Volume 22, Number 3, Fall/Winter 2022, Pages 363–525
Covered by:
Mathematical Reviews
zbMATH (formerly Zentralblatt MATH)
COBISS
SCOPUS
Science Citation Index-Expanded (SCIE)
Web of Science
ISI Alerting Service
Current Contents/Physical, Chemical & Earth Sciences (CC/PC & ES)
dblp computer science bibliography
The University of Primorska
The Society of Mathematicians, Physicists and Astronomers of Slovenia
The Institute of Mathematics, Physics and Mechanics
The Slovenian Discrete and Applied Mathematics Society
The publication is partially supported by the Slovenian Research Agency from the Call for
co-financing of scientific periodical publications.

Contents
A compact presentation for the alternating central extension of the
positive part of Uq(ŝl2)
Paul M. Terwilliger . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363
The adjacency dimension of graphs
Sergio Bermudo, José M. Rodríguez, Juan A. Rodríguez-Velázquez,
José M. Sigarreta . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 387
On the chromatic index of generalized truncations
Brian Alspach, Aditya Joshi . . . . . . . . . . . . . . . . . . . . . . . . . 403
Convex drawings of the complete graph: topology meets geometry
Alan Arroyo, Dan McQuillan, R. Bruce Richter, Gelasio Salazar . . . . . . 415
On the essential annihilating-ideal graph of commutative rings
Mohd Nazim, Nadeem ur Rehman . . . . . . . . . . . . . . . . . . . . . . 443
Cell reducing and the dimension of the C1 bivariate spline space
Gašper Jaklič . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 459
Avoidance in bowtie systems
Mike J. Grannell, Terry S. Griggs, Giovanni Lo Faro, Antoinette Tripodi . . 477
Two families of pseudo metacirculants
Li Cui, Jin-Xin Zhou . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 489
Partial-dual Euler-genus distributions for bouquets with
small Euler genus
Yan Yang, Xiaoya Zha . . . . . . . . . . . . . . . . . . . . . . . . . . . . 505
The (non-)existence of perfect codes in Lucas cubes
Michel Mollard . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 519
Volume 22, Number 3, Fall/Winter 2022, Pages 363–525
xix

ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 22 (2022) #P3.01 / 363–386
https://doi.org/10.26493/1855-3974.2669.58c
(Also available at http://amc-journal.eu)
A compact presentation for the alternating
central extension of the positive part of Uq(ŝl2)
Paul M. Terwilliger *
Department of Mathematics, University of Wisconsin,
480 Lincoln Drive, Madison, WI 53706-1388 USA
Received 6 July 2021, accepted 24 August 2021, published online 9 June 2022
Abstract
This paper concerns the positive part U+q of the quantum group Uq(ŝl2). The algebra
U+q has a presentation involving two generators that satisfy the cubic q-Serre relations.
We recently introduced an algebra U+q called the alternating central extension of U+q . We
presented U+q by generators and relations. The presentation is attractive, but the multitude
of generators and relations makes the presentation unwieldy. In this paper we obtain a
presentation of U+q that involves a small subset of the original set of generators and a very
manageable set of relations. We call this presentation the compact presentation of U+q .
Keywords: q-Onsager algebra, q-Serre relations, q-shuffle algebra, tridiagonal pair.
Math. Subj. Class. (2020): 17B37, 05E14, 81R50
1 Introduction
The algebra Uq(ŝl2) is well known in representation theory [15] and statistical mechanics
[20]. This algebra has a subalgebra U+q called the positive part. The algebra U
+
q has a
presentation involving two generators (said to be standard) and two relations, called the
q-Serre relations. The presentation is given in Definition 2.1 below.
Our interest in U+q is motivated by some applications to linear algebra and combina-
torics; these will be described shortly. Before going into detail, we have a comment about q.
In the applications, either q is not a root of unity, or q is a root of unity with exponent large
enough to not interfere with the rest of the application. To keep things simple, throughout
the paper we will assume that q is not a root of unity.
*The author thanks Pascal Baseilhac for many conversations about U+q and its central extension U+q . The
author thanks Kazumasa Nomura for giving this paper a close reading and offering many valuable comments.
E-mail address: terwilli@math.wisc.edu (Paul M. Terwilliger)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
364 Ars Math. Contemp. 22 (2022) #P3.01 / 363–386
Our first application has to do with tridiagonal pairs [17]. A tridiagonal pair is roughly
described as an ordered pair of diagonalizable linear maps on a nonzero finite-dimensional
vector space, that each act on the eigenspaces of the other one in a block-tridiagonal fashion
[17, Definition 1.1]. There is a type of tridiagonal pair said to be q-geometric [18, Defini-
tion 2.6]; for this type of tridiagonal pair the eigenvalues of each map form a q2-geometric
progression. A finite-dimensional irreducible U+q -module on which the standard genera-
tors are not nilpotent, is essentially the same thing as a tridiagonal pair of q-geometric type
[18, Theorem 2.7]; these U+q -modules are described in [18, Section 1]. See [13, 24] for
more background on tridiagonal pairs.
Our next application has to do with distance-regular graphs [1, 14, 32]. Consider a
distance-regular graph Γ that has diameter d ≥ 3 and classical parameters (d, b, α, β)
[14, p. 193] with b = q2 and α = q2 − 1. The condition on α implies that Γ is formally
self-dual in the sense of [14, p. 49]. Let A denote the adjacency matrix of Γ, and let A∗
denote the dual adjacency matrix with respect to any vertex of Γ [19, Section 7]. Then by
[19, Lemma 9.4], there exist complex numbers r, s, r∗, s∗ with r, r∗ nonzero such that
rA + sI , r∗A∗ + s∗I satisfy the q-Serre relations. As mentioned in [19, Example 8.4],
the above parameter restriction is satisfied by the bilinear forms graph [14, p. 280], the
alternating forms graph [14, p. 282], the Hermitean forms graph [14, p. 285], the quadratic
forms graph [14, p. 290], the affine E6 graph [14, p. 340], and the extended ternary Golay
code graph [14, p. 359].
Our next application has to do with uniform posets [23, 27]. Let GF(b) denote a finite
field with b elements, and let N,M denote positive integers. Let H denote a vector space
over GF(b) that has dimension N +M . Let h denote a subspace of H with dimension M .
Let P denote the set of subspaces of H that have zero intersection with h. For x, y ∈ P
define x ≤ y whenever x ⊆ y. The relation ≤ is a partial order on P , and the poset P
is ranked with rank N . The poset P is called an attenuated space poset, and denoted by
Ab(N,M) [21], [27, Example 3.1]. By [27, Theorem 3.2] the poset Ab(N,M) is uniform
in the sense of [27, Definition 2.2]. It is shown in [21, Lemma 3.3] that for Ab(N,M)
the raising matrix R and the lowering matrix L satisfy the q-Serre relations, provided that
b = q2.
Our last application has to do with q-shuffle algebras. Let F denote a field, and let x, y
denote noncommuting indeterminates. Let V denote the free associative F-algebra with
generators x, y. By a letter in V we mean x or y. For an integer n ≥ 0, by a word of length
n in V we mean a product of letters v1v2 · · · vn. The words in V form a basis for the vector
space V . In [25, 26] M. Rosso introduced an algebra structure on V , called the q-shuffle
algebra. For letters u, v their q-shuffle product is u ⋆ v = uv + q⟨u,v⟩vu, where ⟨u, v⟩ = 2
(resp. ⟨u, v⟩ = −2) if u = v (resp.u ̸= v). By [25, Theorem 13], in the q-shuffle algebra
V the elements x, y satisfy the q-Serre relations. Consequently there exists an algebra
homomorphism ♮ from U+q into the q-shuffle algebra V , that sends the standard generators
of U+q to x, y. By [26, Theorem 15] the map ♮ is injective.
Next we recall the alternating elements in U+q [30]. Let v1v2 · · · vn denote a word in
V . This word is called alternating whenever n ≥ 1 and vi−1 ̸= vi for 2 ≤ i ≤ n. Thus the
alternating words have the form · · ·xyxy · · · . The alternating words are displayed below:
x, xyx, xyxyx, xyxyxyx, . . .
y, yxy, yxyxy, yxyxyxy, . . .
yx, yxyx, yxyxyx, yxyxyxyx, . . .
P. M. Terwilliger: A compact presentation for the alternating central extension of the positive . . . 365
xy, xyxy, xyxyxy, xyxyxyxy, . . .
By [30, Theorem 8.3] each alternating word is contained in the image of ♮. An element
of U+q is called alternating whenever it is the ♮-preimage of an alternating word. For ex-
ample, the standard generators of U+q are alternating because they are the ♮-preimages of
the alternating words x, y. It is shown in [30, Lemma 5.12] that for each row in the above
display, the corresponding alternating elements mutually commute. A naming scheme for
alternating elements is introduced in [30, Definition 5.2].
Next we recall the alternating central extension of U+q [29]. In [30] we displayed two
types of relations among the alternating elements of U+q ; the first type is [30, Proposi-
tions 5.7, 5.10, 5.11] and the second type is [30, Propositions 6.3, 8.1]. The relations
in [30, Proposition 5.11] are redundant; they follow from the relations in [30, Proposi-
tions 5.7, 5.10] as pointed out in [2, Propositions 3.1, 3.2] and [5, Remark 2.5]; see also
Corollary 6.3 below. The relations in [30, Proposition 6.3] are also redundant; they follow
from the relations in [30, Propositions 5.7, 5.10] as shown in the proof of [30, Proposi-
tion 6.3]. By [30, Lemma 8.4] and the previous comments, the algebra U+q is presented by
its alternating elements and the relations in [30, Propositions 5.7, 5.10, 8.1]. For this presen-
tation it is natural to ask what happens if the relations in [30, Proposition 8.1] are removed.
To answer this question, in [29, Definition 3.1] we defined an algebra U+q by generators and
relations in the following way. The generators, said to be alternating, are in bijection with
the alternating elements of U+q . The relations are the ones in [30, Propositions 5.7, 5.10].
By construction there exists a surjective algebra homomorphism U+q → U+q that sends
each alternating generator of U+q to the corresponding alternating element of U+q . In
[29, Lemma 3.6, Theorem 5.17] we adjusted this homomorphism to get an algebra isomor-
phism U+q → U+q ⊗F[z1, z2, . . .], where {zn}∞n=1 are mutually commuting indeterminates.
By [29, Theorem 10.2] the alternating generators form a PBW basis for U+q . The algebra
U+q is called the alternating central extension of U+q .
We mentioned above that the algebra U+q is presented by its alternating generators and
the relations in [30, Propositions 5.7, 5.10]. This presentation is attractive, but the multitude
of generators and relations makes the presentation unwieldy. In this paper we obtain a
presentation of U+q that involves a small subset of the original set of generators and a very
manageable set of relations. This presentation is given in Definition 3.1 below; we call it
the compact presentation of U+q . At first glance, it is not clear that the algebra presented in
Definition 3.1 is equal to U+q . So we denote by U the algebra presented in Definition 3.1,
and eventually prove that U = U+q . After this result is established, we describe some
features of U+q that are illuminated by the presentation in Definition 3.1.
Our investigation of U+q is inspired by some recent developments in statistical mechan-
ics, concerning the q-Onsager algebra Oq . In [9] Baseilhac and Koizumi introduce a current
algebra Aq for Oq , in order to solve boundary integrable systems with hidden symmetries.
In [12, Definition 3.1] Baseilhac and Shigechi give a presentation of Aq by generators and
relations. This presentation and the discussion in [12, Section 4] suggest that Aq is related
to Oq in roughly the same way that U+q is related to U+q . The relationship between Aq and
Oq was conjectured in [7, Conjectures 1, 2] and [28, Conjectures 4.5, 4.6, 4.8], before be-
ing settled in [31, Theorems 9.14, 10.2, 10.3, 10.4]. The articles [3, 4, 6, 7, 8, 9, 10, 11, 12]
contain background information on Oq and Aq .
Earlier in this section, we indicated how U+q has applications to tridiagonal pairs,
distance-regular graphs, and uniform posets. Possibly U+q appears in these applications,
and this possibility should be investigated in the future.
366 Ars Math. Contemp. 22 (2022) #P3.01 / 363–386
This paper is organized as follows. In Section 2 we review some facts about U+q . In Sec-
tion 3, we introduce the algebra U and give an algebra homomorphism U+q → U . In Sec-
tion 4, we introduce the alternating generators for U and establish some formulas involving
these generators. In Sections 5, 6 we use these formulas and generating functions to show
that the alternating generators for U satisfy the relations in [30, Propositions 5.7, 5.10].
Using this result, we prove that U = U+q . Theorem 6.2 and Corollary 6.5 are the main
results of the paper. In Section 7 we describe some features of U+q that are illuminated
by the presentation in Definition 3.1. Appendix A contains a list of relations involving the
generating functions from Section 5.
2 The algebra U+q
We now begin our formal argument. For the rest of the paper, the following notational
conventions are in effect. Recall the natural numbers N = {0, 1, 2, . . .}. Let F denote a
field. Every vector space and tensor product mentioned is over F. Every algebra mentioned
is associative, over F, and has a multiplicative identity. Fix a nonzero q ∈ F that is not a
root of unity. Recall the notation
[n]q =
qn − q−n
q − q−1
n ∈ N.
For elements X,Y in any algebra, define their commutator and q-commutator by
[X,Y ] = XY − Y X, [X,Y ]q = qXY − q−1Y X.
Note that
[X, [X, [X,Y ]q]q−1 ] = X
3Y − [3]qX2Y X + [3]qXYX2 − Y X3.
Definition 2.1 ([22, Corollary 3.2.6]). Define the algebra U+q by generators W0, W1 and
relations
[W0, [W0, [W0,W1]q]q−1 ] = 0, [W1, [W1, [W1,W0]q]q−1 ] = 0. (2.1)
We call U+q the positive part of Uq(ŝl2). The generators W0,W1 are called standard. The
relations (2.1) are called the q-Serre relations.
We will use the following concept.
Definition 2.2 ([16, p. 299]). Let A denote an algebra. A Poincaré-Birkhoff-Witt (or PBW)
basis for A consists of a subset Ω ⊆ A and a linear order < on Ω such that the following is
a basis for the vector space A:
a1a2 · · · an n ∈ N, a1, a2, . . . , an ∈ Ω, a1 ≤ a2 ≤ · · · ≤ an.
We interpret the empty product as the multiplicative identity in A.
In [16, p. 299] Damiani obtains a PBW basis for U+q that involves some elements
{Enδ+α0}∞n=0, {Enδ+α1}∞n=0, {Enδ}∞n=1. (2.2)
P. M. Terwilliger: A compact presentation for the alternating central extension of the positive . . . 367
These elements are defined recursively as follows:
Eα0 = W0, Eα1 = W1, Eδ = q
−2W1W0 −W0W1 (2.3)
and for n ≥ 1,
Enδ+α0 =
[Eδ, E(n−1)δ+α0 ]
q + q−1
, Enδ+α1 =
[E(n−1)δ+α1 , Eδ]
q + q−1
, (2.4)
Enδ = q
−2E(n−1)δ+α1W0 −W0E(n−1)δ+α1 . (2.5)
Proposition 2.3 ([16, p. 308]). A PBW basis for U+q is obtained by the elements (2.2) in
the linear order
Eα0 < Eδ+α0 < E2δ+α0 < · · · < Eδ < E2δ < E3δ < · · · < E2δ+α1 < Eδ+α1 < Eα1 .
The elements (2.2) satisfy many relations [16]. We mention a few for later use.
Lemma 2.4 ([16, p. 300]). For i, j ∈ N with i > j the following hold in U+q .
(i) Assume that i− j = 2r + 1 is odd. Then
Eiδ+α0Ejδ+α0 = q
−2Ejδ+α0Eiδ+α0
− (q2 − q−2)
r∑
ℓ=1
q−2ℓE(j+ℓ)δ+α0E(i−ℓ)δ+α0 ,
Ejδ+α1Eiδ+α1 = q
−2Eiδ+α1Ejδ+α1
− (q2 − q−2)
r∑
ℓ=1
q−2ℓE(i−ℓ)δ+α1E(j+ℓ)δ+α1 .
(ii) Assume that i− j = 2r is even. Then
Eiδ+α0Ejδ+α0 = q
−2Ejδ+α0Eiδ+α0 − qj−i+1(q − q−1)E2(r+j)δ+α0
− (q2 − q−2)
r−1∑
ℓ=1
q−2ℓE(j+ℓ)δ+α0E(i−ℓ)δ+α0 ,
Ejδ+α1Eiδ+α1 = q
−2Eiδ+α1Ejδ+α1 − qj−i+1(q − q−1)E2(r+j)δ+α1
− (q2 − q−2)
r−1∑
ℓ=1
q−2ℓE(i−ℓ)δ+α1E(j+ℓ)δ+α1 .
Lemma 2.5. The following (i) – (iii) hold in U+q .
(i) (See [16, p. 307].) For positive i, j ∈ N,
EiδEjδ = EjδEiδ. (2.6)
(ii) (See [16, p. 307].) For i, j ∈ N,
[Eiδ+α0 , Ejδ+α1 ]q = −qE(i+j+1)δ. (2.7)
368 Ars Math. Contemp. 22 (2022) #P3.01 / 363–386
(iii) For i ∈ N,
[W0, Eiδ+α0 ]q
q − q−1
=
i∑
ℓ=0
Eℓδ+α0E(i−ℓ)δ+α0 , (2.8)
[Eiδ+α1 ,W1]q
q − q−1
=
i∑
ℓ=0
E(i−ℓ)δ+α1Eℓδ+α1 . (2.9)
Proof. (iii) To verify (2.8) and (2.9), use Lemma 2.4 to write each term in the PBW basis
for U+q from Proposition 2.3. We give the details for (2.8). Referring to (2.8), let ∆ denote
the right-hand side minus the left-hand side. We show that ∆ = 0. This is quickly verified
for i = 0, so assume that i ≥ 1. For i even (resp. i odd) write i = 2r (resp. i = 2r + 1).
Using Lemma 2.4 we obtain ∆ =
∑r
ℓ=0 αℓEℓδ+α0E(i−ℓ)δ+α0 , where for i even,
α0 = 1 + q
−2 − q
q − q−1
+
q−3
q − q−1
,
αℓ = 1 + q
−2 − (q2 − q−2)
ℓ∑
k=1
q−2k − (q + q−1)q−2ℓ−1 (1 ≤ ℓ ≤ r − 1),
αr = 1− (q − q−1)
r∑
k=1
q1−2k − q−i
and for i odd,
α0 = 1 + q
−2 − q
q − q−1
+
q−3
q − q−1
,
αℓ = 1 + q
−2 − (q2 − q−2)
ℓ∑
k=1
q−2k − (q + q−1)q−2ℓ−1 (1 ≤ ℓ ≤ r).
For either case αℓ = 0 for 0 ≤ ℓ ≤ r, so ∆ = 0. We have verified (2.8). For (2.9) the
details are similar, and omitted.
3 An extension of U+q
In this section we introduce the algebra U . In Section 6 we will show that U coincides with
the alternating central extension U+q of U+q .
Definition 3.1. Define the algebra U by generators W0, W1, {G̃k+1}k∈N and relations
(i) [W0, [W0, [W0,W1]q]q−1 ] = 0,
(ii) [W1, [W1, [W1,W0]q]q−1 ] = 0,
(iii) [G̃1,W1] = q
[[W0,W1]q,W1]
q2−q−2 ,
(iv) [W0, G̃1] = q
[W0,[W0,W1]q ]
q2−q−2 ,
P. M. Terwilliger: A compact presentation for the alternating central extension of the positive . . . 369
(v) for k ≥ 1,
[G̃k+1,W1] =
[[[G̃k,W0]q,W1]q,W1]
(1− q−2)(q2 − q−2)
,
(vi) for k ≥ 1,
[W0, G̃k+1] =
[W0, [W0, [W1, G̃k]q]q]
(1− q−2)(q2 − q−2)
,
(vii) for k, ℓ ∈ N,
[G̃k+1, G̃ℓ+1] = 0.
For notational convenience define G̃0 = 1.
Note 3.2. Referring to Definition 3.1, the relation (iii) (resp. (iv)) is obtained from (v)
(resp. (vi)) by setting k = 0.
Lemma 3.3. There exists a unique algebra homomorphism ♭ : U+q → U that sends W0 7→
W0 and W1 7→ W1.
Proof. Compare Definitions 2.1, 3.1.
In Corollary 6.7 we will show that ♭ is injective. Let ⟨W0,W1⟩ denote the subalgebra
of U generated by W0,W1. Of course ⟨W0,W1⟩ is the ♭-image of U+q . For the elements
(2.2) of U+q , the same notation will be used for their ♭-images in ⟨W0,W1⟩.
4 Augmenting the generating set for U
Some of the relations in Definition 3.1 are nonlinear. Our next goal is to linearize the
relations by adding more generators.
Definition 4.1. We define some elements in U as follows. For k ∈ N,
W−k =
[G̃k,W0]q
q − q−1
, (4.1)
Wk+1 =
[W1, G̃k]q
q − q−1
, (4.2)
Gk+1 = G̃k+1 +
[W1,W−k]
1− q−2
. (4.3)
For notational convenience define G0 = 1.
Lemma 4.2. For k ∈ N the following hold in U:
G̃kW0 = q
−2W0G̃k + (1− q−2)W−k,
G̃kW1 = q
2W1G̃k + (1− q2)Wk+1.
Proof. These are reformulations of (4.1) and (4.2).
370 Ars Math. Contemp. 22 (2022) #P3.01 / 363–386
The following is a generating set for U :
{W−k}k∈N, {Wk+1}k∈N, {Gk+1}k∈N, {G̃k+1}k∈N. (4.4)
The elements of this set will be called alternating. We seek a presentation of U , that
has the above generating set and all relations linear. We will obtain this presentation in
Theorem 6.2.
Next we obtain some formulas that will help us prove Theorem 6.2. We will show that
for n ∈ N,
Wn+1 =
n∑
k=0
Ekδ+α1G̃n−k(−1)kqk
(q − q−1)2k
, (4.5)
W−n =
n∑
k=0
Ekδ+α0G̃n−k(−1)kq3k
(q − q−1)2k
. (4.6)
We will prove (4.5), (4.6) by induction on n. Note that (4.5), (4.6) hold for n = 0, since
W1 = Eα1 and W0 = Eα0 . We will give the main induction argument after a few lemmas.
For the rest of this section k and ℓ are understood to be in N.
Lemma 4.3. Pick n ∈ N, and assume that (4.5), (4.6) hold for n, n− 1, . . . , 1, 0. Then
[W0,Wn+1] = [W−n,W1]. (4.7)
Proof. The commutator [W0,Wn+1] is equal to
W0Wn+1 −Wn+1W0
=
n∑
k=0
W0Ekδ+α1G̃n−k(−1)kqk
(q − q−1)2k
−
n∑
k=0
Ekδ+α1G̃n−kW0(−1)kqk
(q − q−1)2k
=
n∑
k=0
W0Ekδ+α1G̃n−k(−1)kqk
(q − q−1)2k
−
n∑
k=0
Ekδ+α1
(
q−2W0G̃n−k + (1− q−2)Wk−n
)
(−1)kqk
(q − q−1)2k
=
n∑
k=0
(
W0Ekδ+α1 − q−2Ekδ+α1W0
)
G̃n−k(−1)kqk
(q − q−1)2k
−
n∑
k=0
Ekδ+α1Wk−n(−1)kqk−1
(q − q−1)2k−1
= −
n∑
k=0
E(k+1)δG̃n−k(−1)kqk
(q − q−1)2k
−
n∑
k=0
Ekδ+α1Wk−n(−1)kqk−1
(q − q−1)2k−1
= −
n∑
k=0
E(k+1)δG̃n−k(−1)kqk
(q − q−1)2k
−
n∑
k=0
Ekδ+α1(−1)kqk−1
(q − q−1)2k−1
n−k∑
ℓ=0
Eℓδ+α0G̃n−k−ℓ(−1)ℓq3ℓ
(q − q−1)2ℓ
= −
n∑
p=0
E(p+1)δG̃n−p(−1)pqp
(q − q−1)2p
−
n∑
p=0
( ∑
k+ℓ=p
q2ℓEkδ+α1Eℓδ+α0
)
G̃n−p(−1)pqp−1
(q − q−1)2p−1
.
P. M. Terwilliger: A compact presentation for the alternating central extension of the positive . . . 371
The commutator [W−n,W1] is equal to
W−nW1 −W1W−n
=
n∑
k=0
Ekδ+α0G̃n−kW1(−1)kq3k
(q − q−1)2k
−
n∑
k=0
W1Ekδ+α0G̃n−k(−1)kq3k
(q − q−1)2k
=
n∑
k=0
Ekδ+α0
(
q2W1G̃n−k + (1− q2)Wn−k+1
)
(−1)kq3k
(q − q−1)2k
−
n∑
k=0
W1Ekδ+α0G̃n−k(−1)kq3k
(q − q−1)2k
=
n∑
k=0
(
q2Ekδ+α0W1 −W1Ekδ+α0
)
G̃n−k(−1)kq3k
(q − q−1)2k
−
n∑
k=0
Ekδ+α0Wn−k+1(−1)kq3k+1
(q − q−1)2k−1
= −
n∑
k=0
E(k+1)δG̃n−k(−1)kq3k+2
(q − q−1)2k
−
n∑
k=0
Ekδ+α0Wn−k+1(−1)kq3k+1
(q − q−1)2k−1
= −
n∑
k=0
E(k+1)δG̃n−k(−1)kq3k+2
(q − q−1)2k
−
n∑
k=0
Ekδ+α0(−1)kq3k+1
(q − q−1)2k−1
n−k∑
ℓ=0
Eℓδ+α1G̃n−k−ℓ(−1)ℓqℓ
(q − q−1)2ℓ
= −
n∑
p=0
E(p+1)δG̃n−p(−1)pq3p+2
(q − q−1)2p
−
n∑
p=0
( ∑
k+ℓ=p
q2kEkδ+α0Eℓδ+α1
)
G̃n−p(−1)pqp+1
(q − q−1)2p−1
.
By these comments
[W−n,W1]− [W0,Wn+1] =
n∑
p=0
CpG̃n−p(−1)pqp
(q − q−1)2p
,
where for 0 ≤ p ≤ n,
Cp = E(p+1)δ + q
−1(q − q−1)
∑
k+ℓ=p
q2ℓEkδ+α1Eℓδ+α0
− q2p+2E(p+1)δ − q(q − q−1)
∑
k+ℓ=p
q2kEkδ+α0Eℓδ+α1
= (1− q2p+2)E(p+1)δ − (1− q2)
∑
k+ℓ=p
q2ℓ
(
q−2Ekδ+α1Eℓδ+α0 − Eℓδ+α0Ekδ+α1
)
= (1− q2p+2)E(p+1)δ − (1− q2)
∑
k+ℓ=p
q2ℓE(p+1)δ
372 Ars Math. Contemp. 22 (2022) #P3.01 / 363–386
=
(
1− q2p+2 − (1− q2)
p∑
ℓ=0
q2ℓ
)
E(p+1)δ
= 0.
The result follows.
Lemma 4.4. Pick n ∈ N, and assume that (4.5), (4.6) hold for n, n− 1, . . . , 1, 0. Then
[G̃n, Eδ] = 0. (4.8)
Proof. Using Lemma 4.3,
0 = (q − q−1)
(
[W−n,W1]− [W0,Wn+1]
)
= [[G̃n,W0]q,W1]− [W0, [W1, G̃n]q]
= [G̃n, [W0,W1]q]
= −q[G̃n, Eδ].
Lemma 4.5. Pick n ∈ N, and assume that (4.5), (4.6) hold for n, n− 1, . . . , 1, 0. Then
[W−n,W0] = 0. (4.9)
Proof. The commutator [W−n,W0] is equal to
W−nW0 −W0W−n
=
n∑
k=0
Ekδ+α0G̃n−kW0(−1)kq3k
(q − q−1)2k
−
n∑
k=0
W0Ekδ+α0G̃n−k(−1)kq3k
(q − q−1)2k
=
n∑
k=0
Ekδ+α0
(
q−2W0G̃n−k + (1− q−2)Wk−n
)
(−1)kq3k
(q − q−1)2k
−
n∑
k=0
W0Ekδ+α0G̃n−k(−1)kq3k
(q − q−1)2k
=
n∑
k=0
[W0, Ekδ+α0 ]qG̃n−k(−1)k−1q3k−1
(q − q−1)2k
+
n∑
k=0
Ekδ+α0Wk−n(−1)kq3k−1
(q − q−1)2k−1
=
n∑
k=0
[W0, Ekδ+α0 ]qG̃n−k(−1)k−1q3k−1
(q − q−1)2k
+
n∑
k=0
Ekδ+α0(−1)kq3k−1
(q − q−1)2k−1
n−k∑
ℓ=0
Eℓδ+α0G̃n−k−ℓ(−1)ℓq3ℓ
(q − q−1)2ℓ
=
n∑
p=0
[W0, Epδ+α0 ]qG̃n−p(−1)p−1q3p−1
(q − q−1)2p
+
n∑
p=0
( ∑
k+ℓ=p
Ekδ+α0Eℓδ+α0
)
G̃n−p(−1)pq3p−1
(q − q−1)2p−1
.
P. M. Terwilliger: A compact presentation for the alternating central extension of the positive . . . 373
By these comments
[W−n,W0] =
n∑
p=0
SpG̃n−p(−1)p−1q3p−1
(q − q−1)2p−1
where
Sp =
[W0, Epδ+α0 ]q
q − q−1
−
∑
k+ℓ=p
Ekδ+α0Eℓδ+α0 (0 ≤ p ≤ n).
By (2.8) we have Sp = 0 for 0 ≤ p ≤ n. The result follows.
Lemma 4.6. Pick n ∈ N, and assume that (4.5), (4.6) hold for n, n− 1, . . . , 1, 0. Then
[Wn+1,W1] = 0. (4.10)
Proof. The commutator [Wn+1,W1] is equal to
Wn+1W1 −W1Wn+1
=
n∑
k=0
Ekδ+α1G̃n−kW1(−1)kqk
(q − q−1)2k
−
n∑
k=0
W1Ekδ+α1G̃n−k(−1)kqk
(q − q−1)2k
=
n∑
k=0
Ekδ+α1
(
q2W1G̃n−k + (1− q2)Wn−k+1
)
(−1)kqk
(q − q−1)2k
−
n∑
k=0
W1Ekδ+α1G̃n−k(−1)kqk
(q − q−1)2k
=
n∑
k=0
[Ekδ+α1 ,W1]qG̃n−k(−1)kqk+1
(q − q−1)2k
−
n∑
k=0
Ekδ+α1Wn−k+1(−1)kqk+1
(q − q−1)2k−1
=
n∑
k=0
[Ekδ+α1 ,W1]qG̃n−k(−1)kqk+1
(q − q−1)2k
−
n∑
k=0
Ekδ+α1(−1)kqk+1
(q − q−1)2k−1
n−k∑
ℓ=0
Eℓδ+α1G̃n−k−ℓ(−1)ℓqℓ
(q − q−1)2ℓ
=
n∑
p=0
[Epδ+α1 ,W1]qG̃n−p(−1)pqp+1
(q − q−1)2p
−
n∑
p=0
( ∑
k+ℓ=p
Ekδ+α1Eℓδ+α1
)
G̃n−p(−1)pqp+1
(q − q−1)2p−1
.
By these comments
[Wn+1,W1] =
n∑
p=0
TpG̃n−p(−1)pqp+1
(q − q−1)2p−1
374 Ars Math. Contemp. 22 (2022) #P3.01 / 363–386
where
Tp =
[Epδ+α1 ,W1]q
q − q−1
−
∑
k+ℓ=p
Ekδ+α1Eℓδ+α1 (0 ≤ p ≤ n).
By (2.9) we have Tp = 0 for 0 ≤ p ≤ n. The result follows.
Proposition 4.7. The equations (4.5), (4.6) hold in U for n ∈ N.
Proof. The proof is by induction on n. We assume that (4.5), (4.6) hold for
n, n− 1, . . . , 1, 0, and show that (4.5), (4.6) hold for n+ 1. Concerning (4.5),
Wn+2 =
qW1G̃n+1 − q−1G̃n+1W1
q − q−1
by (4.2)
= W1G̃n+1 − q−1
[G̃n+1,W1]
q − q−1
= W1G̃n+1 −
[[[G̃n,W0]q,W1]q,W1]
(q − q−1)2(q2 − q−2)
by Definition 3.1(v)
= W1G̃n+1 −
[[W−n,W1]q,W1]
(q − q−1)(q2 − q−2)
by (4.1)
= W1G̃n+1 −
[[W−n,W1],W1]q
(q − q−1)(q2 − q−2)
= W1G̃n+1 −
[[W0,Wn+1],W1]q
(q − q−1)(q2 − q−2)
by Lemma 4.3
= W1G̃n+1 −
[[W0,W1]q,Wn+1]
(q − q−1)(q2 − q−2)
by Lemma 4.6
= W1G̃n+1 +
q[Eδ,Wn+1]
(q − q−1)(q2 − q−2)
by (2.3)
= W1G̃n+1 + q
n∑
k=0
[Eδ, Ekδ+α1G̃n−k](−1)kqk
(q − q−1)2k+1(q2 − q−2)
by (4.5) and induction
= W1G̃n+1 + q
n∑
k=0
[Eδ, Ekδ+α1 ]G̃n−k(−1)kqk
(q − q−1)2k+1(q2 − q−2)
by Lemma 4.4
= W1G̃n+1 +
n∑
k=0
E(k+1)δ+α1G̃n−k(−1)k+1qk+1
(q − q−1)2k+2
by (2.4)
= Eα1G̃n+1 +
n+1∑
k=1
Ekδ+α1G̃n+1−k(−1)kqk
(q − q−1)2k
=
n+1∑
k=0
Ekδ+α1G̃n+1−k(−1)kqk
(q − q−1)2k
.
We have shown that (4.5) holds for n+ 1. Concerning (4.6),
W−n−1 =
qG̃n+1W0 − q−1W0G̃n+1
q − q−1
by (4.1)
P. M. Terwilliger: A compact presentation for the alternating central extension of the positive . . . 375
= W0G̃n+1 − q
[W0, G̃n+1]
q − q−1
= W0G̃n+1 − q2
[W0, [W0, [W1, G̃n]q]q]
(q − q−1)2(q2 − q−2)
by Definition 3.1(vi)
= W0G̃n+1 − q2
[W0, [W0,Wn+1]q]
(q − q−1)(q2 − q−2)
by (4.2)
= W0G̃n+1 − q2
[W0, [W0,Wn+1]]q
(q − q−1)(q2 − q−2)
= W0G̃n+1 − q2
[W0, [W−n,W1]]q
(q − q−1)(q2 − q−2)
by Lemma 4.3
= W0G̃n+1 − q2
[W−n, [W0,W1]q]
(q − q−1)(q2 − q−2)
by Lemma 4.5
= W0G̃n+1 + q
3 [W−n, Eδ]
(q − q−1)(q2 − q−2)
by (2.3)
= W0G̃n+1 + q
3
n∑
k=0
[Ekδ+α0G̃n−k, Eδ](−1)kq3k
(q − q−1)2k+1(q2 − q−2)
by (4.6) and induction
= W0G̃n+1 + q
3
n∑
k=0
[Ekδ+α0 , Eδ]G̃n−k(−1)kq3k
(q − q−1)2k+1(q2 − q−2)
by Lemma 4.4
= W0G̃n+1 +
n∑
k=0
E(k+1)δ+α0G̃n−k(−1)k+1q3k+3
(q − q−1)2k+2
by (2.4)
= Eα0G̃n+1 +
n+1∑
k=1
Ekδ+α0G̃n+1−k(−1)kq3k
(q − q−1)2k
=
n+1∑
k=0
Ekδ+α0G̃n+1−k(−1)kq3k
(q − q−1)2k
.
We have shown that (4.6) holds for n+ 1.
Lemma 4.8. The following relations hold in U . For n ∈ N,
[W0,Wn+1] = [W−n,W1], [G̃n, Eδ] = 0,
[W−n,W0] = 0, [Wn+1,W1] = 0.
Proof. By Lemmas 4.3 – 4.6 and Proposition 4.7.
Lemma 4.9. The following relations hold in U . For k ∈ N,
(i) [Gk+1,W1]q = [W1, G̃k+1]q;
(ii) [W0, Gk+1]q = [G̃k+1,W0]q .
Proof. (i) We have
[Gk+1,W1]q − [W1, G̃k+1]q =
[
G̃k+1 +
[W1,W−k]
1− q−2
,W1
]
q
− [W1, G̃k+1]q
376 Ars Math. Contemp. 22 (2022) #P3.01 / 363–386
= (q + q−1)[G̃k+1,W1]−
[[W−k,W1],W1]q
1− q−2
= (q + q−1)[G̃k+1,W1]−
[[W−k,W1]q,W1]
1− q−2
= (q + q−1)[G̃k+1,W1]−
[[[G̃k,W0]q,W1]q,W1]
(1− q−2)(q − q−1)
= 0.
(ii) We have
[W0, Gk+1]q − [G̃k+1,W0]q =
[
W0, G̃k+1 +
[Wk+1,W0]
1− q−2
]
q
− [G̃k+1,W0]q
= (q + q−1)[W0, G̃k+1]−
[W0, [W0,Wk+1]]q
1− q−2
= (q + q−1)[W0, G̃k+1]−
[W0, [W0,Wk+1]q]
1− q−2
= (q + q−1)[W0, G̃k+1]−
[W0, [W0, [W1, G̃k]q]q]
(1− q−2)(q − q−1)
= 0.
5 Generating functions
The alternating generators of U are displayed in (4.4). In the previous section we described
how these generators are related to W0 and W1. Our next goal is to describe how the
alternating generators are related to each other. It is convenient to use generating functions.
Definition 5.1. We define some generating functions in an indeterminate t. Referring to
(4.4),
G(t) =
∑
n∈N
Gnt
n, G̃(t) =
∑
n∈N
G̃nt
n,
W−(t) =
∑
n∈N
W−nt
n, W+(t) =
∑
n∈N
Wn+1t
n.
Lemma 5.2. For the algebra U ,
[W0, G(t)]q
q − q−1
= W−(t),
[G̃(t),W0]q
q − q−1
= W−(t),
[W0,W
−(t)] = 0,
[W0,W
+(t)]
1− q−2
= t−1(G̃(t)−G(t))
and
[G(t),W1]q
q − q−1
= W+(t),
[W1, G̃(t)]q
q − q−1
= W+(t),
[W1,W
+(t)] = 0,
[W1,W
−(t)]
1− q−2
= t−1(G(t)− G̃(t)).
P. M. Terwilliger: A compact presentation for the alternating central extension of the positive . . . 377
Proof. Use Definition 4.1 and Lemmas 4.8, 4.9.
For the rest of this section, let s denote an indeterminate that commutes with t.
Lemma 5.3. For the algebra U ,
[W−(s),W−(t)] = 0, [W+(s),W+(t)] = 0,
[W−(s),W+(t)] + [W+(s),W−(t)] = 0,
s[W−(s), G(t)] + t[G(s),W−(t)] = 0,
s[W−(s), G̃(t)] + t[G̃(s),W−(t)] = 0,
s[W+(s), G(t)] + t[G(s),W+(t)] = 0,
s[W+(s), G̃(t)] + t[G̃(s),W+(t)] = 0,
[G(s), G(t)] = 0, [G̃(s), G̃(t)] = 0,
[G̃(s), G(t)] + [G(s), G̃(t)] = 0
and also
[W−(s), G(t)]q = [W
−(t), G(s)]q, [G(s),W
+(t)]q = [G(t),W
+(s)]q,
[G̃(s),W−(t)]q = [G̃(t),W
−(s)]q, [W
+(s), G̃(t)]q = [W
+(t), G̃(s)]q,
t−1[G(s), G̃(t)]− s−1[G(t), G̃(s)] = q[W−(t),W+(s)]q − q[W−(s),W+(t)]q,
t−1[G̃(s), G(t)]− s−1[G̃(t), G(s)] = q[W+(t),W−(s)]q − q[W+(s),W−(t)]q,
[G(s), G̃(t)]q − [G(t), G̃(s)]q = qt[W−(t),W+(s)]− qs[W−(s),W+(t)],
[G̃(s), G(t)]q − [G̃(t), G(s)]q = qt[W+(t),W−(s)]− qs[W+(s),W−(t)].
Proof. We refer to the generating functions A(s, t), B(s, t), . . . , S(s, t) from Appendix A.
The present lemma asserts that for the algebra U these generating functions are all zero.
To verify this assertion, we refer to the canonical relations in Appendix A. We will use
induction with respect to the linear order
I(s, t),M(s, t), N(s, t), A(s, t), B(s, t), Q(s, t), D(s, t), E(s, t), F (s, t),
G(s, t), R(s, t), S(s, t), H(s, t),K(s, t), L(s, t), P (s, t), C(s, t), J(s, t).
For each element in this linear order besides I(s, t), there exists a canonical relation that
expresses the given element in terms of the previous elements in the linear order. So in
U the given element is zero, provided that in U every previous element is zero. Note that
in U we have I(s, t) = 0 by Definition 3.1(vii). By these comments and induction, in
U every element in the linear order is zero. We have shown that in U each of A(s, t),
B(s, t), . . . , S(s, t) is zero.
6 The main results
In this section we present our main results, which are Theorem 6.2 and Corollary 6.5.
Recall the alternating generators (4.4) for U .
378 Ars Math. Contemp. 22 (2022) #P3.01 / 363–386
Lemma 6.1. The following relations hold in U . For k, ℓ ∈ N we have
[W0,Wk+1] = [W−k,W1] = (1− q−2)(G̃k+1 −Gk+1), (6.1)
[W0, Gk+1]q = [G̃k+1,W0]q = (q − q−1)W−k−1, (6.2)
[Gk+1,W1]q = [W1, G̃k+1]q = (q − q−1)Wk+2, (6.3)
[W−k,W−ℓ] = 0, [Wk+1,Wℓ+1] = 0, (6.4)
[W−k,Wℓ+1] + [Wk+1,W−ℓ] = 0, (6.5)
[W−k, Gℓ+1] + [Gk+1,W−ℓ] = 0, (6.6)
[W−k, G̃ℓ+1] + [G̃k+1,W−ℓ] = 0, (6.7)
[Wk+1, Gℓ+1] + [Gk+1,Wℓ+1] = 0, (6.8)
[Wk+1, G̃ℓ+1] + [G̃k+1,Wℓ+1] = 0, (6.9)
[Gk+1, Gℓ+1] = 0, [G̃k+1, G̃ℓ+1] = 0, (6.10)
[G̃k+1, Gℓ+1] + [Gk+1, G̃ℓ+1] = 0 (6.11)
and also
[W−k, Gℓ]q = [W−ℓ, Gk]q, [Gk,Wℓ+1]q = [Gℓ,Wk+1]q, (6.12)
[G̃k,W−ℓ]q = [G̃ℓ,W−k]q, [Wℓ+1, G̃k]q = [Wk+1, G̃ℓ]q, (6.13)
[Gk, G̃ℓ+1]− [Gℓ, G̃k+1] = q[W−ℓ,Wk+1]q − q[W−k,Wℓ+1]q, (6.14)
[G̃k, Gℓ+1]− [G̃ℓ, Gk+1] = q[Wℓ+1,W−k]q − q[Wk+1,W−ℓ]q, (6.15)
[Gk+1, G̃ℓ+1]q − [Gℓ+1, G̃k+1]q = q[W−ℓ,Wk+2]− q[W−k,Wℓ+2], (6.16)
[G̃k+1, Gℓ+1]q − [G̃ℓ+1, Gk+1]q = q[Wℓ+1,W−k−1]− q[Wk+1,W−ℓ−1]. (6.17)
Proof. The relations (6.1) – (6.3) are from Definition 4.1 and Lemmas 4.8, 4.9. The rela-
tions (6.4) – (6.17) follow from Definition 5.1 and Lemma 5.3.
Theorem 6.2. The algebra U has a presentation by generators
{W−k}k∈N, {Wk+1}k∈N, {Gk+1}k∈N, {G̃k+1}k∈N
and the relations in Lemma 6.1.
Proof. It suffices to show that the relations in Definition 3.1 are implied by the relations in
Lemma 6.1. The relation (iii) in Definition 3.1 is obtained from the equation on the left in
(6.3) at k = 0, by eliminating G1 using [W0,W1] = (1−q−2)(G̃1−G1). The relation (iv)
in Definition 3.1 is obtained from the equation on the left in (6.2) at k = 0, by eliminating
G1 using [W0,W1] = (1 − q−2)(G̃1 − G1). For k ≥ 1 the relation (v) in Definition 3.1
is obtained from the equation on the left in (6.3), by eliminating Gk+1 using [W−k,W1] =
(1 − q−2)(G̃k+1 − Gk+1) and evaluating the result using [G̃k,W0]q = (q − q−1)W−k.
For k ≥ 1 the relation (vi) in Definition 3.1 is obtained from the equation on the left in
(6.2), by eliminating Gk+1 using [W0,Wk+1] = (1− q−2)(G̃k+1 −Gk+1) and evaluating
the result using [W1, G̃k]q = (q − q−1)Wk+1. The relation (vii) in Definition 3.1 is from
(6.10). The relation (i) in Definition 3.1 is obtained from [W0,W−1] = 0, by eliminating
W−1 using [G̃1,W0]q = (q − q−1)W−1 and evaluating the result using Definition 3.1(iv).
The relation (ii) in Definition 3.1 is obtained from [W1,W2] = 0, by eliminating W2 using
[W1, G̃1]q = (q − q−1)W2 and evaluating the result using Definition 3.1(iii).
P. M. Terwilliger: A compact presentation for the alternating central extension of the positive . . . 379
It is apparent from the proof of Theorem 6.2 that the relations in Lemma 6.1 are redundant
in the following sense.
Corollary 6.3. The relations in Lemma 6.1 are implied by the relations listed in (i) – (iii)
below:
(i) (6.1) – (6.3);
(ii) (6.4) with k = 0 and ℓ = 1;
(iii) the relations on the right in (6.10).
Proof. By Lemma 6.1 the relations (6.1) – (6.17) are implied by the relations in Defini-
tions 3.1, 4.1. The relations listed in (i) – (iii) are used in the proof of Theorem 6.2 to
obtain the relations in Definition 3.1. The relations listed in (i) imply the relations in Defi-
nition 4.1. The result follows.
The relations in Lemma 6.1 first appeared in [30, Propositions 5.7, 5.10, 5.11]. It was
observed in [2, Propositions 3.1, 3.2] and [5, Remark 2.5] that the relations (6.1) – (6.11)
imply the relations (6.12) – (6.17). This observation motivated the following definition.
Definition 6.4 ([29, Definition 3.1]). Define the algebra U+q by generators
{W−k}k∈N, {Wk+1}k∈N, {Gk+1}k∈N, {G̃k+1}k∈N
and the relations (6.1) – (6.11). The algebra U+q is called the alternating central extension
of U+q .
Corollary 6.5. We have U = U+q .
Proof. By Theorem 6.2, Corollary 6.3, and Definition 6.4.
Definition 6.6. By the compact presentation of U+q we mean the presentation given in
Definition 3.1. By the expanded presentation of U+q we mean the presentation given in
Theorem 6.2.
Corollary 6.7. The map ♭ from Lemma 3.3 is injective.
Proof. By Corollary 6.5 and [29, Proposition 6.4].
7 The subalgebra of U+q generated by {G̃k+1}k∈N
Let G̃ denote the subalgebra of U+q generated by {G̃k+1}k∈N. In this section we describe
G̃ and its relationship to ⟨W0,W1⟩.
The following notation will be useful. Let z1, z2, . . . denote mutually commuting inde-
terminates. Let F[z1, z2, . . .] denote the algebra consisting of the polynomials in z1, z2, . . .
that have all coefficients in F. For notational convenience define z0 = 1.
Lemma 7.1 ([29, Lemma 3.5]). There exists an algebra homomorphism U+q →
F[z1, z2, . . .] that sends
W−n 7→ 0, Wn+1 7→ 0, Gn 7→ zn, G̃n 7→ zn
for n ∈ N.
380 Ars Math. Contemp. 22 (2022) #P3.01 / 363–386
Proof. By Theorem 6.2 and the nature of the relations in Lemma 6.1.
Corollary 7.2 ([29, Theorem 10.2]). The generators {G̃k+1}k∈N of G̃ are algebraically
independent.
Proof. By Lemma 7.1 and since {zk+1}k∈N are algebraically independent.
The following result will help us describe how G̃ is related to ⟨W0,W1⟩.
Lemma 7.3. For n ∈ N,
G̃nW1 = W1G̃n +
n∑
k=1
Ekδ+α1G̃n−k(−1)k+1qk+1
(q − q−1)2k−1
, (7.1)
G̃nW0 = W0G̃n +
n∑
k=1
Ekδ+α0G̃n−k(−1)kq3k−1
(q − q−1)2k−1
. (7.2)
Proof. To obtain (7.1), eliminate Wn+1 from (4.5) using (4.2), and solve the resulting
equation for G̃nW1. To obtain (7.2), eliminate W−n from (4.6) using (4.1), and solve the
resulting equation for G̃nW0.
Shortly we will describe how G̃ is related to ⟨W0,W1⟩. This description involves the
center Z of U+q . To prepare for this description, we have some comments about Z . In
[29, Sections 5, 6] we introduced some algebraically independent elements Z1, Z2, . . . that
generate the algebra Z . For notational convenience define Z0 = 1. Using {Zn}n∈N we
obtain a basis for Z that is described as follows. For n ∈ N, a partition of n is a sequence
λ = {λi}∞i=1 of natural numbers such that λi ≥ λi+1 for i ≥ 1 and n =
∑∞
i=1 λi. The set
Λn consists of the partitions of n. Define Λ = ∪n∈NΛn. For λ ∈ Λ define Zλ =
∏∞
i=1 Zλi .
The elements {Zλ}λ∈Λ form a basis for the vector space Z . Next we describe a grading
for Z . For n ∈ N let Zn denote the subspace of Z with basis {Zλ}λ∈Λn . For example
Z0 = F1. The sum Z =
∑
n∈N Zn is direct. Moreover ZrZs ⊆ Zr+s for r, s ∈ N.
By these comments the subspaces {Zn}n∈N form a grading of Z . Note that Zn ∈ Zn for
n ∈ N. Next we describe how Z is related to ⟨W0,W1⟩.
Lemma 7.4 ([29, Proposition 6.5]). The multiplication map
⟨W0,W1⟩ ⊗ Z → U+q
w ⊗ z 7→ wz
is an algebra isomorphism.
For n ∈ N let Un denote the image of ⟨W0,W1⟩ ⊗ Zn under the multiplication map. By
construction the sum U+q =
∑
n∈N Un is direct.
In the next two lemmas we describe how G̃ is related to Z .
Lemma 7.5 ([29, Lemmas 3.6, 5.9]). For n ∈ N,
G̃n ∈
n∑
k=0
⟨W0,W1⟩Zk, G̃n − Zn ∈
n−1∑
k=0
⟨W0,W1⟩Zk.
P. M. Terwilliger: A compact presentation for the alternating central extension of the positive . . . 381
For λ ∈ Λ define G̃λ =
∏∞
i=1 G̃λi . By Corollary 7.2 the elements {G̃λ}λ∈Λ form a basis
for the vector space G̃.
Lemma 7.6. For n ∈ N and λ ∈ Λn,
G̃λ ∈
n∑
k=0
Uk, G̃λ − Zλ ∈
n−1∑
k=0
Uk.
Proof. By Lemma 7.5 and our comments above Lemma 7.4 about the grading of Z .
Next we describe how G̃ is related to ⟨W0,W1⟩.
Proposition 7.7. The multiplication map
⟨W0,W1⟩ ⊗ G̃ → U+q
w ⊗ g 7→ wg
is an isomorphism of vector spaces.
Proof. The multiplication map is F-linear. The multiplication map is surjective by
Lemma 7.3 and since U+q is generated by W0, W1, G̃. We now show that the multiplicaton
map is injective. Consider a vector v ∈ ⟨W0,W1⟩ ⊗ G̃ that is sent to zero by the multi-
plication map. We show that v = 0. Write v =
∑
λ∈Λ aλ ⊗ G̃λ, where aλ ∈ ⟨W0,W1⟩
for λ ∈ Λ and aλ = 0 for all but finitely many λ ∈ Λ. To show that v = 0, we must
show that aλ = 0 for all λ ∈ Λ. Suppose that there exists λ ∈ Λ such that aλ ̸= 0. Let
C denote the set of natural numbers m such that Λm contains a partition λ with aλ ̸= 0.
The set C is nonempty and finite. Let n denote the maximal element of C. By construction∑
λ∈Λn aλ ⊗ Zλ is nonzero. By Lemma 7.4,∑
λ∈Λn
aλZλ ̸= 0. (7.3)
By construction
0 =
∑
λ∈Λ
aλG̃λ =
n∑
k=0
∑
λ∈Λk
aλG̃λ =
∑
λ∈Λn
aλG̃λ +
n−1∑
k=0
∑
λ∈Λk
aλG̃λ. (7.4)
Using (7.4),
∑
λ∈Λn
aλZλ =
∑
λ∈Λn
aλ(Zλ − G̃λ)−
n−1∑
k=0
∑
λ∈Λk
aλG̃λ. (7.5)
The left-hand side of (7.5) is contained in Un. By Lemma 7.6 the right-hand side of (7.5)
is contained in
∑n−1
k=0 Uk. The subspaces Un and
∑n−1
k=0 Uk have zero intersection because
the sum
∑n
k=0 Uk is direct. This contradicts (7.3), so aλ = 0 for λ ∈ Λ. Consequently
v = 0, as desired. We have shown that the multiplication map is injective. By the above
comments the multiplication map is an isomorphism of vector spaces.
382 Ars Math. Contemp. 22 (2022) #P3.01 / 363–386
ORCID iDs
Paul M. Terwilliger https://orcid.org/0000-0003-0942-5489
References
[1] E. Bannai and T. Ito, Algebraic Combinatorics. I, Association Schemes, The Ben-
jamin/Cummings Publishing Co., Inc., Menlo Park, CA, 1984.
[2] P. Baseilhac, The positive part of uq(ŝl2) and tensor product representation, preprint.
[3] P. Baseilhac, Deformed Dolan-Grady relations in quantum integrable models, Nuclear Phys. B
709 (2005), 491–521, doi:10.1016/j.nuclphysb.2004.12.016.
[4] P. Baseilhac, An integrable structure related with tridiagonal algebras, Nuclear Phys. B 705
(2005), 605–619, doi:10.1016/j.nuclphysb.2004.11.014.
[5] P. Baseilhac, The alternating presentation of Uq(ĝl2) from Freidel-Maillet algebras, Nuclear
Phys. B 967 (2021), Paper No. 115400, 48, doi:10.1016/j.nuclphysb.2021.115400.
[6] P. Baseilhac and S. Belliard, The half-infinite XXZ chain in Onsager’s approach, Nuclear Phys.
B 873 (2013), 550–584, doi:10.1016/j.nuclphysb.2013.05.003.
[7] P. Baseilhac and S. Belliard, An attractive basis for the q−Onsager algebra, 2017,
arXiv:1704.02950 [math.CO].
[8] P. Baseilhac and K. Koizumi, A deformed analogue of Onsager’s symmetry in the XXZ open
spin chain, J. Stat. Mech. Theory Exp. (2005), P10005, 15, doi:10.1088/1742-5468/2005/10/
p10005.
[9] P. Baseilhac and K. Koizumi, A new (in)finite-dimensional algebra for quantum integrable
models, Nuclear Phys. B 720 (2005), 325–347, doi:10.1016/j.nuclphysb.2005.05.021.
[10] P. Baseilhac and K. Koizumi, Exact spectrum of theXXZ open spin chain from the q-Onsager
algebra representation theory, J. Stat. Mech. Theory Exp. (2007), P09006, 27, doi:10.1088/
1742-5468/2007/09/p09006.
[11] P. Baseilhac and S. Kolb, Braid group action and root vectors for the q-Onsager algebra, Trans-
form. Groups 25 (2020), 363–389, doi:10.1007/s00031-020-09555-7.
[12] P. Baseilhac and K. Shigechi, A new current algebra and the reflection equation, Lett. Math.
Phys. 92 (2010), 47–65, doi:10.1007/s11005-010-0380-x.
[13] S. Bockting-Conrad, Tridiagonal pairs of q-Racah type, the double lowering operator ψ, and
the quantum algebra Uq(sl2), Linear Algebra Appl. 445 (2014), 256–279, doi:10.1016/j.laa.
2013.12.007.
[14] A. E. Brouwer, A. M. Cohen and A. Neumaier, Distance-Regular Graphs, volume 18 of Ergeb-
nisse der Mathematik und ihrer Grenzgebiete (3) [Results in Mathematics and Related Areas
(3)], Springer-Verlag, Berlin, 1989, doi:10.1007/978-3-642-74341-2.
[15] V. Chari and A. Pressley, Quantum affine algebras, Comm. Math. Phys. 142 (1991), 261–283,
http://projecteuclid.org/euclid.cmp/1104248585.
[16] I. Damiani, A basis of type Poincaré-Birkhoff-Witt for the quantum algebra of ŝl(2), J. Algebra
161 (1993), 291–310, doi:10.1006/jabr.1993.1220.
[17] T. Ito, K. Tanabe and P. Terwilliger, Some algebra related to P - and Q-polynomial association
schemes, in: Codes and Association Schemes (Piscataway, NJ, 1999), Amer. Math. Soc., Prov-
idence, RI, volume 56 of DIMACS Ser. Discrete Math. Theoret. Comput. Sci., pp. 167–192,
2001, doi:10.1090/dimacs/056/14.
P. M. Terwilliger: A compact presentation for the alternating central extension of the positive . . . 383
[18] T. Ito and P. Terwilliger, Two non-nilpotent linear transformations that satisfy the cubic q-Serre
relations, J. Algebra Appl. 6 (2007), 477–503, doi:10.1142/s021949880700234x.
[19] T. Ito and P. Terwilliger, Distance-regular graphs and the q-tetrahedron algebra, Eur. J. Comb.
30 (2009), 682–697, doi:10.1016/j.ejc.2008.07.011.
[20] M. Jimbo and T. Miwa, Algebraic Analysis of Solvable Lattice Models, volume 85 of CBMS
Regional Conference Series in Mathematics, Published for the Conference Board of the Math-
ematical Sciences, Washington, DC; by the American Mathematical Society, Providence, RI,
1995.
[21] W. Liu, The attenuated space poset Aq(M,N), Linear Algebra Appl. 506 (2016), 244–273,
doi:10.1016/j.laa.2016.05.014.
[22] G. Lusztig, Introduction to Quantum Groups, volume 110 of Progress in Mathematics,
Birkhäuser Boston, Inc., Boston, MA, 1993, doi:10.1007/978-0-8176-4717-9.
[23] Š. Miklavič and P. Terwilliger, Bipartite Q-polynomial distance-regular graphs and uniform
posets, J. Algebraic Comb. 38 (2013), 225–242, doi:10.1007/s10801-012-0401-1.
[24] K. Nomura and P. Terwilliger, Totally bipartite tridiagonal pairs, Electron. J. Linear Algebra
37 (2021), 434–491, doi:10.13001/ela.2021.5029.
[25] M. Rosso, Groupes quantiques et algèbres de battage quantiques, C. R. Acad. Sci. Paris 320
(1995), 145–148.
[26] M. Rosso, Quantum groups and quantum shuffles, Invent. Math. 133 (1998), 399–416, doi:
10.1007/s002220050249.
[27] P. Terwilliger, The incidence algebra of a uniform poset, in: Coding theory and design theory,
Part I, Springer, New York, volume 20 of IMA Vol. Math. Appl., pp. 193–212, 1990, doi:
10.1007/978-1-4613-8994-1\_15.
[28] P. Terwilliger, An action of the free product Z2 ⋆ Z2 ⋆ Z2 on the q-Onsager algebra and its
current algebra, Nuclear Phys. B 936 (2018), 306–319, doi:10.1016/j.nuclphysb.2018.09.020.
[29] P. Terwilliger, The alternating central extension for the positive part of Uq(ŝl2), Nuclear Phys.
B 947 (2019), 114729, 25, doi:10.1016/j.nuclphysb.2019.114729.
[30] P. Terwilliger, The alternating PBW basis for the positive part of Uq(ŝl2), J. Math. Phys. 60
(2019), 071704, 27, doi:10.1063/1.5091801.
[31] P. Terwilliger, The alternating central extension of the q-Onsager algebra, Comm. Math. Phys.
387 (2021), 1771–1819, doi:10.1007/s00220-021-04171-2.
[32] E. R. van Dam, J. H. Koolen and H. Tanaka, Distance-regular graphs, Electron. J. Comb. DS22
(2016), 156, doi:10.37236/4925.
384 Ars Math. Contemp. 22 (2022) #P3.01 / 363–386
Appendix A
Recall the algebra U from Definition 3.1. In this appendix we list some relations that hold
in U . We will define an algebra U∨ that is a homomorphic preimage of U . All the results
in this appendix are about U∨.
Define the algebra U∨ by generators
{W−k}k∈N, {Wk+1}k∈N, {Gk+1}k∈N, {G̃k+1}k∈N
and the following relations. For k ∈ N,
[W0,Wk+1] = [W−k,W1] = (1− q−2)(G̃k+1 −Gk+1), (A.1)
[W0, Gk+1]q = [G̃k+1,W0]q = (q − q−1)W−k−1, (A.2)
[Gk+1,W1]q = [W1, G̃k+1]q = (q − q−1)Wk+2, (A.3)
[W0,W−k] = 0, [W1,Wk+1] = 0. (A.4)
For notational convenience, define G0 = 1 and G̃0 = 1.
For U∨ we define the generating functions W−(t), W+(t), G(t), G̃(t) as in Defini-
tion 5.1. In terms of these generating functions, the relations (A.1) – (A.4) become the
relations in Lemma 5.2. Let s denote an indeterminate that commutes with t. Define
A(s, t) = [W−(s),W−(t)],
B(s, t) = [W+(s),W+(t)],
C(s, t) = [W−(s),W+(t)] + [W+(s),W−(t)],
D(s, t) = s[W−(s), G(t)] + t[G(s),W−(t)],
E(s, t) = s[W−(s), G̃(t)] + t[G̃(s),W−(t)],
F (s, t) = s[W+(s), G(t)] + t[G(s),W+(t)],
G(s, t) = s[W+(s), G̃(t)] + t[G̃(s),W+(t)],
H(s, t) = [G(s), G(t)],
I(s, t) = [G̃(s), G̃(t)],
J(s, t) = [G̃(s), G(t)] + [G(s), G̃(t)]
and also
K(s, t) = [W−(s), G(t)]q − [W−(t), G(s)]q,
L(s, t) = [G(s),W+(t)]q − [G(t),W+(s)]q,
M(s, t) = [G̃(s),W−(t)]q − [G̃(t),W−(s)]q,
N(s, t) = [W+(s), G̃(t)]q − [W+(t), G̃(s)]q,
P (s, t) = t−1[G(s), G̃(t)]− s−1[G(t), G̃(s)]− q[W−(t),W+(s)]q + q[W−(s),W+(t)]q,
Q(s, t) = t−1[G̃(s), G(t)]− s−1[G̃(t), G(s)]− q[W+(t),W−(s)]q + q[W+(s),W−(t)]q,
R(s, t) = [G(s), G̃(t)]q − [G(t), G̃(s)]q − qt[W−(t),W+(s)] + qs[W−(s),W+(t)],
S(s, t) = [G̃(s), G(t)]q − [G̃(t), G(s)]q − qt[W+(t),W−(s)] + qs[W+(s),W−(t)].
P. M. Terwilliger: A compact presentation for the alternating central extension of the positive . . . 385
By linear algebra,
C(s, t) =
(q + q−1)(P (s, t) +Q(s, t))− (s−1 + t−1)(R(s, t) + S(s, t))
(q2 − s−1t)(q2 − st−1)q−1
, (A.5)
J(s, t) =
(q + q−1)(R(s, t) + S(s, t))− (s+ t)(P (s, t) +Q(s, t))
(q2 − s−1t)(q2 − st−1)q−2
. (A.6)
Using Lemma 5.2 we routinely obtain
[W0, A(s, t)] = 0,
[W0, B(s, t)]
1− q−2
=
G(s, t)− F (s, t)
st
,
[W0, C(s, t)]
1− q−2
=
E(s, t)−D(s, t)
st
,
[W0, D(s, t)]q
q − q−1
= (s+ t)A(s, t),
[E(s, t),W0]q
q − q−1
= (s+ t)A(s, t),
[W0, F (s, t)]q
1− q−2
= S(s, t)− (q + q−1)H(s, t),
[G(s, t),W0]q
1− q−2
= S(s, t)− (q + q−1)I(s, t),
[W0, H(s, t)]q2
q − q−1
= K(s, t),
[I(s, t),W0]q2
q − q−1
= M(s, t),
[W0, J(s, t)]
q − q−1
= M(s, t)−K(s, t)
and
[W0,K(s, t)]q
q2 − q−2
= A(s, t),
[W0, L(s, t)]q
q − q−1
= P (s, t)− (s−1 + t−1)H(s, t),
[M(s, t),W0]q
q2 − q−2
= A(s, t),
[N(s, t),W0]q
q − q−1
= Q(s, t)− (s−1 + t−1)I(s, t),
[P (s, t),W0]
q − q−1
= (s−1 + t−1)K(s, t)− (q + q−1)s−1t−1E(s, t),
[W0, Q(s, t)]
q − q−1
= (s−1 + t−1)M(s, t)− (q + q−1)s−1t−1D(s, t),
[W0, R(s, t)]
q − q−1
= (s−1 + t−1)(E(s, t)−D(s, t)),
[W0, S(s, t)]
q2 − q−2
= M(s, t)−K(s, t)
and
[W1, A(s, t)]
1− q−2
=
D(s, t)− E(s, t)
st
, [W1, B(s, t)] = 0,
[W1, C(s, t)]
1− q−2
=
F (s, t)−G(s, t)
st
,
[D(s, t),W1]q
1− q−2
= R(s, t)− (q + q−1)H(s, t),
[W1, E(s, t)]q
1− q−2
= R(s, t)− (q + q−1)I(s, t), [F (s, t),W1]q
q − q−1
= (s+ t)B(s, t),
[W1, G(s, t)]q
q − q−1
= (s+ t)B(s, t),
[H(s, t),W1]q2
q − q−1
= L(s, t),
[W1, I(s, t)]q2
q − q−1
= N(s, t),
[W1, J(s, t)]
q − q−1
= L(s, t)−N(s, t)
386 Ars Math. Contemp. 22 (2022) #P3.01 / 363–386
and
[K(s, t),W1]q
q − q−1
= P (s, t)− (s−1 + t−1)H(s, t), [L(s, t),W1]q
q2 − q−2
= B(s, t),
[W1,M(s, t)]q
q − q−1
= Q(s, t)− (s−1 + t−1)I(s, t), [W1, N(s, t)]q
q2 − q−2
= B(s, t),
[W1, P (s, t)]
q − q−1
= (s−1 + t−1)L(s, t)− (q + q−1)s−1t−1G(s, t),
[Q(s, t),W1]
q − q−1
= (s−1 + t−1)N(s, t)− (q + q−1)s−1t−1F (s, t),
[W1, R(s, t)]
q2 − q−2
= L(s, t)−N(s, t), [W1, S(s, t)]
q − q−1
= (s−1 + t−1)(F (s, t)−G(s, t)).
We just listed 38 relations, including (A.5), (A.6). These 38 relations are called canonical.
ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 22 (2022) #P3.02 / 387–402
https://doi.org/10.26493/1855-3974.2496.07a
(Also available at http://amc-journal.eu)
The adjacency dimension of graphs*
Sergio Bermudo †
Department of Economics, Quantitative Methods and Economic History,
Pablo de Olavide University, Carretera de Utrera Km. 1, 41013-Sevilla, Spain
José M. Rodrı́guez
Departamento de Matemáticas, Universidad Carlos III de Madrid,
Avenida de la Universidad 30, 28911 Leganés, Madrid, Spain
Juan A. Rodrı́guez-Velázquez
Universitat Rovira i Virgili, Departament d’Enginyeria Informàtica i Matemàtiques,
Av. Paı̈sos Catalans 26, 43007 Tarragona, Spain
José M. Sigarreta
Facultad de Matemáticas, Universidad Autónoma de Guerrero,
Carlos E. Adame No. 54 Col. Garita, 39650 Acalpulco Guerrero, Mexico
Received 1 December 2020, accepted 13 September 2021, published online 9 June 2022
Abstract
It is known that the problem of computing the adjacency dimension of a graph is NP-
hard. This suggests finding the adjacency dimension for special classes of graphs or obtain-
ing good bounds on this invariant. In this work we obtain general bounds on the adjacency
dimension of a graph G in terms of known parameters of G. We discuss the tightness
of these bounds and, for some particular classes of graphs, we obtain closed formulae.
In particular, we show the close relationships that exist between the adjacency dimension
and other parameters, like the domination number, the location-domination number, the
2-domination number, the independent 2-domination number, the vertex cover number, the
independence number and the super domination number.
Keywords: Adjacency dimension, metric dimension, location-domination number, independence num-
ber, super domination number.
Math. Subj. Class. (2020): 05C69, 05C7, 05C12
*This work has been supported in part by three grants from “Ministerio de Economı́a y Competitividad, Agen-
cia Estatal de Investigación (AEI)” and “Fondo Europeo de Desarrollo Regional (FEDER)” (MTM2016-78227-
C2-1-P, MTM2015-70531 and MTM2017-90584-REDT), Spain, and by Junta de Andalucı́a, FEDER-UPO Re-
search and Development Call, reference number UPO-1263769.
†Corresponding author.
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
388 Ars Math. Contemp. 22 (2022) #P3.02 / 387–402
1 Introduction
The metric dimension of a general metric space was introduced in 1953 by Blumenthal [2,
p. 95]. A metric generator for a metric space (X, d) is a set S ⊆ X of points in the
space with the property that every point of X is uniquely determined by the distances from
the elements of S, i.e., S ⊆ X is a metric generator for X if for any pair of distinct
points x, x′ ∈ X there exists s ∈ S such that d(x, s) ̸= d(x′, s). A metric generator of
minimum cardinality in X is called a metric basis, and its cardinality, which is denoted
by dim(X), is called the metric dimension of X . The notion of metric dimension of a
graph was introduced by Slater in [23], where the metric generators were called locating
sets. Harary and Melter [11] independently introduced the same concept, where metric
generators were called resolving sets. Given a simple and connected graph G = (V,E),
we consider the function d : V ×V → N∪{0}, where d(x, y) is the length of a shortest path
in G between u and v and N is the set of positive integers. Since (V, d) is a metric space, a
metric generator for a graph G = (V,E) is simply a metric generator for the metric space
(V, d) and we will use the notation dim(G) instead of dim(V ) for the metric dimension
of G.
Several variations of metric generators have been introduced and studied, namely, re-
solving dominating sets [3], locating-dominating sets [24, 25], independent resolving sets
[5], local metric sets [18], strong resolving sets [22], adjacency generators [15, 16], k-
adjacency generators [6], k-metric generators [1, 7, 8], simultaneous metric generators [21]
etc. In this article, we are interested in the study of adjacency generators.
The notion of adjacency generator was introduced by Jannesari and Omoomi in [16]
as a tool to study the metric dimension of lexicographic product graphs. This concept has
been studied further by Fernau and Rodrı́guez-Velázquez in [9, 10] where they showed that
the (local) metric dimension of the corona product of a graph of order n and some non-
trivial graph H equals n times the (local) adjacency dimension of H . As a consequence
of this strong relation they showed that the problem of computing the adjacency dimension
is NP-hard. This suggests finding the adjacency dimension for special classes of graphs
or obtaining good bounds on this invariant. In this work we obtain general bounds on the
adjacency dimension of a graph G in terms of known parameters of G, while for some
particular cases we obtain closed formulae.
In order to introduce the concept of adjacency generator for a graph G = (V,E), we
define the distance function d2 : V × V → N ∪ {0}, where
d2(x, y) = min{d(x, y), 2}.
An adjacency generator for a graph G = (V,E) is a metric generator for the metric space
(V, d2). Hence, the adjacency dimension of G = (V,E), denoted by adim(G), equals the
metric dimension of (V, d2).
Notice that S ⊆ V is an adjacency generator forG = (V,E) if for every pair of vertices
x, y ∈ V \S there exists s ∈ S which is adjacent to exactly one of these two vertices x and
E-mail addresses: sbernav@upo.es (Sergio Bermudo), jomaro@math.uc3m.es (José M. Rodrı́guez),
juanalberto.rodriguez@urv.cat (Juan A. Rodrı́guez-Velázquez), jsmathguerrero@gmail.com (José M. Sigarreta)
S. Bermudo et al.: The adjacency dimension of graphs 389
y. Therefore, S is an adjacency generator for G if and only if S is an adjacency generator
for its complement G. Consequently,
adim(G) = adim(G). (1.1)
From the definition of adjacency and metric bases, we deduce that S is an adjacency basis
of a graph G of diameter at most two if and only if S is a metric basis of G. In these cases,
adim(G) = dim(G). The reader is referred to [6, 10, 15, 16, 20] for known results on the
adjacency dimension.
The paper is organized as follows. Section 2 is devoted to study the variation of the
adjacency dimension of a graph by removing a set of edges. In particular, we wonder how
far can decrease the adjacency dimension by removing edges from a complete graph and
we obtain a lower bound on the adjacency dimension of any graph in terms of the order.
In Section 3 we show the close relationships that exist between the adjacency dimension
and other parameters, like the domination number, the location-domination number, the
2-domination number, the independent 2-domination number, the vertex cover number, the
independence number and the super domination number.
We will use the notation Kn, Kr,n−r, Cn, Pn and Nn for complete graphs, complete
bipartite graphs, cycle graphs, path graphs and empty graphs of order n, respectively. We
use the notation u ∼ v if u and v are adjacent vertices and G ∼= H if G and H are isomor-
phic graphs. For a vertex v of a graph G, N(v) will denote the set of neighbours or open
neighborhood of v in G, i.e., N(v) = {u ∈ V (G) : u ∼ v}. The closed neighborhood,
denoted by N [v], equals N(v) ∪ {v}. We also define deg(v) = |N(v)| as the degree of
v ∈ V (G), as well as, δ = minv∈V (G){deg(v)} and ∆ = maxv∈V (G){deg(v)}. For the
remainder of the paper, definitions will be introduced whenever a concept is needed.
2 The effect of removing edges and bounds in terms of the order
The following theorem is an important tool to derive some of our results.
Theorem 2.1 ([16]). Let G be a graph of order n. Then the following statements hold.
(i) adim(G) = 1 if and only if n ∈ {1, 2, 3}, G ̸∼= K3 and G ̸∼= N3.
(ii) adim(G) = n− 1 if and only if G ∼= Kn or G ∼= Nn.
(iii) If n ≥ 3 and t ∈ {1, . . . , n− 1}, then adim(Kt,n−t) = n− 2.
(iv) If n ≥ 4, then adim(Pn) = adim(Cn) =
⌊
2n+2
5
⌋
.
In this section we show the effect, on the adjacency dimension, of an operation which
removes a set of edges from a graph. Given a non-empty graph G = (V,E) and an edge
e ∈ E we denote by G− e = (V,E \ {e}) the subgraph obtained by removing the edge e
from G. In general, given a set of edges Ek = {e1, . . . , ek} ⊆ E we denote by G− Ek =
(V,E \Ek) the subgraph obtained by removing the k edges in Ek from G. By analogy we
define the supergraphs G+ e = (V,E ∪ {e}) and G+ Ek = (V,E ∪ Ek), where {e} and
Ek are sets of edges of the complement of G.
Theorem 2.2. LetG = (V,E) be a non-empty graph. For any setEk = {e1, . . . , ek} ⊆ E,
adim(G)− k ≤ adim(G− Ek) ≤ adim(G) + k.
390 Ars Math. Contemp. 22 (2022) #P3.02 / 387–402
Proof. Since (G− Ek−1)− ek = G− Ek, it is enough to prove that, for any e ∈ E,
adim(G)− 1 ≤ adim(G− e) ≤ adim(G) + 1.
Let S be an adjacency basis of G − e, where e = xy. Since S ∪ {y} is an adjacency
generator forG, we have that adim(G) ≤ |S∪{y}| ≤ |S|+1 = adim(G−e)+1. Hence,
adim(G)− 1 ≤ adim(G− e).
Finally, let us observe that adim(G − e) = adim(G− e) = adim(G + e) ≤
adim((G+e)−e)+1 = adim(G)+1 = adim(G)+1. Therefore, the result follows.
Since adim(G − Ek) = adim(G− Ek) = adim(G + Ek), we conclude that
adim(G − Ek) = adim(G) − k if and only if the graph H = G + Ek satisfies
adim(H − Ek) = adim(H) + k. Therefore, in order to show that the inequalities above
are tight, we only need to consider one of them. For instance, adim(Kn − e) = n − 2 =
adim(Kn) − 1. With the aim of showing a more general example, let us consider s stars
Hi ∼= K1,r, r ≥ 4, such that vi is the center and ui1 , . . . , uir are the leaves of Hi, for
i ∈ {1, . . . , s}. Let ei = ui1ui2 , Gi = Hi + ei and M = {vivi+1 : 1 ≤ i < s}, and
define Gr,s = (V,E), where V =
⋃s
i=1 V (Gi) and E = (
⋃s
i=1E(Gi)) ∪M . It is read-
ily seen that adim(Gr,s) = s(r − 1) − 1, while for any k ≤ s and Ek = {e1, . . . , ek},
adim(Gr,s − Ek) = s(r − 1)− 1 + k = adim(Gr,s) + k.
Figure 1: The set of black-colored vertices is an adjacency basis of G4,3.
Figure 2: The set of black-colored vertices is an adjacency basis of G4,3 − E2.
Figure 1 shows the graph G4,3, while Figure 2 shows the graph G4,3−E2. In this case,
adim(G4,3 − E2) = 10 = adim(G4,3) + 2.
All graphs of order n are obtained by successive elimination of edges from a complete
graph (or by successive addition of edges to an empty graph). We know from Theorem 2.1
that for any graph G of order n, adim(G) ≤ n − 1 and the equality holds if and only if
G ∼= Kn or G ∼= Nn. Hence, by Theorem 2.2 we conclude that adim(Kn − e) = n − 2
for every e ∈ E(Kn). Now we wonder how far can decrease the adjacency dimension
by removing edges from Kn, i.e., which is the lower bound for the adjacency dimension
in terms of the order of the graph. This problem is addressed in Propositions 2.3 and 2.4.
Before stating it we need to introduce the following notation.
Given a positive integer s, let G′ be the family of all graphs of order s and G′′ the
family of all graphs of order 2s. We can assume that the graphs in G′ are defined on
S. Bermudo et al.: The adjacency dimension of graphs 391
x1 x3
x2
111
222
122
112
212
121
211
221
Figure 3: A graph G ∈ G3 constructed from G′ ∼= N3 ∈ G′ and G′′ ∼= (K2 ∪N6) ∈ G′′.
S = {x1, . . . , xs} and the graphs in G′′ are defined on the set {1, 2}s of binary words of
length s. Let Gs be the family of graphs constructed from G′ and G′′ as follows. We say that
G ∈ Gs if and only if there exist G′ ∈ G′ and G′′ ∈ G′′ such that V (G) = V (G′)∪ V (G′′)
and E(G) = E(G′)∪E(G′′)∪E∗, where E∗ is the set of edges connecting vertices of G′
with vertices of G′′ in such a way that xi is adjacent to y ∈ {1, 2}s if and only if the i-th
letter of y is 1. Notice that S is an adjacency generator for every G ∈ Gs. Figure 3 shows
a graph G ∈ G3 constructed from G′ ∼= N3 ∈ G′ and G′′ ∼= (K2 ∪N6) ∈ G′′.
The following inequality appeared recently in [15], but we characterize here all graphs
satisfying the equality.
Proposition 2.3. For any graph G of order n,
2adim(G) + adim(G) ≥ n. (2.1)
Furthermore, a graph G of order n = 2s + s satisfies adim(G) = s if and only if G ∈ Gs.
Proof. As we mentioned before, the inequality was proved in [15]. By definition of Gs, if
G ∈ Gs, then {x1, . . . , xs} is an adjacency generator. Now, if adim(G) = r < s, then
Equation (2.1) leads to n = s + 2s > r + 2r ≥ n, which is a contradiction. Therefore,
G ∈ Gs leads to adim(G) = s.
Conversely, suppose that G has order n = 2s + s and adim(G) = s. In this case, for
any adjacency basis S = {x1, . . . , xs} ofG, the function ψ : V (G)\S −→ {1, 2}s defined
by
ψ(x) = (d2(x, x1), . . . , d2(x, xs)),
is bijective, as it is injective and |V (G)\S| = 2s. Hence, takingG′ ∈ G′ as the subgraph of
G induced by S, G′′ ∈ G′′ as the subgraph of G induced by V (G) \ S and E∗ as the set of
edges connecting vertices in S with vertices in V (G)\S, we can conclude thatG ∈ Gs.
Proposition 2.4. For any graph G of order n, adim(G) ≥
⌈
ln( 2n3 )
ln(2)
⌉
.
Proof. If G is a graph with order n and adim(G) = k, since
n ≤ 2k + k ≤ 2k + 2k−1 = 2k
(
1 +
1
2
)
= 2k
(
3
2
)
,
we conclude that k ≥ ln(
2n
3 )
ln(2) .
392 Ars Math. Contemp. 22 (2022) #P3.02 / 387–402
The bound above is tight. It is achieved, for instance, for the family Gs of graphs
constructed prior to Proposition 2.3. These graphs have order n = s + 2s and metric
dimension s. To check the tightness of the bound we only need to observe that 2(s+2
s)
3 >
2s−1, for every positive integer s. Examples of graphs of small order for which the bound
is achieved are the path P3, the cycles Cr (4 ≤ r ≤ 6), and the cube Q3 = K2□K2□K2,
as adim(P3) = 1, adim(C4) = adim(C5) = adim(C6) = 2 and adim(Q3) = 3.
By Theorem 2.1, for any non-complete and non-empty graph of order n, adim(G) ≤
n − 2. The characterization for graphs G such that adim(G) = n − 2 appeared recently
in [15].
Theorem 2.5. Let G be a connected graph of order n ≥ 5. Then adim(G) = n− 2 if and
only if one of the following conditions holds.
(i) G ∼= Kt,n−t, for some t ∈ {1, . . . , n− 1}.
(ii) G ∼= Kn−t +Nt, for some t ∈ {2, . . . , n− 2}.
(iii) G ∼= (K1 ∪Kt) +Kn−t−1, for some t ∈ {2, . . . , n− 2}.
We conclude this section with a characterization of all graphs G satisfying that
adim(G) = 2, which also appeared in [15].
Theorem 2.6. Let G be a connected graph of order n. Then adim(G) = 2 if and only if
one of the following conditions holds for G (or G).
(a) G ∼= K3.
(b) n = 4 and G ̸∼= K4.
(c) n = 5 and G ̸∼= K5, G ̸∼= Kt,5−t for t ∈ {1, . . . , 4}, G ̸∼= K5−t + Nt and
G ̸∼= (K1 ∪Kt) +K4−t for t ∈ {2, 3}.
(d) n = 6 and G ∈ G2.
3 Relationship between the adjacency dimension and other parame-
ters
