some simple inequalities which are analogous to the ones in
other games concerning graph domination and hypergraph transversal, most notably to the
fractional domination game [15].
Proposition 3.1.
(i) For every hypergraph H, it holds that
τ∗(H) ≤ τ∗g (H) < 2τ∗(H) and τ∗(H) ≤ τ∗g
′(H) < 2τ∗(H) + 1.
(ii) There is no universal constant C with τg(H) ≤ C · τ∗g (H), and not even with
τ(H) ≤ C · τ∗g (H). The same holds true for τ∗g
′(H), too.
Proof. No matter which player starts the game, at the end the cover function tq is a frac-
tional transversal. This implies the lower bounds τ∗g (H) ≥ τ∗(H) and τ∗g
′(H) ≥ τ∗(H).
Concerning a fractional transversal game G on H and the upper bounds in (i), we can
write the value of the game in the form |G| = W + W ′, where W and W ′ denote the
total sum of weights assigned by Edge-hitter and Staller, respectively. To keep the claimed
bounds, first Edge-hitter can fix an optimal fractional transversal f , i.e. one with |f | =
Cs. Bujtás et al.: Optimal strategies in fractional games: vertex cover and domination 459
τ∗(H). After that, in his moves he can apply the strategy to play submoves (vij , wj) with
the largest possible weights wj which are not only allowed by (∗) but also respect the
inequalities ti−1(vij ) + wj ≤ f(vij ). If such a legal submove with a positive weight does
not exist anymore, then H is fully covered and the game is finished.
This strategy yields W ≤ τ∗(H), with strict inequality if the game is finished by Staller.
We also have W ′ ≤ W or W ′ ≤ W + 1, depending on whether the first move is made
by Edge-hitter or Staller, both with strict inequalities if the game is finished by Edge-hitter.
Since only one of the players can make the last move, the claimed strict upper bounds
follow.
For the proof of (ii) we apply the following result of Alon [1]: For every ϵ > 0 and
for any sufficiently large k, there is a k-uniform hypergraph H = (V, E) such that τ(H) ≥
(1 − ϵ) ln kk (|V | + |E|). On the other hand, a very simple fractional transversal f with
|f | = |V |/k may be constructed by assigning f(v) = 1/k to each vertex v ∈ V . Therefore,
τ∗(H) ≤ |V |k and we obtain
(1/2− ϵ) ln k < sup
H
τ(H)
2τ∗(H)
< sup
H
τ(H)
τ∗g (H)
≤ sup
H
τg(H)
τ∗g (H)
due to the obvious fact τ ≤ τg and the inequality τ∗g < 2τ∗ from (i). For τ∗g
′(H) the proof
is similar, by the second part of (i).
Proposition 3.2. The upper bounds in Proposition 3.1(i) are tight apart from an additive
constant at most 2.
Proof. Consider the complete bipartite graph G = Kk,k2 on k+k2 vertices as a 2-uniform
hypergraph. Clearly, τ∗(G) = k. In any submove of a fractional transversal game, while
G is not fully covered, Staller can always select a vertex from the bigger partite class.
Following this strategy, during k − 1 moves, Staller increases the sum of the loads by at
most (k − 1)k. As G has maximum degree k2, k − 1 moves of Edge-hitter increase the
loads by at most (k−1)k2. Hence, no matter whether Edge-hitter or Staller starts the game,
after 2k − 2 moves we have∑
E∈E
ℓ2k−2(E) ≤ (k − 1)k + (k − 1)k2 = k3 − k < |E| ,
therefore the game is not over yet. This shows τ∗g (G) > 2k − 2 = 2τ∗(G) − 2 and,
similarly, τ∗g
′(H) ≥ 2τ∗(G)− 1 follows if Staller starts the game.
A monotone property of the game fractional transversal number is expressed in the
following idea, which provides a useful tool in simplifying several arguments. Let a hy-
pergraph H with a pre-defined load function ℓ be given, which we consider as a non-zero
starting configuration. We ask about the value |G| of the game started by Edge-hitter, where
the game is finished when ℓ is completed to a load function under which H is fully covered.
The rules are the same as they were in the case of ℓ0 ≡ 0, but here we have ℓ0 = ℓ, while
the value of the game is still computed by starting with the cover function t0 ≡ 0. Under
these conditions and assuming that Edge-hitter starts the fractional transversal game G on
hypergraph H with the pre-defined ℓ, we consider the sets
UH|ℓ = {a ∈ R : Edge-hitter has a strategy that ensures |G| ≤ a} ,
LH|ℓ = {b ∈ R : Staller has a strategy that ensures |G| ≥ b}.
460 Ars Math. Contemp. 24 (2024) #P3.05 / 453–471
Then the game fractional transversal number with predefined load function ℓ is defined as
τ∗g (H|ℓ) = inf(UH|ℓ). It can be shown with a proof analogous to that of Proposition 2.1
that inf(UH|ℓ) = sup(LH|ℓ) always holds.
The corresponding value τ∗g
′(H|ℓ) is defined analogously for the Staller-start game on
H|ℓ.
The imagination strategy is a useful technique applied in many proofs when domination
and transversal games are considered (see e.g. [3, 6, 9, 15, 17, 20]). It was introduced
and first used in [7]; for a detailed explanation and examples we refer the reader to [5,
Chapter 2.2].
Theorem 3.3 (Continuation Principle). If ℓ and ℓ′ are load functions on the hypergraph
H = (V, E) such that ℓ(E) ≤ ℓ′(E) holds for every E ∈ E , then τ∗g (H|ℓ) ≥ τ∗g (H|ℓ′), and
similarly τ∗g
′(H|ℓ) ≥ τ∗g
′(H|ℓ′).
Proof. Assume for a contradiction that τ∗g (H|ℓ) < τ∗g (H|ℓ′), and choose two reals t1, t2
with τ∗(H|ℓ) < t1 < t2 < τ∗g (H|ℓ′). We use the imagination strategy between the
following two games:
Game 1: Edge-hitter plays on H|ℓ applying a strategy which ensures that the value of
the game is at most t1. Staller’s moves in Game 1 are defined according to her moves in
Game 2.
Game 2: Staller plays on H|ℓ′ applying a strategy which ensures that the value of the
game is at least t2. Edge-hitter’s moves in Game 2 are defined according to his moves in
Game 1.
Assume that Edge-hitter starts the game. He chooses his first move in Game 1 according
to the prescribed strategy. We copy this move into Game 2 if it is a legal move there, or
choose an appropriate replacement move for Edge-hitter in Game 2. In the next turn, Staller
replies with a move according to her prescribed strategy in Game 2 and we copy the same
move into Game 1. (We will see that it is always legal.) Then, Edge-hitter replies in Game 1
and we copy it or make the according move for Edge-hitter in Game 2. The two parallel
games continue this way until at least one of them is finished.
The moves essentially are copied (or interpreted) between Game 1 and Game 2 such
that ℓ(E) ≤ ℓ′(E) remains true for all E ∈ E after every move. If this inequality is valid
before Staller’s move in Game 2, then her next move in H|ℓ′ is legal in H|ℓ as well, so that
we can simply copy it into Game 1. The condition ℓ(E) ≤ ℓ′(E) for all E ∈ E clearly
remains valid for the new load functions.
Suppose now that the inequality ℓ(E) ≤ ℓ′(E) is true for all E ∈ E before Edge-
hitter’s move in Game 1. If it is legal, we simply copy it into Game 2 and the inequality
remains valid. In the other case one or more submoves (vik , wk) made in Game 1 are not
legal in Game 2. We then choose the maximum w′k such that (vik , w
′
k) is a legal submove
in Game 2. The remaining weight wk − w′k can be distributed between arbitrary vertices
such that the submoves are legal. Observe, however, that if this happens, all loads on the
edges incident with vik reach 1 after the submove (vik , w
′
k) in Game 2. We infer that
ℓ(E) ≤ ℓ′(E) remains true for all E ∈ E after Edge-hitter’s move.
It follows that the loads will never become smaller in Game 2 than the corresponding
ones in Game 1. Thus, the values g1 and g2 of Games 1 and 2 satisfy g1 ≥ g2. By the
strategies of the players, it is true that t1 ≥ g1 and g2 ≥ t2. We therefore obtain
t1 ≥ g1 ≥ g2 ≥ t2 > t1
Cs. Bujtás et al.: Optimal strategies in fractional games: vertex cover and domination 461
and this contradiction proves τ∗g (H|ℓ) ≥ τ∗g (H|ℓ′).
The analogous conclusion can be reached in the Staller-start game as well, literally by
the same argument, deriving a contradiction from the assumption τ∗g
′(H|ℓ) < τ∗g
′(H|ℓ′).
We obtain the following immediate consequence.
Theorem 3.4. The game fractional transversal numbers for the Staller-start and for the
Edge-hitter-start games on H may differ by at most 1.
Proof. Consider the Staller-start game. Whatever Staller moves first, she assigns total
weight 1, and creates a situation which is at least as favorable for Edge-hitter as the all-
zero load at the beginning of the original transversal game. Then, due to Theorem 3.3,
Edge-hitter can ensure that the game ends using at most τ∗g (H) further weight. This proves
τ∗g
′(H) ≤ τ∗g (H) + 1.
Similarly, if Edge-hitter starts, after his first move he is in at least as favorable position
as with the all-zero load at the beginning of the Staller-start game. This proves the reverse
inequality τ∗g (H) ≤ τ∗g
′(H) + 1.
4 Infinite moves are not necessary
The definition of a legal move in the transversal game admits the option that a player splits
the value 1 into an infinite number of pieces; e.g., wk = 2−k. It turns out, however, that
each legal move on H = (V, E) is equivalent to a move which consists of at most |V |
submoves.
Theorem 4.1. Every legal move in a fractional transversal game can be replaced with a
legal move such that each vertex occurs in at most one submove of it and the two moves
result in the same load function.
Proof. First, consider a vertex v which occurs in two different submoves (vij , wj) and
(vik , wk) of a move. That is, v = vij = vik and we may assume j < k. By the condi-
tion (∗), there exists an edge E ∈ E such that v ∈ E and the second submove (vik , wk)
increases the load of E by exactly wk. If the submove (vij , wj) is deleted from the se-
quence and the weight wk is replaced by wj +wk in the kth submove, the submove and the
whole move remain legal and result in the same load function as before. Performing this
modification repeatedly we can achieve that every vertex occurs in either zero or exactly
one or infinitely many submoves of the move in question. This already proves the statement
if the move contains only a finite number of submoves.
Now, assume that the move is infinite. Then, the sequence of submoves can be split
into two, such that the first part is finite, and in the second infinite part every vertex
(which is present there) is repeated infinitely many times. Consider this infinite subse-
quence S = (vis , ws), . . . . By renaming the vertices of H if necessary, we may assume
that {v1, . . . , vℓ} is the set of the vertices which are present in S. We prove that the finite
sequence S′ = (v1,
∑
j: ij=1
wj), . . . , (vℓ,
∑
j: ij=ℓ
wj) is equivalent to S. Clearly, S and
S′ yield the same load function after the move. So, it is enough to prove that S′ is legal.
Assume for a contradiction that (vk,
∑
j: ij=k
wj) is not a legal submove in S′, and let k
be the smallest such index. Then, after the (k − 1)st submove of S′, every edge E which
contains vk has a load ℓ(E) > 1−
∑
j: ij=k
wj and, moreover, there is a positive constant
462 Ars Math. Contemp. 24 (2024) #P3.05 / 453–471
ϵ such that minvk∈E ℓ(E) +
∑
j: ij=k
wj = 1 + ϵ. Now, consider S again. There is an
index p = p(ϵ) such that
∑
j≥p wj < ϵ and hence, before the p
th submove of S, each edge
containing vk is fully covered. As vk occurs infinitely often in S, and also the occurrences
after the pth submove are legal, this is a contradiction.
5 Edge-hitter’s moves are transposable
We say that a finite move (vi1 , w1), . . . , (vik , wk) is transposable if for any permutation
β(1), . . . , β(k) of 1, . . . , k, the move (viβ(1) , wβ(1)), . . . , (viβ(k), wβ(k)) is legal. We will
show that from the point of view of Edge-hitter, we can restrict our attention to transposable
moves. Note that every transposable move is legal, but not conversely.
We first give a characterization of transposable moves:
Lemma 5.1. A move (vi1 , w1), . . . , (vik , wk), where wj > 0 for all 1 ≤ j ≤ k and
no vertices are repeated, is transposable if and only if the total weight constraint (∗∗) is
satisfied, and after performing the entire move, for every vertex vij ,
min
E∋vij
∑
vi∈E
t(vi) ≤ 1. (5.1)
Proof. A legal move must satisfy the two constraints (∗) and (∗∗).
The total weight constraint (∗∗) for a legal move is explicitly required in the lemma,
and it is insensitive to permuting the submoves.
Let us turn to (∗). If
∑
vi∈E t(vi) ≤ 1 for an edge E containing vij , then omitting the
submove (vij , wj) from the move we obtain
∑
vi∈E\{vij }
t(vi) ≤ 1−wj . Hence (vij , wj)
is a legal submove no matter when it is performed during the move. This means that the
move is transposable whenever the condition (5.1) is satisfied for all j.
In the other direction, assume that for some vij , the left-hand side of (5.1) is bigger
than 1. Consider a permutation in which (vij , wj) is the last submove. Then (vij , wj) is
not legal because (∗) is violated. Consequently the move is not transposable.
Theorem 5.2. If a finite legal move m = (vi1 , w1), . . . , (vik , wk) is not transposable in
the fractional transversal game, then it can be replaced by a transposable (and legal) move
after which no edge gets smaller load than after m.
Proof. First, consider the legal move m = (vi1 , w1), . . . , (vik , wk) and the move m
′ =
(vi2 , w2), . . . , (vik , wk), (vi1 , w1) that is obtained by the cyclic permutation β =
2, . . . , k, 1. It is clear that condition (∗) in the definition remains true for the first k − 1
submoves of m′ and consequently, these submoves are legal. For the last (and not nec-
essarily legal) submove, determine w∗1 as the maximum weight which results in a legal
submove with vi1 . If w
∗
1 ≥ w1, then m′ is legal and gives exactly the same load function
as m. If w∗1 < w1, then the same load function is obtained after the submove (vi1 , w
∗
1) as
after m, because in both cases every edge incident with vi1 is fully covered and the loads
of the other edges are unchanged.
In the latter case, the sum of the weights is decreased by w1−w∗1 . After this change, the
submove (vi1 , w
∗
1) will be legal in any permutation of (vi1 , w
∗
1), (vi2 , w2), . . . , (vik , wk).
That is, if a permutation is not legal after this replacement, this is due to another vertex.
The same is true if some weights ws are replaced by smaller weights.
Cs. Bujtás et al.: Optimal strategies in fractional games: vertex cover and domination 463
We repeat this step for the modified sequence with permutation β = 3, . . . , k, 1, 2,
then with β = 4, . . . , k, 1, 2, 3, and so on, finally with β = k, 1, 2, . . . , k − 1, keeping
all modifications incrementally. Decreasing the weight of the last submove in each step
if necessary, at the end a legal transposable move m∗ is obtained, which yields the same
load function as m and satisfies
∑k
j=1 w
∗
j ≤
∑k
j=1 wj . If ∆ =
∑k
j=1 wj −
∑k
j=1 w
∗
j is
positive, we use the weight ∆ to increase the loads of some non-fully covered edges in an
arbitrary way. The total absolute change of weights is 2∆, or possibly less if the game is
over. Finally, we normalize the move by eliminating any multiple occurrences of vertices,
using Theorem 4.1.
The redistribution of ∆ and the normalization may lead to a move that is not transpos-
able. If so, we repeat the modification described above.
If the process terminates after a finite number of iterations, then the last version of the
move is transposable by construction, and the proof is complete. Otherwise we obtain an
infinite sequence ∆(1),∆(2), . . . of re-distributions from the total unit weight of the move.
The total load of edges increases by at least
∑
i≥1 ∆
(i); hence this sum converges because
altogether the total load is at most |E|.
On the other hand, the weight of a vertex changes by at most ∆(i) in the ith iteration,
hence the local changes in weight (at least one of which is negative in each iteration) in
absolute value sum up to at most 2
∑
i≥1 ∆
(i), and therefore their sum also converges.
Let m(i) denote the move constructed in the ith iteration, before the re-distribution of
the weight ∆(i). We know that this move is transposable. Let w(i)v denote the weight used
for vertex v in the corresponding submove of m(i), and set w(i)v = 0 if v does not appear in
this move.
We have shown that the limit of these weights w∗v = limi→∞ w
(i)
v = 0 exists. Let p
denote the number of vertices with a positive weight in w∗. Relabeling these vertices in
an arbitrary order, we define a move m∗ = (v1, w∗1), . . . , (vp, w
∗
p) with p submoves. By
continuity, the loads achieved by the moves m(i) converge to the corresponding loads after
the move m∗.
We claim that m∗ is transposable. This will complete the proof because the loads never
decrease, hence under w∗ no edge gets smaller load than by move m.
For showing that m∗ is transposable, we use the characterization of Lemma 5.1. The
inequality (5.1) follows from the fact that its left-hand side is a continuous function of the
weights.
Let us finally check that the total weight constraint (∗∗) is satisfied: Since
∑
v w
(i)
v ≤ 1
for all i, this inequality is satisfied in the limit. If in some iteration, the current move
terminates the game (i.e., each edge gets load at least 1 while the total weight in the move
is at most 1), this will hold in all successive moves, since the loads never decrease, and
hence it holds also for m∗, by continuity. Therefore, we need to show
∑
v w
∗
v = 1 only
when none of the moves terminate the game. In this case, the difference 1 −
∑
v w
(i)
v is
bounded by ∆i, which goes to 0, and hence
∑
v w
∗
v = 1 in the limit.
Remark 5.3. Based on Theorem 5.2, Edge-hitter may restrict his strategy to transposable
moves. On the other hand, the result suggests that Staller is advised to perform moves, if
possible, which are ‘very non-transposable’ in a sense.
464 Ars Math. Contemp. 24 (2024) #P3.05 / 453–471
6 Algorithm for computing the value of the game
We consider an equivalent version of the game, the structured game, which is easier to
analyze.
Each move consists of n + 1 rounds. Each round consists of n submoves, which are
dedicated to the vertices v1, . . . , vn in succession. In each submove, the player whose turn
it is can decide the amount w, the weight spent in the submove, by which the cover value of
t(vi) is increased, subject to the usual rules: The increase must be useful, i.e. each submove
must satisfy the condition (∗), and it must be within the budget constraint of total weight 1
to be spent per move. It is possible to skip a submove by simply choosing w to be zero.
The first n rounds are identical, but the last round is special: In each submove, the
weight is greedily chosen as the largest possible legal weight, hence not allowing any free-
dom in choosing w for the player in those submoves. This ensures that the whole move
spends a total weight of 1 unless a cover is obtained.
There are n moves, alternating between the two players. This is enough to ensure that a
cover is constructed when the game terminates. Every move consists of n2 + n submoves,
and in total, the game consists of N = n3 + n2 submoves. We illustrate this for a small
example with n = 4 vertices, where the Edge-hitter starts. The sequence of submoves is
H1, H2, H3, H4; H1, H2, H3, H4; H1, H2, H3, H4; H1, H2, H3, H4; G1, G2, G3, G4;
S1, S2, S3, S4; S1, S2, S3, S4; S1, S2, S3, S4; S1, S2, S3, S4; G1, G2, G3, G4;
H1, H2, H3, H4; H1, H2, H3, H4; H1, H2, H3, H4; H1, H2, H3, H4; G1, G2, G3, G4;
S1, S2, S3, S4; S1, S2, S3, S4; S1, S2, S3, S4; S1, S2, S3, S4; G1, G2, G3, G4.
Here Hi denotes a move of Edge-hitter for vertex i, and Si denotes a move of Staller for
vertex i. The greedy moves are denoted by Gi.
We do not stipulate as part of the rules that the whole budget of 1 unit must be spent
during a move. This capacity is only an upper bound. It is still true that the whole budget
is spent in each move if the game is played from the beginning. However, this arises as a
consequence of the new setup, due to the greedy moves.
As soon as a cover is found, the rules imply that no more weight can be spent, and thus
the game is effectively over.
Lemma 6.1. The structured game has the same value as the original game.
Proof. According to Theorem 4.1, we can assume that every vertex occurs at most once in
a move. We can realize this in the structured game by selecting one vertex per round and
leaving the weight at 0 otherwise. Thus, the structured game does not restrict the players’
strategies, when compared to the original game. On the other hand, the structured game
does not give the players more power: The greedy moves ensure that the total weight of 1
is used as long as it is possible.
Example for the structured game. Consider an Edge-hitter-start game on C4 with strate-
gies of the players as described in Section 2.1. A corresponding structured game, where
we follow the idea in the proof of Lemma 6.1, can be given as follows.
Cs. Bujtás et al.: Optimal strategies in fractional games: vertex cover and domination 465
Edge-hitter’s first move: (v1, 14 ), (v2, 0), (v3, 0), (v4, 0);
(v1, 0), (v2, 14 ), (v3, 0), (v4, 0);
(v1, 0), (v2, 0), (v3, 14 ), (v4, 0);
(v1, 0), (v2, 0), (v3, 0), (v4, 14 );
(v1, 0), (v2, 0), (v3, 0), (v4, 0).
Staller’s first move: (v1, 0), (v2, 0), (v3, 12 ), (v4, 0);
(v1, 0), (v2, 0), (v3, 0), (v4, 12 );
(v1, 0), (v2, 0), (v3, 0), (v4, 0);
(v1, 0), (v2, 0), (v3, 0), (v4, 0);
(v1, 0), (v2, 0), (v3, 0), (v4, 0).
Edge-hitter’s second move: (v1, 0), (v2, 16 ), (v3, 0), (v4, 0);
(v1, 0), (v2, 0), (v3, 0), (v4, 0);
(v1, 0), (v2, 0), (v3, 0), (v4, 0);
(v1, 0), (v2, 0), (v3, 0), (v4, 0);
(v1,
1
3 ), (v2, 0), (v3, 0), (v4, 0).
For demonstration, the last move makes use of a greedy submove.
By Theorem 5.2, it is not a disadvantage for Edge-hitter restricting himself to
transposable moves. Consequently, his submoves can always be performed in the order
(v1, w1), . . . , (vk, wk), and a single round H1, H2, . . . ,Hn followed by n greedy sub-
moves would be a sufficient model for Edge-hitter’s moves. For simplicity, we have how-
ever chosen to treat the two players uniformly. Note that the greedy submoves are necessary
also in case of Edge-hitter. Otherwise, for example, he might pass on the first move and
transform the game to the Staller-start version, which sometimes admits a smaller game
value (as in the example of C4).
Consider the situation after the jth submove, 0 ≤ j ≤ N . Let x⃗ ∈ [0, 1]E be an arbitrary
load vector, and let r ∈ [0, 1] be the budget for the current move that is still available. If j
is written in the form j = k(n2 + n) + i, i.e. k = ⌊ jn2+n⌋ and 0 ≤ i < n
2 + n, then 1− r
is the total weight spent in the last i submoves.
We define
Tj(x⃗, r)
as the sum of weights spent during the remaining part of the game, if both players play
optimally, starting from the current situation. If j is large and the entries of x⃗ are small,
it may happen that a complete fractional cover is not reached, because the game neces-
sarily ends after the nth round. Nevertheless, we have chosen our definition because it
makes Tj well-defined for arbitrary x⃗ and r. (The definition of Tj(x⃗, r) is related to the
game fractional transversal number with predefined load function used in the proof of the
Continuation Principle (Theorem 3.3)6.)
The value of the original game is T0(⃗0, 1).
We will derive a backward recursion for the functions Tj , and thus show that they are
piecewise linear and continuous.
6In particular, if the jth submove is the last submove in a move of Staller and ℓ is the load function corre-
sponding to x⃗, then we would expect Tj(x⃗, 1) to be τ∗g (H|ℓ). However this does not hold in general because, as
just discussed, the structured game may terminate too early.
466 Ars Math. Contemp. 24 (2024) #P3.05 / 453–471
Given j, we know the type of the jth submove (H , S, or G) and the vertex vi to which
it applies. We denote the maximum permitted weight by
wmaxi (x⃗, r) = min{r,max{ 1− xE | E ∋ vi }} , (6.1)
where xE denotes the entry of x⃗ that corresponds to the edge E ∈ E . For the result of
increasing the cover value of vi by w we write
updatei(x⃗, w) = x⃗
′ with x′E =
{
xE , if vi /∈ E,
min{1, xE + w}, if vi ∈ E.
(6.2)
With these definitions, the recursion for a submove Hi for Edge-hitter can be written easily:
Tj−1(x⃗, r) = min{w + Tj(updatei(x⃗, w), r − w) | 0 ≤ w ≤ wmaxi (x⃗, r) } (6.3)
If the submove is for Staller (Si), the recursion is the same as (6.3), except that min is
replaced by max. In the greedy submoves Gi, we always choose w = wmaxi (x⃗, r):
Tj−1(x⃗, r) = w
max
i (x⃗, r) + Tj(updatei(x⃗, w
max
i (x⃗, r)), r − wmaxi (x⃗, r)) (6.4)
The last greedy submove Gn of each move is an exception: Since a different move is about
to start, the budget r is reset to 1. Thus, when j is a multiple of n2 + n, then
Tj−1(x⃗, r) = w
max
n (x⃗, r) + Tj(updaten(x⃗, w
max
n (x⃗, r)), 1). (6.5)
As the recursion anchor, we use the value after the final move, which is simply
TN (x⃗, r) = 0. (6.6)
Theorem 6.2. Each function Tj(x⃗, r) for 0 ≤ j ≤ N is a piecewise linear continuous
function with finitely many linear pieces defined on [0, 1]E × [0, 1]. Moreover, all Tj are
rational in the sense that each linear piece has rational coefficients and rational constant
part. As a consequence, the boundaries between regions of the domain with different linear
functions can be described by linear equations with rational coefficients.
Proof. We will call a function with all the desired properties — piecewise linear, continu-
ous, and rational, with finitely many linear pieces — a PLCR function.
The proof proceeds by backward recursion from TN down to T0. The function TN
from (6.6) is obviously PLCR.
The sum, difference, maximum, or minimum of two PLCR functions is again PLCR,
and the same holds true when substituting one PLCR function into another. It follows
directly that the functions wmaxi and updatei are PLCR functions on the domain [0, 1]
E ×
[0, 1]. This allows us to perform the induction step in the recursions (6.4)–(6.5) for Gi.
In the recursion (6.3) we additionally have a minimization (or, in the analogous recur-
sion for Staller, a maximization) over some range of values w. It has the form
min{F (x⃗, r, w) | 0 ≤ w ≤ wmaxi (x⃗, r) }
with the PLCR function
F (x⃗, r, w) := w + Tj(updatei(x⃗, w), r − w)
Cs. Bujtás et al.: Optimal strategies in fractional games: vertex cover and domination 467
To get rid of the varying upper bound on w, we rewrite the recursion in terms of another
PLCR function
F̂ (x⃗, r, w) = F (x⃗, r,min{w,wmaxi (x⃗, r)}
as
Tj−1(x⃗, r) = min{ F̂ (x⃗, r, w) | 0 ≤ w ≤ 1 }.
Lemma 6.3 below establishes that Tj−1 is a PLCR function.
The same argument applies to the recursion for Staller (Si), where min is replaced by
max.
Lemma 6.3. Suppose that F̂ (y, w) : [0, 1]m × [0, 1] → R is a PLCR function. Then the
function T (y) : [0, 1]m → R defined by minimizing over w:
T (y) := min{ F̂ (y, w) | 0 ≤ w ≤ 1 } (6.7)
is also a PLCR function.
Proof. We first show that T is continuous. Since F̂ is PLCR, it is Lipschitz-continuous.
Let L be its Lipschitz constant with respect to the ∞-norm. (We can compute L as the
maximum L1-norm of all coefficient vectors of the linear pieces of F̂ .) It follows that the
function T in (6.7) is also Lipschitz-continuous with Lipschitz-constant L. To see this, let
∥y0 − y1∥ ≤ ε, and let T (y0) = F̂ (y0, w0) for some w0. Then T (y1) ≤ F̂ (y1, w0) ≤
F̂ (y0, w0) + Lε = T (y0) + Lε. The converse bound T (y0) ≤ T (y1) + Lε follows in the
same way.
We still need to show that T is piecewise linear. For an intuitive way to see this, one can
interpret the minimization over w geometrically. The graph of F̂ : [0, 1]m× [0, 1] → R is a
subset of Rm+2. Taking the minimum over all w amounts to projecting away the coordinate
corresponding to w and taking the lower envelope (with respect to the last coordinate) in the
projection in Rm+1. Figure 2 shows a two-dimensional illustration. This picture can also be
interpreted as a three-dimensional view of the graph of a bivariate function F̂ (y, w) when
the viewing direction is parallel to the w-axis. (In this hypothetical example, the resulting
minimum is discontinuous; this cannot happen when F̂ is continuous and its domain is the
box [0, 1]m × [0, 1].)
Formally, we conduct the proof as follows. We know that the domain [0, 1]m+1 of F̂
splits into finitely many rational convex (m + 1)-dimensional polytopes P on which F̂ is
linear:
F̂ (y, w) = aP y + bPw + cP , for (y, w) ∈ P
for some rational coefficient vector aP and rational coefficients bP and cP . We can thus
write T (y) as the minimum of finitely many functions TP (y) of the form
TP (y) := min{ aP y + bPw + cP | 0 ≤ w ≤ 1, (y, w) ∈ P }, (6.8)
where the minimum of an empty set is taken as ∞.
For fixed y, the minimum in (6.8) depends on the sign of bP . If bP > 0, the minimum
is achieved on a boundary point that lies on some facet P ′ of P whose outer normal has
negative w-coordinate. On such a facet, w can be expressed as a linear function of y, and
thus, TP can be written as a linear function
TP ′(y) = aP ′y + cP ′ , for y ∈ P̄ ′, (6.9)
468 Ars Math. Contemp. 24 (2024) #P3.05 / 453–471
y
{ (y, F̂ (y, w)) | 0 ≤ w ≤ 1 }
F̂
0 1
Figure 2: The lower envelope of a polyhedral set in 2 dimensions (m = 1).
where P̄ ′ is the projection of the facet P ′ to [0, 1]m. Thus, TP (y) is the minimum of finitely
many functions TP ′(y), with the understanding that TP ′(y) is taken as ∞ when y is outside
its domain P̄ ′.
The situation is similar for bP < 0. When bP = 0, then F̂ does not depend on w and
we can simply write
TP (y) = aP y + cP , for y ∈ P̄ , (6.10)
where P̄ is the projection of P .
In summary, the function T (y) can be written as the minimum of finitely many pieces
TP (y), each of which can in turn be written as the minimum of finitely many linear pieces
(6.9) or (6.10). All these pieces have rational coefficients and rational domain boundaries,
and since continuity of T has already been established, the PLCR property of T follows.
The proof of Theorem 6.2 is constructive and, in principle, it provides an algorithm for
computing the value T0(⃗0, 1) of the game. From this, we obtain the following important
corollary.
Theorem 6.4. For every finite hypergraph H = (V, E), the game fractional transversal
number τ∗g (H) and its Staller-start version τ∗g
′(H) are rational. Moreover, each player in
every step has an optimal move with only rational weights, provided that the weights in all
previous submoves were rational.
Remark 6.5. It is not true in general that every optimal strategy uses only rational weights.
A simple counterexample is the graph C4 (Example 1 from Section 2.1), where Staller can
start by placing x and 1 − x on two vertices with any x ∈ [0, 1], no matter if x is rational
or irrational.
7 Concluding remarks and open problems
Putting the fractional domination game [15] into a more general context, in this paper
we introduced the fractional transversal game on hypergraphs. Among other results, we
Cs. Bujtás et al.: Optimal strategies in fractional games: vertex cover and domination 469
proved that the game value is rational, and both players have optimal strategies using ratio-
nal weights and with a finite number of submoves. Since a dominating set of a graph is a
transversal of the closed neighborhood hypergraph, and a total dominating set is a transver-
sal of the open neighborhood hypergraph, the following consequence is immediate.
Theorem 7.1. The fractional versions of both the domination game and the total domi-
nation game have rational game values (game fractional domination number and game
fractional total domination number) on every graph.
We conclude this paper with some conjectures and open questions.
Conjecture 7.2. If each of the first 2k − 1 (k ≥ 1) moves was an integer move in the
fractional transversal game, i.e. of the form (vi1 , 1), then Staller has an integer move in the
(2k)th turn, which is optimal in the fractional transversal game.
This means that fractional moves would be advantageous for Edge-hitter only. If true
then this conjecture implies the following weaker one.
Conjecture 7.3. For every hypergraph H , τ∗g (H) ≤ τg(H).
Perhaps the following stronger version of Conjecture 7.2 is also true.
Conjecture 7.4. Starting from any cover function, there is an optimal strategy for Staller
where, in every submove, she always spends the largest legal weight.
These conjectures could be approached by implementing the algorithm that is implicit
in the proof of Theorem 6.2 by computer. We have not derived an estimate for the com-
plexity (number of pieces) of the piecewise linear continuous functions Tj(x⃗, r) that are
involved in the construction. If the growth of the complexity is not too steep, there is hope
to solve some examples of moderate size, beyond the range of small examples that we
considered in Section 2.1, and this could shed some light on the conjectures.
One would naturally expect that T0(x⃗, r) is monotonically decreasing in x⃗, for fixed r,
and moreover, that it is Lipschitz-continuous with Lipschitz constant 1. In other words, in
every linear piece, the coefficient of each variable xi is between 0 and −1. We have not
explored these properties.
Acknowledgement
We are grateful to Gábor Tardos for helpful discussions.
ORCID iDs
Csilla Bujtás https://orcid.org/0000-0002-0511-5291
Günter Rote https://orcid.org/0000-0002-0351-5945
Zsolt Tuza https://orcid.org/0000-0003-3235-9221
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ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 24 (2024) #P3.06 / 473–488
https://doi.org/10.26493/1855-3974.2892.f07
(Also available at http://amc-journal.eu)
The Sierpiński domination number*
Michael A. Henning †
Department of Mathematics and Applied Mathematics, University of Johannesburg,
Auckland Park, 2006 South Africa
Sandi Klavžar ‡
Faculty of Mathematics and Physics, University of Ljubljana, Slovenia and
Institute of Mathematics, Physics and Mechanics, Ljubljana, Slovenia and
Faculty of Natural Sciences and Mathematics, University of Maribor, Slovenia
Elżbieta Kleszcz
Department of Mathematics and Applied Mathematics, University of Johannesburg,
Auckland Park, 2006 South Africa and
AGH University, Department of Discrete Mathematics, al. Mickiewicza 30,
30-059 Krakow, Poland
Monika Pilśniak §
AGH University, Department of Discrete Mathematics, al. Mickiewicza 30,
30-059 Krakow, Poland
Received 31 May 2022, accepted 8 November 2023, published online 18 July 2024
Abstract
Let G and H be graphs and let f : V (G) → V (H) be a function. The Sierpiński prod-
uct of G and H with respect to f , denoted by G⊗f H , is defined as the graph on the vertex
set V (G) × V (H), consisting of |V (G)| copies of H; for every edge gg′ of G there is an
edge between copies gH and g′H of H associated with the vertices g and g′ of G, respec-
tively, of the form (g, f(g′))(g′, f(g)). In this paper, we define the Sierpiński domination
number as the minimum of γ(G ⊗f H) over all functions f : V (G) → V (H). The up-
per Sierpiński domination number is defined analogously as the corresponding maximum.
*We would like to thank the reviewer very much for careful reading of the paper and in particular for pointing
to a subtle technical detail that led to the reformulation of one of the main theorems.
†This research was obtained during the sabbatical visit of the first author at the University of Ljubljana and
he thanks them for their kindness in hosting him. The first author also acknowledges that his research visit was
supported in part by the University of Johannesburg and the South African National Research Foundation (grants
132588, 129265).
‡S.K. acknowledges the financial support from the Slovenian Research Agency (research core funding No.
P1-0297 and projects J1-2452, N1-0285).
§Corresponding author.
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
474 Ars Math. Contemp. 24 (2024) #P3.06 / 473–488
After establishing general upper and lower bounds, we determine the upper Sierpiński dom-
ination number of the Sierpiński product of two cycles, and determine the lower Sierpiński
domination number of the Sierpiński product of two cycles in half of the cases and in the
other half cases restrict it to two values.
Keywords: Sierpiński graph, Sierpiński product, domination number, Sierpiński domination number.
Math. Subj. Class. (2020): 05C69, 05C76
1 Introduction
Sierpiński graphs represent a very interesting and widely studied family of graphs. They
were introduced in 1997 in the paper [15], where the primary motivation for their intro-
duction was the intrinsic link to the Tower of Hanoi problem, for the latter problem see the
book [11]. Intensive research of Sierpiński graphs led to a review article [12] in which state
of the art up to 2017 is summarized and unified approach to Sierpiński-type graph families
is also proposed. Later research on Sierpiński graphs includes [2, 3, 6, 19, 23].
Sierpiński graphs have a fractal structure, the basic graphs of which are complete
graphs. In 2011, Gravier, Kovše, and Parreau [7] introduced a generalization in such a
way that any graph can act as a fundamental graph, and called the resulting graphs gener-
alized Sierpiński graphs. We refer to the papers [1, 4, 5, 13, 14, 16, 17, 20, 21, 22, 24] for
investigations of generalized Sierpiński graphs in the last few years.
An interesting generalization of Sierpiński graphs in the other direction has recently
been proposed by Kovič, Pisanski, Zemljič, and Žitnik in [18]. Namely, in the spirit of
classical graph products, where the vertex set of a product graph is the Cartesian product of
the vertex sets of the factors, they introduced the Sierpiński product of graphs as follows.
Let G and H be graphs and let f : V (G) → V (H) be an arbitrary function. The Sierpiński
product of graphs G and H with respect to f , denoted by G⊗f H , is defined as the graph
on the vertex set V (G)× V (H) with edges of two types:
• type-1 edge: (g, h)(g, h′) is an edge of G⊗f H for every vertex g ∈ V (G) and every
edge hh′ ∈ E(H),
• type-2 edge: (g, f(g′))(g′, f(g)) is an edge of G⊗f H for every edge gg′ ∈ E(G).
We observe that the edges of type-1 induce n(G) = |V (G)| copies of the graph H in
the Sierpiński product G ⊗f H . For each vertex g ∈ V (G), we let gH be the copy of H
corresponding to the vertex g. A type-2 edge joins vertices from different copies of H in
G⊗f H , and is called a connecting edges of G⊗f H . A vertex incident with a connecting
edge is called a connecting vertex. We observe that two different copies of H in G ⊗f H
are joined by at most one edge. A copy of the graph H corresponding to a vertex of the
graph G in the Sierpiński product G⊗f H is called an H-layer.
Let G and H be graphs and HG be the family of functions from V (G) to V (H).
We introduce new types of domination, the Sierpiński domination number, denoted by
γS(G,H), as the minimum over all functions f from HG of the domination number of
the Sierpiński product with respect to f , and upper Sierpiński domination number, denoted
E-mail addresses: mahenning@uj.ac.za (Michael A. Henning), sandi.klavzar@fmf.uni-lj.si (Sandi Klavžar),
elzbieta.kleszcz@agh.edu.pl (Elżbieta Kleszcz), pilsniak@agh.edu.pl (Monika Pilśniak)
M. A. Henning et al.: The Sierpiński domination number 475
by ΓS(G,H), as the maximum over all functions f ∈ HG of domination number of the
Sierpiński product with respect to f . That is,
γS(G,H) := min
f∈HG
{γ(G⊗f H)}
and
ΓS(G,H) := max
f∈HG
{γ(G⊗f H)} .
In this paper, we initiate the study of Sierpiński domination in graphs. In Section 1.1
we present the graph theory notation and terminology we follow. In Section 2 we discuss
general lower and upper bounds on the (upper) Sierpiński domination number. Our main
contribution in this introductory paper is to determine the upper Sierpiński domination
number of the Sierpiński product of two cycles, and to determine the lower Sierpiński
domination number of the Sierpiński product of two cycles in half of the cases and in the
other half cases restrict it to two values.
1.1 Notation and terminology
We generally follow the graph theory notation and terminology in the books [8, 9, 10] on
domination in graphs. Specifically, let G be a graph with vertex set V (G) and edge set
E(G), and of order n(G) = |V (G)| and size m(G) = |E(G)|. For a subset S of vertices
of a graph G, we denote by G− S the graph obtained from G by deleting the vertices in S
and all edges incident with vertices in S. If S = {v}, then we simply write G − v rather
than G − {v}. The subgraph induced by the set S is denoted by G[S]. We denote the
path, cycle and complete graph on n vertices by Pn, Cn, and Kn, respectively. For k ≥ 1
an integer, we use the notation [k] = {1, . . . , k} and [k]0 = {0, 1, . . . , k}. We generally
label vertices of the considered graphs by elements of [n]. In this case, the mod function
over the set [n] is to be understood in a natural way, more formally, we apply the following
operation for t ≥ 1: t mod∗ n = (t− 1) mod n+ 1.
A vertex dominates itself and its neighbors, where two vertices are neighbors in a graph
if they are adjacent. A dominating set of a graph G is a set S of vertices of G such that
every vertex in G is dominated by a vertex in S. The domination number, γ(G), of G is
the minimum cardinality of a dominating set of G. A dominating set of cardinality γ(G) is
called a γ-set of G. A thorough treatise on dominating sets can be found in [8, 9].
If S is a set of vertices in a graph G, then we will use the notation G|S to denote that
the vertices in the set S are assumed to be dominated and hence γ(G|S) is the minimum
number of vertices in the graph G needed to dominate V (G) \ S. We note that it could be
that a vertex in S is still a member of a such a minimum dominating set no matter that we
do not need to dominate the vertices in S themselves. If S = {x}, then we simply denote
G|S by G|x rather than G|{x}.
2 General lower and upper bounds
We present in this section general lower and upper bounds on the (upper) Sierpiński domi-
nation number.
Theorem 2.1. If G and H are graphs, then
n(G)γ(H)−m(G) ≤ γS(G,H) ≤ ΓS(G,H) ≤ n(G)γ(H) .
476 Ars Math. Contemp. 24 (2024) #P3.06 / 473–488
Proof. Let G ⊗f H be an arbitrary Sierpiński product of graphs G and H and let X be
a γ-set of G ⊗f H . Assuming for a moment that all the connecting edges are removed
from G⊗f H , we obtain n(G) disjoint copies of H for which we clearly need n(G)γ(H)
vertices in a minimum dominating set. Consider now an arbitrary connecting edge e =
(g, f(g′))(g′, f(g)) of G⊗fH . If no end-vertex of e lies in X , then clearly γ(G⊗fH−e) =
γ(G⊗fH). Similarly, if both end-vertices of e lie in X , then γ(G⊗fH−e) = γ(G⊗fH).
Hence the only situation in which e has an effect on γ(G⊗fH) is when (g, f(g′)) ∈ X and
(g′, f(g)) /∈ X (or the other way around). But in this case, the effect of the presence of the
edge e is that because (g, f(g′)) dominates one vertex of g′H , the edge e might reduce the
domination number by 1. That is, each connecting edge can drop the domination number
of G ⊗f H by at most 1, which proves the left inequality. The other two inequalities are
clear.
To show that the lower bound of Theorem 2.1 is achieved, we show later in Theo-
rem 3.10 that for n ≥ 3 and k ≥ 1, if we take G = Cn and H = C3k+1 where
n ≡ 0 (mod 4), then γS(G,H) = kn = n(G)γ(H) −m(G). The upper bound of Theo-
rem 2.1 is obtained, for example, for the Sierpiński product of two complete graphs. More
generally, to achieve equality in the upper bound of Theorem 2.1 we require the graph H
to have the following property.
Theorem 2.2. The equality in ΓS(G,H) ≤ n(G)γ(H) is achieved if and only if there
exists a vertex x ∈ V (H) such that γ(H|x) = γ(H).
Proof. Suppose that H has a vertex x that satisfies γ(H|x) = γ(H). In this case, we
consider the Sierpiński product G ⊗f H with the function f : V (G) → V (H) defined by
f(v) = x for every vertex v ∈ V (G). Consequently each connecting edge in the product
is of the form (g, x)(g′, x). Thus, if X is a γ-set of G ⊗f H , then |X ∩ V (gH)| = γ(H)
because the only vertex of gH that can be dominated from outside gH is (g, x), but we
have assume that γ(H|x) = γ(H). Therefore, ΓS(G,H) = n(G)γ(H).
For the other implication suppose that γ(H|x) < γ(H) for every vertex x ∈ V (H).
For an arbitrary edge g1g2 ∈ E(G), if D corresponds to a γ-set of the product G ⊗f H ,
then |D ∩ V ((G⊗f H)[V (g1H) ∪ V (g2H)])| ≤ 2γ(H)− 1. Consequently, ΓS(G,H) <
n(G)γ(H).
To conclude this section we describe large classes of graphs for which the second and
the third inequality of Theorem 2.1 are both equality.
Proposition 2.3. If G and H are graphs such that ∆(G) < n(H) and γ(H) = 1, then
ΓS(G,H) = γS(G,H) = n(G).
Proof. Let G and H be graphs such that ∆(G) < n(H) and γ(H) = 1. Thus, by The-
orem 2.1, γS(G,H) ≤ ΓS(G,H) ≤ n(G) and it is straightforward that the Sierpiński
product of graphs G and H can be dominated by taking one dominating vertex from each
H-layer to the dominating set. It remains to show that the inequality γS(G,H) ≥ n(G)
also holds. Suppose that γS(G,H) ≤ n(G) − 1. Let D be a dominating set of G ⊗f H ,
where f is such that it minimizes the domination number. Therefore there is an H-layer,
denote it by H ′, of G ⊗f H such that D ∩ V (H ′) = ∅. Since there are at most ∆(G)
connecting edges incident with vertices from each H-layer and ∆(G) < n(H),then all the
vertices from an H-layer cannot be dominated by the vertices from the neighboring layers.
Therefore we have D∩V (H ′′) ̸= ∅ for each H ′′-layer of G⊗f H . Thus γS(G,H) ≥ n(G)
and the result follows.
M. A. Henning et al.: The Sierpiński domination number 477
3 The Sierpiński domination number of cycles
Let us recall firstly the domination number of a path and a cycle.
Fact 3.1. For n ≥ 3, γ(Pn) = γ(Cn) =
⌈
n
3
⌉
.
In this section, we shall prove the following results.
Theorem 3.2. For n ≥ 3, k ≥ 1, and p ∈ [2]0,
γS(Cn, C3k+p) ∈