A setD ⊆ V (G) is a dominating set ofG ifN(x)∩D ̸= ∅ for every vertex x ∈ V (G)\D.
The domination number, γ(G), is the minimum cardinality among all dominating sets of
G. A dominating set of cardinality γ(G) is called a γ(G)-set. The reader is referred to the
books [12, 13] on the domination theory.
The following result is immediate from Equation (1.1) and the fact that at most one
vertex of G is not dominated by the vertices in an adjacency generator of G.
Remark 3.1 ([10]). For any graph G,
adim(G) ≥ max{γ(G), γ(G)} − 1.
The bound above is tight. For instance, it is attained by the corona graphs Kr ⊙ K1,
r ≥ 3, as adim(Kr ⊙K1) = r − 1 and γ(Kr ⊙K1) = r. Another example is any graph
G ∈ G with γ(G) = s+ 1. A particular case is shown in Figure 3.
S. Bermudo et al.: The adjacency dimension of graphs 393
A locating-dominating set is a dominating set D that locates/distinguishes all the ver-
tices in the sense that every vertex not in D is uniquely determined by its neighbourhood
in D, i.e., N(u) ∩ D ̸= N(v) ∩ D for every pair of vertices u, v ∈ V (G) \ D. The
location-domination number of G, denoted λ(G), is the minimum cardinality among all
locating-dominating sets in G. A locating-dominating set of cardinality λ(G) is called a
λ(G)-set. The concept of a locating-dominating set was introduced and first studied by
Slater [24, 25] and studied, for instance, in [4, 14, 19] and elsewhere.
Since every locating-dominating set is an adjacency generator and any adjacency basis
S dominates at least all but one vertex in V (G) \ S, we deduce the following remark.
Remark 3.2. For any graph G,
adim(G) ≤ λ(G) ≤ adim(G) + 1.
Furthermore, λ(G) = adim(G) + 1 if and only if no adjacency basis of G is a dominating
set.
In general, for non-connected graphs we can state the following remark.
Remark 3.3. Let {G1, . . . , Gk} be the set of components of a graph G. If there exists at
least one component where no adjacency basis is a dominating set, then
adim(G) = −1 +
k∑
i=1
λ(Gi).
Otherwise,
adim(G) =
k∑
i=1
λ(Gi) =
k∑
i=1
adim(Gi).
Furthermore, if there are exactly t ≥ 1 components where no adjacency basis is a dominat-
ing set, then
adim(G) = t− 1 +
k∑
i=1
adim(Gi).
According to the two remarks above, tight bounds on adim(G) impose good bounds
on λ(G). In any case, the problem of obtaining the location-domination number of a graph
G from the adjacency dimension of G, forces us to know whether G has dominating basis
or not. Therefore, we can state the following open problem.
Problem 3.4. Characterize the graphs where no adjacency basis is a dominating set.
In order to show some families of graphs where every adjacency basis is a dominating
set, we proceed to state the following lemma obtained previously in [20].
Lemma 3.5 ([20]). Let G be a connected graph. If has diameter D ≥ 6, or G ∼= Cn with
n ≥ 7, or G is a graph of girth g ≥ 5 and minimum degree δ ≥ 3, then for every adjacency
generator B for G and every v ∈ V (G), B ̸⊆ N(v).
Theorem 3.6. Let G be a connected graph. If G has diameter D ≥ 6, or G ∼= Cn with
n ≥ 7, or G is a graph of girth g ≥ 5 and minimum degree δ ≥ 3, then
adim(G) = λ(G).
394 Ars Math. Contemp. 22 (2022) #P3.02 / 387–402
Proof. Let G be a graph satisfying the hypothesis and let S be an adjacency basis of G. By
Lemma 3.5 we deduce that S is a dominating set of G and, since S is an adjacency basis of
G, we can conclude that S is a locating-dominating set ofG. Therefore, adim(G) = λ(G),
as required.
Theorem 3.7. Let G be a graph of order n and maximum degree ∆. If ∆ ln(2) < ln
(
2n
3
)
,
then adim(G) = λ(G).
Proof. Let S be an adjacency basis of G. If ∆ ln(2) < ln
(
2n
3
)
, then Proposition 2.4 leads
to deg(u) ≤ ∆ < ln(
2n
3 )
ln(2) ≤ |S| for every u ∈ V (G) \ S, concluding that S is a locating-
dominating set of G. Therefore, adim(G) = λ(G).
The following result is a direct consequence of the theorem above.
Corollary 3.8. LetG be a graph of order n and minimum degree δ. If δ > n−
⌈
ln( 2n3 )
ln(2)
⌉
−1,
then adim(G) = λ(G).
Theorem 3.9. Given a graph G of order n, the following assertions hold.
(i) If G has at most one isolated vertex, then adim(G) ≤ n− γ(G).
(ii) If G has at most one vertex of degree n− 1, then adim(G) ≤ n− γ(G).
(iii) If G has no isolated vertices, then λ(G) ≤ n− γ(G).
Proof. In this proof, the number of edges of a graph H will be denoted by m(H). Let
G be a graph having at most one isolated vertex and let S be a γ(G)-set such that for any
γ(G)-set S′ it is satisfiedm(⟨S⟩) ≥ m(⟨S′⟩). We shall show that V (G)\S is an adjacency
generator. Suppose to the contrary that V (G) \ S is not an adjacency generator. In such
a case, there exist x, y ∈ S such that for every z ∈ V (G) \ S, either x, y ∈ N(z) or
x, y /∈ N(z). As a result, neither x nor y has any private neighbour (with respect to S) in
V (G) \ S. We can assume that x is not an isolated vertex. Now, if N(x) ∩ S ̸= ∅, then
S \ {x} is a dominating set, which is a contradiction. If N(x) ∩ S = ∅, then taking any
z ∈ N(x) we have that S′ = (S \ {x}) ∪ {z} is a γ(G)-set such that m(⟨S′⟩) > m(⟨S⟩),
which is a contradiction. Therefore, V (G) \ S is an adjacency generator, and so (i) and (ii)
follow.
Furthermore, if G has no isolated vertices, then the complement of every γ(G)-set is a
dominating set, which implies that V (G) \ S is a locating-dominating set. Therefore, (iii)
follows.
The bounds above are tight. For instance, bounds (i) and (iii) are achieved by G ∼= Kn,
G ∼= P4 and Kp,q (2 ≤ p ≤ q). Bound (i) is also achieved by G ∼= K1 ∪Kr (r ≥ 2), as
adim(K1 ∪ Kr) = r − 1 and γ(K1 ∪ Kr) = 2, and (iii) is also achieved by any corona
graph G ∼= H ⊙K1, as in this case λ(G) = |V (H)| = γ(G) = n2 . Obviously, bound (i) is
achieved by a graph G if and only if bound (ii) is achieved by G.
We now emphasize two well-known bounds on the domination number.
Theorem 3.10 ([26]). For any graph G of order n and maximum degree ∆ ≥ 1,
γ(G) ≥
⌈
n
∆+ 1
⌉
.
S. Bermudo et al.: The adjacency dimension of graphs 395
A graph invariant closely related to the domination number is the 2-packing number. A
set S ⊆ V (G) is a 2-packing if for each pair of vertices u, v ∈ S, N [u] ∩N [v] = ∅. The
2-packing number ρ(G) is the cardinality of a maximum 2-packing.
Theorem 3.11 ([13]). For any graph G,
γ(G) ≥ ρ(G).
The following result is a direct consequence of combining Remark 3.1 and Theo-
rems 3.9, 3.10 and 3.11.
Theorem 3.12. Let G be a non-empty graph of order n, maximum degree ∆ and minimum
degree δ. The following assertions hold.
(a) adim(G) ≥ max
{⌈
δ
n−δ
⌉
,
⌈
n−∆−1
∆+1
⌉}
.
(b) adim(G) ≥ max{ρ(G), ρ(G)} − 1.
(c) If δ ≥ 1, then λ(G) ≤ n−max
{
ρ(G),
⌈
n
∆+1
⌉}
.
(d) If G has at most one isolated vertex, then adim(G) ≤ n−max
{
ρ(G),
⌈
n
∆+1
⌉}
.
(e) If G has at most one vertex of degree n− 1, then
adim(G) ≤ n−max
{
ρ(G),
⌈
n
n− δ
⌉}
.
Bound (a) is achieved by complete graphs, while bounds (b) and (c) are achieved by the
corona graphsKr⊙K1, r ≥ 3, as in this case adim(Kr⊙K1) = r−1 and ρ(Kr⊙K1) =
r = λ(Kr ⊙K1). Bounds (c) and (d) are achieved by G = Kn. Obviously, bound (e) is
achieved by a graph G is and only if bound (d) is achieved by G.
A set S ⊆ V (G) is a k-dominating set if |N(v) ∩ S| ≥ k for every v ∈ V (G) \ S.
The minimum cardinality among all k-dominating sets is called the k-domination number
of G and it is denoted by γk(G). A set S ⊆ V (G) is an independent k-dominating set if
it is both an independent set and a k-dominating set. The minimum cardinality among all
independent k-dominating sets is called the independent k-domination number of G and it
is denoted by ik(G).
Theorem 3.13. If G is a non-trivial graph which does not have cycles of length four, then
λ(G) ≤ γ2(G).
Proof. Let S be a 2-dominating set. If S is not an adjacency basis, then there exist u, v ∈
V \ S such that N(u) ∩ S = N(v) ∩ S. Since |N(v) ∩ S| ≥ 2, there exists a cycle with
four vertices, which is a contradiction.
The inequality above is tight. For instance, for the graph shown in Figure 4 we have
that adim(G) = λ(G) = γ2(G) = 4.
Theorem 3.14. Let G be a graph which does not have cycles of length four, and let S be a
γ2(G)-set. If there exists s ∈ S such that N [s] ∩ S ̸= N(x) ∩ S for every x ∈ N(s) \ S,
then adim(G) ≤ γ2(G)− 1.
396 Ars Math. Contemp. 22 (2022) #P3.02 / 387–402
Figure 4: The set of black-colored vertices is an adjacency basis and a 2-dominating set.
Hence, adim(G) = λ(G) = γ2(G) = 4.
Proof. Let s ∈ S such that N [s] ∩ S ̸= N(x) ∩ S for every x ∈ N(s) \ S. Let us see that
S′ = S \{s} is an adjacency generator. Suppose to the contrary, that S′ is not an adjacency
generator. In such a case, there exist u, v ∈ V (G) \ S′ such that N(u) ∩ S′ = N(v) ∩ S′.
We differentiate three cases for these two vertices.
Case 1: u = s. In this case v ̸∈ N(s) and so |N(v) ∩ S′| ≥ 2. Hence, there exists a cycle
with four vertices, which is a contradiction.
Case 2: u ̸∈ N [s]. Since |N(u) ∩ S′| ≥ 2, there exists a cycle with four vertices, which is
a contradiction.
Case 3: u, v ∈ N(s). Since |N(u) ∩ S| ≥ 2, there exists a cycle with four vertices, which
is a contradiction.
Therefore, the result follows.
In the next result we are assuming that any acyclic graph has girth g = +∞.
Corollary 3.15. LetG be a graph of minimum degree δ ≥ 1. Then the following assertions
hold.
(i) If G has girth g ≥ 5, then adim(G) ≤ γ2(G)− 1.
(ii) If G has an independent 2-dominating set and does not have cycles of length four,
then adim(G) ≤ i2(G)− 1.
The bounds above are tight. For instance, for 3 ≤ k ≤ 7 we have that i2(C2k) =
γ2(C2k) = k and adim(C2k) = k − 1.
Recall that a set S of vertices of G is a vertex cover of G if every edge of G is incident
with at least one vertex of S. The vertex cover number of G, denoted by β(G), is the
smallest cardinality of a vertex cover of G. We refer to a β(G)-set in a graph G as a vertex
cover of cardinality β(G). The largest cardinality of a set of vertices of G, no two of
which are adjacent, is called the independence number of G and it is denoted by α(G).
The following well-known result, due to Gallai, states the relationship between the
independence number and the vertex cover number of a graph.
Theorem 3.16. (Gallai’s theorem) For any graph G of order n,
α(G) + β(G) = n.
A leaf is a vertex of degree one and a strong support vertex is a vertex which is adjacent
to more than one leaf.
Theorem 3.17. Let G be a graph of order n without isolated vertices. If G does not have
neither cycles of four vertices nor strong support vertices, then
λ(G) ≤ β(G) = n− α(G).
S. Bermudo et al.: The adjacency dimension of graphs 397
Proof. Let S be a β(G)-set. Since V (G) \ S is an independent set and G does not have
isolated vertices, S is a dominating set. Suppose to the contrary that S is not an adjacency
generator. In such a case, there exist u, v ∈ V (G) \ S such that N(u) ∩ S = N(v) ∩ S. If
|N(v)∩S| ≥ 2, then there exists a cycle with four vertices, which is a contradiction. Now,
if |N(v) ∩ S| = {w}, then w is a strong support vertex, which is a contradiction again.
Therefore, the results follows.
To see that the above inequality is tight, we can consider the graph shown in Figure 4. In
this case, the set of black-colored vertices is a β(G)-set and adim(G) = λ(G) = β(G) =
n− α(G) = 4.
A set S ⊆ V (G) is called a super dominating set ofG if for every vertex u ∈ V (G)\S,
there exists u′ ∈ S such thatN(u′)\S = {u}. The super domination number ofG, denoted
by γsp(G), is the minimum cardinality among all super dominating sets in G. A super
dominating set of cardinality γsp(G) is called a γsp(G)-set. The study of super domination
in graphs was introduced in [17].
Theorem 3.18. For any graph G,
λ(G) ≤ γsp(G).
Furthermore, if G has minimum degree δ ≥ 3 and does not have cycles of length four, then
λ(G) ≤ γsp(G)− 1.
Proof. Let S be a γsp(G)-set, C = V (G) \ S and the function f : C −→ S where f(u)
is one of the vertices in S satisfying that N(f(u)) \ S = {u}. Since, f(u) distinguishes
u ∈ C from any v ∈ C \{u}, we conclude that S is a locating-dominating set ofG. Hence,
λ(G) ≤ |S| = γsp(G).
Assume that G has minimum degree δ ≥ 3 and does not have cycles of length four. Let
A = f(C) be the image of f and B = S \A. We differentiate the following two cases.
Case 1: There exists u ∈ C such that N(u) ∩ B ̸= ∅. We claim that S′ = S \ {f(u)} is
a locating-dominating set. Since N(f(u)) ∩ C = {u} and deg(f(u)) ≥ 3, we have that
|N(f(u)) ∩ S′| ≥ 2. Hence, S′ is a dominating set. Now, every v ∈ C \ {u}
is distinguished from u by f(v) ∈ S′. Finally, if f(u) and v ∈ C are not distinguished
by some vertex in S′, then v, f(u) and two vertices belonging to N(f(u)) ∩ S′ form a
cycle of length four, which is a contradiction. Therefore, S′ is a locating-dominating set,
and so λ(G) ≤ |S′| = γsp(G)− 1.
Case 2: N(u) ∩B = ∅ for every u ∈ C. Notice that |N(f(u)) ∩ S| ≥ 2 for every u ∈ C.
Let u, v ∈ C be two adjacent vertices. We claim that S′ = (S \ {f(u), f(v)}) ∪ {v}
is a locating-dominating set. Obviously, S′ is a dominating set. Now, u is distinguished
from any u′ ∈ C \ {u, v} by f(u′) ∈ S′, and v distinguishes f(u) from f(v). Notice
also that if x ∈ C \ {u, v}, then |N(x) ∩ (S′ \ {v})| = 1 and, since u ∼ v, we have that
f(u) ̸∼ f(v), which implies that |N(y) ∩ (S′ \ {v})| ≥ 2 for every y ∈ {f(u), f(v)}.
Thus, if x ∈ C \ {v} and y ∈ {f(u), f(v)}, then N(x)∩S′ ̸= N(y)∩S′. In summary, S′
is a locating-dominating set and, as a result, λ(G) ≤ |S′| = γsp(G)− 1.
To show that the inequality λ(G) ≤ γsp(G) is tight we consider the following cases:
λ(Kn) = γsp(Kn) = n − 1, λ(K1,n−1) = γsp(K1,n−1) = n − 1, λ(Kr,n−r) =
γsp(Kr,n−r) = n− 2 for 2 ≤ r ≤ n− 2 and λ(H ⊙Nt) = γsp(H ⊙Nt) = |V (H)|t. For
the Petersen graph, shown in Figure 5, we have that λ(G) = γsp(G)− 1 = 3.
398 Ars Math. Contemp. 22 (2022) #P3.02 / 387–402
Figure 5: This figure shows three copies of the Petersen graph. The set of black-coloured
vertices, on the left forms an adjacency basis, on the center forms a λ(G)-set, while on the
right forms a γsp(G)-set.
Lemma 3.19. Let G be a graph with two adjacent vertices x, y ∈ V (G) such that
deg(x) = 1 and deg(y) = 2. If G′ = G − {x, y}, then adim(G) ≤ adim(G′) + 1
and γsp(G) = γsp(G′) + 1.
Proof. If S is an adjacency basis ofG′, then S∪{x} is an adjacency generator ofG, which
implies that adim(G) ≤ adim(G′) + 1.
Assume that D′ is a γsp(G′)-set and u ∈ V (G′) is adjacent to y in G. If u ∈ D′,
then D′ ∪ {y} is a super dominating set of G, while if u /∈ D′, then D′ ∪ {x} is a super
dominating set of G. Therefore, γsp(G) ≤ γsp(G′) + 1. Now, let D be a γsp(G)-set and
v ∈ N(y) \ {x}. If x, y ∈ D, then v /∈ D and (D ∪ {v}) \ {x, y} is a super dominating set
of G′, which implies that γsp(G′) ≤ γsp(G)− 1. Now, if |D ∩ {x, y}| = 1, D \ {x, y} is
a super dominating set of G′ and so γsp(G′) ≤ γsp(G)− 1.
We know that adim(P4) = λ(P4) = γsp(P4) = 2, adim(Kn) = λ(Kn) = γsp(Kn) =
n− 1, adim(Kp,q) = λ(Kp,q) = γsp(Kp,q) = p+ q− 2 (2 ≤ p ≤ q). We proceed to show
that for the remaining graphs, adim(G) ≤ γsp(G)− 1.
Theorem 3.20. For any connected graph G /∈ {P4,Kn,Kp,q}, with 2 ≤ p ≤ q,
adim(G) ≤ γsp(G)− 1.
Proof of Theorem 3.20. Let G be a connected graph such that G /∈ {P4,Kn,Kp,q} for
2 ≤ p ≤ q. If G′ = G − {x, y}, where deg(x) = 1 and deg(y) = 2, then we have the
following:
S. Bermudo et al.: The adjacency dimension of graphs 399
• If G′ ∼= P4, then adim(G) = 2 < 3 = γsp(G).
• If G′ ∼= K1, then adim(G) = 1 < 2 = γsp(G).
• If G′ ∼= Kn−2 (n ≥ 5), then adim(G) = n− 3 < n− 2 = γsp(G).
• If G′ ∼= Kp,q (2 ≤ p ≤ q), then adim(G) = p+ q − 2 < p+ q − 1 = γsp(G).
Hence, by Lemma 3.19 we only need to consider the case where G does not have vertices
of degree one which are adjacent to vertices of degree two.
Let D be a γsp(G)-set, C = V (G) \D and f : C −→ D a function such that, for every
u ∈ C, f(u) is one of the vertices in S satisfying that N(f(u)) \ S = {u}. Let A = f(C)
be the image of f and B = S \ A. Notice that Dc = C ∪ B is also a γsp(G)-set, so any
condition given on A could be also considered on C.
Suppose to the contrary that adim(G) ≥ γsp(G). With the assumptions above in mind,
we proceed to prove the following eight claims.
Claim 1. For any vertex x ∈ C, |N(x) ∩ C| ≤ 1 and |N(f(x)) ∩A| ≤ 1.
Proof of Claim 1. If there exist y, z ∈ C such that f(y), f(z) ∈ N(f(x)) ∩A, then x and
f(x) are distinguished by f(y); x and any u ∈ C \ {x} are distinguished by f(u); while
f(x) and any u ∈ C \ {x} are distinguished by f(y) or by f(z). Hence, D \ {f(x)} is an
adjacency generator, which is a contradiction.
If |N(x) ∩ C| ≤ 1, then we proceed by analogy to the proof above using Dc instead
of D.
Claim 2. For any vertex x ∈ C, deg(x) ≥ 2 and deg(f(x)) ≥ 2.
Proof of Claim 2. Suppose that there exists x ∈ C such that deg(x) = 1. If N(f(x)) ∩
B = ∅, then (by the connectivity of G) Claim 1 leads to deg(f(x)) = 2, which is a
contradiction with our assumptions. Now, if there exists v ∈ N(f(x)) ∩B, then f(x) and
x are distinguished by v; for any y ∈ C \ {x}, f(y) and f(x) are distinguished by y; while
f(y) and x are distinguished by y. Thus, Dc \ {x} is an adjacency generator, which is a
contradiction.
If deg(f(x)) = 1, then we proceed by analogy to the proof above using D instead of
Dc.
Claim 3. Let x ∈ C. IfN(x)∩C = ∅ orN(f(x))∩A = ∅, thenN(x)∩B = N(f(x))∩B.
Proof of Claim 3. If N(f(x)) ∩ A = ∅, then for any z ∈ C \ {x}, f(x) and z are distin-
guished by f(z). SinceD\{f(x)} is not an adjacency generator,N(f(x))∩B = N(x)∩B.
A similar argument works for the case N(x) ∩ C = ∅.
Claim 4. Let x, y ∈ C. If N(f(x)) ∩ A = {f(y)}, then N(f(x)) ∩ B = N(y) ∩ B and
N(f(y)) ∩B = N(x) ∩B.
Proof of Claim 4. Since D \ {f(x)} is not an adjacency generator, if N(f(x)) ∩ A =
{f(y)}, thenN(f(x))∩B = N(y)∩B. Furthermore, by Claim 1,N(f(x))∩A = {f(y)}
leads to N(f(y)) ∩ A = {f(x)}, and since D \ {f(y)} is not an adjacency generator, we
have that N(f(y)) ∩B = N(x) ∩B.
Claim 5. If v ∈ B, then |N(v) ∩A| = 1 and |N(v) ∩ C| = 1.
400 Ars Math. Contemp. 22 (2022) #P3.02 / 387–402
Proof of Claim 5. If v ∈ B and N(v) ∩ A = ∅, then v and any x ∈ C are distinguished
by f(x). Now, if v ∈ B and and there exist y, z ∈ C such that f(y), f(z) ∈ N(v) ∩ A,
then v and any x ∈ C are distinguished by f(y) or by f(z). In both cases, D \ {v} is an
adjacency generator, which is a contradiction. Therefore, |N(v) ∩A| = 1. By analogy we
deduce that |N(v) ∩ C| = 1.
Claim 6. If v1, v2 ∈ B are adjacent vertices, N(v1)∩A = {f(x)} and N(v1)∩C = {y},
then N(v2) ∩A = {f(y)} and N(v2) ∩ C = {x}.
Proof of Claim 6. Assume that v1, v2 ∈ B are adjacent vertices, N(v1)∩A = {f(x)} and
N(v1)∩C = {y}. Since D \ {v1} is not an adjacency generator and f(x) distinguishes v1
and z for every z ∈ C\{x}, we have that x ∈ N(v2). Thus, by Claim 5,N(v2)∩C = {x}.
Furthermore, since D \ {v2} is not an adjacency generator and v1 distinguishes v2 and z
for every z ∈ C \ {y}, we have that f(y) ∈ N(v2). Hence, by Claim 5 we conclude that
N(v2) ∩A = {f(y)}.
Claim 7. If there exists x ∈ C such that N(x) ∩ C = ∅ and N(f(x)) ∩ A = ∅, then
|C| = 1 and G is a complete graph.
Proof of Claim 7. Assume that there exists a vertex x ∈ C such that N(x) ∩ C = ∅ and
N(f(x)) ∩ A = ∅. By Claim 3, N(x) ∩ B = N(f(x)) ∩ B. Let Bx = N(x) ∩ B,
which is nonempty, as G is connected and G ̸∼= K2. If there exist two nonadjacent vertices
vr, vs ∈ Bx, thenD\{vs} is an adjacency generator because x and vs are distinguished by
vr, and any u ∈ C \{x} is distinguished from vs by f(u). Therefore, X = {x, f(x)}∪Bx
induces a complete graph. Now, by the connectivity of G, if V (G) ̸= X , then there
exist two adjacent vertices b, b′ ∈ B such that b ∈ Bx and b′ ∈ B \ Bx. In such a case,
applying Claim 6 for x = y, we conclude that b′ ∈ Bx, which is a contradiction. Therefore,
V (G) = X , |C| = 1 and G is a complete graph.
Claim 8. If there exist x, y ∈ C such that N(f(x)) ∩A = {f(y)}, then G ∼= Kp,q , where
2 ≤ p ≤ q.
Proof of Claim 8. We differentiate two cases.
Case 1: N(x) ∩ C = {y}. Since the subgraph induced by U = {x, f(x), y, f(y)} is
isomorphic to K2,2, V (G) \ U ̸= ∅. By Claims 1, 4 and 5, every vertex in V (G) \ U
which is adjacent to some vertex in U has to belong to B1 = B ∩ N(f(x)) ∩ N(y) or to
B2 = B ∩ N(x) ∩ N(f(y)). Notice that B1 ∩ B2 = ∅. Let X1 = {x, f(y)} ∪ B1 and
X2 = {f(x), y}∪B2. Let us see that G is a complete bipartite graph. Firstly, if there exist
two adjacent vertices u ∈ V (G) \ (X1 ∪ X2) and v ∈ B1 ∪ B2, by the definition of B1
and B2, we know that u ̸∈ A ∪ C. Hence, if v belongs, for instance, to B1, by Claim 6,
u ∈ B2, which is a contradiction. Consequently, V (G) = X1 ∪ X2. Secondly, if there
exist two adjacent vertices u, v ∈ B, by Claim 6, either u ∈ B1 and v ∈ B2 or u ∈ B2 and
v ∈ B1. Finally, if there exist two nonadjacent vertices u ∈ B1 and v ∈ B2, since u and x
are distinguished by v, while u and any z ∈ C \{x} are distinguished by f(z), we have that
D \ {u} is an adjacency generator, which is a contradiction. Therefore, G = (X1 ∪X2, E)
is a complete bipartite graph with |X1| ≥ 2 and |X2| ≥ 2.
Case 2: For any x, y ∈ C such that N(f(x)) ∩ A = {f(y)}, the subgraph induced by
{x, f(x), y, f(y)} is not isomorphic to K2,2. By Claim 2, for every x ∈ C, deg(x) ≥ 2
S. Bermudo et al.: The adjacency dimension of graphs 401
and deg(f(x)) ≥ 2. Since adim(Cn) =
⌊
2n+2
5
⌋
for n ≥ 4 and γsp(Cn) =
⌈
n
2
⌉
for n ≥ 3,
we have that adim(Cn) ≤ γsp(Cn) − 1 for any n ≥ 5. Hence, G ̸∼= Cn and so B ̸= ∅.
By Claim 7, for every x ∈ C either N(x) ∩ C ̸= ∅ or N(f(x)) ∩ A ̸= ∅. Suppose that
there exist x, y ∈ C such that y /∈ N(x) and N(f(x)) ∩ A = {f(y)}. If there exists
b ∈ B ∩ N(x) ∩ N(f(y)), then D′ = (D ∪ {y}) \ {b}, is also a γsp(G)-set. In such
a case, we define a new function f ′ : (C ∪ {b}) \ {y} −→ D′ where f ′(b) = f(y) and
f ′(w) = f(w) for everyw ∈ C\{y}. Since the subgraph induced by {x, f ′(x), b, f ′(b)} is
isomorphic to K2,2, we can conclude the proof using again Case 1. Analogously, suppose
that there exist x, y ∈ C such that f(y) /∈ N(f(x)) and N(x) ∩ C = {y}. If there exists
b ∈ B ∩N(x)∩N(f(y)), then we can take the γsp(G)-set D′ = (Dc ∪{f(x)}) \ {b} and
f ′ : (A ∪ {b}) \ {f(x)} −→ D′ where f ′(b) = x and f ′(f(z)) = z for every z ∈ C \ {x}
to obtain a subgraph isomorphic to K2,2. Applying again Case 1 we get the result.
End of Proof of Theorem 3.20.
The bound above is tight. For instance, for any graph H , adim(H ⊙ Nt) =
γsp(H ⊙ Nt) − 1 = |V (H)|t − 1 and for the Petersen graph shown in Figure 5 we have
adim(G) = γsp(G)− 1 = 3.
ORCID iDs
Sergio Bermudo https://orcid.org/0000-0003-4838-3170
José M. Rodrı́guez https://orcid.org/0000-0003-2851-7442
Juan A. Rodrı́guez-Velázquez https://orcid.org/0000-0002-9082-7647
José M. Sigarreta https://orcid.org/0000-0002-0863-4695
References
[1] A. F. Beardon and J. A. Rodrı́guez-Velázquez, On the k-metric dimension of metric spaces, Ars
Math. Contemp. 16 (2019), 25–38, doi:10.26493/1855-3974.1281.c7f.
[2] L. M. Blumenthal, Theory and Applications of Distance Geometry, Oxford, at the Clarendon
Press, 1953.
[3] R. C. Brigham, G. Chartrand, R. D. Dutton and P. Zhang, Resolving domination in graphs,
Math. Bohem. 128 (2003), 25–36.
[4] J. Cáceres, C. Hernando, M. Mora, I. M. Pelayo and M. L. Puertas, Locating-dominating codes:
bounds and extremal cardinalities, Appl. Math. Comput. 220 (2013), 38–45, doi:10.1016/j.amc.
2013.05.060.
[5] G. Chartrand, V. Saenpholphat and P. Zhang, The independent resolving number of a graph,
Math. Bohem. 128 (2003), 379–393.
[6] A. Estrada-Moreno, Y. Ramı́rez-Cruz and J. A. Rodrı́guez-Velázquez, On the adjacency dimen-
sion of graphs, Appl. Anal. Discrete Math. 10 (2016), 102–127, doi:10.2298/aadm151109022e.
[7] A. Estrada-Moreno, J. A. Rodrı́guez-Velázquez and I. G. Yero, The k-metric dimension of a
graph, Appl. Math. Inf. Sci. 9 (2015), 2829–2840, doi:10.12785/amis.
[8] A. Estrada-Moreno, I. G. Yero and J. A. Rodrı́guez-Velázquez, The k-metric dimension of
Corona product graphs, Bull. Malays. Math. Sci. Soc. 39 (2016), S135–S156, doi:10.1007/
s40840-015-0282-2.
402 Ars Math. Contemp. 22 (2022) #P3.02 / 387–402
[9] H. Fernau and J. A. Rodrı́guez-Velázquez, Notions of metric dimension of corona products:
combinatorial and computational results, in: Computer science—theory and applications,
Springer, Cham, volume 8476 of Lecture Notes in Comput. Sci., pp. 153–166, 2014, doi:
10.1007/978-3-319-06686-8\ 12.
[10] H. Fernau and J. A. Rodrı́guez-Velázquez, On the (adjacency) metric dimension of corona
and strong product graphs and their local variants: combinatorial and computational results,
Discrete Appl. Math. 236 (2018), 183–202, doi:10.1016/j.dam.2017.11.019.
[11] F. Harary and R. A. Melter, On the metric dimension of a graph, Ars Combin. 2 (1976), 191–
195.
[12] T. W. Haynes, S. T. Hedetniemi and P. J. Slater, Domination in Graphs: Volume 2: Advanced
Topics, Taylor & Francis, 1998.
[13] T. W. Haynes, S. T. Hedetniemi and P. J. Slater, Fundamentals of Domination in Graphs, vol-
ume 208 of Monographs and Textbooks in Pure and Applied Mathematics, Marcel Dekker, Inc.,
New York, 1998.
[14] C. Hernando, M. Mora and I. M. Pelayo, On global location-domination in graphs, Ars Math.
Contemp. 8 (2015), 365–379, doi:10.26493/1855-3974.591.5d0.
[15] M. Jannesari, Graphs with constant adjacency dimension, Discrete Math. Algorithms Appl.
(2021), doi:10.1142/s1793830921501342.
[16] M. Jannesari and B. Omoomi, The metric dimension of the lexicographic product of graphs,
Discrete Math. 312 (2012), 3349–3356, doi:10.1016/j.disc.2012.07.025.
[17] M. Lemańska, V. Swaminathan, Y. B. Venkatakrishnan and R. Zuazua, Super dominat-
ing sets in graphs, Proc. Nat. Acad. Sci. India Sect. A 85 (2015), 353–357, doi:10.1007/
s40010-015-0208-2.
[18] F. Okamoto, B. Phinezy and P. Zhang, The local metric dimension of a graph, Math. Bohem.
135 (2010), 239–255.
[19] B. S. Panda and A. Pandey, Algorithmic aspects of open neighborhood location-domination in
graphs, Discrete Appl. Math. 216 (2017), 290–306, doi:10.1016/j.dam.2015.03.002.
[20] Y. Ramı́rez-Cruz, A. Estrada-Moreno and J. A. Rodrı́guez-Velázquez, The simultaneous metric
dimension of families composed by lexicographic product graphs, Graphs Comb. 32 (2016),
2093–2120, doi:10.1007/s00373-016-1675-1.
[21] Y. Ramı́rez-Cruz, O. R. Oellermann and J. A. Rodrı́guez-Velázquez, The simultaneous metric
dimension of graph families, Discrete Appl. Math. 198 (2016), 241–250, doi:10.1016/j.dam.
2015.06.012.
[22] A. Sebő and E. Tannier, On metric generators of graphs, Math. Oper. Res. 29 (2004), 383–393,
doi:10.1287/moor.1030.0070.
[23] P. J. Slater, Leaves of trees, in: Proceedings of the Sixth Southeastern Conference on Com-
binatorics, Graph Theory, and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1975),
Congressus Numerantium, No. XIV, 1975 pp. 549–559.
[24] P. J. Slater, Domination and location in acyclic graphs, Networks 17 (1987), 55–64, doi:10.
1002/net.3230170105.
[25] P. J. Slater, Dominating and reference sets in a graph, J. Math. Phys. Sci. 22 (1988), 445–455.
[26] H. Walikar, B. Acharya and E. Sampathkumar, Recent Developments in the Theory of Dom-
ination in Graphs and Its Applications, MRI Lecture Notes in mathematics, 1979, https:
//books.google.si/books?id=b4i-PgAACAAJ.
ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 22 (2022) #P3.03 / 403–413
https://doi.org/10.26493/1855-3974.2638.d0b
(Also available at http://amc-journal.eu)
On the chromatic index
of generalized truncations
Brian Alspach * , Aditya Joshi †
School of Mathematical and Physical Sciences, University of Newcastle, Callaghan,
NSW 2308, Australia
Received 21 May 2021, accepted 17 March 2022, published online 24 June 2022
Abstract
We examine the chromatic index of generalized truncations of graphs and multigraphs.
The insertion graphs considered are complete graphs, cycles, regular graphs and forests.
Keywords: generalized truncation, chromatic index
Math. Subj. Class. (2020): 05C15
1 Introduction
A broad definition of generalized truncations of graphs was introduced in [1]. We give
this definition now for completeness, but first a brief word about terminology is in order.
The term multigraph is used if multiple edges are allowed. Thus, a graph does not have
multiple edges. We use V (X) to denote the set of vertices of a multigraph X and E(X) to
denote the set of edges. The order of X is |V (X)| and the size of X is |E(X)|. Finally,
the valency of a vertex u, denoted val(u), is the number of edges incident with u. The term
k-valent multigraph is used for regular multigraphs of valency k. Throughout the paper,
∆(X) denotes the maximum valency of the multigraph X and ∆ often is used when the
graph X involved is apparent.
Given a multigraph X , a generalized truncation of X is obtained as follows via a two-
step operation. The first step is the excision step. Let M denote an auxiliary matching (no
two edges have a vertex in common) of size |E(X)|. Let F : E(X) → M be a bijective
function and for [u, v] ∈ E(X), label the ends of the edge F ([u, v]) with u and v. Let
*Corresponding author.
†Author wishes to thank the University of Newcastle for support from a 2020 - 2021 Summer Research Schol-
arship during which time this research was begun.
E-mail addresses: brian.alspach@newcastle.edu.au (Brian Alspach), aditya.joshi@uon.edu.au (Aditya
Joshi)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
404 Ars Math. Contemp. 22 (2022) #P3.03 / 403–413
MF denote the vertex-labelled matching thus obtained. So MF represents the edges of X
completely disassembled.
The second step is the assemblage step. For each v ∈ V (X), the set of vertices of
MF labelled with v is called the cluster at v and is denoted cl(v). Insert an arbitrary graph
on cl(v). The inserted graph on cl(v) is called the constituent graph at v and is denoted
con(v). The resulting graph
MF ∪v∈V (X) con(v)
is a generalized truncation of X . We usually think of the labels on the vertices of MF as
being removed following the assemblage stage, but there are many times when the labels
are useful in the exposition. We use TR(X) to denote a generalized truncation of the
multigraph X .
Truncations arise via action involving the edges incident with a vertex. Consequently,
isolated vertices are useless and we make the important convention that the multigraphs
from which we are forming generalized truncations do not have isolated vertices. This
will not be mentioned in the subsequent material, but is required for the validity of a few
statements.
Recall that a proper edge coloring of a multigraph X is a coloring of the edges so that
adjacent edges do not have the same color. The chromatic index of X , denoted χ′(X), is
the fewest number of colors for which a proper edge coloring exists. We now state Vizing’s
well-known theorem [4] which plays a fundamental role in studying chromatic index.
Theorem 1.1. If X is a graph, then its chromatic index is either ∆(X) or ∆(X) + 1.
The preceding theorem leads to a classification of graphs as follows. A graph is class I if
its chromatic index is equal to its maximum valency ∆ and is class II otherwise. There is a
generalization for multigraphs and to ease subsequent exposition we shall say a multigraph
is class I if its chromatic index equals its maximum valency.
The following result was proved in [1]. A generalized truncation TR(X) is called
complete when every constituent is a complete subgraph.
Theorem 1.2. If X is a class I graph, then its complete truncation also is class I. If X is a
class II graph and its maximum valency is even, then its complete truncation is class I.
The purpose of this paper is to extend the preceding result and explore edge colorings
of generalized truncations in more detail.
2 Some useful results
In the following material we shall be considering the relationship between multigraphs and
their generalized truncations that are class I. Doing so requires the use of some well-known
edge coloring results which we now give for completeness.
Let σ be a permutation of the vertex set of the complete graph Kn of order n. If Y is a
subgraph of Kn, then σ(Y ) denotes the subgraph of Kn whose edge set is
{[σ(u), σ(v)] : [u, v] ∈ E(Y )}. Let n be even and ρ be the permutation defined by
ρ = (u0)(u1 u2 u3 · · · un−1). So ρ fixes one vertex and cyclically rotates the remaining
vertices. Let Y be the perfect matching of Kn consisting of the edges
[u0, u1], [u2, un−1], [u3, un−2], . . . , [un/2, u(n+2)/2].
B. Alspach and A. Joshi: On the chromatic index of generalized truncations 405
t t
t t
t tttu0
u1
u2
u3
u4u5
u6
u7
Figure 1: Canonical edge coloring.
We then obtain a proper edge coloring of Kn by letting the color classes be
Y, ρ(Y ), ρ2(Y ), . . . , ρn−2(Y ). We call this proper edge coloring the canonical edge col-
oring of Kn. Figure 1 shows Y for K8. We obtain a proper edge coloring of Kn with n
colors, n odd, by starting with the canonical edge coloring of Kn+1 and then removing the
central vertex u0 and the edges incident with it. Note that each vertex is missing an edge
of one color and the set of missing colors has cardinality n. This fact is true for any proper
edge coloring of Kn, n odd.
Because we use the preceding canonical edge coloring for both even and odd orders, we
describe it by referring to the edge of length 1 as the anchor. When n is even, the anchor
is the edge [un/2, u1+n/2]. When n is odd, the anchor is the edge [u(n+1)/2, u(n+3)/2]. To
obtain proper edge colorings of complete graphs, we cyclically rotate the canonical edge
coloring which, of course, cyclically rotates the anchor. So we may determine a color class
by specifying the anchor.
Lemma 2.1. Every proper edge coloring of Kn with n colors, n odd, has the property that
there is one color missing on the edges incident with a given vertex, and the set of missing
colors at the vertices has cardinality n.
Proof. There are at most (n− 1)/2 edges of a fixed color in a proper edge coloring of Kn
because n is odd. Thus, there is no proper edge coloring using just n − 1 colors because
(n − 1)2/2 <
(
n
2
)
. There is a proper edge coloring using n colors by Vizing’s Theorem.
Hence, each color class contains precisely (n − 1)/2 edges from which the conclusion
follows.
We use a result about list chromatic index so we discuss it briefly here and present the
result. Given a graph X and for each edge [u, v] a list L[u,v] of colors we may use for the
edge [u, v], a proper list edge coloring is a proper edge coloring of X so that the color on
each edge [u, v] belongs to L[u,v]. The list chromatic index of X is the smallest N such
that if every list L[u,v] has cardinality N , then X admits a proper list edge coloring. We
denote this value by χ′L(X). The following important result was proved by Häggkvist and
Jansen in [2].
Theorem 2.2. The complete graph Kn satisfies χ′L(Kn) ≤ n.
406 Ars Math. Contemp. 22 (2022) #P3.03 / 403–413
3 A general result
We are interested in determining which generalized truncations of a given multigraph X
are class I. Theorem 3.1 below provides a general answer and in subsequent sections we use
the theorem to describe class I generalized truncations of specific types. Some definitions
and notation are required before stating the theorem.
Given a generalized truncation TR(X), for convenience we call the subgraph induced
by the edges of con(v) together with the edges of MF incident with the vertices of con(v)
the sun centered at con(v). Given that con(v) is regular of valency d− 1, when is the sun
centered at con(v) class I? If it is class I, then there is a proper edge coloring using d colors.
We use a vector of length d to describe the number of edges of MF of the various colors in
the sun centered at con(v).
Given a vertex v ∈ X of valency r, we say that a vector (x1, x2, . . . , xd) is admissible
if
∑
i xi = r, the edges of MF are colored according to the vector, that is, xi edges have
color i, and there is a (d− 1)-regular graph on con(v) such that the sun centered at con(v)
is class I. We say the vector is totally inadmissible if there is no (d − 1)-regular graph on
con(v) which makes the sun class I.
Theorem 3.1. If (x1, x2, . . . , xd) satisfies x1+x2+· · ·+xd = r ≥ d, then (x1, x2, . . . , xd)
is admissible if and only if all of x1, x2, . . . , xd and r have the same parity and when the
parity is odd, d also is odd. Otherwise, the vector is totally inadmissible.
Proof. Suppose that val(v) = r in a multigraph X and that there is a class I sun centered
at con(v) which is regular of valency d. Let (x1, x2, . . . , xd) be the vector for the colors
of MF . For an arbitrary color c(i), if the number of edges of color c(i) in con(v) is
k, then the number of edges of color c(i) in MF must be r − 2k. It then follows that
x1, x2, . . . , xd and r all have the same parity. Moreover, if r is odd, then the graph con(v)
is regular of valency d− 1 and order r which implies that d− 1 is even. This completes the
necessity.
Now suppose that all coordinates of (x1, x2, . . . , xd) are odd, r is odd and x1 + x2 +
· · ·+ xd = r. In this case we also have that d is odd. Label the r vertices of the constituent
u1, u2, u3, . . . , ur so that the first x1 vertices are incident with the x1 edges of MF of color
c(1), the next x2 vertices are incident with the x2 edges of MF of color c(2) and continue
in this way. Also carry out subscript arithmetic modulo r on the residues 1, 2, 3, . . . , r.
Consider the xi successive vertices incident with the xi edges of MF of color c(i).
Because xi is odd, there is a central edge of MF of color c(i) and let it be incident with
ua. Letting α = (xi − 1)/2, the vertices incident with edges of MF of color c(i) are
ua−α, ua−α+1, . . . , ua, ua+1, . . . , ua+α.
Consider the canonical color class whose anchor is [ua+(r−1)/2, ua+(r+1)/2]. Color
the edges of this class with color c(i) starting with the anchor and stopping with the edge
[ua−α−1, ua+α+1]. This yields edges colored with c(i) so that each vertex of the con-
stituent is incident with one edge of color c(i). After doing this for each of the d colors, the
resulting sun centered at the constituent is class I and the constituent is regular of valency
d− 1.
This leaves us with the case that all the coordinates of (x1, x2, . . . , xd) are even so that
their sum r also is even. There is no restriction on the parity of d in this case. Also note
that xi = 0 is possible for various values of i. Without loss of generality, let x1, x2, . . . , xa
be the non-zero entries of the vector.
B. Alspach and A. Joshi: On the chromatic index of generalized truncations 407
We label the vertices of the constituent a little differently using the canonical edge
coloring of Figure 1 as a template. The central vertex is labelled u0 and the others are
labelled u1, u2, . . . , ur−1.
Let the x1 edges of MF of color c(1) be incident with u0, u1, . . . , ux1−1. Let the
remaining edges of MF be incident with vertices of the constituent as in the preceding
case, that is, edges of the same color are incident with successively labelled vertices.
The edges of color c(1) have a different form than the other colored edges making
the completion of the coloring a little different for them. Vertices u1, u2, . . . , ux1−1 are
incident with edges of MF of color c(1). This corresponds to the color class with anchor
[u(x1+r−4)/2, u(x1+r−2)/2]. So color the edges of that class with color c(1) starting with
the anchor until finishing with [ur−1, ux1 ]. Every vertex of the constituent now is incident
with an edge of color c(1).
For all the other colors c(j), 1 < j ≤ a, there are an even number of successive
vertices, say uj , uj+1, . . . , uj+t, incident with edges of MF of color c(j). Then the edge
[uj+(t−1)/2, uj+(t+1)/2] is the anchor of a color class. So complete the coloring of the
edges of this color class starting with [uj−1, uj+t+1] moving away from the anchor. For
xi = 0, simply include an entire color class from the canonical coloring scheme using an
unused anchor.
Doing the preceding yields a class I coloring of the sun centered at the constituent so
that the constituent graph is regular of valency d− 1 and completes the proof.
We now obtain three corollaries from Theorem 3.1, but first require a couple of defini-
tions. An edge coloring of a multigraph X is said to be parity-balanced if for each vertex
v of X , the parity of the number of edges of each color incident with v is the same as the
parity of val(v). A regular truncation is a generalized truncation for which every vertex
has the same valency. A generalized truncation is said to be semiregular if each sun cen-
tered at a constituent con(v) is class I and the subgraph con(v) is regular. The distinction
between a semiregular truncation and a regular truncation is that valencies of regularity for
a semiregular truncation may differ over the constituents. Note that the source multigraph
need not be regular in order to have a regular truncation.
Corollary 3.2. Let every entry of the feasible vector (x1, x2, . . . , xd) be even and x1 +
x2 + · · · + xd = r. If d′ is the number of non-zero entries of the vector, then there are
class I suns centered at the appropriate constituent such that the constituents are regular
of every valency from d′ through r − 1.
Proof. This result follows from the proof of Theorem 3.1 rather than the statement. Be-
cause the number of colors on edges of MF is d′, the scheme used in the proof of Theorem
3.1 introduces no new colors so that the valency of regularity is d′. Because r is even, we
may add canonical coloring classes one at a time until reaching Kr. This completes the
proof.
Corollary 3.3. A multigraph X has a class I semiregular truncation if and only if it has a
parity-balanced edge coloring.
Proof. If X has a parity-balanced edge coloring, it follows immediately from Theorem 3.1
that it has a semiregular truncation. On the other hand, if X has a semiregular truncation,
then performing the standard contraction and retention of the colors on the edges of MF
produces a parity-balanced edge coloring of X .
408 Ars Math. Contemp. 22 (2022) #P3.03 / 403–413
Corollary 3.4. A multigraph X has a class I regular truncation of valency d if and only if
it has a parity-balanced coloring and one of the following conditions holds:
(i) When d is odd, every vertex of odd valency has precisely d colors on its incident
edges, and every vertex of even valency has valency at least d + 1 and at most d
colors on its incident edges ; or
(ii) When d is even, every vertex of X has even valency at least d.
Proof. Let X be a mutigraph and suppose it has a generalized truncation Y which is regular
of valency d. Consider the case that d is even. This means that a constituent con(v) is
regular of valency d − 1 and the latter is odd. Thus, the constituent has even order. This
implies that every vertex of X has even valency. Clearly every valency is at least d.
When d is odd, each constituent must itself be regular of valency d− 1. Because d− 1
is even, there is no restriction on the order N of the constituent other than it must be at
least d. If N is odd, then for each of the d distinct colors, there must be at least one edge
of that color belonging to the edges of MF incident with vertices of the constituent. So
the corresponding vertex of X is incident with edges of precisely d different colors. When
N is even there is no restriction on the number of colors on edges of MF incident with
vertices of the constituent other than it is at most d.
For the other direction, when d is even, every constituent has even order and by Corol-
lary 3.2 it is easy to obtain regular valency of d− 1 on the constituent. When d is odd, the
result follows from Theorem 3.1.
4 Complete truncations
Let X be a multigraph with maximum valency ∆. If ∆ is odd and X has an edge coloring
with ∆ colors such that the edges incident with every vertex of valency ∆ have distinct
colors, then we say that X is edge-feasible. The next result characterizes graphs whose
complete truncations are class I. Of course, a complete truncation is a semi-regular trunca-
tion.
Theorem 4.1. The complete truncation of a multigraph X is class I if and only if either
the maximum valency ∆ of X is even, or ∆ is odd and X is edge-feasible.
Proof. This theorem is an improvement on Theorem 1.2 as the latter does not include
multigraphs as part of the hypotheses. Because a complete truncation is a semiregular
truncation, Corollary 3.3 implies that the coloring of the edges in MF must correspond to
a parity-balanced coloring of X . When ∆ is even, coloring all the edges of MF with a
single color corresponds to a parity-balanced coloring of X . Any constituent of order ∆
can be properly edge-colored with ∆ − 1 colors because ∆ is even. Any constituent of
order less than ∆ can be properly edge-colored with at most ∆ − 1 colors. Hence, the
complete truncation of X is class I.
For the remainder of the proof we assume that ∆ is odd. First suppose that the complete
truncation TR(X) is class I. If u ∈ V (X) has valency ∆, then con(u) = K∆ the complete
graph of order ∆. We conclude that the edges of MF incident with con(u) all have different
colors by Lemma 2.1. So if we contract each constituent to a single vertex, delete all the
loops formed and keep the colors of the edges of MF , we obtain an edge coloring of X .
Clearly, all the edges incident to any vertex of valency ∆ in X have distinct colors. Thus,
X is edge-feasible.
B. Alspach and A. Joshi: On the chromatic index of generalized truncations 409
Now let X be edge-feasible and choose an edge coloring of X which is edge-feasible.
Let TR(X) be the complete truncation of X . Color the edges of MF with the same color
they had in the edge coloring of X . These are the edges between the constituents all of
which are complete subgraphs. At this point we have used ∆ colors. It suffices to show
that we can color the edges of the constituents without introducing any new colors so that
the resulting edge coloring of TR(X) is proper.
Any constituent con(u) of order ∆ corresponds to a vertex u of X of valency ∆. This
implies that the edges of MF incident with the vertices of con(u) all have distinct colors
because the coloring of X was edge-feasible. From Lemma 2.1 it is clear that we may color
the edges of con(u) with ∆ colors so that the missing color at each vertex is the color of
the edge of MF at the vertex. This coloring of the edges does not violate the definition of
a proper edge coloring.
Now consider any constituent con(u) of order r ≤ ∆−2. Because each edge of con(u)
has one edge of MF at each of its end vertices, the number of possible colors for the edge
that do not violate the proper edge coloring condition is ∆ − 2. So each edge has a list
of ∆− 2 possible colors, and Theorem 2.2 implies that we may color the edges of con(u)
without violating the proper edge coloring condition because r ≤ ∆− 2.
This leaves us with the case that con(u) has order ∆−1 and this is the most complicated
case. The first observation we make is that the coloring pattern of the edges of MF incident
with the vertices of con(u) can vary all the way from having ∆ − 1 distinct colors to
every color being the same. So we introduce a sequence describing the color pattern. Let
(s1, s2, . . . , st) satisfy s1 ≤ s2 ≤ · · · ≤ st and s1 + s2 + · · ·+ st = ∆− 1. The sequence
means there are t distinct colors on the edges of MF incident with vertices of con(u) and
si such edges have the same color c(i) for i = 1, 2, . . . , t.
We need to show that no matter what the color sequence is we may color the edges of
con(u) so that the sun centered at con(u) is properly edge colored with ∆ colors. As a first
step, label the vertices of con(u) with v0, v1, . . . , va, where a = ∆ − 2. Let the vertices
v1, v2, . . . , vs1 be incident with the s1 edges of MF of color c(1). Let the next s2 vertices
be incident with the s2 edges of MF of color c(2). Continue labelling the vertices in the
obvious manner and let the edge of MF incident with v0 have color c(t).
Carry out an initial coloring of the edges of con(u) using the canonical edge coloring
described in Section 2 with v0 acting as the fixed vertex and ρ = (v0)(v1 v2 · · · va).
The strategy now is to choose the colors for the color classes in such a way that we may
re-color some edges to obtain a proper edge coloring for the sun centered at con(u).
The first observation we make is that only ∆ − 2 colors are required for a canonical
edge coloring of con(u). So we do not use the colors c(t−1) or c(t) for the canonical edge
coloring of con(u). Hence, if t = 1 or t = 2, we already have an edge coloring of con(u)
that does not violate the conditions for a proper edge coloring. Thus, we assume that t ≥ 3.
The anchor for a color class plays an active role as follows. Suppose we require two
edges of a color class so that the two edges use four successive vertices under the cyclic
labelling of {v1, v2, . . . , va}. If the anchor is [vi, vi+1], then adding the edge [vi−1, vi+2]
from the same color class easily does the job. It is now easy to see how we may obtain
k edges from the same color class so that they cover 2k successive vertices. With this
in mind, we now describe an iterative process for determining the colors of certain color
classes.
If s1 is odd, then we color the color class for which [v(s1+1)/2, v(s1+3)/2] is the anchor
with c(1). If s1 is even, then we color the color class for which [vs1/2, v(s1+2)/2] is the
410 Ars Math. Contemp. 22 (2022) #P3.03 / 403–413
anchor with c(1). There are two points to observe about the preceding choices. When s1 is
odd, the edge from v1 to vs1+1 has color c(1) and note that the edge of MF incident with
vs1+1 has color c(2). When s1 is even, the edge from v1 to vs1 has color c(1) and vertex
vs1 is the last vertex for which the edge of MF incident with it has color c(1).
We continue in the manner suggested by the preceding paragraph, but discuss it further
to make it clearer. When si is odd and the first vertex incident with an edge of MF of color
c(i) is vd, then the anchor is the edge [vd+(si−1)/2, vd+(si+1)/2] and we color this color
class with c(i). Note that the edge [vd, vd+si ] is in this color class, but the edge of MF
incident with vd+si has color c(i+ 1).
When si is even and the first vertex incident with an edge of MF of color c(i) is vd,
then the anchor is [vd+(si−2)/2, vd+si/2] and we color this color class with c(i). In this
case, the edge from vd to vd+si−1 is colored c(i) and vd+si−1 is the last vertex incident
with an edge of MF of color c(i).
The preceding procedure is carried out for the colors c(1) through c(t − 2) at which
point it stops because colors c(t− 1) and c(t) are not used for the canonical edge coloring
of con(u). The edges of con(u) that need to be re-colored are those that are adjacent with
edges of MF having the same color. We have seen that when si is odd, there is an edge
in con(u) of color c(i) with one end vertex incident with an edge of MF of color c(i) and
the other end vertex incident with an edge of MF of color c(i + 1). So at the end vertex
incident with an edge of MF of color c(i + 1), the procedure gives another edge of color
c(i+ 1) which also needs to be re-colored.
Thus, the edges that need to be recolored are isolated and possibly paths. Luckily we
have two colors to use for the re-coloring and we arbitrarily re-color isolated edges with
either c(t− 1) or c(t), and alternately re-color the edges of the paths making certain that if
one of the paths terminates with a vertex whose incident edge from MF has color c(t− 1),
we color the edge of the path terminating there with c(t). This removes all potential color
conflicts for con(u) and completes the proof of the theorem.
Corollary 4.2. Let X be a class I graph and TR(X) a generalized truncation.
If ∆(X) = ∆(TR(X)), then TR(X) is class I.
Proof. Let TR(X) be a generalized truncation of X satisfying ∆(TR(X)) = ∆(X).
Then a proper edge coloring of TR(X) requires at least ∆ colors. We know the complete
truncation Y of X is class I by Theorem 4.1, that is, it has a proper edge coloring using
∆ colors. Remove any edges of Y not belonging to TR(X) and we are left with a proper
edge coloring of TR(X) using ∆ colors. So TR(X) is class I.
It is well known that both the Petersen graph and its complete truncation are class II
graphs. The extension to a regular multigraph is an immediate corollary of Theorem 4.1.
Corollary 4.3. A regular multigraph of odd valency is class I if and only if its complete
truncation is class I.
5 Cyclic truncations
Complete truncations have been used by various authors, and there is another generalized
truncation that has been employed frequently. Namely, the generalized truncation obtained
by letting each constituent graph be a cycle. We shall call these cyclic truncations. Probably
the best known example of this is the cube-connected cycles graph first introduced in [3].
B. Alspach and A. Joshi: On the chromatic index of generalized truncations 411
Of course, the ancient Greeks studied truncations of Platonic and Archimedian solids, and
these resulted in cyclic truncations.
When moving from a multigraph X to a generalized truncation Y , we frequently wish
Y to inherit properties of X . This can be a problem for cyclic truncations. For example,
poorly chosen cyclic truncations can play havoc with automorphisms.
We are interested in determining conditions for which a cyclic truncation is class I. Of
course, a cyclic truncation is a regular truncation so that we may use Theorem 3.1. We use
the same vector describing the numbers of colors used on the edges of MF belonging to a
sun centered at a constituent. In this case the vector has length 3 as we are looking at cyclic
truncations all of which are 3-valent. There is a new term not used before, namely, a vector
(x1, x2, x3) is universal if the sun centered at the corresponding constituent is class I for
every cycle on the vertices of the constituent.
Corollary 5.1. If (x1, x2, x3) satisfies x1 + x2 + x3 = d ≥ 3, then (x1, x2, x3) is ad-
missible if and only x1, x2, x3 and d all have the same parity. Otherwise, the vector is
totally inadmissible. Moreover, if just one of x1, x2 and x3 is non-zero, then (x1, x2, x3) is
universal when d is even.
Proof. The portion of the statement prior to the sentence about universality is simply The-
orem 3.1 restricted to d = 3. When the vector has the form (x1, 0, 0) and x1 = d is even,
this corresponds to all the edges of MF incident with the vertices of the constituent having
the same color. Clearly, no matter which even length cycle we form on the vertices of the
constituent, the resulting sun is class I.
Corollary 5.1 gives us an easy way to construct a class I cyclic truncation of a multi-
graph X based on parity conditions for the colors on the edges of MF . Thus, we need to
concentrate on coloring the edges in the source multigraph.
Corollary 5.2. If every vertex of the multigraph X has even valency, then every cyclic
truncation of X is class I.
Proof. This follows from Corollary 5.1 by coloring all edges of MF with a single color.
Corollary 5.3. If X is a regular multigraph of odd valency d > 1 and is class I, then there
are class I cyclic truncations of X .
Proof. Let X be a class I regular multigraph of odd valency d ≥ 3. Choose a proper edge
coloring of X using d colors. In forming a cyclic truncation Y of X , color the edges of
MF according to the colors c(1), c(2), . . . , c(d) the edges have in the proper edge coloring
of X . Now change the color of any edge of colors c(4), c(5), . . . , c(d) to c(3). The vector
for each constituent becomes (1, 1, d − 2) all of which are odd. We now may find a cycle
for each constituent such that the cyclic truncation is class I by Corollary 5.1.
We now give a sufficient condition for a multigraph to have a class I cyclic truncation. A
definition is required. Let X be an arbitrary multigraph and V0, V1, V2 and V3 be a partition
of V (X), where Vi contains the vertices whose valencies are congruent to i modulo 4.
Given a submultigraph Y of X , let X \ Y be the submultigraph obtained by removing the
edges of Y from X . Moreover, let V ′0 , V
′
1 , V
′
2 and V
′
3 denote the vertices of V (X) whose
valencies in X \ Y are congruent to i modulo 4. A submultigraph Y of X is called an
enabling submultigraph if it satisfies:
412 Ars Math. Contemp. 22 (2022) #P3.03 / 403–413
• if v ∈ V0, then v ∈ V ′0 ;
• if v ∈ V1, then v ∈ V ′2 ;
• if v ∈ V2, then v ∈ V ′0 ; and
• if v ∈ V3, then v ∈ V ′2 .
The preceding conditions are not as complicated as they may look at first. The first
thing to notice is that the components of X \ Y are Eulerian. Second, for many graphs
the conditions are simple. For example, if X is 3-regular, then a perfect matching is an
enabling subgraph.
Corollary 5.4. Let X be a multigraph with an enabling submultigraph Y . If every compo-
nent of X \ Y has even size, then there is a class I cyclic truncation of X .
Proof. Because each component of X\Y has even size, we may alternately color the edges
of an Euler tour of the component with two colors so that we finish with same number of
colors on the edges incident with every vertex. Doing this for each component yields a
2-coloring so that each vertex of X \ Y has the same number of colors incident with it.
If we now color all the edges of Y with a third color, it is easy to verify that the con-
ditions of Corollary 5.1 are met for X . We check one possibility for illustrative purposes.
If v ∈ V3, it belongs to V ′2 in X \ Y . Thus, the number of edges of Y incident with v is
congruent to 1 modulo 4, that is, it is incident to an odd number of edges of the third color
in X . Because it is in V ′2 , it is incident with an odd number of edges of each of the first two
colors. Thus, the vector for v has three odd components.
Proposition 5.5. If a trivalent graph X has a cut edge, then X is class II.
The proof of the preceding proposition is immediate and leads to examples such as the
following. Let X be the graph of order 10 obtained by taking two vertex-disjoint copies of
K5 and joining the two copies with a single edge from a vertex of one copy to a vertex of the
other copy. Lemma 2.1 implies that X is class I. If we take any cyclic truncation TR(X)
of X , then TR(X) is class II by Proposition 5.5 because the edge of MF connecting the
two copies still is a cut edge in TR(X).
6 Arboreal truncations
We now examine another generalized truncation. An arboreal truncation is a generalized
truncation for which every constituent graph is a forest. The following result is useful for
arboreal truncations.
Theorem 6.1. Let TR(X) be a generalized truncation of a multigraph X . If the maximum
valency of TR(X) is ∆ and every constituent of TR(X) having a vertex of valency ∆ is
class I, then TR(X) is class I.
Proof. Let con(v) have a vertex of valency ∆ in TR(X). Then the maximum valency in
the subgraph con(v) is ∆ − 1. Its edges may be properly edge-colored with ∆ − 1 colors
because it is class I. Any constituent not having a vertex of valency ∆ may have its edges
properly colored with at most ∆−1 colors because its maximum valency is ∆−2. Then the
edges of MF may be colored with a new color yielding a proper edge coloring of TR(X)
with ∆ colors. The conclusion now follows.
B. Alspach and A. Joshi: On the chromatic index of generalized truncations 413
Corollary 6.2. Every arboreal truncation of a multigraph X is class I.
Proof. This follows immediately from Theorem 6.1 because forests are class I.
ORCID iDs
Brian Alspach https://orcid.org/0000-0002-1034-3993
References
[1] B. Alspach and J. B. Connor, Some graph theoretical aspects of generalized truncations, Aus-
tralas. J. Comb. 79 (2021), 476–494, https://ajc.maths.uq.edu.au/?page=get_
volumes&volume=79.
[2] R. Häggkvist and J. Janssen, New bounds on the list-chromatic index of the complete
graph and other simple graphs, Comb. Probab. Comput. 6 (1997), 295–313, doi:10.1017/
s0963548397002927.
[3] F. P. Preparata and J. Vuillemin, The cube-connected cycles: a versatile network for parallel
computation, Comm. ACM 24 (1981), 300–309, doi:10.1145/358645.358660.
[4] V. G. Vizing, On an estimate of the chromatic class of a p-graph, Diskret. Analiz (1964), 25–30.

ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 22 (2022) #P3.04 / 415–441
https://doi.org/10.26493/1855-3974.2134.ac9
(Also available at http://amc-journal.eu)
Convex drawings of the complete graph:
topology meets geometry*
Alan Arroyo †
University of Waterloo, Waterloo, Ontario, Canada,
current address: Vienna, Austria
Dan McQuillan
Norwich University, Northfield, Vermont, United States
R. Bruce Richter ‡
University of Waterloo, Waterloo, Ontario, Canada
Gelasio Salazar §
Universidad Autónoma de San Luis Potosı́, Mexico
Received 1 October 2019, accepted 8 September 2021, published online 24 June 2022
Abstract
In a geometric drawing of Kn, trivially each 3-cycle bounds a convex region: if two
vertices are in that region, then so is the (geometric) edge between them. We define a
topological drawing D of Kn in the sphere to be convex if each 3-cycle bounds a closed
region R (either of the two sides of the 3-cycle) such that any two vertices in R have the
(topological) edge between them contained in R.
While convex drawings generalize geometric drawings, they specialize topological ones.
Therefore it might be surprising if all optimal (that is, crossing-minimal) topological draw-
ings of Kn were convex. However, we take a first step to showing that they are convex:
we show that if D has a non-convex K5 all of whose extensions to a K7 have no other
non-convex K5, then D is not optimal (without reference to the conjecture for the crossing
number of Kn). This is the first example of non-trivial local considerations providing suffi-
cient conditions for suboptimality. At our request, Aichholzer has computationally verified
that, up to n = 12, every optimal drawing of Kn is convex.
*We are very grateful to the referees for their detailed reading and useful suggestions for some simplifications
and improving presentation. We also appreciate the contributions of Matthew Sullivan and Kasper Szabo Lyngsie
for simplifying some of our original arguments.
†Supported by CONACYT.
‡Corresponding author. Supported by NSERC grant #50503-10940-500.
§Supported by CONACYT.
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
416 Ars Math. Contemp. 22 (2022) #P3.04 / 415–441
Convexity naturally lends itself to refinements, including hereditarily convex (h-convex)
and face convex (f-convex). The hierarchy rectilinear ⊆ f-convex ⊆ h-convex ⊆ convex ⊆
topological provides links between geometric and topological drawings. It is known that
f-convex is equivalent to pseudolinear (generalizing rectilinear) and h-convex is equivalent
to pseudospherical (generalizing spherical geodesic). We characterize h-convexity by three
forbidden (topological) subdrawings.
This hierarchy provides a framework to consider generalizations of other geometric
questions for point sets in the plane. We provide two examples of such questions, namely
numbers of empty triangles and existence of convex k-gons.
Keywords: Simple drawings, complete graphs, convex drawings.
Math. Subj. Class. (2020): 05C10
1 Introduction
Hill’s long-standing conjecture [14, 10] asserts that the crossing number cr(Kn) of Kn is
equal to
H(n) :=
1
4
⌊
n
2
⌋⌊
n− 1
2
⌋⌊
n− 2
2
⌋⌊
n− 3
2
⌋
.
To date, Hill’s conjecture has only been verified up to n ≤ 12. Moreover, current proofs for
n = 11, 12 rely on extensive computer searches, therefore providing limited explanation for
the elegance of the expression in Hill’s conjecture. Guy’s [13] original proof for n = 9, 10
also relied on an extensive case analysis, with most details left to the reader, similar to a
computer proof.
The main point of this work is the introduction of the class of convex drawings of
Kn. It turns out that, of the (up to spherical homeomorphisms) five drawings of K5 in the
sphere, the drawings K̃35 and K̃55 in Figure 1 are not convex in our sense. Furthermore, an
elementary but principal result of this work is to characterize (rather than define) a spherical
drawing of Kn as convex if and only if neither K̃35 nor K̃55 occurs as a subdrawing.
Our study of these drawings was motivated by a couple of specific events. One was
the computer-free proof by two of the authors that the crossing number of K9 is 36 [21].
As part of that proof, the two drawings K̃35 and K̃55 were both shown not to occur in any
optimal (that is, fewest crossings) drawing of K7. Another was the question, “Is there, for
some n, an optimal drawing (or even one with the conjectured fewest crossings) of Kn that
contains K̃55?”
One of us asked Tilo Wiedera to check by computer if any optimal drawing of K9
contains a K̃55. Not only was the answer negative as expected, but Wiedera also found that
the smallest number of crossings in a drawing of K9 that contains K̃55 has 40 crossings –
a surprising 4 more than optimal! At the Crossing Number Workshop in Osnäbruck (May
2017), the authors then asked Aichholzer for the smallest n for which an optimal drawing
of Kn could contain it. After checking for both K̃35 and K̃55 among all optimal drawings
for Kn with n ≤ 12, he announced at the workshop his findings, implying that if such n
exists, it must be at least 13. Theorem 5.1 below gives further evidence that the answer to
the question is “no”.
E-mail addresses: alanarroyoguevara@gmail.com (Alan Arroyo), dmcquill@norwich.edu (Dan McQuillan),
brichter@uwaterloo.ca (R. Bruce Richter), gsalazar@ifisica.uaslp.mx (Gelasio Salazar)
A. Arroyo et al.: Convex drawings of the complete graph: topology meets geometry 417
Figure 1: Drawings of interest: K̃35, K̃55, K116 , and TC8.
Thus, we were quite naturally led to the class of simple (i.e., no edge crosses itself and
no two closed edges intersect twice; often referred to as “good”) drawings of Kn in which
neither K̃35 nor K̃55 occurs; we first thought of these as “locally rectilinear”, as, of the (up to
spherical homeomorphisms) five drawings of K5 in the sphere, these are the two that are
not isomorphic (via a homeomorphism from the sphere with an appropriate point deleted
to the plane) to rectilinear drawings of K5. We became especially interested in them when
we realized they have a topological characterization (Theorem 2.6, below), which we have
since taken to be the definition of convex drawings in Definition 1.1 below.
If D is a drawing of a graph G, and H is a subgraph of G (or even a set of vertices and
edges of G), then we let D[H] denote the drawing of H induced by D. In a simple drawing
D of a graph G, for a 3-cycle T of G, D[T ] is a simple closed curve.
Definition 1.1. Let D be a simple drawing of Kn in the sphere.
1. If T is a 3-cycle in Kn, then a closed disc ∆ bounded by D[T ] is a convex side of T
if, for any distinct vertices x and y of Kn such that D[x] and D[y] are both contained
in ∆, then D[xy] is also contained in ∆.
2. The drawing D is convex if every 3-cycle of Kn has a convex side.
Evidently every rectilinear or spherical geodesic drawing is convex. Therefore, the “tin
can” drawing TC8 of K8 shown in Figure 1 is convex (see Section 2); it is not homeomor-
phic to any rectilinear drawing. Indeed, K8 is known to have rectilinear crossing number
of 19, while TC8 is optimal with the minimum of 18 crossings.
Wagner [25] showed that establishing the Hill Conjecture for spherical geodesic draw-
ings is equivalent to the special case of the Spherical Generalized Upper Bound Con-
jecture for arrangements of n hemispheres in an (n − 4)-dimensional sphere. In par-
ticular, proving the convex or h-convex crossing number of Kn is H(n) would estab-
lish this case of the SGUBC. Flag algebras are used by Balogh et al. [7] to show that
limn→∞ cr(Kn)/H(n) > 0.985. Restricted to convex drawings, the same technique gets
a lower bound of 0.996. The 0.996 is also the lower bound for spherical geodesic drawings
(these are all necessarily convex).
There are two natural refinements of Definition 1.1; the first is satisfied by spherical
geodesic drawings, while the second holds for rectilinear and pseudolinear drawings.
Definition 1.2. Let D be a convex drawing of Kn.
1. Then D is hereditarily convex (abbreviated to h-convex) if, for every 3-cycle T ,
there is a choice ∆T of a convex side, such that, if T1 and T2 are 3-cycles with
D[T2] ⊆ ∆T1 , then ∆T2 ⊆ ∆T1 .
418 Ars Math. Contemp. 22 (2022) #P3.04 / 415–441
2. Then D is face convex (abbreviated to f-convex) if there is a face Γ of D such that,
for every 3-cycle T of Kn, the side of D[T ] disjoint from Γ is convex.
The drawing K116 of K6 shown in Figure 1 is convex but not h-convex. The (optimal)
drawing TC8 of K8 is h-convex but not f-convex.
It is an easy exercise to prove that an f-convex drawing is also h-convex. Moreover, ev-
ery rectilinear (or, more generally, pseudolinear) drawing of Kn is f-convex, with (in both
cases) Γ being the unbounded face. In fact, Aichholzer et al. [4] and, independently, the
current authors [5], have shown that f-convex is equivalent to pseudolinear. Generalizing
spherical geodesic drawings, Arroyo et al. [6] have introduced a natural notion of “pseu-
dospherical drawings” of Kn in the sphere; surprisingly, they are exactly the h-convex
drawings.
It would be very interesting to obtain an analogous “geometric-style” generalization
that characterizes convexity. At this time, we have no suggestion for what this might be.
Thus, our definitions regarding drawings of complete graphs correspond precisely to
geometric descriptions of point-sets. These geometric connections open up new possibil-
ities for studying geometric questions to see how the results differ for convex drawings.
For complete graphs, we now have a geometrically meaningful hierarchy of drawings. It
is, from most to least restrictive: rectilinear ⊆ f-convex (= pseudolinear) ⊆ h-convex (=
pseudospherical) ⊆ convex ⊆ topological.
One question of long-standing interest is: given n points in general position in the plane,
how many of the 3-tuples (that is, triangles) have none of the other points inside the triangle
(empty triangle)? Currently, we know that there can be as few as about 1.6n2+o(n2) empty
triangles [9] and every set of n points has at least n2+O(n) empty triangles (n2+o(n2) first
proved in [8]). In [5], we proved the n2 + o(n2) bound also holds for f-convex drawings.
At the other extreme, Harborth [15] presented an example of a topological drawing of Kn
having only 2n−4 empty triangles, while Aichholzer et al. [3] show that every topological
drawing of Kn has at least n empty triangles. We have shown in [5] that every convex
drawing of Kn has at least 13n
2 + O(n) empty triangles. For h-convex, it is shown in [6],
using the f-convex result and other facts about h-convex drawings, that there are at least
3
4n
2+o(n2) empty triangles. We would be interested in progress related to the coefficients
1
3 and
3
4 .
Another question of interest is: given n points in general position in the plane, what
is the largest k so that k of the n points are the corners of a k-gon in convex position? In
Theorem 3.4, we generalize to convex drawings the Erdős-Szekeres theorem [12] that, for
every k, there is an n such that every set of n points in the plane in general position has a
set of k points that are the corners of a convex k-gon. Finding the least such n is of current
interest. Suk [24] has shown that 2k+o(k) points suffices in the geometric case. For k = 5,
9 points is best possible in the rectilinear case (see Bonnice [11] for a short proof).
For a general drawing D of Kn, we can ask whether there is a subdrawing D[Kk] such
that one face is bounded by a k-cycle: this is a natural drawing of Kk. (In [21], these
drawings are quite appropriately labelled “convex”. We think convex is very descriptive of
the drawings considered in this work and expect there to be no confusion with the two quite
different uses of the term “convex”.) Bonnice’s proof adapts easily to the pseudolinear case
(that is, the f-convex case). Aichholzer (personal communication) has verified by computer
that 11 points is best possible for k = 5 for the convex drawings considered in this article.
A trivial consequence of our Theorem 4.5 characterizing h-convex drawings is that any
convex, but not h-convex, drawing has a natural K5; therefore, 11 is also best possible
A. Arroyo et al.: Convex drawings of the complete graph: topology meets geometry 419
for h-convex drawings. For general drawings of Kn, there need not be a natural K5. In
Harborth’s example [15] (originally from [16]), every K5 is isomorphic to K̃55.
We remark that Pach et al. [23] show that, for any fixed positive integer r, there is
a large enough integer N = N(r) such that, for n ≥ N , every simple drawing of Kn
contains either a natural Kr or Harborth’s generalization of K̃55 on r vertices. In Harborth’s
generalization, every K5 is isomorphic to K̃55. These are the “twisted” Krs in [23].
Section 2 introduces many fundamental properties of a convex drawing D of Kn, in-
cluding showing convexity of D is equivalent to not containing either of the two non-
rectilinear drawings K̃35 and K̃55 of K5 (the first two drawings in Figure 1). Another equiv-
alence is that every 3-cycle T has a side such that every vertex v on that side is such that
D[T + v] is a non-crossing K4.
Section 3 proves that a convex drawing D of Kn has a particularly nice structure: there
is a natural Kr such that D[Kr] has a face Γ bounded by an r-cycle C; if D[v] is in Γ and
D[w] is in the closure of Γ, then D[vw] is in Γ ∪ {D[w]}; and if D[v] and D[w] are in
the closure of the complement of Γ, then D[vw] is in the complement of Γ. This structure
theorem may provide a strategy for showing that a convex drawing of Kn has at least H(n)
crossings.
Section 4 treats h-convex drawings. The main result here is that a convex drawing D
is h-convex if and only if there is no K6 such that D[K6] is K116 in Figure 1. We do not
know a comparable result distinguishing f-convex drawings from h-convex. The tin can
drawing TC8 of K8 in Figure 1 is one such (as are the larger tin can drawings). However it
is not clear to us whether TC8 is the only minimal one or, indeed, if there are only finitely
many minimal distinguishing examples. The final result of the section is that testing a set
of convex sides for h-convexity is also a “Four Point Property”, which is to say that it can
be verified by checking all sets of four points.
Finally, in Section 5, we prove the principal result Theorem 5.1, showing that some
non-convex drawings of Kn are not optimal.
Table 1: The convexity hierarchy.
level characterization distinguish
general edges share ≤ 1 point
convex general, no K̃35, K̃55 K̃35
h-convex convex, no K116 K116
f-convex h-convex + ?? TC8
rectilinear unlikely Pappus
This work will provide characterizations of the different kinds of convexity and distin-
guishing between them by examples and theorems. A summary is given in Table 1.
Matoušek [19] gives a nice exposition of the theorem of Mnëv [22] that testing the
stretchability of an arrangement of pseudolines in the plane is ∃R-complete. A straightfor-
420 Ars Math. Contemp. 22 (2022) #P3.04 / 415–441
ward application of Levi’s Enlargement Lemma [18] (see also [5]) turns such an arrange-
ment with n pseudolines into a pseudolinear drawing of K2n that is not stretchable. Hence,
unless P=NP=∃R, there are infinitely many non-stretchable f-convex drawings of Kn.
2 Convex drawings
In this section we introduce the basics of convexity. We already mentioned in the intro-
duction that the two drawings K̃35 and K̃55 of K5 in Figure 1 are not convex. In fact, their
absence characterizes convexity. We first prove some intermediate results that make this
completely clear. Our first observation is immediate from the definition of convex side and
is surprisingly useful.
Observation 2.1. If J is such that D[J ] is a crossing K4, and T is a 3-cycle in J , then the
side of D[T ] containing the fourth vertex in J is not convex. 2
We had some difficulty deciding on the right definition of convexity. At the level of
individual 3-cycles, the definition given in the introduction makes more sense. At the level
of a drawing being convex, there is a simpler one, as shown in the next lemma and, more
particularly, its Corollary 2.4: we only need to test single points in the closed disc ∆ and
how they connect to the three corners.
Definition 2.2. Let D be a drawing of Kn, let T be a 3-cycle in Kn, and let ∆ be a closed
disc bounded by D[T ]. Then ∆ has the Four Point Property if, for every vertex v of Kn
not in T such that D[v] ∈ ∆, D[T + v] is a non-crossing K4.
Lemma 2.3. Let D be a drawing of K5 such that the side ∆ of a 3-cycle T has the Four
Point Property. Suppose u and v are vertices of K5 such that D[u], D[v] ∈ ∆. If D[uv] is
not contained in ∆, then there is a vertex b of T such that neither side of the 3-cycle induced
by u, v, and b satisfies the Four Point Property; in particular, neither side is convex.
Proof. Since ∆T has the Four Point Property, neither u nor v is in T . Because D[u] and
D[v] are both on the same side of D[T ], D[uv] crosses D[T ] an even number of times.
However, D[uv] crosses each of the three sides of D[T ] at most once, so D[uv] crosses
D[T ] at most three times. Thus, D[uv] crosses D[T ] either 0 or 2 times.
As D[uv] is not contained in ∆T , D[uv] crosses D[T ] a positive number of times. We
conclude they cross exactly twice. Label the vertices of T as a, b, and c so that D[uv]
crosses both D[ab] and D[ac].
Since D[T +u] is a non-crossing K4, the three edges of T +u incident with u partition
∆T into three faces, each incident with a different two of a, b, and c. Because D[uv]
crosses D[ab] and D[ac], but not any of the three edges of T + u incident with u, v must
be in one of the faces of D[T + u] incident with a. We choose the labelling so that v is in
the face of D[T + u] incident with both a and c.
The Four Point Property implies D[vb] is contained in ∆T . It must cross either D[ua]
or D[uc]. To show that it crosses D[uc], we assume by way of contradiction that it crosses
D[ua]. Let × be the point where D[ab] crosses D[uv]. Then D[vb] must exit the region
incident with a, u, and ×, but it cannot cross either D[ab] or D[uv], and it cannot cross
D[au] a second time. This contradiction shows D[vb] crosses D[uc].
Therefore D[T + {u, v}] is isomorphic to K̃35, with u and v being the upper left and
upper right vertices, respectively, and a the peak in Figure 1. Letting b and c be the lower
A. Arroyo et al.: Convex drawings of the complete graph: topology meets geometry 421
left and lower right vertices, respectively, the 3-cycles uvb and uvc have no convex side.
Obviously, if ∆ is a convex side of D[T ], then ∆ has the Four Point Property. The
following converse is an immediate consequence of Lemma 2.3.
Corollary 2.4. Let D be a drawing of Kn and, for each 3-cycle T in Kn, let ∆T be a
closed disc bounded by D[T ]. Suppose, for each T , ∆T has the Four Point Property. Then
each ∆T is convex; in particular, D is convex.
Our next two corollaries yield additional characterizations of convexity.
Corollary 2.5. Let D be a drawing of Kn.
(a) Then D is not convex if and only if there exists a 3-cycle T of Kn and vertices u, w of
Kn, one in the interior of each side of D[T ], such that both D[T + u] and D[T +w]
are crossing K4’s.
(b) If D is convex and T is a 3-cycle in Kn, then a side ∆ of D[T ] is convex if and only
if it satisfies the Four Point Property.
Proof. For Item (a), Observation 2.1 shows that if D is convex, then no such 3-cycle can
exist. Conversely, Corollary 2.4 implies that some 3-cycle T of Kn does not have a side
that satisfies the Four Point Property. This implies that, for each side ∆ of D[T ], there is a
vertex v∆ such that D[T + v∆] is a crossing K4, as required.
The proof of Item (b) is direct from the definition of convex side and Lemma 2.3.
We came to the concept of convexity by considering drawings of Kn without the two
drawings K̃35 and K̃55 (see Figure 1) of K5 for reasons that have been subsumed by some of
the developments described in this article. Since the remaining drawings of K5 are rectilin-
ear, we think of such drawings of Kn as locally rectilinear. Our next result is the surprising
equivalence with convexity and this led us to consider convexity and its strengthenings to
h- and f-convex.
Theorem 2.6. A drawing D of Kn is convex if and only if, for every subgraph J of Kn
isomorphic to K5, D[J ] is not isomorphic to either K̃35 or K̃55.
Proof. In the drawing of K̃35 in Figure 1, we see that a 3-cycle consisting of one of the
edges that is not a straight segment together with the longer horizontal edge has no convex
side. In the drawing of K̃55, there are two 3-cycles that have the “interior vertex” in their
interiors. Neither of these 3-cycles is convex. Thus, these two drawings of K5 cannot occur
in a convex drawing of Kn.
Conversely, in a rectilinear drawing of K5, the bounded side of each 3-cycle has the
Four Point Property. Thus, Corollary 2.4 shows a rectilinear drawing of K5 is convex.
On the other hand, Corollary 2.5 shows every non-convex drawing of Kn contains a non-
convex drawing of K5. Such a drawing is either K̃35 or K̃55.
Theorem 2.6 has the following simple corollary. (To help with phrasing, we refer to an
isomorph J of Kr to mean J is a specific one of the complete subgraphs of Kn having r
vertices.)
422 Ars Math. Contemp. 22 (2022) #P3.04 / 415–441
Corollary 2.7. Let D be a drawing of Kn. Suppose, for every isomorph J of K5, there
is, for some m ≤ 12, an isomorph L of Km containing J such that D[L] is an optimal
drawing of Km. Then D is convex.
Proof. Aichholzer [2] has verified that D[L] is convex, so D[J ] is a convex K5.
As illustration of the utility of Corollary 2.7, Ábrego et al. [1] exhibit, for odd n, a
drawing Nn,n,1 of K2n+1 having H(2n + 1) crossings, with every edge involved in at
least one crossing. We argue here that it is convex; furthermore, we believe that analogous
arguments apply to any of the known examples of drawings of Kn having H(n) crossings.
We start with the tin can drawing TC2n of K2n (TC8 is illustrated in Figure 1; in
this figure, the labelling described below is counterclockwise). This has concentric regular
n-gons, the outside one labelled u0, . . . , un−1 and the inside one labelled v0, . . . , vn−1.
For each i, ui and vi are roughly “antipodal”, so ui is closest to vi+⌊n/2⌋. We use the
non-self-crossing perfect matching consisting of the edges uivi+⌊n/2⌋ to work from (the
indices are always read modulo n). We join ui to vi+⌊n/2⌋+1, vi+⌊n/2⌋+2, . . . , vi−1 in one
direction from uivi+⌊n/2⌋ around the cylinder, and to vi+⌊n/2⌋−1, vi+⌊n/2⌋−2, . . . , vi+1, vi
in the other direction. The two sides are as equal as possible. Thus, the rotation at ui is
ui+1, ui+2, . . . , ui−1, vi, vi+1, . . . vi−1 .
We mentioned in the introduction that this is homeomorphic to a spherical drawing and
therefore convex. However, an earlier referee requested more detail for proofs of convexity,
so we provide the argument for TC2n and adapt it to prove the convexity of Nn,n,1.
To argue that TC2n is convex, let V be any set of five vertices in TC2n. These vertices
are covered by at most five edges of the matching M consisting of the antipodal edges uivi.
Add one or two more edges from M as needed to get up to five edges, and use the induced
drawing on these 10 vertices. This is TC10 because the induced rotations of the edges at
the 10 vertices satisfy the rotation described in the preceding paragraph. This is an optimal
drawing of K10, so Corollary 2.7 shows TC2n is convex.
To get the Ábrego et al. drawing Nn,n,1, add a vertex w to TC2n near the centre of the
concentric n-gons. This is joined to the 2n vertices by straight line segments, creating the
drawing TC+2n. The drawing Nn,n,1 is completed by redrawing the edges uivi of TC
+
2n as
illustrated in Figure 2; these are the green edges in their Figure 4. Their labelling and ours
coincide except that our indices on ui and vi are one less than theirs.
v0
v1
v2
v3
u0u1
u2 u3
w
Figure 2: N4,4,1.
Many of the K5’s in Nn,n,1 are exactly the same in TC+2n; all of these are convex.
This is even true for a K5 that contains just one green edge, say u0v0 and no vertex from
A. Arroyo et al.: Convex drawings of the complete graph: topology meets geometry 423
u1, u2, . . . , u⌊n/2⌋: the drawings of this K5 in Nn,n,1 and TC
+
2n are isomorphic. This is
helpful in dealing with the case there is only one green edge: we may assume such a K5
also has at least one of the uj listed in the preceding paragraph.
In case of a single green edge and the green edge is not involved in a crossing, we make
use of the following simple observation.
Observation 2.8. Let D be a convex drawing of K5 and let D′ be another (simple) drawing
of K5 obtained from D by rerouting a single edge e. If e is uncrossed in D′, then D′ is
convex.
Proof. If e is uncrossed in D, then D and D′ are homeomorphic drawings and the result is
trivial. Otherwise, the edge e is crossed in D, so cr(D′) < cr(D). These numbers are both
odd (Kleitman [17]) and at most 5. Since the spherical drawing of K5 with one crossing is
convex, the only non-trivial case is D is the natural drawing of K5.
In this case, e is one of the crossed edges; these are all the same. Only the face of G−e
bounded by the 5-cycle is incident with both ends of e. Thus, D′ is a homeomorph of the
rectilinear drawing whose outer face is bounded by a 4-cycle.
Next, consider a K5 having a unique green edge crossed in the K5. With 4 vertices de-
termined, if the fifth vertex is w or vj , then the K5 is optimal and hence convex. Otherwise,
there are four “u vertices” and it is easy to see that it is one of the rectilinear K5’s.
Lastly, there may be two green edges. Except for one exceptional case when n is even,
these edges must cross. Moreover, they give us four of the vertices of the K5 and it is not
difficult to check the different possible locations for the remaining vertex.
The following definition and corollary to Lemma 2.3 will be used in the next section
for the structural description of a convex drawing.
Definition 2.9. Let D be a drawing of Kn, let u be a vertex of Kn, and let J be a complete
subgraph of Kn − u. If D[J ] is natural and D[u] is in the face of D[J ] bounded by a
|V (J)|-cycle, then u is planarly joined to J if no edge from D[u] to D[J ] crosses any edge
of J .
Corollary 2.10. Let D be a convex drawing of K5 with vertices u, v such that D − {u, v}
is the 3-cycle T . If D[u] and D[v] are in the same face of D[T ] and u and v are both
planarly joined to T , then D[uv] is in the same face of D[T ] as D[u] and D[v].
A further perusal of the five drawings of K5 shows that the following further refinement
of forbidden substructures is possible. This configuration was mentioned at Crossing Num-
ber Workshop 2015 (Rio de Janeiro) in the context of being one forbidden configuration
for a drawing of an arbitrary graph to be pseudolinear. The proof is quite straightforward.
Lemma 2.11. Let D be a drawing of Kn. Then D is convex if and only if, for every path
P of length 4, D[P ] is not isomorphic to P̃4.
We will use the following observation in Section 5. Its proof, left to the reader, is a
good exercise in using the fact that no two closed edges can have two points in common.
Observation 2.12. Let D be a drawing of K5 in which some 3-cycle is crossed three times
by a single edge. Then D is K̃55 (as in Figure 1).
424 Ars Math. Contemp. 22 (2022) #P3.04 / 415–441
Figure 3: The drawing P̃4.
3 Convexity and natural drawings of Kn
We recall from Section 1 that a natural drawing of Kn is a drawing in which an n-cycle
bounds a face Γ. It is easy to see that, in any natural drawing of Kn, every 3-cycle T has a
side ∆T that is disjoint from Γ and there is no vertex of Kn in the interior of ∆T . Thus, Γ
and the ∆T show that a natural drawing of Kn is f-convex.
In this section, we show that if D is a convex drawing of Kn with the maximum number(
n
4
)
of crossings, then D is a natural drawing of Kn. This leads us to a structure theorem
for convex drawings of Kn whose central piece is, for some r ≥ 4, a natural drawing of
Kr. It also leads to the Erdős-Szekeres Theorem for convex drawings: for every r ≥ 5, if
n is sufficiently large, then every convex drawing of Kn contains a natural Kr.
Lemma 3.1. Let D be a drawing of Kn. Then D is a convex drawing of Kn with
(
n
4
)
crossings if and only if D is a natural drawing of Kn.
Proof. One direction is trivial; for the other, let D be a convex drawing of Kn with
(
n
4
)
crossings; all K4’s are crossing in D. We proceed by induction on n, the cases n ≤ 5 being
trivial. Let v0 be any vertex of Kn and let H be the Hamilton cycle of Kn − v0 such that
D[H] bounds a face F of D[Kn − v0].
Claim 3.2. If T is a 3-cycle in Kn, then one side of D[T ] has no vertex drawn in its interior.
Proof. Otherwise, the convex side ∆ of D[T ] has a vertex v with D[v] in the interior of ∆.
This yields the contradiction that D[T + v] is a non-crossing K4.
In particular, Claim 3.2 applied to all the 3-cycles in Kn − v0 shows v0 is in F .
Claim 3.3. Suppose x, y, z are distinct vertices of Kn − v0 such that yz ∈ E(H) and the
edge D[v0x] crosses the edge D[yz]. Then, for any vertex w of Kn − {v0, y, z}:
(a) D[v0w] crosses D[yz]; and
(b) yz is the only edge of H crossed by v0w.
Proof. For (a), if D[v0w] does not cross D[yz], then the 3-cycle D[v0wx] has D[y] and
D[z] on different sides, contradicting Claim 3.2.
For (b), in traversing v0w, let ab be the first edge of Kn − v0 such that D[v0w] crosses
D[ab]. Then ab is in H and v0w does not cross either aw or bw. That is, the portion of
D[v0w] from the crossing with D[ab] to D[w] is contained inside the 3-cycle D[abw].
A. Arroyo et al.: Convex drawings of the complete graph: topology meets geometry 425
Since every K4 is crossing, v0 is not planarly joined to D − v0. Therefore, there is
an edge v0x that crosses an edge yz of H . Claim 3.3(a) shows that, for any vertex w of
D − {v0, y, z}, v0w also crosses yz.
If v0y crosses some edge y′z′ of H , then Claim 3.3(a) shows that, for any vertex w
of D − {v0, y′, z′} also crosses y′z′. Note that y /∈ {y′, z′}. Since n ≥ 6, there is a
w0 different from all of v0, y, z, y′, z′, so v0w0 crosses both yz and y′z′, contradicting
Claim 3.3(b).
Replacing the edge yz of H with the path y, v0, z yields a Hamilton cycle in D that
proves D is a natural drawing.
The convex version of the Erdős–Szekeres Theorem is an immediate consequence of the
main result of Pach et al. [23]. Their bounds are better than those coming from our Ramsey
argument. Lemma 3.1 and Ramsey’s Theorem also easily prove the convex version of the
Erdős–Szekeres Theorem. We suppose r ≥ 5 is an integer and choose n large enough so
that some subset of V (Kn) of size r is such that every K4 in the Kr is crossing. (For r ≥ 5,
they cannot all be non-crossing.) If the drawing D of Kn is convex, Lemma 3.1 implies
D[Kr] is natural. We state the theorem here for reference.
Theorem 3.4. Let r ≥ 5 be an integer. Then there is an integer N = N(r) such that, if
n ≥ N and D is a convex drawing of Kn, then there is a subgraph J of Kn isomorphic to
Kr such that D[J ] is a natural Kr. 2
The remainder of this section is devoted to a structure theorem for convex drawings.
Let D be a convex drawing of Kn and, for some r ≥ 4, let J be a Kr in Kn such that D[J ]
is natural. We set CJ to be the facial r-cycle in D[J ]. We refer to the face of D[J ] bounded
by CJ as the outside of J and the other side of CJ as the inside of J .
The proof uses the following elementary observations that are somewhat interesting
and otherwise useful in their own right.
Lemma 3.5. Let D be a convex drawing of Kn and, for some r ≥ 4, let J be a Kr such
that D[J ] is natural, with facial r-cycle CJ .
(a) If u is inside J , then, for each v ∈ V (J), D[uv] is inside J .
(b) If u and v are both inside J , then D[uv] is inside J .
(c) If u and v are both outside J and planarly joined to J , then D[uv] is contained in
the outside of J .
(d) Let u be outside of J and suppose there is a vertex v of J such that D[uv] crosses
CJ . Then D[uv] crosses CJ exactly once.
(e) Suppose u is outside of J but, for vertices v and w of J , D[uv] and D[uw] both cross
CJ . Let e and f be the edges of CJ crossed by D[uv] and D[uw]. Then v and w are
in the same component of CJ − {e, f}.
(f) Suppose u is outside of J , v is a vertex of J , and D[uv] crosses CJ on the edge ab.
Then D[ua] and D[ub] are contained in the outside of J .
Proof. We start with (a). If we consider the edges of J incident with v, they partition the
inside of J into discs bounded by 3-cycles. As |V (J)| ≥ 4, the disc containing u is the
convex side of its bounding 3-cycle. Thus, D[uv] is inside this disc and so is inside J .
426 Ars Math. Contemp. 22 (2022) #P3.04 / 415–441
For (b), we present an argument suggested by Kasper Szabo Lyngsie that simplifies our
original. There is an edge xy in CJ such that v is in the side ∆ of D[uxy] that has no
vertices of J − {x, y}. If there is an edge of J incident with either x or y that crosses the
3-cycle uxy, then v is in the crossing side of a natural K4 containing u, x, and y. In this
case, ∆ is the convex side of uxy, so D[uv] is inside ∆.
In the other case, let x′ and y′ be the neighbours of x and y, respectively, in CJ − xy.
Then ∆ is contained in the convex side ∆′ of the 3-cycle x′xy, and again D[uv] is contained
in ∆′ and consequently inside J . (We remark that, in fact, D[uv] is contained inside ∆, but
vx or vy might cross uxy, so ∆ need not be the convex side of uxy.)
Moving on to (c), let x, y, z be any three vertices of J and let L be the K5 induced by
u, v, x, y, z. Then D[L] is a convex drawing of K5. Let T be the 3-cycle (x, y, z). The
assumption that u and v are planarly joined to T in D shows that the side ∆T of T that
contains u and v satisfies the Four Point Property in D[L].
Corollary 2.10 implies that D[uv] is contained in ∆T . This is true for every three
vertices of J , so D[uv] is contained in the intersection of all the ∆T ’s; this is precisely the
closure of the face of D[J ] containing D[u] and D[v], as required.
In the proof of (d), we suppose the first crossing of uv is with the edge xy of CJ . The
3-cycle xyv is inside J and, by the definition of drawing, uv cannot cross xyv a second
time.
Turning to (e), we suppose that v and w are in different components of CJ −{e, f} and
that uv crosses e, while uw crosses f . Let x be the end of e in the component of CJ−{e, f}
containing v and let y be the end of f in the component of CJ − {e, f} containing w. By
the definition of drawing, x ̸= v and y ̸= w.
The edge xw crosses uv and the edge yv crosses uw. Moreover, x and y are on different
sides of the 3-cycle uvw, so uvw has no convex side, a contradiction.
For (f), it suffices by symmetry to show D[ua] is outside J . In the alternative, ua
crosses CJ . Since it cannot cross uv by goodness, it must cross the av-subpath of CJ −ab.
But now ua and uv violate (e).
We now turn to the basic ingredient in the structure theorem. Let D be a convex drawing
of Kn and, for some r ≥ 4, let J be a Kr such that D[J ] is natural. The J-induced drawing
J̄ consists of the subdrawing induced by D[J ] and all vertices inside of J . The following
is the main point in the proof of the structure theorem.
Lemma 3.6. Let D be a convex drawing of Kn and, for some r ≥ 4, let J be a Kr such
that D[J ] is natural. If there is a vertex u outside J and a vertex v of J such that D[uv]
crosses CJ , then there is, for some s ≥ 4, a Ks-subgraph J ′ including u such that D[J ′]
is natural and J̄ ⊂ J̄ ′.
Proof. Let ab be the edge of CJ crossed by uv. Lemma 3.5(f) implies that D[ua] and
D[ub] are contained in the outside of J . It follows that, in the av-subpath of CJ −ab, there
is a vertex wa nearest v such that D[uwa] is contained in the outside of J . Likewise, there
is a nearest such vertex wb in the bv-subpath.
For any internal vertex x in the wawb-subpath P of CJ − ab, the edge ux must cross
CJ ; Lemma 3.5(d) and (f) imply ux must cross wawb. It follows that ux does not cross P .
Thus, the cycle consisting of u, together with P , makes the facial cycle for a natural Ks
(s = 1 + |V (P )| ≥ 4) and all the points of J̄ are on or inside this Ks.
Our structure theorem is an immediate consequence of Lemmas 3.5(c) and 3.6.
A. Arroyo et al.: Convex drawings of the complete graph: topology meets geometry 427
Theorem 3.7 (Structure Theorem). Let n ≥ 5 and let D be a convex drawing of Kn. Then,
for some r ≥ 4, there is a Kr-subgraph J such that D[J ] is natural, every vertex outside
of J is planarly joined to J , and any two vertices outside J are joined outside J . 2
As a consequence of the Structure Theorem, we have the following straightforward,
though not trivial, observation. As it is not of immediate interest, we give only a proof
sketch.
Theorem 3.8. Let n ≥ 5 and let D be a drawing of Kn. Suppose that, for every subgraph
J of Kn that is isomorphic to a K4 and D[J ] has a crossing, there are no vertices of Kn
inside D[J ]. Then D[Kn] is convex and either:
1. a natural Kn; or
2. a natural Kn−1 with one vertex outside that is planarly joined to the Kn−1; or
3. the unique drawing of K6 with three crossings.
Proof sketch. The hypothesis on the crossing K4’s implies the drawing is convex: in both
K̃35 and K̃55, there is a crossing K4 with a vertex inside the K4.
Apply the Structure Theorem 3.7 to D to get a subgraph J of Kn such that D[J ] is a
natural Kr, with r ≥ 4 and every other vertex of Kn is either inside D[J ] or is outside J
and planarly joined to J .
Any vertex inside D[J ] is in a face that is incident with a crossing of some crossing K4
involving four vertices in J . Since this is forbidden, there is no vertex inside D[J ].
If there are three vertices of Kn outside D[J ], then there is a crossing K4 with a vertex
inside.
If there are two vertices u, v of Kn outside D[J ] and some edge from u to J crosses
two edges from v to J , then there is a crossing K4 with a vertex inside. In particular, if
r ≥ 5, then there is at most one vertex outside J .
The remaining case is r = 4 and no uJ-edge crosses two vJ-edges and no vJ-edge
crosses two uJ-edges. This is the unique drawing of K6 with three crossings.
In general, if, in a convex drawing of Kn, we bound by a non-negative integer p the
number of vertices allowed inside any natural K4, there is a theorem in the spirit of Theo-
rem 3.8. There are more special cases with n small, but if n is large enough (on the order
of 3p), the structure is: a natural Kr, with r at least roughly p/3, and at most one of the
remaining points is outside the natural Kr.
4 h-convex drawings
In this section, we investigate h-convex drawings. Our main result is a characterization of
h-convex drawings; this immediately yields a polynomial time algorithm for determining
if a drawing is h-convex.
Consider the drawing K116 . It is convex, but not h-convex. To see that it is convex, it
suffices to check the six K5’s and observe that none of them is either K̃35 or K̃55. To see
that it is not h-convex, consider the dashed K4 (including the thick edge) highlighted in
Figure 1. For this K4, either of the 3-cycles T containing the thick edge has its bounded (in
the figure) side convex, while its unbounded side is not convex. A similar statement holds
428 Ars Math. Contemp. 22 (2022) #P3.04 / 415–441
for the a 3-cycle in the dotted K4 that contains the thick edge (bounded and unbounded
sides reversing roles). These 3-cycles show that D is not h-convex.
Definition 4.1. Let D be a drawing of Kn and let J and J ′ be distinct K4’s in D such that
both D[J ] and D[J ′] are crossing K4’s. For 3-cycles T and T ′ in J and J ′, respectively,
let ∆T and ∆T ′ be the sides of T and T ′, respectively, not containing the fourth vertex of
J and J ′, respectively. Then J and J ′ are inverted K4’s in D if there are 3-cycles T in J
and T ′ in J ′ such that D[T ] ⊆ ∆T ′ but ∆T ̸⊆ ∆T ′ .
The following observation is immediate from the definition.
Observation 4.2. Let J and J ′ be inverted K4’s in a drawing D of Kn and let T and T ′ be
3-cycles in J and J ′, respectively. Let ∆T and ∆T ′ be the sides of T and T ′, respectively,
not containing the fourth vertex of J and J ′, respectively. If D[T ] ⊆ ∆T ′ but ∆T ̸⊆ ∆T ′ ,
then D[T ′] ⊆ ∆T but ∆T ′ ̸⊆ ∆T .
We are ready for our first characterization of h-convex drawings.
Lemma 4.3. Let D be a convex drawing of Kn. Then D is h-convex if and only if there
are no inverted K4’s.
Proof. It is clear that if D is h-convex, then there are no inverted K4’s.
For the converse, we shall inductively obtain a list C of convex sides, one for each 3-
cycle of Kn. Along the way, the list C will have convex sides for some, but not all, of the
3-cycles of Kn. Such a partial list is hereditary if, for any 3-cycles T and T ′ having convex
sides ∆T and ∆T ′ , respectively, in C, if D[T ] ⊆ ∆T ′ , then ∆T ⊆ ∆T ′ .
Our initial list C0 consists of the convex sides for every 3-cycle that is in a crossing K4.
The assumption that there are no inverted K4’s immediately implies C0 is hereditary.
Let T1, . . . , Tr be the 3-cycles in Kn such that, for i = 1, 2, . . . , r, Ti is not in any
crossing K4. For j ≥ 1, suppose that Cj−1 is a hereditary list of convex sides that includes
C0 and a convex side for each of T1, . . . , Tj−1.
If there is a convex side ∆T ∈ Cj−1 such that D[Tj ] ⊆ ∆T , then we choose ∆Tj so
that ∆Tj ⊆ ∆T . Otherwise, we choose ∆Tj arbitrarily from the two sides of D[Tj ]. Set
Cj = Cj−1 ∪ {∆Tj}.
We show that Cj is hereditary. If not, then, since Cj−1 is hereditary, there is a 3-
cycle T with a convex side ∆T ∈ Cj−1 such that either D[T ] ⊆ ∆Tj and ∆T ̸⊆ ∆Tj or
D[Tj ] ⊆ ∆T and ∆Tj ̸⊆ ∆T . The second case implies that D[T ] ⊆ ∆Tj and ∆T ̸⊆ ∆Tj ,
which is the first case.
Thus, in both cases, we have that D[T ] ⊆ ∆Tj and ∆T ̸⊆ ∆Tj . By the choice of ∆Tj ,
there is a second already considered triangle T ′ such that D[T ′] is contained in the other
side ∆Tj of D[Tj ] but ∆T ′ ̸⊆ ∆Tj .
Evidently, D[T ] ⊆ ∆T ′ and ∆T ̸⊆ ∆T ′ , yielding the contradiction that Cj−1 is not
hereditary.
Lemma 4.3 yields an O(n8) algorithm for determining whether a drawing is h-convex.
Also, a similar argument proves the following analogous fact for f-convexity. This is es-
sentially the characterization of pseudolinearity due to Aichholzer et al. [4].
Theorem 4.4 ([4]). Let D be a drawing of Kn. Then D is f-convex if and only if there
is a face Γ such that, for every isomorph J of K4 for which D[J ] is a crossing K4, Γ is
contained in the face of D[J ] bounded by the 4-cycle. 2
A. Arroyo et al.: Convex drawings of the complete graph: topology meets geometry 429
There is a colourful way to understand this theorem. For each isomorph J of K4 for
which D[J ] is a crossing K4, let CJ be the 4-cycle in J that bounds a face of D[J ]. Paint
the side of D[CJ ] that contains the crossing of D[J ]. If the whole sphere is painted, then
D is not f-convex. Otherwise, with respect to any face F of D[Kn] that is not painted, F
witnesses that D is f-convex.
Our next result gives a surprising characterization of h-convex drawings of Kn by a
single additional forbidden configuration.
Theorem 4.5. Let D be a convex drawing of Kn. Then D is h-convex if and only if, for
each isomorph J of K6 in Kn, D[J ] is not isomorphic to K116 .
Proof. Since h-convexity is evidently inherited by induced subgraphs, no h-convex drawing
of Kn can contain K116 . Conversely, suppose D is not h-convex; we show D contains K116 .
By Lemma 4.3, there exist isomorphs J1 and J2 of K4 that are inverted in D. For
i = 1, 2, let Ti be a 3-cycle in Ji with convex side ∆Ti such that D[T1] ⊆ ∆T2 and
∆T1 ̸⊆ ∆T2 .
Let w be the vertex of J1 not in T1; D[w] is separated from D[T2] by D[T1]. Let x be
the vertex of T1 such that D[wx] crosses D[T1]. Complete D[wx] to a simple closed curve
γ by adding a segment on the non-convex side of D[T1] joining D[w] and D[x]. Clearly
γ separates the two vertices of T1 − x. Moreover, D[T1] and, therefore, D[w] as well, are
all contained in ∆2. Convexity implies D[J1] ⊆ ∆2. Thus, γ also separates one of the
vertices of T1 − x from D[T2]; let z be the one separated from T2 by γ and let y be the
other.
Since D[T1] ⊆ ∆T2 , D[T2 + z] is a non-crossing K4. If any of the edges from z to T2
crosses T1, then we have proof that the side ∆T1 of D[T1] is not convex, a contradiction.
Therefore, D[T1] is contained in a face Γ of D[T2 + z] that is incident with z. It follows
that w is also in Γ.
Let a be the vertex of T2 not incident with Γ. The edge wx has both its ends in Γ. Since
γ separates z from T2, γ must cross za and, therefore, is not contained in Γ. It follows that
Γ is not the convex side of the 3-cycle T3 that bounds Γ.
Evidently, D[T3] ⊆ ∆T2 and ∆T3 ̸⊆ ∆T2 . Corollary 2.5(b) implies that there is a vertex
v3 such that v3 ∈ Γ and D[T3 + v3] is a crossing K4. Because T2 is in the isomorph J2 of
K4, there is a vertex v2 in J2 that is not in T2. Since D[J2] is a crossing K4, D[v2] /∈ ∆T2 .
We now consider the isomorph of K6 consisting of (T2 ∪ T3) + {v2, v3}. Because
D[(T2 ∪ T3) + v2] is contained in ∆T3 , no edge from v2 to a vertex in T2 ∪ T3 can cross
D[T3]. In particular, (recall that a is the vertex of T2 not in T3) D[v2a] does not cross
D[T2 ∪ T3]. Let b be the vertex of T2 such that D[v2b] crosses D[T2] and let c be the third
vertex of T2.
Symmetrically, z is the vertex of T3 that is not in T2, so D[v3z] does not cross
D[T2∪T3]. As both D[z] and D[v2] are in ∆T3 , the edge zv2 cannot cross T3. Since D[v2b]
crosses D[ac] but not D[T3], it must also cross D[az]. It follows that D[v2z] crosses only
D[ac].
Let b′ be the one of b and c such that D[v3b′] crosses D[T3] and let c′ be the other.
There are two cases to consider: b = b′ and c = c′; or b = c′ and c = b′. Note that, in
each case, convexity and the definitions of b and b′ determine the routings of all the edges
except v2v3 and v3a.
Let T4 be the 3-cycle induced by b, v2 and c. Since D[ac] crosses D[v2b], the convex
side of D[T4] is the side that contains v3. Thus, D[v2v3] must be contained in this side of
430 Ars Math. Contemp. 22 (2022) #P3.04 / 415–441
D[T4]. In the case b = b′, D[v3b] crosses D[cz], D[zv2], and D[az]. Thus, the only routing
for D[v2v3] is across D[ac] and D[zc]. In the case b = c′, the only routing for D[v2v3]
is across D[ac], D[az], and D[zb]. In both cases there is only one routing available for
D[v3a].
To see in each case that these drawings are both K116 , focus on the face-bounding 4-
cycles induced by b, a′, v3, c and b, a, v2, c.
The previous results were about a drawing being h-convex. The following result char-
acterizes when a collection of sides, one for each 3-cycle of Kn, is a set of h-convex sides.
Its proof is similar to the proof of Theorem 4.5.
Lemma 4.6. Let D be a drawing of Kn and, for each 3-cycle T of Kn, let ∆T be one
of the closed discs bounded by D[T ]. Let C be the set of all these ∆T . Then C is a set of
h-convex sides if and only if both of the following hold:
(1) each ∆T has the Four Point Property; and
(2) for each non-crossing K4, at least three of the four (closed) faces of the non-crossing
K4 are in C.
Proof. Let C be a set of h-convex sides. They are also convex sides, so obviously satisfy
(1). If J is a non-crossing K4 and T is a 3-cycle in J , then the side ∆T of T that is in C is
either empty or contains J ; in the latter case all the empty sides of the other 3-cycles in J
are in C by heredity. Thus, at most one of the four 3-cycles in J has non-empty side in C,
which is (2).
Conversely, let C satisfy (1) and (2). Then C is a set of convex sides by Corollary 2.4.
Now let T, T ′ be 3-cycles such that D[T ′] ⊆ ∆T .
Suppose there is a t ∈ T \ T ′ such that T ′ + t is a crossing K4. Then the convex
side ∆T ′ of T ′ is the side that does not contain t. Because D[T ′] ⊆ ∆(T ), this implies
∆T ′ ⊆ ∆(T ), as required. Thus, we may assume T ′ is not crossed in D.
Let v be any vertex of T \T ′. Then D[T + v] is a non-crossing K4. By (2), each of the
three faces of D[T + v] incident with v is the side of the bounding 3-cycle that is in C.
We proceed by induction on the number of vertices of T ′ that are not in T ; this number
being at least 1. Because T ′ is not crossed in D, the remaining vertices of T ′ are either in
or incident with the same face F of T + v. Let T ′′ be the 3-cycle bounding F .
The base of the induction is that T ′− v ⊆ T . In this case, T ′ = T ′′ and ∆T ′ is F ; thus,
∆T ′ ⊆ ∆T , as required.
For the induction step, suppose |T ′ \ T | ≥ 2. We have already seen that the 3-cycle
T ′′ bounding F has ∆T ′′ = F . For this induction step, there is a vertex w of T ′ in the
interior of F , so the side of ∆T ′′ in C is not a face of T ′′ + w. Now T ′ ⊆ ∆T ′′ and
|T ′ \ T ′′| < |T ′ \ T |. By induction, ∆T ′ ⊆ ∆T ′′ . Since ∆T ′′ ⊆ ∆T , we have shown C is
the set of h-convex sides, as required.
Although Lemma 4.6 shows that h-convexity is determined by considering all sets of
four points, it is not evident that there is an O(n4) algorithm to test whether a drawing is
h-convex. Theorem 4.5 makes it clear that there is an O(n6) algorithm to determine if a
drawing of Kn is h-convex.
It is O(n4) to check that a drawing is convex. To see this, we use the Four Point Property
(Corollary 2.4): for each 3-cycle T and each vertex v, we determine which side of T v is
on and whether T + v is a planar K4.
A. Arroyo et al.: Convex drawings of the complete graph: topology meets geometry 431
We conclude this section with an observation related to the Structure Theorem 3.7.
Lemma 4.7. Let D be an h-convex drawing of Kn consisting of a natural Kr (with r ≥ 4)
and all other points inside the natural Kr. Then D is f-convex.
Proof. Any triangle of the Kr has its convex side determined; it is the side that avoids the
face F bounded by the cycle Cr. Let T be any other triangle. If an edge of Kr crosses T ,
then the convex side of T must also avoid F . If this does not happen, then T is inside a
triangle of Kr and the result follows from the hereditary property.
5 Suboptimal drawings of Kn having either K̃35 or K̃
5
5
In this section, we prove that a broad class of “locally determined” drawings of Kn are
suboptimal. This is the first theorem of its type. The theorem requires the presence of
either K̃35 or K̃55 in the drawing, but, for at least one such K5, the occurrence is restricted.
This might be a first step towards showing that all optimal drawings of Kn are convex.
This line of research was stimulated by Tilo Wiedera’s computation (personal commu-
nication) showing that any drawing of K9 that contains a K̃55 has at least 40 crossings. This
is in line with Aichholzer’s later computations (see the remark following the statement of
Theorem 5.1 below).
We also rethink the approach in [21] that cr(K9) = 36. This was done before convexity
became known to us. Using the fact that cr(K7) = 9, it is easy to see that cr(K9) ≥ 34. At
the end of this section, we show easily by hand that there is no non-convex drawing D of
K9 such that cr(D) = 34. Thus, to prove that cr(K9) = 36, it suffices to consider convex
drawings of K9.
A principal result of this work is the following, which shows that if D is a drawing of
Kn such that some non-convex K5 intersects every other non-convex K5 in at most two
vertices, then D is not optimal.
Theorem 5.1. Let D be a drawing of Kn such that there is an isomorph J of K5 with D[J ]
either K̃35 or K̃55. Suppose, for every isomorph H of K7 in Kn containing J , D[J ] is the
only non-convex K5 in D[H].
1. If J is K̃35, then there is a drawing D′ of Kn such that cr(D′) ≤ cr(D)− 2.
2. If J is K̃55, there is a drawing D′ of Kn such that cr(D′) ≤ cr(D)−4. If, in addition,
n is even, then cr(D′) ≤ cr(D)− 5.
We remark that the lower bounds 2, 4, and 5 for cr(D) − cr(D′) exhibited in Theo-
rem 5.1 are precisely the smallest differences found by Aichholzer (private communica-
tion) between any drawing, for n ≤ 12, of Kn that has either a K̃35 or a K̃55 and an optimal
drawing of Kn.
Proof of Theorem 5.1. We use the labelling of J as shown in Figure 4. We first deal with
the case J = K̃35.
(I) J = K̃35 .
Claim 5.2. There is no vertex of D[Kn] in the side of any of the 3-cycles D[stw], D[suv],
and D[tuv] that has no vertex of D[J ].
432 Ars Math. Contemp. 22 (2022) #P3.04 / 415–441
Figure 4: Labelled K̃35 and K̃55 for the proof of Theorem 5.1.
Proof of Claim 5.2. We start with D[stw]. Similar arguments apply to D[suv]. Finally,
symmetry shows that D[tuv] also does not have a vertex on the side empty in D[J ].
Suppose to the contrary that there is a vertex x of Kn such that D[x] is in the side of
D[stw] that is empty in D[J ]. By hypothesis, the K5 consisting of J −w plus x is convex
in D. Since D[x] is incident with a face of D[J − w] that is incident with the crossing of
D[J −w], Observation 2.1 and convexity imply D[xu] does not cross the 4-cycle D[stuv].
Likewise, the K5 consisting of J − v together with x is convex in D. Again, D[x] is in
a face of D[J − v] incident with a crossing, so D[xu] does not cross the 4-cycle D[wtus].
However, D[x] and D[u] are in different faces of D[stuv]∪D[wtus], so D[xu] must cross
at least one of the two 4-cycles.
The same deletions show that any vertex in the empty side of D[suv] cannot connect to
t.
There are two remaining regions of interest. Let × be the crossing of su with tv. Let
R1 be the region bounded by D[wt×sw] that does not contain D[u] and D[v]; R2 is the
region bounded by D[stuv] that does not contain D[w].
The following observations follow immediately from the convexity of the correspond-
ing K5 and the knowledge of the crossings.
(Obs. 1) If x ∈ R1, then J − w + x is convex, so the routings of x to s, t, u, v are
determined.
(Obs. 2) If y ∈ R2, then J−u+y and J − v + y are convex, so the routings of y to
J are determined.2
These observations yield all of (1) – (5) in the following assertions except for (3).
Claim 5.3. If D[x] ∈ R1 and D[y] ∈ R2, then:
(1) D[xu] crosses D[J ] only on D[tv] and D[xv] crosses D[J ] only on D[su];
(2) D[xs] and D[xt] do not cross D[J ];
(3) D[xw] either does not cross D[J ], or crosses D[st] and at least one of D[su] and
D[tv];
(4) D[ys], D[yt], D[yu], and D[yv] cross D[J ] at most in either D[uw] or D[vw] (or
both);
(5) D[yw] crosses only D[st].
A. Arroyo et al.: Convex drawings of the complete graph: topology meets geometry 433
Moreover, if zz′ is an edge of G with neither z nor z′ in J and T is one of the 3-cycles
stw, suv, and tuv, then either D[zz′] does not cross D[T ] or it crosses the one of D[st],
D[su], and D[tv] that is in D[T ].
Proof of Claim 5.3. For xw, the following argument is due to Matthew Sullivan, simplify-
ing our original. Consider the isomorph L of K2,4 with x and w on one side and s, t, u, v
on the other side. Then D[xw] does not cross (the planar drawing) D[L] and so is contained
in one of the four faces of D[L]. The face of D[L] bounded by swtx is disjoint from D[J ].
In each of the other three faces, D[xw] must cross D[st]. In two of these three faces, it also
crosses exactly one of D[su] and D[tv]. In the third, it crosses both D[su] and D[tv].
For the moreover, we consider the remaining three types of edges z1z2: D[z1] and
D[z2] can both be in R1; both in R2; or one in each. In all three cases for z1, z2 and
all three cases for the three-cycle T , D[z1] and D[z2] are on the same side of D[T ]. In
the event that D[z1] and D[z2] are both planarly joined to D[T ], Corollary 2.10 applies to
show D[z1z2] does not cross the 3-cycle.
In the remaining cases, we assume that D[z1] is not planarly joined to D[T ]. If T =
stw, then Claim 5.3(3) and (5) imply the only possible crossing with D[T ] is D[z1w]
crossing D[st]. As D[z1z2] has either 0 or 2 crossings with D[stw], but does not cross
D[z1w], the two crossings of D[z1z2] and D[T ] cannot be on D[ws] and D[wt]. For
T = suv and T = tuv, the edges z1u and z1v, respectively, produce analogous results.
We are now prepared for the final part of the proof. For i = 1, 2, let ri be the number
of vertices of D[Kn] that are in (the interior of) Ri.
Let D′ be the drawing of Kn obtained from D by making the following two changes:
(C1) reroute st alongside the path D[swt], so as to not cross D[wu] and D[wv]; and
(C2) reroute su alongside the path D[svu] so as to cross D[vw].
We first consider the changes in crossings arising from rerouting st. There are at least
2+ r2 crossing pairs of edges in D that do not cross in D′: two from D[st] crossing D[wv]
and D[wu], plus all the crossings of D[st] from those edges incident with D[w] that cross
D[st]. There are at least r2 of these latter crossings, as every vertex z such that D[z] is in
R2 has D[zw] crossing D[st].
On the other hand, there is a set of at most r1 crossing pairs in D′ that do not cross
in D. These arise from the the edges joining a vertex drawn in R1 to D[w]; these might
not intersect D[J ]. Those that do intersect D[J ] cross D[st] and, therefore, yield further
savings.
We show that every other edge z1z2 has no more crossings in D′ than it has in D.
Case 1: z1, say, is in J .
In this case, we use Claim 5.3. Items (1), (2), (4), and (5) show that no such edge has
more crossings in D′ than in D, except possibly xw.
If D[xw] does not cross D[J ], then D′[xw] also does not cross D′[J − st], as required.
If D[xw] crosses D[J ], then Claim 5.3(3) implies that D[xw] crosses D[st]. Thus, D[xw]
crosses both D[su] and one of D[sv] and D[tu]; in this case, the same is true of D′[xw] in
D′, as required.
Case 2: neither z1 nor z2 is in J .
434 Ars Math. Contemp. 22 (2022) #P3.04 / 415–441
If D[z1z2] crosses the 3-cycle D[stw], then the moreover part of Claim 5.3 shows it
crosses D[st]. Therefore, it crosses exactly one of D[sw] and D[wt], showing that D′[z1z2]
crosses D′[st] and the same one of D′[sw] and D′[wt]. That is, z1z2 crosses the same two
edges in both drawings, and we are done.
The net result is that the rerouting of st contributes at least 2 + (r2 − r1) to cr(D) −
cr(D′).
Now we turn our attention to the changes in crossings from rerouting su. The crossing
of D[su] with D[tv] is replaced by a crossing of D′[su] with D′[vw]. In addition, r2 edges
incident with v do not cross D[su], but cross D′[su], while r1 edges incident with v cross
D[su], but do not cross D′[su].
The only additional remark special to this rerouting is the observation that, for z in
R1, Claim 5.3 implies that if D[zw] crosses the 3-cycle D[suv], then zw crosses su. This
shows that D′[zw] also crosses D′[suv] twice, so there are no other “new” crossings.
Therefore, the su rerouting contributes at least r1 − r2 to the difference cr(D) −
cr(D′). Combining this with the contribution from st, we get that cr(D) − cr(D′) ≥
(2 + r2 − r1) + (r1 − r2) = 2. That is, cr(D′) ≤ cr(D)− 2, as required.
(II) J = K̃55 .
In this case, there is a homeomorphism Θ of the sphere to itself that is an involution
that restricts to J as, using the labelling in Figure 4: s ↔ w; t ↔ v; and u is fixed. This
will be helpful at several points in the following discussion. The outline of the argument is
the same as for K̃35, but there are some interesting differences.
Let R1 be the face of D[J ] incident with all three points in D[{s, t, u}] (the unbounded
face in the diagram) and let R2 be the face of D[J ] incident with all three points in
D[{u, v, w}] (note that R2 = Θ(R1)).
Claim 5.4. If z is a vertex of Kn not in J , then D[z] ∈ R1 ∪R2.
Proof of Claim 5.4. Suppose x is a vertex of Kn−V (J) such that D[x] is not in R1 ∪R2.
Suppose first that D[x] is in the region bounded by the 4-cycle D[wtsv].
The convexity of D[(J − s) + x] and of D[(J − w) + x] imply that D[xu] does not
cross the 4-cycles D[twvu] and D[stuv], respectively. However, D[x] is not in a face of
D[twvu] ∪D[stuv] incident with D[u], a contradiction.
The remaining possibility is that D[x] is in the face F that is both distinct from R1 and
either incident with D[ut] or, symmetrically, incident with uv; we assume the former. The
convexity of D[(J−t)+x] and D[(J−v)+x] show that D[xw] does not cross the 4-cycles
D[swvu] and D[swut], respectively. However, D[x] is not in a face of D[swvu]∪D[swut]
incident with D[w], a contradiction.
We next move to the routings of the edges from a vertex D[x] in R1 ∪R2 to D[J ].
Claim 5.5. If D[x] ∈ R1, then:
(1) D[xu] and D[xs] do not cross D[J ];
(2) D[xv] crosses D[J ] only on D[uw], and D[xw] crosses D[J ] only on D[sv] and
D[tv]; and
A. Arroyo et al.: Convex drawings of the complete graph: topology meets geometry 435
(3) D[xt] either does not cross D[J ] or it crosses D[J ] precisely on D[sv], D[sw], and
D[su].
Furthermore, if D[x], D[x′] ∈ R1, then D[xx′] ⊆ R1.
Proof of Claim 5.5. The convexity of D[(J−t)+x] and D[(J−u)+x] shows D[xs] does
not cross the 4-cycles D[suvw] and D[stwv], respectively. The convexity of J − s + x
shows xu does not cross J − s. Also xu does not cross su by simplicity, so xu does not
cross any edge incident with s.
The convexity of J − t+ x shows xv, respectively xw, crosses J − t only uw, respec-
tively tv. Also xv, respectively xw, does not cross vt, respectively wt, by simplicity, so
xv, respectively xw, does not cross any edge incident with t.
The convexity of D[(J − s) + x] determines the routing of D[xt], except with respect
to D[s], leaving the two options described.
For the furthermore conclusion, D[x] and D[x′] are planarly joined to the 3-cycle
D[svu]. Corollary 2.10 shows that D[xx′] is disjoint from D[svu]. In the same way,
D[xx′] is disjoint from D[swu], and D[tvu]. Thus, D[xx′] can only cross D[J ] on D[st].
However, letting × denote the crossing of D[su] with D[tv], D[xx′] must cross the 3-cycle
D[st×] an even number of times and it can only cross it on D[st], which is impossible.
The homeomorphism Θ implies a completely symmetric statement when x ∈ R2. We
provide it here for ease of reference.
Claim 5.6. If D[x] ∈ R2, then, in D[J + x]:
(1) D[xu] and D[xw] do not cross D[J ];
(2) D[xt] crosses D[J ] only on D[us], and D[xs] crosses D[J ] only on D[tw] and
D[tv]; and
(3) D[xv] either does not cross D[J ] or it crosses D[J ] precisely on D[wt], D[ws], and
D[wu].
Furthermore, if D[x], D[x′] ∈ R2, then D[xx′] ⊆ R2.
Using the homeomorphism Θ, we may choose the labelling of J so that the number r1
of vertices of D[Kn] drawn in R1 is at most the number r2 drawn in R2.
Our next claim was somewhat surprising to us in the strength of its conclusion.
Claim 5.7. If there is a vertex x of Kn − V (J) such that D[x] ∈ R1 and D[xt] crosses
D[sv], D[sw], and D[su], then there is a drawing D′ of Kn such that cr(D′) ≤ cr(D)− 4
and, if n is even, cr(D′) ≤ cr(D)− 5.
Proof of Claim 5.7. Choose such an x so that D[xt] crosses D[sv], D[sw], and D[su] and
such that, among all such x, the crossing of D[xt] with D[sv] is as close to D[s] on D[sv]
as possible. Let ∆ be the closed disc bounded by the 3-cycle D[sxt] that does not contain
the vertices D[{v, u, w}].
If there is a vertex y of Kn such that D[y] is in the interior of ∆, then D[y] is in the
face of D[J + x] contained in ∆ and incident with D[sx]. However, the convexity in D of
(J − {u,w}) + {x, y} implies D[yt] crosses D[sv] closer to s in D[sv] than D[xt] does,
contradicting the choice of x. Therefore, no vertex of D[Kn] is in ∆.
436 Ars Math. Contemp. 22 (2022) #P3.04 / 415–441
The drawing D′ is obtained from D by rerouting xt to go alongside the path D[xst],
on the side not in ∆. (That is, D[xt] is pushed to the other side of D[s].)
The hardest part of the analysis of the crossings of D′ compared to D is determining
what happens to an edge of D[Kn] that crosses D[st]. No edge of D[J ] crosses D[st].
Claims 5.5 and 5.6 imply that: no edge from a vertex in R1 ∪ R2 to a vertex in D[J ]
crosses D[st]; and no edge with both incident vertices in the same one of R1 or R2 crosses
D[st]. Thus, the only possible crossing of D[st] is by an edge D[yz], with D[y] ∈ R1 and
D[z] ∈ R2.
Because of the routing of D[sz], D[yz] cannot also cross D[xs]. Therefore, D[yz] also
crosses D[xt]. It follows that such an edge has the same number of crossings of xt in
both D and D′. Therefore, any edge that crosses D[xs] crosses D[xt] and so has the same
number of crossings with D[xt] and D′[xt].
The only changes then are in the number of crossings of D[xt] with edges incident
with D[s] and the number of crossings with D[J ]. There are 3 fewer of the latter. From
R1 to D[s], there are at most r1 − 1 crossings of D′[xt]. From R2 to D[s], we have lost
r2 crossings of D[xt]. Thus, D′ has at least (r2 − (r1 − 1)) + 3 = (r2 − r1) + 4 fewer
crossings than D. This proves the first conclusion.
Since n = 5 + r1 + r2, if n is even, then r1 ̸= r2 and, therefore, r2 − r1 ≥ 1. In this
case D′ has at least 5 fewer crossings, as claimed.
It follows from Claim 5.7 that we may assume that, for D[x] ∈ R1, D[xt] is disjoint
from D[J ]. Combining this with the other information from Claim 5.5, we may assume the
following property.
R1 Assumption: If D[x] ∈ R1, then D[x] is planarly joined to D[J − w].
Let D′ be obtained from D by rerouting D[tv] on the other side of the path D[tsv].
There are two claims that complete the proof of Theorem 5.1. The first, similar to Claim 5.7,
shows that there are at least 2 fewer crossings in D′ (3 if n is even). The second shows that
D′ satisfies the hypotheses of Theorem 5.1. Therefore, there is a third drawing D′′ with at
least two fewer crossings than D′, as required.
Claim 5.8. cr(D′) ≤ cr(D)− ((r2 − r1)− 2).
Proof of Claim 5.8. The proof is very similar to that of Claim 5.7. The main point is to
see that no edge e can have D[e] cross both D[ts] and D[sv]. Let x be incident with e.
If D[x] ∈ R2, then the routing of D[xs] is known; since D[xs] does not cross D[e], the
crossings of D[e] with D[ts] and D[sv] are joined by an arc of D[e] outside D[stuv]. But
then D[e] has both ends in J ∪R2. This contradicts Claim 5.6, so D[x] /∈ R2.
Likewise, Claim 5.5 shows D[x] /∈ R1 and clearly e is not in J . Therefore, there is no
such e.
It is now easy to see that there are (r2 − r1) + 2 fewer crossings of D′[tv] with edges
incident with s than there are of D[tv]. All other crossings of D′[tv] pair off with crossings
of D[tv].
Finally, we show that the drawing D′ satisfies the hypotheses of Theorem 5.1. It is
routine to verify that D′[J ] is K̃35. Now let N be a K5 in Kn such that N ∩ J has 3 or 4
vertices.
If any of s, t, v is not in N , then D′[N ] is homeomorphic to D[N ] and so is convex.
Thus, we may assume s, t, v are all in N .
A. Arroyo et al.: Convex drawings of the complete graph: topology meets geometry 437
Case 1: N ∩ J has four vertices.
In this case, there is a vertex x not in J such that N is either (J−w)+x or (J−u)+x.
If D[x] is in R1, then the routings are determined and we can see by inspection that D′[N ]
is, respectively, the K5 with 1 crossing or the convex K5 with 3 crossings.
If D[x] ∈ R2, then Claim 5.6 shows that D′[(J −w) + x] is the K5 with one crossing.
However, Claim 5.6(3) shows D′[(J − u) + x] has two possibilities for xv, depending on
which way around w it goes. If it crosses both wt and ws, then the drawing is the natural
one. The alternative is to reroute it to not cross D′[N − x]. In this case, Observation 2.8
shows D′[N ] is convex.
Case 2: N ∩ J has 3 vertices.
In this case N = (J − {u,w}) + {x, y}. Since D[x], D[y] ∈ R1 ∪ R2, they are both
on the same side of D[stv]. The routings from either to D[J ] are determined by Claims 5.5
and 5.6 and the assumption following the proof of Claim 5.7. Only when D[x] and D[y]
are in different ones of R1 and R2 is it possible that D[xy] crosses D[stv].
We consider the three possibilities for D[x] and D[y].
Subcase 2.1: D[x] ∈ R1 and D[y] ∈ R2.
All routings in D′[N ] are determined except for D[xy]. The 4-cycle D[xvyt] is un-
crossed in D[N−xy]. As D is a drawing, D[xy] does not cross D[xvyt]. Therefore, either
D[xvyt] or D[xvy] is a face of D′[N ], showing D′[N ] is convex.
Subcase 2.2: D[x] and D[y] are both in R2.
Since D[x] and D[y] are both planarly joined to D′[stv] and D[xy] does not cross
D′[stv], D′[stv] bounds a face of D′[N ]. Thus, D′[N ] is convex.
Subcase 2.3: D[x], D[y] are both in R1.
Suppose D′[N ] is not convex. Then Corollary 2.5(a) implies there is a 3-cycle T in
N such that the two vertices z, z′ of N not in T are in different faces of D′[T ] and both
D′[T + z] and D′[T + z′] are crossing K4’s.
Since both x and y are in the same face of D′[stv], T ̸= stv. If a ∈ {s, t, v}, then the
routings of the edges from x and y to stv show that the two vertices in {s, t, v} \ {a} are
on the same side of D′[xya], so xya ̸= T . The only remaining possibility is that T has x,
say, and two of s, t, v.
Claim 5.9. The 3-cycle D′[tvx] has no convex side.
Proof of Claim 5.9. In the alternative, T is either stx or svx. These two situations are
very similar, so we treat only stx, leaving the completely analogous argument for svx to
the reader. Our strategy is to show that assuming that stx has no convex side in D′ implies
that tvx has no convex side in D′ either.
The vertices v and y are on different sides of D′[stx] and D[vt] crosses D[sx], showing
that the side of D′[stx] containing D[v] is not convex. The edge D[sx] also shows that the
side of D′[tvx] containing D[s] is not convex.
Likewise, there is an edge e incident with y to one of s, t, and x such that D[e] crosses
D′[stx]. Notice that D[xy] does not cross D[xs] and D[xt] by definition of drawing and
D[xy] does not cross D[st] by the R1 Assumption. Therefore, D[xy] does not cross D[st]
and we conclude that D[xy] does not cross D[stx].
438 Ars Math. Contemp. 22 (2022) #P3.04 / 415–441
Next suppose that that D[yt] crosses D[xs]. The R1 Assumption shows that D[yt] does
not cross D[stv] and so D[yt] crosses D[vx]. Therefore, this side of D′[tvx] is also not
convex. Combined with the second paragraph of this proof, D′[tvx] is not convex.
In the final case, D[ys] crosses D[xt]. As we traverse D[ys] from D[y], there is the
crossing with D[xt]. A point of D[ys] just beyond this crossing is on the other side of
D[tvx] from both y and s.
The edge D[yv] is contained on the same side of the 3-cycle D[sty] as D[v]. Therefore,
D[yv] must also cross D[xt], showing that the D[y]-side of D[tvx] is also not convex, as
required.
Notice that D[y] is in one side of D′[tvx] and D[s] is on the other. Since s /∈ {t, v, x, y},
D[{t, v, x, y}] and D′[{t, v, x, y}] are homeomorphic. Thus, the side of D[tvx] that con-
tains D[y] is not convex in D.
On the other hand, we know that, in D, D[w] is on the other side of D[tvx] from
D[y]. However, Claim 5.5(2) shows that D[wx] crosses D[tv]. This shows that the side of
D[tvx] containing D[w] is not convex. Combined with the preceding paragraph, the K5
induced by t, v, w, x, y is not convex in D, contradicting the hypothesis of the theorem.
This completes the proof of Subcase 2.3 and the theorem.
It would be significant progress to prove some analogue of Theorem 5.1 with a weaker
hypothesis on extensions J . Indeed, one might expect that no hypothesis beyond the exis-
tence of J is required, as is easily verified for n = 7 (Lemmas 7.5 and 7.7 [21] prove this
for K7 as a simple consequence of the theory developed).
Suppose a drawing D of K8 has a non-convex K5. This K5 is in three different K7’s,
each having at least 11 crossings. Lemma 5.10 below shows D has at least 20 crossings, in
agreement with Aichholzers’s computations.
Lemma 5.10. Let n be an integer, n ≥ 4, and let D be a drawing of Kn. Then
(n− 4) cr(D) =
∑
v∈V (Kn) cr(D − v).
A similar argument shows that a non-convex drawing of K9 cannot have 34 crossings.
Let J be any non-convex K5 in a drawing D of K9 having 34 crossings. Then J is con-
tained in four K8’s in the K9. The preceding paragraph shows each of these K8’s has at
least 20 crossings. Lemma 5.10 and the assumption that D has only 34 crossings shows
that the five remaining K8’s are optimal and hence convex. Thus, J is the only non-convex
K5 in D and so the hypothesis of Theorem 5.1 trivially holds.
The hypothesis of Theorem 5.1 is stronger than we would like and stronger than needed
for the preceding argument for K8. It is not so strong, however, as to force a single non-
convex K5 in a drawing. For example, we have a drawing of K8—a modified TC8—having
two different non-convex K5’s and satisfying the hypotheses of Theorem 5.1.
6 Questions and conjectures
We conclude with a few questions and conjectures.
1. In Section 1 we presented a table with the convexity hierarchy. One obvious omission
is a forbidden drawing characterization of when an h-convex drawing is f-convex.
We pointed out that TC8 is one example of h-convex that is not f-convex. Rerouting
some of the edges between the central and outer crossing K4’s produces a few more
examples.
A. Arroyo et al.: Convex drawings of the complete graph: topology meets geometry 439
Question 6.1. Is there a characterization by forbidden subdrawings of those h-convex
drawings of Kn that are f-convex?
2. The deficiency δ(D) of a drawing D of Kn is the number cr(D)−H(n). The drawing
D has the natural deficiency property if, for every vertex v of Kn, δ(D−v) ≤ 2δ(D).
If the Hill Conjecture is true for n = 2k + 1, then every drawing of K2k has the
natural deficiency property. We prove this in the Appendix.
Conjecture 6.2. For every k ≥ 2, every simple drawing of K2k has the natural
deficiency property.
This seems to be an interesting weakening of the Hill Conjecture; it came up tangen-
tially in the proof that cr(K13) > 217 [20].
3. Pach, Solymosi, and Tóth [23] proved that, for each positive integer r, there is an
N(r) = O(2r
8
) such that, for every n ≥ N(r), every drawing D of Kn contains
either the natural Kr or the Harborth Kr [15]. If D is convex, then it must be the
natural Kr.
Question 6.3. Can the O(2r
8
) bound be improved for convex drawings?
4. The big question is: Is every optimal drawing of Kn convex? While we believe this
is quite conceivable, Ramsey theory suggests other possibilities. Hedging our bets,
we have the following conjecture.
Conjecture 6.4. Exactly one of the following holds:
(a) for all n ≥ 5, no optimal drawing of Kn contains K̃55; and
(b) for any p ≥ 1 and any drawing D of Kp, there is some n ≥ p and an optimal
drawing of Kn (or at least one with at most H(n) crossings) that contains
D[Kp].
ORCID iDs
Alan Arroyo https://orcid.org/0000-0003-2401-8670
R. Bruce Richter https://orcid.org/0000-0002-1444-9699
References
[1] B. Ábrego, O. Aichholzer, S. Fernández-Merchant, P. Ramos, and B. Vogtenhuber, Non-
shellable drawings of kn with few crossings, in: 26th Annual Canadian Conference
on Computational Geometry CCCG 2014, 2014 pp. 46–51, https://www.cccg.ca/
proceedings/2014/.
[2] O. Aichholzer, Rotation systems with specific drawings of K5, in preparation.
[3] O. Aichholzer, T. Hackl, A. Pilz, P. Ramos, V. Sacristán and B. Vogtenhuber, Empty triangles
in good drawings of the complete graph, Graphs Comb. 31 (2015), 335–345, doi:10.1007/
s00373-015-1550-5.
[4] O. Aichholzer, T. Hackl, A. Pilz, G. Salazar, and B. Vogtenhuber, Deciding monotonicity of
good drawings of the complete graph, in: XVI Spanish Meeting on Computational Geometry
(EGC 2015), 2015 pp. 33–36, http://www.ist.tugraz.at/cpgg/publications.
php.
440 Ars Math. Contemp. 22 (2022) #P3.04 / 415–441
[5] A. Arroyo, D. McQuillan, R. B. Richter and G. Salazar, Levi’s lemma, pseudolinear drawings
of Kn, and empty triangles, J. Graph Theory 87 (2018), 443–459, doi:10.1002/jgt.22167.
[6] A. Arroyo, R. B. Richter and M. Sunohara, Extending drawings of complete graphs into ar-
rangements of pseudocircles, SIAM J. Discrete Math. 35 (2021), 1050–1076, doi:10.1137/
20m1313234.
[7] J. Balogh, B. Lidický and G. Salazar, Closing in on Hill’s conjecture, SIAM J. Discrete Math.
33 (2019), 1261–1276, doi:10.1137/17m1158859.
[8] I. Bárány and Z. Füredi, Empty simplices in Euclidean space, Can. Math. Bull. 30 (1987),
436–445, doi:10.4153/cmb-1987-064-1.
[9] I. Bárány and P. Valtr, Planar point sets with a small number of empty convex polygons, Stud.
Sci. Math. Hung. 41 (2004), 243–266, doi:10.1556/sscmath.41.2004.2.4.
[10] L. Beineke and R. Wilson, The early history of the brick factory problem, Math. Intell. 32
(2010), 41–48, doi:10.1007/s00283-009-9120-4.
[11] W. E. Bonnice, On convex polygons determined by a finite planar set, Am. Math. Mon. 81
(1974), 749–752, doi:10.2307/2319566.
[12] P. Erdős and G. Szekeres, On some extremum problems in elementary geometry, Ann. Univ.
Sci. Budap. Rolando Eötvös, Sect. Math. 3-4 (1961), 53–62, https://users.renyi.hu/
˜p_erdos/Erdos.html.
[13] R. K. Guy, Crossing numbers of graphs, in: Y. Alavi, D. R. Lick and A. T. White (eds.), Graph
Theory and Applications, Springer Berlin Heidelberg, Berlin, Heidelberg, 1972 pp. 111–124,
doi:10.1007/bfb0067363.
[14] F. Harary and A. Hill, On the number of crossings in a complete graph, Proc. Edinb. Math.
Soc., II. Ser. 13 (1963), 333–338, doi:10.1017/s0013091500025645.
[15] H. Harborth, Empty triangles in drawings of the complete graph, Discrete Math. 191 (1998),
109–111, doi:10.1016/s0012-365x(98)00098-3.
[16] H. Harborth and I. Mengersen, Drawings of the complete graph with maximum number of
crossings, in: Proceedings of the Twenty-Third Southeastern International Conference on Com-
binatorics, Graph Theory, and Computing (Boca Raton, FL, 1992), Congr. Numer. 88, pp.
225–228, 1992.
[17] D. J. Kleitman, A note on the parity of the number of crossings of a graph, J. Comb. Theory,
Ser. B 21 (1976), 88–89, doi:10.1016/0095-8956(76)90032-0.
[18] F. Levi, Die Teilung der projektiven Ebene durch Gerade oder Pseudogerade, Ber. Math.-Phys.
Kl. Sächs. Akad. Wiss. Leipzig 78 (1926), 256–267.
[19] J. Matousek, Intersection graphs of segments and ∃R2, 2014, arXiv:1406.2636
[math.CO].
[20] D. McQuillan, S. Pan and R. B. Richter, On the crossing number of K13, J. Comb. Theory, Ser.
B 115 (2015), 224–235, doi:10.1016/j.jctb.2015.06.002.
[21] D. McQuillan and R. B. Richter, On the crossing number of Kn without computer assistance,
J. Graph Theory 82 (2016), 387–432, doi:10.1002/jgt.21908.
[22] N. E. Mnëv, The universality theorems on the classification problem of configuration varieties
and convex polytopes varieties, in: Topology and Geometry—Rohlin Seminar, Springer, Berlin,
volume 1346 of Lecture Notes in Math., pp. 527–543, 1988, doi:10.1007/bfb0082792.
[23] J. Pach, J. Solymosi and G. Tóth, Unavoidable configurations in complete topological graphs,
Discrete Comput. Geom. 30 (2003), 311–320, doi:10.1007/s00454-003-0012-9.
A. Arroyo et al.: Convex drawings of the complete graph: topology meets geometry 441
[24] A. Suk, On the Erdős-Szekeres convex polygon problem, J. Am. Math. Soc. 30 (2017), 1047–
1053, doi:10.1090/jams/869.
[25] U. Wagner, On a geometric generalization of the upper bound theorem, in: 2006 47th Annual
IEEE Symposium on Foundations of Computer Science (FOCS’06), 2006 pp. 635–645, doi:
10.1109/focs.2006.53.
Appendix
Here we prove the claim made in the discussion about the natural deficiency property,
related to Conjecture 6.2: if the Hill Conjecture is true for n = 2k+ 1, then every drawing
of K2n has the natural deficiency property.
Let D be a simple drawing of K2n, let v be any vertex of K2n, and let r(v) denote
the total number of crossings in D involving edges incident with v. Duplicating v to get
the drawing D + v of K2n+1, we get cr(D + v) = cr(D) + W (2n − 1) + r(v), where
W (m) = ⌊m2 ⌋⌊
m−1
2 ⌋ is the smallest number of crossings in a drawing of K2,m such that
the vertices on the 2-side have the same clockwise rotation of the m remaining vertices.
By assumption, cr(D+v) ≥ H(2n+1), so H(2n+1) ≤ cr(D)+W (2n−1)+ r(v).
On the other hand, arithmetic shows that H(2n+ 1) = H(2n) +W (2n− 1) + (H(2n)−
H(2n− 1)). Therefore,
2H(2n) +W (2n− 1)−H(2n− 1) = H(2n+ 1)
≤ cr(D) +W (2n− 1) + r(v) .
Cancelling the common W (2n− 1) and rearranging yields
r(v) +H(2n− 1) ≥ 2H(2n)− cr(D) . (A.1)
Inequality (A.1) implies
δ(D − v) = cr(D − v)−H(2n− 1)
= cr(D)− r(v)−H(2n− 1)
≤ 2 cr(D)− 2H(2n)
= 2δ(D) ,
as claimed.

ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 22 (2022) #P3.05 / 443–458
https://doi.org/10.26493/1855-3974.2645.8fc
(Also available at http://amc-journal.eu)
On the essential annihilating-ideal graph
of commutative rings*
Mohd Nazim , Nadeem ur Rehman †
Department of Mathematics, Aligarh Muslim University, Aligarh-202002, India
Received 30 May 2021, accepted 8 September 2021, published online 9 June 2022
Abstract
Let R be a commutative ring with unity, A(R) be the set of annihilating-ideals of R and
A∗(R) = A(R) \ {0}. In this paper, we introduced and studied the essential annihilating-
ideal graph of R, denoted by EG(R), with vertex set A∗(R) and two distinct vertices I1 and
I2 are adjacent if and only if Ann(I1I2) is an essential ideal of R. We prove that EG(R) is
a connected graph with diameter at most three and girth at most four if EG(R) contains a
cycle. Furthermore, the rings R are characterized for which EG(R) is a star or a complete
graph. Finally, we classify all the Artinian rings R for which EG(R) is isomorphic to some
well-known graphs.
Keywords: Annihilating-ideal graph, zero-divisor graph, complete graph, planar graph, genus of a
graph.
Math. Subj. Class. (2020): 13A15, 05C10, 05C12, 05C25
1 Introduction
Throughout this paper all rings are commutative rings (not a field) with unit element such
that 1 ̸= 0. For a commutative ring R, we use I(R) to denote the set of ideals of R and
I∗(R) = I(R) \ {0}. An ideal I of R is said to be non-trivial if it is nonzero and proper
both. An ideal I of R is said to be annihilator ideal if there is a nonzero ideal J of R such
that IJ = 0. For X ⊆ R, we define annihilator of X as Ann(X) = {r ∈ R : rX = 0}.
We use A(R) to denote the set of annihilator ideas of R and A∗(R) = A(R) \ {0}. We
denote the set of zero-divisors, the set of nilpotent elements, the set of maximal ideals, the
set of minimal prime ideals, and the set of Jacobson radical of a ring R by Z(R), Nil(R),
*The authors are greatly indebted to the referee for his/her constructive comments and suggestion, which
improves the quality of the paper a lot.
†Corresponding author.
E-mail addresses: mnazim1882@gmail.com (Mohd Nazim), nu.rehman.mm@amu.ac.in (Nadeem ur
Rehman)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
444 Ars Math. Contemp. 22 (2022) #P3.05 / 443–458
Max (R), Min(R) and J(R), respectively. A nonzero ideal I of R is called essential,
denoted by I ≤e R, if I has a nonzero intersection with every nonzero ideal of R. Also,
if I is not an essential ideal of R then, it is denoted by I ̸≤e R. A ring R is said to be
reduced, if it has no nonzero nilpotent element. For a nonzero nilpotent element x of R,
we use η to denote the index of nilpotency of x. If S is any subset of R, then S∗ denote the
set S \ {0}. For any undefined notation or terminology in ring theory, we refer the reader
to see [9].
Let G be a graph with vertex set V (G). The distance between two vertices u and v
of G denoted by d(u, v), is the smallest path from u to v. If there is no such path, then
d(u, v) = ∞. The diameter of G is defined as diam(G) = sup{d(u, v) : u, v ∈ V (G)}.
A cycle is a closed path in G. The girth of G denoted by gr(G) is the length of a shortest
cycle in G (gr(G) = ∞ if G contains no cycle). A graph is said to be complete if all its
vertices are adjacent to each other. A complete graph with n vertices is denoted by Kn. If
G is a graph such that the vertices of G can be partitioned into two nonempty disjoint sets
U1 and U2 such that vertices u and v are adjacent if and only if u ∈ U1 and v ∈ U2, then
G is called a complete bipartite graph. A complete bipartite graph with disjoint vertex sets
of size m and n, respectively, is denoted by Km,n. We write Kn,∞ (respectively, K∞,∞)
if one (respectively, both) of the disjoint vertex sets is infinite. A complete bipartite graph
of the form K1,n is called a star graph. A graph G is said to be planar if it can be drawn in
the plane so that its edges intersect only at their ends. A subdivision of a graph is a graph
obtained from it by replacing edges with pairwise internally-disjoint paths. A remarkably
simple characterization of planar graphs was given by Kuratowski in 1930. Kuratowski’s
Theorem says that a graph G is planar if and only if it contains no subdivision of K5 or
K3,3. The genus of a graph G, denoted by γ(G), is the minimum integer k such that the
graph can be drawn without crossing itself on a sphere with k handles (i.e. an oriented
surface of genus k). Thus, a planar graph has genus 0, because it can be drawn on a sphere
without self-crossing. For more details on graph theory, we refer to reader to see [21, 22].
The concept of zero-divisor graph of a commutative ring R, denoted by Γ(R), was
introduced by I. Beck [10]. The vertex set of Γ(R) is Z∗(R) = Z(R) \ {0} (set of nonzero
zero-divisors of R) and two distinct vertices x and y are adjacent if and only if xy =
0, for details see [5, 8, 7]. In [14], Dolžan and Oblak also obtained several interesting
results related with zero-divisor graph of rings and semirings. The zero-divisor graph of a
noncommutative ring has been introduced and studied by Redmond [18], whereas the same
concept for semigroup by Demeyer et al. [13].
In [11], Behboodi et al. generalized the zero-divisor graph to ideals by defining the
annihilating-ideal graph AG(R), with vertex set is A∗(R) and two distinct vertices I1 and
I2 are adjacent if and only if I1I2 = 0. For more details on annihilating-ideal graph, we
refer the reader to see [1, 2, 3, 4, 6, 12, 16].
In [17], M. Nikmehr et al. introduced the essential graph EG(R) with vertex set
Z∗(R) = Z(R)\{0} and two distinct vertices x and y are adjacent if and only if annR(xy)
is an essential ideal of R.
Motivated by [17], we define the essential annihilating-ideal graph of R denoted by
EG(R) with vertex set A∗(R) and two distinct vertices I1 and I2 adjacent if and only if
Ann(I1I2) is an essential ideal of R. In this paper we first prove that AG(R) is a sub-
graph of EG(R) and then studied some basic properties of EG(R) such as connectedness,
diameter, girth and shows that EG(R) is a connected graph with diam(EG(R)) ≤ 3 and
gr(EG(R)) ≤ 4, if EG(R) contains a cycle. In the third section, we determine some condi-
M. Nazim and N. Rehman: On the essential annihilating-ideal graph of commutative rings 445
tions on R under which EG(R) is a star graph or a complete graph. In the last, we identify
all the Artinian rings R for which EG(R) is isomorphic to some well-known graphs.
2 Basic properties of essential annihilating-ideal graph
We begin this section with the following lemma given by [17].
Lemma 2.1 ([17, Lemma 2.1]). Let R be a commutative ring and I be an ideal of R. Then
(1) I +Ann(I) is an essential ideal of R.
(2) If I2 = (0), then Ann(I) is an essential ideal of R.
(3) If R contains no proper essential ideals, then J(R) = (0).
The following lemma is analogue of [17, Lemma 2.2].
Lemma 2.2. Let R be a commutative ring. Then
(1) If I1 and I2 are adjacent in AG(R), then I1 and I2 are also adjacent in EG(R).
(2) If I2 = 0 for some I ∈ A∗(R), then I is adjacent to every other vertex in EG(R).
Proof. (1) Suppose I1 and I2 are adjacent in AG(R), then I1I2 = 0 and so
Ann(I1I2) = R, is an essential ideal of R. Thus I1 and I2 are also adjacent in EG(R).
(2) Suppose that I2 = 0 for some I ∈ A∗(R). Then by Lemma 2.1(2), Ann(I) is an
essential ideal of R. Since Ann(I) ⊆ Ann(IJ) for every J ∈ A∗(R), therefore Ann(IJ)
is also an essential ideal of R. Thus I is adjacent to every other vertex of EG(R).
Let R be a commutative ring. By [11, Theorem 2.1], the annihilating ideal graph
AG(R) is a connected graph with diam(AG(R)) ≤ 3. Moreover, if AG(R) contains
a cycle, then gr(AG(R)) ≤ 4.
In view of part (1) of Lemma 2.2, we have the following result.
Theorem 2.3. Let R be a commutative ring. Then EG(R) is connected with diam(EG(R)) ≤
3. Moreover, if EG(R) contain a cycle, then gr(EG(R)) ≤ 4.
In Lemma 2.2(1), we proved that AG(R) is a spanning subgraph of EG(R) but this
containment may be proper. The following examples shows that AG(R) and EG(R) are
not identical.
Example 2.4.
1. If R = Z16, then AG(R) is P3 and EG(R) is K3.
2. If R = Zp5 , where p is a prime number. Then AG(R) is the following graph and
EG(R) is K4.
Theorem 2.5. Let R be a commutative reduced ring. Then EG(R) = AG(R).
Proof. Clearly, AG(R) ⊆ EG(R). We just have to prove that EG(R) is a subgraph of
AG(R). Suppose on contrary that I1 ∼ I2 is an edge of EG(R) such that I1I2 ̸= 0. Since
R is a reduced ring, then I1I2 ∩ Ann(I1I2) = 0, which implies that Ann(I1I2) is not an
essential ideal of R, a contradiction. Thus I1I2 = 0 and EG(R) = AG(R).
446 Ars Math. Contemp. 22 (2022) #P3.05 / 443–458
Figure 1: The graph AG(Zp5).
Theorem 2.6 ([12, Theorem 1.9(3)]). Let R be a commutative ring with finitely many
minimal primes. Then diam(AG(R)) = 2 if and only if either R is reduced with exactly
two minimal primes and at least three nonzero annihilating-ideals, or R is not reduced,
Z(R) is an ideal whose square is not (0) and for each pair of annihilating-ideals I1 and
I2, I1 + I2 is an annihilating-ideal.
Theorem 2.7. Let R be a commutative ring with |Min(R)| < ∞. Then
(1) If R is reduced ring, then diam(EG(R)) = 2 if and only if |Min(R)| = 2 and R has
at least three nonzero annihilating-ideals. Moreover, in this case
gr(EG(R)) ∈ {4,∞}.
(2) If R is non-reduced, then diam(EG(R)) ≤ 2. Moreover, in this case
gr(EG(R)) ∈ {3,∞}.
Proof. (1) First part is clear from Theorems 2.5 and 2.6. Now, let Min(R) = {P1, P2},
then EG(R) is a complete bipartite graph with partitions V1 = {I ∈ V (EG) : I ⊆ P1} and
V2 = {I ∈ V (EG) : I ⊆ P2} by [12, Theorem 1.2]. Hence gr(EG(R)) ∈ {4,∞}.
(2) Since R is a non-reduced ring, then there is I1 ∈ A∗(R) such that I21 = 0. Thus by
Lemma 2.2(2), I1 is adjacent to every other vertex of EG(R). Hence diam(EG(R)) ≤ 2.
Also, if there are I, J ∈ V (EG(R)) \ {I1} such that I ∼ J is an edge of EG(R),
then I1 ∼ I ∼ J ∼ I1 is a triangle in EG(R). Thus, gr(EG(R)) = 3, otherwise
gr(EG(R)) = ∞.
3 Completeness of essential annihilating-ideal graph
In this section, we characterize commutative rings R for which EG(R) is a star graph or a
complete graph. We begin with the following lemma.
Lemma 3.1. Let R be a commutative nonreduced ring. Then
(1) For every nilpotent ideal I1 of R, I1 is adjacent to every other vertex of EG(R).
(2) The subgraph induced by the nilpotent ideals of R is a complete subgraph of EG(R).
Proof. (1) Suppose that I1 be any nilpotent ideal of R. Let I2 ∈ A∗(R). We show
that Ann(I1I2) ≤e R. Since Ann(I1) ⊆ Ann(I1I2), then it is enough to show that
Ann(I1) ≤e R. Suppose on contrary that Ann(I1) ̸≤e R, then there exists I3 ∈ I∗(R)
such that Ann(I1) ∩ I3 = 0, which implies that rI1 ̸= 0 for every r ∈ I∗3 . Since
0 ̸= rI1 ⊆ I3, then I1 · rI1 = rI21 ̸= 0. Continuing this process, we get rIn1 ̸= 0,
for every positive integer n, which is a contradiction. This complete the proof.
(2) It is clear from (1).
M. Nazim and N. Rehman: On the essential annihilating-ideal graph of commutative rings 447
Lemma 3.2. Let (R,m) be a commutative Artinian local ring. Then EG(R) is a complete
graph.
Proof. Follows from Lemma 3.1.
Lemma 3.3. Let R be a commutative decomposable ring. Then EG(R) is a star graph if
and only if R = F ×D, where F is a field and D is an integral domain.
Proof. (⇒) Suppose that EG(R) is a star graph and let R = R1 × R2, where R1 and R2
are commutative rings. If R1 and R2 both are not fields and I1 ∈ I∗(R1), I2 ∈ I∗(R2),
then (R1 × (0)) ∼ ((0) × R2) ∼ (I1 × (0)) ∼ ((0) × I2) ∼ (R1 × (0)) is a cycle of
length 4 in EG(R), a contradiction. Thus, without loss of generality we can assume that
R1 is a field. We claim that R2 is an integral domain. Suppose on contrary that R2 is not
an integral domain, then there exists I3, I4 ∈ I∗(R1) such that I3I4 = 0. If I3 ̸= I4, then
(R1× (0)) ∼ ((0)×I3) ∼ ((0)×I4) ∼ (R1× (0)) is a triangle in EG(R), a contradiction.
Also, if I3 = I4, then by Lemma 3.1, (R1× (0)) ∼ ((0)× I3) ∼ ((0)×R2) ∼ (R1× (0))
is a triangle in EG(R), again a contradiction. This complete the proof.
(⇐) is clear.
Theorem 3.4. Let R be an Artinian commutative ring with atleast two non-trivial ideals.
Then EG(R) is a star graph if and only if EG(R) ∼= K2.
Proof. (⇒) Suppose EG(R) is a star graph. If R is a local ring, then from Lemma 3.2,
EG(R) is a complete graph. Since EG(R) is a star graph, therefore EG(R) ∼= K2. If R is
non-local ring, then it is decomposable. Thus by Lemma 3.3, R = F × D, where F is a
field and D is an integral domain. Since R is Artinian ring, then D is Artinian and hence
is a field. Thus EG(R) ∼= K2.
(⇐) is evident.
Theorem 3.5. Let R be a commutative ring with at least two non-trivial ideals. Then
EG(R) is a star graph if and only if one of the following holds:
(1) R has exactly two non-trivial ideals.
(2) R = F ×D, where F is a field and D is an integral domain which is not a field.
(3) R has a minimal ideal I1 such that I1 is not an essential ideal of R, I21 = 0 and for
any nonzero annihilating ideal I2 of R, Ann(I2) = I1.
Proof. (⇒) Suppose EG(R) is a star graph. If |A∗(R)| < ∞, then from [11, Theorem 1.1],
R is an Artinian ring. Thus, by Theorem 3.4, EG(R) ∼= K2 and hence (1) hold.
Now, let |A∗(R)| = ∞ and I1 is adjacent to every other vertex of EG(R). We show that I1
is minimal ideal of R. Suppose on contrary that there exists I2 ∈ I∗(R) such that I2 ⊂ I1.
Let I3 ∈ A∗(R) \ {I1, I2}, then Ann(I1I3) ≤e R. Since I2I3 ⊆ I1I3, then Ann(I2I3)
is also essential ideal of R. This implies that I2 is also adjacent to every other vertex of
EG(R), a contradiction. Now, following two cases occur:
Case I: I21 ̸= 0. Then I21 = I1, thus by Brauer’s Lemma [15, p. 172, Lemma 10.22], R
is decomposable. Since |A∗(R)| = ∞ and EG(R) is a star graph. Then from Lemma 3.3,
R = F ×D, where F is a field and D is an integral domain which is not a field. Hence (2)
hold.
Case II: I21 = 0. Let I2 ∈ A∗(R) \ {I1}. Then I2 ̸= Ann(I2), otherwise I22 = 0
448 Ars Math. Contemp. 22 (2022) #P3.05 / 443–458
implies that I2 is also adjacent to every other vertex of EG(R), a contradiction. Now, since
I2 ∼ Ann(I2), then Ann(I2) = I1. If I1 is an essential ideal of R, then Ann(I2) is
also an essential ideal of R. This shows that I2 is also adjacent with every other vertex of
EG(R), which is a contradiction to our assumption that EG(R) is a star graph because we
are assuming that I1 is adjacent with every other vertex of EG(R) and I1 ̸= I2. Hence I1
is not an essential ideal of R.
(⇐) If R has exactly two non-trivial ideals, then R is Artinian ring with |A∗(R)| = 2.
Since EG(R) is connected, therefore EG(R) ∼= K2. If R = F ×D, where F is a field and
D is an integral domain which is not a field, then from Lemma 3.3, EG(R) is a star graph.
Now, suppose that R has a minimal ideal I1 such that I1 is not an essential ideal of R,
I21 = 0 and for any nonzero annihilating ideal I2 of R, Ann(I2) = I1. Let I2, I3 ∈
A∗(R) \ {I1} such that I2 ∼ I3 in EG(R). This implies that Ann(I2I3) ≤e R and
Ann(I2) = I1 = Ann(I3). Since Ann(I2) = Ann(I3) is not an essential of R, there
exists a nonzero ideal I4 of R such that Ann(I2) ∩ I4 = Ann(I3) ∩ I4 = 0. This shows
that rI2 ̸= 0 and rI3 ̸= 0 for every r ∈ I∗4 . On the other hand, since Ann(I2I3) ≤e R, then
Ann(I2I3) ∩ I4 ̸= 0. That is there exists s ∈ I∗4 such that sI2I3 = 0. Now, observe that
sI2 ⊆ I∗4 satisfies sI2 ⊆ Ann(I3), which implies that Ann(I3) ∩ I4 ̸= 0, a contradiction.
This complete the proof.
Theorem 3.6. Let R be a commutative Artinian ring. Then EG(R) is a complete graph if
and only if one of the following holds:
(1) R = F1 × F2, where F1 and F2 are fields.
(2) R is a local ring.
Proof. (⇒) Suppose that EG(R) is a complete graph. Since R is Artinian, then
R ∼= R1 × R2 × · · · × Rn, where Ri is Artinian local ring for each 1 ≤ i ≤ n. The
following cases occur:
Case I: n ≥ 3. Then R1 × (0) × · · · × (0) and R1 × (0) × R3 × · · · × (0) are nonzero
annihilating ideals of R such that (R1 × (0)× · · · × (0)) ̸∼ (R1 × (0)×R3 × · · · × (0))
in EG(R), a contradiction.
Case II: n = 2. We show that R1 and R2 are fields. Suppose on contrary that R1 is not a
field with non-trivial maximal ideal m. Then Ann(((0)× R2) · (m× R2)) = Ann((0)×
R2) = R1 × (0), which is not an essential ideal of R. Thus ((0) × R2) ̸∼ (m × R2) in
EG(R), a contradiction. Hence (2) holds.
Case III: n = 1. Then R is Artinian local ring and (1) holds.
(⇐) If R is local, then from Lemma 3.2, EG(R) is a complete graph. If R = F1 × F2,
where F1 and F2 are fields, then EG(R) ∼= K2.
Theorem 3.7. Let R be a commutative ring with at least one minimal ideal. Then EG(R) ∼=
Km,n, where m,n ≥ 2 if and only if R = D × S, where D and S are integral domains
which are not fields.
Proof. (⇒) Suppose that EG(R) ∼= Km,n, where m,n ≥ 2. Let I1 be minimal ideal of
R. If I21 = 0, then from Lemma 2.2, I1 is adjacent to every other vertex, a contradic-
tion. Thus I21 ̸= 0. Since I1 is minimal, therefore I21 = I1. Therefore, Brauer’s Lemma
[15, p. 172, Lemma 10.22], R = R1×R2, where R1 and R2 are commutative rings. Now,
our objective is to show that R1 and R2 are integral domains. Suppose on contrary that
M. Nazim and N. Rehman: On the essential annihilating-ideal graph of commutative rings 449
R1 is not an integral domain with nonzero annihilating ideal I2. As above, I22 ̸= 0 which
implies that I2 /∈ Ann(I2). Thus (I2×(0)) ∼ ((0)×R2) ∼ (Ann(I2)×(0)) ∼ (I2×(0))
is a triangle in EG(R), a contradiction. Hence R1 is an integral domain. Similarly, one can
prove that R2 is an integral domain. Since m,n ≥ 2, therefore R1 and R2 are not fields.
(⇐) Suppose that R = D × S, where D and S are integral domains which are not fields.
Let U = {I1×(0) : I1 ∈ I∗(D)} and V = {(0)×I2 : I2 ∈ I∗(S)}. Then A∗(R) = U ∪V
such that no two vertices of U or V are adjacent in EG(R). Also, every vertex of U is
adjacent to every vertex of V in EG(R). Thus, EG(R) ∼= Km,n. Since D and S are not
fields, therefore m,n ≥ 2.
Lemma 3.8. Let R be a commutative ring. Then
(1) Let I1, I2, I3 ∈ A∗(R) such that Ann(I1) = Ann(I2). Then I1 ∼ I3 is an edge of
EG(R) if and only if I2 ∼ I3 is an edge of EG(R).
(2) Let I ∈ A∗(R). Then Ann(I) ≤e R if and only if Ann(In) ≤e R for every n ≥ 2.
In particular, if Ann(I3) ≤e R, then Ann(In) ≤e R for every n ≥ 1.
Proof. (1) (⇒) Suppose that I1 ∼ I3 is an edge of EG(R), then Ann(I1I3) ≤e R. We
have to show that Ann(I2I3) ≤e R. Suppose on contrary that Ann(I2I3) is not an essential
ideal of R, then there exits I4 ∈ I∗(R) such that Ann(I2I3) ∩ I4 = 0. This implies that
rI2I3 ̸= 0 for all r ∈ I∗4 . On the other hand, since Ann(I1I3) is an essential ideal of R,
then Ann(I1I3) ∩ I4 ̸= 0. That is there exists some s ∈ I∗4 such that sI1I3 = 0. Now,
observe that sI3 ⊆ I∗4 satisfies sI3 ⊆ Ann(I1) = Ann(I2), which implies that sI2I3 = 0,
a contradiction.
(⇐) Using similar argument as above we get the required result.
(2) (⇒) is clear.
(⇐) Suppose on contrary that Ann(I) is not an essential ideal of R, then there exists
nonzero ideal I1 of R such that Ann(I) ∩ I1 = 0. This implies that rI ̸= 0 for all r ∈ I∗1 .
On the other hand, since Ann(I2) ≤e R, then Ann(I2) ∩ I1 ̸= 0. That is there exists
some s ∈ I∗1 such that sI2 = 0. Now, observe that r = sI ⊆ I∗1 such that rI = 0, a
contradiction.
For the particular case, we need to show that Ann(I2) ≤e R. Suppose on contrary that
there is some I1 ∈ I∗(R) such that Ann(I2) ∩ I1 = 0, which implies that rI2 ̸= 0 for
all r ∈ I∗1 . On the other hand, since Ann(I3) ≤e R, then Ann(I3) ∩ I1 ̸= 0. That is
there exists some s ∈ I∗1 such that sI3 = 0. Now, observe that r = sI2 ⊆ I∗1 such that
rI = 0, which implies that Ann(I) ∩ I1 ̸= 0. Since Ann(I) is a subset of Ann(I2), then
Ann(I2) ∩ I1 ̸= 0, a contradiction.
Theorem 3.9. Let R be a commutative non-reduced ring. Then EG(R) is a complete graph
if and only if Ann(I) ≤e R for every I ∈ A∗(R).
Proof. (⇒) Suppose that EG(R) is a complete graph. We claim that R is indecompos-
able ring. Suppose on contrary that R = R1 × R2, where R1 and R2 are commu-
tative rings. Since R is non-reduced ring, without loss of generality, we can assume
that R1 is non-reduced ring with nonzero nilpotent element x. Let I1 = xR1. Then
Ann((I1 × R2) · ((0) × R2)) = Ann((0) × R2) = R1 × (0), is not an essential ideal
of R, a contradiction to the completeness of EG(R). Let I ∈ A∗(R) be arbitrary. If I is
nilpotent ideal, then from Lemma 3.1(1), Ann(I) ≤e R. Suppose I is not nilpotent ideal.
450 Ars Math. Contemp. 22 (2022) #P3.05 / 443–458
Since R is indecomposable, then I2 ̸= I , which implies that Ann(I3) ≤e R. Hence by
Lemma 3.8(2), Ann(I) ≤e R.
(⇐) is evident.
4 Essential annihilating-ideal graph as some special type of graphs
In this section, we characterize all the Artinian rings R for which EG(R) is a tree, a unicycle
graph, a split graph, a outerplanar graph, a planar graph and a toroidal graph.
Theorem 4.1. Let R be a commutative Artinian ring (not a field). Then EG(R) is a tree
if and only if either R ∼= F1 × F2, where F1 and F2 are fields or R is a local ring with at
most two non-trivial ideals.
Proof. Suppose that EG(R) is a tree. Since R is an Artinain ring, then
R ∼= R1×R2×· · ·×Rn, where each Ri is an Artinian local ring. If n ≥ 3. Consider I1 =
R1×(0)×· · ·×(0), I2 = (0)×R2×(0)×· · ·×(0) and I3 = (0)×(0)×R3×(0)×· · ·×(0).
Then I1 ∼ I2 ∼ I3 ∼ I1 is a cycle of in EG(R), a contradiction.
Suppose n = 2, then we show that R1 and R2 both are fields. Suppose on contrary that
R1 is not a field with nonzero maximal ideal m. Consider J1 = (0) × R2, J2 = m × (0),
J3 = m× R2 and J4 = R1 × (0). Then J1 ∼ J2 ∼ J3 ∼ J4 ∼ J1 is a cycle in EG(R), a
contradiction.
If n = 1, then R is Artinian local ring. Thus by Lemma 3.2, EG(R) is a complete graph.
Since EG(R) is a tree, therefore R has at most two non-trivial ideal.
Converse is clear.
Theorem 4.2. Let R be a commutative Artinian ring (not a field). Then EG(R) is unicycle
if and only if either R ∼= F1 × F2 × F3, where Fi is a field for each 1 ≤ i ≤ 3 or R is an
Artinain local ring with exactly three non-trivial ideals.
Proof. Suppose that EG(R) is unicycle. Since R is Artinian ring, then
R ∼= R1 × R2 × · · · × Rn, where Ri is Artinian local ring for each 1 ≤ i ≤ n. Let
n ≥ 4. Consider I1 = R1 × (0) × · · · × (0), I2 = (0) × R2 × (0) × · · · × (0),
I3 = (0) × (0) × R3 × (0) × · · · × (0) and J1 = (0) × (0) × R3 × (0) × · · · × (0),
J2 = R1 × R2 × (0) × · · · × (0), J3 = (0) × (0) × (0) × R4 × (0) × · · · × (0). Then
I1 ∼ I2 ∼ I3 ∼ I1 as well as J1 ∼ J2 ∼ J3 ∼ J1 are two different cycles in EG(R), a
contradiction. Hence n ≤ 3.
First, let n = 3 and suppose on contrary that R2 is not a field with nonzero maximal ideal
m. Consider I1 = R1 × (0) × (0), I2 = (0) × R2 × (0), I3 = (0) × (0) × R3 and
J1 = R1 × (0)× (0), J2 = (0)×m× (0), J3 = (0)× (0)×R3. Then I1 ∼ I2 ∼ I3 ∼ I1
and J1 ∼ J2 ∼ J3 ∼ J1 are two different cycles in EG(R), a contradiction. Hence Ri is a
field for each 1 ≤ i ≤ 3.
Now, let n = 2. If R1 and R2 both are fields then EG(R) ∼= K2, a contradiction. Thus one
of Ri say R2 is not a field with nonzero maximal ideal m. Then (R1× (0)) ∼ ((0)×m) ∼
((0)×R2) ∼ (R1 × (0)) as well as (R1 ×m ∼ ((0)×m) ∼ ((0)×R2) ∼ (R1 ×m) are
two different cycles in EG(R), again a contradiction.
If n = 1, then R is an Artinian local ring. Thus, by Lemma 3.2, EG(R) is a complete
graph. Since EG(R) is unicycle, R have exactly three non-trivial ideals.
Theorem 4.3 ([21]). Let G be a connected graph. Then G is a split graph if and only if G
contains no induced subgraph isomorphic to 2K2, C4, C5.
M. Nazim and N. Rehman: On the essential annihilating-ideal graph of commutative rings 451
Theorem 4.4. Let R be a commutative Artinian non-local ring. Then EG(R) is split graph
if and only if either R ∼= F1 × F2 × F3 or R ∼= F1 × F2, where Fi is a field for each
1 ≤ i ≤ 3.
Proof. Suppose that EG(R) is a split graph. Since R is Artinian non-local ring, then
R ∼= R1 × R2 × · · · × Rn, where each Ri is an Artinian local ring and n ≥ 2. If n ≥ 4,
then I1 = R1 ×R2 × (0)× · · · × (0) ∼ J1 = (0)× (0)×R3 ×R4 × (0)× · · · × (0) and
I2 = R1 × (0) × R3 × (0) × · · · × (0) ∼ J2 = (0) × R2 × (0) × R4 × (0) × · · · × (0)
induces 2K2 in EG(R), a contradiction. Hence n = 2 or 3. We have following cases:
Case I: If n = 3, then we show that each Ri ia a field. Suppose on contrary that R1 is
not a field with nonzero maximal ideal m. Then (R1 × (0) × (0)) ∼ ((0) × R2 × R3) ∼
(m× (0)× (0)) ∼ ((0)×R2 × (0)) ∼ (R1 × (0)× (0)) is C4 in EG(R), a contradiction.
Hence Ri is a field for each 1 ≤ i ≤ 3.
Case II: Let n = 2 and suppose that R2 is not a field with nonzero maximal ideal m′. Then
(R1 × (0)) ∼ ((0) × R2) ∼ (R1 × m′) ∼ ((0) × m′) ∼ (R1 × (0)) is C4 in EG(R), a
contradiction. Hence R1 and R2 both are fields.
Converse is clear.
Theorem 4.5 ([22]). A graph G is outerplanar if and only if it does not contain a subdivi-
sion of K4 or K2,3.
Theorem 4.6. Let R be a commutative Artinian ring. Then EG(R) is outerplanar if and
only if one of the following holds:
(1) R = F1 × F2 × F3, where Fi is a field for each 1 ≤ i ≤ 3.
(2) R = F1 × F2, where F1 and F2 are fields.
(3) R = F × R1, where F is a field and (R1,m) is a local ring with m is the only
non-trivial ideal of R1.
(4) R is a local ring with at most three non-trivial ideals.
Proof. Suppose that EG(R) is outerplanar. Since R is Artinian ring, then
R ∼= R1×R2×· · ·×Rn, where each Ri is Artinian local ring. If n ≥ 4, then the set {I1 =
R1×(0)×· · ·×(0), I2 = (0)×R2×(0)×· · ·×(0), I3 = (0)×(0)×R3×(0)×· · ·×(0),
I4 = (0)× (0)× (0)×R4× (0)×· · ·× (0)} induces K4 in EG(R), a contradiction. Hence
n ≤ 3. The following cases occur:
Case I: n = 3. We claim that Ri is a field for each 1 ≤ i ≤ 3. Suppose on contrary that R2
is not a field with nonzero maximal ideal m. Then the set
{R1 × (0) × (0), R1 × m × (0), (0) × m × (0), (0) × (0) × R3, (0) × R2 × R3} in-
duces a copy of K2,3 with partition sets A = {(0) × (0) × R3, (0) × R2 × R3} and
B = {R1 × (0) × (0), R1 × m × (0), (0) × m × (0)}, a contradiction. Therefore Ri is a
field for each 1 ≤ i ≤ 3.
Case II: n = 2 and let Ri is not a field with nonzero maximal ideal mi for each i =
1, 2. Then the set {R1 × (0), (0) × R2,m1 × (0), (0) × m2} induces a copy of K4 in
EG(R), a contradiction. Hence one of Ri (say R1) must be a field. Let I be a non-trivial
ideal of R2 other than maximal ideal m2. Then the set {R1 × (0), R1 × m2, (0) × R2,
(0) × m2, (0) × I} induces a copy of K2,3 with partition sets A = {R1 × (0), R1 × m2}
and B = {(0)×R2, (0)×m2, (0)× I} in EG(R), a contradiction. Hence R2 is a field or
452 Ars Math. Contemp. 22 (2022) #P3.05 / 443–458
has unique non-trivial ideal.
Case III: n = 1, then R is an Artinian local ring. Thus by Lemma 3.2, EG(R) is a complete
graph. Since EG(R) is outerplanar, R have at most three non-trivial ideals.
Converse follows from Lemma 3.2, Theorem 4.5, Figures 2 and 3.
F1 × (0)× F3
(0)× F2 × F3
(0)× (0)× F3
F1 × F2 × (0)
F1 × (0)× (0)
(0)× F2 × (0)
Figure 2: The graph EG(F1 × F2 × F3).
0×m F ×m
F × (0) (0)×R
Figure 3: The graph EG(F ×R1), where m is the only non-trivial ideal of R1.
Lemma 4.7 ([20, Proposition 2.7]). If (R,m) is an Artinian local ring and there is an ideal
I of R such that I ̸= mi for every i, then R has at least three distinct non-trivial ideals
J,K and L such that J,K,L ̸= mi for each i.
Theorem 4.8 (Kuratowski’s Theorem). A graph G is planar if and only if it contains no
subdivision of K5 or K3,3.
Lemma 4.9. Let (R,m) be a commutative Artinian local ring. Then EG(R) is planar if
and only if R have at most four non-trivial ideals.
Proof. It is clear from Lemma 3.2 and Theorem 4.8.
Theorem 4.10. Let R be a commutative Artinian ring. Then EG(R) is planar graph if and
only if one of the following hold:
(1) R = F1 × F2 × F3, where Fi is a field for each 1 ≤ i ≤ 3.
M. Nazim and N. Rehman: On the essential annihilating-ideal graph of commutative rings 453
(2) R has at most four non-trivial ideals.
Proof. Suppose that EG(R) is a planar graph. If |A∗(R)| ≤ 4, then (2) holds. Thus, we
assume that |A∗(R)| ≥ 5. Since R is Artinian ring, then R ∼= R1 ×R2 × · · · ×Rn, where
each Ri is Artinian local ring. If n ≥ 4, then the set {R1 × (0) × · · · × (0), R1 × R2 ×
(0)×· · ·× (0), (0)×R2× (0)×· · ·× (0)} ∪ {(0)× (0)×R3×R4× (0)×· · ·× (0), (0)×
(0) × R3 × (0) × · · · × (0), (0) × (0) × (0) × R4 × (0) × · · · × (0)} induces a copy of
K3,3 in EG(R), a contradiction. Hence n ≤ 3. The following cases occur:
Case I: n = 3. We claim that Ri is a field for each 1 ≤ i ≤ 3. Suppose on con-
trary that one of Ri say R2 is not a field with nonzero maximal ideal m. Then the set
{R1×(0)×(0), R1×m×(0), (0)×m×(0), (0)×m×R3, (0)×(0)×R3, (0)×R2×R3}
induces a copy of K3,3 with partition sets A = {R1×(0)×(0), R1×m×(0), (0)×m×(0)}
and B = {(0)×(0)×R3, (0)×m×R3, (0)×R2×R3} in EG(R), a contradiction. Hence,
(1) satisfied.
Case II: n = 2. Since |A∗(R)| ≥ 5, then one of Ri is not a field for some i = 1, 2.
Suppose that R1 is not a field with nonzero maximal ideal m1. If R2 is a field, then
|A∗(R)| ≥ 5 shows that R1 have at least two non-trivial ideals. Let I be a non-trivial ideal
of R1 other than the maximal ideal. Then the set {R1×(0),m1×(0), I×(0)}∪{(0)×R2,
m1 ×R2, I ×R2} induces a copy of K3,3 in EG(R), a contradiction.
Now, if R2 is not a field with nonzero maximal ideal m2, then the set
{R1 × (0), (0) × m2, R1 × m2} ∪ {(0) × R2,m1 × (0),m1 × R2} induces a copy of
K3,3 in EG(R), again a contradiction.
Case III: n = 1. Then R is an Artinian local ring. Thus, by Lemma 3.2, EG(R) is
a complete graph. Since |A∗(R)| ≥ 5, then EG(R) contains a copy of K5, which is a
contradiction.
Conversely, If R is an Artinian ring with at most four non-trivial ideals, then by Theo-
rem 4.8, EG(R) is planar. Also, if R = F1×F2×F3, where Fi is a field for each 1 ≤ i ≤ 3,
then from Figure 2, EG(R) is planar.
Lemma 4.11 ([22]). γ(Kn) = ⌈ 112 (n − 3)(n − 4)⌉, where ⌈x⌉ is the least integer that is
greater than or equal to x. In particular, γ(Kn) = 1 if n = 5, 6, 7.
Lemma 4.12 ([22]). γ(Km,n) = ⌈ 14 (m − 2)(n − 2)⌉, where ⌈x⌉ is the least integer that
is greater than or equal to x. In particular, γ(K4,4) = γ(K3,n) = 1 if n = 3, 4, 5, 6.
Theorem 4.13. Let (R,m) be a commutative Artinian local ring. Then γ(EG(R)) = 1 if
and only if R have at least five and at most seven non-trivial ideals.
Proof. Since (R,m) is an Artinian local ring, then from Lemma 3.2, EG(R) is a complete
graph. Thus, by Lemma 4.11, 5 ≤ r ≤ 7, where r is the number of non-trivial ideals of
R.
Theorem 4.14. Let R be a commutative Artinian ring such that R = F1 × F2 × · · · × Fn,
where n ≥ 4 and Fi is a field for each 1 ≤ i ≤ n. Then γ(EG(R)) = 1 if and only if
n = 4.
Proof. Since R is a reduced ring, EG(R) = AG(R) by Theorem 2.5. Hence the result
follows from [19, Theorem 2].
454 Ars Math. Contemp. 22 (2022) #P3.05 / 443–458
Theorem 4.15. Let R be a commutative Artinian ring such that R = R1×R2×· · ·×Rn,
where n ≥ 2 and each (Ri,mi) is an Artinian local ring with mi ̸= 0. Let ηi be the
nilpotency of mi. Then γ(EG(R)) = 1 if and only if n = 2 and m1 and m2 are the only
non-trivial ideals of R1 and R2 respectively.
Proof. Suppose that γ(EG(R)) = 1. If n ≥ 3, then the set {mη1−11 × (0)×· · ·× (0), (0)×
mη2−12 × (0)×· · ·× (0),m
η1−1
1 ×m
η2−1
2 × (0)×· · ·× (0)}∪{(0)× (0)×R3× (0)×· · ·×
(0), (0)× (0)×m3 × (0)× · · · × (0),m1 ×m2 ×m3 × (0)× · · · × (0),m1 × (0)×m3 ×
(0)× · · · × (0), (0)×m2 ×m3 × (0)× · · · × (0),m1 × (0)×R3 × (0)× · · · × (0), (0)×
m2 × R3 × (0) × · · · × (0)} induces a copy of K3,7 in EG(R). Thus, from Lemma 4.12,
γ(EG(R)) > 1, a contradiction. Hence n = 2.
Suppose I is non-trivial ideal of R1 such that I ̸= m1. Then the set {R1 × (0),
m1 × (0), R1 × m2, I × (0), I × m2} ∪ {(0) × R2, (0) × m2,m1 × R2,m1 × m2} in-
duces a copy of K4,5 in EG(R). By Lemma 4.12, γ(EG(R)) > 1, a contradiction. Hence
R1 has unique non-trivial ideal m1. Similarly, we can show that R2 has unique non-trivial
ideal m2.
Conversely, let R = R1×R2, where m1 and m2 are the only non-trivial ideals of R1 and R2
respectively, then |A∗(R)| = 7. It is easy to see that the set {R1 × (0),
m1 × (0), R1 × m2} ∪ {(0) × R2, (0) × m2,m1 × R2} induces a copy of K3,3, which
implies that K3,3 ≤ EG(R) ≤ K7. Hence, by Lemma 4.11 and 4.12, γ(EG(R)) = 1.
Theorem 4.16 ([19, Theorem 4]). Let R = R1 × R2 × F be a commutative ring, where
each (Ri,mi) is a local ring with mi ̸= 0 and F is a field. Let ηi be the nilpotency of mi.
Then γ(AG(R)) > 1.
Theorem 4.17 ([19, Theorem 5]). Let R = R1 × F1 × F2 × · · · × Fm be a commutative
ring, where each (R1,m1) is a local ring with m1 ̸= 0 and each Fj is a field. Let η1 be the
nilpotency of m1 and m ≥ 3. Then γ(AG(R)) > 1.
Theorem 4.18. Let R be a commutative Artinian ring such that R = R1×R2×· · ·×Rn×
F1 × F2 × · · · × Fm, where each (Ri,mi) is an Artinian local ring with mi ̸= 0 and each
Fj is a field. Let ηi be the nilpotency of mi and n ≥ 2 or m ≥ 3. Then γ(EG(R)) > 1.
Proof. Follows from Theorems 4.16 and 4.17.
Theorem 4.19. Let R be a commutative Artinian ring such that R = R1×F1×F2, where
(R1,m) is an Artinian local ring and F1 and F2 are fields. Let η be the nilpotency of m.
Then γ(EG(R)) = 1 if and only if η = 2 and m is the only non-trivial ideal of R1.
Proof. Suppose that η = 2 and m is the only non-trivial ideal of R1. Then from Figure 5,
we get γ(EG(R)) = 1, where a = m× (0)× (0), b = R1 × (0)× (0), c = m× F1 × F2,
d = (0) × F1 × F2, e = m × (0) × F2, f = (0) × F1 × (0), g = R1 × F1 × (0),
h = R1 × (0)× F2, i = (0)× (0)× F2, j = m× F1 × (0).
Conversely, assume that γ(EG(R)) = 1. Let J be a non-trivial ideal of R1 such that J ̸= m.
Then the set {m× (0)× (0),m× F1 × (0), J × F1 × (0), (0)× F1 × (0)} ∪ {J × (0)×
(0),m× (0)× F2, J × (0)× F2, (0)× (0)× F2, R1 × (0)× (0)} induces a copy of K4,5
in EG(R), which is a contradiction. Hence m is the only non-trivial ideal of R1.
Theorem 4.20. Let R be a commutative Artinian ring such that R = R1 × F , where
(R1,m) is an Artinian local ring and F is a field. Let η be the nilpotency of m. Then
γ(EG(R)) = 1 if and only if one of the following holds:
M. Nazim and N. Rehman: On the essential annihilating-ideal graph of commutative rings 455
m1 ×m2 m1 ×m2
m1 ×m2 m1 ×m2
(0)×m2
m1 ×R2
m1 ×R2
(0)×R2 (0)×R2
R1 ×m2
R1 ×m2
R1 × (0) R1 × (0)
m1 × (0)
Figure 4: Toroidal embedding of EG(R1 × R2), where mi is the only non-trivial ideal of
Ri for i = 1, 2.
a a
aa
b
f f
e e
i
i
g
h
j
j
d c
Figure 5: Toroidal embedding of EG(R1 × F1 × F2), where m is the only non-trivial ideal
of R1.
(1) η = 3 and m and m2 are the only non-trivial ideals of R1.
(2) η = 4 and m, m2 and m3 are the only non-trivial ideals of R1.
Proof. Suppose that γ(EG(R)) = 1. If η ≥ 5, then the set {mη−1 × (0),mη−2 ×
(0),mη−3 × (0)}∪{R1 × (0),m× (0), (0)×F,mη−1 ×F,mη−2 ×F,mη−3 ×F,m×F}
456 Ars Math. Contemp. 22 (2022) #P3.05 / 443–458
m2 × Fm2 × (0)
m× (0)
(0)× F
(0)× F
R1 × (0) R1 × (0)
R1 × (0)R1 × (0)
m× F
Figure 6: Toroidal embedding of EG(R1 × F ), where m and m2 are only non-trivial ideals
of R1.
induces a copy of K3,7. Thus, by Lemma 4.12, γ(EG(R)) > 1, a contradiction. Hence
η ≤ 4. We have following cases:
Case I: η = 2. Let J be a non-trivial ideal of R1 such that J ̸= m. Then by Lemma 4.7,
R1 has at least three non-trivial ideals I1, I2 and I3 such that I1, I2, I3 ̸= m. We can see
that the set {R1×(0), J×(0), I1×(0), I2×(0)}∪{(0)×F, J×F, I1×F, I2×F,m×F}
induces a copy of K3,7 in EG(R), a contradiction. Hence m is the only non-trivial ideal of
R1. It follows from Theorem 4.10 that EG(R) is a planar graph, a contradiction.
Case II: η = 3. Let I be a non-trivial ideal of R1 such that I ̸= m,m2. Then by Lemma 4.7,
R1 has at least three non-trivial ideals I1, I2 and I3 such that I1, I2, I3 ̸= m,m2. It is easy
to see that the set {R1 × (0),m × (0),m2 × (0)} ∪ {I × (0), I1 × (0), I2 × (0), 0 × F,
m × F,m2 × F, I × F} induces a copy of K3,7 in EG(R), a contradiction. Hence m and
m2 are the only non-trivial ideals of R1.
Case III: η = 4. Let I be a non-trivial ideal of R1 such that I ̸= mi for each
i = 1, 2, 3. Then by Lemma 4.7, R1 has at least three non-trivial ideals I1, I2 and I3
such that I1, I2, I3 ̸= mi for each i = 1, 2, 3. It is easy to see that the set {m × (0),
m2 × (0),m3 × (0)} ∪ {R1 × (0), I × (0), I1 × (0), I2 × (0),m×F,m2 × (0),m3 × (0)}
induces a copy of K3,7 in EG(R), a contradiction. Hence m, m2 and m3 are the only non-
trivial ideals of R1.
Conversely, if m and m2 are the only non-trivial ideals of R1, then |A∗(R)| = 6 and the
set {R1 × (0),m× (0),m2 × (0)} ∪ {(0)× F,m× F,m2 × F} induces a copy of K3,3 in
EG(R). Thus, K3,3 ≤ EG(R) ≤ K6, which implies that γ(EG(R)) = 1.
Now, if m, m2 and m3 are the only non-trivial ideals of R1. Then from Figure 7,
γ(EG(R)) = 1.
M. Nazim and N. Rehman: On the essential annihilating-ideal graph of commutative rings 457
m2 × F
m2 × F
m2 × (0)
m2 × (0)
(0)× F (0)× F
m3 × (0)
m× (0) m3 × F
R1 × (0)
R1 × (0) R1 × (0)
R1 × (0)
m× F m× F
Figure 7: Toroidal embedding of EG(R1 × F ), where m, m2 and m3 are non-trivial ideals
of R1.
ORCID iDs
Mohd Nazim https://orcid.org/0000-0001-8817-4336
Nadeem ur Rehman https://orcid.org/0000-0003-3955-7941
References
[1] G. Aalipour, S. Akbari, M. Behboodi, R. Nikandish, M. J. Nikmehr and F. Shaveisi, The clas-
sification of the annihilating-ideal graphs of commutative rings, Algebra Colloq. 21 (2014),
249–256, doi:10.1142/s1005386714000200.
[2] G. Aalipour, S. Akbari, R. Nikandish, M. Nikmehr and F. Shaveisi, On the coloring of the
annihilating-ideal graph of a commutative ring, Discrete Math. 312 (2012), 2620–2626, doi:
10.1016/j.disc.2011.10.020.
[3] G. Aalipour, S. Akbari, R. Nikandish, M. J. Nikmehr and F. Shaveisi, Minimal prime ideals
and cycles in annihilating-ideal graphs, Rocky Mountain J. Math. 43 (2013), 1415–1425, doi:
10.1216/rmj-2013-43-5-1415.
[4] M. Ahrari, S. A. S. Sabet and B. Amini, On the girth of the annihilating-ideal graph of a
commutative ring, J. Linear Topol. Algebra 4 (2015), 209–216.
[5] S. Akbari and A. Mohammadian, On the zero-divisor graph of a commutative ring, J. Algebra
274 (2004), 847–855, doi:10.1016/s0021-8693(03)00435-6.
[6] F. Aliniaeifard, M. Behboodi, E. Mehdi-Nezhad and A. M. Rahimi, On the diameter and girth
of an annihilating-ideal graph, 2014, arXiv:1411.4163v1 [math.CO].
[7] D. D. Anderson and M. Naseer, Beck’s coloring of a commutative ring, J. Algebra 159 (1993),
500–514, doi:10.1006/jabr.1993.1171.
[8] D. F. Anderson and P. S. Livingston, The zero-divisor graph of a commutative ring, J. Algebra
217 (1999), 434–447, doi:10.1006/jabr.1998.7840.
458 Ars Math. Contemp. 22 (2022) #P3.05 / 443–458
[9] M. F. Atiyah and I. G. Macdonald, Introduction to Commutative Algebra, Addison-Wesley
Publishing Co., Reading, Mass.-London-Don Mills, Ont., 1969.
[10] I. Beck, Coloring of commutative rings, J. Algebra 116 (1988), 208–226, doi:10.1016/
0021-8693(88)90202-5.
[11] M. Behboodi and Z. Rakeei, The annihilating-ideal graph of commutative rings I, J. Algebra
Appl. 10 (2011), 727–739, doi:10.1142/s0219498811004896.
[12] M. Behboodi and Z. Rakeei, The annihilating-ideal graph of commutative rings II, J. Algebra
Appl. 10 (2011), 741–753, doi:10.1142/s0219498811004902.
[13] F. R. DeMeyer, T. McKenzie and K. Schneider, The zero-divisor graph of a commutative semi-
group, Semigroup Forum 65 (2002), 206–214, doi:10.1007/s002330010128.
[14] D. Dolžan and P. Oblak, The zero-divisor graphs of rings and semirings, Internat. J. Algebra
Comput. 22 (2012), 1250033, 20, doi:10.1142/s0218196712500336.
[15] T. Y. Lam, A First Course in Non-Commutative Rings, Springer-Verlag, New York, 1991, doi:
10.1007/978-1-4684-0406-7.
[16] M. J. Nikmehr and S. M. Hosseini, More on the annihilator-ideal graph of a commutative ring,
J. Algebra Appl. 18 (2019), 1950160, 14, doi:10.1142/s0219498819501603.
[17] M. J. Nikmehr, R. Nikandish and M. Bakhtyiari, On the essential graph of a commutative ring,
J. Algebra Appl. 16 (2017), 1750132, 14, doi:10.1142/s0219498817501328.
[18] S. P. Redmond, The zero-divisor graph of a non-commutative ring, in: Internat. J. Commutative
Rings, Nova Sci. Publ., Hauppauge, NY, volume 1, 2002.
[19] K. Selvakumar and P. Subbulakshmi, Classification of rings with toroidal annihilating-ideal
graph, Commun. Comb. Optim. 3 (2018), 93–119, doi:10.22049/cco.2018.26060.1072.
[20] K. Selvakumar, P. Subbulakshmi and J. Amjadi, On the genus of the graph associated to
a commutative ring, Discrete Math. Algorithms Appl. 9 (2017), 1750058, 11, doi:10.1142/
s1793830917500586.
[21] D. B. West, Introduction to Graph Theory, Prentice-Hall of India, New Delhi, 2002.
[22] A. T. White, Graphs, Groups and Surfaces, North-Holland Mathematics Studies, No. 8, North-
Holland Publishing Co., Amsterdam, 1973.
ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 22 (2022) #P3.06 / 459–476
https://doi.org/10.26493/1855-3974.2646.c07
(Also available at http://amc-journal.eu)
Cell reducing and the dimension
of the C1 bivariate spline space
Gašper Jaklič
FGG and IMFM, University of Ljubljana, and
IAM, University of Primorska
Received 1 June 2021, accepted 8 November 2021, published online 9 June 2022
Abstract
In this paper, the problem of determining the dimension of the space S1n(△), n ≥ 3 of
bivariate C1 splines of degree ≤ n over a triangulation △ is considered. The piecewise
polynomials are represented as blossoms, and the smoothness conditions are written as
a system of linear equations. The rank of the system matrix is analysed by repeatedly
reducing small subtriangulations (cells) at the boundary of a triangulation. It is shown that
the dimension of the bivariate spline space S1n(△), n ≥ 3 is equal to Schumaker’s lower
bound for a large class of triangulations.
Keywords: Dimension, spline space, triangulation, cell.
Math. Subj. Class. (2020): 65D05, 65D07, 65D17, 15A03
1 Introduction
In the last 40 years the problem of determining the dimension of the bivariate spline space
has received a considerable attention. For a given triangulation △ of a polygonal region
Ω ⊂ R2 with N triangles Ωi, the bivariate spline space of degree n and smoothness r is
defined as
Srn(△) := {f ∈ Cr(Ω); f |Ωi ∈ Πn(R2), i = 1, 2, . . . , N},
where Πn(R2) denotes the space of bivariate polynomials of total degree ≤ n. In contrast
to the univariate case, the bivariate spline space has a much more complex structure and
even such basic problems as determining its dimension or construction of its basis are
surprisingly hard to tackle. Even more surprising is the fact that the “simplest” spaces of
splines of the lowest degrees are the most complex. For example, for the most interesting
E-mail address: gasper.jaklic@fgg.uni-lj.si (Gašper Jaklič)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
460 Ars Math. Contemp. 22 (2022) #P3.06 / 459–476
case - the space of cubic C1 splines S13(△), quite frequently used in practical applications,
the dimension is still unknown in general, even though a great deal of research has been
done on the topic.
But it is essential that the dimension is known in advance in some important applica-
tions, in particular for Lagrange interpolation by bivariate splines.
In general, the problem has been solved for a spline space of degree n and smoothness
r over a regular triangulation △, Srn(△), where the degree n is large in comparison to the
smoothness r (n ≥ 3r + 2 ([6]), n = 4, r = 1 ([1])). Recall that a triangulation is regular,
if two adjacent triangles Ωi,Ωj can have only one vertex or the whole edge in common.
The dimension of the spline space S13(△) is known for particular classes of triangula-
tions only [4, 7, 5], etc. It has been conjectured that the dimension is equal to Schumaker’s
lower bound ([12, 13])
dimS13(△) ≥ 3VB(△) + 2VI(△) + σ(△) + 1, (1.1)
where VB(△) denotes the number of boundary vertices, VI(△) the number of internal
vertices, and σ(△) =
∑VI(△)
i=1 σi,
σi =
{
1, if vertex is singular,
0, otherwise.
A vertex is singular if it is obtained as an intersection of exactly two lines.
Suppose that a triangulation △ consists of a set of triangles that all have one common
vertex v. Suppose every triangle in △ has at least one neighbour with which it shares a
common edge. Then we call △ a cell. If v is an interior vertex, then △ is an interior cell,
otherwise it is a boundary cell (see [11]). Cell degree is the degree (valency) of the vertex
v.
The main obstacle in the study of the dimension problem is the fact that the dimension
depends not only on the topology of the triangulation △ but also on its geometry. It has
been conjectured (see [14]) that the dimension is equal to Schumaker’s lower bound for
n ≥ 2r + 1 and that the dimension jump occurs only for singular vertices.
Various approaches have been applied to tackle the dimension problem (tools from
linear algebra, algebraic topology, graph theory, symbolic computation and computer aided
design), but the problem is very hard. It has been compared even with the well-known Four
Color Map Problem.
For the space of cubic C1 splines, the dimension has been determined for special tri-
angulations only (triangulations of type 1 and 2 ([12, 13]), nested polygon triangulations
([4]), reducible triangulations ([7]), etc.). Dimension is known also for generic triangula-
tions, i.e., such triangulations, that if the dimension of the spline space exceeds the lower
bound, a small perturbation of vertices of the triangulation causes the dimension to match
the lower bound (see [2, 15], and more elementary proof in [11]). In all the cases the
dimension equals the lower bound (1.1).
In this paper, the blossoming approach is used (see [3, 7]). The idea is to study the
smoothness conditions between polynomial patches, written as their blossoms ([10]). This
is a dual approach to the well known classical approach (see [11], e.g.) and brings a new
insight to the dimension problem. An overview of cell reduction at the boundary of the
triangulation is given. Thus sufficient conditions for an inductive approach for determin-
ing whether the dimension of S1n(△), n ≥ 3, is equal to Schumaker’s lower bound for
G. Jaklič: Cell reducing and the dimension of the C1 bivariate spline space 461
a large class of triangulations △ are obtained. It is shown that interior cells of degrees
k = 4, 5, . . . , 8 can be tackled, but the reduction can be applied only in the case k = 4
and for special cases for k = 5. For k = 6, 7, 8, a negative result is proven. Furthermore,
interior cells with more than 2 free boundary edges are studied. Since it is possible to re-
duce most of the cases by methods for boundary cells, we focus the study to the cases with
collinearities. It is proven that a cell of degree 4 with 1 common edge and a cell of degree
5 with 2 common edges with the rest of the triangulation can be reduced.
An algorithm that extends the results of [7] is presented, and it is proven that the results
can be generalized to S1n(△), n > 3.
The structure of the paper is as follows. In Section 2, an overview of the blossoming
approach to the dimension problem is given. In Section 3, the cell reduction is studied
and the main results of the paper are presented. In Section 4, an efficient algorithm for
determining whether a given triangulation belongs to the observed class of triangulations is
presented. Section 5 extends the results from the cubic (S13(△)) to the general case n ≥ 3
(S1n(△)). The paper is concluded by a proof of one of the main results.
2 Blossoming approach to the dimension problem
First, let us recall the blossoming approach. It is well known that there exists a bijective
correspondence between a bivariate polynomial and its blossom: for every polynomial
p ∈ Πn(Rm) there exists a unique symmetric n-affine polynomial
Bn(p)
(
x(1), x(2), . . . , x(n)
)
, x(i) ∈ Rm,
with a diagonal property
Bn(p)
(
x, x, . . . , x︸ ︷︷ ︸
n
)
= p(x), x ∈ Rm.
The polynomial Bn(p) is called the blossom of the polynomial p.
For example, the blossom of the polynomial
p2(x, y) = a0 + a1x+ a2y + a3x
2 + a4xy + a5y
2
is
B2(p2)
(
(u1, v1), (u2, v2)
)
=
a0 + a1
u1 + u2
2
+ a2
v1 + v2
2
+ a3u1u2 + a4
u1v2 + u2v1
2
+ a5v1v2.
The blossom generalizes a given polynomial. One can study the smoothness conditions
between polynomial patches over adjacent triangles, expressed in the blossoming form. In
order to determine the dimension of the spline space, smoothness conditions over all inner
edges of the triangulation need to be studied.
By studying the dual representation of a triangulated graph [3], it can be seen that some
smoothness conditions are independent and can be omitted. The study of the rest results
into the dimension equation
dimSrn(△) = N
(
n+ 2
2
)
−
(
r + 2
2
)
(EI − VI)− rankMn. (2.1)
462 Ars Math. Contemp. 22 (2022) #P3.06 / 459–476
Here N denotes the number of triangles, EI the number of inner edges, and VI the number
of inner vertices of the triangulation. The matrix Mn describes the smoothness relations
that need to be studied. It has a particular structure:
Mn := (Mkm)
n−r;n−r
k=1;m=1 := (Mr,km)
n−r;n−r
k=1;m=1 (2.2)
is a block upper triangular matrix with blocks Mkm of size (r+1)EI×(m+r+1)(N−1).
The matrix Mkm is also a block matrix with EI block rows and N − 1 block columns: ℓ-
th block row corresponds to smoothness conditions across the edge eℓ = (i, j) between
triangles Ωi and Ωj , and it has at most two nonzero blocks Qℓi := Qkm,ℓi, Qℓj :=
Qkm,ℓj , with Qkm,ℓi+Qkm,ℓj = 0. Blocks Qkm,ℓi are circulant matrices of size (r+1)×
(m+ r + 1). Theirs first row is defined by(
n− r
k
) ∑
|γ|=k
(
k
γ
)(
n− r − k
β − γ
)
vγℓ u
β−γ
ℓ = f ·
∑
|γ|=k
(
k
γ
)(
m− k
β − γ
)
vγℓ u
β−γ
ℓ ,
β = (b1,m− b1), b1 = m,m− 1, . . . , 0,
with
f := fn,r,k,m =
(
n− r − k
m− k
)
.
The rest r elements of the row are zeros. Here uℓ denotes an arbitrary point on the edge
eℓ with the normalized directional vector vℓ. The standard multiindex notation is used. For
more details, see [3]. In the following section, the matrix Mn for the case n = 3 will be
described in detail.
Thus the main problem is how to determine the rank of a large symbolic matrix Mn that
depends on the geometry and the topology of a triangulation. Such a problem is very hard
to tackle in general. A natural idea is to reduce the problem to a smaller one, if particular
assumptions are satisfied.
3 Cell reduction
The idea of the blossoming approach to the dimension problem is to inductively reduce the
problem from the given triangulation △ to its subtriangulation △\△1 for a proper choice
of the subtriangulation △1 (see Figure 1).
Let B denote the intersection of △1 and △\△1. We call the subtriangulation △1 proper,
if it is simply connected, has a vertex T0 on the outer face of △ and contains all the triangles
in △ with the vertex T0, no triangle in △1 has two edges on B, there are no consecutive
pairs of collinear edges at the vertices of degree 3 in △1 on B, and every singular vertex in
△1 lies in the interior of △1.
Let vℓ = (αℓ, βℓ) denote a normalized directional vector of the edge eℓ between trian-
gles Ωi and Ωj . Further, let
vi × vj := αiβj − αjβi
be the planar vector product. The matrix M := Mn could be written as
M = M(△) =
 M(△1) 0M(△1,△) M(△,△1)
0 M(△\△1)
 ,
G. Jaklič: Cell reducing and the dimension of the C1 bivariate spline space 463
Figure 1: Reduction of a triangulation.
where the matrices M(△1,△),M(△,△1) represent the common part of the smoothness
conditions between △1 and △\△1, M(△1) represents the conditions inside △1, and
M(△\△1) the conditions inside △\△1. Let
M̃(△1,△) :=
[
M(△1)
M(△1,△)
]
∈ Rr×c.
Theorem 3.1 ([7, Theorem 1]). Suppose that △1 is a proper subtriangulation of △ that
satisfies VB(△1) ≤ 2VI(△1) + 6. If
rank M̃(△1,△) = r − σ(△1) (3.1)
and dimS13(△\△1) is equal to the lower bound, then the dimension dimS13(△) is equal
to the lower bound (1.1) too.
The matrices M̃(△1,△) that have to be studied, have a block structure, based on the
incidence matrix of the underlying graph of the triangulation. They consist of blocks
Q11,ℓi = −Q11,ℓj =
[
αℓ βℓ 0
0 αℓ βℓ
]
,
Q22,ℓi = −Q22,ℓj =
[
α2ℓ 2αℓβℓ β
2
ℓ 0
0 α2ℓ 2αℓβℓ β
2
ℓ
]
,
and blocks Q12,ℓj that depend not only on the directions but also on the vertices of the
triangulation and some arbitrary additional points. More precisely, by using [3, Lemma 3.1]
it is possible to simplify some of the blocks in M12 without changing the rank of M by
choosing the points tℓ := (cℓ, dℓ) as the inner vertices of the triangulation, and by choosing
some arbitrary additional points zk := (xk, yk) ∈ R2, k = 1, 2, . . . , N for the faces Ωk.
The block Q12,ℓi is of the form[
αℓ(cℓ − xi) αℓ(dℓ − yi) + βℓ(cℓ − xi) βℓ(dℓ − yi) 0
0 αℓ(cℓ − xi) αℓ(dℓ − yi) + βℓ(cℓ − xi) βℓ(dℓ − yi)
]
, (3.2)
and the block Q12,ℓj reads
−
[
αℓ(cℓ − xj) αℓ(dℓ − yj) + βℓ(cℓ − xj) βℓ(dℓ − yj) 0
0 αℓ(cℓ − xj) αℓ(dℓ − yj) + βℓ(cℓ − xj) βℓ(dℓ − yj)
]
,
(3.3)
464 Ars Math. Contemp. 22 (2022) #P3.06 / 459–476
so the choice zk = tℓ, k ∈ {i, j} reduces (3.2) or (3.3) to zero block. Of course, not all
blocks can be simplified in this way.
The following theorem gives conditions on boundary cells △1 where the reduction can
be applied.
Theorem 3.2 ([7, Theorem 2]). Let △1 be a proper subtriangulation of △ with a vertex
T0 of degree s+ 1 on the outer face of △. If
(1) s ≤ 4, |△1| = s, or
(2) s = 2, 3, |△1| = s + 2, and △1 includes an interior cell, and dimS13(△\△1) is
equal to the lower bound, then the dimension dimS13(△) is equal to the lower bound
(1.1) too.
Therefore, as a natural extension, it is interesting to study interior cell reduction at the
boundary of the triangulation △. Usually, the main obstacle are collinear edges. Those
are the only special cases for n ≥ 3r + 2 and for known results for n = 2r + 1. For
n = 2r, there exist other geometric configurations, that result in a change of the dimension
(Morgan-Scott triangulation [3], e.g.).
By studying interior cells, the collinear edges can be included in the subtriangulation
△1. This will enable us to study the dimension problem on a wider class of triangulations.
The conditions of Theorem 3.1 allow us to study cells of degree k, k = 4, 5, . . . , 8. In
light of the previous discussion, the study will be limited to particularly interesting cells
where there is a collinearity of edges between the inner vertex and the vertices, adjacent to
the boundary vertex.
Theorem 3.3. Let △1 be an interior cell of degree k, k = 5, 6, 7, 8, and let △1 be a proper
subtriangulation of the triangulation △ with a vertex T0 of degree 3 at the outer face of
the triangulation △ and an inner vertex T1. Let the edges adjacent to T1 be denoted in
clockwise direction as e1 = T0T1, e2, . . . , ek (see Figures 3, 5, 6, 7). Let there be the
collinearity e2||ek.
(1) If k = 5 and there is a collinearity of edges e1||e3 or e1||e4 (Figure 3), and
dimS13(△\△1) is equal to Schumaker’s lower bound, then also the dimension
dimS13(△) is equal to the lower bound (1.1).
(2) If k = 6, 7, 8, then Theorem 3.1 can not be applied (Figures 5, 6, 7).
Proof. Proof of the theorem is given as the last section of the paper.
By Theorem 3.1 we have to study ranks of certain matrices M̃(△1,△) that belong
to cells considered. Ranks will be obtained by studying appropriate minors. Since the
matrices are large and symbolic, symbolic computer algebra tools and [9] will be used for
the computation of determinants and the simplification of huge symbolic expressions. Here,
properties of triangulation’s topology and geometry have to be used very carefully, since
otherwise computations are futile because of the huge time and memory requirements and
enormous symbolic expressions. Note that the determinants obtained can be easily verified
by evaluating the polynomials in enough number of points, exact numerical determinant
calculation, and the well-known results on multivariate Lagrange polynomial interpolation
(see [8, Lemma 1]).
G. Jaklič: Cell reducing and the dimension of the C1 bivariate spline space 465
Remark 3.4. Interior cells of degree 4 and boundary cells of degree ≤ 5 are covered by
Theorem 3.2.
Remark 3.5. The study of interior cells in the general position is much more difficult.
1. For k = 5, 6, the appropriate minors can be computed, but the obtained polynomial
expressions are difficult to simplify enough to be able to prove that the polynomial
is nonzero for all geometrically admissible edge directions and vertex positions.
2. For k = 7 and k = 8 also in the general case the appropriate matrices are not of the
full rank and the considered cell reducing approach can not be applied. The proof of
Theorem 3.3 holds in general, since the collinearity of the edges is not used.
Remark 3.6. In general, it is interesting to consider interior cells of degree k at the bound-
ary of triangulation, where there are ℓ = 1, 2, . . . , k − 2 edges in the common border
B. It can be easily seen from (3.1), that the matrix dimensions limit the consideration to
ℓ ≤ min{k − 2, ⌊3/4k⌋}. It is important to notice that most of the cases can be reduced
by methods for boundary cells (Theorem 3.3). As expected, the most interesting cases with
collinearities are left for the study.
Theorem 3.7. For interior cells of small degrees k = 4, 5, 6 there is only one problematic
configuration, namely, an interior cell with ℓ = k − 3 edges in B and two collinearities in
the interior. For k = 4, 5 such configurations can be reduced.
Proof. For k = 4, by [7] the matrix obtained is not of full rank because of the interior
singular vertex. Consider Figure 2. Let T1 = (c1, d1) be the interior point of the cell,
and T2 = (c2, d2) be an additional point on B, the join of e5 and e6. Further, let w6 :=
(β6,−α6). The minor, obtained by omitting rows, corresponding to the boundary edge e5
and additionally row 1 and columns 1, 2, 3, 4, 13, 14, 15, 16, 26, is
α3α6β
2
4β
2
6(v3 × v4)9(v6 × v3)2(β6(−c1 + c2) + α6(d1 − d2)),
which is nonzero, since (β6(−c1 + c2) +α6(d1 − d2)) = ⟨w6, T1 − T2⟩ and w6 is perpen-
dicular to v6 = (α6, β6).
Figure 2: An interior cell of degree 4 at the boundary of the triangulation.
For k = 5 we similarly first omit the block rows and columns belonging to e8 (see
Figure 3), apply the same simplification as in the proof of [7, Theorem 2], and then omit
466 Ars Math. Contemp. 22 (2022) #P3.06 / 459–476
Figure 3: A cell of degree 5 at the boundary of the triangulation.
the columns 15, 16, 19, 20 and further the columns 3, 14, 15, 16, 18, 19, 24 in the obtained
matrix. The resulting minor is
6g2α1α
3
2α
2
7β
2
1β
2
2(v2 × v1)2(v6 × v3)5(v7 × v3)3(v7 × v6)(v1 × v8)(v2 × v8). (3.4)
Here g denotes the length of T2 − T1 = gv3. Since by assumption the cell is proper,
v7 × v6 ̸= 0, thus the expression (3.4) is nonzero.
Remark 3.8. For interior cells of higher degrees k ≥ 6 the matrices obtained are unfortu-
nately too large to study with current computational facilities.
Numerical computations suggest that in the case k = 5 the interior cell reduction can
be applied for all but one configuration. Unfortunately this seems hard to prove, and we
will state it as the following conjecture with a discussion on its possible proof.
Conjecture. Let k = 5 and let the assumptions of Theorem 3.3 be fulfilled. Then the cell
reduction can be applied for almost every position of edges of the interior cell. There exists
a unique configuration of edges where the reduction can not be used.
A computation of all possible
(
35
32
)
= 6545 minors of size 32 and a careful simplifica-
tion of nonzero expressions reveals that all contain the same linear expression in α1 and β1.
This results in a unique condition on the positions of the edges of the cell, where the matrix
considered, M̃(△1,△), is not of full rank. This configuration is geometrically admissible.
Thus the reduction of the interior cell can not be applied in this special case. Unfortu-
nately, the symbolic polynomial expressions are huge and it is not feasible to write them
down. Thus it is difficult to find a proper minor and prove that it is nonzero for all possible
geometrically admissible edge directions, with the exception of the considered one.
4 An algorithm for the reduction of the triangulation
By using Theorem 3.2, Theorem 3.3 and Theorem 3.7, we can construct an algorithm that
determines if the triangulation △ belongs to the class of triangulations where the dimension
dimS13(△) can be obtained by sequential reductions of the triangulation △. This algorithm
improves the algorithm in [7].
We are given a triangulation △. First, it is rotated to a general position, such that no
edge lies on the coordinate axes. Then the reduction step can be used on every boundary
G. Jaklič: Cell reducing and the dimension of the C1 bivariate spline space 467
vertex that satisfies one of the conditions of Theorem 3.2, Theorem 3.3, item 1 or Theo-
rem 3.7. In the algorithm, let B denote the intersection of a current boundary or interior
cell △1 and △\△1. Let |B| denote the number of edges in B.
An algorithm for the reduction of the triangulation
// T - a given triangulation
// deg(v) - degree of the vertex v
list<vertex> L = {list of all vertices of degree <=5 on the
outer face of the triangulation T};
while (!L.empty()) {
if (T= interior cell or T=triangle) break(success);
//check for possible reductions using L
v=L.pop();
check for collinearities between neighbours of v in B;
//boundary cells
if (no collinearities && all neighbours of v are in B){
T.delete_vertex(v);
if any neighbour(v) has new degree <=5, add it to L;
}
//interior cells
else if ((collinearity at the inner vertex z, deg(z)=4,
deg(v)<5), |B|=1 or 2 ||
(collinearity at the inner vertex z, deg(v)=3, deg(z)=5,
edge vz is collinear with another edge from z,
|B|=2 or 3)){
T.delete_vertex(v);
T.delete_vertex(z);
add vertices in B with new degree <=5 to L;
}
else {
L.append(v);
}
}
if (success) {
print("Dimension equals Schumaker’s lower bound");
}
else {
print("Algorithm can not be used");
}
}
If all the vertices in the list L were checked, and no reduction could be applied, the list
L remains the same, and the algorithm stops. In such a case (because of too large degrees
of the boundary vertices or because of the collinearities) this method can not be applied
for the study of the dimension. If the while loop successfully terminates, the dimension
dimS13(△) is equal to Schumaker’s lower bound (1.1). The answer of the algorithm is
quite clearly independent of the enumeration of the vertices, i.e., on a particular sequence
468 Ars Math. Contemp. 22 (2022) #P3.06 / 459–476
of reductions.
As an example, consider the triangulation in Figure 4. It is derived from a nested
polygonal configuration (see [4] and an example in [7]). Because of a slight modification
(deletion of some vertices and edges, and perturbation of vertices), the tools in [4] can not
be applied. Similarly, the algorithm from [7] can not tackle it because of collinearities
along reduction boundaries at all boundary vertices. But the algorithm, presented in this
paper, enables the reduction of the triangulation. Thus the dimension of the C1 cubic spline
space on the triangulation in Figure 4 is equal to Schumaker’s lower bound.
0 1 2
3 4 5
6 7 8
10 11 12
141516
Figure 4: A triangulation △, where the algorithm determines dimS13(△).
5 Generalization to n ≥ 3
In [3], it has been shown how particular triangulations can be tackled under the assumption,
that no inner edges that share a common vertex have the same slope. We will show that this
assumption can be omitted, and apply this to generalize our results to n ≥ 3. For the sake
of completeness, we will include the relevant results from [3].
First, let us recall Schumaker’s lower bound in general,
dimSrn(△) ≥ LBrn(△) :=
(
n+ 2
2
)
+
(
n− r + 1
2
)
EI− (5.1)((
n+ 2
2
)
−
(
r + 2
2
))
VI +
VI∑
i=1
σi,
where
σi =
n−r∑
j=1
(r + j + 1− j ei)+, i = 1, 2, . . . , VI , (5.2)
and EI denotes the number of interior edges, VI the number of inner vertices, and ei the
number of edges with distinct slopes in an inner vertex vi.
Let σi(n) denote the number σi, defined in (5.2), that belong to the vertex i and the
considered spline space Srn(△).
Lemma 5.1. Let n > 2r. Then σi(n) = σi(2r).
G. Jaklič: Cell reducing and the dimension of the C1 bivariate spline space 469
Proof. The number σi(n) can be written as
σi(n) =
n−r∑
j=1
(r + j + 1− j ei)+ = σi(2r) +
n−r∑
j=r+1
(r + j + 1− j ei)+,
where ei is the number of edges with different edge slopes at the inner vertex i. Note that
in the second term j ≥ r + 1. Now let us consider the expression
r + j + 1− j ei = r + 1− j(ei − 1) ≤ r + 1− (r + 1)(ei − 1) = (r + 1)(2− ei).
Since for the inner vertices of the triangulation in general position ei ≥ 3 and for the
singular vertices ei = 2,
(r + 1)(2− ei) ≤ 0.
Thus σi does not change if the polynomial degree increases.
Now we can show that [3, Theorem 3.2] and [3, Theorem 3.3] hold also for triangula-
tions that contain collinearities.
Recall the definition of the matrix Mn in (2.2).
Theorem 5.2 ([3, Theorem 3.2]). Let △ be a regular triangulation and n ≥ n0 ≥ 2r.
Then
rankMn ≥ rankMn0 + (r + 1)EI(n− n0). (5.3)
Proof. The matrix Mn can be written as
Mn =
[
Mn−1 X
0 Mn−k,n−k
]
.
Clearly, rankMn ≥ rankMn−1 + rankMn−k,n−k. We need to prove that
rankMn−k,n−k = (r + 1)EI ,
i.e., the rank of the matrix Mn−k,n−k is equal to the number of rows. The number of
columns of Mn−k,n−k is equal to
(n+ 1)(N − 1) ≥ 2(r + 1)(N − 1) ≥ 4
3
(r + 1)EI > (r + 1)EI ,
since 3N ≥ 2EI + 3, thus it suffices to prove, that the rows are linearly independent.
Suppose that the rows are not independent. Then there exists a vector x ∈ R(r+1)EI ,
x ̸= 0, such that xTMn−k,n−k = 0. Since the rows that correspond to boundary triangles
with only one inner edge are clearly independent, the corresponding components of the
vector x are zero. Thus we can assume, that all boundary triangles have only one outer
edge. Let Ωi be such a boundary triangle in △. Then eℓ1 = (j1, i), eℓ2 = (i, j2) for some
j1, j2, and the block matrices Qℓ1i, Qℓ2i are the only nonzero blocks in the block column
i. Let x|ℓj denote the ℓj-th block row in x. Then
[x|ℓ1 , x|ℓ2 ]T
[
Qℓ1i
Qℓ2i
]
= 0.
470 Ars Math. Contemp. 22 (2022) #P3.06 / 459–476
It can easily be seen that the matrix[
Qℓ1i
Qℓ2i
]
∈ R2(r+1)×(n+1) (5.4)
where n+ 1 ≥ 2r + 1 + 1 = 2(r + 1), is of full rank. Indeed, the rank of the matrix (5.4)
stays unchanged, if n− 2r − 1 zero columns are added. If we append another n− 2r − 1
rows, such that they cyclically continue Qℓ1i, and n− 2r− 1 rows by cyclical continuation
of Qℓ2i, the rank of the matrix (5.4) increases at most by 2(n − 2r − 1). The resulting
matrix is of dimension 2(n − r) × 2(n − r). Its determinant is equal to the resultant of
polynomials
pℓ1(x) := (v1,ℓ1x+ v2,ℓ1)
n−r, pℓ2(x) := (v1,ℓ2x+ v2,ℓ2)
n−r,
where vℓi := (v1,ℓi , v2,ℓi) is the direction of eℓi . Since the directions vℓ1 and vℓ2 are
different, the polynomials pℓ1 and pℓ2 can not have common zeros. Thus theirs resultant
(v1,ℓ1v2,ℓ2 − v2,ℓ1v1,ℓ2)(n−r)
2
is nonzero, hence the matrix (5.4) is of full rank. Thus
x|ℓ1 = x|ℓ2 = 0. So one can study △\{Ωi} only, and use the previous result on the
reduced triangulation. The procedure can be repeated until only one triangle is left. This
concludes the proof.
Theorem 5.3 ([3, Theorem 3.3]). Let △ be a triangulation, and n ≥ n0 ≥ 2r. The
function
θ(n, n0, r) := dimS
r
n(△)−N
((
n+ 2
2
)
−
(
n0 + 2
2
))
+ (r + 1)EI(n− n0)
is nonincreasing function of n, and
LBrn0(△) ≤ θ(n, n0, r) ≤ dimS
r
n0(△). (5.5)
Proof. It is enough to consider n > n0. The first claim follows from (2.1) and Theorem 5.2.
Recall Schumaker’s lower bound (5.1). By Lemma 5.1, σi stays unchanged for any n > n0.
Now it is straightforward to apply (2.1) and (5.3) to obtain (5.5).
This approach has been used in [3] in order to determine the dimension of S2n(△MS),
where △MS denotes the Morgan-Scott triangulation. The key observation is the fact, that
if in (5.5) the right inequality reduces to equality, so does the left. Of course, the main
problem is how to determine rankMn.
From Theorem 5.3 it follows that the results of Theorem 3.2 and Theorem 3.3 hold not
only for the cubic case, but also for spline spaces of higher degrees.
Remark 5.4. The algorithm, given in Section 4, determines whether the dimension
dimS1n(△), n ≥ 3, equals Schumaker’s lower bound for a large class of triangulations
△. If the answer is in the affirmative, dimS13(△) agrees with Schumaker’s lower bound,
and we can apply Theorem 5.3. Of course, for n ≥ 4, the dimension S1n(△) is known
for any triangulation △ (see [1] and [6]). In this case, the blossoming approach does not
yield anything new. It is promising for use on the spline spaces with higher degrees of
smoothness r ≥ 1, as observed for the special case Sr2r(△) in [3].
G. Jaklič: Cell reducing and the dimension of the C1 bivariate spline space 471
6 Proof of Theorem 3.3
Proof. Let us shorten the notation by Qkℓij := Qkℓ,ij . It turns out that instead of considering
all the cells at once, it is more reasonable to choose different points for each case k =
5, 6, . . . , 8 in order to simplify non-diagonal blocks as much as possible. Thus each case
will be studied separately.
First, let us consider the case k = 5. The points on the edges are chosen as e1 : T1,
e2 : T1, e3 : T1, e4 : T1, e5 : T1, e6 : T2, e7 : T2, e8 : T3, and the points for the faces
as Ω1 : T1, Ω2 : T1, Ω3 : T1, Ω4 : T1, Ω5 : T1 (Figure 3). This implies that in the matrix
M12 only three nonzero blocks remain: Q1262, Q
12
73, Q
12
84. The matrix M̃(△1,△) is