{kn}; p = 0,
{kn, kn+ 1}; p = 1,
{kn+
⌊n
2
⌋
, kn+
⌊n
2
⌋
+ 1}; p = 2.
Moreover, if n ≡ 0 mod 4, then γS(Cn, C3k+1) = kn and γS(Cn, C3k+2) = kn+
⌊n
2
⌋
.
Theorem 3.3. For n ≥ 3, k ≥ 1, and p ∈ [2]0,
ΓS(Cn, C3k+p) =

kn; p = 0,
kn+
⌈n
3
⌉
; p = 1,
(k + 1)n; p = 2.
In order to prove Theorems 3.2 and 3.3, we consider three cases, depending on the
value of p.
3.1 The cycle Cn and cycles C3k+1
To determine ΓS(Cn, C3k+1), we prove a slightly more general result. For this purpose,
we define a class of graphs Hk as follows.
Definition 3.4. For k ≥ 1, let Hk be the class of all graphs H that have the following
properties.
(a) γ(H) = k + 1 and γ(H − v) = k for every vertex v ∈ V (H).
(b) If x, y ∈ V (H), then there exists a γ-set of H that contains x and y, where x = y is
allowed.
We show, for example, that for every k ≥ 1, the cycle C3k+1 belongs to the class Hk.
Proposition 3.5. For k ≥ 1, the class Hk of graphs contains the cycle C3k+1.
Proof. For k ≥ 1, let H ∼= C3k+1. Since γ(Cn) = γ(Pn) = ⌈n/3⌉, property (a) in Defi-
nition 3.4 holds. To prove that property (b) in Definition 3.4 holds, let x, y ∈ V (H). Since
H is vertex-transitive, every specified vertex belongs to some γ-set of H . In particular, if
x = y, then property (b) is immediate. Hence, we may assume that x ̸= y. Let H be the
cycle v1v2 . . . v3k+1v1, where renaming vertices if necessary, we may assume that x = v1.
Let y = vi, and so i ∈ [3k + 1] \ {1}.
Let H ′ = H − N [{x, y}], that is, H ′ is obtained from H by removing x and y, and
removing all neighbors of x and y. If H ′ is connected, then H ′ is a path P3(k−2)+j for
478 Ars Math. Contemp. 24 (2024) #P3.06 / 473–488
some j where j ∈ [3]. In this case, γ(H ′) = k − 1. If H ′ is disconnected, then H ′ is the
disjoint union of two paths Pk1 and Pk2 , where k1+ k2 = 3(k− 2)+1. Thus renaming k1
and k2 if necessary, we may assume that either k1 = 3j1 and k2 = 3j2 + 1 where j1 ≥ 1,
j2 ≥ 0, and j1 + j2 = k − 2 or k1 = 3j1 + 2 and k2 = 3j2 + 2 where j1, j2 ≥ 0 and
j1 + j2 = k − 3. In both cases, γ(H ′) = ⌈k1/3⌉+ ⌈k2/3⌉ = k − 1. Letting D′ be a γ-set
of H ′, the set D = D′ ∪ {x, y} is a dominating set of H of cardinality k + 1 = γ(H),
implying that D is a γ-set of H that contains both x and y. Hence, property (b) holds.
For n ≥ 3 an integer, a circulant graph Cn⟨L⟩ with a given list L ⊆ {1, . . . , ⌊ 12n⌋}
is a graph on n vertices in which the ith vertex is adjacent to the (i + j)th and (i − j)th
vertices for each j in the list L and where addition is taken modulo n. For example, for
n = 3k + 1 where k ≥ 1 and L = {1}, the circulant graph Cn⟨L⟩ is the cycle C3k+1,
which, by Proposition 3.5, belongs to the class Hk. More generally, for n = k(2p+1)+ 1
where k ≥ 1, p ≥ 1, and L = [p], the circulant graph Cn⟨L⟩ belongs to the class Hk.
We omit the relatively straightforward proof. These examples of circulant graphs serve to
illustrate that for each k ≥ 1, one can construct infinitely many graphs in the class Hk. We
determine next the upper Sierpiński domination number ΓS(Cn, H) of a cycle Cn and a
graph H in the family Hk.
Theorem 3.6. For n ≥ 3 and k ≥ 1, if H ∈ Hk, then
ΓS(Cn, H) = kn+
⌈n
3
⌉
.
Proof. For n ≥ 3 and k ≥ 1, let G ∼= Cn and let H ∈ Hk. Let G be the cycle given
by g1g2 . . . gng1. In what follows, we adopt the following notation. For each i ∈ [n], we
denote the copy giH of H corresponding to the vertex gi simply by Hi. We proceed further
with two claims. The first claim establishes a lower bound on ΓS(Cn, H), and the second
claim establishes an upper bound on ΓS(Cn, H).
Claim 3.7. ΓS(Cn, H) ≥ kn+
⌈n
3
⌉
.
Proof. Let f : V (G) → V (H) be a constant function, that is, we select h ∈ V (H) and
for every vertex g ∈ V (G), we set f(g) = h. Let DG be a γ-set of G. Thus, |DG| =
γ(Cn) = ⌈n/3⌉. By property (a) in Definition 3.4, for every vertex g ∈ V (G), there
exists a γ-set of gH that contains the vertex (g, f(g)) = (g, h). If g ∈ DG, let Dg be a
γ-set of gH that contains the vertex (g, f(g)) = (g, h), and so |Dg| = γ(H) = k + 1. If
g ∈ V (G) \DG, let Dg be a γ-set of gH − (g, f(g)) = gH − (g, h), and so in this case
|Dg| = γ(H − h) = γ(H)− 1 = k. Let
D =
⋃
g∈V (G)
Dg.
The set D is a dominating set of G⊗f H , and so
γ(G⊗f H) ≤ |D| = γ(G)(k + 1) + (n− γ(G))k = kn+ γ(G) = kn+
⌈n
3
⌉
. (3.1)
For the fixed vertex h chosen earlier, we note that the set of vertices (g, h) for all
g ∈ V (G) induces a subgraph of G ⊗f H that is isomorphic to G ∼= Cn. We denote this
M. A. Henning et al.: The Sierpiński domination number 479
copy of G by Gh. Among all γ-sets of G ⊗f H , let D∗ be chosen to contain as many
vertices of Gh as possible. Let D∗g = D
∗ ∩ V (gH) for every g ∈ V (G). Further let
D∗G = {(g, h) ∈ D∗ : g ∈ V (G)}, that is, D∗G is the restriction of D∗ to the copy of
G. If a vertex (g, h) /∈ D∗G and (g, h) is not dominated by D∗G, then D∗g is a γ-set of
gH by the minimality of the set D∗. However in this case, we could replace the set D∗g
be a γ-set of gH that contains the vertex (g, h) to produce a new γ-set of G ⊗f H that
contains more vertices from the copy of G than does D∗, a contradiction. Hence, the set
D∗G is a dominating set in the copy of G, and so |D∗g(G)| ≥ γ(G). By the minimality
of the set D∗ and by property (a) in Definition 3.4, for each vertex g ∈ V (G), we have
|D∗g | = γ(H) = k + 1 if the vertex (g, h) ∈ D∗G and |D∗g | = γ(H − h) = k if the vertex
(g, h) /∈ D∗G. Therefore,
γ(G⊗f H) = |D∗| = |D∗G|(k + 1) + (n− |D∗G|)k = kn+ |D∗G|
≥ kn+ γ(G) = kn+
⌈n
3
⌉
. (3.2)
By inequalities (3.1) and (3.2), we have
γ(G⊗f H) = kn+
⌈n
3
⌉
. (3.3)
By equation (3.3), we have ΓS(Cn, H) ≥ γ(G⊗f H) = kn+ ⌈n/3⌉. This completes
the proof of Claim 3.7. (2)
Claim 3.8. ΓS(Cn, H) ≤ kn+
⌈n
3
⌉
.
Proof. Let f : V (G) → V (H) be an arbitrary function. Let Hi be the ith copy of H
corresponding to the vertex gi of G for all i ∈ [n]. Let D be the dominating set of G⊗f H
constructed as follows. Let xiyi+1 be the connecting edge from Hi to Hi+1 for all i ∈ [n],
where addition is taken modulo n. Thus, the vertex xi ∈ V (Hi) is adjacent to the vertex
yi+1 ∈ V (Hi+1) in the graph G⊗fH , that is, xi = (gi, f(gi+1)) and yi+1 = (gi+1, f(gi)).
We note that possibly xi = yi. By property (b) in Definition 3.4, there exists a γ-set of Hi
that contains both xi and yi. For i ∈ [n], we define the sets Di,1, Di,2, and Di,3 as follows.
Let Di,1 be a γ-set of Hi − xi. Let Di,2 be a γ-set of Hi that contains both xi and yi. Let
Di,3 be a γ-set of Hi − yi. We note that |Di,1| = |Di,3| = k and |Di,2| = k + 1. For
i ∈ [n], we define the set Di as follows.
Di =

Di,1; i ≡ 1 (mod 3) and i ̸= n,
Di,2; i ≡ 2 (mod 3) or i ≡ 1 (mod 3) and i = n,
Di,3; i ≡ 0 (mod 3).
For example, the set D1 dominates all vertices of H1 − x1. The set D2 contains the
vertex y2, which is adjacent to the vertex x1 of H1, and contains the vertex x2, which
is adjacent to the vertex y3 of H3, implying that D2 dominates the vertex x1 of H1, all
vertices of H2, and the vertex y3 of H3. The set D3 dominates all vertices of H3 − y3.
Thus, D1 ∪D2 ∪D3 dominates all vertices in V (H1)∪ V (H2)∪ V (H3) in the Sierpiński
product G⊗fH . Moreover, |D1|+|D2|+|D3| = k+(k+1)+k = 3k+1. More generally,
the set D3j−2∪D3j−1∪D3j dominates all vertices in V (H3j−2)∪V (H3j−1)∪V (H3j) in
480 Ars Math. Contemp. 24 (2024) #P3.06 / 473–488
the Sierpiński product G⊗f H for all j ∈ {1, . . . , ⌊n/3⌋}. Moreover, |D3j−2|+ |D3j−1|+
|D3j | = k + (k + 1) + k = 3k + 1. If n ≡ 1 (mod 3), then the set Dn is a γ-set of Hn,
and in this case |Dn| = k + 1. If n ≡ 2 (mod 3), then the set Dn−1 ∪Dn dominates all
vertices in V (Hn−1) ∪ V (Hn), and in this case |Dn−1|+ |Dn| = k + (k + 1) = 2k + 1.
The set
D =
n⋃
i=1
Di
is therefore a dominating set of G⊗f H , implying that
γ(G⊗f H) ≤ |D| =
n∑
i=1
|Di| = kn+
⌈n
3
⌉
.
This completes the proof of Claim 3.8. (2)
The proof of Theorem 3.6 follows as an immediate consequence of Claims 3.7 and 3.8.
As a consequence of Proposition 3.5, we have the following special case of Theo-
rem 3.6.
Corollary 3.9. For n ≥ 3 and k ≥ 1,
ΓS(Cn, C3k+1) = kn+
⌈n
3
⌉
.
We consider next the Sierpiński domination number of Cn and C3k+1, and show that
γS(Cn, C3k+1) = kn if n ≡ 0 (mod 4) and γS(Cn, C3k+1) ∈ {kn, kn+ 1}, otherwise.
Theorem 3.10. For n ≥ 3 and k ≥ 1,
γS(Cn, C3k+1) ∈ {kn, kn+ 1}.
Moreover, if n ≡ 0 mod 4, then γS(Cn, C3k+1) = kn.
Proof. For n ≥ 3 and k ≥ 1, let G = Cn and let H = C3k+1. Let G be the cycle given
by g1g2 . . . gng1. We adopt our notation employed in our earlier proofs. For notational
convenience, we let V (H) = {1, 2, . . . , 3k+1} where vertices i and i+1 are consecutive
on the cycle H for all i ∈ [3k + 1] (and where addition is taken modulo 3k + 1, and so
vertex 1 and vertex 3k + 1 are adjacent).
As before, we denote the copy giH of H corresponding to the vertex gi simply by Hi
for each i ∈ [n]. Thus, Hi = C3k+1 is the cycle (gi, 1), (gi, 2), . . . , (gi, 3k + 1), (gi, 1)
for all i ∈ [n]. Recall that we denote the connecting edge from Hi to Hi+1 by xiyi+1 for
all i ∈ [n], where xi ∈ V (Hi), yi+1 ∈ V (Hi+1), and addition is taken modulo n. Thus,
yi = (gi, f(gi−1)) and xi = (gi, f(gi+1)) for all i ∈ [n].
By Proposition 3.5, the graph H belongs to the class Hk. Thus, γ(H) = k + 1 and
γ(H − v) = k for every vertex v ∈ V (H). Furthermore, if x, y ∈ V (H) where x = y is
allowed, then there exists a γ-set of H that contains x and y.
By the elementary lower bound on the Sierpiński domination number given in Theo-
rem 2.1, γS(G,H) ≥ n(G)γ(H)−m(G) = kn, noting that here n(G) = m(G) = n and
γ(H) = k + 1. It follows that γS(Cn, H) ≥ kn.
M. A. Henning et al.: The Sierpiński domination number 481
To complete the proof we are going to prove that
γS(Cn, H) ≤ kn+
⌈n
4
⌉
−
⌊n
4
⌋
.
Let f : V (G) → V (H) be the function defined by
f(gi) =
{
1; i mod 4 ∈ {1, 2},
3; otherwise.
for all i ∈ [n] where addition is taken modulo n. Adopting our earlier notation, recall
that yi = (gi, f(gi−1)) and xi = (gi, f(gi+1)) for all i ∈ [n]. Let n = 4ℓ + j where
j ∈ [3]0 = {0, 1, 2, 3}. We note that f(g4i−3) = f(g4i−2) = 1 and f(g4i−1) = f(g4i) = 3
for all i ∈ [ℓ]. Let Di be the unique γ-set of Hi − yi ∼= P3k which consists of all vertices
at distance 2 modulo 3 from yi in the graph Hi for all i ∈ [n], and let
D =
n⋃
i=1
Di.
We note that |Di| = k for all i ∈ [n], and so |D| = kn. For all i ∈ {2, 3, . . . , ℓ − 1}, the
following four properties hold.
(P1) y4i−3 = (g4i−3, 3) and x4i−3 = (g4i−3, 1).
(P2) y4i−2 = (g4i−2, 1) and x4i−2 = (g4i−2, 3).
(P3) y4i−1 = (g4i−1, 1) and x4i−1 = (g4i−1, 3).
(P4) y4i = (g4i, 3) and x4i = (g4i, 1).
Hence for all i ∈ {2, 3, . . . , ℓ − 1}, the vertices xi and yi are at distance 2 in Hi,
implying that xi ∈ Di. We consider four cases to determine which properties hold for
the boundary conditions (that is for i ∈ {1, ℓ}) and finally to set the upper bound on the
domination number in each case.
Case 1. n ≡ 0 (mod 4), that is n = 4ℓ.
In this case, properties (P1) and (P4) also hold for i = 1 and i = ℓ, respectively. Thus,
y1 = (g1, 3) and x1 = (g1, 1), and y4ℓ = (g4ℓ, 3) and x4ℓ = (g4ℓ, 1), implying that
x1, x4ℓ ∈ D. The set D is therefore a dominating set of G ⊗f H , and so γ(G ⊗f H) ≤
|D| = kn = kn+ ⌈n/4⌉ − ⌊n/4⌋.
Case 2. n ≡ 1 (mod 4), that is n = 4ℓ+ 1.
In this case, y1 = x1 = (g1, 1), and y4ℓ+1 = (g4ℓ+1, 3) and x4ℓ+1 = (g4ℓ+1, 1). In
particular, property (P4) also holds for i = ℓ, and so x4ℓ+1 ∈ D. The set D ∪ {x1}
is therefore a dominating set of G ⊗f H , and so γ(G ⊗f H) ≤ |D| + 1 = kn + 1 =
kn+ ⌈n/4⌉ − ⌊n/4⌋.
Case 3. n ≡ 2 (mod 4), that is n = 4ℓ+ 2.
In this case, y1 = x1 = (g1, 1), and y4ℓ+2 = x4ℓ+2 = (g4ℓ+2, 1). We note that neither x1
nor x4ℓ+2 belong to the set D. The set D ∪ {x1} is a dominating set of G ⊗f H , and so
γ(G⊗f H) ≤ |D|+ 1 = kn+ 1 = kn+ ⌈n/4⌉ − ⌊n/4⌋.
Case 4. n ≡ 3 (mod 4), that is n = 4ℓ+ 3.
482 Ars Math. Contemp. 24 (2024) #P3.06 / 473–488
In this case, y1 = (g1, 3) and x1 = (g1, 1), and y4ℓ+3 = x4ℓ+3 = (g4ℓ+3, 1). In particular,
property (P1) also holds for i = 1, and so x1 ∈ D. However, x4ℓ+3 /∈ D. The set
D ∪ {x4ℓ+3} is therefore a dominating set of G ⊗f H , and so γ(G ⊗f H) ≤ |D| + 1 =
kn+ 1 = kn+ ⌈n/4⌉ − ⌊n/4⌋.
In all four cases, γ(G⊗f H) ≤ kn+ ⌈n/4⌉ − ⌊n/4⌋.
3.2 The cycle Cn and cycles C3k+2
In this section, we determine the Sierpiński domination number γS(Cn, C3k+2) and the
upper Sierpiński domination number ΓS(Cn, C3k+2).
Theorem 3.11. For n ≥ 3 and k ≥ 1, we have ΓS(Cn, C3k+2) = (k + 1)n.
Proof. For n ≥ 3 and k ≥ 1, let G ∼= Cn and let H ∼= C3k+2. Let f : V (G) → V (H)
be a constant function, that is, we select h ∈ V (H) and for every vertex g ∈ V (G),
we set f(g) = h. For each vertex g ∈ V (G), let Hg denote the copy of H associated
with the vertex g. Let D be a dominating set of G ⊗f H , and let Dg = D ∩ V (Hg),
and so Dg is the restriction of D to the copy Hg of H . If the vertex (g, h) does not
belong to Dg , then Dg dominates all vertices on the path Hg − (g, h) ∼= P3k+1, and so
|Dg| ≥ γ(P3k+1) = k + 1. If the vertex (g, h) does belong to Dg , then Dg dominates
all vertices on the cycle Hg ∼= C3k+2, and so |Dg| ≥ γ(C3k+2) = k + 1. In both cases,
|Dg| ≥ k + 1. Therefore,
γ(G⊗f H) = |D| =
∑
g∈V (G)
|Dg| ≥ (k + 1)n,
implying that ΓS(Cn, C3k+2) ≥ (k + 1)n. By the upper bound in Theorem 2.1, we have
ΓS(G,H) ≤ n(G)γ(H) = (k + 1)n, noting that in this case γ(H) = γ(C3k+2) = k + 1.
Consequently, ΓS(Cn, C3k+2) = (k + 1)n.
Theorem 3.12. For n ≥ 3 and k ≥ 1,
γS(Cn, C3k+2) ∈ {kn+
⌊n
2
⌋
, kn+
⌊n
2
⌋
+ 1}.
Moreover, if n ≡ 0 mod 4, then γS(Cn, C3k+2) = kn+
⌊n
2
⌋
.
Proof. For n ≥ 3 and k ≥ 1, let G ∼= Cn and let H ∼= C3k+2. We adopt our notation
employed in our earlier proofs. Thus, the cycle G is given by g1g2 . . . gng1, and V (H) =
{1, 2, . . . , 3k + 2} where vertices i and i + 1 are consecutive on the cycle H for all i ∈
[3k + 2] (and where addition is taken modulo 3k + 2, and so vertex 1 and vertex 3k + 2
are adjacent). As before, we denote the copy giH of H corresponding to the vertex gi
simply by Hi for each i ∈ [n]. Thus, Hi = C3k+2 is the cycle (gi, 1), (gi, 2), . . . , (gi, 3k+
2), (gi, 1) for all i ∈ [n].
We adopt our notation from the proof of Theorem 3.6. Thus, we denote the connecting
edge from Hi to Hi+1 by xiyi+1 for all i ∈ [n], where xi ∈ V (Hi), yi+1 ∈ V (Hi+1),
and addition is taken modulo n. Thus, yi = (gi, f(gi−1)) and xi = (gi, f(gi+1)) for all
i ∈ [n].
We proceed further with two claims. The first claim establishes a lower bound on
γS(G,H), and the second claim an upper bound on γS(G,H). Combining these two
bounds yields the desired result in the statement of the theorem.
M. A. Henning et al.: The Sierpiński domination number 483
Claim 3.13. γS(Cn, H) ≥ kn+
⌊n
2
⌋
.
Proof. Let f : V (G) → V (H) be an arbitrary function. We show that
γ(G⊗f H) ≥ kn+
⌊n
2
⌋
. (3.4)
Let D be a γ-set of G ⊗f H constructed, and let Di = D ∩ V (Hi) for i ∈ [n]. If the
vertex xi is not dominated by Di, then either xi ̸= yi, in which case xi is dominated by
the vertex yi+1 ∈ D, or xi = yi, in which case xi is dominated by the vertex xi−1 ∈ D
or the vertex yi+1 ∈ D. Analogously, if the vertex yi is not dominated by Di, then either
xi ̸= yi, in which case yi is dominated by the vertex xi−1 ∈ D, or xi = yi, in which
case yi is dominated by the vertex xi−1 ∈ D or the vertex yi+1 ∈ D. If a vertex is not
dominated by Di, then such a vertex is xi or yi, and we say that such a vertex is dominated
from outside Hi.
Similarly as before, we proceed with a claim that delivers properties of sets Di leading
to the desired lower bound on the Sierpiński domination number.
Claim 3.14. The following properties hold in the graph Hi.
(a) If d(xi, yi) ≡ 1 (mod 3), then |Di| = k. Further, both xi and yi are dominated from
outside Hi.
(b) If d(xi, yi) ̸≡ 1 (mod 3), then |Di| = k + 1.
Proof. Suppose that Di contains a vertex wi that dominates xi. Possibly, wi = xi. In
order to dominate the 3(k−1)+2 vertices in Hi not dominated by wi, at least k additional
vertices are needed even if the vertex yi is dominated outside the cycle Hi. Thus in this
case, |Di| ≥ k+1, implying by the minimality of the set D that |Di| = k+1. Analogously,
if Di contains a vertex that dominates yi, then |Di| = k + 1. Hence, if xi or yi (or both xi
and yi) are dominated by Di, then |Di| = k + 1.
Suppose that neither xi nor yi is dominated by Di, implying that both xi and yi are
dominated from outside the cycle Hi. Thus, Di is a dominating set of H ′i = Hi−xi−yi. If
xi = yi, then H ′i = P3k+1, and by the minimality of D we have |Di| = γ(P3k+1) = k+1.
Hence, we may assume that xi ̸= yi. If xi and yi are adjacent, then H ′i = P3k, and by the
minimality of D we have |Di| = γ(P3k) = k. Suppose that xi and yi are not adjacent, and
so H ′ is the disjoint union of two paths Pk1 and Pk2 , where k1 + k2 = 3k. If k1 = 3j1 +1
and k2 = 3j2 + 2 (or if k1 = 3j1 + 2 and k2 = 3j2 + 1) for some integers ji and j2 where
j1+ j2 = k− 1, then |Di| = ⌈k1/3⌉+ ⌈k2/3⌉ = (j1+1)+ (j2+1) = k+1. If k1 = 3j1
and k2 = 3j2 where j1 + j2 = k, then |Di| = ⌈k1/3⌉ + ⌈k2/3⌉ = j1 + j2 = k. Hence
if neither xi nor yi is dominated by Di, then either d(xi, yi) ≡ 1 (mod 3), in which case
|Di| = k, or d(xi, yi) ̸≡ 1 (mod 3), in which case |Di| = k+1. This proves properties (a)
and (b) of the claim. (2)
By Claim 3.14, if |Di| = k for some i ∈ [n], then |Di−1| = |Di+1| = k + 1 where
addition is taken modulo n. Furthermore in this case when |Di| = k, the vertices xi and
yi are distinct and are both dominated from outside Hi, implying that yi+1 ∈ Di+1 and
xi−1 ∈ Di−1. This implies that if n is even, then |D| ≥ kn + n/2, and if n is odd, then
|D| ≥ kn+ (n+ 1)/2. This proves inequality (3.4).
Claim 3.15. γS(Cn, H) ≤ kn+
⌊n
2
⌋
+
⌈n
4
⌉
−
⌊n
4
⌋
.
484 Ars Math. Contemp. 24 (2024) #P3.06 / 473–488
Proof. Let f : V (G) → V (H) be the function defined by
f(gi) =