Q1111 0 0 0 Q
11
15 0 0 0 0 0
Q1121 Q
11
22 0 0 0 0 0 0 0 0
0 Q1132 Q
11
33 0 0 0 0 0 0 0
0 0 Q1143 Q
11
44 0 0 0 0 0 0
0 0 0 Q1154 Q
11
55 0 0 0 0 0
0 Q1162 0 0 0 0 Q
12
62 0 0 0
0 0 Q1173 0 0 0 0 Q
12
73 0 0
0 0 0 Q1184 0 0 0 0 Q
12
84 0
0 0 0 0 0 Q2211 0 0 0 Q
22
15
0 0 0 0 0 Q2221 Q
22
22 0 0 0
0 0 0 0 0 0 Q2232 Q
22
33 0 0
0 0 0 0 0 0 0 Q2243 Q
22
44 0
0 0 0 0 0 0 0 0 Q2254 Q
22
55
0 0 0 0 0 0 Q2262 0 0 0
0 0 0 0 0 0 0 Q2273 0 0
0 0 0 0 0 0 0 0 Q2284 0

32×35
.
Since
det
[
Q2211
Q2221
]
4×4
= (v2 × v1)4 ̸= 0,
the rows 17, 18, 19, 20, and columns 16, 17, 18, 19, can be omitted without changing the
rank of the matrix M̃(△1,△). If the columns 3, 30 and 31 in the new matrix are omitted,
in the case of the collinearity e1||e3 we obtain the minor
∥e4∥2α62(v2 × v3)(v2 × v4)(v4 × v3)2(v3 × v6)5·
(v7 × v4)2(v6 × v7)(v8 × v4)3(v8 × v7)3 ̸= 0,
and in the case of the collinearity e1||e4 the minor
∥e4∥2α62(v2 × v3)(v2 × v4)(v4 × v3)2(v3 × v6)3·
(v7 × v4)2(v6 × v7)3(v4 × v8)5(v7 × v8) ̸= 0.
Therefore the matrix M̃(△1,△) is of full rank in the considered special cases. The condi-
tions of Theorem 3.1 are fulfilled.
In the case k = 6 we pick the points on the edges as e1 : T1, e2 : T1, e3 : T1,
e4 : T1, e5 : T1, e6 : T1, e7 : T2, e8 : T2, e9 : T3, e10 : T3, and the points for the faces
Ω1 : T1, Ω2 : T1, Ω3 : T1, Ω4 : T1, Ω5 : T1, Ω6 : T1 (Figure 5). This choice implies
that in the matrix M12 only 4 nonzero blocks remain: Q1272, Q
12
83, Q
12
94, Q
12
10,5. The matrix
M̃(△1,△) is
472 Ars Math. Contemp. 22 (2022) #P3.06 / 459–476
Figure 5: A cell of degree 6 at the boundary of the triangulation.

Q1111 0 0 0 0 Q
11
16 0 0 0 0 0 0
Q1121 Q
11
22 0 0 0 0 0 0 0 0 0 0
0 Q1132 Q
11
33 0 0 0 0 0 0 0 0 0
0 0 Q1143 Q
11
44 0 0 0 0 0 0 0 0
0 0 0 Q1154 Q
11
55 0 0 0 0 0 0 0
0 0 0 0 Q1165 Q
11
66 0 0 0 0 0 0
0 Q1172 0 0 0 0 0 Q
12
72 0 0 0 0
0 0 Q1183 0 0 0 0 0 Q
12
83 0 0 0
0 0 0 Q1194 0 0 0 0 0 Q
12
94 0 0
0 0 0 0 Q1110,5 0 0 0 0 0 Q
12
10,5 0
0 0 0 0 0 0 Q2211 0 0 0 0 Q
22
16
0 0 0 0 0 0 Q2221 Q
22
22 0 0 0 0
0 0 0 0 0 0 0 Q2232 Q
22
33 0 0 0
0 0 0 0 0 0 0 0 Q2243 Q
22
44 0 0
0 0 0 0 0 0 0 0 0 Q2254 Q
22
55 0
0 0 0 0 0 0 0 0 0 0 Q2265 Q
22
66
0 0 0 0 0 0 0 Q2272 0 0 0 0
0 0 0 0 0 0 0 0 Q2283 0 0 0
0 0 0 0 0 0 0 0 0 Q2294 0 0
0 0 0 0 0 0 0 0 0 0 Q2210,5 0

40×42
.
Since the submatrix [
Q2211
Q2221
]
,
that belong to the block column for Ω1 in M22, is nonsingular, the block rows for e1, e2 and
the block column for Ω1 in M12 and M22 can be omitted. A matrix of dimension 36× 38
remains. It is enough to compute 6 minors, where 2 of the columns that belong to the last
block column of the matrix are omitted (the rest of the minors are 0 because of the linearly
dependent columns in the last block column). A quick computation shows that all minors
are 0. Therefore the matrix M̃(△1,△) is not of full rank.
In the case k = 7 we choose the points on the edges as e1 : T1, e2 : T1, e3 : T1,
e4 : T1, e5 : T1, e6 : T1, e7 : T1, e8 : T2, e9 : T2, e10 : T3, e11 : T3, e12 :
T4, and the points for the faces Ω1 : T1, Ω2 : T1, Ω3 : T1, Ω4 : T1, Ω5 : T1,
Ω6 : T1, Ω7 : T1 (Figure 6). Therefore, in the matrix M12 only 5 nonzero blocks re-
G. Jaklič: Cell reducing and the dimension of the C1 bivariate spline space 473
Figure 6: A cell of degree 7 at the boundary of the triangulation.
main: Q1282, Q
12
93, Q
12
10,4, Q
12
11,5, Q
12
12,6. The matrix M̃(△1,△) is

Q1111 0 0 0 0 0 Q
11
17 0 0 0 0 0 0 0
Q1121 Q
11
22 0 0 0 0 0 0 0 0 0 0 0 0
0 Q1132 Q
11
33 0 0 0 0 0 0 0 0 0 0 0
0 0 Q1143 Q
11
44 0 0 0 0 0 0 0 0 0 0
0 0 0 Q1154 Q
11
55 0 0 0 0 0 0 0 0 0
0 0 0 0 Q1165 Q
11
66 0 0 0 0 0 0 0 0
0 0 0 0 0 Q1176 Q
11
77 0 0 0 0 0 0 0
0 Q1182 0 0 0 0 0 0 Q
12
82 0 0 0 0 0
0 0 Q1193 0 0 0 0 0 0 Q
12
93 0 0 0 0
0 0 0 Q1110,4 0 0 0 0 0 0 Q
12
10,4 0 0 0
0 0 0 0 Q1111,5 0 0 0 0 0 0 Q
12
11,5 0 0
0 0 0 0 0 Q1112,6 0 0 0 0 0 0 Q
12
12,6 0
0 0 0 0 0 0 0 Q2211 0 0 0 0 0 Q
22
17
0 0 0 0 0 0 0 Q2221 Q
22
22 0 0 0 0 0
0 0 0 0 0 0 0 0 Q2232 Q
22
33 0 0 0 0
0 0 0 0 0 0 0 0 0 Q2243 Q
22
44 0 0 0
0 0 0 0 0 0 0 0 0 0 Q2254 Q
22
55 0 0
0 0 0 0 0 0 0 0 0 0 0 Q2265 Q
22
66 0
0 0 0 0 0 0 0 0 0 0 0 0 Q2276 Q
22
77
0 0 0 0 0 0 0 0 Q2282 0 0 0 0 0
0 0 0 0 0 0 0 0 0 Q2293 0 0 0 0
0 0 0 0 0 0 0 0 0 0 Q2210,4 0 0 0
0 0 0 0 0 0 0 0 0 0 0 Q2211,5 0 0
0 0 0 0 0 0 0 0 0 0 0 0 Q2212,6 0

48×49
.
Figure 7: A cell of degree 8 at the boundary of the triangulation.
In the case k = 8 we choose the points on the edges as e1 : T1, e2 : T1, e3 : T1,
474 Ars Math. Contemp. 22 (2022) #P3.06 / 459–476
e4 : T1, e5 : T1, e6 : T1, e7 : T1, e8 : T1, e9 : T2, e10 : T2, e11 : T3, e12 : T3,
e13 : T4, e14 : T4, and the points for the faces as Ω1 : T1, Ω2 : T1, Ω3 : T1,
Ω4 : T1, Ω5 : T1, Ω6 : T1, Ω7 : T1, Ω8 : T1 (Figure 7). Therefore, in the matrix
M12 only 6 nonzero blocks remain: Q1292, Q
12
10,3, Q
12
11,4, Q
12
12,5, Q
12
13,6, Q
12
14,7. The matrix
M̃(△1,△) is

Q1111 0 0 0 0 0 0 Q
11
18 0 0 0 0 0 0 0 0
Q1121 Q
11
22 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 Q1132 Q
11
33 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 Q1143 Q
11
44 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 Q1154 Q
11
55 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 Q1165 Q
11
66 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 Q1176 Q
11
77 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 Q1187 Q
11
88 0 0 0 0 0 0 0 0
0 Q1192 0 0 0 0 0 0 0 Q
12
92 0 0 0 0 0 0
0 0 Q1110,3 0 0 0 0 0 0 0 Q
12
10,3 0 0 0 0 0
0 0 0 Q1111,4 0 0 0 0 0 0 0 Q
12
11,4 0 0 0 0
0 0 0 0 Q1112,5 0 0 0 0 0 0 0 Q
12
12,5 0 0 0
0 0 0 0 0 Q1113,6 0 0 0 0 0 0 0 Q
12
13,6 0 0
0 0 0 0 0 0 Q1114,7 0 0 0 0 0 0 0 Q
12
14,7 0
0 0 0 0 0 0 0 0 Q2211 0 0 0 0 0 0 Q
22
18
0 0 0 0 0 0 0 0 Q2221 Q
22
22 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 Q2232 Q
22
33 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 Q2243 Q
22
44 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 Q2254 Q
22
55 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 Q2265 Q
22
66 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 Q2276 Q
22
77 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 Q2287 Q
22
88
0 0 0 0 0 0 0 0 0 Q2292 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 Q2210,3 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 Q2211,4 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 Q2212,5 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 Q2213,6 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 Q2214,7 0