1; i ≡ 1 (mod 4),
2; i ≡ 2 (mod 4),
3; otherwise.
for all i ∈ [n] where addition is taken modulo n. Adopting our earlier notation, recall that
yi = (gi, f(gi−1)) and xi = (gi, f(gi+1)) for all i ∈ [n]. Let n = 4ℓ+ j where j ∈ [3]0 =
{0, 1, 2, 3}. We note that f(g4i−3) = 1, f(g4i−2) = 2, and f(g4i−1) = f(g4i) = 3 for all
i ∈ [ℓ].
Case 1. n ≡ 0 (mod 4).
Thus, n = 4ℓ. We note that y4i−3 = (g4i−3, 3) and x4i−3 = (g4i−3, 2) for all i ∈ [n], and
so in the graph H4i−3 the vertices x4i−3 and y4i−3 are at distance 1. Moreover, y4i−1 =
(g4i−1, 2) and x4i−1 = (g4i−1, 3) for all i ∈ [n], and so in the graph H4i−1 the vertices
x4i−1 and y4i−1 are at distance 1. This implies that H4i−j − {x4i−j , y4i−j} ∼= C3k for
j ∈ {1, 3}. Let D4i−j be a γ-set of H4i−j − {x4i−j , y4i−j} for j ∈ {1, 3}, and so
|D4i−j | = k.
We also note that y4i−2 = (g4i−2, 1) and x4i−2 = (g4i−2, 3) for all i ∈ [n], and so in
the graph H4i−2 the vertices x4i−2 and y4i−2 are at distance 2. Moreover, y4i = (g4i, 3)
and x4i = (g4i, 1) for all i ∈ [n], and so in the graph H4i the vertices x4i and y4i are at
distance 2. This implies that H4i−j − N [{x4i−j , y4i−j}] ∼= C3(k−1) for j ∈ {0, 2}. Let
D4i−j be a γ-set of H4i−j that contains both vertices x4i−j and y4i−j for j ∈ {0, 2}, and
so |D4i−j | = k + 1. The set
D =
4ℓ⋃
i=1
Di
is a dominating set of G⊗f H , and so γ(G⊗f H) ≤ |D| = 4kℓ+ 2ℓ = kn+ n/2.
Case 2. n ≡ 2 (mod 4).
Thus, n = 4ℓ + 2 and in this case, f(g4ℓ+1) = 1 and f(g4ℓ+2) = 2. We note that in the
graph H4ℓ+1, the vertices x4ℓ+1 and y4ℓ+1 are at distance 1 and in the graph H4ℓ+2 we
have x4ℓ+2 = y4ℓ+2. For i ∈ [4ℓ], we define the set Di exactly as in the previous case.
Further, let D4ℓ+1 be a γ-set of H4ℓ+1 − {x4ℓ+1, y4ℓ+1} ∼= C3k, and let D4ℓ+2 be a γ-set
of H4ℓ+2 containing x4ℓ+2. We note that |D4ℓ+1| = k and |D4ℓ+2| = k + 1. The set
D =
4ℓ+2⋃
i=1
Di
is a dominating set of G⊗f H , and so γ(G⊗f H) ≤ |D| = 4kℓ+2k+2ℓ+1 = kn+n/2.
Case 3. n ≡ 1 (mod 4).
Thus, n = 4ℓ+ 1, and in this case, f(g4ℓ+1) = 1. Thus, y4ℓ+1 = (g4ℓ+1, 3) and x4ℓ+1 =
(g4ℓ+1, 1), and so in the graph H4ℓ+1, the vertices x4ℓ+1 and y4ℓ+1 are at distance 2. For
i ∈ [4ℓ], we define the set Di exactly as in the previous cases. Further, let D4ℓ+1 be a γ-set
of H4ℓ+1 that contains the vertex x4ℓ+1. We note that |D4ℓ+1| = k + 1. The set
D =
4ℓ+1⋃
i=1
Di
M. A. Henning et al.: The Sierpiński domination number 485
is a dominating set of G⊗fH , and so γ(G⊗fH) ≤ |D| = 4kℓ+k+2ℓ+1 = kn+(n+1)/2.
Case 4. n ≡ 3 (mod 4).
Thus, n = 4ℓ + 3, and in this case, f(g4ℓ+1) = 1, f(g4ℓ+2) = 2, and f(g4ℓ+3) = 3.
In particular, y4ℓ+3 = (g4ℓ+3, 2) and x4ℓ+3 = (g4ℓ+3, 1), and so in the graph H4ℓ+3, the
vertices x4ℓ+3 and y4ℓ+3 are at distance 1. For i ∈ [4ℓ + 2], we define the set Di exactly
as in Case 2. Further, let D4ℓ+3 be a γ-set of H4ℓ+3 containing the vertex x4ℓ+3. We note
that |D4ℓ+3| = k + 1. The set
D =
4ℓ+3⋃
i=1
Di
is a dominating set of G ⊗f H , and so γ(G ⊗f H) ≤ |D| = 4kℓ + 3k + 2ℓ + 2 =
kn+ (n+ 1)/2. The desired result of the claim now follows from the four cases above. (2)
The proof of Theorem 3.12 follows as an immediate consequence of Claim 3.13 and
Claim 3.15.
3.3 The cycle Cn and cycles C3k
In this section, we determine the Sierpiński domination number γS(Cn, C3k) and the upper
Sierpiński domination number ΓS(Cn, C3k).
Theorem 3.16. For n ≥ 3 and k ≥ 1,
γS(Cn, C3k) = ΓS(Cn, C3k) = kn.
Proof. We adopt our notation from the earlier sections. Let G ∼= Cn be the cycle
g1g2 . . . gng1, and let Hi be the ith copy of C3k corresponding to the vertices gi of G
for i ∈ [n]. As before, we denote the connecting edge from Hi to Hi+1 by xiyi+1 for all
i ∈ [n].
Let f : V (G) → V (H) be an arbitrary function. Let D be a γ-set of G ⊗f H , and let
Di = D∩V (Hi) for i ∈ [n]. We show that |Di| = k for all i ∈ [n]. If both vertices xi and
yi are dominated by Di, then Di is a γ-set of Hi ∼= C3k, and so |Di| = k. If exactly one
of xi and yi is dominated by Di, say xi, then by the minimality of the set D, the set Di is
a γ-set of Hi − yi ∼= P3k−1, and so |Di| = k. Hence, we may assume that neither xi nor
yi is dominated by Di, for otherwise, |Di| = k and the desired bound follows.
With our assumption that neither xi nor yi is dominated by Di, the set Di is a γ-set of
H ′i = Hi − {xi, yi}. If xi = yi, then H ′i = P3k−1, and by the minimality of D we have
|Di| = γ(P3k−1) = k. Hence, we may assume that xi ̸= yi. If xi and yi are adjacent,
then H ′i = P3k−2, and by the minimality of D we have |Di| = γ(P3k−2) = k. Suppose
that xi and yi are not adjacent, and so H ′ is the disjoint union of two paths Pk1 and Pk2 ,
where k1 + k2 = 3k − 2. If k1 = 3j1 + 1 and k2 = 3j2 for some integers ji and j2 where
j1+j2 = k−1, then |Di| = ⌈k1/3⌉+⌈k2/3⌉ = (j1+1)+j2 = k. Analogously, if k1 = 3j1
and k2 = 3j2 + 1, then |Di| = k. If k1 = 3j1 + 2 and k2 = 3j2 + 2 for some integers ji
and j2 where j1 + j2 = k − 2, then |Di| = ⌈k1/3⌉+ ⌈k2/3⌉ = (j1 + 1) + (j2 + 1) = k.
In all cases, |Di| = k, implying that
γ(G⊗f H) = |D| =
n∑
i=1
|Di| = kn.
Since f : V (G) → V (H) was chosen as an arbitrary function, and D as an arbitrary
γ-set of G⊗f H , we deduce that γS(Cn, C3k) = ΓS(Cn, C3k) = γ(G⊗f H) = kn.
486 Ars Math. Contemp. 24 (2024) #P3.06 / 473–488
4 Concluding remarks
It seems to us that in the vast majority of cases where the lower Sierpiński domination num-
ber of the Sierpiński product of two cycles is specified to two values exactly, the larger of
the two is the correct value. However, the following example, which surprised us, demon-
strates that there are also cases where the exact value is the smaller of the two possible
values.
Let G ∼= C18 with V (G) = [18] and let H ∼= C7 with V (H) = [7] and let the function
f : V (G) → V (H) be defined as follows:
f(1) = f(4) = f(5) = f(18) = 4,
f(2) = f(3) = f(6) = f(7) = 2,
f(8) = f(9) = 7,
f(10) = f(11) = 5,
f(12) = f(13) = 3,
f(14) = f(15) = 1,
f(16) = f(17) = 6.
Then Theorem 3.2 asserts that γ(G⊗f H) ∈ {36, 37} and it is straightforward to check
that the exact value is γ(G⊗f H) = 36.
ORCID iDs
Michael A. Henning https://orcid.org/0000-0001-8185-067X
Sandi Klavžar https://orcid.org/0000-0002-1556-4744
Elżbieta Kleszcz https://orcid.org/0000-0002-1413-2413
Monika Pilśniak https://orcid.org/0000-0002-3734-7230
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ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 24 (2024) #P3.07 / 489–507
https://doi.org/10.26493/1855-3974.2835.8f0
(Also available at http://amc-journal.eu)
On the minisymposium problem*
Peter Danziger †
Department of Mathematics, Toronto Metropolitan University, Toronto,
ON M5B 2K3, Canada
Eric Mendelsohn
Department of Mathematics, University of Toronto, Toronto, ON M5S 2E4, Canada
Brett Stevens ‡
School of Mathematics and Statistics - Carleton University, Ottawa, ON K1S 5B6, Canada
Tommaso Traetta §
DICATAM, Università degli Studi di Brescia, Via Branze 43, 25123 Brescia, Italy
Received 26 February 2022, accepted 25 July 2023, published online 29 July 2024
Abstract
The generalized Oberwolfach problem asks for a factorization of the complete graph
Kv into prescribed 2-factors and at most a 1-factor. When all 2-factors are pairwise isomor-
phic and v is odd, we have the classic Oberwolfach problem, which was originally stated
as a seating problem: given v attendees at a conference with t circular tables such that the
ith table seats ai people and
∑t
i=1 ai = v, find a seating arrangement over the
v−1
2 days of
the conference, so that every person sits next to each other person exactly once.
In this paper we introduce the related minisymposium problem, which requires a solu-
tion to the generalized Oberwolfach problem on v vertices that contains a subsystem on m
vertices. That is, the decomposition restricted to the required m vertices is a solution to the
generalized Oberwolfach problem on m vertices. In the seating context above, the larger
conference contains a minisymposium of m participants, and we also require that pairs of
these m participants be seated next to each other for
⌊
m−1
2
⌋
of the days.
When the cycles are as long as possible, i.e. v, m and v−m, a flexible method of Hilton
and Johnson provides a solution. We use this result to provide further solutions when
v ≡ m ≡ 2 (mod 4) and all cycle lengths are even. In addition, we provide extensive
*We thank the anonymous referees for their many useful suggestions that helped strongly improve this paper.
†P. Danziger has received support from NSERC Discovery Grants RGPIN-2022-03816.
‡B. Stevens recieved support from NSERC Discovery Grant RGPIN-2017-06392.
§Corresponding author. T. Traetta has received support from GNSAGA of Istituto Nazionale di Alta Matem-
atica.
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
490 Ars Math. Contemp. 24 (2024) #P3.07 / 489–507
results in the case where all cycle lengths are equal to k, solving all cases when m | v,
except possibly when k is odd and v is even.
Keywords: Minisymposium problem, (generalized) Oberwolfach problem, 2-factorizations, subsys-
tems.
Math. Subj. Class. (2020): 05C51, 05B30
1 Introduction
We assume that the reader is familiar with the fundamentals of graph theory and of design
theory and refer them to [41] and [14], respectively. In particular, a factor is a spanning
subgraph and an r-factor is a factor which is r-regular, so in a 1-factor every vertex has
degree one and a 2-factor is a disjoint union of cycles. Given a collection of factors, F , an
F-factorization of a graph G is a decomposition of the edges of G into subgraphs, each of
which is isomorphic to some F ∈ F . If F = {F} we speak of an F -factorization.
We use Kn to denote the complete graph on n vertices and K∗n to denote the graph
Kn when n is odd and Kn − I , where I is a 1-factor, when n is even. Similarly, Km,n
denotes the complete bipartite graph with parts of sizes m and n. If the parts are X and
Y , respectively, we may also speak of KX,Y . A 2-factor is called uniform if all of its con-
stituent cycles are of the same length; it is called Hamiltonian* if its cycles have longest
possible lengths given the requirements of the factorization. If a 2-factor, F , consists en-
tirely of cycles of a particular length, k say, we refer to an F -factor and an F -factorization
as a Ck-factor and Ck-factorization, respectively. Given a graph G, we denote by G[n]
the lexicographic product of G with the empty graph on n vertices. Specifically, the
vertex set of G[n] is V (G) × Zn (where Zn denotes the cyclic group of order n) and
(x, i)(y, j) ∈ E(G[n]) if and only if xy ∈ E(G), i, j ∈ Zn.
The well known Oberwolfach problem asks for a 2-factorization of K∗n into 2-factors
all of which are isomorphic to a given 2-factor F . A summary of results up until 2006 can
be found in [14, Section VI.12], in particular the case of uniform factors has been solved
[1, 2, 26].
Theorem 1.1 ([1, 2, 26]). Given integers v, k ≥ 3, there is a Ck-factorization of K∗v if and
only if k | v, except that there is no C3-factorization of K∗6 or K∗12.
The case when all cycles in F have even length has been completely solved in [6].
The case with exactly two cycles is solved in [38]. The case of the complete graph Kℵ,
where ℵ is any infinite cardinal has been completely solved in [15]. In the related Hamilton-
Waterloo problem two 2-factors F1 and F2 are specified and we are asked for a factorization
of K∗n into a given number of each of the factors. There has been much recent progress
in this problem, see [3, 5, 7, 10, 11, 12, 17, 27, 28, 29, 32, 39, 40]. More generally,
in the generalized Oberwolfach problem we are given a set of 2-factors F1, F2, . . . , Ft
of Kn and positive integers α1, α2, . . . , αt, where
∑
αi = ⌊n−12 ⌋, and are asked for a
factorization of K∗n which contains αi copies of the 2-factor Fi, see [6, 13, 21]. A major
recent development gives a non-constructive asymptotic existence result for the generalized
Oberwolfach problem [23].
E-mail addresses: danziger@torontomu.ca (Peter Danziger), mendelso@math.utoronto.ca (Eric
Mendelsohn), brett@math.carleton.ca (Brett Stevens), tommaso.traetta@unibs.it (Tommaso Traetta)
P. Danziger et al.: On the minisymposium problem 491
Other graphs have also been considered. In particular, Liu has shown the following for
the complete multipartite graph.
Theorem 1.2 ([30, 31]). Let k, t and u be positive integers with k ≥ 3. There exists a
Ck-factorization of Kt[u] if and only if k | tu, (t − 1)u is even, further k is even if t = 2,
and (k, t, u) ̸∈ {(3, 3, 2), (3, 6, 2), (3, 3, 6), (6, 2, 6)}.
Originally the Oberwolfach problem was stated as a seating problem:
Given an odd number v of attendees at a conference with t circular tables such
that the ith table seats ai people and
∑t
i=1 ai = v, find a seating arrangement
over the v−12 days of the conference, so that every person sits next to each
other person exactly once.
In this paper we introduce the related minisymposium problem. In this case we require
a solution to the generalized Oberwolfach problem on v vertices such that its restriction to
a subset of m vertices constitutes a solution to the generalized Oberwolfach problem on
m vertices. Another way of considering the problem asks for a solution to the generalized
Oberwolfach problem on v vertices which contains a subsystem on m vertices. In the
seating context above, the larger conference contains a minisymposium of m participants,
and we also require that pairs of these m participants be seated next to each other for⌊
m−1
2
⌋
of the days. A similar problem has been considered, for example, in [9] for whist
tournaments.
Section 2 gives the formal definition of a minisymposium factorization and some nec-
essary conditions for its existence, as well as introduces some special cases. In Section 3
we show how to use a flexible theorem by Hilton and Johnson [25] to solve the case of
Hamiltonian* 2-factors (where the cycles are as long as possible). The same section con-
siders the case where all cycles are of even length and v ≡ m ≡ 2 (mod 4). Section 4
considers the uniform case, where all cycles have the same length. We completely solve
the case where all cycles are of length m when (v − 1)m is even. In Section 5 we discuss
and give some preliminary results on factorizations that contain more than one subsystem.
We provide some concluding remarks in the final section.
2 Preliminaries
We begin by giving a formal definition of a minisymposium factorization. The minisym-
posium problem is equivalent to the original Oberwolfach problem when v = m. Hence
we will generally assume that v > m.
Definition 2.1. Given positive integers v and m with v ≥ m, let
F =
{
Fi : i = 1, . . . ,
⌊
v − 1
2
⌋
−
⌊
m− 1
2
⌋}
,
be a collection of 2-factors on v vertices and let
G =
{
(Ti, Ui) : i = 1, . . . ,
⌊
m− 1
2
⌋}
,
where the Ti are 2-factors on m vertices and the Ui are 2-factors on v − m vertices. We
define a minisymposium factorization MSF(F ,G) as a factorization of K∗v into 2-factors
F ∈ F and Gi = Ti ∪Ui, where (Ti, Ui) ∈ G, such that T = {Ti : i = 1, . . . ,
⌊
m−1
2
⌋
} is
a factorization of a subgraph of K∗v isomorphic to K
∗
m.
492 Ars Math. Contemp. 24 (2024) #P3.07 / 489–507
Note that with the notation MSF(F ,G), we assume that the parameters v, m, T and
U = {Ui : i = 1, . . . ,
⌊
m−1
2
⌋
} are defined implicitly. We may also use the notation
MSF(F , (T ,U)), if we wish to explicitly refer to the factorizations T and U .
An MSF(F ,G) can be thought of as a 2-factorization of K∗v with a subsystem of size
m. When m = 1 or 2, this is just a factorization of K∗v into 2-factors in F , which is
equivalent to a solution of the generalized Oberwolfach problem, and so we will assume
m ≥ 3. Similarly, when m = v, this is a factorization into the Ti, so we assume that
m < v.
Removing the subsystem, we can talk about a 2-factorization of K∗v with a “hole” of
size m. However, care must be taken when either v or m is even as the placement of the
various 1-factors must be considered, as noted below.
We note that the size of F is⌊
v − 1
2
⌋
−
⌊
m− 1
2
⌋
=