56×56
.
The cases k = 7 and k = 8 can be considered simultaneously. Since the submatrix[
Q2211
Q2221
]
,
that belong to the block column for Ω1 in M22, is nonsingular, Gaussian eliminations on
the rows transform it to the identity matrix. Then the eliminations, applied on the columns,
can be used to set all the rest of the elements in the block rows for e1 and e2 in M22 to zero.
This simplifies the matrix, the considered block rows and the block column for Ω1 can be
omitted. In the last block column only the nonzero block Q2277 (Q
22
88) of dimension 2 × 4
remains. In order to choose minors to prove full rank of the matrix, we have to omit two of
the columns in the last block column, otherwise the columns are linearly dependent. But
this is not possible because of the matrix dimensions. For k = 7 one and for k = 8 none of
the critical columns can be removed. Therefore the matrices M̃(△1,△) for k = 7, 8 are
not of full rank.
7 Conclusion
The problem of determining the dimension of the cubic C1 bivariate spline space over
triangulations may seem easy. But for over 40 years it remains unsolved. Various math-
ematical tools were applied, from numerical mathematics, algebra and graph theory. A
G. Jaklič: Cell reducing and the dimension of the C1 bivariate spline space 475
possible way in practice is by introducing triangle subdivision. Unfortunately this modifies
the triangulation and significantly increases the number of triangles, and thus influences
further numerical computations. In this paper an approach by using cell reduction at the
boundary of a triangulation is presented. If the triangulation is reduced to a single triangle,
this proves the dimension result. Otherwise it could be combined with some other method,
that would yield the result for the remaining subtriangulation. Larger interior cells remain
to be analysed, since the applied technique has a very large memory and processor power
requirements.
An another interesting research topic is the study of the space S12(△) and its general-
ization Sr2r(△), r ≥ 1, where very little is known in general. In a more general setting, not
much is known on very complex spline spaces on higher dimensional simplical complexes,
where tools from commutative algebra show a lot of promise.
ORCID iDs
Gašper Jaklič https://orcid.org/0000-0002-9837-7177
References
[1] P. Alfeld, B. Piper and L. L. Schumaker, An explicit basis for C1 quartic bivariate splines,
SIAM J. Numer. Anal. 24 (1987), 891–911, doi:10.1137/0724058.
[2] L. J. Billera, Homology of smooth splines: Generic triangulations and a conjecture of Strang,
Trans. Am. Math. Soc. 310 (1988), 325–340, doi:10.2307/2001125.
[3] Z. Chen, Y. Feng and J. Kozak, The blossom approach to the dimension of the bivariate spline
space, J. Comput. Math. 18 (2000), 183–198, doi:10.4208/jcm.1310-fe1.
[4] O. Davydov, G. Nürnberger and F. Zeilfelder, Cubic spline interpolation on nested polygon
triangulations, in: Curve and Surface Fitting: Saint-Malo 1999, Vanderbilt University Press,
2000 pp. 161–170, https://apps.dtic.mil/sti/citations/ADP011983.
[5] G. Farin, Dimensions of spline spaces over unconstricted triangulations, J. Comput. Appl. Math.
192 (2006), 320–327, doi:10.1016/j.cam.2005.05.010.
[6] A. K. Ibrahim and L. L. Schumaker, Super spline spaces of smoothness r and degree d ≥ 3r+2,
Constr. Approx. 7 (1991), 401–423.
[7] G. Jaklič, On the dimension of the bivariate spline space S13(∆), Int. J. Comput. Math. 82
(2005), 1355–1369, doi:10.1080/00207160412331336035.
[8] G. Jaklič and J. Modic, On properties of cell matrices, Appl. Math. Comput. 216 (2010), 2016–
2023, doi:10.1016/j.amc.2010.03.032.
[9] C. Krattenthaler, Advanced determinant calculus, Sémin. Lothar. Comb. 42 (1999), b42q, 67,
doi:10.1007/978-3-642-56513-7 17.
[10] M.-J. Lai, A characterization theorem of multivariate splines in blossoming form, Comput.
Aided Geom. Des. 8 (1991), 513–521, doi:10.1016/0167-8396(91)90034-9.
[11] M.-J. Lai and L. L. Schumaker, Spline functions on triangulations, volume 110 of Encycl.
Math. Appl., Cambridge University Press, Cambridge, 2007, doi:10.1017/cbo9780511721588.
[12] L. L. Schumaker, On the Dimension of Spaces Of Piecewise Polynomials in Two Variables,
Birkhäuser Basel, Basel, pp. 396–412, 1979, doi:10.1007/978-3-0348-6289-9 26.
[13] L. L. Schumaker, Bounds on the dimension of spaces of multivariate piecewise polynomials,
Rocky Mt. J. Math. 14 (1984), 251–264, doi:10.1216/rmj-1984-14-1-251.
476 Ars Math. Contemp. 22 (2022) #P3.06 / 459–476
[14] Ş. O. Tohǎneanu, Smooth planar r-splines of degree 2r, J. Approx. Theory 132 (2005), 72–76,
doi:10.1016/j.jat.2004.10.011.
[15] W. Whiteley, A matrix for splines, in: P. Nevai and A. Pinkus (eds.), Progress in Approximation
Theory, Academic Press, Boston, pp. 821–828, 1991.
ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 22 (2022) #P3.07 / 477–488
https://doi.org/10.26493/1855-3974.2341.af2
(Also available at http://amc-journal.eu)
Avoidance in bowtie systems
Mike J. Grannell , Terry S. Griggs
The Open University, School of Mathematics and Statistics,
Walton Hall, Milton Keynes MK7 6AA, United Kingdom
Giovanni Lo Faro , Antoinette Tripodi *
Università di Messina, Dipartimento di Scienze Matematiche e Informatiche,
Scienze Fisiche e Scienze della Terra, Viale Ferdinando Stagno d’Alcontres 31,
98166 Messina, Italy
Received 20 May 2020, accepted 5 November 2021, published online 9 June 2022
Abstract
There are ten configurations of two bowties that can arise in a bowtie system. The
avoidance spectrum for three of these was determined in a previous paper (Aequat. Math.
85 (2013), 347–358). In this paper the avoidance spectrum for a further five configurations
is determined.
Keywords: Bowtie system, configuration, avoidance, Steiner triple system.
Math. Subj. Class. (2020): 05B30, 05B05, 05B070
1 Introduction
Let X = (V,E) be the graph with vertex set V = {x, a, b, c, d} and edge set
E = {xa, xb, xc, xd, ab, cd}. Such a graph is called a bowtie and will be represented
throughout this paper by the notation a, b − x − c, d. The vertex x is called the centre of
the bowtie and the other vertices are called endpoints. A decomposition of the complete
graph Kn into subgraphs isomorphic to X is called a bowtie system of order n and denoted
by BTS(n). An elementary counting argument shows that a necessary condition for the
existence of a BTS(n) is n ≡ 1 or 9 (mod 12). In a BTS(n), if every vertex of the com-
plete graph Kn occurs the same number of times as the centre of a bowtie, then the bowtie
*Corresponding author. G. Lo Faro and A. Tripodi were supported by INDAM (GNSAGA) and A. Tripodi
was supported by FFABR Unime 2019.
E-mail addresses: m.j.grannell@open.ac.uk (Mike J. Grannell), t.s.griggs@open.ac.uk (Terry S. Griggs),
lofaro@unime.it (Giovanni Lo Faro), atripodi@unime.it (Antoinette Tripodi)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
478 Ars Math. Contemp. 22 (2022) #P3.07 / 477–488
system is said to be balanced, otherwise the system is said to be unbalanced. A necessary
condition for the existence of a balanced BTS(n) is n ≡ 1 (mod 12).
It is easy to see that, given a BTS(n), by regarding each of the two triangles of every
bowtie as separate entities, we have a Steiner triple system STS(n). We call this the asso-
ciated Steiner triple system of the bowtie system. Conversely, if n ≡ 1 or 9 (mod 12), it
is also true that the triangles of every STS(n) can be amalgamated to form bowties. This
is a consequence of the fact that the block intersection graph of every Steiner triple system
is Hamiltonian, see for example [2, Section 13.6]. If n ≡ 1 (mod 12), there exists a cyclic
STS(n), see also [2, Section 7.2], and this system will have an even number of full orbits.
It is then immediate that we can amalgamate triangles from pairs of orbits to form a bal-
anced BTS(n). Hence the necessary conditions for both BTS(n) and balanced BTS(n)
given above are also sufficient.
A configuration in a bowtie system (resp. Steiner triple system) is a small collection of
bowties (resp. triangles) which may occur in the system. The study of configurations in
STS(n) is now well established and the whole of Chapter 13 of [2] is devoted to various
results about them and in particular includes formulae for the number of occurrences of
all possible configurations of four or fewer triangles. Those for configurations of one, two
or three triangles are functions of n. Such configurations are called constant because the
number of occurrences is independent of the structure of the STS(n). Other configura-
tions are variable. There are 16 non-isomorphic configurations of four triangles of which
5 are constant and 11 are variable. An important concept is that of avoidance; given any
particular configuration in a bowtie system (resp. Steiner triple system), to determine the
spectrum of n for which there exists a BTS(n) (resp. STS(n)) which does not contain that
configuration. Avoidance sets for all configurations of four or fewer triangles in Steiner
triple systems are known. Most, particularly those for constant configurations, are easy
to determine but that for the so-called Pasch configuration (four triangles isomorphic to
{a, b, c}, {a, y, z}, {x, b, z}, {x, y, c}) was more challenging. It is n ≡ 1 or 3 (mod 6),
n ̸= 7, 13 and a complete solution appears in the two papers [7] and [6].
In this paper we will be concerned with the avoidance sets of configurations of two
bowties in a BTS(n). There are ten such configurations which were determined in [3] and
are illustrated in Figure 1. In this figure each triangle of a bowtie is represented by a path
on three vertices and, in each case, one bowtie is represented by solid lines and the second
by dashed lines. The intersection of two solid lines or two dashed lines is the centre of the
bowtie and the other four points are the endpoints. The ten configurations are each labelled
Ĉi for some value of i, 1 ≤ i ≤ 16, to reflect the fact that the bowtie configuration with that
label gives the configuration Ci in the standard listing of configurations of four triangles in
Steiner triple systems as given in [5] or [2, Section 13.1]. Indeed it was by examining all
16 possible configurations of four triangles in a Steiner triple system and identifying which
could be obtained from two bowties that the ten possible configurations of two bowties
were obtained.
There are four equations which connect the number of occurrences of the various con-
figurations of two bowties and these were proved in [3]. Denoting the number of occur-
rences of the configuration Ĉi by ci, the equations are the following.
4c7 + c8 + c11 + c15 = n(n− 1)(n− 5)/24. (1.1)
c11 + c12 + 2c14 + 3c15 + 4c16 = n(n− 1)/3. (1.2)
M. J. Grannell et al.: Avoidance in bowtie systems 479
Ĉ3 Ĉ7 Ĉ8 Ĉ9
Ĉ10 Ĉ11 Ĉ12 Ĉ14
Ĉ15 Ĉ16
Figure 1: Configurations of two bowties.
480 Ars Math. Contemp. 22 (2022) #P3.07 / 477–488
c8 + c9 + 2c10 + c11 + c12 + c14 = n(n− 1)(n− 7)/12. (1.3)
4c3 + c8 + 2c9 + c12 = n(n− 1)(n− 7)(n− 9)/72. (1.4)
If the bowtie system is balanced, there is a further equation.
c7 = n(n− 1)(n− 13)/288. (1.5)
All configurations are variable except that Ĉ7 is constant in balanced bowtie systems.
Avoidance sets for the three most compact configurations, Ĉ14, Ĉ15 and Ĉ16 have al-
ready been determined in [3]. The following theorem was proved.
Theorem 1.1. For each n ≡ 1 (mod 12) there exists both a balanced and an unbalanced
BTS(n) simultaneously avoiding Ĉ14, Ĉ15 and Ĉ16. For each n ≡ 9 (mod 12), n ̸=
9 there exists a (necessarily unbalanced) BTS(n) simultaneously avoiding Ĉ14, Ĉ15 and
Ĉ16.
Thus not only can each of these three configurations be avoided for all values of n for
which both balanced BTS(n) and unbalanced BTS(n) exist except for n = 9, they can all
be avoided simultaneously. There are precisely 12 non-isomorphic BTS(9)s which were
enumerated in [4]. All avoid Ĉ16, none avoid Ĉ15 and just one avoids Ĉ14. The details are
in [3].
In this paper, we consider five further configurations. In particular we show that BTS(n)
avoiding three of the least compact configurations Ĉ3, Ĉ7 and Ĉ8 do not exist if n > 13.
Our main results are that for each of the configurations Ĉ11 and Ĉ12, and for all admissible
values of n, there exists a BTS(n) avoiding that configuration, with the single exception of
Ĉ11 when n = 13. The situation for the two configurations Ĉ9 and Ĉ10 remains unresolved.
2 Avoiding Ĉ3, Ĉ7 and Ĉ8
We begin with Ĉ7. The number of bowties in a BTS(n) is n(n− 1)/12. Hence if n > 13,
there will be two bowties with a common centre. So the only possible systems which may
avoid Ĉ7 are balanced BTS(13)s, and indeed all such systems do avoid Ĉ7, and BTS(9)s.
Checking the data of the 12 non-isomorphic BTS(9)s from [3] shows that six of these do
avoid Ĉ7 and the other six do not. We state this formally as a theorem.
Theorem 2.1. The only bowtie systems to avoid Ĉ7 are six of the twelve non-isomorphic
BTS(9)s and all balanced BTS(13)s.
Next we consider Ĉ8 and begin with some observations. First, if a, b − x − c, d is a
bowtie in a BTS(n) which has no Ĉ8 configurations, then there are at most two bowties
whose centre is a. This is because any such bowtie must intersect the bowtie a, b−x− c, d
in a further point which can only be c or d. Similarly, there are at most two bowties whose
centre is b, c or d.
Secondly, in any BTS(n), a point x can be the centre of at most (n − 1)/4 bowties.
Thus if the BTS(n) has no Ĉ8 configurations and x is the centre of less then (n − 1)/4
bowties, then it is an endpoint of at least one other bowtie and so, by the above, there are
at most two bowties whose centre is x. As a consequence, in a BTS(n) which has no Ĉ8
configurations, each point x is the centre of 0, 1, 2 or (n− 1)/4 bowties. Furthermore, if a
point is the centre of (n− 1)/4 bowties, then all remaining points are the centre of at most
two bowties. We can now prove the following theorem.
M. J. Grannell et al.: Avoidance in bowtie systems 481
Theorem 2.2. A BTS(n) avoiding Ĉ8 can only exist if n ≤ 13.
Proof. Subtracting equation 1.2 from equation 1.1 and re-arranging terms gives
c8 = n(n− 1)(n− 13)/24− 4c7 + c12 + 2c14 + 2c15 + 4c16.
Hence c8 ≥ n(n− 1)(n− 13)/24− 4c7.
Now let ax be the number of bowties in a BTS(n) whose centre is x. Then c7 =∑
x∈V
(
ax
2
)
where V denotes the set of n points in the design. Suppose that n > 13 and
that the BTS(n) has no Ĉ8 configurations. Let m be the maximum number of bowties
centred on any point in the BTS(n). Then from the argument above either m = 2 or
m = (n− 1)/4 and all but one point is the centre of at most two bowties.
In either case
c7 =
∑
x∈V
(
ax
2
)
≤
(
(n− 1)/4
2
)
+ (n− 1) = (n− 1)(n+ 27)/32.
Hence
c8 ≥ n(n− 1)(n− 13)/24− (n− 1)(n+ 27)/8 = (n− 1)(n2 − 16n− 81)/24.
The right hand side of this expression is strictly positive for n ≥ 21, and the result follows.
In order to complete the avoidance spectrum for the configuration Ĉ8, we have the
following result.
Theorem 2.3. All BTS(9)s avoid Ĉ8 but no balanced BTS(13) avoids Ĉ8. There exist
unbalanced BTS(13)s which avoid Ĉ8.
Proof. Checking the data of the 12 non-isomorphic BTS(9)s from [3] shows that all avoid
Ĉ8. The fact that no balanced BTS(13) avoids Ĉ8 follows from an exhaustive com-
puter search of all 1, 411, 422 non-isomorphic systems identified in [4]. Two unbalanced
BTS(13)s on the point set {0, 1, 2, . . . , 12} which avoid Ĉ8 are given below. In the first
case the associated STS(13) is cyclic and in the second case it is non-cyclic.
(1) 0, 4− 1− 2, 5; 0, 7− 2− 3, 6; 2, 9− 4− 3, 7;
0, 6− 8− 1, 3; 4, 5− 8− 9, 12; 1, 7− 9− 5, 6;
0, 9− 10− 6, 7; 2, 8− 10− 3, 5; 0, 5− 11− 1, 10;
2, 12− 11− 4, 6; 3, 9− 11− 7, 8; 0, 3− 12− 4, 10;
1, 6− 12− 5, 7.
(2) 1, 4− 0− 2, 7; 6, 8− 0− 9, 10; 0, 12− 3− 1, 8;
2, 6− 3− 4, 7; 2, 9− 4− 5, 8; 1, 2− 5− 3, 10;
1, 7− 9− 5, 6; 2, 8− 10− 6, 7; 0, 5− 11− 1, 6;
2, 12− 11− 4, 10; 3, 9− 11− 7, 8; 1, 10− 12− 5, 7;
4, 6− 12− 8, 9.
Finally in this section we consider Ĉ3. We have a parallel result to Theorem 2.2 for the
configuration Ĉ8.
482 Ars Math. Contemp. 22 (2022) #P3.07 / 477–488
Theorem 2.4. A BTS(n) avoiding Ĉ3 can only exist if n ≤ 13.
Proof. Assume that n > 13, so that from Theorem 2.2, c8 > 0. From equation 1.3,
c9 < n(n − 1)(n − 7)/12 and c8 + c9 + c12 ≤ n(n − 1)(n − 7)/12. So by addition
c8 + 2c9 + c12 < n(n− 1)(n− 7)/6. From equation 1.4,
4c3 = n(n− 1)(n− 7)(n− 9)/72− (c8 + 2c9 + c12).
Therefore 4c3 > n(n − 1)(n − 7)(n − 21)/72. Throughout this proof all inequalities are
strict and since n > 13, i.e. n ≥ 21, we have that c3 > 0.
Again, to complete the avoidance spectrum for the configuration Ĉ3, we have the fol-
lowing result.
Theorem 2.5. The avoidance spectrum of the configuration Ĉ3 is the set {9, 13}.
Proof. The configuration Ĉ3 has ten points so all BTS(9)s avoid Ĉ3. A balanced BTS(13)
on the set Z13 which avoid Ĉ3 is the set of bowties (i+ 1), (i+ 4)− i− (i+ 2), (i+ 7),
0 ≤ i ≤ 12, with arithmetic modulo 13. An unbalanced BTS(13) can be obtained by
replacing the bowties 1, 4 − 0 − 2, 7 and 7, 10 − 6 − 8, 0 with the bowties 1, 4 − 0 − 6, 8
and 6, 10− 7− 0, 2.
3 Avoiding Ĉ11 and Ĉ12
The method we use to construct bowtie systems which avoid the configurations Ĉ11 and
Ĉ12 is similar to how we proved Theorem 1.1 on avoiding Ĉ14, Ĉ15 and Ĉ16 and uses stan-
dard techniques involving group divisible designs. It is however more intricate. We note
that all GDDs used in this paper exist (see [1, Section IV 4.1]). An essential component of
the construction is the following BTS(9) which is System (a)(I) in [4] and avoids both Ĉ11
and Ĉ12.
1, 2− 0− 3, 6; 4, 8− 0− 5, 7; 3, 5− 4− 1, 7;
6, 7− 8− 2, 5; 5, 6− 1− 3, 8; 3, 7− 2− 4, 6.
We begin by proving the following result.
Theorem 3.1. For each n ≡ 1, 9 (mod 24), there exists a BTS(n) avoiding Ĉ12.
Proof. Take a 3-GDD of type 4t, where t = 3s or 3s+ 1 and s ≥ 1. Denote the points of
the ith group, 1 ≤ i ≤ t, by (i, 1), (i, 2), (i, 3) and (i, 4). Inflate each point to two points,
i.e. a point (i, j) becomes two points (i, j) and (i, j′). Add a single new point ∞. On each
inflated group of 8 points augmented with the ∞ point place a copy of the BTS(9) above,
identifying the points as follows.
∞ = 0, (i, 1) = 1, (i, 1′) = 3, (i, 2) = 2, (i, 2′) = 6,
(i, 3) = 4, (i, 3′) = 5, (i, 4) = 8, (i, 4′) = 7.
On each of the original blocks of the GDD, say {(i1, j1), (i2, j2), (i3, j3)}, where i1 ̸=
i2 ̸= i3 ̸= i1, place the two bowties (i2, j2), (i3, j3) − (i1, j1) − (i2, j′2), (i3, j′3) and
(i2, j2), (i3, j
′
3)− (i1, j′1)− (i2, j′2), (i3, j3). The bowties in the resulting BTS(8t+1) can
be thought of as being of two types; (i) those resulting from a BTS(9) which we will call
M. J. Grannell et al.: Avoidance in bowtie systems 483
BTS bowties and (ii) those resulting from the blocks of the GDD which we will call GDD
bowties. We need to consider pairs of bowties which arise from all possibilities. There are
five cases to consider.
(1) Two GDD bowties which come from the same block of the GDD. By the construc-
tion these form a configuration Ĉ16.
(2) Two GDD bowties which come from different blocks of the GDD. There are four
possible scenarios.
(a) If the two bowties are disjoint then they form a configuration Ĉ3.
(b) If the centres of the two bowties are the same, then they have no further points
in common and we have a configuration Ĉ7.
(c) If the centre of one of the bowties is an endpoint of the other bowtie, then again
they have no further points in common and we have a configuration Ĉ8.
(d) If the two bowties have an endpoint in common, then they also have a further
endpoint in common and they form a configuration Ĉ10.
(3) Two BTS bowties which come from the same BTS(9). The configuration they form
is completely determined by the structure of the BTS(9) and so avoids Ĉ12 (and
Ĉ11).
(4) Two BTS bowties which come from different BTS(9)s. If the two bowties are dis-
joint then they form a configuration Ĉ3. Otherwise they can only intersect in the
point ∞ which will be the centre of both bowties and we have a configuration Ĉ7.
(5) A BTS bowtie and a GDD bowtie. If the two bowties are disjoint then they form a
configuration Ĉ3. If they have just one point in common then they also avoid Ĉ12.
Otherwise they have two points in common and these points will both be endpoints
of the GDD bowtie. Further, the two points will be (i, j) and (i, j′) for some i, j
such that 1 ≤ i ≤ t and 1 ≤ j ≤ 4. If either of these points is the centre of the
BTS bowtie, then the other point is an endpoint and we have a configuration Ĉ11.
Otherwise both points are endpoints of the BTS bowtie and, because of the way in
which the points of the BTS(9) were assigned to the points ∞, (i, j) and (i, j′), they
are in different triangles. Hence we have a configuration Ĉ10.
We now prove a parallel result for the configuration Ĉ11.
Theorem 3.2. For each n ≡ 1, 9 (mod 24), there exists a BTS(n) avoiding Ĉ11.
Proof. This follows the same steps as the previous theorem. However the way in which
each inflated group of 8 points augmented with the ∞ point is identified with the points of
the BTS(9) is different. In this case it is as follows.
∞ = 0, (i, 1) = 1, (i, 1′) = 2, (i, 2) = 3, (i, 2′) = 6,
(i, 3) = 4, (i, 3′) = 8, (i, 4) = 5, (i, 4′) = 7.
The construction of the GDD bowties is the same. Also, in the analysis of pairs of bowties,
the first four cases are the same. So we only need to consider case (5) of a BTS bowtie and
a GDD bowtie. Again, if the two bowties are disjoint then they form a configuration Ĉ3.
484 Ars Math. Contemp. 22 (2022) #P3.07 / 477–488
If they have just one point in common then they also avoid Ĉ11. Otherwise they have two
points in common and they are (i, j) and (i, j′) as before. Because of the way in which the
points of the BTS(9) were assigned to the points ∞, (i, j) and (i, j′), no BTS bowtie has
its centre at a point (i, j) (resp. (i, j′)) and an endpoint at the point (i, j′) (resp. (i, j)). So
both points are endpoints of the BTS bowtie. If they are in the same triangle then we have
a configuration Ĉ12. If they are in different triangles then we have a configuration Ĉ10.
We next consider the cases n ≡ 13, 21 (mod 24). In order to deal with bowtie systems
in these residue classes avoiding Ĉ12, the following further BTS(13) is used.
1, 4− 0− 9, 10; 2, 7− 0− 6, 8; 3, 12− 0− 5, 11;
1, 5− 2− 3, 6; 1, 8− 3− 5, 10; 2, 10− 8− 4, 5;
7, 9− 1− 10, 11; 1, 12− 6− 7, 10; 2, 4− 9− 5, 6;
4, 7− 3− 9, 11; 2, 12− 11− 4, 6; 4, 10− 12− 8, 9;
5, 12− 7− 8, 11.
This system avoids the configuration Ĉ12 and has the property that one point, namely 0,
is at the centre of three bowties and never appears as an endpoint. We can now prove the
following result.
Theorem 3.3. For each n ≡ 13, 21 (mod 24), there exists a BTS(n) avoiding Ĉ12.
Proof. Take a 3-GDD of type 4t61, where t = 3s or 3s + 1 and s ≥ 1. Proceed as
in Theorem 3.1 where in addition the points of the long group are denoted by (t + 1, j),
1 ≤ j ≤ 6. On this inflated group of 12 points augmented with the ∞ point place a copy
of the BTS(13) above, identifying the points as follows.
∞ = 0,
(t+ 1, 1) = 1, (t+ 1, 1′) = 10, (t+ 1, 2) = 4, (t+ 1, 2′) = 9,
(t+ 1, 3) = 2, (t+ 1, 3′) = 6, (t+ 1, 4) = 7, (t+ 1, 4′) = 8,
(t+ 1, 5) = 3, (t+ 1, 5′) = 5, (t+ 1, 6) = 12, (t+ 1, 6′) = 11.
The proof now follows that of Theorem 3.1 . This proves the result for all stated values of
n except n = 21. A solution for this value is the following.
15, 9− 3− 11, 17; 17, 9− 5− 11, 15; 18, 10− 3− 14, 19;
19, 10− 5− 14, 18; 16, 12− 3− 13, 20; 20, 12− 5− 13, 16;
18, 9− 4− 11, 19; 19, 9− 8− 11, 18; 16, 10− 4− 14, 20;
20, 10− 8− 14, 16; 15, 12− 4− 13, 17; 17, 12− 8− 13, 15;
16, 9− 6− 11, 20; 20, 9− 7− 11, 16; 15, 10− 6− 14, 17;
17, 10− 7− 14, 15; 18, 12− 6− 13, 19; 19, 12− 7− 13, 18;
0, 7− 3− 1, 5; 2, 3− 6− 5, 7; 2, 7− 8− 3, 4;
6, 8− 1− 4, 7; 0, 6− 4− 2, 5; 2, 9− 12− 11, 13;
2, 13− 14− 9, 10; 12, 14− 1− 10, 13; 0, 14− 11− 1, 9;
0, 12− 10− 2, 11; 0, 19− 15− 1, 17; 2, 15− 18− 17, 19;
2, 19− 20− 15, 16; 18, 20− 1− 16, 19; 0, 18− 16− 2, 17;
1, 2− 0− 9, 13; 8, 5− 0− 20, 17.
M. J. Grannell et al.: Avoidance in bowtie systems 485
Turning our attention to avoiding Ĉ11, we have shown by an exhaustive computer
search that there is no BTS(13) that avoids this configuration. So for the residue classes
13 and 21 (mod 24) we use the modified constructions given in Theorems 3.4 and 3.5.
For balanced BTS(13)s the minimum number of Ĉ11 configurations is 10 for both associ-
ated cyclic and non-cyclic STS(13)s. For unbalanced systems with the associated cyclic
STS(13), we find that the minimum is 5, but for unbalanced systems with the associated
non-cyclic STS(13), we find that the minimum is 4 and an example is given below.
0, 12− 3− 1, 8; 2, 6− 3− 4, 7; 3, 5− 10− 6, 7;
2, 4− 9− 3, 11; 2, 7− 0− 5, 11; 0, 10− 9− 8, 12;
0, 8− 6− 4, 12; 1, 7− 9− 5, 6; 0, 4− 1− 10, 12;
2, 5− 1− 6, 11; 2, 11− 12− 5, 7; 2, 8− 10− 4, 11;
4, 5− 8− 7, 11.
Theorem 3.4. For each n ≡ 21 (mod 24), there exists a BTS(n) avoiding Ĉ11.
Proof. Take a 3-GDD of type 3t, where t = 4s+3 and s ≥ 0. Denote the points of the ith
group, 1 ≤ i ≤ t, by (i, 1), (i, 2) and (i, 3). As before inflate each point to two points, i.e.
a point (i, j) becomes two points (i, j) and (i, j′). Add three new points ∞0, ∞1 and ∞2.
On each inflated group of 6 points augmented with the three ∞ points first place a copy of
the BTS(9) at the beginning of this Section, identifying the points as follows.
∞0 = 0, ∞1 = 1, ∞2 = 2,
(i, 1) = 3, (i, 1′) = 6, (i, 2) = 4, (i, 2′) = 8,
(i, 3) = 5, (i, 3′) = 7.
The triangle {∞0,∞1,∞2} now occurs 4s+ 3 times. Remove the bowties
∞1,∞2 −∞0 − (i, 1), (i, 1′)
for all i such that 2 ≤ i ≤ 4s+ 3 and replace them by the bowties
(2i, 1), (2i, 1′)−∞0 − (2i+ 1, 1), (2i+ 1, 1′), 1 ≤ i ≤ 2s+ 1.
We call these BTS⋆ bowties. The construction of the GDD bowties is as in the previous
three theorems.
We need to prove that a bowtie system constructed in this way avoids configuration
Ĉ11. The proof for the five cases involving just BTS bowties and GDD bowties is as
in Theorem 3.2. So any putative configuration Ĉ11 must contain a BTS⋆ bowtie. We
show that this is impossible. A configuration Ĉ11 consists of two bowties isomorphic to
c, y − x − b, z and a, z − y − d, e. The centre of every BTS⋆ bowtie is ∞0; however this
point never occurs as the endpoint of any bowtie. So y ̸= ∞0. Now suppose that x = ∞0
and that c, y − x− b, z is a BTS⋆ bowtie. Then without loss of generality y = (2i, 1) and
z = (2i + 1, 1) for some i such that 1 ≤ i ≤ 2s + 1, say i = q. Therefore the bowtie
a, z − y − d, e is a GDD bowtie and either d or e = (2q + 1, 1′) = b which means that we
do not have a configuration Ĉ11.
We note that, by using a 3-GDD of type 3t where t = 4s+1, s ≥ 1, the above theorem
can also be used to provide an alternative proof of the existence of a BTS(n) avoiding Ĉ11
for the residue class 9 (mod 24).
486 Ars Math. Contemp. 22 (2022) #P3.07 / 477–488
A BTS(21) avoiding Ĉ11 from the above theorem is given below. This will be needed
in the proof of the final theorem. It has the crucial property that one point, again namely 0,
is at the centre of five bowties and never appears as an endpoint.
1, 2− 0− 3, 6; 4, 8− 0− 5, 7; 10, 14− 0− 11, 13;
16, 20− 0− 17, 19; 9, 12− 0− 15, 18;
3, 5− 4− 1, 7; 9, 11− 10− 1, 13; 15, 17− 16− 1, 19;
6, 7− 8− 2, 5; 12, 13− 14− 2, 11; 18, 19− 20− 2, 17;
5, 6− 1− 3, 8; 11, 12− 1− 9, 14; 17, 18− 1− 15, 20;
3, 7− 2− 4, 6; 9, 13− 2− 10, 12; 15, 19− 2− 16, 18;
9, 15− 3− 12, 18; 10, 17− 3− 14, 19; 11, 16− 3− 13, 20;
9, 18− 6− 12, 15; 10, 19− 6− 14, 17; 11, 20− 6− 13, 16;
10, 16− 4− 14, 20; 11, 15− 4− 13, 18; 9, 17− 4− 12, 19;
10, 20− 8− 14, 16; 11, 18− 8− 13, 15; 9, 19− 8− 12, 17;
11, 17− 5− 13, 19; 9, 16− 5− 12, 20; 10, 15− 5− 14, 18;
11, 19− 7− 13, 17; 9, 20− 7− 12, 16; 10, 18− 7− 14, 15.
Theorem 3.5. For each n ≡ 13 (mod 24), except for n = 13, there exists a BTS(n)
avoiding Ĉ11.
Proof. Take a 3-GDD of type 4t101, where t = 3s+ 2, s ≥ 1. Proceed as in Theorem 3.2
where the points of the long group are denoted by (t + 1, j), 1 ≤ j ≤ 10. On this
inflated group of 20 points augmented with the ∞ point place a copy of the BTS(21)
above, identifying the points as follows.
∞ = 0,
(t+ 1, 1) = 1, (t+ 1, 1′) = 2, (t+ 1, 2) = 3, (t+ 1, 2′) = 6,
(t+ 1, 3) = 4, (t+ 1, 3′) = 8, (t+ 1, 4) = 5, (t+ 1, 4′) = 7,
(t+ 1, 5) = 10, (t+ 1, 5′) = 14, (t+ 1, 6) = 11, (t+ 1, 6′) = 13,
(t+ 1, 7) = 16, (t+ 1, 7′) = 20, (t+ 1, 8) = 17, (t+ 1, 8′) = 19,
(t+ 1, 9) = 9, (t+ 1, 9′) = 12, (t+ 1, 10) = 15, (t+ 1, 10′) = 18.
The proof now follows that of Theorem 3.2. This proves the result for all stated values of
n except n = 37. A solution for this value is given in Table 1 below.
Finally, we again note that, by using a 3-GDD of type 4t101 where t = 3s, s ≥ 1,
the above theorem can also be used to provide an alternative proof of the existence of a
BTS(n) avoiding Ĉ11 for the residue class 21 (mod 24).
M. J. Grannell et al.: Avoidance in bowtie systems 487
Table 1: A BTS(37) avoiding Ĉ11.
16, 34− 0− 17, 35; 18, 36− 0− 1, 19; 8, 26− 0− 9, 27;
10, 28− 0− 11, 29; 12, 30− 0− 13, 31; 14, 32− 0− 15, 33;
2, 20− 0− 7, 25; 3, 21− 0− 4, 22; 5, 23− 0− 6, 24;
7, 18− 6− 36, 1; 25, 36− 24− 18, 19; 24, 1− 7− 19, 36;
6, 19− 25− 1, 18; 9, 8− 1− 26, 27; 27, 8− 19− 26, 9;
7, 5− 2− 23, 25; 25, 5− 20− 23, 7; 14, 13− 3− 31, 32;
32, 13− 21− 31, 14; 15, 11− 4− 29, 33; 33, 11− 22− 29, 15;
12, 10− 6− 28, 30; 30, 10− 24− 28, 12; 11, 10− 1− 28, 29;
29, 10− 19− 28, 11; 15, 13− 2− 31, 33; 33, 13− 20− 31, 15;
6, 5− 3− 23, 24; 24, 5− 21− 23, 6; 12, 8− 4− 26, 30;
30, 8− 22− 26, 12; 14, 9− 7− 27, 32; 32, 9− 25− 27, 14;
13, 12− 1− 30, 31; 31, 12− 19− 30, 13; 11, 9− 2− 27, 29;
29, 9− 20− 27, 11; 7, 4− 3− 22, 25; 25, 4− 21− 22, 7;
15, 10− 5− 28, 33; 33, 10− 23− 28, 15; 14, 8− 6− 26, 32;
32, 8− 24− 26, 14; 15, 14− 1− 32, 33; 33, 14− 19− 32, 15;
6, 4− 2− 22, 24; 24, 4− 20− 22, 6; 11, 8− 3− 26, 29;
29, 8− 21− 26, 11; 12, 9− 5− 27, 30; 30, 9− 23− 27, 12;
13, 10− 7− 28, 31; 31, 10− 25− 28, 13; 18, 17− 16− 35, 36;
36, 17− 34− 35, 18; 2, 3− 1− 21, 16; 20, 21− 19− 3, 34;
19, 16− 2− 34, 21; 1, 34− 20− 16, 3; 10, 14− 4− 32, 16;
28, 32− 22− 14, 34; 22, 16− 10− 34, 32; 4, 34− 28− 16, 14;
8, 13− 5− 31, 16; 26, 31− 23− 13, 34; 23, 16− 8− 34, 31;
5, 34− 26− 16, 13; 9, 15− 6− 33, 16; 27, 33− 24− 15, 34;
24, 16− 9− 34, 33; 6, 34− 27− 16, 15; 11, 12− 7− 30, 16;
29, 30− 25− 12, 34; 25, 16− 11− 34, 30; 7, 34− 29− 16, 12;
4, 5− 1− 23, 17; 22, 23− 19− 5, 35; 19, 17− 4− 35, 23;
1, 35− 22− 17, 5; 12, 14− 2− 32, 17; 30, 32− 20− 14, 35;
20, 17− 12− 35, 32; 2, 35− 30− 17, 14; 9, 10− 3− 28, 17;
27, 28− 21− 10, 35; 21, 17− 9− 35, 28; 3, 35− 27− 17, 10;
11, 13− 6− 31, 17; 29, 31− 24− 13, 35; 24, 17− 11− 35, 31;
6, 35− 29− 17, 13; 8, 15− 7− 33, 17; 26, 33− 25− 15, 35;
25, 17− 8− 35, 33; 7, 35− 26− 17, 15; 8, 10− 2− 28, 18;
26, 28− 20− 10, 36; 20, 18− 8− 36, 28; 2, 36− 26− 18, 10;
12, 15− 3− 33, 18; 30, 33− 21− 15, 36; 21, 18− 12− 36, 33;
3, 36− 30− 18, 15; 9, 13− 4− 31, 18; 27, 31− 22− 13, 36;
22, 18− 9− 36, 31; 4, 36− 27− 18, 13; 11, 14− 5− 32, 18;
29, 32− 23− 14, 36; 23, 18− 11− 36, 32; 5, 36− 29− 18, 14.
488 Ars Math. Contemp. 22 (2022) #P3.07 / 477–488
ORCID iDs
Mike Grannell https://orcid.org/0000-0002-0429-0493
Giovanni Lo Faro https://orcid.org/0000-0001-6174-8627
Antoinette Tripodi https://orcid.org/0000-0003-0767-4457
References
[1] C. J. Colbourn and J. H. Dinitz (eds.), The Handbook of Combinatorial Designs, Discrete Math.
Appl. (Boca Raton), CRC Press, Boca Raton, 2nd edition, 2007, doi:10.1201/9781420010541.
[2] C. J. Colbourn and A. Rosa, Triple Systems, Oxford Mathematical Monographs, The Clarendon
Press, Oxford University Press, New York, 1999.
[3] M. J. Grannell, T. S. Griggs, G. Lo Faro and A. Tripodi, Configurations in bowtie systems,
Aequationes Math. 85 (2013), 347–358, doi:10.1007/s00010-013-0199-5.
[4] M. J. Grannell, T. S. Griggs, G. LoFaro and A. Tripodi, Small bowtie systems: an enumeration,
J. Comb. Math. Comb. Comput. 70 (2009), 149–159.
[5] M. J. Grannell, T. S. Griggs and E. Mendelsohn, A small basis for four-line configurations in
Steiner triple systems, J. Comb. Des. 3 (1995), 51–59, doi:10.1002/jcd.3180030107.
[6] M. J. Grannell, T. S. Griggs and C. A. Whitehead, The resolution of the anti-Pasch conjecture,
J. Comb. Des. 8 (2000), 300–309, doi:10.1002/1520-6610(2000)8:4⟨300::aid-jcd7⟩3.3.co;2-i.
[7] A. C. H. Ling, C. J. Colbourn, M. J. Grannell and T. S. Griggs, Construction techniques for
anti-Pasch Steiner triple systems, J. London Math. Soc. (2) 61 (2000), 641–657, doi:10.1112/
s0024610700008838.
ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 22 (2022) #P3.08 / 489–504
https://doi.org/10.26493/1855-3974.2147.da8
(Also available at http://amc-journal.eu)
Two families of pseudo metacirculants*
Li Cui , Jin-Xin Zhou †
Department of Mathematics, Beijing Jiaotong University, Beijing 100044, P. R. China
Received 12 October 2019, accepted 31 December 2021, published online 9 June 2022
Abstract
A split weak metacirculant which is not metacirculant is simply called a pseudo metacir-
culant. In this paper, two infinite families of pseudo metacirculants are constructed.
Keywords: Metacirculant, metacyclic, split weak metacirculant.
Math. Subj. Class. (2020): 20B25, 05C25
1 Introduction
Metacirculant graphs were introduced by Alspach and Parsons [1]. In 2008 Marušič and
Šparl [11] gave an equivalent definition of metacirculant graphs as follows. Let m ≥ 1 and
n ≥ 2 be integers. A graph Γ of order mn is called [11] an (m,n)-metacirculant graph
(in short (m,n)-metacirculant) if it has an automorphism σ of order n such that ⟨σ⟩ is
semiregular on the vertex set of Γ, and an automorphism τ normalizing ⟨σ⟩ and cyclically
permuting the m orbits of ⟨σ⟩ such that τ has a cycle of size m in its cycle decomposition.
A graph is called a metacirculant if it is an (m,n)-metacirculant for some m and n.
It follows from the definition above that a metacirculant Γ has a vertex-transitive au-
tomorphism group ⟨σ, τ⟩ which is metacyclic. If we, instead, require that the graph has a
vertex-transitive metacyclic group of automorphisms, then we obtained the so-called weak
metacirculants, which were introduced by Marušič and Šparl [11] in 2008. In [10] Li et
al. initiated the study of relationship between metacirculants and weak metacirculants, and
they divided the weak metacirculants into the following two subclasses: A weak metacircu-
lant which has a vertex-transitive split metacyclic automorphism group is called split weak
metacirculant. Otherwise, a weak metacirculant Γ is called a non-split weak metacirculant
if its full automorphism group does not contain any split metacyclic subgroup which is
vertex-transitive. In [10, Lemma 2.2] it was proved that every metacirculant is a split weak
*This work was partially supported by the National Natural Science Foundation of China (12071023,
1211101360).
†Corresponding author.
E-mail addresses: 16118417@bjtu.edu.cn (Li Cui), jxzhou@bjtu.edu.cn (Jin-Xin Zhou)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
490 Ars Math. Contemp. 22 (2022) #P3.08 / 489–504
metacirculant, but it was unknown whether the converse of this statement is true. In [14]
Sanming Zhou and the second author asked the following question:
Question 1.1 ([14, Question A]). Is it true that any split weak metacirculant is a metacir-
culant?
If the graph under consideration is of prime-power order, it was shown by the authors
[5, 14] that the answer to the above question is positive. However, in [6] we show that
there do exist infinitely many split weak metacirculants which are not metacirculants. For
convenience, we shall say that a split metacirculant is a pseudo metacirculant if it is not
metacirculant. To the best of our knowledge, up to now the only known pseudo metacircu-
lants are the graphs constructed in [6]. So it might be interesting to find some other families
of pseudo metacirculants. Furthermore, it seems that the existence of pseudo metacircu-
lants is closely related to the orders of graphs. Motivated by this, it is natural to consider
the following problem.
Problem 1.2. Characterize those integers n for which there is a pseudo metacirculant of
order n.
There are some partial answers to Problem 1.2. By [5, 14], there do not exist a pseudo
metacirculant with a prime-power order. The construction of pseudo metacirculants in [6]
shows that for two any primes p, q with q | p − 1, there exists a pseudo metacirculant of
order pmq for each m ≥ 3. In this paper, two new infinite families of pseudo metacirculants
are constructed. Our construction implies that for any primes p, q, if either p⌊
m
2 ⌋+1 | q − 1
or p = 2 and 4 | q − 1, then there exists a pseudo metacirculant of order pmq with m ≥ 3.
Our research also shows that the three families of pseudo metacirculants constructed
in [6] and this paper are crucial for solving Problem 1.2, and we shall use them to give a
complete solution of Problem 1.2 in our subsequent paper [4].
2 Preliminaries
2.1 Definitions and notation
For a positive integer n, we denote by Cn the cyclic group of order n, by Zn the ring of
integers modulo n, by Z∗n the multiplicative group of Zn consisting of numbers coprime to
n, and by D2n the dihedral group of order 2n. For two groups M and N , N : M denotes
a semidirect product of N by M . Given a group G, denote by 1, Aut(G), and Z(G) the
identity element, full automorphism group and center of G, respectively. Denote by o(x)
the order of an element x of G. For a subgroup H of G, denote by CG(H) the centralizer
of H in G. A group G is called metacyclic if it contains a normal cyclic subgroup N such
that G/N is cyclic. In other words, a metacyclic group G is an extension of a cyclic group
N ∼= Cn by a cyclic group G/N ∼= Cm, written as G ∼= Cn.Cm. If this extension is split,
namely G ∼= Cn : Cm, then G is called a split metacyclic group.
Let G be a permutation group on a set Ω and α ∈ Ω. Denote by Gα the stabilizer of
α in G, that is, the subgroup of G fixing the point α. We say that G is semiregular on Ω
if Gα = 1 for every α ∈ Ω and regular if G is transitive and semiregular. For any subset
∆ of Ω, use G∆ and G(∆) to denote the subgroups of G fixing ∆ setwise and pointwise,
respectively. A block of imprimitivity of G on Ω is a subset ∆ of Ω with 1 < |∆| < |Ω|
such that for any g ∈ G, either ∆g = ∆ or ∆g ∩∆ = ∅. In this case the blocks ∆g, g ∈ G
form a G-invariant partition of Ω.
L. Cui and J.-X. Zhou: Two families of pseudo metacirculants 491
All graphs in this paper are finite, simple and undirected. For a graph Γ, we denote
its vertex set and edge set by V (Γ) and E(Γ), respectively. Given two adjacent vertices
u, v of Γ, denote by {u, v} the edge between u and v. Denote by Γ(v) the neighbourhood
of v, and by Γ[B] the subgraph of Γ induced by a subset B of V (Γ). An s-cycle in Γ,
denoted by Cs, is an (s + 1)-tuple of pairwise distinct vertices (v0, v1, . . . , vs) such that
{vi−1, vi} ∈ E(Γ) for 1 ≤ i ≤ s and {vs, v0} ∈ E(Γ). Denote by Kn the complete graph
of order n, and by Kn,n the complete bipartite graph with biparts of cardinality n. The full
automorphism group of Γ is denoted by Aut(Γ).
2.2 Quotient graph
Let Γ be a connected vertex-transitive graph, and let G ≤ Aut(Γ) be vertex-transitive on
Γ. A partition B of V (Γ) is said to be G-invariant if for any B ∈ B and g ∈ G we have
Bg ∈ B. For a G-invariant partition B of V (Γ), the quotient graph ΓB is defined as the
graph with vertex set B such that, for any two different vertices B,C ∈ B, B is adjacent to
C if and only if there exist u ∈ B and v ∈ C which are adjacent in Γ. Let N be a normal
subgroup of G. Then the set B of orbits of N on V (Γ) is a G-invariant partition of V (Γ).
In this case, the symbol ΓB will be replaced by ΓN , and the original graph Γ is said to be a
cover of ΓN if Γ and ΓN have the same valency.
2.3 Cayley graph
Given a finite group G and an inverse closed subset S ⊆ G \ {1}, the Cayley graph
Cay(G,S) on G with respect to S is a graph with vertex set G and edge set {{g, sg} |
g ∈ G, s ∈ S}. For any g ∈ G, R(g) is the permutation of G defined by R(g) : x 7→ xg
for x ∈ G. Set R(G) := {R(g) | g ∈ G}. It is well-known that R(G) is a subgroup of
Aut(Cay(G,S)). A Cayley graph Cay(G,S) is said to be normal if R(G) is normal in
Aut(Cay(G,S)). This concept was introduced by Xu in [13], and for more results about
normal Cayley graphs, we refer the reader to [7].
The following proposition determines the normalizer of R(G) in the full automorphism
group of Cay(G,S).
Proposition 2.1 ([8, Lemma 2.1]). Let Γ = Cay(G,S) be a Cayley graph on G with
respect to S. Then NAut(Γ)(R(G)) = R(G) : Aut(G,S), where Aut(G,S) is the group
of automorphisms of G fixing the set S setwise.
2.4 Coset graph
Let G be a group and for a subgroup H of G, let Ω = [G : H] = {Hx | x ∈ G}, the
set of right cosets of H in G. For g ∈ G, define RH(g) : Hx 7→ Hxg, x ∈ G, and set
RH(G) = {RH(g) | g ∈ G}. The map g 7→ RH(g), g ∈ G, is a homomorphism from G
to SΩ and it is called the coset action of G relative to H . The kernel of the coset action is
HG = ∩g∈GHg , the largest normal subgroup of G contained in H , and G/HG ∼= RH(G).
It is well-known that any transitive action of G on Ω is equivalent to the coset action of G
relative the subgroup Gα for any given α ∈ Ω. If HG = 1, we say that H is core-free in G.
Let D be a union of several double-cosets of the form HgH with g /∈ H such that
D = D−1. The coset graph Γ = Cos(G,H,D) of G with respect to H and D is defined as
the graph with vertex set V (Γ) = [G : H], and edge set E(Γ) = {{Hg,Hdg} | g ∈ G, d ∈
D}. It is easy to see that Γ is well defined and has valency |D|/|H|, and Γ is connected if
492 Ars Math. Contemp. 22 (2022) #P3.08 / 489–504
and only if D generates G. Further, RH(G) ≤ Aut(Γ), and hence Γ is vertex-transitive.
Let Aut(G,H,D) = {α ∈ Aut(G) | Hα = H,Dα = D}. For any α ∈
Aut(G,H,D), define αH : Hx 7→ Hxα, x ∈ G, and consider the action of Aut(G,H,D)
on [G : H] induced by α 7→ αH . It follows that Aut(G,H,D)/L ∼= Aut(G,H,D)H ,
where Aut(G,H,D)H = {αH | α ∈ Aut(G,H,D)} and L is the kernel of the action.
Furthermore, it is easy to see that Aut(G,H,D)H ≤ Aut(Γ) and Aut(G,H,D)H fixes
the vertex H . For h ∈ H , let σ(h) be the inner automorphism of G induced by h, that is,
σ(h) : g 7→ h−1gh, g ∈ G. One may show that σ(H) = {σ(h) | h ∈ H} is a subgroup of
Aut(G,H,D) and hence RH(H) = {RH(h) | h ∈ H} is a subgroup of Aut(G,H,D)H .
The following proposition determines the normalizer of RH(G) in the full automor-
phism group of Cos(G,H,D).
Proposition 2.2 ([12, Lemma 2.10]). Let G be a finite group, H a core-free subgroup of G
and D a union of several double-cosets HgH such that H ⊈ D. Let Γ = Cos(G,H,D)
and A = Aut(Γ). Then RH(G) ∼= G, Aut(G,H,D)H ∼= Aut(G,H,D), RH(H) ∼=
σ(H), and NA(RH(G)) = RH(G)Aut(G,H,D)H with RH(G) ∩ Aut(G,H,D)H =
RH(H).
Below, we prove a technical lemma.
Lemma 2.3. Let G be a finite group and let H be a core-free subgroup of G. Let S be a
set of non-identity elements of G such that S is self-inverse, and let T be any self-inverse
subset of S. Let D = HSH and let C = HTH . Set
Γ = Cos(G,H,D) and Σ = Cos(G,H,C).
If the vertex-stabilizer Aut(Γ)H fixes Σ(H) = {Hd | d ∈ C} setwise, then Aut(Γ) ≤
Aut(Σ).
Proof. Let A = Aut(Γ). Suppose that AH fixes Σ(H) = {Hd | d ∈ C} setwise. Then for
any g ∈ G, we have AHg = (AH)RH(g), and so AHg fixes the following set setwise:
{Hd | d ∈ C}RH(g) = {Hdg | d ∈ C} = Σ(Hg).
Take x ∈ A and take any edge e = {Hg,Hdg} of Σ. To show A ≤ Aut(Σ), it suffices
to show that ex ∈ E(Σ). Since G acts transitively on V (Γ) by right multiplication, there
exists g′ ∈ G such that (Hg)x = Hgg′, and then (Hg)xRH((g′)−1) = Hg. It follows
that (Hdg)xRH((g
′)−1) ∈ Σ(Hg) and so (Hdg)x ∈ (Σ(Hg))RH((g′)) = Σ(Hgg′) =
Σ((Hg)x). Hence, we have ex ∈ E(Σ), and consequently, A ≤ Aut(Σ).
2.5 Circulants
A circulant of order n is a Cayley graph over a cyclic group of order n. The following
proposition gives a classification of arc-transitive circulants. Before stating this result, we
introduce several concepts.
If a graph Γ has n > 1 connected components, each of which is isomorphic to a graph
Σ, then we shall write Γ = nΣ. The lexicographic (or wreath) product of graphs Γ1 and Γ2
is a graph Γ1◦Γ2 with vertex set V (Γ1)×V (Γ2) such that {(x1, x2), (y1, y2)} ∈ E(Γ1◦Γ2)
if and only if either x1 = y1 and {x2, y2} ∈ E(Γ2), or {x1, y1} ∈ E(Γ1). If Γ2 is of
order m with V (Γ2) = {y1, y2, . . . , ym}, then we have a natural embedding of mΓ1 in
L. Cui and J.-X. Zhou: Two families of pseudo metacirculants 493
Γ1 ◦ Γ2, where, for 1 ≤ i ≤ m, the ith copy of Γ1 is the subgraph induced by the subset
of vertices of Γ1 ◦ Γ2. The deleted lexicographic product of graphs Γ1 and Γ2, denoted by
Γ1 ◦ Γ2 −mΓ1, is the graph obtained by deleting from Γ1 ◦ Γ2 the edges of mΓ1.
Proposition 2.4 ([9, Theorem 1]). Let Γ be a connected arc-transitive circulant of order
n. Then one of the following holds:
(a) Γ = Kn;
(b) Γ is a normal circulant;
(c) Γ = Σ ◦ K̄d, where n = md and Σ is a connected arc-transitive circulant of order
m;
(d) Γ = Σ ◦ K̄d − dΣ, where n = md, d > 3, gcd(d,m) = 1 and Σ is a connected
arc-transitive circulant of order m.
3 Pseudo metacirculants–Family A
Consruction A. Let p be a prime such that 4 | p− 1 and let n ≥ 2 be an integer. If p > 5,
then let r ∈ Z∗p be such that r2 ≡ −1 (mod p), and if p = 5, then let r = 2. Let
G2,n,p,r = ⟨a, b, c | a2
n
= bp = c4 = 1, ab = ba, c−1ac = a−1, c−1bc = br⟩.
Let
Γ2,n,p,r = Cos(G2,n,p,r, H,H{(ab)±1, (abc)±1, (bc)±1, c±1}H),
where H = ⟨a2n−1c2⟩.
We first prove a lemma.
Lemma 3.1. Let G = G2,n,p,r, H = ⟨a2
n−1
c2⟩, and let D = H{(ab)±1, (abc)±1,
(bc)±1, c±1}H . If n = 2 and p > 5, then Aut(G,H,D) contains exactly one involu-
tion.
Proof. Since H ∼= C2, by Proposition 2.2, Aut(G,H,D) has an involution. Take an
involution α ∈ Aut(G,H,D). Since ⟨b⟩ is a normal Sylow p-subgroup of G, ⟨b⟩ is char-
acteristic in G, implying that bα ∈ ⟨b⟩. Since α has order 2, one has bα = b or b−1. By a
direct computation, we see that CG(b) = ⟨a⟩ × ⟨b⟩. Then
aα ∈ CG(bα) = CG(b) = ⟨a⟩ × ⟨b⟩.
It follows that aα ∈ ⟨a⟩. Since α has order 2 and n = 2, one has aα = a or a−1, and hence
(a2)α = a2.
Assume that cα = aibjck for some i ∈ Z2n , j ∈ Zp and k ∈ Z∗4. Considering the
image of the equality c−1bc = br under α, we obtain that (aibjck)−1(b±1)aibjck = b±r,
and hence br
k
= br. It follow that k = 1 in Z4, and so cα = aibjc. Moreover, (c2)α =
bj(1−r)c2. From Hα = H we obtain that (a2c2)α = a2c2. As (a2)α = a2, one has
(c2)α = c2. Thus, j = 0 in Zp, and so cα = aic.
Now by a direct computation, we have
ac, a−1c, ba2c, b−1a2c, a−1bc, a−1b−1c /∈ D. (3.1)
494 Ars Math. Contemp. 22 (2022) #P3.08 / 489–504
Remember that Dα = D. Since c ∈ D, cα = aic implies that aic ∈ D. So the only
possibility is either cα = c or cα = a2c. If the latter happens, then (bc)α = ba2c or
b−1a2c, and since bc ∈ D, either ba2c or b−1a2c belongs to D, contrary to Equation (3.1).
Thus, cα = c.
Recall that aα = a−1 or a. For the former, we have (abc)α = a−1bc or a−1b−1c,
and since abc ∈ D, either a−1bc or a−1b−1c belongs to D. This is again impossible by
Equation (3.1). Thus, aα = a, and hence we have that
α : a 7→ a, b 7→ b−1, c 7→ c.
This implies that Aut(G,H,D) has exactly one involution.
Below we shall determine the full automorphism group of Γ2,n,p,r.
Lemma 3.2. Let Γ = Γ2,n,p,r and let G = G2,n,p,r. Then Aut(Γ) = RH(G) ∼= G.
Proof. Let A = Aut(Γ). It is easy to see that H is a non-normal subgroup of G, and
so H is core-free in G. It follows that G acts faithfully and transitively on V (Γ) by right
multiplication, and so we may view G as a transitive subgroup of A. If n = 2 and p = 5,
then by Magma [3], we obtain that Aut(Γ) = G. In what follows, we shall always assume
that either p > 5 or n > 2.
Noting that ab = ba, we have ⟨ab⟩ ∼= C2np. Clearly, ⟨ab⟩ ⊴ G, so ⟨ab⟩ is semiregular
on V (Γ). Since |V (Γ)| = |G : H| = 2n+1p, ⟨ab⟩ has two orbits on V (Γ) which are listed
as follows:
V0 = {Haibj | i ∈ Z2n , j ∈ Zp} and V1 = {Haibjc | i ∈ Z2n , j ∈ Zp}.
The kernel of G acting on {V0, V1} is ⟨ab⟩ : ⟨c2⟩. We can also easily obtain the following
two observations:
∀i ∈ Z2n , j ∈ Zp, Haibjc2 = Hc2ai(c2bjc2) = Ha2
n−1+ib−j ,
∀i1, i2 ∈ Z2n , j1, j2 ∈ Zp, k ∈ Z2 Hai1bj1ck = Hai2bj2ck ⇔
{
i1 ≡ i2 (mod 2n),
j1 ≡ j2 (mod p).
(3.2)
Set ∆1 = {Hd | d ∈ H{(ab)±1}H} and ∆2 = {Hd | d ∈ H{(abc)±1, (bc)±1, c±1}H}.
Then Γ(H) = ∆1 ∪∆2. Furthermore, an easy computation shows that
∆1 = {Hab,H(ab)−1, Hab−1, Ha−1b},
∆2 = {Habc,Ha1+2
n−1
brc,Hab−1c,Ha1+2
n−1
b−rc,
Hbc,Ha2
n−1
brc,Hb−1c,Ha2
n−1
b−rc,Hc,Ha2
n−1
c}.
L. Cui and J.-X. Zhou: Two families of pseudo metacirculants 495
q
Habc
qHc q
Hab−1c
qHab−1 qHa2n−1c qHab
q
Ha1+2
n−1
brc
qHa−1b−1 qHa−1b q
Ha1+2
n−1
b−rc
q
Ha2
n−1
brc
q
Hbc
q
Hb−1c
q
Ha2
n−1
b−rc
Figure 1: The subgraph induced by Γ(H) when p > 5 and n > 2.
So for any i ∈ Z2n , j ∈ Zp, we have
Γ(Haibj) = {Hai+1bj+1, Hai−1bj−1, Hai+1bj−1, Hai−1bj+1, Ha1−ib1−jrc,
Ha2
n−1+1−ibr−jrc,Ha1−ib−1−jrc,Ha1+2
n−1−ib−r−jrc,
Ha−ib1−jrc,Ha2
n−1−ibr−jrc,Ha−ib−1−jrc,Ha2
n−1−ib−r−jrc,
Ha−ib−jrc,Ha2
n−1−ib−jrc},
Γ(Haibjc) = {Hai+1bj+1c,Hai−1bj−1c,Hai+1bj−1c,Hai−1bj+1c,Ha2
n−1+1−ibjr−1,
Ha1−ibjr−r, Ha2
n−1+1−ib1+jr, Ha1−ibr+jr, Ha2
n−1−ibjr−1,
Ha−ibjr−r, Ha2
n−1−ib1+jr, Ha−ibr+jr, Ha2
n−1−ibjr, Ha−ibjr}.
We shall finish the proof by the following four steps.
Step 1: Let Σ = Cos(G,H,H{(bc)±1, c±1}H) and let M = ⟨bc, c,H⟩. Then A ≤
Aut(Σ). In particular, the orbit HM = {Hg | g ∈ M} of M on V (Γ) containing H is a
block of imprimitivity of A on V (Γ).
By direct computations, we may depict the subgraph induced by Γ(H) as in
Figures 1 – 3. From these three figures one may see that the vertex-stabilizer AH fixes the
following set setwise:
Σ(H) = {Hd | d ∈ H{(bc)±1, c±1}H}
= {Hbc,Ha2
n−1
brc,Hb−1c,Ha2
n−1
b−rc,Hc,Ha2
n−1
c}.
From Lemma 2.3 it follows that A ≤ Aut(Σ).
Since M = ⟨bc, c,H⟩ = ⟨a2n−1⟩ × ⟨b, c⟩ ∼= C2 × (Cp : C4), the coset graph
∆ = Cos(M,H,H{(bc)±1, c±1}H})
is just a component of Σ, and since A is transitive on V (Σ) = V (Γ), the orbit of M con-
taining H is a block of imprimitivity of A on V (Γ).
496 Ars Math. Contemp. 22 (2022) #P3.08 / 489–504
q
Ha−1b−1
qHc q
Ha−1b
qHa1+2n−1brc q