v−m
2 v ≡ m (mod 2)
v−m+1
2 v + 1 ≡ m ≡ 0 (mod 2)
v−m−1
2 v ≡ m+ 1 ≡ 0 (mod 2).
• In the case when both v and m are even, we are considering a factorization of K∗v =
Kv − I , where I is a 1-factor, containing a factorization of a subgraph G− J of K∗v
where G ∼= Km and J is a 1-factor of G contained in I .
• When v is even and m is odd, we are considering a factorization of Kv − I , where
I is a 1-factor, containing a factorization of a subgraph G ∼= Km. Note that none of
the edges of I are contained in G.
• When v is odd and m is even, we are considering a factorization of K∗v = Kv
containing a factorization of G − J , where G ∼= Km and J is a 1-factor of G. We
note that the edges of J are not covered by factors in T , and hence must be covered
by factors in F .
• When both v and m are odd there is no 1-factor to consider.
Since in all cases G ∼= Km, we will henceforth refer to it as the Km. We note that none
of the edges of the Km are covered by F , except in the case of m even and v odd; in this
case, the edges of the 1-factor J of the Km are covered by F . We use this observation in
the proofs of the lemmas below, where for a given 2-factor F , we define ak(F ) to be the
number of cycles of length k in F .
Lemma 2.2. For a given v and m, if there is a minisymposium factorization, MSF(F ,G),
then for each factor F ∈ F using cF edges in the Km,
v −m ≥ −cF +
v∑
i=3
ai(F )
⌈
i
2
⌉
, (2.1)
m ≤ cF +
v∑
i=3
ai(F )
⌊
i
2
⌋
. (2.2)
In particular, if v is even, or m is odd, all of the cF = 0 and therefore,∑v
i=3 ai(F )
⌈
i
2
⌉∑v
i=3 ai(F )
⌊
i
2
⌋ ≤ v −m
m
. (2.3)
P. Danziger et al.: On the minisymposium problem 493
Proof. We first deal with the case when v is even, or m is odd. In this case none of the
edges of the Km appear in any F ∈ F . Therefore, for each F ∈ F , at most
⌊
i
2
⌋
vertices
of any cycle of length i in F are inside the Km, hence
m ≤
v∑
i=3
ai(F )
⌊
i
2
⌋
. (2.4)
Similarly, for any cycle of length i in F , at least ⌈ i2⌉ vertices of the cycle are not in the
Km, thus
v∑
i=3
ai(F )
⌈
i
2
⌉
≤ v −m. (2.5)
Thus Inequality (2.3) follows.
Now, if v is odd and m is even the m2 edges in the 1-factor J of the Km must be used
in factors from F . Suppose that F ∈ F uses cF of these edges. Each edge of the Km used
can increase the right hand side of Inequality (2.4) by no more than one and decrease the
left hand side of Inequality (2.5) by no more than one.
Theorem 2.3. For a given v and m, if there is a minisymposium factorization, MSF(F ,G),
then v ≥ 2m, unless v is odd and m is even, in which case v ≥ 2m− 1.
Proof. When v is even or m is odd, the left hand side of Inequality (2.3) is at least 1 and
therefore v ≥ 2m.
When v is odd and m is even,
∑
F∈F cF =
m
2 . Since v is odd, each F ∈ F must
contain at least one odd cycle, therefore
∑v
i=3 ai(F )⌈
i
2⌉ ≥ ⌈
v
2⌉ =
v+1
2 . Also note that
the number of factors F ∈ F is v−m+12 . Summing Inequality (2.1) over all of the F ∈ F
twice gives
(v −m+ 1)(v −m) ≥ 2
∑
F∈F
(
−cF +
v∑
i=3
ai(F )
⌈
i
2
⌉)
= −m+ 2
∑
F∈F
v∑
i=3
ai(F )
⌈
i
2
⌉
≥ −m+ 2
∑
F∈F
v + 1
2
= (v −m+ 1)v + 1
2
−m
Hence,
0 ≤ (v −m+ 1)(v −m)− (v −m+ 1)v + 1
2
+m
=
(v −m+ 1)(v − 2m− 1) + 2m
2
=
(v − (2m− 1))(v − (m+ 1))
2
.
494 Ars Math. Contemp. 24 (2024) #P3.07 / 489–507
When v = m + 1, the factors in U are required to be 2-regular graphs on a single vertex,
which is not possible, so v − (m + 1) > 0. Thus v − (2m − 1) ≥ 0 and the result
follows.
In the case where v is odd and m is even and v = 2m − 1, we have that U is a
factorization of Kv−m. So we can interchange both the roles of m and v −m, as well as
those of T and U . Thus, without loss of generality, we may assume that v ≥ 2m in all
cases.
There are two cases of initial special interest. Firstly, the case of uniform cycle lengths
(when all cycles in a factor are of the same length), which we consider in detail in Section 4.
Secondly, the case where the cycles are as long as possible, which in correspondence with
the definition of K∗n and the Hamiltonian-like nature of such factorizations we will call
Hamiltonian* factorizations. We formally define Hamiltonian* factorizations in Section 3.
There we show that a method of Hilton and Johnson completely settles their existence.
An MSF(F ,G) in which all of the factors in F , T and U are uniform with the same
cycle length k is called uniform and we refer to it as a UMSF(v,m, k). In this case we have
the following necessary conditions.
Theorem 2.4. If v > m and a UMSF(v,m, k) exists, then k ≥ 3, k | m and k | v.
Furthermore,
• if k is even, then v ≥ 2m;
• if k is odd, then v ≥ 2mkk−1 .
Proof. Since we are forming 2-factors with cycles of length k, we require k ≥ 3. The
divisibility conditions follow directly from the requirement for a factorization of K∗v and
K∗m into k-cycles. If k is even, then v is even, since it is a multiple of k, and Theorem 2.3
gives v ≥ 2m.
If k is odd, we note that for a Ck-factor F ∈ F , we have ai(F ) = vk when i = k and 0
otherwise, ⌊k2 ⌋ =
k−1
2 and ⌈
k
2 ⌉ =
k+1
2 . Thus, when v is even or m is odd, Inequality (2.3)
implies that v ≥ 2mkk−1 and the result follows.
This leaves the case when v and k are odd and m is even. We sum Inequality (2.1) over
all the F ∈ F to obtain∑
F∈F (v −m) ≥
∑
F∈F
(
−cF +
∑v
i=3 ai(F )
⌈
i
2
⌉)
1
2 (v −m+ 1)(v −m) ≥ −m/2 +
∑
F∈F
v
k
k+1
2
2k(v −m+ 1)(v −m) ≥ v(v −m+ 1)(k + 1)− 2mk.
Rearranging and expanding in v gives
(k − 1)v2 + (−3km+m+ k − 1)v + 2km2 ≥ 0. (2.6)
Let
f(v) = (k − 1)v2 + (−3km+m+ k − 1)v + 2km2.
By Theorem 2.3, v ≥ 2m− 1, but
f (2m− 1) = m(1 + k − 2m) < 0,
P. Danziger et al.: On the minisymposium problem 495
so v is at least as large as the larger of the two roots of f . Now
f
(
2mk
k − 1
− 2
)
= −2(m+ 1− k) < 0, and f
(
2mk
k − 1
− 1
)
= (k − 1)m > 0.
Thus f has its larger root between 2mkk−1 − 2 and
2mk
k−1 − 1.
It is left to check that v ̸= ⌊ 2mkk−1 −1⌋, ⌈
2mk
k−1 −1⌉. Recalling that k | v, if v = ⌊
2mk
k−1 −1⌋
or v = ⌈ 2mkk−1 − 1⌉, then there exist a rational number 0 ≤ ϵ < 1 and a positive integer a
such that
v =
2mk
k − 1
− 1± ϵ = ak.
Multiplying both sides by k−12k and rearranging we have that
m− ak − 1
2
= (1± ϵ)k − 1
2k
.
Since k is odd, the left side is an integer. However, 0 < (1 ± ϵ)k−12k < 1, a contradiction.
We conclude that v ≥ 2mkk−1 .
One interesting case in light of these necessary conditions is when m = k, i.e. a
UMSF(v, k, k). For these parameters, since k < v and k | v, we must have v ≥ 2k,
so the necessary conditions in Theorem 2.3 are satisfied. We consider these types of fac-
torizations in Section 4.
3 Hamiltonian* and bipartite factors
Considering non-uniform factors, an obvious case to consider is a Hamiltonian* minisym-
posium factorization, which is one in which the cycles have the longest possible lengths.
Specifically, the factors in F are all v-cycles, factors in T are all m-cycles and the factors in
U are all (v −m)-cycles. Such an MSF(F ,G) is denoted by HMSF(v,m). We sometimes
refer to the cycles in T and U as ‘short’ cycles. Because of the lengths of these cycles there
are no further necessary conditions beyond those of Theorem 2.3.
In a paper on the Oberwolfach problem, Hilton and Johnson prove the following theo-
rem on a flexible construction technique.
Theorem 3.1 ([25]). Let m and n be integers, 1 ≤ m < n. Let (s1, . . . , st), si ∈ 1, 2, 1 ≤
i ≤ t, be a composition of n−1. Let Km be edge coloured with t colours c1, . . . , ct. Let fi
be the number of edges coloured ci and Km(ci) be the ith colour class. This colouring can
be extended to an edge-colouring of Kn in which the colour class Kn(ci) is an si-factor,
1 ≤ i ≤ t, and when si = 2, Kn(ci) contains exactly one more cycle than Km(ci) if and
only if for all i = 1, 2, . . . , t:
fi ≥ si(m− n/2),
sin is even,
∆(Km(ci)) ≤ si.
This theorem is sufficient to provide a solution to the Hamiltonian* minisymposium
factorization.
496 Ars Math. Contemp. 24 (2024) #P3.07 / 489–507
Corollary 3.2. An HMSF(v,m) exists if and only if m ≥ 3, v ≥ 2m−1 in case m is even,
and v ≥ 2m in case m is odd.
Proof. The given conditions are necessary by Theorem 2.3. To prove sufficiency, we will
define an edge colouring of the Km from a decomposition of the Km into Hamiltonian
cycles and possibly a single 1-factor using Theorem 1.1. If m is odd, this defines edge
colours ci, 1 ≤ i ≤ (m− 1)/2. If v is also odd, extend this to a (v − 1)/2-edge colouring
of the Km by including (v −m)/2 empty colour classes ci, (m+ 1)/2 ≤ i ≤ (v − 1)/2.
Let si = 2 for all 1 ≤ i ≤ (v − 1)/2. If v is even, extend the colouring to a v/2-edge
colouring of the Km by adding empty colour classes. Let si = 2 for all 1 ≤ i < v/2 and
sv/2 = 1. In both cases, it can be verified that Theorem 3.1 now gives an HMSF(v,m) as
desired.
If m is even, define m/2 edge colour classes of the Km from a decomposition into
Hamiltonian cycles and one 1-factor. Let c1 be the colour class of the 1-factor. If v is also
even, extend this to a v/2-edge colouring by adding empty colour classes. Let s1 = 1 and
si = 2 for 2 ≤ i ≤ v/2. If v is odd, extend this to a (v − 1)/2-edge colouring by adding
empty colour classes. Let si = 2 for 1 ≤ i ≤ v/2. In both cases, it can be verified that
Theorem 3.1 now gives an HMSF(v,m) as desired.
Theorem 3.1 is more than just an existence result; a recursive procedure can be extracted
from the proof to algorithmically build the edge decompositions. We have a more direct
construction of all HMSF(v,m) which uses difference methods and decompositions of
Cayley graphs [16].
Theorem 3.1 can be used much more generally to build minisymposium factorizations.
Essentially it shows that it is possible to extend any 2-factorization of the Km to one of Kv ,
provided that the necessary conditions hold, where the additional 2-factors are Hamiltonian,
with an additional 1-factor when v is even.
When all of the cycles of the factors in F , U and T are bipartite (i.e. contain only even
cycles), we apply the Theorem of Häggkvist [24] (given below) to HMSF(v,m) to give us
a solution to the minisymposium problem when v ≡ m ≡ 2 (mod 4).
Theorem 3.3 ([24]). If F is a bipartite 2-regular graph of order 2n, then there is a factor-
ization of Cn[2] into 2 isomorphic copies of F .
We note that in the case where the factors are bipartite, so all cycle lengths are even, v
and m are both even and Theorem 2.3 gives v ≥ 2m.
Theorem 3.4. If v ≡ m ≡ 2 (mod 4), and F = {Fi : 1 ≤ i ≤ (v −m)/2}, U = {Uj :
1 ≤ j ≤ (m−2)/2} and T = {Tj : 1 ≤ j ≤ (m−2)/2} are sets of bipartite factors with
Fi = Fi+1, Uj = Uj+1 and Tj = Tj+1 for every odd i and j, then an MSF(F , (T ,U))
exists if and only v ≥ 2m.
Proof. We note that if v ≡ m ≡ 2 (mod 4), the number of the Fi is (v − m)/2 and the
number of the Uj and the Tj is (m− 2)/2, so the number of the Fi, Uj and Tj are all even.
We take an HMSF(v/2,m/2), which exists by Corollary 3.2, with factors F ′i of order v/2,
U ′i of order (v − m)/2 and T ′i of order m/2. We blow up each vertex by 2 and apply
Theorem 3.3 to factor each F ′i [2] into 2 copies of Fi, each U
′
j [2] into 2 copies of Uj and
each T ′j [2] into 2 copies of Tj .
P. Danziger et al.: On the minisymposium problem 497
If v ≡ 0 (mod 4) or m ≡ 0 (mod 4), then v/2 or m/2 would be even and the
HMSF(v/2,m/2) would contain 1-factors either in Kv/2 or the Km/2. When a 1-factor is
blown up as done in Theorem 3.3, it results in a C4-factor, which prevents constructing the
desired MSF unless F , T , and U already contain this kind of factor.
An immediate consequence of Theorem 3.4 is the following relating to uniform factors.
Corollary 3.5. If k > 3, k ≡ 2 (mod 4), v ≡ m ≡ k (mod 2k), then a UMSF(v,m, k)
exists if and only if v ≥ 2m.
4 Uniform factors
In this section we consider the case of uniform factors, i.e. when all cycles are of the same
length, k. We recall from Theorem 2.4 that in order for a UMSF(v,m, k) to exist, we
require that k ≥ 3, which we will assume throughout this section. We also require k | m
and k | v. Additionally, if k is even, then v ≥ 2m and if k is odd, then v ≥ 2mkk−1 .
Corollary 3.5 gives a powerful result in the case when k ≡ 2 (mod 4) and v ≡ m ≡ k
(mod 2k). The case where k = 3 has been considered in [34, 35, 36] when v and m are
both odd, and [18, 19, 20, 22, 37] when they are both even. However, the case when m and
v have opposite parities appears to be completely open. We summarize these results in the
following theorem.
Theorem 4.1 ([20, 35]). If v ≡ m (mod 2), there exists a UMSF(v,m, 3) if and only if
v ≥ 3m, v ≡ m ≡ 0 (mod 3), and if v,m are even, then v,m > 12.
We will find the following results useful. A corollary of a result in [4] yields the fol-
lowing.
Theorem 4.2 ([4]). If G is a Hamiltonian decomposable graph, then G[n] is also Hamil-
tonian decomposable. In particular, Cm[n] has a Cmn-factorization for every m ≥ 3.
Piotrowski [33] has shown the following result for m ≥ 4. The case m = 3 is covered
by Theorem 1.2.
Theorem 4.3 ([33]). There exists a Cm-factorization of Cm[n], except if n = 2 and m is
odd, or when (m,n) = (3, 6).
Piotrowski [33] has also shown the following result.
Theorem 4.4 ([33]). Let F be a bipartite 2-regular graph of order 2n. The complete
bipartite graph K2[n] has an F -factorization if and only if n is even, except when n = 6
and F consists of two 6-cycles.
We now give some recursive constructions for uniform minisymposium factorizations.
Theorem 4.5. Let m ≥ k ≥ 3 and t ≥ 2 be integers. If (t − 1)m is even and k | m,
then there is a UMSF(mt,m, k), except that there is no UMSF(6t, 6, 3) UMSF(12t, 12, 3),
UMSF(12, 6, 6), or UMSF(2m,m, k) when k is odd.
Proof. The non-existence of a UMSF(6t, 6, 3) and UMSF(12t, 12, 3) are covered by The-
orem 4.1. Since a UMSF(2m,m, k) is equivalent to a Ck-factorization of the complete
bipartite graph K2[m], it clearly does not exist when the cycle length k is odd, or when
k = m = 6 by Theorem 4.4. In all remaining cases, the following conditions simultane-
ously hold:
498 Ars Math. Contemp. 24 (2024) #P3.07 / 489–507
1. (m, k) ̸∈ {(6, 3), (12, 3)},
2. (t,m, k) ̸= (2, 6, 6),
3. if k is odd, then t > 2.
The assumptions of Theorems 1.1 and 1.2 are then satisfied. Hence there is a Ck-factorization
of Kt[m] and a Ck-factorization of K∗m, which we use to fill in the parts of size m in Kt[m].
This completes the proof.
Considering the necessary conditions in Theorem 2.4, we get the following corollaries.
Corollary 4.6. Suppose that either k is even or v is odd, and m | v. Then there exists a
UMSF(v,m, k) if and only if k | m, v ≥ 2m when k is even, and v ≥ 3m when k is odd,
except that UMSF(v, 6, 3), UMSF(v, 12, 3) and UMSF(12, 6, 6) do not exist.
Proof. Taking v = mt, Theorem 2.4 gives the necessary conditions k | m and v ≥ 2m
when k is even. When k is odd, the necessary condition from Theorem 2.4 is v ≥ 2mkk−1 , but
since m | v, this implies k ≥ 3m. Given the conditions of k and v, the sufficiency comes
from Theorem 4.5.
We note that if m | v this corollary completely solves all cases except when k is odd
and v is even. One case of particular interest is when k = m, in this case m | v is necessary.
Corollary 4.7. Let m(t− 1) be even. Then a UMSF(tm,m,m) exists if and only if t ≥ 2
when m is even, t ≥ 3 when m is odd, and (t,m) ̸= (2, 6).
The previous results all require m | v, however the next theorem allows us to recur-
sively construct solutions to cases where m does not divide v.
Theorem 4.8. Assume there is a UMSF(v,m, k) and let t ≥ 1. Then there exists a
UMSF(vtk,mtk, ℓ), with ℓ ∈ {k, kt}, in each of the following cases:
(1) v and m have the same parity;
(2) v and t are even, ℓ = tk, and m and k are both odd, except possibly when (k, t) =
(3, 2).
Proof. Letting w ∈ {m, v}, we factorize K∗wtk into Γw = K∗w[tk] and Γw = K∗wtk − Γw.
Note that Γw is the vertex disjoint union of
(1) w copies of K∗tk when w is odd, or
(2) w/2 copies of K∗2tk when w is even.
Without loss of generality, we can assume that Γm ⊆ Γv and Γm ⊆ Γv , except when v is
odd and m is even. In this case the components of Γm are copies of K∗2tk, while those of
Γv are isomorphic to K∗tk, therefore Γm ⊆ Γv cannot hold. We proceed by constructing
(a) a Cℓ-factorization of Γv containing a Cℓ-factorization of Γm, and
(b) a Cℓ-factorization of Γv containing a Cℓ-factorization of Γm,
P. Danziger et al.: On the minisymposium problem 499
which together will provide the desired UMSF (vtk,mtk, ℓ).
We blow up each vertex of the UMSF(v,m, k) by tk, to obtain a Ck[tk]-factorization
of Γv containing a Ck[tk]-factorization of Γm. To construct (a) it is therefore enough to
factorize Ck[tk] into Cℓ-factors, ℓ ∈ {k, tk}. By Theorem 4.3 there is a Ck-factorization
of Ck[tk], except when (t, k) = (2, 3). In this case, the desired UMSF(6v, 6m, 3) exists
by Theorem 4.1. Considering that Ck[tk] = Ck[t][k], by Theorem 4.2 there exists a Ckt-
factorization of Ck[t] which we blow up by k to obtain Ckt[k]-factorization of Ck[tk]. By
Theorem 4.3, each Ckt[k]-factor can be further decomposed into Ckt-factors yielding a
Ckt-factorization of Ck[tk].
It is left to construct (b). If m and v have the same parity, the components of Γm and
Γv are pairwise isomorphic: they are copies of K∗tk or K
∗
2tk. It is then enough to build a
Cℓ-factorization of K∗tk and K
∗
2tk for ℓ ∈ {k, kt}. They exist by Theorem 1.1 except when
ℓ = k = 3 and one of the following two conditions hold,
1. mv is odd and t ∈ {2, 4}, or
2. m and v are even, and t ∈ {1, 2}.
In each of these cases, the existence of the desired UMSF (3vt, 3mt, 3) is guaranteed by
Theorem 4.1.
If v and t are even, ℓ = tk, and both m and k are odd, the components of Γm are
isomorphic to K∗tk, while those of Γv are isomorphic to K
∗
2tk. Since we can factorize
K∗2tk into K2[tk] and two copies of K
∗
tk, it is enough to decompose both K2[tk] and K
∗
tk
into Ctk-factors. These factorizations exist by Theorem 4.4 and Theorem 1.1, respectively,
except possibly when (k, t) = (3, 2).
We may now use the result on triples (Theorem 4.1) to obtain the following.
Corollary 4.9. Let v ≡ m ≡ 0, 3 (mod 6), with v ≥ 3m and m ̸∈ {0, 6, 12}. Then there
exists a UMSF(3tv, 3tm, 3t) for all t > 0.
Additionally, we may use Theorem 3.5 to obtain the following result.
Corollary 4.10. Let 3 < k, k ≡ 2 (mod 4), v ≡ m ≡ k (mod 2k) and v ≥ 2m. Then
there exists a UMSF(vtk,mtk, k) and a UMSF(vtk,mtk, tk) for all t > 0.
We note that the above result can be used to obtain UMSF’s with cycle length, subsys-
tem size or number of vertices congruent to 0 (mod 4) by taking t even. However, in all
cases, the number of vertices and subsystem size will be divisible by k2.
5 Multiple subsystems
A natural question to ask is if a system can have multiple subsystems. In general, it seems
likely to be hard to navigate through the lattice of subsystems and all the possible ways the
subsystems can be distributed across the main system. However, when the subsystems are
disjoint, have small common intersections or are nested, the problem is more tractable. We
give some preliminary results in the next three subsections.
500 Ars Math. Contemp. 24 (2024) #P3.07 / 489–507
5.1 Disjoint subsystems
In the uniform case, the flexibility of Theorem 1.2 allows us to create a large number of
disjoint subsystems. We refer to a factorization of K∗v into k-cycles with subsystems on
disjoint vertex sets of sizes mj for 1 ≤ j ≤ n as a UMSF(v, {mj}, k).
Lemma 5.1. Let k | mj for 1 ≤ j ≤ n. Let m be an integer such that there is a
UMSF(m,mj , k) for each 1 ≤ j ≤ n. Then there exists a UMSF(ms, {mj}, k) for all
s ≥ max{2, n} if k is even, and for all s ≥ max{3, n} such that (s − 1)m is even if k is
odd, except when (k, s,m) = (6, 2, 6).
Proof. Theorem 1.2 guarantees the existence of a Ck-factorization of Ks[m]. For each
1 ≤ j ≤ n, place a copy of the ingredient UMSF(m,mj , k) on the jth part of size m
of Ks[m], and any Ck-factorization of K∗m on each of the remaining parts. The definite
exception (k, s,m) = (6, 2, 6) follows from the non-existence of a UMSF(12, 6, 6) (see
Theorem 4.5).
As with the uniform factorizations containing a single subsystem in this paper, the
easiest case is when mj | m for all 1 ≤ j ≤ n and either k is even or m is odd.
Corollary 5.2. Let m = lcm{mj : j = 1, . . . , n}, and assume the following conditions
are all satisfied:
(1) k | mj for all j ∈ {1, . . . , n} and m | v;
(2) k(m− 1) is even;
(3) if k = 3, then mj ̸∈ {6, 12} for all j;
(4) if (k,m) = (6, 12), then mj ̸= 6 for all j;
(5) (v,m, k) ̸= (12, 6, 6);
(6) v/m ≥ max{2, n} if k is even;
(7) v/m ≥ max{3, n} and v is odd if k is odd.
Then there exists a UMSF(v, {mj}, k).
Proof. Since each mj is a divisor of m and conditions (1)–(4) hold, we can apply either
Theorem 1.1 or Corollary 4.6, as needed, to ensure the existence of a UMSF(m,mj , k) for
every j ∈ {1, . . . , n}. These are the ingredient designs needed in Lemma 5.1, which can
be applied in view of conditions (5)–(7).
We note that for any fixed multiset of mj , this corollary constructs UMSF(v, {mj}, k)
for all but a finite number of v permitted by the necessary conditions when mj | v and
either k is even or m is odd.
P. Danziger et al.: On the minisymposium problem 501
5.2 Scattered subsystems
The proof of Lemma 5.1 builds systems whose factors intersect either all of the subsystems
or none of them. A balancing of the sizes of these intersections could be an interesting
property. For instance, we could ask for systems whose factors do not intersect more than
one subsystem. In other words, we ask for a Ck-factorization F of K∗v that contains n
subsystems of sizes m1,m2, . . . ,mn, such that no two factors of any of the subsystems are
contained in the same factor of F . We denote such a factorization by UMSF(v, [mj ], k)
and say that the subsystems are scattered.
Partial results in this direction can be easily obtained by making use of cycle frames.
We recall that a k-cycle frame (k-CF) of Ks[m] is a decomposition of Ks[m] into holey
Ck-factors; a holey Ck-factor is a vertex-disjoint union of k-cycles covering all vertices
Ks[m] except those belonging to one part. The following result, proven in [8], provides
necessary and sufficient conditions for the existence of k-cycle frames.
Theorem 5.3 ([8]). Let m ≥ 2 and k, s ≥ 3. There exists a k-cycle frame of Ks[m] if and
only if m is even, m(s− 1) ≡ 0 (mod k), k is even when s = 3, and (k,m, s) ̸= (6, 6, 3).
By making use of Theorem 5.3, we obtain the following.
Lemma 5.4. Let u ∈ {1, 2}. If there exists a UMSF(2m+ u,mj , k) for each 1 ≤ j ≤ n,
then there exists a UMSF(2ms+ u, [mj ], k) whenever 2s ≡ 2 (mod k) and s ≥ n.
Proof. Let n ≥ 1, 2s ≡ 2 (mod k) and s ≥ n. It follows that 2m(s − 1) ≡ 0 (mod k),
k = 4 when s = 3, and (k, 2m, s) ̸= (6, 6, 3). Therefore, Theorem 5.3 guarantees the
existence of a k-cycle frame F of Ks[2m]. Let Pi denote the i-th part of Ks[2m], for
1 ≤ i ≤ s. Also, let
F = {Fij : 1 ≤ i ≤ s, 1 ≤ j ≤ m},
where the Fijs are the holey Ck-factors of F missing Pi, for 1 ≤ i ≤ s. By assumption,
there is a UMSF(2m + u,mj , k) on Pi ∪ {∞1,∞u}, say Hi = {Hij : 1 ≤ j ≤ m}. It
follows that F∗ = {Fij ∪Hij : 1 ≤ i ≤ s, 1 ≤ j ≤ m} is a Ck-factorization of K2ms+u
with scattered subsystems of sizes m1,m2, . . . ,mn. Indeed, the factors of the subsystems
belong to the Hijs, each of which belongs to exactly one factor of F∗.
In the UMSF(2ms+u, [mj ], k) constructed in the proof of Lemma 5.4, two subsystems
may intersect in 0, 1, or 2 vertices, which are necessarily in the set {∞1,∞u}.
Theorem 4.5 provides sufficient conditions for the existence of a UMSF(v,m, k) if m
is a divisor of v. From that, we easily obtain the following corollary.
Corollary 5.5. Let u ∈ {1, 2}, k ≥ 3, and let k | mj | (2m + u) for each 1 ≤ j ≤ n.
Then there exists a UMSF(2ms+ u, [mj ], k) whenever the following conditions hold:
(1) mj is even or (2m+ u)/mj is odd,
(2) 2n ≤ 2s and 2s ≡ 2 (mod k),
except when (mj , k) ∈ {(6, 3), (12, 3)}, and except possibly when (2m + u,mj , k) =
(12, 6, 6), or k is odd and 2m+ u = 2mj , for some j ∈ {1, . . . , n}.
Note that for values of the triple (2m + u,mj , k) determining a possible exception in
Corollary 5.5 it is possible for a UMSF(2ms + u, [mj ], k) to exist. However, our method
cannot construct them because the UMSF(2m+u,mj , k) to use in the construction does not
exist. It is possible that other construction methods would build a UMSF(2ms+u, [mj ], k).
502 Ars Math. Contemp. 24 (2024) #P3.07 / 489–507
5.3 Nested subsystems
A scenario complementary to the susbsystems being all disjoint is when the subsystems
are completely nested, on vertex sets M1 ⊃ M2 ⊃ · · · ⊃ Mn−1. We modify our notation
slightly for this section to make it less cumbersome in this specific context.
Definition 5.6. Let v = m0 > m1 > · · · > mn−1 > mn = 0 be non-negative integers.
For 1 ≤ i ≤ ⌊m0−12 ⌋ and 0 ≤ j < n, let Ui,j be a 2-regular graph of order
|V (Ui,j)| =

mj −mj+1, if 1 ≤ i ≤ ⌊mj+1−12 ⌋,
mj , if ⌊mj+1+12 ⌋ ≤ i ≤ ⌊
mj−1
2 ⌋,
0, if ⌊mj+12 ⌋ ≤ i ≤ ⌊
m0−1
2 ⌋.
A nested minisymposium factorization nMSF({Ui,j : 1 ≤ i ≤ ⌊m0−12 ⌋, 0 ≤ j < n}) is a
2-factorization F = {Fi : 1 ≤ i ≤ ⌊m0−12 ⌋} of K
∗
v such that
• V (K∗v ) = M0 ⊃ M1 ⊃ · · · ⊃ Mn−1 ⊃ Mn = ∅ are nested sets with |Mj | = mj ;
• each 2-factor Fi =
⋃n−1
j=0 Fi,j , where
V (Fi,j) =

Mj \Mj+1 if 1 ≤ i ≤ ⌊mj+1−12 ⌋
Mj if ⌊mj+1+12 ⌋ ≤ i ≤ ⌊
mj−1
2 ⌋
∅ otherwise,
and each Fi,j is isomorphic to Ui,j ;
• for every 0 ≤ ℓ < n, {
⋃n−1
j=ℓ Fi,j : 1 ≤ i ≤ ⌊
mℓ−1
2 ⌋} is a 2-factorization of a graph
isomorphic to K∗mℓ .
In other words, the factorization F of K∗v restricted to vertex set Mℓ factorizes a graph
isomorphic to K∗mℓ into 2-factors whose structure is determined by the Ui,js.
Our construction of nested minisymposium factorizations is most tidily expressed by
defining holey factorizations.
Definition 5.7. Given positive integers v and m with v ≥ m, let
U =
{
Ui : 1 ≤ i ≤
⌊
v − 1
2
⌋}
,
be a collection of 2-regular graphs on v−m vertices for 1 ≤ i ≤ ⌊m−12 ⌋, and on v vertices
for ⌊m+12 ⌋ ≤ i ≤ ⌊
v−1
2 ⌋. A holey factorization HF(U) is a decomposition F = {Fi : 1 ≤
i ≤
⌊
v−1
2
⌋
} of K∗v −G (i.e., K∗v minus the edges of G) where each Fi ∼= Ui and G ∼= K∗m.
If v is even, then there is a 1-factor, Iv , on the vertices of K∗v whose edges are not
present in K∗v − G. If v and m are both even there is a 1-factor, Jm, on the vertices of G
which is a subgraph of Iv . If v is even and m odd, then no edges of Iv are induced on the
vertices of G. If v is odd and m is even, then there is a 1-factor Jm on the vertices of G
whose edges are present in K∗v −G.
By removing the 2-factors of a subsystem or “filling the hole” with them (making the
Jm in the hole coincide with the Im of the subsystem as required by the parities of v and
m) we have an equivalence between the existence of minisymposium factorizations and
holey factorizations.
P. Danziger et al.: On the minisymposium problem 503
Theorem 5.8. Let T = {Ti : 1 ≤ i ≤ ⌊m−12 ⌋} be a 2-factorization of K
∗
m and
U =
{
Ui : 1 ≤ i ≤
⌊
v − 1
2
⌋}
,
be a collection of 2-regular graphs on v−m vertices for 1 ≤ i ≤ ⌊m−12 ⌋ and on v vertices
for ⌊m+12 ⌋ ≤ i ≤ ⌊
v−1
2 ⌋. Then a HF(U) exists if and only if a
MSF({Ui : ⌊m+12 ⌋ ≤ i ≤ ⌊
v−1
2 ⌋}, {(Ti, Ui) : 1 ≤ i ≤ ⌊
m−1
2 ⌋})
exists.
Because in a nested minisymposium factorization the holes are nested and emptying or
filling them does not affect the edges outside the hole, this equivalence extends to nested
minisymposium factorizations and shows that they can be constructed exactly when the
various holey factorizations of K∗mj with holes of size mj+1 exist.
Theorem 5.9. Let v = m0 > m1 > · · · > mn−1 > mn = 0 be positive integers. For
1 ≤ i ≤ ⌊m0−12 ⌋ and 0 ≤ j < n, let Ui,j be a 2-regular graph of order
|V (Ui,j)| =

mj −mj+1, if 1 ≤ i ≤ ⌊mj+1−12 ⌋,
mj , if ⌊mj+1+12 ⌋ ≤ i ≤ ⌊
mj−1
2 ⌋,
0, if ⌊mj+12 ⌋ ≤ i ≤ ⌊
m0−1
2 ⌋.
A nested minisymposium factorization nMSF({Ui,j : 1 ≤ i ≤ ⌊m0−12 ⌋, 0 ≤ j < n})
exists if and only if a HF({Ui,j : 1 ≤ i ≤ ⌊mj−12 ⌋}) exists for each 0 ≤ j < n.
Proof. The forward direction is proved simply by restricting the system to Mj and re-
moving the subsystem on Mj+1. The converse is proved by a recursive construction
starting with j = n − 1: in this case, a HF({Ui,n−1 : 1 ≤ i ≤ ⌊mn−1−12 ⌋}) is an
nMSF({Ui,n−1 : 1 ≤ i ≤ ⌊mn−1−12 ⌋}), say Fn−1.
At stage j < n − 1, use Theorem 5.8 to construct an nMSF({Ui,ℓ : 1 ≤ i ≤
⌊mj−12 ⌋, j ≤ ℓ < n}), say Fj , by filling the hole in the HF({Ui,j : 1 ≤ i ≤ ⌊
mj−1
2 ⌋})
with the nMSF({Ui,ℓ : 1 ≤ i ≤ ⌊mj+1−12 ⌋, j + 1 ≤ ℓ < n}), denoted by Fj+1, built at
stage j + 1.
Between the extremes of disjoint and nested subsystems, there are factorizations with
multiple subsystems with arbitrary intersections. Some structured instances of this much
more general problem may be amenable to solution but we leave this to future work.
6 Conclusions and further work
We have introduced the minisymposium problem: a subsystem variant of the generalized
Oberwolfach problem. This variant asks for a solution to a generalized Oberwolfach prob-
lem that contains a subsystem of a given size. When v, the number of vertices, is even,
it is traditional in 2-factor decomposition problems to ask for decompositions of Kv − I
where I is a 1-factor. When the number of vertices in the system and the subsystem are
both even, then we require that the 1-factor in the subsystem be a subgraph of the 1-factor
in the full system. Therefore when the parities of the system and the subsystem agree, the
504 Ars Math. Contemp. 24 (2024) #P3.07 / 489–507
problem becomes more tractable. When the parities are opposite, either the 1-factor of the
full system must avoid the subsystem, or the edges of the 1-factor in the subsystem must
be in 2-factors of the whole system.
Clearly, this is a very broad statement and we identify some particularly interesting
cases, Hamiltonian* and uniform. In the Hamiltonian* minisymposium problem there are
as few cycles as possible and in the uniform minisymposium problem all cycles are of the
same length. We have shown that the work of Hilton and Johnson provides a complete
solution for the Hamiltonian* minisymposium problem in Corollary 3.2. In the case when
v ≡ m ≡ 2 (mod 4), Theorem 3.4 uses this Hamiltonian* construction to provide a
wide range of solutions when the resulting factors are all bipartite. In particular, a uniform
factorization with k ≡ 2 (mod 4), v ≥ 2m and v ≡ m ≡ k (mod 2k) always exists.
Corollary 4.10 can be used to extend this to uniform factorizations where k ≡ 0 (mod 4)
or v ≡ m ≡ 0 (mod 2k).
In Section 4 we considered the uniform case. We have solved a large part of the spec-
trum. In particular, when k is even or v is odd, Corollary 4.6 gives all cases when m | v
and Corollary 4.7 completely solves all cases when k = m has the same parity as v. Theo-
rem 4.8 gives a powerful recursive construction which is applicable in cases where m does
not divide v. By applying it to the case when k = 3, we obtain uniform factorizations
with cycle lengths divisible by 3. The case when m is odd and v is even seems to be the
hardest. Even in the simplest case when k = 3, which has been well studied otherwise
[18, 19, 20, 22, 34, 35, 36, 37], the case with v and m having opposite parities has not been
previously considered and remains open.
While the Hamiltonian* problem is solved and we have made significant inroads into
the uniform case, the general problem remains wide open. We expect that when m and
v have the same parity solutions will be easier to find. When m and v have opposite
parity we expect that v odd with m even is more tractable than the reverse. Considering
2-factorizations where a solution to the Oberwolfach problem is known might be a good
starting point. A natural case to consider is the case when all factors are isomorphic i.e.
Fi ∼= Tj ∪ Uj for all i and j. Uniform factorizations are an example of this, but other
variations are possible, for example, requiring all factors to be isomorphic to Cv−m ∪Cm.
Indeed, Theorem 3.4 solves all these cases when the factors are bipartite and v ≡ m ≡ 2
(mod 4), but this broader variant remains open.
More complex variants can also be considered. We have briefly considered systems
with multiple subsystems. When these subsystems are completely nested the problem es-
sentially reduces to the existence of the necessary ingredients as described in Theorem 5.9.
Let {mj} be a multiset of subsystem sizes. When the subsystems are pairwise disjoint, v is
divisible by each mj and either v is odd or at least one subsystem is even, then Lemma 5.1
and Corollary 5.2 use Theorem 1.2 to construct a UMSF(v, {mj}, k) for all but a finite
number of admissible v. Even in the seemingly simple case when the subsystems are all
disjoint the problem remains generally open even for the uniform case. Further partial re-
sults are obtained when the subsystems are scattered, that is, when no two minisymposia
have meetings taking place on the same day. Cycle frames in Theorem 5.3 allow us to
construct uniform factorizations as described in Lemma 5.4 and Theorem 5.5. The more
general case when the intersections of multiple subsystems are arbitrary is completely open.
P. Danziger et al.: On the minisymposium problem 505
ORCID iDs
Tommaso Traetta https://orcid.org/0000-0001-8141-0535
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ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 24 (2024) #P3.08 / 509–534
https://doi.org/10.26493/1855-3974.3035.6ac
(Also available at http://amc-journal.eu)
Two kinds of partial Motzkin paths
with air pockets*
Jean-Luc Baril †
LIB, Université de Bourgogne, B.P. 47 870, 21078 Dijon Cedex, France
Paul Barry
School of Science, South East Technological University (SETU), Ireland
Received 23 December 2022, accepted 23 October 2023, published online 16 August 2024
Abstract
Motzkin paths with air pockets (MAP) are defined as a generalization of Dyck paths
with air pockets by allowing some horizontal steps with certain conditions. In this paper,
we introduce two generalizations. The first one consists of lattice paths in N2 starting at
the origin, made of steps U = (1, 1), Dk = (1,−k), k ⩾ 1 and H = (1, 0), where two
down steps cannot be consecutive, while the second one are lattice paths in N2 starting at
the origin, made of steps U , Dk and H , where each step Dk and H is necessarily followed
by an up step, except for the last step of the path. We provide enumerative results for these
paths according to the length, the type of the last step, and the height of its end-point. A
similar study is made for these paths read from right to left. As a byproduct, we obtain new
classes of paths counted by the Motzkin numbers. Finally, we express our results using
Riordan arrays.
Keywords: Enumeration, Motzkin paths, kernel method, Riordan array.
Math. Subj. Class. (2020): 05A05, 05A15, 05A15
1 Introduction
In a recent paper [2], the authors introduce, study and enumerate special classes of lattice
paths, called Dyck paths with air pockets (DAP for short). Such paths are non empty lattice
paths in the first quadrant of Z2 starting at the origin, and consisting of up-steps U = (1, 1)
and down-steps Dk = (1,−k), k ⩾ 1, where two down steps cannot be consecutive.
*The authors would like to thank the anonymous referees for useful remarks and comments.
†Corresponding author.
E-mail addresses: barjl@u-bourgogne.fr (Jean-Luc Baril), pbarry@wit.ie (Paul Barry)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
510 Ars Math. Contemp. 24 (2024) #P3.08 / 509–534
The length of a path is the number of its steps. These paths can be viewed as ordinary
Dyck paths where each maximal run of down-steps is condensed into one large down step.
As mentioned in [2], they also correspond to a stack evolution with (partial) reset opera-
tions that cannot be consecutive (see for instance [7]). Whenever the last point is on the
x-axis, they prove that DAP of length n are in one-to-one correspondence with the peakless
Motzkin paths of length n − 1. They also investigate the popularity of many patterns in
these paths and they give asymptotic approximations. We also refer to [10] for the enu-
meration of DAP with respect to the length, the type (up or down) of the last step and the
height of the end-point (i.e., the ordinate of the end-point). In a second work [3], the au-
thors make a study for a generalization of these paths by allowing them to go below the
x-axis. They call these paths Grand Dyck paths with air pockets (GDAP), and they also
yield enumerative results for these paths according to the length and several restrictions on
the height.
In this paper, we introduce two generalizations of partial Dyck paths of air pockets by
allowing some possible horizontal steps H = (1, 0) with some conditions. These two kinds
of paths can be viewed as special partial Motzkin paths (lattice paths in N2 starting at the
origin and made of U = (1, 1), D = (1,−1), and H = (1, 0)), where each maximal run of
down-steps is condensed into one large down step.
Definition 1.1. A partial Motzkin path with air pocket of the first kind is a lattice path in
N2 starting at the origin, consisting of steps U , Dk and H , where two down steps cannot
be consecutive. Let M1 be the set of these paths.
Definition 1.2. A partial Motzkin path with air pocket of second kind is a lattice path in N2
starting at the origin, consisting of steps U , Dk and H , where any step H and Dk (except
the last step of the path) is immediately followed by an up step U . Let M2 be the set of
these paths.
Whenever these paths end on the x-axis, we call them Motzkin paths with air pockets
of the first and second kinds, respectively. For short, we denote by PMAP all paths in Mi,
i ∈ {1, 2}, and we denote by MAP all paths ending on the x-axis.
For instance, UUUDHUD2 ∈ M1 and UUUDUHUD2 ∈ M2 are two PMAP of the
first and second kinds, respectively. The paths UUUDHUD3UD and UUUDUHUD4
UD are MAP of the first and second kinds, respectively. See also Figure 1 and Figure 5 for
other examples of PMAP and MAP.
We also consider lattice paths obtained from MAP in M1 and M2 by reading them
from right to left and by replacing down-steps with up-steps and vice versa. More precisely,
we define the two sets M′1 and M′2 as follows.
Definition 1.3. Let M′1 be the set of paths in N2 starting at the origin, consisting of up
steps Uk = (1, k), k ⩾ 1, horizontal step H and down step D, where two up steps cannot
be consecutive.
Definition 1.4. Let M′2 be the set of paths in N2 starting at the origin, consisting of up
steps Uk = (1, k), k ⩾ 1, horizontal step H and down step D, where any H and Uk,
k ⩾ 1, (except the first step of the path) is preceded by a down step D.
All paths in M′1 and M′2 will be be called PMAP from right to left, and in the case
where they end on the x-axis, we call them MAP from right to left. For instance, we
have UDU3DHUD ∈ M′1, UDU4DHDU ∈ M′2, and the path UDU3DHUDDD
J.-L. Baril and P. Barry: Two kinds of partial Motzkin paths with air pockets 511
(resp. UDU4DHDUDDD) is a MAP of the first (resp. second) kind from right to left.
See also Figure 3 and Figure 7 for other examples of PMAP and MAP from right to left.
Notice that D1 = D and U1 = U , and we will use both notations.
Throughout the paper, and for each class of paths Mi and M′i, i ∈ {1, 2}, described
above, we will use the following notations. For k ⩾ 0, we consider the generating function
fk = fk(z) (resp. gk = gk(z), resp. hk = hk(z)), where the coefficient of zn in the series
expansion is the number of partial Motzkin paths with air pockets of length n ending at
height k with an up-step, (resp. with a down-step, resp. with a horizontal step H).
We introduce the bivariate generating functions
F (z, u) =
∑
k⩾0
ukfk(z), G(z, u) =
∑
k⩾0
ukgk(z), and H(z, u) =
∑
k⩾0
ukhk(z).
For short, we also use the notation F (u), G(u) and H(u) for these functions, and we
introduce the generating function
Total(z, u) = F (u) +G(u) +H(u).
Finally, for any bivariate generating function H(z, u) = H(u), we will use the notation
[uk]H(u) for the coefficient of uk in H(u).
The outline of this paper is the following. In Section 2, we present enumerative results
for partial Motzkin paths with air pockets of the first kind, and for these paths when we read
them from right to left. We provide bivariate generating functions that count these paths
with respect to the length, the type of the last step (up, down or horizontal step) and the
height of the end-point. In Section 3, we make a similar study for PMAP of second kind,
and we present new classes of lattice paths counted by the well known Motzkin numbers.
All these results are obtained algebraically by using the famous kernel method for solving
several systems of functional equations. More precisely, Sections 2.1, 2.2, 3.1 and 3.2 have
the same structure: we show how to obtain a system of equations involving fk, gk and hk,
and we apply the kernel method in order to provide some expressions of F (u), G(u) and
H(u). Finally, in Section 4 we express our results using Riordan arrays and we deduce
closed forms for PMAP of length n ending at height k.
2 PMAP of the first kind
In this section, we focus on PMAP of the first kind, i.e. lattices paths in N2 starting at the
origin, made of steps U , Dk and H , such that two down steps cannot be consecutive. The
first subsection considers the paths in M1, while the second subsection handles the paths in
M′1 (see Introduction for the definition of these two sets). We yield enumerative results for
these paths according to the length, the type of the last step, and the height of its end-point.
2.1 PMAP in M1 - From left to right
In this part, we consider PMAP in M1. Figure 1 shows two examples of such paths.
Let P be a length n PMAP in M1 ending at height k ⩾ 0. If the last step of P is U ,
then k ⩾ 1 and P can be written P = QU where Q is a length (n − 1) PMAP ending at
height k − 1. So, we obtain the first relation fk = zfk−1 + zgk−1 + zhk−1 for k ⩾ 1,
anchored with f0 = 1 by considering the empty path. If the last step of P is a down step
Da, a ⩾ 1, then we have P = QDa where Q is a length (n − 1) PMAP ending at height
512 Ars Math. Contemp. 24 (2024) #P3.08 / 509–534
Figure 1: The left drawing shows a Motzkin path with air pockets of length 18 in M1. The
right drawing shows a partial Motzkin path with air pockets of length 18 ending at height
3 in M1.
ℓ = a + k ⩾ k + 1 with no down step at its end. So, we obtain the second relation
gk = z
∑
ℓ⩾k+1 fℓ + z
∑
ℓ⩾k+1 hℓ. If the last step of P is a horizontal step H , then we
have P = QH where Q is a length (n − 1) PMAP ending at height k, which implies
hk = zfk + zgk + zhk.
Therefore, we have to solve the following system of equations.
f0 = 1, and fk = zfk−1 + zgk−1 + zhk−1, k ⩾ 1,
gk = z
∑
ℓ⩾k+1
fℓ + z
∑
ℓ⩾k+1
hℓ, k ⩾ 0,
hk = zfk + zgk + zhk, k ⩾ 0.
(2.1)
Summing the recursions in (2.1), we have:
F (u) = 1 + z
∑
k⩾1
ukfk−1 + z
∑
k⩾1
ukgk−1 + z
∑
k⩾1
ukhk−1
= 1 + zuF (u) + zuG(u) + zuH(u),
G(u) = z
∑
k⩾0
uk
( ∑
ℓ⩾k+1
fℓ
)
+ z
∑
k⩾0
uk
( ∑
ℓ⩾k+1
hℓ
)
= z
∑
k⩾1
fk(1 + u+ . . .+ u
k−1) + z
∑
k⩾1
hk(1 + u+ . . .+ u
k−1)
= z
∑
k⩾1
uk − 1
u− 1
fk + z
∑
k⩾1
uk − 1
u− 1
hk
=
z
u− 1
(F (u)− F (1) +H(u)−H(1)),
H(u) = zF (u) + zG(u) + zH(u).
Notice that we have F (1) − H(1) = 1 by considering the difference of the first and
third equations. Now, setting a := F (1) and solving these functional equations, we obtain
F (u) =
2au z2 − u z2 + uz + z2 − u− z + 1
u2z + u z2 + z2 − u− z + 1
,
G(u) = −z (2auz + 2az − uz − 2a− z + 2)
u2z + u z2 + z2 − u− z + 1
,
H(u) =
z (2az − u− 2z + 1)
u2z + u z2 + z2 − u− z + 1
.
J.-L. Baril and P. Barry: Two kinds of partial Motzkin paths with air pockets 513
In order to compute a = F (1), we use the kernel method (see [1, 9]) on F (u). This
method consists in cancelling the denominator by finding u as an algebraic function of z,
s(z). So, if we substitute u with s(z) in the numerator, then it necessarily equals zero (in
order to counterbalance the cancellation of the denominator), which allows to find a =
F (1). Finally, we can deduce the generating function F (u).
We can write the denominator (which is a polynomial in u of degree 2), as z(u − r)
(u− s) with
r =
1− z2 +
√
z4 − 4z3 + 2z2 − 4z + 1
2z
,
s =
1− z2 −
√
z4 − 4z3 + 2z2 − 4z + 1
2z
.
Replacing u with s (which have a Taylor expansion at z = 0) in order to cancel the numer-
ator of F (u), we obtain the equation
2as z2 − s z2 + sz + z2 − s− z + 1 = 0.
Using zrs = z2 − z + 1, a straightforward calculation provides
a = F (1) = H(1) + 1 =
r(s− 1)
2z
.
Finally, after simplifying by the factor (u − s) in the numerators and denominators, we
obtain
F (u) =
r
r − u
, G(u) =
r(s− 1)− z
r − u
, and H(u) =
1
r − u
.
Extracting the coefficient of uk in the series expansion of each generating function, we
obtain
[uk]F (u) := fk =
1
rk
,
[uk]G(u) := gk =
s− 1
rk
− z
rk+1
,
[uk]H(u) := hk =
1
rk+1
.
Theorem 2.1. The bivariate generating function for the total number of PMAP in M1 with
respect to the length and the height of the end-point is given by
Total(z, u) =
1
z(r − u)
,
and we have
[uk]Total(z, u) =
1
zrk+1
.
Finally, setting t(n, k) := [zn][uk]Total(z, u), we have for n ⩾ 2, k ⩾ 1,
t(n, k) = t(n, k − 1) + t(n− 1, k)− t(n− 1, k − 2)− t(n− 2, k)− t(n− 2, k − 1),
and setting tn := t(n, 0), then we have t0 = t1 = 1, and for n ⩾ 2,
tn = tn−1 + tn−2 +
n−3∑
k=0
tktn−k−3 +
n−1∑
k=2
(tk − tk−1) tn−k−1.
514 Ars Math. Contemp. 24 (2024) #P3.08 / 509–534
Proof. Since Total(z, u) = F (u)+G(u)+H(u), the first two equalities are immediately
deduced from the previous results. The third equality is obtained using the Mathematica
package Guess.m (written by Manuel Kauers [6]) for guessing recurrence relations. After
this, it suffices to check algebraically:
Total(z, u) = (u+ z − u2z − z2 − uz2)Total(z, u)− u+ (z
2 − z + 1)(1 + rs− z)
r
.
Now, let us prove the last equality. Any length n MAP of the first kind is of the form
(i) HP , or (ii) UDP , or (iii) UPHDQ where P,Q are some MAP so that the length k of
P lies into [0, n− 3], and the length of Q is n− k − 3, or (iv) P ♯Q where P ♯ = UP ′Dℓ,
ℓ ⩾ 1, and P ′ is a PMAP such that P ′Dℓ−1 is a MAP of length k lying into [2, n− 1], and
the length of Q is n− k − 1. Taking into account all these cases, we obtain the result.
Let T be the infinite matrix T := [t(n, k)]n⩾0,k⩾0. The first few rows of the matrix T
are
T =

1 0 0 0 0 0 0 0 · · ·
1 1 0 0 0 0 0 0 · · ·
2 2 1 0 0 0 0 0 · · ·
5 5 3 1 0 0 0 0 · · ·
13 14 9 4 1 0 0 0 · · ·
36 40 28 14 5 1 0 0 · · ·
105 118 87 48 20 6 1 0 · · ·
317 359 273 161 75 27 7 1 · · ·
...
...
...
...
...
...
...
...
. . .

.
Corollary 2.2. The generating function that counts the PMAP in M1 with respect to the
length is given by
Total(z, 1) =
1
z(r − 1)
.
The first few terms of the series expansion of Total(z, 1) are 1 + 2z + 5z2 + 14z3 +
41z4 + 124z5 + 385z6 + 1220z7 + 3929z8 + 12822z9 + O(z10), which correspond
to the sequence A159771 in [13] counting the n-leaf binary trees that do not contain
(()((()())((()())()))) as a subtree (see [11]). See Figure 2 for an illustration of the 14
PMAP of length 3.
Corollary 2.3. The generating function that counts the MAP in M1 with respect to the
length is given by
Total(z, 0) =
1
zr
.
The first few terms of the series expansion of Total(z, 0) are 1 + z + 2z2 + 5z3 +
13z4+36z5+105z6+317z7+982z8+3105z9+O(z10) which correspond to the sequence
A114465 in [13] counting Dyck paths of length 2n having no ascents of length 2 that start
at an odd level. We leave open the question of finding a constructive bijection between
these sets.
J.-L. Baril and P. Barry: Two kinds of partial Motzkin paths with air pockets 515
Figure 2: The 14 PMAP of length three in M1. Notice that five paths end on the x-axis,
five paths end at height 1, three paths end at height 2, and one path ends at height 3, which
correspond to the fourth row of T .
2.2 PMAP in M′1 - From right to left
Here, we consider the paths of the previous section, but we read them from right to left. This
means that down steps become up steps and vice versa, and horizontal steps are unchanged,
which implies that two up steps cannot be consecutive now. See Definition 1.3 and Figure 3
for two examples of such paths.
Figure 3: The left drawing shows a Motzkin path with air pockets of length 18 in M′1. The
right drawing shows a partial Motzkin path with air pockets of length 18 ending at height
2 in M′1.
Let P be a length n PMAP in M′1 ending at height k ⩾ 0. If the last step of P is Ua,
a ⩾ 1, then k ⩾ a and we have P = QUa where Q is a length (n − 1) PMAP ending
at height ℓ = k − a with a horizontal step or a down step. So, we obtain the first relation
fk = z + z(g0 + g1 + . . . + gk−1) + z(h0 + h1 + . . . + hk−1) for k ⩾ 1, anchored with
f0 = 1 by considering the empty path. If the last step of P is a down step D, then we have
P = QD where Q is a length (n − 1) PMAP ending at height k + 1. So, we obtain the
second relation gk = zfk+1 + zgk+1 + zhk+1. If the last step of P is a horizontal step
H , then we have P = QH where Q is a length (n − 1) PMAP ending at height k, which
implies hk = zfk + zgk + zhk.
Therefore, we have to solve the following system of equations.

f0 = 1,
fk = z + z(g0 + g1 + . . .+ gk−1) + z(h0 + h1 + . . .+ hk−1), k ⩾ 1,
gk = zfk+1 + zgk+1 + zhk+1, k ⩾ 0,
hk = zfk + zgk + zhk, k ⩾ 0.
(2.2)
516 Ars Math. Contemp. 24 (2024) #P3.08 / 509–534
Summing the recursions in (2.2), we have:
F (u) = 1 +
zu
1− u
+ z
∑
k⩾1
(g0 + . . .+ gk−1)u
k + z
∑
k⩾1
(h0 + . . .+ hk−1)u
k
= 1 +
zu
1− u
+ z
∑
k⩾0
gk
uk+1
1− u
+ z
∑
k⩾0
hk
uk+1
1− u
= 1 +
zu
1− u
(1 +G(u) +H(u)),
G(u) = z
∑
k⩾0
fk+1u
k + z
∑
k⩾0
gk+1u
k + z
∑
k⩾0
hk+1u
k
=
z
u
(F (u)− F (0) +G(u)−G(0) +H(u)−H(0)),
H(u) = zF (u) + zG(u) + zH(u).
Notice that we have H(0) = z(1+G(0))1−z by considering the third equation and F (0) = 1.
Solving these functional equations using a := G(0), we obtain
F (u) =
−u2z3 + a uz2 + 3u2z2 − uz3 − 3u2z + 2uz2 + u2 + uz − z2 − u+ z
(1− z) (u2z2 − u2z + uz2 + u2 − u+ z)
,
G(u) =
z
(
a uz2 − a uz + a u+ a z + uz − a
)
(−1 + z) (u2z2 − u2z + uz2 + u2 − u+ z)
,
H(u) =
(
a uz2 + u2z2 − a uz − 2u2z + uz2 + a z + u2 − u+ z
)
z
(1− z) (u2z2 − u2z + uz2 + u2 − u+ z)
.
In order to compute a = G(0), we use the kernel method on G(u). We can write the
denominator (which is a polynomial in u of degree 2), as (z−1)(z2−z+1)(u−r)(u−s)
with
r =
1− z2 +
√
z4 − 4 z3 + 2 z2 − 4 z + 1
2(z2 − z + 1)
,
s =
1− z2 −
√
z4 − 4 z3 + 2 z2 − 4 z + 1
2(z2 − z + 1)
.
Replacing u with s (which have a Taylor expansion at z = 0) in order to cancel the numer-
ator of G(u), we obtain the equation
a(sz2 − sz + s+ z − 1) + sz = 0.
Using sr(z2 − z + 1) = z, we deduce
a = G(0) =
1− r
r
− sz.
Finally, after simplifying by the factor (u− s) in the numerators and denominators, we
obtain
F (u) = 1 +
sru
r − u
, G(u) =
s(1− r + rz)
r − u
, and H(u) =
sr − sru(1− z)
r − u
,
J.-L. Baril and P. Barry: Two kinds of partial Motzkin paths with air pockets 517
which implies that
[uk]F (u) := fk = [k = 0] + [k ̸= 0] ·
s
rk−1
,
[uk]G(u) := gk =
s(1− r + rz)
rk+1
,
[uk]H(u) := hk =
s
rk
− [k ̸= 0] · (1− z)s
rk−1
,
where [k = 0] (resp. [k ̸= 0]) equals 1 whenever k = 0 (resp. k ̸= 0), and 0 otherwise.
Theorem 2.4. The bivariate generating function for the total number of PMAP (read from
right to left) with respect to the length and the height of the end-point is given by
Total(z, u) = 1 +
s(1 + rz + ruz)
r − u
,
and we have
[uk]Total(z, u) = [k = 0] +
s(rz + 1)
rk+1
+ [k ̸= 0] · sz
rk−1
.
Finally, setting t(n, k) := [zn][uk]Total(z, u), we have for n ⩾ 2, k ⩾ 1,
t(n, k) = t(n− 2, k− 1) + t(n− 2, k)− t(n− 1, k− 1) + t(n− 1, k+ 1) + t(n, k− 1),
and setting tn := t(n, 0), we have t0 = t1 = 1, and for n ⩾ 2,
tn = tn−1 + tn−2 +
n−3∑
k=0
tktn−k−3 +
n−1∑
k=2
(tk − tk−1) tn−k−1.
Proof. The first two equalities are directly deduced from the previous results. Since we
have
(u− u2z2 − uz2 + zu2 − z − u2)Total(z, u) + u2(1− z)− (1 + s− sz)u+ s = 0,
we deduce the third relation. The last equality is already given in Theorem 2.1, since the
number of MAP in M′1 is obviously equal to the number of MAP in M1.
Let T be the infinite matrix T := [t(n, k)]n⩾0,k⩾0. The first few rows of the matrix T
are
T =

1 0 0 0 0 0 · · ·
1 1 1 1 1 1 · · ·
2 3 3 3 3 3 · · ·
5 8 10 12 14 16 · · ·
13 23 33 43 53 63 · · ·
36 69 107 149 195 245 · · ·
105 212 348 512 704 924 · · ·
317 665 1141 1753 2509 3417 · · ·
...
...
...
...
...
...
. . .

.
518 Ars Math. Contemp. 24 (2024) #P3.08 / 509–534
Since there is an infinite number of PMAP of length n, we do not provide an ordinary
generating function (with respect to the length) for these paths. So, we get around this by
counting PMAP ending on a point (x, n− x) for a given n ⩾ 0.
Corollary 2.5. The generating function that counts the partial PMAP ending on the line
y = n− x is given by
Total(z, z) = 1 +
s(1 + rz + rz2)
r − z
.
The first few terms of the series expansion of Total(z, z) are 1 + z + 3z2 + 9z3 +
25z4 + 73z5 + 223z6 + 697z7 + 2217z8 + 7161z9 + O(z10), which correspond to the
sequence A101499 in [13], which is a Chebyshev transform of the Catalan number that
counts peakless Motzkin paths of length n where horizontal steps at level at least one come
in 2 colors. See Figure 4 for an illustration of the 9 PMAP in M′1 ending on the line
y = 3− x.
Notice that, from Theorem 2.4 we have
Total(z, 0) = 1 +
s(1 + rz)
r
,
which is obviously equal to the expression derived in Corollary 2.3 that counts MAP of the
first kind with respect to the length.
Figure 4: The 9 PMAP ending on the line y = 3− x in M′1. Notice that five paths end on
the x-axis, three paths end at height 1, and one path ends at height 2, which correspond to
the fourth diagonal of T .
3 PMAP of the second kind
In this section, we focus on PMAP of the second kind. See Figure 5 for an illustration of
such paths. The first subsection considers paths in M2, while the second handles paths in
M′2. We yield enumerative results for these paths according to the length, the type of the
last step, and the height of the end-point.
3.1 PMAP in M2 - From left to right
In this part, we consider PMAP in M2, i.e. lattice paths in N2 starting at the origin,
consisting of steps U , Dk and H , and where any down step or horizontal step (except
for the last step of the path) is immediately followed by an up step. Figure 5 shows two
examples of such paths.
J.-L. Baril and P. Barry: Two kinds of partial Motzkin paths with air pockets 519
Figure 5: The left drawing shows a MAP of length 18 in M2. The right drawing shows a
PMAP of length 18 ending at height 3 in M2.
Let P be a length n PMAP in M2 ending at height k ⩾ 0. If the last step of P is U ,
then k ⩾ 1 and we have P = QU where Q is a length (n−1) PMAP ending at height k−1.
So, we obtain the first relation fk = zfk−1 + zgk−1 + zhk−1 for k ⩾ 1, anchored with
f0 = 1 by considering the empty path. If the last step of P is a down step Da, a ⩾ 1, then
we have P = QDa where Q is a length (n− 1) PMAP ending at height ℓ = a+ k ⩾ k+1
with an up step. So, we obtain the second relation gk = z
∑
ℓ⩾k+1 fℓ. If the last step of P
is a horizontal step H , then we have P = QH where Q is a length (n− 1) PMAP ending
at height k with an up step, which implies hk = zfk.
So, we have to solve the following system of equations.
f0 = 1, and fk = zfk−1 + zgk−1 + zhk−1, k ⩾ 1,
gk = z
∑
ℓ⩾k+1
fℓ, k ⩾ 0,
hk = zfk, k ⩾ 0.
(3.1)
Summing the recursions in (3.1), we have:
F (u) = 1 + z
∑
k⩾1
ukfk−1 + z
∑
k⩾1
ukgk−1 + z
∑
k⩾1
ukhk−1
= 1 + zuF (u) + zuG(u) + zuH(u),
G(u) = z
∑
k⩾0
uk
( ∑
ℓ⩾k+1
fℓ
)
= z
∑
k⩾1
uk − 1
u− 1
fk
=
z
u− 1
(F (u)− F (1)),
H(u) = zF (u).
Now, setting a := F (1) and solving these functional equations, we deduce
F (u) =
au z2 − u+ 1
u2z2 + u2z − uz − u+ 1
,
G(u) = −
z
(
au z2 + auz − a+ 1
)
u2z2 + u2z − uz − u+ 1
, H(u) =
z
(
au z2 − u+ 1
)
u2z2 + u2z − uz − u+ 1
.
In order to compute a = F (1), we use the kernel method on F (u). We can write the
denominator (which is a polynomial in u of degree 2), as (z2 + z)(u− r)(u− s) with
r =
1 + z +
√
−3z2 − 2z + 1
2z (z + 1)
, and s =
1 + z −
√
−3z2 − 2z + 1
2z (z + 1)
.
520 Ars Math. Contemp. 24 (2024) #P3.08 / 509–534
Replacing u with s (which have a Taylor expansion at z = 0) in order to cancel the numer-
ator of F (u), we obtain the equation
as z2 − s+ 1 = 0,
and thus
a = F (1) =
s− 1
sz2
.
Finally using z(1 + z)rs = 1 and simplifying by the factor (u − s) in the numerators
and denominators, we obtain
F (u) =
r
r − u
, G(u) =
s− 1
sz(r − u)
, and H(u) =
zr
r − u
,
which implies that
[uk]F (u) := fk =
1
rk
,
[uk]G(u) := gk = (1 + z) ·
s− 1
rk
,
[uk]H(u) := hk =
z
rk
.
Theorem 3.1. The bivariate generating function for the total number of PMAP with respect
to the length and the height of the end-point is given by
Total(z, u) =
1
z(r − u)
,
and we have
[uk]Total(z, u) =
1
zrk+1
.
Finally, setting t(n, k) = [zn][uk]Total(z, u), we have for n ⩾ 2, k ⩾ 1,
t(n, k) = t(n, k − 1) + t(n− 1, k − 1)− t(n− 1, k − 2)− t(n− 2, k − 2),
and setting tn := t(n, 0), we have t0 = 1, and for n ⩾ 1,
tn = tn−1 +
n−2∑
k=1
tktn−1−k.
Proof. The first two equalities are immediately deduced from the previous results. The
third equality is obtained using the Mathematica package Guess.m ([6]) for guessing
recurrence relations. After this, it suffices to check algebraically:
Total(z, u) = (u+ uz − u2z − u2z2)Total(z, u)− u(1 + z) + 1
zr
.
For the last equality, it suffices to remark that the generating function of the sequence
(tn)n⩾0, that is 1/(zr), generates a shift of the well known Motzkin sequence A001006
in [13].
J.-L. Baril and P. Barry: Two kinds of partial Motzkin paths with air pockets 521
Let T be the infinite matrix T := [t(n, k)]n⩾0,k⩾0. The first few rows of the matrix T
are
T =

1 0 0 0 0 0 0 0 · · ·
1 1 0 0 0 0 0 0 · · ·
1 2 1 0 0 0 0 0 · · ·
2 3 3 1 0 0 0 0 · · ·
4 6 6 4 1 0 0 0 · · ·
9 13 13 10 5 1 0 0 · · ·
21 30 30 24 15 6 1 0 · · ·
51 72 72 59 40 21 7 1 · · ·
...
...
...
...
...
...
...
...
. . .

.
Corollary 3.2. The generating function that counts the PMAP with respect to the length is
given by
Total(z, 1) =
1
z(r − 1)
.
The first few terms of the series expansion of Total(z, 1) are 1 + 2z + 4z2 + 9z3 +
21z4+51z5+127z6+323z7+835z8+2188z9+O(z10), which correspond to a shift of
the sequence A001006 in [13] that counts the Motzkin paths of a given length. See Figure 6
for an illustration of the 9 paths of length 3.
Corollary 3.3. The generating function that counts the MAP with respect to the length is
given by
Total(z, 0) =
1
zr
.
The first few terms of the series expansion of Total(z, 0) are 1+ z+ z2+2z3+4z4+
9z5+21z6+51z7+127z8+323z9+O(z10) which correspond to a shift of the sequence
A001006 in [13] that counts Motzkin paths of a given length.
Figure 6: The 9 PMAP of length three in M2. Notice that two paths end on the x-axis,
three paths end at height 1, three paths end at height 2, and one path ends at height 3, which
correspond to the fourth row of T .
522 Ars Math. Contemp. 24 (2024) #P3.08 / 509–534
3.2 PMAP in M′2 - From right to left
Here, we consider the paths of the previous section, but we read them from right to left.
This means that down steps become up steps and vice versa, and horizontal steps are un-
changed, which implies that any up step or horizontal step (except the first step of the path)
is preceded by a down step. See Definition 1.4 and Figure 7 for two examples of such
paths.
Figure 7: The left drawing shows a Motzkin path with air pockets of length 18 in M′2
(read from right to left). The right drawing shows a partial Motzkin path with air pockets
of length 18 ending at height 2 in M′2.
Let P be a length n PMAP in M′2 ending at height k ⩾ 0. If the last step of P is Ua, a ⩾
1, then k ⩾ a and we have P = QUa where Q is a length (n− 1) PMAP ending at height
ℓ = k−a with a down step. So, we obtain the first relation fk = z+z(g0+g1+. . .+gk−1)
for k ⩾ 1, anchored with f0 = 1 by considering the empty path. If the last step of P is a
down step D, then we have P = QD where Q is a length (n− 1) PMAP ending at height
k+1. So, we obtain the second relation gk = zfk+1+zgk+1+zhk+1. If the last step of P
is a horizontal step H , then we have P = QH where Q is a length (n− 1) PMAP ending
at height k with a down step whenever it is nonempty. If k ⩾ 1, then we have hk = zgk; if
k = 0 then we have h0 = z + zg0 where the monomial z corresponds to the path P = H .
So, we have to solve the following system of equations.
f0 = 1, and fk = z(1 + g0 + g1 + . . .+ gk−1), k ⩾ 1,
gk = zfk+1 + zgk+1 + zhk+1, k ⩾ 0,
h0 = z + zg0, and hk = zgk, k ⩾ 1.
(3.2)
Using the same notations as in the previous sections, and summing the recursions in
(3.2), we have:
F (u) = 1 +
∑
k⩾1
ukfk = 1 + z
∑
k⩾1
(1 + g0 + . . .+ gk−1)u
k
= 1 +
zu
1− u
(1 +G(u)),
G(u) = z
∑
k⩾0
(fk+1 + gk+1 + hk+1)u
k
=
z
u
(F (u)− F (0) +G(u)−G(0) +H(u)−H(0)),
H(u) = z + zG(u).
Notice that F (0) = 1 and H(0) = z + zG(0) by the third relation. Now, setting
J.-L. Baril and P. Barry: Two kinds of partial Motzkin paths with air pockets 523
a := G(0) and solving these functional equations, we deduce
F (u) =
a uz3 + a uz2 + uz3 − u2z + uz2 + u2 − uz + z2 − u+ z
u2 − uz + z2 − u+ z
,
G(u) = −z (a uz + a u− a z + uz − a)
u2 − uz + z2 − u+ z
,
H(u) = −
z
(
a uz2 + a uz − a z2 + uz2 − a z − u2 + uz − z2 + u− z
)
u2 − uz + z2 − u+ z
.
In order to compute a = G(0), we use the kernel method on F (u). We can write the
denominator (which is a polynomial in u of degree 2), as (u− r)(u− s) with
r =
1 + z +
√
−3 z2 − 2 z + 1
2
, and s =
1 + z −
√
−3 z2 − 2 z + 1
2
.
Replacing u with s (which have a Taylor expansion at z = 0) in oder to cancel the numer-
ator of F (u), we obtain the equation
a sz3 + a sz2 + sz3 − s2z + sz2 + s2 − sz + z2 − s+ z = 0.
Using rs = z(1 + z), we deduce
a = G(0) =
1− r
r
.
Finally, after simplifying by the factor (u − s) in the numerators and denominators, we
obtain
F (u) =
u(z − 1) + r
r − u
, G(u) =
1− r
r − u
, and H(u) =
z(1− u)
r − u
,
which implies that
[uk]F (u) := fk =
1
rk
+ [k ̸= 0] · z − 1
rk
,
[uk]G(u) := gk =
1− r
rk+1
,
[uk]H(u) := hk =
z
rk+1
− [k ̸= 0] · z
rk
.
Theorem 3.4. The bivariate generating function for the total number of PMAP with respect
to the length and the height of the end-point is given by
Total(z, u) =
1− u+ z
r − u
.
[uk]Total(z, u) =
z + 1
rk+1
− [k ̸= 0] · 1
rk
.
Finally, setting t(n, k) = [zn][uk]Total(z, u), we have for n ⩾ 1, k ⩾ 1,
t(n, k) = t(n, k − 1)− t(n− 1, k) + t(n− 2, k + 1) + t(n− 1, k + 1),
and setting tn := t(n, 0), we have t0 = 1, and for n ⩾ 2,
tn = tn−1 +
n−2∑
k=1
tktn−1−k.
524 Ars Math. Contemp. 24 (2024) #P3.08 / 509–534
Proof. The proof are obtained mutatis mutandis as for the previous theorems.
Let T be the infinite matrix T := [t(n, k)]n⩾0,k⩾0. The first few rows of the matrix T
are
T =

1 0 0 0 0 0 0 · · ·
1 1 1 1 1 1 1 · · ·
1 1 1 1 1 1 1 · · ·
2 3 4 5 6 7 8 · · ·
4 6 8 10 12 14 16 · · ·
9 15 22 30 39 49 60 · · ·
21 36 54 75 99 126 156 · · ·
51 91 142 205 281 371 476 · · ·
...
...
...
...
...
...
...
. . .

.
Since there is an infinite number of PMAP of length n, we do not provide an ordinary
generating function (with respect to the length) for these paths. So, we get around this by
counting PMAP ending on a point (x, n− x) for a given n ⩾ 0.
Corollary 3.5. The generating function that counts the partial PMAP ending on the line
y = n− x is given by
Total(z, z) =
1
r − z
.
The first few terms of the series expansion of Total(z, z) are 1 + z + 2z2 + 4z3 +
9z4+21z5+51z6+127z7+323z8+835z9+O(z10), which correspond to the sequence
A001006 in [13] that counts the Motzkin paths with respect to the length. See Figure 8 for
the illustration of the 9 PMAP in M′2 ending on the line y = 4− x.
Notice that we obviously retrieve the results of Corollary 3.3, i.e., the generating func-
tion Total(z, 0) that counts the MAP with respect to the length is also a shift of the
Motzkin sequence A001006 in [13].
Figure 8: The 9 PMAP ending on the line y = 4 − x in M′2. Notice that four paths end
on the x-axis, three paths end at height 1, one path ends at height 2, and one path ends at
height 3, which correspond to the fifth diagonal of T .
J.-L. Baril and P. Barry: Two kinds of partial Motzkin paths with air pockets 525
4 A Riordan array point of view
In this section, we make links between the previous matrices T = [tn,k]n⩾0,k⩾0 and some
Riordan arrays or almost Riordan arrays. We first give a short background on Riordan
arrays [4, 5, 12].
An infinite column vector (a0, a1, . . . )T has generating function f(z) if f(z) =∑
n⩾0 anz
n. A Riordan array is an infinite lower triangular matrix whose k-th column
has generating function g(z)f(z)k for all k ⩾ 0, for some formal power series g(z) and
f(z), with g(0) ̸= 0, f(0) = 0, and f ′(0) ̸= 0. Such a Riordan array is denoted by
(g(z), f(z)). If we multiply this matrix by a column vector (b0, b1, . . . )T having generat-
ing function h(z) =
∑
n⩾0 bnz
n, then the resulting column vector has generating function
g(z)h(f(z)). This property is known as the fundamental theorem of Riordan arrays.
The product of two Riordan arrays (g(z), f(z)) and (h(z), l(z)) is defined by
(g(z), f(z)) ∗ (h(z), l(z)) = (g(z)h(f(z)), l(f(z))) .
Under the operation “∗”, the set of all Riordan arrays is a group [12]. The identity element
is I = (1, z), and the inverse of (g(z), f(z)) is
(g(z), f(z))−1 =
(
1/
(
g ◦ f
)
(z), f(z)
)
,
where f(z) denotes the compositional inverse of f(z).
Moreover, if a matrix T = [tn,k]n⩾0,k⩾0 is a Riordan array (g(z), f(z)) then tn,k
equals the coefficient of znuk in the series expansion of the bivariate generating function
g(z)
1− uf(z)
,
and we say that this is the bivariate generating function of the matrix T .
An almost Riordan array T ′ is a matrix that consists of an initial column vector
(d0, d1, . . . )
T with generating function g0(z) =
∑
n⩾0 dnz
n, followed by a vertically
shifted Riordan array (g(z), f(z)) as illustrated below.
T ′ =

d0 0 · · · 0 0 · · ·
d1
d2
d3 (g(z), f(z))
d4
d5
...
...
...
...
...
...

Therefore, the bivariate generating function for this matrix is given by
g0(z) + zu
g(z)
1− uf(z)
.
Finally, we will say that a matrix M = [mn,k]n⩾0,k⩾0 is the rectification of the Riordan
array (g(z), f(z)) whenever mn,k equals the coefficient of znuk in the series expansion of
526 Ars Math. Contemp. 24 (2024) #P3.08 / 509–534
the bivariate generating function
g(z)
1− u f(z)z
.
In this section, C(z) = 1−
√
1−4z
2z will be the generating function where the coefficient
of zn in its series expansion is the Catalan number cn = 1n+1
(
2n
n
)
.
4.1 Comment on Section 2.1
Let T = [t(n, k)]n⩾0,k⩾0 be the matrix given in Section 2.1 where t(n, k) is the number
of length n PMAP in M1 ending at height k.
Proposition 4.1. The matrix T = [t(n, k)]n⩾0,k⩾0 is a Riordan array defined by
(
1
1− z2
C
(
z(1− z + z2)
(1− z2)2
)
,
z
1− z2
C
(
z(1− z + z2)
(1− z2)2
))
.
Proof. Considering r defined in Section 2.1, we have
[uk]Total(z, u) =
1
zrk+1
=
(
1
zr
)(
1
r
)k
=
2
1− z2 +
√
1− 4z + 2z2 − 4z3 + z4
(
2z
1− z2 +
√
1− 4z + 2z2 − 4z3 + z4
)k
.
Therefore, the array T satisfies
T =
(
2
1− z2 +
√
1− 4z + 2z2 − 4z3 + z4
,
2z
1− z2 +
√
1− 4z + 2z2 − 4z3 + z4
)
=
(
1− z2 −
√
1− 4z + 2z2 − 4z3 + z4
2z(1− z + z2)
,
1− z2 −
√
1− 4z + 2z2 − 4z3 + z4
2(1− z + z2)
)
=
(
1
1− z2
C
(
z(1− z + z2)
(1− z2)2
)
,
z
1− z2
C
(
z(1− z + z2)
(1− z2)2
))
.
Proposition 4.2. The general term t(n, k) equals
n−k∑
i=0
(
k + n−k−i2
n−k−i
2
)
1 + (−1)n−k−i
2
i∑
j=0
Ck+j,k
j∑
m=0
(
j
m
)
(−1)m
m∑
p=0
(
m
p
)
(−1)p
(
2j − 1 + i−j−m−p2
i−j−m−p
2
)
1 + (−1)i−j−m−p
2
,
where Cn,k = k+1n+1
(
2n−k
n−k
)
is the general term of the Catalan matrix (A033184 in [13]).
J.-L. Baril and P. Barry: Two kinds of partial Motzkin paths with air pockets 527
Proof. Setting Z = z(1−z+z
2)
(1−z2)2 , we have
t(n, k) = [zn]zk
1
(1− z2)k+1
C(Z)k+1
= [zn−k]
1
(1− z2)k+1
C(Z)k+1
=
n−k∑
i=0
[zn−k−i]
1
(1− z2)k+1
[zi]C(Z)k+1 (product rule [8])
=
n−k∑
i=0
[zn−k−i]
1
(1− z2)k+1
i∑
j=0
[zj ]C(z)k+1[zi]Zj (composition rule [8])
=
n−k∑
i=0
[zn−k−i]
1
(1− z2)k+1
i∑
j=0
[zj ]
1
zk
C(z)(zC(z))k[zi]Zj
=
n−k∑
i=0
(
k + n−k−i2
n−k−i
2
)
1 + (−1)n−k−i
2
i∑
j=0
Ck+j,k[z
i]Zj (see [8]).
Since we have
[zi]Zj = [zi]
(
z(1− z + z2)
(1− z2)2
)j
=
j∑
m=0
(
j
m
)
(−1)m
m∑
p=0
(
m
p
)
(−1)p
(
2j − 1 + i−j−m−p2
i−j−m−p
2
)
1 + (−1)i−j−m−p
2
,
the result follows.
4.2 Comment on Section 2.2
Let T = [t(n, k)]n⩾0,k⩾0 be the matrix given in Section 2.2 where t(n, k) is the number
of length n PMAP in M′1 ending at height k.
Proposition 4.3. The matrix T = [t(n, k)]n⩾0,k⩾0 can be written
T =

1 0 0 0 0 0 · · ·
1 1 1 1 1 1 · · ·
2 3 3 3 3 3 · · ·
5 8 10 12 14 16 · · ·
13 23 33 43 53 63 · · ·
36 69 107 149 195 245 · · ·
...
...
...
...
...
...
. . .

= A ·B
528 Ars Math. Contemp. 24 (2024) #P3.08 / 509–534
where
A =

1 0 0 0 0 0 · · ·
1 1 0 0 0 0 · · ·
2 3 0 0 0 0 · · ·
5 8 2 0 0 0 · · ·
13 23 10 0 0 0 · · ·
36 69 38 4 0 0 · · ·
...
...
...
...
...
...
. . .

and B =

1 0 0 0 0 0 · · ·
0 1 1 1 1 1 · · ·
0 0 1 2 3 4 · · ·
0 0 0 1 3 6 · · ·
0 0 0 0 1 4 · · ·
0 0 0 0 0 1 · · ·
...
...
...
...
...
...
. . .

are defined as follows:
• The matrix B = [bn,k]n,k⩾0 is defined by b0,0 = 1, and bn,0 = b0,n = 0 if n ⩾ 1,
and bn,k =
(
k−1
n−1
)
otherwise, which is a kind of Pascal matrix.
• The matrix A = [an,k]n,k⩾0 is the almost Riordan array with initial column of gen-
erating function
g0(z) =
1− z2 −
√
1− 4z + 2z2 − 4z3 + z4
2z(1− z + z2)
,
followed by the shifted Riordan array(
1− 3z + z2 − z3 − (1− z)
√
1− 4z + 2z2 − 4z3 + z4
2z3(1− z + z2)
,
1− 2z − z2 −
√
1− 4z + 2z2 − 4z3 + z4
2z
)
.
The second column of T 1, 3, 8, 23, 69, . . . with generating function
1− 3z + z2 − z3 − (1− z)
√
1− 4z + 2z2 − 4z3 + z4
2z3(1− z + z2)
is the convolution of the first column of T 1, 1, 2, 5, 13, 36, . . . (A114465 in [13]) and the
sequence 1, 2, 4, 10, 28, . . . (A187256 in [13]).
Proof. An almost Riordan array is represented by an initial column vector with generating
function g0(z), followed by a vertically shifted Riordan array (g(z), f(z)). The bivariate
generating function of this matrix is then given by g0(z) + zu
g(z)
1−uf(z) . In our case, for the
almost Riordan array A, we have
g0(z) =
1− z2 −
√
1− 4z + 2z2 − 4z3 + z4
2z(1− z + z2)
,
g(z) =
1− 3z + z2 − z3 − (1− z)
√
1− 4z + 2z2 − 4z3 + z4
2z3(1− z + z2)
,
f(z) =
1− 2z − z2 −
√
1− 4z + 2z2 − 4z3 + z4
2z
.
J.-L. Baril and P. Barry: Two kinds of partial Motzkin paths with air pockets 529
We let G(z, u) = g0(z) + zu
g(z)
1−uf(z) , the bivariate generating function of the almost Rior-
dan array. Now, it suffices to check that the generating function corresponding to the matrix
A ·B, that is
G(z,
u
1− u
) =
(1− u+ uz)(1− 2z + 2uz − (1 + 2u)z2 −
√
1− 4z + 2z2 − 4z3 + z4
2(1− z + z2)(u(u− 1) + (1− u2)z + u(u+ 1)z2)
,
coincides with the generating function Total(z, u) of the matrix T .
Proposition 4.4. The matrix
[t(n, k)]n⩾1,k⩾1 =

1 1 1 1 1 · · ·
3 3 3 3 3 · · ·
8 10 12 14 16 · · ·
23 33 43 53 63 · · ·
69 107 149 195 245 · · ·
...
...
...
...
...
. . .

is the rectification of the Riordan array (g(z), h(z)) with
h(z) =
1− z2 −
√
1− 4z + 2z2 − 4z3 + z4
2
, and
g(z) =
1− 3z + z2 − z3 − (1− z)
√
1− 4z + 2z2 − 4z3 + z4
2z3(1− z + z2)
.
Proof. It suffices to check that the generating function of T , i.e. Total(z, u), given in
Theorem 2.4, equals to
Total(z, 0) + zu
g(z)
1− uh(z)z
.
We can express h(z) and g(z), respectively, in the following form
h(z) =
z(1− z + z2)
1− z2
C
(
z(1− z + z2)
(1− z2)2
)
,
g(z) =
1
1− 3z + z2 − z3
C
(
z3(1− z + z2)
(1− 3z + z2 − z3)2
)
.
Then g(z) expands to give the first column 1, 3, 8, 23, . . ., whose n-th term vn can be
expressed
vn =
n∑
k=0
k∑
j=0
(
k
j
)
(−1)j
j∑
i=0
(
j
i
)
(−1)i
n−3k−j−i∑
ℓ=0
(
2k + ℓ
ℓ
) ℓ∑
m=0
(
ℓ
m
)
3ℓ−m(−1)m
(
m
n− 3k − j − i− ℓ−m
)
(−1)n−3k−j−i−ℓ−mck.
Using vn, we can deduce the following.
530 Ars Math. Contemp. 24 (2024) #P3.08 / 509–534
Proposition 4.5. The general term t(n, k) equals
n+k∑
i=0
vn+k−i
i∑
j=0
j∑
m=0
Mm,k
m∑
p=0
(
m
p
)
(−1)p
p∑
q=0
(
p
q
)
(−1)q
(
2m− 1 + j−m−p−q2
j−m−p−q
2
)
1 + (−1)j−m−p−q
2
(
k
i−j
2
)
(−1)
i−j
2
1 + (−1)i−j
2
,
where
Mn,k =
{
[k = 0] if n = 0,
n
k
(
2n−k−1
n−k
)
otherwise,
is the general term of Riordan array (1, zC(z)) (see A106566 in [13]).
4.3 Comment on Section 3.1
Let T = [t(n, k)]n⩾0,k⩾0 be the matrix given in Section 3.1 where t(n, k) is the number
of length n PMAP in M2 ending at height k.
Proposition 4.6. The matrix T = [t(n, k)]n⩾0,k⩾0 is the Riordan array
T = (1 + zM(z), z(1 + zM(z)))
=
(
C
(
z
1 + z
)
, zC
(
z
1 + z
))
,
where M(z) = 1−z−
√
1−2z−3z2
2z2 is the generating function of the Motzkin numbers (see
A001006 in [13]).
Proof. It suffices to check that Total(z, u) given in Theorem 3.1 satisfies
Total(z, u) =
C
(
z
1+z
)
1− uzC
(
z
1+z
) ,
and that 1 + zM(z) = C
(
z
1+z
)
.
This triangle T corresponds to A091836 in [13] where the coefficient of row n− 1 and
column k is the number of Motzkin paths of length n having k points on the horizontal axis
(besides the first and last point). As mentioned in [13], we obtain
t(n, k) =

1, if n = k,
k+1
n+1
n−k∑
j=1
j(−1)n−k−j
(
n+j
j
) n−k∑
i=0
1
n−k
(
i
n−k−i+j
)(
n−k
i
)
, otherwise.
A second expression for t(n, k) is given by the following proposition.
Proposition 4.7. The general term t(n, k) of the Riordan array (1+zM(z), z(1+zM(z)))
is given by
t(n, k) =
{
1 if n = k,
k+1
n−k
∑k
j=0
(
k
j
)∑n−k
i=0
(
n−k
i
)(
i
n−k−i−j−1
)
otherwise.
J.-L. Baril and P. Barry: Two kinds of partial Motzkin paths with air pockets 531
Proof. We prove this using Lagrange inversion, using the fact that
z
1 + z + z2
M
(
z
1 + z + z2
)
= z,
which means that the compositional inverse (zM(z))−1 of zM(z) is
(zM(z))−1 =
z
1 + z + z2
.
Thus we have
t(n, k) = [zn](1 + zM(z))(z(1 + zM(z)))k
= [zn−k](1 + zM(z))k+1
= [zn−k]G(zM(z)), with G(z) = (1 + z)k+1
=
1
n− k
[zn−k−1]G′(z)
(
z
(zM(z))−1
)n−k
(Lagrange inversion)
=
1
n− k
[zn−k−1](k + 1)(1 + z)k(1 + z + z2)n−k
=
k + 1
n− k
[zn−k]
k∑
j=0
(
k
j
)
zj
n−k∑
i=0
(
n− k
i
)
zi(1 + z)i
=
k + 1
n− k
[zn−k]
k∑
j=0
(
k
j
)
zj
n−k∑
i=0
(
n− k
i
)
zi
i∑
ℓ=0
(
i
ℓ
)
zℓ
=
k + 1
n− k
k∑
j=0
(
k
j
) n−k∑
i=0
(
n− k
i
)(
i
n− k − i− j − 1
)
.
Remark 4.8. The Riordan array (1 + zM(z), z(1 + zM(z))) is a pseudo-involution in
the Riordan group (see [5, Example 8]), that is, the matrix [(−1)ktn,k]n,k⩾0 is idempotent.
Thus, this work yields a significant lattice path interpretation of this array.
4.4 Comment on Section 3.2
Let T = [t(n, k)]n⩾0,k⩾0 be the matrix given in Section 3.2 where t(n, k) is the number
of length n PMAP in M′2 ending at height k.
Proposition 4.9. The matrix T = [t(n, k)]n⩾0,k⩾0 can be written
T =

1 0 0 0 0 0 · · ·
1 1 1 1 1 1 · · ·
1 1 1 1 1 1 · · ·
2 3 4 5 6 7 · · ·
4 6 8 10 12 14 · · ·
9 15 22 30 39 49 · · ·
...
...
...
...
...
...
. . .

= A ·B
532 Ars Math. Contemp. 24 (2024) #P3.08 / 509–534
where
A =

1 0 0 0 0 0 · · ·
1 1 0 0 0 0 · · ·
1 1 0 0 0 0 · · ·
2 3 1 0 0 0 · · ·
4 6 2 0 0 0 · · ·
9 15 7 1 0 0 · · ·
...
...
...
...
...
...
. . .

and B =

1 0 0 0 0 0 · · ·
0 1 1 1 1 1 · · ·
0 0 1 2 3 4 · · ·
0 0 0 1 3 6 · · ·
0 0 0 0 1 4 · · ·
0 0 0 0 0 1 · · ·
...
...
...
...
...
...
. . .

are defined as follows:
• The matrix B = [bn,k]n,k⩾0 is defined by b0,0 = 1, and bn,0 = b0,n = 0 if n ⩾ 1,
and bn,k =
(
k−1
n−1
)
otherwise, which is the same as in Proposition 4.3.
• The matrix A = [an,k]n,k⩾0 is the almost Riordan array with initial column whose
generating function is
g0(z) = 1 + zM(z) =
1 + z −
√
1− 2z − 3z2
2z
,
which is followed by the shifted Riordan array (g(z), z2g(z)) where
g(z) =
1− z − 2z2 −
√
1− 2z − 3z2
2z3(1 + z)
.
Proof. The almost Riordan array A has generating function
g0(z) + zu
g(z)
1− z2ug(z)
.
Setting G(z, u) = g0(z) + zu
g(z)
1−z2ug(z) , it suffices to check that the generating function
corresponding to the matrix A·B, that is G(z, u1−u ), coincides with the generating function
Total(z, u) of the matrix T given in Theorem 3.4.
Proposition 4.10. The matrix
[t(n, k)]n⩾1,k⩾1 =

1 1 1 1 1 · · ·
1 1 1 1 1 · · ·
3 4 5 6 7 · · ·
6 8 10 12 14 · · ·
15 22 30 39 49 · · ·
36 54 75 99 126 · · ·
...
...
...
...
...
. . .

is the rectification of the Riordan array (M(z), zR(z)) where M(z) = 1−z−
√
1−2z−3z2
2z2
is the generating function of the Motzkin numbers, and R(z) = 1+z−
√
1−2z−3z2
2z(1+z) is the
generating function of the Riordan numbers (A005043 in [13]).
J.-L. Baril and P. Barry: Two kinds of partial Motzkin paths with air pockets 533
Proof. It suffices to check that the generating function of T , i.e. Total(z, u), given in
Theorem 3.4, equals to g0(z) + zu
M(z)
1−u zR(z)z
.
We let mn denote the n-th Motzkin number mn =
∑⌊n2 ⌋
k=0
(
n
2k
)
ck where ck is the k-th
Catalan defined above.
Corollary 4.11. We have
t(n, k) =
{
[k = 0] if n = 0,
r(n− 1, k) otherwise,
where
r(n, k) =
n∑
i=0
mi · (k + [n = i])
n+ k − i+ [n = i]
n−i∑
j=0
(−1)n−i−j
(
n+ k − i+ j − 1
j
) n+k−i∑
ℓ=0
(
n+ k − i
ℓ
)(
ℓ
n− i− j − ℓ
)
.
Proof. We have (M(z), zR(z))−1 =
(
(1−x)2
1−x+x2 ,
x(1−x)
1−x+x2
)
. If we denote by (v(z), u(z))
this inverse Riordan array, then we obtain
(M(z), zR(z)) =
(
1
v(ū(z))
, ū(z)
)
,
where ū is the compositional inverse [8] of u. Using the definition of a Riordan array, and
Lagrange inversion, we find that the Riordan array (M(z), zR(z)) has general term r̃(n, k)
given by
n∑
i=0
mi · (k + [n = k + i])
n− i+ [n = k + i]
n−k−i∑
j=0
(−1)n−k−i−j
(
n− i+ j − 1
j
) n−i∑
ℓ=0
(
n− i
ℓ
)(
ℓ
n− k − i− j − ℓ
)
.
This array begins as follows:
1 0 0 0 0 0 · · ·
1 1 0 0 0 0 · · ·
2 1 1 0 0 0 · · ·
4 3 1 1 0 0 · · ·
9 6 4 1 1 0 · · ·
21 15 8 5 1 1 · · ·
...
...
...
...
...
...
. . .

.
To rectify this array and thus to obtain the array (M(z), R(z)), we change n to n+ k, and
we obtain r(n, k) above. To this array we must now prepend the row (1, 0, 0, 0, . . .), and
the result follows.
ORCID iDs
Jean-Luc Baril https://orcid.org/0000-0002-9811-8028
Paul Barry https://orcid.org/0000-0003-1909-8725
534 Ars Math. Contemp. 24 (2024) #P3.08 / 509–534
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ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 24 (2024) #P3.09 / 535–551
https://doi.org/10.26493/1855-3974.3113.6f7
(Also available at http://amc-journal.eu)
Independent coalition in graphs:
existence and characterization
Mohammad Reza Samadzadeh , Doost Ali Mojdeh *
Department of Mathematics, Faculty of Mathematical Sciences,
University of Mazandaran, Babolsar, Iran
Received 2 May 2023, accepted 3 November 2023, published online 23 August 2024
Abstract
An independent coalition in a graph G consists of two disjoint sets of vertices V1 and
V2 neither of which is an independent dominating set but whose union V1 ∪ V2 is an inde-
pendent dominating set. An independent coalition partition, abbreviated, ic-partition, in a
graph G is a vertex partition π = {V1, V2, . . . , Vk} such that each set Vi of π either is a
singleton dominating set, or is not an independent dominating set but forms an independent
coalition with another set Vj ∈ π. The maximum number of classes of an ic-partition of
G is the independent coalition number of G, denoted by IC(G). In this paper, we study
the concept of ic-partition. In particular, we discuss the possibility of the existence of ic-
partitions in graphs and introduce a family of graphs for which no ic-partition exists. We
also determine the independent coalition number of some classes of graphs and investi-
gate graphs G of order n with IC(G) ∈ {1, 2, 3, 4, n} and the trees T of order n with
IC(T ) = n− 1.
Keywords: Independent coalition, independent coalition partition, independent dominating set, ido-
matic partition.
Math. Subj. Class. (2020): 05C69
1 Introduction
Let G = (V,E) denote a simple graph of order n with vertex set V = V (G) and edge set
E = E(G). The open neighborhood of a vertex v ∈ V is the set N(v) = {u|{u, v} ∈ E},
and its closed neighborhood is the set N [v] = N(v)∪{v}. Each vertex of N(v) is called a
neighbor of v, and the cardinality of N(v) is called the degree of v, denoted by deg(v) or
*Corresponding author.
E-mail addresses: m.samadzadeh02@umail.umz.ac.ir (Mohammad Reza Samadzadeh),
damojdeh@umz.ac.ir (Doost Ali Mojdeh)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
536 Ars Math. Contemp. 24 (2024) #P3.09 / 535–551
degG(v). A vertex v of degree 1 is called a pendant vertex or leaf, and its neighbor is called
a support vertex. A vertex of degree n− 1 is called a full vertex while a vertex of degree 0
is called an isolated vertex. The minimum and maximum degree of G are denoted by δ(G)
and ∆(G), respectively. For a set S of vertices of G, the subgraph induced by S is denoted
by G[S]. For two sets X and Y of vertices, let [X,Y ] denote the set of edges between X
and Y . If every vertex of X is adjacent to every vertex of Y , we say that [X,Y ] is full,
while if there are no edges between them, we say that [X,Y ] is empty. A subset Vi ⊆ V
is called a singleton set if |Vi| = 1, and is called a non-singleton set if |Vi| ≥ 2. The join
G+H of two disjoint graphs G and H is the graph obtained from the union of G and H by
adding every possible edge between the vertices of G and the vertices of H . We denote the
family of paths, cycles, complete graphs and stars of order n by Pn, Cn, Kn and K1,n−1,
respectively, and the complete k-partite graph with partite sets of order n1, n2, . . . , nk, by
Kn1,...,nk . A double star with respectively p and q leaves connected to each support vertex
is denoted by Sp,q . The complete graph K3 is called a triangle, and a graph is triangle-free
if it has no K3 as an induced subgraph. The girth of a graph with a cycle is the length of its
shortest cycle. For a graph G, the girth of G is denoted by g(G). For a graph G of order n,
let G denote the complement of G with V (G) = V (G) and E(G) = E(Kn)−E(G) [13].
A set S ⊆ V in a graph G = (V,E) is called a dominating set if every vertex v ∈ V
is either an element of S or is adjacent to an element of S. A set S ⊆ V is called an
independent set if its vertices are pairwise nonadjacent. The vertex independence number,
denoted by α(G), is the maximum cardinality of an independent set of G. An independent
dominating set in a graph G is a set which is both independent and dominating.
A partition of the vertices of G into dominating sets (independent dominating sets)
is called a domatic partition (idomatic partition). The maximum number of classes of a
domatic partition (idomatic partition) of G is called the domatic number (idomatic number)
of G, denoted by d(G) (id(G)). The concepts of domination and domatic partition and their
variations have been studied widely in the literature. See, for example, [1, 2, 3, 4, 5, 6, 12].
The term coalition was introduced by Haynes et al, [7] and has been studied further
in [8, 9, 10, 11].
Definition 1.1 ([7]). A coalition in a graph G consists of two disjoint sets of vertices
V1, V2 ⊂ V , neither of which is a dominating set but whose union V1 ∪ V2 is a dominating
set. We say that the sets V1 and V2 form a coalition, and are coalition partners.
Definition 1.2 ([7]). A coalition partition, henceforth called a c-partition, in a graph G
is a vertex partition π = {V1, V2, . . . , Vk} such that every set Vi of π is either a singleton
dominating set, or is not a dominating set but forms a coalition with another set Vj in π. The
coalition number C(G) equals the maximum order k of a c-partition of G, and a c-partition
of G having order C(G) is called a C(G)-partition.
Herein we will focus on coalitions involving independent dominating sets in graphs. In
other words, we will study the concepts of independent coalition and independent coalition
partition which have been introduced in [7] as an area for future research. We begin with
the following definitions.
Definition 1.3. An independent coalition in a graph G consists of two disjoint sets of
independent vertices V1 and V2, neither of which is an independent dominating set but
whose union V1 ∪ V2 is an independent dominating set. We say the sets V1 and V2 form an
independent coalition, and are independent coalition partners (or ic-partners).
M. R. Samadzadeh and D. A. Mojdeh: Independent coalition in graphs: existence and . . . 537
Definition 1.4. An independent coalition partition, abbreviated ic-partition, in a graph G
is a vertex partition π = {V1, V2, . . . , Vk} such that every set Vi of π is either a singleton
dominating set, or is not an independent dominating set but forms an independent coalition
with another set Vj ∈ π. The independent coalition number IC(G) equals the maximum
number of classes of an ic-partition of G, and an ic-partition of G having order IC(G) is
called an IC(G)-partition.
Definition 1.5 ([8]). Let G be a graph of order n with vertex set V = {v1, v2, . . . , vn}.
The singleton partition, denoted π1, of G is the partition of V into n singleton sets, that is,
π1 = {{v1}, {v2}, . . . , {vn}}.
This paper is organized as follows. Section 1 is devoted to terminology and definitions.
We discuss the possibility of the existence of ic-partitions in graphs and derive some bounds
on independent coalition number in Section 2. In Section 3, we determine the independent
coalition number of some classes of graphs. The graphs G with IC(G) ∈ {1, 2, 3, 4}
are investigated in Section 4. In Section 5, we characterize triangle-free graphs G with
IC(G) = n and trees T with IC(T ) = n − 1. Finally, we end the paper with some
research problems.
2 Independent coalition partition: existence and bound
This section is divided into two subsections. In the first subsection, we show that not all
graphs admit an ic-partition, and in the second subsection, we present some bounds on
IC(G) whenever the graph G admits an ic-partition.
2.1 Existence
In the following definition, we construct graphs with arbitrarily large order for which no
ic-partition exists.
Definition 2.1. Let B be the set of all graphs obtained from the complete graph Kn, (n ≥ 4)
with the vertices vi, (1 ≤ i ≤ n), and two additional vertices vn+1, vn+2 such that vn+1
and vn+2 are adjacent to vn, and vn+1 is adjacent to vn−1. Figure 1 illustrates such a graph
for n = 4.
v1
v2 v3
v4
v5 v6
Figure 1: The graph G in B for n = 4.
Proposition 2.2. Let G be a graph. If G ∈ B, then G has no ic-partition.
Proof. Suppose, to the contrary, that G has an ic-partition π. The vertices v1, v2, . . . , vn−1
are pairwise adjacent, so they must be in different classes. Further, vn is a full vertex, so
it must be in a singleton class. Since vn−1 is adjacent to all vertices except vn+2, and
538 Ars Math. Contemp. 24 (2024) #P3.09 / 535–551
{vn−1, vn+2} dominates G, it follows that {vn−1} ∈ π. Further, since {vn−1} can only
form an independent coalition with {vn+2}, it follows that {vn+2} ∈ π. If {vn+1} ∈ π,
then π is a singleton partition. In this case, {vn+1} has no ic-partner, a contradiction.
Hence, {vn+1} /∈ π. It follows that π consists of a non-singleton set {vn+1, vi} such
that vi ∈ {v1, v2, . . . , vn−2}, and n singleton sets. Assume, without loss of generality,
that {vn+1, v1} ∈ π. Now for each 2 ≤ i ≤ n − 2, the set {vi} has no ic-partner, a
contradiction.
2.2 Bounds
Definition 1.4 implies that an ic-partition of a graph G is also a c-partition. Further, we
note that an ic-partition of G is a proper coloring as well. Hence, we have the following
two sharp bounds on IC(G). To see the sharpness of them, consider the complete graph
Kn.
Observation 2.3. Let G be a graph. If G has an ic-partition, then IC(G) ≤ C(G).
Furthermore, this bound is sharp.
Observation 2.4. Let G be a graph. If G has an ic-partition, then IC(G) ≥ χ(G). Fur-
thermore, this bound is sharp.
Given a connected graph G and an ic-partition π of it, the following theorem shows
that each set in π admits at most ∆(G) ic-partners.
Theorem 2.5. Let G be a connected graph with maximum degree ∆(G), and let π be an ic-
partition of G. If X ∈ π, then X is in at most ∆(G) independent coalitions. Furthermore,
this bound is sharp.
Proof. Let π be an ic-partition of G, and let X be a set in π. If X is a dominating set,
then it has no ic-partner. Hence, we may assume that X does not dominate G. Let x
be a vertex that is not dominated by X . Now every ic-partner of X must dominate x,
that is, it must contain a vertex in N [x]. Hence, there are at most |N [x]| ≤ ∆(G) + 1
sets in π that can form an independent coalition with X . Now we show that X cannot
form an independent coalition with ∆(G) + 1 sets. Suppose, to the contrary, that X has
∆(G) + 1 ic-partners (name V1, V2, . . . , V∆+1). Consequently, [X,Vi] is empty for each
1 ≤ i ≤ ∆(G) + 1. Let U =
⋃∆(G)+1
i=1 Vi, and G
′ = G[U ]. Consider an arbitrary vertex
v ∈ U (say v ∈ Vi) and an arbitrary set Vj such that 1 ≤ j ≤ ∆(G) + 1 and j ̸= i.
Since X ∪ Vj dominates G and [X,Vi] is empty, it follows that v has a neighbor in Vj .
Choosing Vj arbitrarily, we conclude that degG′(v) ≥ ∆(G), and so degG′(v) = ∆(G).
Hence, for each v ∈ U , we have degG′(v) = ∆(G). Now since G is connected, there
is a path P = (v0, v1, . . . , vk) connecting U to X such that v0 ∈ U and vk ∈ X . Note
that [U,X] is empty, and so V (P ) \ (U ∪ X) ̸= ∅. Let i be the smallest index for which
vi /∈ U ∪ X . It follows that vi−1 ∈ U , and so degG′(vi−1) = ∆(G). Thus, we have
degG(vi−1) ≥ degG′(vi−1) + 1 = ∆(G) + 1, a contradiction.
To prove the sharpness, let G be the graph that is obtained from the complete graph
Kn with vertices vi, (1 ≤ i ≤ n), and a path P2 = (a, b), where b is adjacent to v1. Let
A = {a}, B = {b} and Vi = {vi}, for 1 ≤ i ≤ n. One can observe that ∆(G) = n and
that the singleton partition π1 = {V1, V2, . . . , Vn, A,B} is an ic-partition of G such that A
forms an independent coalition with Vi, for each 1 ≤ i ≤ n. This completes the proof.
M. R. Samadzadeh and D. A. Mojdeh: Independent coalition in graphs: existence and . . . 539
Note that the bound presented in Theorem 2.5 does not hold for disconnected graphs.
As a counterexample, consider the graph G = K2 ∪ K2 and the singleton partition π1 of
it. On can verify that π1 is an ic-partition of G such that each set in π1 has two ic-partners,
while ∆(G) = 1.
The next bound relates independent coalition number of a graph to its idomatic number.
As we will see in the proof of Theorem 2.6, any graph admitting an idomatic partition has
an ic-partition. However, the converse is not necessarily true. For example, the singleton
partition of the cycle C5 is an ic-partition of it, while C5 has no idomatic partition. Or the
cycle C7 has the ic-partition π = {{v1, v5}, {v2, v4}, {v3}, {v6}, {v7}}, while it has no
idomatic partition.
Theorem 2.6. Let G be a connected graph, and let r ≥ 0 be the number of full vertices of
G. If G admits an idomatic partition, then IC(G) ≥ 2id(G)− r.
Proof. Let F = {v1, v2, . . . , vr} be the set of full vertices of G, and let π = {V1, V2, . . . ,
Vid(G)} be an idomatic partition of G of order id(G). Note that each full vertex must
be in a singleton set of π. Without loss of generality, assume that vi ∈ Vi, for each
1 ≤ i ≤ r. It follows that for each r + 1 ≤ i ≤ k, we have |Vi| ≥ 2. Now for
each r + 1 ≤ i ≤ k, we partition Vi into two nonempty subsets Vi,1 and Vi,2. Note
that no proper subset of Vi is a dominating set. Thus, neither Vi,1 nor Vi,2 is an in-
dependent dominating set, and so Vi,1 and Vi,2 are ic-partners. It follows that the par-
tition π′ = {V1, V2, . . . , Vr, Vr+1,1, Vr+1,2, Vr+2,1, Vr+2,2, . . . , Vid(G),1, Vid(G),2} is an
ic-partition of G of order 2id(G)− r. Hence, IC(G) ≥ 2id(G)− r.
3 Independent coalition number for some classes of graphs
Let us begin this section with some routine results.
Observation 3.1. For n ≥ 1, we have IC(Kn) = n.
Observation 3.2. For n ≥ 3, we have IC(K1,n−1) = 3.
Observation 3.3. For p, q ≥ 1, we have IC(Sp,q) = 4.
For complete multipartite graphs, the following result is obtained.
Proposition 3.4. Let G = Kn1,n2,...,nk be a complete k-partite graph with m ≥ 0 full
vertices (m partite sets of cardinality 1). Then IC(G) = 2k −m.
Proof. Let π = {V1, V2, . . . , Vk} be the partition of G into its partite sets. Assume, without
loss of generality, that the sets Vi, for 1 ≤ i ≤ m, are those containing full vertices. Now
for each m+ 1 ≤ i ≤ k, we partition Vi into two sets Vi,1 and Vi,2. Observe that Vi,1 and
Vi,2 are ic-partners, and so the partition π′ = {{V1}, {V2}, . . . , {Vm}, {Vm+1,1, Vm+1,2},
{Vm+2,1, Vm+2,2}, . . . , {Vk,1, Vk,2}} is an ic-partition of G of order 2k − m. Thus,
IC(G) ≥ 2k −m. Now let π′′ be an ic-partition of G. We note that π′′ has the following
properties:
• For any set S ∈ π′′, all vertices in S are in the same partite set of G.
• For any set Vi ∈ π, the vertices in Vi are in at most two sets of π′′.
Hence, we have IC(G) ≤ 2k −m, and so IC(G) = 2k −m.
540 Ars Math. Contemp. 24 (2024) #P3.09 / 535–551
Next we determine the independent coalition number of all paths and cycles.
Lemma 3.5 ([7]). For any path Pn, C(Pn) ≤ 6.
Theorem 3.6. For the path Pn,
IC(Pn) =

n if n ≤ 4;
4 if n = 5;
5 if n = 6, 7, 8, 9;
6 if n ≥ 10.
Proof. It is clear that for 1 ≤ n ≤ 4, we have IC(Pn) = n. Now let n = 5. Con-
sider the path P5 with V (P5) = {v1, v2, v3, v4, v5}. It is easily seen that IC(P5) ̸= 5.
Thus, IC(P5) ≤ 4. The partition {{v1, v3}, {v2}, {v4}, {v5}} is an ic-partition of P5, so
IC(P5) = 4. Now assume n = 6. Consider the path P6 with V (P6) = {v1, v2, v3, v4,
v5, v6}. It is clear that IC(P6) ̸= 6. The partition {{v1, v6}, {v2}, {v3}, {v4}, {v5}}
is an ic-partition of P6, so IC(P6) = 5. Next assume n = 7. Consider the path P7
with V (P7) = {v1, v2, v3, v4, v5, v6, v7}. By Lemma 3.5 and Observation 2.3, we have
IC(Pn) ≤ 6. Now we show that IC(P7) ̸= 6. Suppose that IC(P7) = 6. Let π be an
IC(P7)-partition. We note that π consists of a set (name A) of cardinality 2 and five sin-
gleton sets. Since γi(P7) = 3, each singleton set must be an ic-partner of A. On the other
hand, Theorem 2.5 implies that A has at most two ic-partners, a contradiction. The parti-
tion {{v1, v6}, {v2, v7}, {v3}, {v4}, {v5}} is an ic-partition of P7. Therefore, IC(P7) = 5.
Next we assume n = 8. Consider the path P8 with V (P8) = {v1, v2, v3, v4, v5, v6, v7, v8}.
By Lemma 3.5 and Observation 2.3, we have IC(Pn) ≤ 6. Now we show that IC(P8) ̸=
6. Suppose that IC(P8) = 6. Let π be an IC(P8)-partition. We consider two cases.
Case 1. π consists of a set (name A) of cardinality 3 and five singleton sets. Since
γi(P8) = 3, each singleton set must be an ic-partner of A. On the other hand, Theorem 2.5
implies that A has at most two ic-partners, a contradiction.
Case 2. π consists of two sets of cardinality 2 and four singleton sets. Since γi(P8) =
3, each singleton set must be an ic-partner of a set of cardinality 2. Therefore, using
Theorem 2.5, we deduce that for any two ic-partners C and D, it holds that |C∪D| = 3. On
the other hand, v3 and v6 are not present in any independent dominating set of cardinality
3, a contradiction.
The partition {{v1, v3, v6}, {v2, v7}, {v8}, {v4}, {v5}} is an ic-partition of P8. Therefore,
IC(P8) = 5.
Now let n = 9. Consider the path P9 with V (P9) = {v1, v2, v3, v4, v5, v6, v7, v8, v9}.
By Lemma 3.5 and Observation 2.3, we have IC(Pn) ≤ 6. Now we show that IC(P9) ̸=
6. Suppose that IC(P9) = 6. Let π be an IC(P9)-partition. There exist three cases.
Case 1. π consists of a set (name A) of cardinality 4 and five singleton sets. Since
γi(P9) = 3, each singleton set must be an ic-partner of A. On the other hand, by Theo-
rem 2.5, A has at most two ic-partners, a contradiction.
Case 2. π consists of a set (name A) of cardinality 3, a set (name B) of cardinality
2 and four singleton sets. Since γi(P9) = 3, no two singleton sets in π are ic-partners.
Furthermore, by Theorem 2.5, A has at most two ic-partners, so at least two singleton
sets of π must be ic-partners of B, which is impossible, as P9 has a unique independent
dominating set of cardinality 3.
M. R. Samadzadeh and D. A. Mojdeh: Independent coalition in graphs: existence and . . . 541
Case 3. π consists of three sets of cardinality 2, and three singleton sets. We note that
each singleton set in π must be an ic-partner of a set of cardinality 2, which is impossible,
as P9 has a unique independent dominating set of cardinality 3.
The partition {{v1, v3, v5}, {v2, v4, v9}, {v6}, {v7}, {v8}} is an ic-partition of P9.
Therefore, IC(P9) = 5.
Finally, let n ≥ 10. Consider the path Pn with V (Pn) = {v1, v2, . . . , vn}. By
Lemma 3.5 and Observation 2.3, we have IC(Pn) ≤ 6. Now we consider the sets
V1 = {v1, v6} ∪ {v2n−1 : n ≥ 5}, V2 = {v2, v5} ∪ {v2n : n ≥ 5}, V3 = {v3},
V4 = {v4}, V5 = {v7}, V6 = {v8}. Then π = {V1, V2, V3, V4, V5, V6} is an ic-partition
of Pn, where V3 and V4 are ic-partners of V1, and V5 and V6 are ic-partners of V2. So the
proof is complete.
Lemma 3.7 ([7]). For any cycle Cn, C(Cn) ≤ 6.
Lemma 3.7 and Observation 2.3 imply the following result.
Lemma 3.8. For any cycle Cn, IC(Cn) ≤ 6.
Lemma 3.9. For any cycle Cn with n ≥ 8 and n ≡ 0 (mod 2), it holds that IC(Cn) = 6.
Proof. Let V (Cn) = {v1, v2, . . . , vn}. Consider the sets V1 = {v1, v6}∪{v2n−1 : n ≥ 5},
V2 = {v2, v5} ∪ {v2n : n ≥ 5}, V3 = {v3}, V4 = {v4}, V5 = {v7}, V6 = {v8}.
Then π = {V1, V2, V3, V4, V5, V6} is an ic-partition of Cn, for n ≥ 8, where V3 and V4
are ic-partners of V1, and V5 and V6 are ic-partners of V2. Hence, by Lemma 3.8 and
Observation 2.3, we have IC(Cn) = 6.
Lemma 3.10. For any cycle Cn with n ≥ 8 and n ≡ 0 (mod 3), it holds that
IC(Cn) = 6.
Proof. Let V (Cn) = {v1, v2, . . . , v3k}. Consider the sets V1 = {v3i+1}, V2 = {v3i+2}
and V3 = {v3i+3}, for 0 ≤ i ≤ k − 1. Now for each 1 ≤ i ≤ 3, we partition Vi into
two nonempty sets Vi,1 and Vi,2. Observe that Vi,1 and Vi,2 are ic-partners. Hence, by
Lemma 3.8 and Observation 2.3, we have IC(Cn) = 6.
Lemma 3.11. For any cycle Cn with n ≥ 8 and n ≡ 5 (mod 6), it holds that
IC(Cn) = 6.
Proof. Assume n = 6k − 1, (k ≥ 2). Let V (Cn) = {v1, v2, . . . , vn}. Consider the sets
A =
k−1⋃
i=0
{v3i+1}, A1 =
2k−1⋃
i=k
{v3i+1}, A2 =
2k−1⋃
i=k
{v3i},
B =
2k⋃
i=k
{v3i−1}, B1 =
k−1⋃
i=1
{v3i−1}, B2 =
k−1⋃
i=1
{v3i}.
Let π = {A,A1, A2, B,B1, B2}. One can observe that π is an ic-partition of Cn,
where A1 and A2 are ic-partners of A, and B1 and B2 are ic-partners of B. Now using
Lemma 3.8 and Observation 2.3, we have IC(Cn) = 6.
Lemma 3.12. For any cycle Cn with n ≥ 8 and n ≡ 1 (mod 6), it holds that
IC(Cn) = 6.
542 Ars Math. Contemp. 24 (2024) #P3.09 / 535–551
Proof. Assume n = 6k + 1, (k ≥ 2). Let V (Cn) = {v1, v2, . . . , vn}. Consider the sets
A =
(
k⋃
i=0
{v3i+1}
)
∪ {v3k+3}, A1 =
2k⋃
i=k+2
{v3i}, A2 =
2k⋃
i=k+2
{v3i−1},
B =
(
2k⋃
i=k+1
{v3i+1}
)
∪ {v3k+2}, B1 =
k⋃
i=1
{v3i−1}, B2 =
k⋃
i=1
{v3i}.
Let π = {A,A1, A2, B,B1, B2}. One can observe that π is an ic-partition of Cn, where
A1 and A2 are ic-partners of A, and B1 and B2 are ic-partners of B. Now using Lemma 3.8
and Observation 2.3, we have IC(Cn) = 6.
Theorem 3.13. For the cycle Cn,
IC(Cn) =

n if n ≤ 6;
5 if n = 7;
6 if n ≥ 8.
Proof. If 1 ≤ n ≤ 6, then it is easy to check that IC(Cn) = n. Now assume n = 7. Con-
sider the cycle C7 with V (C7) = {v1, v2, v3, v4, v5, v6, v7}. First we show that IC(C7) ̸=
6. Suppose, to the contrary, that IC(C7) = 6. Let π be an IC(C7)-partition. We note that
π consists of five singleton sets and a set of cardinality 2 (name A). By Theorem 2.5, A has
at most two ic-partners. Hence, π contains two singleton sets that are ic-partners, which
contradicts the fact that γi(C7) = 3. The partition {{v1, v3}, {v5}, {v6}, {v4, v7}, {v2}}
is an ic-partition of C7, so IC(C7) = 5. Furthermore, by Lemmas 3.9, 3.10, 3.11 and 3.12
we have IC(Cn) = 6, for n ≥ 8.
4 Graphs with small independent coalition number
In this section we investigate graphs G with IC(G) ∈ {1, 2, 3, 4}. We will make use of the
following two lemmas.
Lemma 4.1. Let G be a graph of order n containing r ≥ 1 full vertices, and let
F = {v1, v2, . . . , vr} be the set of full vertices of G. Then IC(G) = k, if and only if
IC(G[V \ F ]) = k − r, where r < k ≤ n.
Proof. Assume first that IC(G[V \ F ]) = k − r. Let π = {V1, V2, . . . , Vk−r} be an
IC(G[V \F ])-partition. Now the partition π′ = {V1, V2, . . . , Vk−r, {v1}, {v2}, . . . , {vr}},
is an ic-partition of G, so IC(G) ≥ k. Now we prove that IC(G) = k. Suppose, to
the contrary, that IC(G) > k. Let π be an IC(G)-partition. Now the partition π′ = π \
{{v1}, {v2}, . . . , {vr}} is an ic-partition of G[V \F ] such that |π′| > k−r, a contradiction.
Hence, IC(G) = k. Conversely, assume that IC(G) = k. Let π be an IC(G)-partition.
Now the partition π′ = π \ {{v1}, {v2}, . . . , {vr}} is an ic-partition of G[V \ F ], so
IC(G[V \ F ]) ≥ k − r. Now we prove that IC(G[V \ F ]) = k − r. Suppose, to the
contrary, that IC(G[V \ F ]) > k − r. Let π be an IC(G[V \ F ])-partition. Now the
partition π′ = π ∪ {{v1}, {v2}, . . . , {vr}} is an ic-partition of G such that |π′| > k, a
contradiction. Hence, IC(G[V \ F ]) = k − r.
M. R. Samadzadeh and D. A. Mojdeh: Independent coalition in graphs: existence and . . . 543
Lemma 4.2. Let G be a graph containing a nonempty set of isolated vertices I . If IC(G) ≥
3, then for any IC(G)-partition π, there is a set Vr ∈ π such that Vr = I .
Proof. First we show that all vertices in I are in the same set of π. Suppose, to the contrary,
that there are sets Vi ∈ π and Vj ∈ π such that both Vi and Vj contain isolated vertices.
Let Vk ∈ π be an arbitrary set in π such that Vk /∈ {Vi, Vj}. (Since IC(G) ≥ 3, such a set
exists). Then Vk has no ic-partner, a contradiction. Now let Vr be the set in π containing
isolated vertices. Further, let v be an arbitrary vertex in Vr, and let u ∈ V (G) be an
arbitrary vertex such that u ̸= v. If u ∈ Vr, then u is not adjacent to v. Otherwise, the
set in π containing u is an ic-partner of Vr, which again implies that u is not adjacent to v.
Hence, we have deg(v) = 0. Choosing v arbitrarily, we conclude that Vr = I .
Proposition 4.3. Let G be a graph of order n. Then
(1) IC(G) = 1 if and only if G ≃ K1.
(2) IC(G) = 2 if and only if G ≃ K2 or G ≃ Kn, for some n ≥ 2.
Proof. (1) It is clear that IC(G) = 1 if and only if G ≃ K1.
(2) If G ≃ K2, then we clearly have IC(G) = 2. Now assume G ≃ Kn, for some n ≥ 2.
Let π be an ic-partition of G. Note that no more than two sets in π contain isolated vertices,
for otherwise, no two sets in π are ic-partners. Thus, |π| ≤ 2. Partitioning vertices of G into
two nonempty sets yields an ic-partition of G. Hence, IC(G) = 2. Conversely, suppose
that IC(G) = 2. Let π = {V1, V2} be an IC(G)-partition. If both V1 and V2 are singleton
dominating sets, then G ≃ K2. Hence, we may assume that at least one of them (say V1)
is not a singleton dominating set. It follows that V2 is not a singleton dominating set either,
for otherwise, G is a star, and so by Observation 3.2, we have IC(G) = 3. Hence, V1 and
V2 are ic-partners, and so V = V1 ∪ V2 is an independent set. Hence, G ≃ Kn, for some
n ≥ 2.
Definition 4.4. Let B1 represent the family of bipartite graphs H with partite sets H1 and
H2 such that |H1| ≥ 2, |H2| ≥ 2, δ(H) ≥ 1 and id(H) = 2.
Definition 4.5. For m ≥ 1, let B2 represent the family of graphs H ∪Km, where H is a
bipartite graph with δ(H) ≥ 1 and id(H) = 2.
Definition 4.6. For m ≥ 1, let B3 represent the family of graphs H ∪Km, where H is a
3-partite graph with δ(H) ≥ 1 and id(H) = 3.
Proposition 4.7. Let G be a graph. Then IC(G) = 3, if and only if G ∈ {K3,K1,n−1} ∪
B2.
Proof. Observations 3.1 and 3.2 imply that IC(K3) = 3 and that IC(K1,n−1) = 3, re-
spectively. Now let G ∈ B2. Let I be the set of isolated vertices of G, and let {H1, H2}
be a partition of G− I into its partite sets. We observe that the partition {I,H1, H2} is an
ic-partition of G, so IC(G) ≥ 3. Now we show that IC(G) = 3. Suppose, to the con-
trary, that IC(G) ≥ 4. Let π = {V1, V2, . . . , Vk} be an IC(G)-partition. By Lemma 4.2,
we have I ∈ {V1, V2, . . . , Vk}. Assume, without loss of generality, that I = V1. Now
for each 2 ≤ i ≤ k, Vi forms an independent coalition with V1, and so Vi dominates H .
Hence, the partition {V2, . . . , Vk} is an idomatic partition of H , which contradicts the as-
sumption. Hence, IC(G) = 3. Conversely, let G be a graph with IC(G) = 3, and let
544 Ars Math. Contemp. 24 (2024) #P3.09 / 535–551
π = {V1, V2, V3} be an IC(G)-partition. We consider four cases depending on the number
of full vertices of G.
Case 1. G has three full vertices. In this case, the sets V1, V2 and V3 are all singleton
dominating sets, so G ≃ K3.
Case 2. G has two full vertices. Note that this case never occurs.
Case 3. G has one full vertex. Let v1 be the full vertex of G. Lemma 4.1 implies that
IC(G − v1) = 2. Thus, by Proposition 4.3, either G − v1 ≃ K2, implying that G ≃ K3,
or G− v1 ≃ Kn, for some n ≥ 2, which implies that G ≃ K1,n−1, for some n ≥ 3.
Case 4. G has no full vertex. Let I be the set of isolated vertices of G. First we
note that V1, V2 and V3 are not pairwise ic-partners, for otherwise, we have G ≃ Kn,
and so by Proposition 4.3, we have IC(G) = 2, a contradiction. Hence, π contains a set
(say V1) that forms an independent coalition with V2 and V3, while V2 and V3 are not ic-
partners. Therefore, each vertex in V1 is an isolated vertex, so it follows from Lemma 4.2
that I = V1. Further, the sets V2 and V3 are independent dominating sets of G[V2 ∪ V3],
implying that id(G[V2 ∪ V3]) ≥ 2. It remains to show that id(G[V2 ∪ V3]) = 2. Suppose,
to the contrary, that id(G[V2 ∪ V3]) ≥ 3. Let π′ = {U1, U2, . . . , Uk}, (k ≥ 3), be an
idomatic partition of G[V2 ∪ V3]. Then the partition π′′ = {U1, U2, . . . , Uk, V1} is clearly
an ic-partition of G, implying that IC(G) ≥ 4, a contradiction. Hence, G ∈ B2.
Proposition 4.8. Let G be a graph. If IC(G) = 4, then G ∈ {K4,K2 +Kn,K1 +B} ∪
B1 ∪ B3, where n ≥ 2 and B ∈ B2.
Proof. Let π = {V1, V2, V3, V4} be an ic-partition of G. We consider two cases.
Case 1. G has a full vertex. Let v1 be a full vertex of G. Lemma 4.1 implies that
IC(G−v1) = 3. Thus, by Proposition 4.7, we have G−v1 ≃ K3, implying that G ≃ K4,
or G − v1 ≃ K1,n, for some n ≥ 2, implying that G ≃ K2 + Kn, for some n ≥ 2, or
G− v1 ∈ B2, which implies that G ≃ K1 +B, where B ∈ B2.
Case 2. G has no full vertex. First assume that G contains a nonempty set I of isolated
vertices. Then by Lemma 4.2, we have I ∈ π. Without loss of generality, assume that
I = V4. Now for each 1 ≤ i ≤ 3, Vi must form an independent coalition with I . Thus,
U = G[V1 ∪ V2 ∪ V3] is a 3-partite graph with id(U) ≥ 3. Since IC(G) = 4, the case
id(U) > 3 is impossible. Hence, id(U) = 3, and so G ∈ B3. Now assume that G contains
no isolated vertex. Since G has neither full vertices nor isolated vertices, each set of π has
either one or two ic-partners. If there is a set of π, (say V1) having one ic-partner, (say V2),
then it follows that V3 and V4 are ic-partners, and so G is a bipartitie graph with partite sets
V1 ∪ V2 and V3 ∪ V4. Otherwise, assume, without loss of generality, that V2 and V3 are
ic-partners of V1. It follows that V4 has an ic-partner in {V2, V3}. By symmetry, we may
assume that V4 and V3 are ic-partners. Then G is again a bipartitie graph with partite sets
V1 ∪ V2 and V3 ∪ V4. Now using Theorem 2.6, we have id(G) = 2, and so G ∈ B1.
5 Graphs with large independent coalition number
Our main goal in this section is to investigate structure of graphs G of order n with
IC(G) = n, under specified conditions. In addition, we will characterize all trees T of
order n with IC(T ) = n − 1. Let us begin with an observation that characterizes all
disconnected graphs G of order n with IC(G) = n.
Observation 5.1. Let G be a disconnected graph of order n. Then IC(G) = n if and only
if G ≃ Ks ∪Kr, for some s ≥ 1, and r ≥ 1.
M. R. Samadzadeh and D. A. Mojdeh: Independent coalition in graphs: existence and . . . 545
Now we introduce two sufficient conditions for a graph G of order n to have indepen-
dent coalition number n.
Observation 5.2. If G is a graph of order n with α(G) = 2, then IC(G) = n.
Proof. Let G be a graph of order n such that α(G) = 2. Consider the singleton partition
π1 of G. Note that for any two non-adjacent vertices v and u in V (G), the sets {v} and
{u} in π1, are ic-partners. Hence, π1 is an ic-partition of G, and so IC(G) = n.
Observation 5.3. Let G be a graph of order n. If G admits a partition of its vertices into
two maximal cliques, then IC(G) = n.
5.1 Graphs G with δ(G) = 1 and IC(G) = n
In this subsection, we characterize graphs G of order n with δ(G) = 1 and IC(G) = n.
We need the following definition.
Definition 5.4. Let G be a graph of order n, (n ≥ 3), and let δ(G) = 1. Furthermore, let
x be a pendant vertex of G, and let y be the support vertex of x. Then G ∈ F if and only
if V (G) \ {x, y} induces a clique.
Theorem 5.5. Let G be a graph of order n with δ(G) = 1. Then IC(G) = n if and only
if either G ≃ K2, or G ∈ F .
Proof. Obviously, IC(K2) = 2. Now assume that G ∈ F . Let x be a pendant vertex of G,
and let y be the support vertex of x. Further, let U = V (G) \ {x, y}. Note that U contains
no full vertex. If y is a full vertex, then G is obtained from the complete graph Kn−1, where
one of its vertices is adjacent to a leaf. In this case, we clearly have IC(G) = n. Thus,
we may assume that y is not a full vertex, that is, there is a vertex u ∈ U such that u is not
adjacent to y. Then it is easy to verify that the sets {y} and {u} are ic-partners, and that
each vertex in U \ {u} forms an independent coalition with {x}. Therefore, IC(G) = n.
Conversely, suppose that G is a graph with δ(G) = 1 and IC(G) = n. Let x be a leaf
of G, and let y be the support vertex of x. Consider the singleton partition π1 of G. Note
that each set in π1\{{x}, {y}} must be an ic-partner of {x} or {y}, to dominate x. Let
A = N(y) \ {x}, and B = V (G) \ ({x, y} ∪A). We consider four cases.
Case 1. A = ∅ and B = ∅. In this case, we have G ≃ K2.
Case 2. A = ∅ and B ̸= ∅. By Observation 5.1, we have G ≃ K2 ∪ Kr, for some
r ≥ 1. Thus, G ∈ F .
Case 3. A ̸= ∅ and B = ∅. For each v ∈ A, the set {v} cannot be an ic-partner of {y},
so it must be an ic-partner of {x}. This implies that A induces a clique. Hence, G ∈ F .
Case 4. A ̸= ∅ and B ̸= ∅. For each v ∈ A, the set {v} cannot be an ic-partner of
{y}, so it must be an ic-partner of {x}. This implies that [A,B] is full and that A induces
a clique. Now for each vertex u ∈ B, in order for the set {u} to be an ic-partner of {x}
or {y}, u must be adjacent to all other vertices in B. Hence, B induces a clique, and so
G ∈ F , which completes the proof.
As an immediate result from Theorem 5.5 we have:
Corollary 5.6. Let T be a tree of order n. Then IC(T ) = n if and only if T ∈ {P1, P2,
P3, P4}.
546 Ars Math. Contemp. 24 (2024) #P3.09 / 535–551
5.2 Triangle-free graphs G with IC(G) = n
In this subsection, we characterize graphs G of order n with g(G) = 4 and IC(G) = n.
This will lead to characterization of all triangle-free graphs G of order n with IC(G) = n.
We will make use the following lemmas.
Lemma 5.7. Let G be a triangle-free graph of order n with IC(G) = n. Then g(G) ≤ 6.
Proof. Let G be a graph of order n with IC(G) = n, and suppose, to the contrary, that
g(G) ≥ 7. Let C ⊆ G be a cycle of order g(G). Consider an arbitrary vertex v ∈ V (C).
Note that γi(C) ≥ 3, and so {v} is not an ic-partner of any set {u} ⊂ V (C). Therefore,
it must be an ic-partner of a set {u} ⊆ V (G) \ V (C). It follows that, {u} dominates
V (C) \Nc[v], which implies that G contains triangles, a contradiction.
Lemma 5.8. Let G be a graph of order n with g(G) = 6. Then IC(G) = n if and only if
G ≃ C6.
Proof. Let G be a graph of order n with g(G) = 6. If G ≃ C6, then by Theorem 3.13,
we have IC(G) = 6. Conversely, assume that IC(G) = n. Let C ⊆ G be a cycle of
order 6, and suppose, to the contrary, that V (G) \ V (C) ̸= ∅. Consider an arbitrary vertex
v ∈ V (G) \ V (C). If {v} is an ic-partner of a set {u} ⊂ V (C), then {v} must dominate
V (C) \ Nc[u], which implies that G contains triangles, a contradiction. Otherwise, {v}
must be an ic-partner of a set {u} ⊂ V (G) \ V (C). Now since {u, v} dominates C, it
follows that G contains triangles, or induces cycles of order 4, a contradiction.
Our next result can be established almost the same way as Lemma 5.8, so we state it
without proof.
Lemma 5.9. Let G be a graph of order n with g(G) = 5. Then IC(G) = n if and only if
G ≃ C5.
In order to characterize graphs G of order n with IC(G) = n and g(G) = 4, we need
the following definitions.
Definition 5.10. Let K0 represent a bipartite graph with partite sets H1 = {v1, v2, v3, v4}
and H2 = {u1, u2, u3, u4} such that for each 1 ≤ i ≤ 4, vi is adjacent to all vertices in
H2, except ui (see Figure 2).
Definition 5.11. Let K represent a family of 4-partite graphs with partite sets
H1 = {v1, v2, v3, v4}, H2 = {u1, u2, u3, u4}, H3 = {n1, n2, . . . , nk} and
H4 = {m1,m2, . . . ,mk}, for k ≥ 1, with the following properties:
• [H1, H3] is full and [H2, H4] is full,
• [H1, H4] is empty and [H2, H3] is empty,
• For each 1 ≤ i ≤ 4, vi is adjacent to all vertices in H2, except ui,
• For each 1 ≤ i ≤ k, ni is adjacent to all vertices in H4, except mi.
Figure 3 illustrates such a graph for k = 3.
M. R. Samadzadeh and D. A. Mojdeh: Independent coalition in graphs: existence and . . . 547
v1 v2 v3 v4
u1 u2 u3 u4
Figure 2: The graph K0.
v1
v2
v3
v4 u4
u3
u2
u1
n1
n2
n3 m3
m2
m1
Figure 3: The graph in K for k = 3.
Theorem 5.12. Let G be a graph of order n with g(G) = 4. Then IC(G) = n if and only
if G ∈ {C4,K0} ∪ K.
Proof. It is easy to check that IC(C4) = 4 and that IC(K0) = 8. Now let G ∈ K.
We observe that for each 1 ≤ i ≤ 4, {vi} and {ui} are ic-partners, and that for each
1 ≤ i ≤ k, {ni} and {mi} are ic-partners. Thus, IC(G) = n. Conversely, let G be a
graph of order n with g(G) = 4 and IC(G) = n, and let C be a cycle of G of order 4
with V (C) = {x, y, z, t} and E(C) = {xy, yz, zt, tx}. If G = C, then the desired result
follows. Hence, we assume that G ̸= C. Since x is adjacent to y and t, neither {y} nor {t}
is an ic-partner of {x}. Now consider two cases.
Case 1. {x} and {z} are ic-partners. In this case, G is dominated by {x, z}. Let
A = N(x) \ {y, t} and B = N(z) \ {y, t}. If A ̸= ∅, (say v ∈ A), then it is not difficult
to check that {v} has no ic-partner. Thus, A = ∅, and so by symmetry, we have B = ∅.
Hence, G ≃ C4.
Case 2. {x} and {z} are not ic-partners. Let {e} be an ic-partner of {x}. Since {x, e}
dominates G and z is not adjacent to x, it must be adjacent to e. Let A = N(x) \ {y, t}
and B = N(e) \ {z}. It is not difficult to verify that A ∩ B = ∅. Now if A = ∅, then
{z} cannot form an independent coalition with any other set, so A ̸= ∅. Let {f} ⊆ A be
an ic-partner of {z}. We note that if a set {g} forms an independent coalition with {y},
then g ∈ B. Further, if a set {h} forms an independent coalition with {t} then h ∈ B. Let
{g} and {h} be ic-partners of {y} and {t}, respectively. Observe that {g} ≠ {h}. Now let
A′ = A \ {f} and B′ = B \ {g, h}. There exist the following subcases.
548 Ars Math. Contemp. 24 (2024) #P3.09 / 535–551
Subcase 2.1. A′ = ∅ and B′ = ∅. In this case, we have G ≃ K0.
Subcase 2.2. A′ = ∅ and B′ ̸= ∅. Let v ∈ B′. One can verify that {v} cannot form an
independent coalition with any other set. Thus, this case is impossible.
Subcase 2.3. A′ ̸= ∅ and B′ = ∅. Let v ∈ A′. One can verify that {v} cannot form an
independent coalition with any other set. Thus, this case is impossible.
Subcase 2.4. A′ ̸= ∅ and B′ ̸= ∅. Let v ∈ A′. If a set {u} forms an independent
coalition with {v}, then u ∈ B′. Furthermore, for each vertex u ∈ B′, {u} cannot form
an independent coalition with more than one sets {v} ⊆ A′. Thus, |A′| ≤ |B′|. Using
a similar argument, we deduce that |B′| ≤ |A′|, and so |A′| = |B′|. Consequently, the
following statements hold in the graph G:
• G[{x, y, z, t, e, f, g, h}] is a bipartite graph with partite sets V1 = {x, z, g, h} and
V2 = {y, t, e, f}, which is isomorphic to K0,
• [V1, A′] is full and [V2, B′] is full,
• [V1, B′] is empty and [V2, A′] is empty,
• G[A′∪B′] is a bipartite graph with partite sets A′ and B′ such that degG[A′∪B′](v) =
|A′| − 1 = |B′| − 1, for each v ∈ A′ ∪B′.
Hence, G ∈ K and the proof is complete.
Using Observation 5.1, Corollary 5.6, Lemmas 5.7, 5.8 and 5.9, and Theorem 5.12, we
infer the following result.
Corollary 5.13. Let G be a triangle-free graph of order n. Then IC(G) = n if and only if
G ∈ {C4, C5, C6, P1, P2, P3, P4,K2,K1 ∪K2,K2 ∪K2,K0} ∪ K.
5.3 Trees T with IC(T ) = n − 1
The following theorem characterizes all trees T of order n with IC(T ) = n− 1.
Theorem 5.14. Let T be a tree of order n. Then IC(T ) = n − 1 if and only if T ∈
{P5, P6, S1,2,K1,3}.
Proof. By Theorem 3.6, we have IC(P5) = 4 and IC(P6) = 5. Further, by Observa-
tion 3.2, we have IC(K1,3) = 3 and by Observation 3.3, we have IC(S1,2) = 4. Con-
versely, let T be a tree of order n with IC(T ) = n − 1, where x is a leaf, and y is the
support vertex of x. Define A = N(y) \ {x} and B = V (G) \ ({x, y} ∪ A). Further,
let π be an IC(T )-partition. Note that π contains a set of cardinality 2 (say V1 = {u, v})
and n − 2 singleton sets. Since x and y are adjacent, we have V1 ̸= {x, y}. Note as well
that any set in π must be an ic-partner of the set containing x, or the set containing y, to
dominate x. We consider two cases.
Case 1. B = ∅. If A = ∅, then we have T ≃ K2, and so IC(T ) = 2 ̸= n− 1. Hence,
A ̸= ∅, and so T ≃ K1,n−1, for some n ≥ 3. Now by Lemma 3.7, we have IC(T ) = 3.
Hence, T ≃ K1,3.
Case 2. B ̸= ∅. Since T is connected, we have A ̸= ∅. We divide this case into some
subcases.
Subcase 2.1. u ∈ A and v ∈ B. We first show that |A| = 1. Suppose, to the contrary,
that |A| ≥ 2. Then there is a vertex z ∈ A such that z ̸= u. Since z and y are adjacent,
M. R. Samadzadeh and D. A. Mojdeh: Independent coalition in graphs: existence and . . . 549
{z} cannot be an ic-partner of {y}, so it must be an ic-partner of {x}. Since {x} does not
dominate u, u must be adjacent to z, which is a contradiction, since y, z and u induce a
triangle. Now {y} cannot be an ic-partner of {x} or {u, v}, so it must have an ic-partner
in B. This implies that |B| ≥ 2. Let {t} ⊂ B be an ic-partner of {y}. we show that
B \ {v, t} = ∅. Suppose that B \ {v, t} ̸= ∅. Let z ∈ B \ {v, t}. Note that v and t are
adjacent. Now {z} must be an ic-partner of {x} or {y}, so z must be adjacent to t and
v, which is a contradiction, since z, t and v induce a triangle. Hence, B = {v, t} and so
T ≃ P5.
Subcase 2.2. {u, v} ⊆ B. An argument similar to the one presented above implies
that |A| = 1. Now we show that B \ {u, v} = ∅. Suppose that B \ {u, v} ≠ ∅. Let
z ∈ B \ {u, v}, and let A = {t}. Since t and y are adjacent, {t} cannot be an ic-partner
of {y}, so it must be an ic-partner of {x}. Thus, t must be adjacent to u, v and z. Now
{z} must be an ic-partner of {x} or {y}, so z must be adjacent to u and v, which is a
contradiction, since z, u and t induce a triangle. Hence, T ≃ S1,2.
Subcase 2.3. {u, v} ⊆ A. We first show that A\{u, v} = ∅. Suppose that A\{u, v} ≠
∅. Let z ∈ A \ {u, v}. Since z is adjacent to y, {z} must be an ic-partner of {x}, so z must
be adjacent to u and v, which is a contradiction, since z, u and y induce a triangle. Now
we show that |B| = 1. Suppose that |B| ≠ 1. First assume |B| ≥ 3. Let z, t, w ∈ B. Now
z, t and w induce a triangle, since the sets containing each of them, must be an ic-partner
of {x} or {y}, a contradiction. Now assume |B| = 2. Let B = {z, t}. Each of the sets
{z} and {t} must be an ic-partner of {x} or {y}. Thus, z must be adjacent to t. Now
{u, v} must be an ic-partner of {x}, so z and t must be dominated by {u, v}. Now the
induced subgraph T [{u, v, z, t}] contains at least one cycle, a contradiction. Hence, we
have T ≃ S1,2.
Subcase 2.4. u = y and v ∈ B. we first show that |A| = 1. Suppose that |A| ≥ 2.
Let z, t ∈ A. Since z and t are adjacent to y, {z} and {t} cannot be an ic-partner of y,
so each of them must be an ic-partner of {x}. Thus, z must be adjacent to t, which is a
contradiction, since z, t and y induce a triangle. Now we show that |B| = 2. Suppose that
|B| ̸= 2 . If |B| = 1, then T ≃ P4, a contradiction. Otherwise, let {v, z, t} ⊆ B and
A = {w}. Now {w} cannot be an ic-partner of {u, v}, so it must be an ic-partner of {x}.
Thus, w must be adjacent to v, z and t. Now observe that {u, v} must have an ic-partner
in B. Assume, without loss of generality, that {u, v} and {z} are ic-partners. This implies
that t is adjacent to z or v, which is impossible, since both cases lead to existence of an
induced triangle. Hence, T ≃ S1,2.
Subcase 2.5. u = x and v ∈ A. We first show that |B| ≤ 2. Suppose that |B| ≥ 3.
Let {z, t, w} ⊆ B. Note that {y} must have an ic-partner in B. Assume, without loss of
generality, that {y} and {z} are ic-partners. It follows that z is adjacent to t and w. Now if
{y} is an ic-partner of {t} or {w}, then t must be adjacent to w, which is impossible, since
z, t and w induce a triangle. Hence, both t and w must be ic-partners of {u, v}, which
implies that t is adjacent to w. Now z, t and w induce a triangle, a contradiction. Now we
show that |A| ≤ 2. Suppose that |A| ≥ 3. Let {z, t, v} ⊆ A. The sets {z} and {t} must be
ic-partners of {u, v}. This implies that z is adjacent to t. Now z, t and y induce a triangle,
a contradiction. Further, we observe that the case |A| = |B| = 2 is impossible. Hence,
either |A| = 2 and |B| = 1, which implies that T ≃ S1,2, or |A| = 1 and |B| = 2, which
implies that T ≃ P5.
Subcase 2.6. u = x and v ∈ B. We first show that |A| = 1. Suppose that |A| ≥ 2.
Let {z, t} ⊆ A. Now each of the sets {z} and {t} must be an ic-partner of {u, v}. This
550 Ars Math. Contemp. 24 (2024) #P3.09 / 535–551
implies that z is adjacent to t, which is a contradiction, since y, z and t induce a triangle.
Now we show that |B| ≤ 3. Suppose that |B| ≥ 4. Let {v, t, w, h} ⊆ B and A = {z}.
Note that {y} must have an ic-partner in B. Assume, without loss of generality, that {y}
and {t} are ic-partners. It follows that t is adjacent to w, h and v. Now {w} must be an
ic-partner of {y} or {u, v}. One can observe that both cases lead to contradiction. Hence,
either |B| = 2, which implies that T ≃ P5, or |B| = 3, which implies that T ≃ P6.
6 Discussion and conclusions
In Proposition 2.2, we introduced a family of graphs admitting no ic-partition. This result
motivates the following problem:
Problem 6.1. Characterize graphs admitting an ic-partition.
In Observations 2.3 and 2.4, we presented the sharp inequalities IC(G) ≤ C(G) and
IC(G) ≥ χ(G). This raises the following problems:
Problem 6.2. Characterize graphs G in which the equality IC(G) = C(G) holds.
Problem 6.3. Characterize graphs G in which the equality IC(G) = χ(G) holds.
In Theorem 5.14, trees T of order n, with IC(T ) = n − 1 have been characterized.
This raises the following problem:
Problem 6.4. Characterize graphs G of order n with IC(G) = n− 1.
ORCID iDs
Mohammad Reza Samadzadeh https://orcid.org/0009-0004-4370-5905
Doost Ali Mojdeh https://orcid.org/0000-0001-9373-3390
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ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 24 (2024) #P3.10 / 553–566
https://doi.org/10.26493/1855-3974.3227.fd5
(Also available at http://amc-journal.eu)
Spectra of signed graphs and
related oriented graphs
Zoran Stanić *
Faculty of Mathematics, University of Belgrade,
Studentski trg 16, 11 000 Belgrade, Serbia
Received 14 September 2023, accepted 23 April 2024, published online 6 September 2024
Abstract
For every oriented graph G′, there exists a bipartite signed graph Ḣ such that the spec-
trum of Ḣ contains the full information about the spectrum of the skew adjacency ma-
trix of G′. This allows us to transfer some problems concerning the skew eigenvalues of
oriented graphs to the framework of signed graphs, where the theory of real symmetric
matrices can be employed. In this paper, we continue the previous research by relating
the characteristic polynomials, eigenspaces and the energy of G′ to those of Ḣ . Simul-
taneously, we address some open problems concerning the skew eigenvalues of oriented
graphs.
Keywords: Oriented graph, signed graph, eigenvalues, characteristic polynomial, eigenspaces, en-
ergy.
Math. Subj. Class. (2020): 05C20, 05C22, 05C50
1 Introduction
For a finite simple undirected graph G = (V,E), an oriented graph G′ is a pair (G, σ′),
where σ′ is the edge orientation satisfying σ′(ij) ∈ {i, j}, for every ij ∈ E. Similarly, a
signed graph Ġ is a pair (G, σ̇), where σ̇ is the edge signature satisfying σ̇(ij) ∈ {+1,−1},
for every ij ∈ E. In both cases, G is referred to as the underlying graph. The order n is the
number of vertices of G. The edge set of G′ consists of oriented edges, where the edge ij
is oriented from i to j if σ′(ij) = j; this is designated by i → j (or j ← i). The edge set
of Ġ consists of positive and negative edges.
*The research is supported by the Science Fund of the Republic of Serbia; grant number 7749676: Spectrally
Constrained Signed Graphs with Applications in Coding Theory and Control Theory – SCSG-ctct.
E-mail address: zstanic@matf.bg.ac.rs (Zoran Stanić)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
554 Ars Math. Contemp. 24 (2024) #P3.10 / 553–566
The skew adjacency matrix SG′ = (sij) of G′ is the n × n matrix such that sij = 0
if ij is not an edge of G′, sij = 1 if i → j, and sij = −1 otherwise. This matrix
is skew symmetric and differs from the adjacency matrix of G′ whose (i, j)-entry is 0
whenever i ← j. The characteristic polynomial, the eigenvalues and the spectrum of SG′
are known as the skew characteristic polynomial, the skew eigenvalues, the skew spectrum
of G′, respectively. To easy language and be consistent with the forthcoming terminology
for signed graphs, in this article we omit the prefix ‘skew’. The spectrum of G′ consists
of purely imaginary numbers, the non-zero eigenvalues come as complex conjugates, and
thus the rank of SG′ is even. Also, S
⊺
G′SG′ = −S2G′ .
The adjacency matrix AĠ of Ġ is obtained from the standard (0, 1)-adjacency matrix
of G by reversing the sign of all edges mapped to −1 by σ̇. By the characteristic poly-
nomial, the eigenvalues and the spectrum of Ġ we mean the characteristic polynomial, the
eigenvalues and the spectrum of AĠ, respectively. Since AĠ is symmetric, its eigenvalues
are real.
An r-cube or a hypercube Qr is the r-regular graph of order 2r with vertex set {0, 1}r
(all possible binary r-tuples) in which two vertices are adjacent if and only if they differ
in exactly one coordinate. Accordingly, an oriented (resp. signed) r-cube is an oriented
(signed) graph underlined by Qr. If Γ is either an oriented graph or a signed graph, then its
energy E(Γ) is the sum of modulus of its eigenvalues.
It follows from [19] that every oriented graph G′ is related to a bipartite signed graph Ḣ
in such a way that the spectrum of Ḣ contains the full information about the spectrum of G′.
All necessary details about this relation are given in the next section. This means that the
theory of spectra of oriented graphs is strongly connected to the theory of spectra of signed
graphs, and that many problems concerning spectral parameters of oriented graphs can be
transferred to the framework of signed graphs, where the entire theory of real symmetric
matrices can be employed. In this way, some known results on oriented graphs are proved
in an elementary way [19], some open problems are resolved [15] and some known results
concerning signed graphs are transferred to the context of oriented graphs [18].
In this article we continue the research by expressing the coefficients of the character-
istic polynomial of G′ in terms of the coefficients of the characteristic polynomial of Ḣ .
We also generate the eigenspaces of G′ on the basis of the eigenvectors of Ḣ and relate
the energies of both graphs. The results on characteristic polynomials and energies are of
particular interest since they address some open problems posed in literature. The results
on eigenspaces are followed by an immediate application in the engineering domain.
Section 2 contains additional terminology and notation, along with a short review of re-
sults of [19]. In particular, oriented graphs are related to signed graphs in the forthcoming
Theorem 2.1. Some comments and results that arise directly from this theorem are given
in Section 3. Characteristic polynomials, eigenspaces and energies are considered in Sec-
tions 4–6, respectively. Some notes on particular (oriented or signed) hypercubes are given
in Section 7.
2 Preliminaries
We say that an oriented or a signed graph is connected, regular, or bipartite if the same
holds for its underlying graph. Similarly, a matching (or a perfect matching) refers to the
matching in the underlying graph.
Two oriented (resp. signed) graphs with the same underlying graph are switching equiv-
Z. Stanić: Spectra of signed graphs and related oriented graphs 555
alent if there is a subset U of the vertex set V , such that one of them is obtained by reversing
the orientation (sign) of every edge located between U and V \ U . In matrix terminology,
G′1 and G
′
2 are switching equivalent if there exists a diagonal matrix S with±1 on the main
diagonal, such that SG′2 = S
−1SG′1S, and similarly for signed graphs. S is referred to as
the switching matrix. Observe that the spectrum remains unchanged under the switching
operation.
We say that an oriented even cycle C ′2ℓ is oriented uniformly if by traversing along the
cycle we pass through an odd (resp. even) number of edges oriented in the route direction
for ℓ odd (even), where the ‘route direction’ refers to any of two possible directions: clock-
wise or counterclockwise. A canonical orientation in a bipartite graph G is the orientation
which orients all the edges from one colour class to the other. Clearly, in this orientation
every cycle is oriented uniformly.
A cycle Ċ in a signed graph is positive if the product of its edge signs σ̇(Ċ) is 1.
Otherwise, it is negative. A signed graph is said to be homogeneous if all edges have the
same sign, e.g., if its edge set is empty. It is balanced if it switches to its underlying graph;
equivalently, it does not contain negative cycles. The negation −Ġ of a signed graph is
obtained by reversing the sign of every edge of Ġ. Observe that if Ġ is bipartite, then it is
switching equivalent to −Ġ and they share the same spectrum.
We proceed with results of [19]. The signature σ̇ is associated with the orientation σ′
(and also Ġ is associated with G′) if
σ̇(ik)σ̇(jk) = siksjk holds for every pair of edges ik and jk. (2.1)
Being associated is a symmetric relation. We write σ̇ ∼ σ′ to indicate that σ̇ and σ′ are
mutually associated. The following results hold: If σ̇ ∼ σ′, then −S2G′ = A2Ġ. For a
graph G and an orientation σ′, there exists a signature σ̇ associated with σ′ if and only if G
is bipartite.
The next result gives a crucial relation between the spectrum of an oriented graph and
the spectrum of a related signed graph. If G′ is an oriented graph, then its bipartite dou-
ble bd(G′) is the oriented graph whose skew adjacency matrix is the Kronecker product
Sbd(G′) = AK2 ⊗ SG′ , where K2 is the complete graph with 2 vertices. This defini-
tion extends the definition of a bipartite double of an ordinary graph, where bd(G) has
Abd(G) = AK2 ⊗ AG as the adjacency matrix. Evidently, if G underlies G′, then bd(G)
underlies bd(G′). In our notation, exponents denote multiplicities of the eigenvalues.
Theorem 2.1. Let G′ = (G, σ′) be an oriented graph with rank(SG′) = 2k. The following
statements hold true:
(i) If G′ is bipartite and σ̇ ∼ σ′, then ±iλ1,±iλ2, . . . ,±iλk, 0(n−2k) are the eigen-
values of G′ if and only if ±λ1,±λ2, . . . ,±λk, 0(n−2k) are the eigenvalues of Ġ =
(G, σ̇).
(ii) If G′ is non-bipartite, H ′ = (bd(G), σ′′) is a bipartite double of G′ and σ̇ ∼
σ′′, then ±iλ1,±iλ2, . . . ,±iλk, 0(n−2k) are the eigenvalues of G′ if and only if
(±λ1)(2), (±λ2)(2), . . . , (±λk)(2), 0(2n−4k) are the eigenvalues of Ḣ = (bd(G), σ̇).
Therefore, G′ is related to a bipartite signed graph whose spectrum gives the full infor-
mation on the spectrum of G′. If G′ is bipartite, a required signed graph is its associate.
If G′ is non-bipartite, then a required signed graph is associated with bd(G′). One may
556 Ars Math. Contemp. 24 (2024) #P3.10 / 553–566
Figure 1: An infinite family of oriented graphs G′ and signed graphs Ġ such that −S2G′ =
A2
Ġ
. Negative edges are dashed.
notice that item (ii) of the previous theorem covers the bipartite case in the sense that, if G′
is bipartite, then bd(G′) consists of two copies of G′ and Ḣ also has two identical copies,
each associated with G′. However, this would lead to unnecessary complicating, as there
is no need to deal with a bipartite double if G′ is already bipartite. Thus, the bipartite case
is separated in item (i) of the same theorem.
3 Comments to Theorem 2.1
In the bipartite case, G′ is associated with Ġ if and only if it is associated with −Ġ.
Hence, G′ actually has two associates (with the same spectrum). Similarly, Ġ is associ-
ated with G′ if and only if it is associated with the oriented graph obtained by reversing the
orientation of every edge of G′. Again, Ġ has two associates which share the same spec-
trum. Henceforth, when we say ‘let Ġ be a signed graph associated with G′’ (or something
similar), we always mean that Ġ is allowed to take any of the two options (that differ up to
negation).
We have pointed out in the previous section that−S2G′ = A2Ġ holds whenever G
′ and Ġ
are associated. However, this identity is not exclusively reserved for associated graphs. For
example, Figure 1 illustrates infinite families of graphs that are not mutually associated
in the sense of the equality (2.1) (since they are non-bipartite), but satisfy the previous
matrix identity. In this case, they share the same underlying graph, but the identity can
occur even if they do not. In fact, the identity occurs if and only if for every pair i, j of
vertices, −s(2)ij = a
(2)
ij holds, where an exponent indicates that we deal with the entry of
matrix square. This leads to the following result.
Theorem 3.1. If an oriented graph G′ and a signed graph Ġ are defined on the same vertex
set, then −S2G′ = A2Ġ holds if and only if for every pair of vertices i, j∣∣{k : (i→ k∧j → k)∨(i← k∧j ← k)}∣∣−∣∣{k : (i→ k∧j ← k)∨(i← k∧j → k)}∣∣
in G′ is equal to ∣∣{k : σ̇(ik)σ̇(jk) = 1}∣∣− ∣∣{k : σ̇(ik)σ̇(jk) = −1}∣∣
in Ġ.
Z. Stanić: Spectra of signed graphs and related oriented graphs 557
To demonstrate an application of Theorem 2.1, we deduce a known result. Observe
that a bipartite canonically oriented graph is associated with a homogeneous signed graph,
necessarily switching equivalent to its underlying graph, and so sharing the spectrum with
it. Since bipartite oriented graphs are switching equivalent if and only if associated signed
graphs are switching equivalent (where the equivalence is realized by the same switch-
ing matrix), we deduce the following result (conjectured in [7], and proved in [3]): If
±iλ1,±iλ2, . . . ,±iλn are the skew eigenvalues of a bipartite oriented graph G′ = (G, σ′),
then the eigenvalues of G are ±λ1,±λ2, . . . ,±λn if and only if G′ switches to a canoni-
cally oriented graph.
Here are more details on the cycle structure of Ḣ of Theorem 2.1(ii). It will be used in
the forthcoming sections.
Theorem 3.2. Let G′ and Ḣ be as in Theorem 2.1(ii). For every odd cycle C ′ of G′, the
signed cycle Ċ of Ḣ associated with bd(C ′) is negative.
Proof. If the vertices of C ′, labelled in the natural order, are 1, 2, . . . , ℓ, then the vertices
of its bipartite double bd(C ′) are divided into the colour classes, say A = {a1, a2, . . . , aℓ}
and B{b1, b2, . . . , bℓ}, the edges of bd(C ′) are a1b2, b2a3, a3b4, . . . , bℓ−1aℓ, aℓb1,
b1a2, . . . , aℓ−1bℓ, bℓa1, and these edges inherit the orientation from G′ in the sense that
i→ j implies ai → bj and bi → aj .
Now, if the edges of G′ are oriented in the route direction, so are the edges of bd(G′).
In this case, the edges of the associated signed cycle Ċ alternate in sign, which in particular
means that Ċ is negative (as it counts exactly ℓ negative edges and ℓ is odd) and the edges
aibj and biaj differ in sign (again, since ℓ is odd). If the edges of G′ are not oriented in
the route direction, then C ′ is obtained from the previous cycle by reversing the orientation
of some edges. The desired conclusion follows since changing the orientation of a single
edge ij changes the orientation of both aibj and biaj (in bd(C ′)) and changes the sign of
both aibj and biaj (in Ċ), so it does not change the signature of Ċ.
Corollary 3.3. Let G′ and Ḣ be as in Theorem 2.1(ii). Then Ḣ is unbalanced and bd(G′)
contains a cycle that is not oriented uniformly.
Proof. The first statement follows from the previous theorem. The second one follows
from an easy observation that a negative cycle in Ḣ corresponds to a cycle in bd(G′) that
is not oriented uniformly.
4 Characteristic polynomials
Here are relations between the coefficients of the characteristic polynomials.
Theorem 4.1. If
∑n
i=0 six
n−i is the characteristic polynomial of an oriented graph G′,
then si = 0 for i odd and si ≥ 0 for i even. If G′ is bipartite and
∑⌊n/2⌋
i=0 a2ix
n−2i is the
characteristic polynomial of an associated signed graph, then
a2i = (−1)is2i. (4.1)
If G′ is non-bipartite and
∑n
i=0 a2ix
n−2i is the characteristic polynomial of Ḣ (where Ḣ
is as in the formulation of Theorem 2.1(ii)), then
558 Ars Math. Contemp. 24 (2024) #P3.10 / 553–566
a2i = (−1)i
min{i,n−i}∑
ℓ = −min{i, n − i}
i ≡ ℓ (mod 2)
si−ℓsi+ℓ. (4.2)
Proof. Under the assumptions of Theorem 2.1, the characteristic polynomial of the skew
adjacency matrix of G′ is
xn + s1x
n−1 + · · ·+ sn−1x+ sn = xn−2k(x2 + λ21)(x2 + λ22) · · · (x2 + λ2k). (4.3)
We immediately obtain si = 0 for i odd (since si is the ith elementary symmetric polyno-
mial in eigenvalues), and si ≥ 0 for i even.
If G′ is bipartite, then the characteristic polynomial of an associated signed graph reads∑⌊n/2⌋
i=0 a2ix
n−2i = xn−2k(x2 − λ21)(x2 − λ22) · · · (x2 − λ2k), which yields that even coef-
ficients alternate in sign, and |a2i| = s2i. This implies the equality (4.1).
If G′ is non-bipartite, then the characteristic polynomial of Ḣ is
n∑
i=0
a2ix
n−2i =
(
xn−2k(x2 − λ21)(x2 − λ22) · · · (x2 − λ2k)
)2
.
Comparing it with (4.3), we arrive at
a2i = (−1)i
min{i,n−i}∑
ℓ=−min{i,n−i}
si−ℓsi+ℓ,
but we know that odd coefficients under the sum are zero, so the summation reduces to
even coefficients, which leads to (4.2).
For 2i ≤ n, we note that the formula (4.2) is simplified to
a2i = (−1)i
i∑
ℓ=0
s2ℓs2(i−ℓ).
We visualize this in the following example.
Example 4.2. Let G′ be the oriented graph with 10 vertices illustrated in Figure 1. The
coefficients of its characteristic polynomial are (s0, s2, . . . , s10) = (1, 15, 60, 92, 48, 0).
The coefficients of the characteristic polynomial of Ḣ of Theorem 2.1(ii) are
(a0, a2, . . . , a20) = (1,−30, 345,−1984, 6456,−12480, 14224,−8832, 2304, 0, 0).
Say, 345 = a4 = s0s4 + s22 + s4s0 = 60 + 225 + 60 or −8832 = a14 = (−1)7(s10s4 +
s8s6 + s6s8 + s4s10) = 0− 2 · 92 · 48 + 0, as in Theorem 4.1.
We recall that a basic figure in a graph is a disjoint union of edges and cycles. If∑n
i=0 aix
i is the characteristic polynomial of the adjacency matrix of a signed graph Ġ =
(G, σ̇), then from [4]
ai =
∑
B∈Bi
(−1)p(B)σ̇(B)2|c(B)|,
Z. Stanić: Spectra of signed graphs and related oriented graphs 559
where Bi is the set of basic figures on i vertices in G, p(B) is the number of components of
B, c(B) is the set of cycles in B and σ̇(B) =
∏
Ċ∈c(B) σ̇(Ċ). This result can be extended
to oriented graphs on the basis of Theorems 2.1 and 4.1. However, this has already been
performed in [5, pages 4516–4517] and [9, Theorem 2.3], and so we just refer the reader to
these references. It is worth mentioning that these results address the research problem of
[1, Section 6], asking for an interpretation of the coefficients si in terms of G′.
We conclude this section by the following observation. If the characteristic polynomials
of G (the underlying graph), Ġ = (G, σ̇) (a signed graph) and G′ = (G, σ′) (an oriented
graph) are
∑n
i=0 ai(G)x
n−i,
∑n
i=0 ai(Ġ)x
n−i and
∑n
i=0 si(G
′)xn−i, respectively, then
ai(G) = ai(Ġ) = si(G
′) (mod 2), for 1 ≤ i ≤ n, as AG = AĠ = SG′ (mod 2). Since
si(G
′) = 0 for i odd, this in particular means that ai(G) and ai(Ġ) are even, whenever i is
odd. Moreover, we have the following consequence.
Theorem 4.3. For a graph G, let Ġ (resp. G′) consist of all signed graphs (oriented graphs)
having G as the underlying graph. If there is at least one Γ ∈ {G} ∪ Ġ ∪ G′, such that the
determinant of Γ is odd, then G, all signed graphs of Ġ and all oriented graphs of G′ are
non-singular (i.e. their determinant is non-zero).
Proof. This result follows from the previous observation applied to a0(G), a0(Ġ) and
s0(G
′). Indeed, if for example a0(Ġ) is odd, then a0(Ġ) = a0(G) = 1 (mod 2), and
the same holds for the elements of Ġ ∪ G′ in the role of G.
5 Eigenspaces
Let E(λ) and E(−λ) be the eigenspaces of eigenvalues λ ( ̸= 0) and−λ of a bipartite signed
graph Ġ, and E(iλ), E(−iλ) the eigenspaces of iλ and −iλ belonging to the spectrum of
an associated oriented graph G′. Then E(iλ) ∪ E(−iλ) is spanned (in Cn) by the union
of bases of E(λ) and E(−λ). Indeed, since −S2G′ = A2Ġ, the eigenspace of λ
2 for A2
Ġ
coincides with the eigenspace of−λ2 for S2, and thus the former eigenspace is spanned by
the union of eigenbases of λ and −λ (for AĠ), and the latter one is spanned by the union
of eigenbases of iλ and −iλ (for SG′ ). By the same reasoning, if 0 is an eigenvalue of Ġ,
then it has the same eigenspace for Ġ and G′. However, we can say more.
We first note the following, probably known, result. If x is an eigenvector associated
with the eigenvalue λ (̸= 0) of a bipartite signed graph Ġ, then by negating the entries on
one colour class, we get an eigenvector associated with −λ. Indeed, with an appropriate
vertex labelling, the adjacency matrix has the form
AĠ =
(
O B
B⊺ O
)
(5.1)
If x =
(
x1
x2
)
, then AĠ
(
−x1
x2
)
=
(
Bx2
−B⊺x1
)
= −λ
(
−x1
x2
)
, as desired. If AĠ has
no the previous form, the eigenvectors are permutationally equal, and the result follows.
Henceforth, we assume that the adjacency matrix of a bipartite signed graph is as in (5.1).
Theorem 5.1. Let xj =
(
x1
x2
)
and x−j =
(
−x1
x2
)
be eigenvectors associated with dis-
tinct eigenvalues λj and −λj of a bipartite signed graph Ġ, and let, with a consistent
vertex labelling, G′ be an associated oriented graph. Then xj + ix−j and xj − ix−j are
eigenvectors associated with iλj and −iλj .
560 Ars Math. Contemp. 24 (2024) #P3.10 / 553–566
Proof. If the adjacency matrix of Ġ has the form (5.1), then the skew adjacency matrix
of G′ is SG′ =
(
O B
−B⊺ O
)
. Indeed, in the adjacency matrix of the underlying graph
G, B has no negative entries, and reversing the sign of an edge uv changes the sign of both
auv, avu in AĠ and both suv, svu in SG′ .
We compute
SG′(xj ± ix−j) =
(
Bx2
−B⊺x1
)
± i
(
Bx2
B⊺x1
)
=
(
λjx1
−λjx2
)
± i
(
λjx1
λjx2
)
= ± iλj
((
x1
x2
)
± i
(
−x1
x2
))
= ±iλj(xj ± ix−j),
as desired.
The previous theorem tells us that, in the bipartite case, the eigenspaces of iλj and−iλj
of G′ contain vectors of the form
yij =
(
x1
x2
)
+ i
(
−x1
x2
)
and y−ij = yij =
(
x1
x2
)
− i
(
−x1
x2
)
,
respectively. Conversely, if these eigenvectors are given, then the eigenvectors associated
with λj and −λj of Ġ are easily computed. We proceed with the non-bipartite case.
Theorem 5.2. Let G′ be a non-bipartite oriented graph and y and y eigenvectors as-
sociated with the eigenvalues iλj and −iλj , respectively. Then the pair
(
y
y
)
,
(
y
−y
)
(resp.
(
y
y
)
,
(
y
−y
)
) is associated with iλj (resp. −iλj) in the bipartite double bd(G′),
where Sbd(G′) =
(
0 1
1 0
)
⊗ SG′ . Let Ḣ be a signed graph associated with bd(G′), and
k the multiplicity of λj in the spectrum of Ḣ . The first (resp. second) pair is spanned by
xjℓ + ix−jℓ (resp. xjℓ − ix−jℓ), for 1 ≤ ℓ ≤ k, where xjℓ and x−jℓ span the eigenspaces
of λj and −λj , respectively.
Proof. Since Sbd(G′) is the tensor product, the eiegnvectors of iλj for bd(G′) are (1, 1)⊺⊗
y and (1,−1)⊺ ⊗ y, and similarly for −iλj . This establishes the proof of the first part of
the statement.
By Theorem 5.1, the eigenvectors of iλj for bd(G′) are xjℓ + ix−jℓ, for 1 ≤ ℓ ≤ k.
They are obviously linearly independent (because xjℓ and x−jℓ are). Since the multiplicity
of iλj in the spectrum of bd(G′) is k (cf. Theorem 2.1(ii)), its geometric multiplicity is at
most k and the desired conclusion follows.
We proceed with an application in control theory. The following differential equation
is the standard model of multi-agent single-input linear control systems:
dx
dt
= Mx+ bu, (5.2)
where the scalar u = u(t) is the control input, M ∈ Rn × Rn and x,b ∈ Rn. The system
is controllable if for any x⋆ and time t⋆, there exists a control function u(t), 0 ≤ t ≤ t⋆,
Z. Stanić: Spectra of signed graphs and related oriented graphs 561
such that the solution of the differential equation gives x⋆ = x(t⋆) irrespective of x(0).
There are many controllable criteria, one of which is the Popov-Belevitch-Hautus Test to
be found in any of [6, 11]. Accordingly, the system is controllable if and only if there is no
z ∈ Cn \ {0} such that z∗M = λz∗ and z∗b = 0, where ∗ denotes the complex conjugate.
Theorem 5.3. Let A be the adjacency matrix of a bipartite signed graph and S the skew
adjacency matrix of an associated oriented graph. If the system (5.2) with M = A is
controllable, then it is controllable with M = S.
Proof. Since the system with M = A is controllable, we have x⊺b ̸= 0 for every eigen-
vector of A. Moreover A has no repeated eigenvalues. Indeed, if x1 and x2 are linearly
independent eigenvectors associated with the same eigenvalue, with x⊺1b = b1 ̸= 0 and
x⊺2b = b2 ̸= 0, then (b2x1 − b1x2)⊺b = 0.
As z∗S = λz∗ is equivalent to Sz = λz, and z is an eigenvector for S if and only if z∗
is an eigenvector for the same matrix, we have to show that for every eigenvector y of S,
y∗b = 0 holds.
Since A and S share the same eigenspace for 0, the claim holds for eigenvectors asso-
ciated with 0 (if any). Every eigenvector associated with iλj has the form z(xj + ix−j),
where z = z1 + iz2 ̸= 0, and xj and x−j are as in the formulation of Theorem 5.1. As
before, we may take x⊺jb = b1 ̸= 0 and x
⊺
−jb = b2 ̸= 0. Then
(z(xj + ix−j))
∗b =
(
z1xj − z2x−j + i(z1x−j + z2xj)
)∗
b
=(z1xj − z2x−j)⊺b− i(z1x−j + z2xj)⊺b
= z1b1 − z2b2 + i(z1b2 + z2b1).
Equating with 0, we obtain z1 = b2b1 z2 and z2 = −
b2
b1
z1, which leads to the impossible
scenario z1 = z2 = 0. Hence, the system with M = S is controllable.
6 Energy
We start with the following result concerning signed graphs. There is a similar result for
oriented graphs reported in [1].
Theorem 6.1. Let E(Ġ) be the energy of a signed graph Ġ having n vertices, m edges,
average vertex degree d = 2mn and eigenvalues λ1, λ2, . . . , λn. Then√√√√2(m+ (n
2
)( n∏
i=1
|λi|
) 2
n
)
≤ E(Ġ) ≤ n
√
d. (6.1)
Both equalities hold if and only if Ġ is either edgeless or has exactly two distinct eigenval-
ues and these eigenvalues are equal in absolute value.
Proof. The proof of inequalities of (6.1) is an imitation of the proof of [1, Theorem 2.5]
(concerning oriented graphs). Namely, both follow from the next chain of inequalities and
562 Ars Math. Contemp. 24 (2024) #P3.10 / 553–566
equalities:
2
(
m+
(
n
2
)( n∏
i=1
|λi|
) 2
n
)
≤
n∑
i=1
|λi|2 + 2
∑
1≤i<j≤n
|λi||λj | =
( n∑
i=1
|λi|
)2
≤n
n∑
i=1
|λi|2 = 2mn = n2d.
To clarify, the first inequality in above chain follows from the inequality between the geo-
metric mean and the arithmetic mean, in our case:
( n∏
i=1
|λi|
) 2
n
= n(n−1)
√√√√ n∏
i=1
|λi|2(n−1) = n(n−1)
√ ∏
1≤i<j≤n
(|λi||λj |)2
≤
2
∑
1≤i<j≤n |λi||λj |
n(n− 1)
.
The second inequality in the chain is a consequence of the Cauchy-Schwarz inequality.
Consider now the equality cases. The first equality in (6.1) holds if and only if |λi||λj |
is a constant for every i ̸= j. This occurs if either λi = 0, for all i, or λi = ±λj ̸= 0,
for all i ̸= j. The second equality holds if and only if (|λ1|, |λ2|, . . . , |λn|) is a constant
vector (possibly zero). Hence, both inequalities hold if and only if Ġ is as in the statement
formulation.
We observe that, in the particular case of graphs, the equalities in (6.1) are attained if
and only if G a disjoint union of either isolated vertices or isolated edges. In the ‘signed’
case, there are many other examples. All of them are regular with an even number of
vertices, say 2n, and a symmetric spectrum of the form [λn, (−λ)n]. These signed graphs
belong to the class of strongly regular signed graphs in the sense of definition of [13];
we note in passing that this definition generalizes the concept of strongly regular graphs.
Considering the minimal polynomial, we deduce that the common vertex degree is equal to
λ2. Therefore, λ is the square root of an integer. Disconnected examples are not of interest,
since each of them is a disjoint union of connected ones. We know from [14] that a signed
r-cube without positive quadrangles has the spectrum
[√
r
2r−1
, (−
√
r)2
r−1]
. There are
no other examples for λ ≤
√
3. Signed graphs with spectrum [2n, (−2)n] are completely
determined in [10, 16] (see also [12]). For those with [
√
5
n
, (−
√
5)n], see [17]. Some
other constructions can be found in [8]. We also remark that all signed line graphs with the
required spectrum are constructed in [16].
Observe next that the upper bound (6.1) does not depend on the eigenvalues of Ġ. This,
in particular, means that it simultaneously holds for a signed graph Ġ and its underlying
graph G. According to the previous discussion, in the connected case, this bound is simul-
taneously attained for Ġ and G if and only if G ∼= K1 or G ∼= K2; so, in exactly two simple
cases, and in both Ġ switches to G. For the remaining connected signed graphs that attain
this bound, their underlying graphs do not attain it. This leads to an assumption that the en-
ergy of a signed graph could often be larger than the energy of its underlying graph. In this
context we have experimented with a large number of connected graphs having small order,
or small number of edges (i.e., obtained by inserting few edges to a tree), or large number
of edges (i.e., obtained by deleting few edges of a complete graph). Our conclusions are
Z. Stanić: Spectra of signed graphs and related oriented graphs 563
Figure 2: The signed graph Ġ with 7.814 ≈ E(Ġ) < E(G) ≈ 7.996.
summarized as follows: E(G) = E(Ġ) holds if Ġ switches to G, or G is non-bipartite
and Ġ switches to −G. (For example, the latter occurs for every unicyclic graph with an
odd cycle). We have encountered hundreds of examples in which E(G) < E(Ġ). An ex-
ample in which E(G) > E(Ġ) is illustrated in Figure 2. Motivated by these experiments,
we formulate the following problems.
Problem 6.2. Determine graphs G such that E(G) ≤ E((G, σ̇)) holds for every signature σ̇
defined on the edge set of G.
Problem 6.3. Determine (or, at least, characterize) signed graphs Ġ with E(Ġ) < E(G).
We proceed with oriented graphs. If G′ is associated with Ġ, then E(G′) = E(Ġ), by
Theorem 2.1(i). If Ḣ is as in Theorem 2.1(ii), then E(G′) = E(Ḣ)2 . Accordingly, if the
underlying graph G is regular of degree r, then E(G′) = n
√
r (the upper bound (6.1))
if and only if G′ has exactly two eigenvalues (complex conjugates); this is established
in [1], as well. Some examples are constructed in the mentioned reference. Here we note
that the search on regular oriented graphs attaining the upper bound (6.1) is reduced to
the search on signed graphs with the same spectral property: An associate of a bipartite
signed graph with E(Ġ) = n
√
r has the required spectral property, and if it figures as a
bipartite double, then a corresponding constituent has the same property. In this context it
is worth mentioning that, on the basis of the results of [10, 16], all oriented graphs with n
vertices and spectrum [i2n/2, (−i2)n/2] are determined in [15] (their energy attains 2n);
they include infinite families of both bipartite and non-bipartite oriented graphs. More
examples can be extracted from known signed graphs with two eigenvalues obtained in the
foregoing references.
We also observe that an oriented graph shares the energy with its underlying graph G
whenever it is associated with a balanced signed graph. In this context, we point out that,
due to Corollary 3.3, Ḣ of Theorem 2.1(ii) is never balanced. It also shares the energy
with G whenever it is associated with a signed graph that switches to −G.
This section is concluded with the following result. A conference matrix A is an n× n
matrix with diagonal entries 0 and off-diagonal entries ±1, satisfying A⊺A = (n− 1)I .
Theorem 6.4. An oriented graph G′ (resp. a signed graph Ġ) with n vertices attains
E(G′) = n
√
n− 1 (E(Ġ) = n
√
n− 1) if and only if its adjacency matrix is the skew-
symmetric conference matrix (symmetric conference matrix).
Proof. If E(G′) = n
√
n− 1, by Theorems 2.1(ii) and 6.1, G′ is regular of degree n − 1
and its spectrum is
[
i
√
n− 1n/2,−(i
√
n− 1)n/2
]
. Considering the minimal polynomial
of AG′ , we obtain A2G′ = (n−1)I , which means that AG′ is a skew-symmetric conference
564 Ars Math. Contemp. 24 (2024) #P3.10 / 553–566
matrix. Conversely, if AG′ is a skew-symmetric conference matrix then, by definition, we
have A2G′ = (n− 1)I , which gives E(G′) = n
√
n− 1.
The proof for Ġ is analogous.
The previous result addresses the open problem (6) of [1, Section 6] related to the
existence of skew-symmetric conference matrices that do not give the maximum energy of
the corresponding oriented graph. It also partially addresses the problem (2) of the same
reference related to determination of oriented graphs with maximum energy, as it gives their
characterization via the matrix structure. However, their determination remains a difficult
research problem, since it is equivalent to the complete determination of skew-symmetric
conference matrices.
7 Notes on hypercubes
Due to [22, Theorem 2.1(iv)] (see also [21, Theorem 2]) a signed graph is balanced if its
cycle basis has all positive cycles. Since the induced quadrangles of a signed r-cube contain
a cycle basis, we arrive at the following result.
Lemma 7.1 (cf. [22, Theorem 2.1(iv)]). For r ≥ 2, a signed r-cube Q̇r is balanced if and
only if it has no negative quadrangles.
In [2, 3, 20], the authors gave several algorithms which iteratively construct an oriented
r-cube such that its energy is either equal to the energy of the underlying graph or attains
the upper bound (6.1). (This upper bound reduces to n
√
r.) These algorithms are useful
since they give explicit orientations for which the previous settings occur. In this context
we offer the following contribution (see also the text below the theorem).
Theorem 7.2. The following statements hold true:
(i) A signed r-cube without negative quadrangle is associated with an oriented r-cube
whose energy is equal to the energy of its underlying r-cube.
(ii) A signed r-cube is associated with an oriented r-cube whose energy attains the upper
bound of (6.1) if and only if it has no positive quadrangle.
Proof. (i): If every quadrangle (if any) in Q̇r is positive then Q̇r is balanced, by Lemma 7.1,
and therefore it switches to the underlying graph Qr. Consequently, Q̇r and Qr have the
same energy. By Theorem 2.1, Q̇r shares the energy with an associated oriented cube, and
the result follows.
(ii): Since a signed graph and an associated oriented graph share the same energy, it
is sufficient to show that the energy n
√
r is exclusive to a signed cube without positive
quadrangles. First, such a cube has this energy (as mentioned in the previous section).
Conversely, assume that, for r ≥ 3, a signed r-cube with the adjacency matrix A has a
positive quadrangle, say uvwz (where the vertices are in the natural order). It follows that
the (u,w)-entry of A2 is 2, and so A2 is not a multiple of the identity matrix, but then the
spectrum of A deviates the equality condition of Theorem 6.1, as follows by considering
the minimal polynomial.
In other words, to construct an oriented r-cube whose energy is E(Qr) (resp. n
√
r),
it is sufficient to take any signed r-cube without negative quadrangle (without positive
quadrangle), and construct an oriented associate.
Z. Stanić: Spectra of signed graphs and related oriented graphs 565
ORCID iDs
Zoran Stanić https://orcid.org/0000-0002-4949-4203
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