C
C
C
C
C
Habc qCCC
C
C





Hab−1c qHa1+2n−1b−rc
qHab−1 qHa2n−1c qHab
q
Ha2
n−1
brc
q
Hbc
q
Hb−1c
q
Ha2
n−1
b−rc
Figure 2: The subgraph induced by Γ(H) when p > 5 and n = 2.
qHa2n−1b2c q
Ha1+2
n−1
b−2c
qHa−1b q
Hab−1c
qHc q
Habc
qHa−1b−1 q
Ha1+2
n−1
b2c
qHa2n−1b−2c
qHbc qHab−1 qHa2n−1c qHab qHb−1c
Figure 3: The subgraph induced by Γ(H) when p = 5, r = 2 and n > 2.
Step 2: Set N = ⟨a2n−1⟩ × ⟨b⟩. Then each orbit of N is a block of imprimitivity of A
on V (Γ).
Let O be the orbit of M on V (Γ) containing H . Then O = {Hg | g ∈ M} = V (∆).
By Step 1, O is a block of imprimitivity of A on V (Γ). Let S = {Og | g ∈ G}. Then
S is an imprimitivity block system of A on V (Γ). Let K be the kernel of A acting on S.
Note that N ⊴ G. Since N ≤ M , N fixes every block Og in S. It follows that N ≤ K.
Note that the subgraph Σ[Og] of Σ is just a component which is isomorphic to ∆. It is
easy to see that N acts on each Og semiregularly with two orbits {Hng | n ∈ N} and
{Htcg | t ∈ N}. As
Σ(Hg) = {Hbcg,Ha2
n−1
brcg,Hb−1cg,Ha2
n−1
b−rcg,Hcg,Ha2
n−1
cg},
the component Σ[Og] is a bipartite graph with the two orbits of N on Og as its two parts.
This implies that every orbit of N on V (Γ) is a block of imprimitivity of A on V (Γ).
Step 3: Take two adjacent orbits B0, B1 of N on V (Γ) such that B0 ⊆ V0 and B1 ⊆ V1.
Then we have A(B0) = A(B1). In particular, AH fixes ∆1 = {Hab,H(ab)−1, Hab−1,
Ha−1b} = {Hd | d ∈ H{(ab)±1}H} setwise. Since G acts transitively on V (Γ), we may
let B0 = B = {Hn | n ∈ N}. Since N ⊴G, one has B = NH ≤ G. Recall that
Γ(H) = {Hab,H(ab)−1, Hab−1, Ha−1b,
Habc,Ha1+2
n−1
brc,Hab−1c,Ha1+2
n−1
b−rc,
Hbc,Ha2
n−1
brc,Hb−1c,Ha2
n−1
b−rc,Hc,Ha2
n−1
c}.
L. Cui and J.-X. Zhou: Two families of pseudo metacirculants 497
Each orbit of N on V (Γ) is also an independent subset of Γ. Consider the orbits Ba,Ba−1, Bc
and Bac of N . Since B = NH , one has
Hab,Hab−1 ∈ Ba,
Ha−1b,Ha−1b−1 ∈ Ba−1,
Habc,Ha1+2
n−1
brc,Hab−1c,Ha1+2
n−1
b−rc ∈ Bac,
Hbc,Ha2
n−1
brc,Hb−1c,Ha2
n−1
b−rc,Hc,Ha2
n−1
c ∈ Bc.
So Ba,Ba−1, Bc and Bac are all orbits of N adjacent to B. Furthermore, it is easy to
check that if n > 2, then Ba,Ba−1, Bc and Bac are four pair-wise different orbits of
N , and if n = 2, then Ba = Ba−1, Bc and Bac are three pair-wise different orbits of
N . Clearly, Bc,Bac ⊆ V1. So B1 = Bc or Bac. Note that Γ[B,Bc] has valency 6 and
Γ[B ∪Bac] has valency 4.
If n > 2, then both Γ[B ∪Ba] and Γ[B ∪Ba−1] have valency 2. This implies that AB
also fixes Bc and Bac. If n = 2, then Γ[B∪Ba] has valency 4, and from Figure 2 one may
see that AH fixes {Hab,Ha−1b−1, Hab−1, Ha−1b} setwise and so AH fixes Ba. Again,
we have AB also fixes each of Bc and Bac.
Thus, we always have that AB also fixes each of Bc and Bac. Clearly, the subgraph
Γ[B∪B1] is bipartite, where B1 is either Bc or Bac. Let K be the subgroup of Aut(Γ[B∪
B1]) fixing B setwise. Then B and B1 are two orbits of K. Let K(B) be the kernel of K
acting on B. If K(B) does not fix every vertex of B1, then each orbit of K(B) on B1 has
length 2p, p or 2. Take two vertices u, v of B1 such that u, v are in the same orbit of K(B).
Then u, v will share the common neighborhood in Γ[B ∪ B1]. Without loss of generality,
we may assume that H is one of their common neighbors.
If B1 = Bac, then u, v ∈ {Habc,Ha1+2
n−1
brc,Hab−1c,Ha1+2
n−1
b−rc}. Note that
Γ(Habc) ∩B = {Ha2
n−1
br−1, H,Ha2
n−1
b1+r, Hb2r},
Γ(Ha1+2
n−1
brc) ∩B = {Hb−2, Ha2
n−1
b−1−r, H,Ha2
n−1
br−1},
Γ(Hab−1c) ∩B = {Ha2
n−1
b−r−1, Hb−2r, Ha2
n−1
b1−r, H},
Γ(Ha1+2
n−1
b−rc) ∩B = {H,Ha2
n−1
b1−r, Hb2, Ha2
n−1
br+1}.
It is easy to see no two of the above four sets are the same, and so no two vertices
in {Habc,Ha1+2n−1brc,Hab−1c,Ha1+2n−1b−rc} share the common neighborhood in
Γ[B,Bac], a contradiction.
If B1 = Bc, then u, v ∈ {Hbc,Ha2
n−1
brc,Hb−1c,Ha2
n−1
b−rc,Hc,Ha2
n−1
c}.
Note that
Γ(Hbc) ∩B = {Ha2
n−1
br−1, H,Ha2
n−1
b1+r, Hb2r, Ha2
n−1
br, Hbr},
Γ(Ha2
n−1
brc) ∩B = {Hb−2, Ha−2
n−1
b−1−r, H,Ha−2
n−1
br−1, Hb−1, Ha−2
n−1
b−1},
Γ(Hb−1c) ∩B = {Ha2
n−1
b−r−1, Hb−2r, Ha2
n−1
b1−r, H,Ha2
n−1
b−r, Hb−r},
Γ(Ha2
n−1
b−rc) ∩B = {H,Ha−2
n−1
b1−r, Hb2, Ha−2
n−1
br+1, Hb,Ha−2
n−1
b},
Γ(Hc) ∩B = {Ha2
n−1
b−1, Hb−r, Ha2
n−1
b,Hbr, Ha2
n−1
, H},
Γ(Ha2
n−1
c) ∩B = {Hb−1, Ha2
n−1
b−r, Hb,Ha−2
n−1
br, H,Ha−2
n−1
}.
498 Ars Math. Contemp. 22 (2022) #P3.08 / 489–504
It is easily checked that the above six sets are pair-wise different, and no two vertices in
{Hbc,Ha2n−1brc,Hb−1c,Ha2n−1b−rc,Hc,Ha2n−1c} share the common neighborhood
in Γ[B,Bc], a contradiction. Thus, K(B) also fixes every vertex of B1. Consequently, we
have A(B0) = A(B) = A(B1) with B1 = Bc or Bac.
Step 4: A = G.
By Step 3, AH fixes ∆1 setwise and so fixes ∆2 setwise. By Lemma 2.3, we have
A ≤ Aut(Λ), where Λ = Cos(G,H,D2) with D2 = H{(abc)±1, (bc)±1, c±1}H . It is
easy to see that Λ is a connected bipartite graph with V0 and V1 as its two parts. Let K be
the kernel of A acting on {V0, V1}. Again, by Step 3, we obtain that K acts faithfully on
V0. It is easy to see that Γ[V0] ∼= Θ = Cay(⟨ab⟩, {ab, (ab)−1, ab−1, a−1b}). By [2, Corol-
lary 1.3], Θ is a normal Cayley graph on ⟨ab⟩. Then Aut(⟨ab⟩, {ab, (ab)−1, ab−1, a−1b}) ∼=
C2 × C2 is regular on {ab, (ab)−1, ab−1, a−1b}. So, |K| ≤ 4|V0|. It follows that |A| ≤
4|V (Γ)| and hence |A : G| ≤ 2. Consequently, we have G ⊴ A. By Proposition 2.2, we
have AH = Aut(G,H,D) ≤ Aut(⟨ab⟩, {ab, (ab)−1, ab−1, a−1b}). If n > 2, then from
Figures 1 and 3, we can see that AH is intransitive on the set of four neighbors of H con-
tained in V0. It follows that AH ∼= C2 and hence A = G. If n = 2 and p > 5, then by
Lemma 3.1, we must have Aut(G,H,D) ∼= C2, implying that A = G.
Corollary 3.3. The graph Γ2,n,p,r is non-Cayley.
Proof. Let Γ = Γ2,n,p,r, and let A = Aut(Γ). Suppose on the contrary that Γ is a Cay-
ley graph. Then A has a regular subgroup, say T , and then A = T : AH . Since A is
metacyclic, every Sylow 2-subgroup of T must be cyclic. It follows that every Sylow 2-
subgroup of A has a cyclic maximal subgroup. However, this is impossible because from
the Construction B we know that every Sylow 2-subgroup of A is isomorphic to C2n : C4,
a contradiction.
Theorem 3.4. The graph Γ2,n,p,r is a pseudo metacirculant.
Proof. Let Γ = Γ2,n,p,r, and let G = G2,n,p,r. Note that G acts faithfully and transitively
on V (Γ) by right multiplication. Since G = ⟨ab⟩ : ⟨c⟩ ∼= C2n·p : C4, Γ is a split weak
metacirculant.
Suppose that Γ is also a metacirculant. Then by the definition of metacirculant, Γ has
two automorphisms σ, τ satisfying the following conditions:
(1) ⟨σ⟩ is semiregular and has m orbits on V (Γ),
(2) τ normalizes ⟨σ⟩ and cyclically permutes the m orbits of ⟨σ⟩,
(3) τ has a cycle of size m in its cycle decomposition.
By Lemma 3.2, we have Aut(Γ) = G. By Corollary 3.3, Γ is a non-Cayley graph, and
then we have G = ⟨σ, τ⟩. Thus, τm ̸= 1 and hence ⟨τm⟩ ∼= C2. Since G is transitive on
V (Γ), we may assume that ⟨τm⟩ = GH = ⟨a2
n−1
c2⟩. Then there would exist an element
x of G of order 4 such that x2 = τm = a2
n−1
c2. By a direct computation, we have
CG(a
2n−1c2) = ⟨a, c⟩. Then x ∈ ⟨a, c⟩ and so x = aicj for some integers i, j. However,
x2 = c2j due to c−1ac = a−1. A contradiction occurs. Thus, Γ is not a metacirculant.
L. Cui and J.-X. Zhou: Two families of pseudo metacirculants 499
4 Pseudo metacirculants—Family B
Consruction B. Let m > n > 1 be positive integers and let p, q be primes such that
pm | q − 1. Let r ∈ Z∗q be of order pm, and let
Gp,q,m,n,r = ⟨a, b, c | ap
n
= bq = cp
m
= 1, ab = ba, ac = ca, c−1bc = br⟩.
Let
Γp,q,m,n,r = Cos(Gp,q,m,n,r, H,H{(ab)±1, c±1}H),
where H = ⟨cpm−1apn−1⟩.
We shall first give some basic properties of G.
Lemma 4.1. Let G = Gp,q,m,n,r, and let D = H{(ab)±1, c±1}H . Then the following
hold.
(1) |H| = p, H is non-normal in G and CG(H) = ⟨a, c⟩.
(2) CG(b) = ⟨a, b⟩.
(3) For any g ∈ G, if ⟨g⟩⊴G, then g ∈ ⟨a, b⟩.
(4) D = (
⋃
k∈Zp H(ab
rkp
m−1
)) ∪ (
⋃
k∈Zp H(ab
rkp
m−1
)−1) ∪Hc ∪Hc−1.
(5) Aut(G,H,D) ∼= Cp.
Proof. Note that G = ⟨a⟩ × (⟨b⟩ : ⟨c⟩) ∼= Zpn × (Zq : Zpm). Let P = ⟨a, c⟩. Clearly,
P = ⟨a⟩ × ⟨c⟩ ∼= Cpn × Cpm , so H = ⟨cp
m−1
ap
n−1⟩ has order p. If H ⊴ G, then b
centralizes cp
m−1
ap
n−1
and then centralizes cp
m−1
since a ∈ Z(G). This is impossible
because c−p
m−1
bcp
m−1
= br
pm−1 ̸= b. Thus, H is non-normal in G. It follows that
⟨a, c⟩ ≤ CG(H) < G. Observing that |G : ⟨a, c⟩| = q, we have CG(H) = ⟨a, c⟩.
Therefore, item (1) holds.
For (2) , it is easy to see that ⟨a, b⟩ ≤ CG(b) and G = ⟨a, b⟩ : ⟨c⟩. Recall that
c−1bc = br with r an element of Z∗q of order pm. This implies that CG(b) = ⟨a, b⟩.
For (3) , recall that G/⟨a⟩ ∼= ⟨b⟩ : ⟨c⟩ ∼= Zq : Zpm , and ⟨b⟩ is self-centralized in
⟨b⟩ : ⟨c⟩. This implies that ⟨b⟩ is the unique non-trivial normal cyclic subgroup of ⟨b⟩ : ⟨c⟩.
For any g ∈ G, if ⟨g⟩ ⊴ G, then ⟨g⟩⟨a⟩/⟨a⟩ is normal in G/⟨a⟩, and then ⟨g⟩⟨a⟩/⟨a⟩ ≤
⟨b⟩⟨a⟩/⟨a⟩. So g ∈ ⟨a, b⟩, as desired.
For (4) , we have D = H{(ab)±1, c±1}H = H{ab, (ab)−1}H ∪H{c, c−1}H . Since
c centralizes H , one has H{c, c−1}H = Hc ∪Hc−1. Clearly,
H{ab, (ab)−1}H = HabH ∪H(ab)−1H.
Then HabH =
⋃
k∈Zp Hab(c
pm−1ap
n−1
)k =
⋃
k∈Zp Habc
kpm−1akp
n−1
. Since c−1bc =
br, one has bckp
m−1
= ckp
m−1
br
kpm−1
. As a ∈ Z(G), it follows that
HabH =
⋃
k∈Zp
Habckp
m−1
akp
n−1
=
⋃
k∈Zp
Habr
kpm−1
.
500 Ars Math. Contemp. 22 (2022) #P3.08 / 489–504
Similarly, H(ab)−1H =
⋃
k∈Zp H(ab
rkp
m−1
)−1. (4) is proved.
Finally, we shall prove (5) . Take α ∈ Aut(G,H,D). Then Hα = H and Dα = D.
Observe that ⟨b⟩ is a normal Sylow q-subgroup of G and ⟨a⟩ is just the center of G. It
follows that bα ∈ ⟨b⟩ and aα ∈ ⟨a⟩. Since Hα = H , one has cα ∈ CG(H) = ⟨a, c⟩.
Assume that
aα = ai, bα = bj , cα = asct, with i ∈ Z∗pn , j ∈ Z∗q , s ∈ Zpn , t ∈ Z∗pm .
Considering the image of the equality c−1bc = br under α, we obtain that
(asct)−1bj(asct) = bjr,
and hence bjr
t
= bjr. It follows that rt ≡ r (mod q). Since r is an element of Z∗pm of
order pm and t ∈ Z∗pm , the equality rt ≡ r (mod q) implies that t = 1 in Zpm , and so
cα = asc. Consequently, asc = cα ∈ Dα = D. By a direct computation, we have aℓc /∈ D
for any ℓ ̸= 0 (in Zpn ). So we must have cα = c.
Since ab ∈ D, one has aibj = (ab)α ∈ D. Since j ∈ Z∗q , one has aibj /∈ Hc ∪Hc−1
because (Hc ∪Hc−1) ⊆ ⟨a, c⟩. Then by (4) we have
aibj ∈ (
⋃
k∈Zp
H(abr
kpm−1
)) ∪ (
⋃
k∈Zp
H(abr
kpm−1
)−1).
Note that ⟨a, b⟩ ∩H = 1. It follows that (ab)α = aibj ∈ {(abrkp
m−1
)±1 | k ∈ Zp}, and so
aibj = abr
kpm−1
or (abr
kpm−1
)−1 for some k ∈ Zp. Since ⟨a, b⟩ = ⟨a⟩× ⟨b⟩ ∼= Zpn ×Zq ,
one has
(a, b)α = (ai, bj) = (a, br
kpm−1
) or (a−1, b−r
kpm−1
).
Since Hα = H , one has cp
m−1
(ai)p
n−1
= (cp
m−1
ap
n−1
)α ∈ H , and hence aα = a. It
follows that (a, b)α = (ai, bj) = (a, br
kpm−1
). Hence |Aut(G,H,D)| ≤ p. Note that
H ≤ Aut(G,H,D). Then Aut(G,H,D) ∼= Cp.
Next we shall determine the full automorphism group of Γp,q,n,m,r.
Lemma 4.2. Let Γ = Γp,q,m,n,r and let G = Gp,q,m,n,r. Then Aut(Γ) = RH(G) ∼= G.
Moreover, Γ is a non-Cayley graph.
Proof. We shall first prove three claims.
Claim 1. Let Λ = Cos(G,H,H{ab, (ab)−1}H). Then Aut(Γ) ≤ Aut(Λ).
Proof of Claim 1. By Lemma 4.1(4), the neighborhood of H in Γ is
Γ(H) = {H(abr
kpm−1
)±1, Hc±1 | k ∈ Zp},
and the neighborhood of H is Λ is
Λ(H) = {H(abr
kpm−1
)±1 | k ∈ Zp}.
L. Cui and J.-X. Zhou: Two families of pseudo metacirculants 501
By direct computations, we see that for any k ∈ Zp,
{H,Ha2b1+r
kpm−1
| k ∈ Zp} ⊆ Γ(Habr
kpm−1
) ∩ Γ(Hab),
{H,H(a2b1+r
kpm−1
)−1 | k ∈ Zp} ⊆ Γ(H(abr
kpm−1
)−1) ∩ Γ(H(ab)−1),
and moreover, for any Hx ∈ {H(abrkp
m−1
)±1 | k ∈ Zp} and Hcℓ ∈ {Hc,Hc−1}, we
have
Γ(Hx) ∩ Γ(Hcℓ) = {H}.
It then follows that the vertex-stabilizer Aut(Γ)H fixes the following set setwise:
Λ(H) = {H(abr
kpm−1
)±1 | k ∈ Zp}.
By Lemma 2.3, we have Aut(Γ) ≤ Aut(Λ).
Claim 2. Let Vi = {Hajbkci | j ∈ Zpn , k ∈ Zq} with i ∈ Zpm−1 . Then the following
hold.
(1) Vi is a block of imprimitivity of Aut(Γ) on V (Γ).
(2) The edges of Γ between Vi and Vi+1 form a perfect matching, where the subscripts
are modulo pm−1.
(3) Let B = {V0, V1, . . . , Vpm−1−1}. Then the quotient graph ΓB is isomorphic to
Cpm−1 .
Proof of Claim 2. By Claim 1, we have Aut(Γ) ≤ Aut(Λ). Recall that
Λ = Cos(G,H,H{ab, (ab)−1}H).
Then Λ is disconnected, and the coset graph
∆ = Cos(⟨ab,H⟩, H,H{ab, (ab)−1}H)
is a component of Λ. Consequently, V0 = V (∆) = {Hg | g ∈ ⟨ab⟩} is a block of
imprimitivity of Aut(Γ) on V (Λ) = V (Γ). Clearly, each Vi is an orbit of ⟨ab⟩ on V (Γ).
Since ⟨ab⟩ ⊴ G and G is transitive on V (Γ), Vi is a block of imprimitivity of Aut(Γ) on
V (Γ).
For (2), observing that V1 ∩ Γ(H) = {Hc}, the edges between V0 and V1 form a
perfect matching. As c cyclically permutates the orbits V0, V1, . . . , Vpm−1−1 of ⟨ab⟩, for
every i ∈ Zpm−1 , the subgraph of Γ induces by the edges between Vi and Vi+1 form a
perfect matching, where the subscripts are modulo pm−1.
For (3), noting that Γ has valency 2p + 2 while ∆ has valency 2p, the quotient graph
ΓB must be a cycle of length pm−1.
Claim 3. Let ∆ = Cos(⟨ab,H⟩, H,H{ab, (ab)−1H}). Then |Aut(∆)| = 2pn+1q.
Proof of Claim 3. It is easy to see that ∆ is isomorphic to the following Cayley graph
Θ = Cay(⟨ab⟩, {abr
kpm−1
, (abr
kpm−1
)−1 | k ∈ Zp}).
502 Ars Math. Contemp. 22 (2022) #P3.08 / 489–504
Let M = ⟨a, b⟩(∼= Zpnq) and S = {abr
kpm−1
, (abr
kpm−1
)−1 | k ∈ Zp}. The maps
α : a 7→ a, b 7→ brp
m−1
and :
¯
a 7→ a−1, b 7→ b−1 induce two automorphisms of M which
fix S setwise. So, ⟨α, β⟩ ≤ Aut(M,S). It is easy to see that ⟨α, β⟩ acts transitively on S.
Hence Θ is a connected arc-transitive Cayley graph on M . Suppose that Θ is not normal.
It is obvious that Θ ≇ Kpnq . By Proposition 2.4, there exists a connected arc-transitive
circulant Σ of order t such that one of the following happens:
(i) Θ ∼= Σ ◦ K̄d with pnq = td,
(ii) Θ ∼= Σ ◦ K̄d − dΣ, where pnq = td, d > 3 and gcd(d, t) = 1.
Suppose that (i) happens. Let k be the valency of Σ. Then Σ[K̄d] has valency kd. Noting
that Θ has valency 2p, Θ ∼= Σ ◦ K̄d implies that kd = 2p. As pnq = td, one has d = p,
and hence k = 2. It follows that Σ is a cycle of length t. Also, Θ ∼= Σ ◦ K̄p implies
that there exist t independent subsets, say D0, D1, D2, . . . , Dt−1 of V (Θ) of cardinality p
such that the subgraph induced by Di ∪Di+1 is isomorphic to Kp,p, where the subscripts
are modulo t. Furthermore, these t subsets are also blocks of imprimitivity of Aut(Θ)
on V (Θ). Assume that D0 contains the identity of M . Since M acts on V (Θ) by right
multiplication, D0 will be a subgroup of M of order p, and then D0 = ⟨ap
n−1⟩ and then
each Di is a coset of D0 in M . Recall that the only two blocks adjacent to D0 are D1
and D2, and that any two adjacent blocks induce a subgraph isomorphic to Kp,p. Then
D1∪Dt−1 = S. We may assume that ab ∈ D1. Then D1 = D0ab and Dt−1 = D0(ab)−1.
This is contrary to the fact that S = {abrkp
m−1
, (abr
kpm−1
)−1 | k ∈ Zp}.
Suppose now that (ii) holds. Observing that the valency of Σ◦K̄d−dΣ is a multiple of
d− 1, one has d− 1 | 2p, and hence d− 1 = p due to d > 3. As pnq = td and d = p+ 1,
one has p+ 1 | pnq, implying p+ 1 = q. This is contrary to the fact that p, q are odd.
Thus, Θ is a normal Cayley graph on M . By Proposition 2.1, we have Aut(Θ) =
R(M) : Aut(M,S). As M is cyclic, Aut(M,S) is abelian, and so Aut(M,S) acts regu-
larly on S. It follows that |Aut(M,S)| = 2p, and so |Aut(Θ)| = 2pn+1q, as claimed.
Proof of Lemma 4.2, continued: Now we are ready to finish the proof. Let A = Aut(Γ).
By Claim 2(1), B = {V0, V1, . . . , Vpm−1−1} is a system of blocks of A. Let K be the
kernel of A acting on B. By Claim 2(3), we have A/K ≤ Aut(ΓB) ∼= D2pm−1 . By
Claim 2(2), one may see that K acts faithfully on V0. So K can be viewed as a subgroup of
the full automorphism group of the subgraph ∆ of Γ induced by V0. By Claim 3, we have
|Aut(∆)| = 2pn+1q, and so |K| ≤ 2pn+1q.
From Lemma 4.1(1) we know that H is non-normal in G, and so G acts faithfully
on V (Γ) by right multiplication. Therefore, we may identify G with RH(G), and then
G is a vertex-transitive subgroup of A. Then GK/K ∼= Cpm−1 which is normal in
A/K ≤ D2pm−1 . In particular, GK ⊴ A. Furthermore, |GK/K| = pm−1 implies that
|GK| = pm−1|K| ≤ 2pm+nq. So |GK : G| ≤ 2, and hence G is the unique Hall {p, q}-
subgroup of GK. Thus, G is characteristic in GK, and so normal in A since GK ⊴A. By
Proposition 2.2, the stabilizer of H in A is AH = Aut(G,H,D). By Lemma 4.1(5), we
have Aut(G,H,D) ∼= Cp. This implies that G = A.
Finally, suppose that Γ is a Cayley graph. Then A has a regular subgroup, say T , and
then A = T : AH . Since A = G is metacyclic, every Sylow p-subgroup of T must
be cyclic because AH ∼= Cp. It follows that every Sylow p-subgroup of A has a cyclic
L. Cui and J.-X. Zhou: Two families of pseudo metacirculants 503
maximal subgroup. However, this is impossible because from the Construction B we know
that every Sylow p-subgroup of A is isomorphic to Cpn : Cpm with m > n > 1.
Theorem 4.3. The graph Γp,q,m,n,r is a pseudo metacirculant.
Proof. Let Γ = Γp,q,n,m,r, and let G = Gp,q,n,m,r. Note that G acts faithfully and transi-
tively on V (Γ) by right multiplication. Since G = ⟨ab⟩ : ⟨c⟩ ∼= Cpnq : Cpm , Γ is a split
weak metacirculant.
Suppose that Γ is also a metacirculant. Then by the definition of metacirculant, Γ has
two automorphisms σ, τ satisfying the following conditions:
(1) ⟨σ⟩ is semiregular and has t orbits on V (Γ),
(2) τ normalizes ⟨σ⟩ and cyclically permutes the t orbits of ⟨σ⟩,
(3) τ has a cycle of size t in its cycle decomposition.
By Lemma 4.2, we have A = G and Γ is a non-Cayley graph. Since |G| = |V (Γ)|p, we
must have G = ⟨σ, τ⟩. Thus, τ t ̸= 1 and hence ⟨τ t⟩ ∼= Cp. Since G is transitive on V (Γ),
we may assume that ⟨τ t⟩ = GH = H = ⟨ap
n−1
cp
m−1⟩. By Lemma 4.1(3), we have σ ∈
⟨a, b⟩, and so o(σ) | pnq. Consequently, pm | o(τ). Let x ∈ ⟨τ⟩ be of order pm such that
xp
m−1
= τ t = ap
n−1
cp
m−1
. Then x ∈ CG(H), and by Lemma 4.1(1), CG(H) = ⟨a, c⟩.
So x = aicj for some integers i, j. However, xp
m−1
= aip
m−1
cjp
m−1
= cjp
m−1
due to
m > n. A contradiction occurs. Thus, Γ is not a metacirculant.
ORCID iDs
Li Cui https://orcid.org/0000-0002-7470-6511
Jin-Xin Zhou https://orcid.org/0000-0002-8353-896X
References
[1] B. Alspach and T. D. Parsons, A construction for vertex-transitive graphs, Can. J. Math. 34
(1982), 307–318, doi:10.4153/cjm-1982-020-8.
[2] Y.-G. Baik, Y. Feng, H.-S. Sim and M. Xu, On the normality of Cayley graphs of abelian
groups, Algebra Colloq. 5 (1998), 297–304.
[3] W. Bosma, J. Cannon and C. Playoust, The Magma algebra system. I: The user language, J.
Symb. Comput. 24 (1997), 235–265, doi:10.1006/jsco.1996.0125.
[4] L. Cui and J.-X. Zhou, Split metacyclic groups and pseudo metacirculants, submitted.
[5] L. Cui and J.-X. Zhou, Absolutely split metacyclic groups and weak metacirculants, J. Algebra
Appl. 18 (2019), 13, doi:10.1142/s0219498819501172, id/No 1950117.
[6] L. Cui and J.-X. Zhou, A construction of pseudo metacirculants, Discrete Math. 343 (2020), 8,
doi:10.1016/j.disc.2020.111830, id/No 111830.
[7] Y.-Q. Feng, Z.-P. Lu and M.-Y. Xu, Automorphism groups of Cayley digraphs, in: Applications
of Group Theory to Combinatorics, Taylor & Francis Group, London, pp. 13–25, 2008, doi:
10.1201/9780203885765-5.
[8] C. Godsil, On the full automorphism group of a graph, Combinatorica 1 (1981), 243–256,
doi:10.1007/bf02579330.
504 Ars Math. Contemp. 22 (2022) #P3.08 / 489–504
[9] I. Kovács, Classifying arc-transitive circulants, J. Algebr. Comb. 20 (2004), 353–358, doi:10.
1023/b:jaco.0000048519.27295.3b.
[10] C. H. Li, S. J. Song and D. J. Wang, A characterization of metacirculants, J. Comb. Theory, Ser.
A 120 (2013), 39–48, doi:10.1016/j.jcta.2012.06.010.
[11] D. Marušič and P. Šparl, On quartic half-arc-transitive metacirculants, J. Algebr. Comb. 28
(2008), 365–395, doi:10.1007/s10801-007-0107-y.
[12] Y. Wang, Y.-Q. Feng and J.-X. Zhou, Cayley digraphs of 2-genetic groups of odd prime-power
order, J. Comb. Theory, Ser. A 143 (2016), 88–106, doi:10.1016/j.jcta.2016.05.001.
[13] M. Xu, Automorphism groups and isomorphisms of Cayley digraphs, Discrete Math. 182
(1998), 309–319, doi:10.1016/s0012-365x(97)00152-0.
[14] J.-X. Zhou and S. Zhou, Weak metacirculants of odd prime power order, J. Comb. Theory, Ser.
A 155 (2018), 225–243, doi:10.1016/j.jcta.2017.11.007.
ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 22 (2022) #P3.09 / 505–517
https://doi.org/10.26493/1855-3974.2603.376
(Also available at http://amc-journal.eu)
Partial-dual Euler-genus distributions
for bouquets with small Euler genus*
Yan Yang †
School of Mathematics, Tianjin University, Tianjin, China
Xiaoya Zha
Department of Mathematical Sciences, Middle Tennessee State University,
Murfreesboro, USA
Received 13 April 2021, accepted 22 December 2021, published online 9 June 2022
Abstract
For an arbitrary ribbon graph G, the partial-dual Euler-genus polynomial of G is a gen-
erating function that enumerates partial duals of G by Euler genus. When G is an orientable
ribbon graph, the partial-dual orientable genus polynomial of G is a generating function that
enumerates partial duals of G by orientable genus. Gross, Mansour, and Tucker inaugu-
rated these partial-dual Euler-genus and orientable genus distribution problems in 2020. A
bouquet is a one-vertex ribbon graph. Given a ribbon graph G, its partial-dual Euler-genus
polynomial is the same as that of some bouquet; this motivates our focus on bouquets. We
obtain the partial-dual Euler-genus polynomials for all the bouquets with Euler genus at
most two.
Keywords: Ribbon graph, partial dual, Euler-genus polynomial, orientable genus polynomial.
Math. Subj. Class. (2020): 05C10, 05C30, 05C31, 57M15
1 Introduction
Ribbon graphs are well-known to be equivalent to 2-cell embeddings of graphs. Let G be a
ribbon graph, V (G) and E(G) denote its vertex-disk set and edge-ribbon (or simply ribbon)
set, respectively. In 2009, Chmutov [1] defined the partial dual of G when he was studying
signed Bollobás–Riordan polynomials. For any A ⊆ E(G), the partial dual ribbon graph
with respect to A is denoted by GA. As a generalization of the geometric duality, partial
*The authors would like to thank the anonymous referees for their careful reading and helpful comments.
†Corresponding author. Supported by NNSF of China under Grant NSFC-11971346.
E-mail addresses: yanyang@tju.edu.cn (Yan Yang), xzha@mtsu.edu, xiaoyazha@yahoo.com (Xiaoya Zha)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
506 Ars Math. Contemp. 22 (2022) #P3.09 / 505–517
duality has developed into a topic of independent interest for its applications in graph theory
and topology. The reader is referred to [1, 3] for more background about ribbon graphs and
partial duals.
Given a ribbon graph G, there are 2|E(G)| partial duals of G in total, the problem of
the enumeration of its partial duals GA according to their Euler-genus ε(GA) or orientable
genus γ(GA) was inaugurated by Gross, Mansour, and Tucker [5] in 2020. They defined
the partial-dual Euler-genus (and orientable genus) polynomial of a ribbon graph G, which
was motivated by the Euler-genus (and orientable genus) polynomial [4] of a graph.
Definition 1.1 ([5]). The partial-dual Euler-genus polynomial of a ribbon graph G is the
generating function
∂εG(z) =
∑
A⊆E(G)
zε(G
A)
that enumerates partial duals by Euler genus. The partial-dual orientable genus polynomial
of a ribbon graph G is the generating function
∂γG(z) =
∑
A⊆E(G)
zγ(G
A)
that enumerates partial duals by orientable genus.
In [5], the authors discussed some properties of each polynomial, gave a recursion for
subdivision of an edge, and obtained the partial-dual Euler-genus (or orientable genus)
polynomials for some infinite families of ribbon graphs. In [8], Yan and Jin found some
bouquets having a non-constant partial-dual orientable genus polynomial with only one
non-zero coefficient, which disproved one of conjectures provided in [5], they also ob-
tained the partial-dual Euler-genus polynomials for all bouquets with the number of loops
at most 3 and the partial-dual orientable genus polynomials for all bouquets with the num-
ber of loops at most 4. They also found two infinite families of counterexamples to another
conjecture provided in [5] about the interpolating property of partial-dual Euler-genus poly-
nomials of non-orientable ribbon graphs in [10]. Then Chmutov and Vignes-Tourneret [2]
and Yan and Jin [9] proved that the family of counterexamples given in [8] are the only
counterexamples for that conjecture, independently.
A bouquet is a one-vertex ribbon graph, we denote the bouquet with n loops by Bn. The
bouquets with orientable genus 0 and 1 are called plane bouquets and toroidal bouquets,
respectively; the bouquets with nonorientable genus 1 and 2 are called projective planar
bouquets and Klein bottle bouquets, respectively. In this paper we focus on the partial-
dual Euler-genus polynomial for these bouquets. The following two lemmas motivate our
interest in bouquets.
Lemma 1.2 ([8]). If G is a ribbon graph and A ⊆ E(G), then ∂εG(z) = ∂εGA(z). When
G is orientable, ∂γG(z) = ∂γGA(z).
Lemma 1.3 ([3, 5]). If A is a spanning tree for G, then the partial dual GA has only one
vertex.
From Lemmas 1.2 and 1.3, given a ribbon graph G, its partial-dual Euler-genus poly-
nomial and partial-dual orientable genus polynomial when G is orientable will be the same
as that of some bouquet. From this point of view, for the partial-dual Euler-genus (and
orientable genus) distribution problems, we only need to focus on bouquets.
Y. Yang and X. Zha: Partial-dual Euler-genus distributions for bouquets with small . . . 507
Given a bouquet Bn, we label each loop by distinct letter. By reading these letters we
meet when travelling around the boundary of the vertex-disk of Bn, we obtain a cyclic
order of 2n letters, call it the rotation of Bn. A loop in Bn is called a twisted loop if its
neighbourhood is homeomorphic to a Möbius band, and untwisted loop if its neighbour-
hood is homeomorphic to an annulus. To distinguish between untwisted loops and twisted
loops, we give the same signs + to the two same letters which correspond to a common
untwisted loop, and give two different signs (one +, the other −) to the two same letters
which correspond to a common twisted loop. For simplicity, we usually omit the sign +.
We call the cyclic order of 2n signed letters the signed rotation of Bn. It is easy to check
that there is a 1-to-1 correspondence between the signed rotations and bouquets.
Two loops a and b in a bouquet are interlaced if the rotation of this bouquet is of form
a · · · b · · · a · · · b · · · , otherwise parallel. A loop e in a bouquet is trivial if e is untwisted
and not interlaced with any other loops; otherwise nontrivial. Notice that, all the twisted
loops are nontrivial in this paper, which is different with that in [8] (where a loop in a
bouquet is trivial if it is not interlaced with any other loops, so both trivial untwisted loops
and trivial twisted loops may exist). An essential bouquet is a bouquet whose loops are all
nontrivial. A ribbon graph H is a ribbon subgraph of G if H can be obtained by deleting
vertices and ribbons of G. Given a bouquet B, by deleting all trivial loops, we obtain
an essential bouquet B′ which is a ribbon subgraph of B, and we call B′ the maximum
essential subgraph of B.
The join of two disjoint ribbon graphs G1 and G2, denoted by G1 ∨ G2, is a ribbon
graph which can be obtained by the following way. Firstly we choose an arc p1 on the
boundary of a vertex-disk of G1 that lies between two consecutive ribbon ends, and choose
another such arc p2 on the boundary of a vertex-disk of G2. Next by identifying the arcs
p1 and p2, we merge the two vertex-disks from G1 and G2 into one new vertex-disk, the
ribbon graph obtained is the graph G1∨G2. For the partial-dual Euler-genus polynomial of
the graph G1 ∨G2, the following result had been obtained by Gross, Mansour, and Tucker.
Lemma 1.4 ([5]). If G1 and G2 are disjoint ribbon graphs, then
∂εG1∨G2(z) =
∂εG1(z) · ∂εG2(z).
For a bouquet, it can be obtained by the join of its maximum essential subgraph with
each trivial loops successively. From Lemma 1.4, the following result can be deduced.
Lemma 1.5 ([8]). Let G be a bouquet and G′ be the maximum essential subgraph of G. If
|E(G)| − |E(G′)| = i, then
∂εG(z) = 2
i · ∂εG′(z).
From Lemma 1.5, we can reduce the problem of partial-dual Euler-genus polynomial
of a bouquet to that of its maximum essential subgraph. We also note that the Euler genus
of a bouquet is equal to that of its maximum essential subgraph, due to the additivity of
Euler genus over the join operation.
If A ⊆ E(G), the induced ribbon subgraph G|A is a ribbon subgraph of G whose
ribbon set is A and whose the vertex-disk set consists of all ends of ribbons in A. We use
Ac := E(G) \ A to denote the complement of A ⊆ E(G). By the following lemma, the
Euler genus of a partial dual of a bouquet can be computed from the Euler genus of its two
ribbon subgraphs.
508 Ars Math. Contemp. 22 (2022) #P3.09 / 505–517
Lemma 1.6 ([5]). If G is a bouquet and A ⊆ E(G), then
ε(GA) = ε(G|A) + ε(G|AC ).
We note that ε(G) = 2γ(G) for orientable ribbon graphs, the following two lemmas
can be deduced easily.
Lemma 1.7 ([5]). If G is an orientable bouquet and A ⊆ E(G), then
γ(GA) = γ(G|A) + γ(G|AC ).
Lemma 1.8 ([5]). If G is an orientable ribbon graph, then ∂γG(z) = ∂εG(z2).
In this paper we deduce the forms of the signed rotations of maximum essential sub-
graphs for all projective planar bouquets, Klein bottle bouquets, plane bouquets, and toroidal
bouquets, respectively, and obtain the partial-dual Euler-genus polynomials for these bou-
quets.
2 Partial-dual Euler-genus polynomials of projective planar bouquets
and Klein bottle bouquets
Recall that a ribbon graph is equivalent to a 2-cell embedding of a graph. From the Propo-
sition 4.1.5 in [6], we get the following fact easily.
Lemma 2.1 ([6]). If H is a subgraph of a ribbon graph G, then ε(H) ≤ ε(G).
Lemma 2.2. The bouquet Bn is a plane bouquet if and only if all of its loops are trivial.
Proof. This lemma follows from the definition of trivial loops in Bn.
Lemma 2.3. For a nonorientable bouquet Bn,
(i) if there exists a pair of parallel twisted loops, then ε(Bn) ≥ 2;
(ii) if there exists a pair of interlaced loops, in which one is twisted and the other one is
untwisted, then ε(Bn) ≥ 2;
(iii) if there exists a pair of interlaced untwisted loops, then ε(Bn) ≥ 3.
Proof. For (i) and (ii), the bouquet Bn has ribbon subgraphs B2 and B′2, as shown in
Figure 1(a) and (b), respectively. It is easy to check that ε(B2) = ε(B′2) = 2, then ε(Bn) ≥
2 by Lemma 2.1.
For (iii), there exists a ribbon subgraph with three ribbons e1, e2 and e3 in which e1 and
e2 are interlaced untwisted loops and e3 is a twisted loop (for any nonorientable bouquet,
a twisted loop must exist). The loop e3 can be interlaced with the other i (0 ≤ i ≤ 2)
loops in this ribbon subgraph, by a case analysis, there are three nonequivalent such ribbon
subgraphs B3, B′3 and B
′′
3 , as shown in Figure 1(c)-(e). One can check that ε(B3) =
ε(B′3) = ε(B
′′
3 ) = 3, then ε(Bn) ≥ 3 follows from Lemma 2.1.
The following is easy to prove by induction and simple facial walk counting. We present
it as a technical lemma for the convenience of our later proof. It in fact reveals some
important differences between twisted and untwisted loops in a bouquet.
Y. Yang and X. Zha: Partial-dual Euler-genus distributions for bouquets with small . . . 509
(a) B2 (b) B
′
2 (c) B3
(d) B′3 (e) B
′′
3 (f) B
′′′
3
Figure 1: Some bouquets.
Lemma 2.4. If G is a bouquet and A ⊆ E(G), then
(i) If A (AC , respectively) consists of k untwisted and pairwise parallel loops, then
ε(G|A) = 0 (ε(G|AC ) = 0, respectively);
(ii) If A (AC , respectively) consists of k twisted and pairwise parallel loops, then
ε(G|A) = k (ε(G|AC ) = k, respectively);
(iii) If A (AC , respectively) consists of k twisted and pairwise interlaced loops, then
ε(G|A) = 1 (ε(G|AC ) = 1, respectively).
We first consider the partial-dual Euler-genus polynomials for projective planar bou-
quets.
Lemma 2.5. If Bn is a projective planar bouquet and Bk is the maximum essential sub-
graph of Bn, then the signed rotation of Bk has the form
a1 . . . ak(−a1) . . . (−ak), (1 ≤ k ≤ n). (1)
Proof. Firstly a bouquet with signed rotation (1) has Euler genus 1, from Lemma 2.4(iii).
From items (ii) and (iii) in Lemma 2.3, all the untwisted loops in Bn are trivial loops, so all
the loops in Bk are twisted loops. From item (i) in Lemma 2.3, any pair of twisted loops
in Bk cannot be paralleled to each other, so its signed rotation must have the form (1). The
lemma follows.
Lemma 2.6. If G is a bouquet with the signed rotation a1 . . . at(−a1) . . . (−at), then
∂εG(z) = 2z + (2
t − 2)z2.
Proof. Let X ⊆ E(G) and x = |X|. When x = 0 or x = t, ε(GX) = ε(G) = 1.
When 0 < x < t, the signed rotations of both G|X and G|XC have the form (1), then
510 Ars Math. Contemp. 22 (2022) #P3.09 / 505–517
ε(G|X) = ε(G|XC ) = 1. By Lemma 1.6, ε(GX) = 2. Since there are
(
t
x
)
ways to choose
the x edges,
∂εG(z) = 2z +
t−1∑
x=1
(
t
x
)
z2 = 2z + (2t − 2)z2.
Theorem 2.7. If Bn is a projective planar bouquet with k (1 ≤ k ≤ n) twisted loops, then
∂εBn(z) = 2
n−k(2z + (2k − 2)z2).
Proof. From Lemma 2.5, the k twisted loops induce the maximum essential subgraph of
Bn whose signed rotation has form (1). Combining it with Lemmas 1.5 and 2.6, the theo-
rem follows.
We now consider the partial-dual Euler-genus polynomials for Klein bottle bouquets.
Firstly, we deduce the form of the signed rotation of maximum essential subgraphs for
Klein bottle bouquets. We use the model of disk sum of two projective planes for the Klein
bottle. The fundamental group of Klein bottle is generated by two topologically disjoint
orientation reversing loops a and b (called twisted in this paper). There are three non-
separating loops a, b and d up to homeomorphism, where d = ab is orientation preserving
(called untwisted in this paper).
Lemma 2.8. If Bn is a Klein bottle bouquet and Bk is the maximum essential subgraph of
Bn, then the signed rotation of Bk has the form
a1 . . . aid1 . . . dq(−a1) . . . (−ai)b1 . . . bjdq . . . d1(−b1) . . . (−bj) (2)
in which k = i+ j + q; and
(i) i, j, q ≥ 1, or
(ii) i, j ≥ 1 and q = 0, or
(iii) i, q ≥ 1 and j = 0, or
(iv) j, q ≥ 1 and i = 0.
Proof. It is routine check that the bouquet whose signed rotation with form (2) has Euler
genus two. Notice that Case (iii) and Case (iv) are symmetric and we only need to consider
Case (iii). We have the following observations:
(α): From Lemma 2.4(ii), there are no three twisted loops that are paralleled to each other,
otherwise the Euler genus of Bk is at least three. And there is no subgraph B′′′3 as shown in
Figure 1(f), because ε(B′′′3 ) = 3. Let B
′
k be the ribbon subgraph induced by all the twisted
loops in Bk, the signed rotation of B′k is of form
a1 . . . ai(−a1) . . . (−ai)b1 . . . bj(−b1) . . . (−bj),
in which i+ j ≥ 1.
(β): From Lemma 2.3(iii), there exist no pairs of interlaced untwisted loops in a Klein
bottle bouquet. Each untwisted loop must be interlaced with some twisted loops, otherwise
the untwisted loop is trivial.
(γ): Each nontrivial and untwisted loop must be interlaced with all twisted loops, other-
wise there exits a ribbon subgraph B̂ whose signed rotation is dad(−a)b(−b), and ε(B̂) =
3.
Combining observations (α), (β), and (γ), the lemma follows.
Y. Yang and X. Zha: Partial-dual Euler-genus distributions for bouquets with small . . . 511
According to the four cases listed in Lemma 2.8, the signed rotation of Bk (the maxi-
mum essential subgraph of a Klein bottle bouquet) has one of the following three forms,
a1 . . . ai(−a1) . . . (−ai)b1 . . . bj(−b1) . . . (−bj), (3)
a1 . . . aid1 . . . dq(−a1) . . . (−ai)dq . . . d1, (4)
a1 . . . aid1 . . . dq(−a1) . . . (−ai)b1 . . . bjdq . . . d1(−b1) . . . (−bj) (5)
in which k = i+ j, i, j ≥ 1 in form (3); k = i+ q, i, q ≥ 1 in form (4); and k = i+ j+ q,
i, j, q ≥ 1 in form (5).
Next we compute the partial-dual Euler-genus polynomials for bouquets whose signed
rotation has form (3), (4) and (5) respectively. Let A = {a1, . . . , at}, B = {b1, . . . ,
bm}, D = {d1, . . . , ds}, X ⊆ E(G) and x = |X|.
Lemma 2.9. If G is a bouquet with the signed rotation
a1 . . . at(−a1) . . . (−at)b1 . . . bm(−b1) . . . (−bm), (t,m ≥ 1),
then
∂εG(z) = 4z
2 + 2(2t + 2m − 4)z3 + (2t − 2)(2m − 2)z4.
Proof. Let Bt and Bm be projective planar bouquets whose signed rotation have forms
a1 . . . at(−a1) . . . (−at) and b1 . . . bm(−b1) . . . (−bm), respectively. Then we have G =
Bt ∨Bm. From Lemmas 1.4 and 2.6, the result follows.
Lemma 2.10. If G is a bouquet with the signed rotation
a1 . . . atd1 . . . ds(−a1) . . . (−at)ds . . . d1, (t, s ≥ 1),
then
∂εG(z) = 2z + 2(2
s − 1)z2 + 2(2t − 2)z3 + (2t − 2)(2s − 2)z4.
Proof. When x = 0 or t + s, ε(GX) = ε(G) = 2. When 0 < x < t + s, we have the
following three cases.
Case 1: Assume X ⊆ A. If X = A, then X consists of t twisted and pairwise interlaced
loops and XC consists of s untwisted and pairwise parallel loops. From Lemma 2.4, we
have ε(G|X) = 1 and ε(G|XC ) = 0. Then from Lemma 1.6, we have ε(GX) = 1.
If X ⊂ A, then the signed rotations of the induced ribbon subgraphs G|X and G|XC
have form (1) and (4) respectively, so we have ε(G|X) = 1, ε(G|XC ) = 2 and ε(GX) = 3.
There are
∑t−1
x=1
(
t
x
)
= 2t− 2 ways to choose the x edges. Hence in this case, we have one
partial dual with Euler genus one, and 2t − 2 partial duals with Euler genus three.
Case 2: Assume X ⊆ D. The argument is similar to that for Case 1, by using Lemmas 2.4,
2.8 and 1.6, we get ε(G|X), ε(G|XC ), and ε(GX). Then we calculate the number of ways
to choose the x edges, as shown in Table 1.
Table 1: Euler genus distribution in Case 2.
X ε(G|X) ε(G|XC ) ε(GX) number
X = D 0 1 1 1
X ⊂ D 0 2 2 2s − 2
512 Ars Math. Contemp. 22 (2022) #P3.09 / 505–517
Table 2: Euler genus distribution in Case 3.
X ε(G|X) ε(G|XC ) ε(GX) number
A ⊂ X and D ̸⊂ X 2 0 2 2s − 2
D ⊂ X and A ̸⊂ X 2 1 3 2t − 2
A ̸⊂ X and D ̸⊂ X 2 2 4 (2t − 2)(2s − 2)
Case 3: Assume X ∩ A ̸= ∅ and X ∩ D ̸= ∅. We discuss three subcases as shown in
Table 2, in which we use Lemmas 2.4, 2.5 and 2.8 to compute ε(G|X) and ε(G|XC ).
Summarizing the above, we have ∂εG(z) = 2z + 2(2s − 1)z2 + 2(2t − 2)z3 +
(2t − 2)(2s − 2)z4.
Lemma 2.11. If G is a bouquet with the signed rotation
a1 . . . atd1 . . . ds(−a1) . . . (−at)b1 . . . bmds . . . d1(−b1) . . . (−bm), (t,m, s ≥ 1),
then
∂εG(z) = 2
s+1z2 + (2t+1 + 2m+1 − 4)z3 + (2t+m+s − 2t+1 − 2m+1 − 2s+1 + 4)z4.
Proof. We consider the Euler genus of GX with the following three cases.
Case 1: Ribbons in X come from one of the three ribbon sets A, B and D. We discuss two
subcases as shown in Table 3. And notice that, subcase X ⊆ D includes X = ∅.
Table 3: Euler genus distribution in Case 1.
X ε(G|X) ε(G|XC ) ε(GX) number
X ⊆ D 0 2 2 2s
X ̸= ∅; and X ⊆ A or X ⊆ B 1 2 3 2t + 2m − 2
Case 2: Ribbons in X come from two of the three ribbon sets A, B and D. We discuss
three subcases: X ⊆ A∪B, X ⊆ A∪D and X ⊆ B∪D respectively, as shown in Table 4.
Table 4: Euler genus distribution in Case 2.
X ε(G|X) ε(G|XC ) ε(GX) number
X = A ∪B 2 0 2 1
X ⊂ A ∪B 2 2 4 (2t − 1)(2m − 1)− 1
X = A ∪D or X = B ∪D 2 1 3 2
X ⊂ A ∪D or X ⊂ B ∪D 2 2 4 (2t − 1)(2s − 1) + (2m − 1)(2s − 1)− 2
Case 3: Ribbons in X come from all of three ribbon sets A, B and D. We discuss three
subcases as shown in Table 5. Notice that, subcase XC ⊂ D includes XC = ∅. And when
ribbons in XC come from at least two of A, B and D, the signed rotations of both G|X
Y. Yang and X. Zha: Partial-dual Euler-genus distributions for bouquets with small . . . 513
and G|XC have form (2), thus ε(GX) = ε(G|X) + ε(G|XC ) = 2 + 2 = 4. There are
t−1∑
x=1
(
t
x
)m−1∑
x=1
(
m
x
) s−1∑
x=1
(
s
x
)
+
m−1∑
x=1
(
m
x
) s−1∑
x=1
(
s
x
)
+
t−1∑
x=1
(
t
x
) s−1∑
x=1
(
s
x
)
+
t−1∑
x=1
(
t
x
)m−1∑
x=1
(
m
x
)
= (2t − 2)(2m − 2)(2s − 2) + (2m − 2)(2s − 2)
+ (2t − 2)(2s − 2) + (2t − 2)(2m − 2)
= 2t+m+s − 2t+m − 2t+s − 2m+s + 4
ways to choose X .
Table 5: Euler genus distribution in Case 3.
X ε(G|X) ε(G|XC ) ε(GX) number
XC ⊂ D 2 0 2 2s − 1
XC ̸= ∅;
and XC ⊆ A or XC ⊆ B 2 1 3 2
t + 2m − 4
Ribbons in XC come from
at least two of A, B and D
2 2 4 2t+m+s − 2t+m − 2t+s − 2m+s + 4
Summarizing the above, we have
∂εG(z) = (2
s + 1 + 2s − 1)z2 + (2t + 2m − 2 + 2 + 2t + 2m − 4)z3
+ ((2t − 1)(2m − 1) + (2t − 1)(2s − 1) + (2m − 1)(2s − 1)− 3
+ 2t+m+s − 2t+m − 2t+s − 2m+s + 4)z4
= 2s+1z2 + (2t+1 + 2m+1 − 4)z3
+ (2t+m+s − 2t+1 − 2m+1 − 2s+1 + 4)z4.
From Lemmas 1.5, 2.9, 2.10 and 2.11 the following three theorems can be deduced.
Theorem 2.12. If Bn is a Klein bottle bouquet and the signed rotation of its maximum
essential subgraph is of form a1 . . . ai(−a1) . . . (−ai)b1 . . . bj(−b1) . . . (−bj), (i, j ≥ 1),
then
∂εBn(z) = 2
n−i−j(4z2 + 2(2i + 2j − 4)z3 + (2i − 2)(2j − 2)z4).
Theorem 2.13. If Bn is a Klein bottle bouquet and the signed rotation of its maximum
essential subgraph is of form a1 . . . aid1 . . . dq(−a1) . . . (−ai)dq . . . d1, (i, q ≥ 1), then
∂εBn(z) = 2
n−i−q(2z + 2(2q − 1)z2 + 2(2i − 2)z3 + (2i − 2)(2q − 2)z4).
Theorem 2.14. If Bn is a Klein bottle bouquet and the signed rotation of its maximum
essential subgraph is of form
a1 . . . aid1 . . . dq(−a1) . . . (−ai)b1 . . . bjdq . . . d1(−b1) . . . (−bj), (i, j, q ≥ 1),
then
∂εBn(z) = 2
n−i−j−q(2q+1z2+(2i+1+2j+1−4)z3+(2i+j+q−2i+1−2j+1−2q+1+4)z4).
514 Ars Math. Contemp. 22 (2022) #P3.09 / 505–517
3 Partial-dual orientable genus polynomials of plane bouquets and
toroidal bouquets
In this section, we will deduce the partial-dual orientable genus polynomials of plane bou-
quets and toroidal bouquets, then their partial-dual Euler-genus polynomials follow.
Theorem 3.1. If Bn is a plane bouquet, then ∂γBn(z) = ∂εBn(z) = 2n.
Proof. From Lemma 2.2, the maximum essential subgraph of any plane bouquet is a bou-
quet with no loop. Combining it with Lemmas 1.5 and 1.8, the theorem follows.
For toroidal bouquets, in order to obtain the signed rotation of their maximum essential
subgraphs, we will use the structure of the maximal nonhomotopic loop system of the torus.
Assume x is a point on surface S. A nonhomotopic loop system, denoted by L = {li : i =
1, . . . , t}, is a collection of nonocontractible loops with base point x such that li and lj only
intersect at x and are not homotopic to each other for 1 ≤ i < j ≤ t. A nonhomotopic
loop system L is maximal if adding any noncontractible loop l with x as the base point to
L then l will either be homotopic to some loop li in L or intersect li at some point other
than x. Let ρ(S) = max{|L|}, where L is a maximal nonhomotopic loop system of S, |L|
is the number of loops in L, and the maximality is taken over all such system of S. For the
torus, the following result can be found in both [6] (in Chapter 4) and [7].
Lemma 3.2 ([6, 7]). If S1 is the torus, then ρ(S1) = 3.
Lemma 3.3. If Bn is a toroidal bouquet, then all the nontrivial loops can be partitioned
into at least two homotopy classes and at most three homotopy classes. Furthermore, any
pair of nonhomotopic nontrivial loops interlace mutually.
Proof. For Bn, viewing it as an embedding of one-vertex graph on torus, all the nontrivial
loops are nonocontractible and all the trivial loops are contractible. By Lemma 3.2 all
the nontrivial loops of Bn can be partitioned into at most three homotopy classes. The
fundamental group of torus is Z ⊕ Z, generated by two elements which are not homotopic
and intersect each other transversely, so all the nontrivial loops of Bn can be partitioned
into at least two homotopy classes. On the torus, two nontrivial loops are homotopic if and
only if they are homotopically disjoint. Therefore any pair of nonhomotopic and nontrivial
loops interlace mutually.
Lemma 3.4. If Bn is a toroidal bouquet, then the signed rotation of its maximum essential
subgraph has the form
a1 . . . aib1 . . . bjd1 . . . dqai . . . a1bj . . . b1dq . . . d1, (6)
in which i, j ≥ 1, q ≥ 0 and i+ j + q ≤ n.
Proof. From Lemma 3.3, loops in the maximum essential subgraph are partitioned into
two or three homotopy classes, the loops in the same homotopy class are paralleled with
each other, and loops from different homotopy classes are interlaced with each other, the
theorem follows.
In the following, we will discuss the partial dual distributions of toroidal bouquets in
two cases depending on whether q = 0 or not in Lemma 3.4.
Y. Yang and X. Zha: Partial-dual Euler-genus distributions for bouquets with small . . . 515
Lemma 3.5. If G is a bouquet with the signed rotation
a1 . . . atb1 . . . bmat . . . a1bm . . . b1, (t,m ≥ 1), (7)
then
∂γG(z) = 2 + 2(2
t + 2m − 3)z + (2t − 2)(2m − 2)z2.
Proof. When x = 0 or t + m, γ(GX) = γ(G) = 1. When 0 < x < t + m, we have
the following three cases as shown in Table 6, in which we get γ(G|X), γ(G|XC ) by
Lemmas 2.4 and 3.4, and get γ(GX) by Lemma 1.7.
Table 6: Orientable genus distribution when 0 < x < t+m.
X γ(G|X) γ(G|XC ) γ(GX) number
X = A or X = B 0 0 0 2
X ⊂ A, or X ⊂ B,
or XC ⊂ A, or XC ⊂ B 0 1 1
2
t−1∑
x=1
(
t
x
)
+ 2
m−1∑
x=1
(
m
x
)
= 2(2t + 2m − 4)
X ∩A ̸= ∅, X ∩B ̸= ∅,
XC ∩A ̸= ∅, and XC ∩B ̸= ∅ 1 1 2
t−1∑
x=1
(
t
x
)m−1∑
x=1
(
m
x
)
= (2t − 2)(2m − 2)
Summarizing the above, we have ∂γG(z) = 2 + 2(2t + 2m − 3)z +
(2t − 2)(2m − 2)z2.
Lemma 3.6. If G is a bouquet with the signed rotation
a1 . . . atb1 . . . bmd1 . . . dsat . . . a1bm . . . b1ds . . . d1, (t,m, s ≥ 1), (8)
then
∂γG(z) = 2(2
t + 2m + 2s − 2)z + (2t+m+s − 2t+1 − 2m+1 − 2s+1 + 4)z2.
Proof. We will consider the orientable genus of GX into the following three cases.
Case 1: Ribbons in X come from one of the three ribbon sets A, B and D.
Case 2: Ribbons in X come from two of the three ribbon sets A, B and D.
In this case, we could only consider that ribbons in X come from both A and B, by sym-
metry. Then we have the following two subcases, i.e.,
Subcase 2a: X = A ∪B and Subcase 2b: X ̸= A ∪B.
Case 3: Ribbons in X come from all of the three ribbon sets A, B and D.
In this case, we have the following two subcase, i.e.,
Subcase 3a: XC ⊂ A, or B, or D and Subcase 3b: Otherwise.
In each case, we discuss γ(G|X), γ(G|XC ) and γ(GX), and compute the number of ways
to choose X , as shown in Table 7.
Notice that, in Case 1, because the case X = ∅ has been counted three times in the
three summations, we subtract 2 from the counting. So we have 2t + 2m + 2s − 2 partial
duals with orientable genus one in this case. In Case 2, by symmetry, we have three partial
duals with orientable genus one and
(2t − 1)(2m − 1)− 1 + (2t − 1)(2s − 1)− 1 + (2m − 1)(2s − 1)− 1
= 2t+m + 2m+s + 2t+s − 2t+1 − 2m+1 − 2s+1
516 Ars Math. Contemp. 22 (2022) #P3.09 / 505–517
Table 7: Some orientable genus distribution of GX .
X γ(G|X) γ(G|XC ) γ(GX) number
Case 1 0 1 1
t∑
x=0
(
t
x
)
+
m∑
x=0
(
m
x
)
+
s∑
x=0
(
s
x
)
− 2
= 2t + 2m + 2s − 2
Subcase 2a 1 0 1 1
Subcase 2b 1 1 2
t∑
x=1
(
t
x
) m∑
x=1
(
m
x
)
− 1
= (2t − 1)(2m − 1)− 1
Subcase 3a 1 0 1
t−1∑
x=1
(
t
x
)
+
m−1∑
x=1
(
m
x
)
+
s−1∑
x=1
(
s
x
)
+ 1
= 2t + 2m + 2s − 5
Subcase 3b 1 1 2 2t+m+s − 2t+m − 2t+s − 2m+s + 4
partial duals with orientable genus two. In Subcase 3a, notice that, we haven’t include
XC = ∅ in any of the three summations, so we add one to the counting. In Subcase 3b,
there are
t−1∑
x=1
(
t
x
)m−1∑
x=1
(
m
x
) s−1∑
x=1
(
s
x
)
+
m−1∑
x=1
(
m
x
) s−1∑
x=1
(
s
x
)
+
t−1∑
x=1
(
t
x
) s−1∑
x=1
(
s
x
)
+
t−1∑
x=1
(
t
x
)m−1∑
x=1
(
m
x
)
= (2t − 2)(2m − 2)(2s − 2) + (2m − 2)(2s − 2)
+ (2t − 2)(2s − 2) + (2t − 2)(2m − 2)
= 2t+m+s − 2t+m − 2t+s − 2m+s + 4
ways to choose the x edges. Hence, in Case 3, we have 2t +2m +2s − 5 partial duals with
orientable genus one and 2t+m+s − 2t+m − 2t+s − 2m+s +4 partial duals with orientable
genus two.
Summarizing the above, we have
∂γG(z) = (2
t + 2m + 2s − 2 + 3 + 2t + 2m + 2s − 5)z
+ (2t+m + 2m+s + 2t+s − 2t+1 − 2m+1
− 2s+1 + 2t+m+s − 2t+m − 2t+s − 2m+s + 4)z2
= 2(2t + 2m + 2s − 2)z + (2t+m+s − 2t+1 − 2m+1 − 2s+1 + 4)z2.
From Lemmas 1.5, 1.8, 3.5 and 3.6, the following two theorems can be deduced.
Theorem 3.7. If Bn is a toroidal bouquet and the signed rotation of its maximum essential
subgraph has the form a1 . . . aib1 . . . bjai . . . a1bj . . . b1 in which i, j ≥ 1 and i + j ≤ n,
then
∂γBn(z) = 2
n−i−j(2 + 2(2i + 2j − 3)z + (2i − 2)(2j − 2)z2),
and
∂εBn(z) = 2
n−i−j(2 + 2(2i + 2j − 3)z2 + (2i − 2)(2j − 2)z4).
Y. Yang and X. Zha: Partial-dual Euler-genus distributions for bouquets with small . . . 517
Theorem 3.8. If Bn is a toroidal bouquet and the signed rotation of its maximum es-
sential subgraph has the form a1 . . . aib1 . . . bjd1 . . . dqai . . . a1bj . . . b1dq . . . d1, in which
i, j, q ≥ 1 and i+ j + q ≤ n, then
∂γBn(z) = 2
n−i−j−q(2(2i + 2j + 2q − 2)z + (2i+j+q − 2i+1 − 2j+1 − 2q+1 + 4)z2),
and
∂εBn(z) = 2
n−i−j−q(2(2i + 2j + 2q − 2)z2 + (2i+j+q − 2i+1 − 2j+1 − 2q+1 + 4)z4).
ORCID iDs
Yan Yang https://orcid.org/0000-0002-9666-5167
Xiaoya Zha https://orcid.org/0000-0003-1940-7428
References
[1] S. Chmutov, Generalized duality for graphs on surfaces and the signed Bollobás-Riordan poly-
nomial, J. Comb. Theory, Ser. B 99 (2009), 617–638, doi:10.1016/j.jctb.2008.09.007.
[2] S. Chmutov and F. Vignes-Tourneret, On a conjecture of Gross, Mansour and Tucker, Eur. J.
Comb. 97 (2021), 7, doi:10.1016/j.ejc.2021.103368, id/No 103368.
[3] J. A. Ellis-Monaghan and I. Moffatt, Graphs on Surfaces. Dualities, Polynomials, and Knots,
SpringerBriefs Math., Springer, New York, NY, 2013, doi:10.1007/978-1-4614-6971-1.
[4] J. L. Gross and M. L. Furst, Hierarchy for imbedding-distribution invariants of a graph, J.
Graph Theory 11 (1987), 205–220, doi:10.1002/jgt.3190110211.
[5] J. L. Gross, T. Mansour and T. W. Tucker, Partial duality for ribbon graphs. I: Distributions,
Eur. J. Comb. 86 (2020), 20, doi:10.1016/j.ejc.2020.103084, id/No 103084.
[6] B. Mohar and C. Thomassen, Graphs on Surfaces, Johns Hopkins University Press, Baltimore,
MD, 2001, https://www.sfu.ca/˜mohar/Book.html.
[7] B. Xu and X. Zha, Thickness and outerthickness for embedded graphs, Discrete Math. 341
(2018), 1688–1695, doi:10.1016/j.disc.2018.02.024.
[8] Q. Yan and X. Jin, Counterexamples to a conjecture by Gross, Mansour and Tucker on partial-
dual genus polynomials of ribbon graphs, Eur. J. Comb. 93 (2021), 13, doi:10.1016/j.ejc.2020.
103285, id/No 103285.
[9] Q. Yan and X. Jin, Partial-dual genus polynomials and signed intersection graphs, 2021,
arXiv:2102.01823v1 [math.CO].
[10] Q. Yan and X. Jin, Counterexamples to the interpolating conjecture on partial-dual genus poly-
nomials of ribbon graphs, Eur. J. Comb. 102 (2022), 7, doi:10.1016/j.ejc.2021.103493, id/No
103493.

ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 22 (2022) #P3.10 / 519–525
https://doi.org/10.26493/1855-3974.2308.de6
(Also available at http://amc-journal.eu)
The (non-)existence of perfect codes
in Lucas cubes*
Michel Mollard
Institut Fourier, CNRS, Université Grenoble Alpes, France
Received 10 April 2020, accepted 10 December 2021, published online 9 June 2022
Abstract
The Fibonacci cube of dimension n, denoted as Γn, is the subgraph of the n-cube Qn
induced by vertices with no consecutive 1’s. Ashrafi and his co-authors proved the non-
existence of perfect codes in Γn for n ≥ 4. As an open problem the authors suggest to
consider the existence of perfect codes in generalizations of Fibonacci cubes. The most
direct generalization is the family Γn(1s) of subgraphs induced by strings without 1s as a
substring where s ≥ 2 is a given integer. In a precedent work we proved the existence of a
perfect code in Γn(1s) for n = 2p − 1 and s ≥ 3.2p−2 for any integer p ≥ 2. The Lucas
cube Λn is obtained from Γn by removing vertices that start and end with 1. Very often the
same problems are studied on Fibonacci cubes and Lucas cube. In this note we prove the
non-existence of perfect codes in Λn for n ≥ 4 and prove the existence of perfect codes in
some generalized Lucas cube Λn(1s).
Keywords: Error correcting codes, perfect code, Fibonacci cube.
Math. Subj. Class. (2020): 94B50, 05C69
1 Introduction and notations
An interconnection topology can be represented by a graph G = (V,E), where V denotes
the processors and E the communication links. The hypercube Qn is a popular intercon-
nection network because of its structural properties.
The Fibonacci cube was introduced in [8] as a new interconnection network. This graph
is an isometric subgraph of the hypercube which is inspired in the Fibonacci numbers. It has
attractive recurrent structures such as its decomposition into two subgraphs which are also
Fibonacci cubes by themselves. Structural properties of these graphs were more extensively
studied afterwards. See [12] for a survey.
*Footnote on the title.
E-mail address: michel.mollard@univ-grenoble-alpes.fr (Michel Mollard)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
520 Ars Math. Contemp. 22 (2022) #P3.10 / 519–525
Lucas cubes, introduced in [17], have attracted the attention as well due to the fact that
these cubes are closely related to the Fibonacci cubes. They have also been widely studied
[3, 4, 6, 11, 13, 14, 18, 20, 23].
We will next define some concepts needed in this paper. Let G be a connected graph.
The open neighbourhood of a vertex u is NG(u) the set of vertices adjacent to u. The
closed neighbourhood of u is NG[u] = NG(u) ∪ {u}. The distance between two vertices
denoted dG(x, y) is the length of a shortest path between x and y. We have thus NG[u] =
{v ∈ V (G); dG(u, v) ≤ 1}. We will use the notations d(x, y) and N [u] when the graph is
unambiguous.
A dominating set D of G is a set of vertices such that every vertex of G belongs to
the closed neighbourhood of at least one vertex of D. In [2], Biggs initiated the study of
perfect codes in graphs as a generalization of classical 1-error perfect correcting codes. A
code C in G is a set of vertices C such that for every pair of distinct vertices c, c′ of C we
have NG[c] ∩NG[c′] = ∅ or equivalently such that dG(c, c′) ≥ 3.
A perfect code of a graph G is both a dominating set and a code. It is thus a set of
vertices C such that every vertex of G belongs to the closed neighbourhood of exactly one
vertex of C. A perfect code is also known as an efficient dominating set. The existence or
non-existence of perfect codes have been considered for many graphs. See the introduction
of [1] for some references.
The vertex set of the n-cube Qn is the set Bn of binary strings of length n, two vertices
being adjacent if they differ in precisely one position. Classical 1-error correcting codes
and perfect codes are codes and perfect codes in the graph Qn. The weight of a binary
string is the number of 1s. The concatenation of strings x and y is denoted x||y or just xy
when there is no ambiguity. A string f is a substring of a string s if there exist strings x
and y, may be empty, such that s = xfy.
A Fibonacci string of length n is a binary string b = b1 . . . bn with bi · bi+1 = 0 for
1 ≤ i < n. In other words a Fibonacci string is a binary string without 11 as substring.
The Fibonacci cube Γn (n ≥ 1) is the subgraph of Qn induced by the Fibonacci strings of
length n. Adjacent vertices in Γn differ in one bit. Because of the empty string, Γ0 = K1.
A Fibonacci string of length n is a Lucas string if b1 · bn ̸= 1. That is, a Lucas string
has no two consecutive 1’s including the first and the last elements of the string. The
Lucas cube Λn is the subgraph of Qn induced by the Lucas strings of length n. We have
Λ0 = Λ1 = K1.
Let Fn and Ln be the set of strings of Fibonacci strings and Lucas strings of length n.
By Γn,k and Λn,k we denote the vertices of of weight k in respectively Γn and Λn.
Since
Ln = {0s; s ∈ Fn−1} ∪ {10s0; s ∈ Fn−3}
and
|Γn,k| =
(
n− k + 1
k
)
it is immediate to derive the following classical result.
Proposition 1.1. Let n ≥ 1. The number of vertices of weight k ≤ n in Λn is
|Λn,k| =
(
n− k
k
)
+
(
n− k − 1
k − 1
)
.
M. Mollard: The (non-)existence of perfect codes in Lucas cubes 521
u
u
u
01
00
10
u
u
u
u
001
000
010
100 u
u
u
u
u
u
u
0001
0000
0010
01001000
1010
0101 u
u
u
u
u
u
u
uu
u
u



00001
00000
00010
00100
01001
10000
01000
01010
00101
10100
10010
Figure 1: Γ2 = Λ2, Λ3, Λ4 and Λ5.
It will be convenient to consider the binary strings of length n as vectors of Fn the
vector space of dimension n over the field F = Z2 thus to associate to a string x1x2 . . . xn
the vector θ(x1x2 . . . xn) = (x1, x2, . . . , xn). The Hamming distance between two vectors
x,y ∈ Fn, d(x,y) is the number of coordinates in which they differ. By the correspon-
dence θ we can define the binary sum x+ y and the Hamming distance d(x,y) of strings
in Bn. Note that the Hamming distance is the usual graph distance in Qn.
We will first recall some basic results about perfect codes in Qn. Since Qn is a regular
graph of degree n the existence of a perfect code of cardinality |C| implies |C|(n+1) = 2n
thus a necessary condition of existence is that n + 1 is a power of 2 thus that n = 2p − 1
for some integer p.
For any integer p Hamming [7] constructed, a linear subspace of F2p−1 which is a
perfect code. It is easy to prove that all linear perfect codes are Hamming codes. Notice
that 1n belongs to the Hamming code of length n.
In 1961 Vasilev [22], and later many authors, see [5, 21] for a survey, constructed
perfect codes which are not linear codes.
In a recent work [1] Ashrafi and his co-authors proved the non-existence of perfect
codes in Γn for n ≥ 4. As an open problem the authors suggest to consider the existence
of perfect codes in generalizations of Fibonacci cubes. The most complete generalization
proposed in [9] is, for a given string f , to consider Γn(f) the subgraph of Qn induced
by strings that do not contain f as substring. Since Fibonacci cubes are Γn(11) the most
immediate generalization [15, 19] is to consider Γn(1s) for a given integer s. In [16] we
proved the existence of a perfect code in Γn(1s) for n = 2p − 1 and s ≥ 3.2p−2 for any
integer p ≥ 2.
In the next section we will prove the main result of this note.
Theorem 1.2. The Lucas cube Λn, n ≥ 0, admits a perfect code if and only if n ≤ 3 .
2 Perfect codes in Lucas cube
It can be easily checked by hand that {0n} is a perfect code of Λn for n ≤ 3 and that Λ4
and Λ5 do not contain a perfect code (Figure 1).
Assume thus n ≥ 6.
Note first that from Proposition 1.1 we have
|Λn,2| =
n(n− 3)
2
and |Λn,3| =
n(n− 4)(n− 5)
6
.
Therefore Λn,2 and Λn,3 are none empty.
522 Ars Math. Contemp. 22 (2022) #P3.10 / 519–525
Let Λ1n,k be the vertices of Λn,k that start with 1. Since Ln = {0s; s ∈ Fn−1} ∪
{10s0; s ∈ Fn−3} the number of vertices in Λ1n,k is
|Λ1n,k| = |Γn−3,k−1| =
(
n− 1− k
k − 1
)
.
Lemma 2.1. If n ≥ 6 and C is a perfect code of Λn then 0n ∈ C.
Proof. Suppose on the contrary that 0n /∈ C. Since 0n must be dominated there exists a
vertex in Λn,1 ∩ C. This vertex is unique and because of the circular symmetry of Λn we
can assume 10n−1 ∈ C.
Since 0n /∈ C the other vertices of Λn,1 must be dominated by vertices in Λn,2. But a
vertex in Λn,2 has precisely two neighbors in Λn,1 thus n must be odd and
|Λn,2 ∩ C| =
n− 1
2
.
The unique vertex 10n−1 in Λn,1∩C has exactly n−3 neighbors in Λn,2. Let D be the
vertices of Λn,2 not in C and not dominated by 10n−1. Vertices in D must be dominated
by vertices in Λn,3 ∩ C. Each vertex of Λn,3 ∩ C has exactly exacty three neighbors in
Λn,2. Thus 3 divides the number of vertices in D. This number is
|D| = |Λn,2| − (n− 3)−
n− 1
2
=
n2 − 6n+ 7
2
.
This is not possible since there exists no odd integer n such that 6 divides n2 + 1. Indeed
since n is odd, 6 does not divide n thus divides (n+1)(n−1) = n2−1 or (n+2)(n−2) =
n2 − 4 or (n+ 3)(n− 3) = n2 − 9 thus cannot divide n2 + 1.
We are now going to prove Theorem 1.2.
Proof of Theorem 1.2. Let n ≥ 6 and C be a perfect code. Since 0n ∈ C all vertices of
Λn,1 are dominated by 0n and thus Λn,2 ∩ C = Λn,1 ∩ C = ∅. Consequently, each vertex
of Λn,2 must be dominated by a vertex in Λn,3. Since each vertex in Λn,3 has precisely
three neighbors in Λn,2 we obtain that
|Λn,3 ∩ C| =
|Λn,2|
3
.
This number must be an integer thus 3 divides |Λn,2| = n(n−3)2 and therefore 3 divides
n(n− 3). This is only possible if n is a multiple of 3.
Each vertex of Λ1n,2 must be dominated by a vertex in Λ
1
n,3. Furthermore a vertex in
Λ1n,3 has precisely two neighbors in Λ
1
n,2. Therefore |Λ1n,2| = n− 3 must be even and thus
n = 6p+ 3 for some integer p ≥ 1.
Let E be the set of vertices of Λn,3 not in C. Vertices in E must be dominated by a
vertex in Λn,4. Furthermore each vertex in Λn,4 has precisely four neighbors in Λn,3.
Therefore 4 divides |E| with
|E| = |Λn,3| − |Λn,3 ∩ C| =
n(n− 4)(n− 5)
6
− n(n− 3)
6
=
n(n2 − 10n+ 23)
6
.
Replacing n by 6p+3 we obtain that 4 divides the odd number (2p+1)(18p2 − 12p+1).
This contradiction proves Theorem 1.2.
M. Mollard: The (non-)existence of perfect codes in Lucas cubes 523
3 Perfect codes in generalized Lucas cube
The analogous of the generalisation of Fibonacci cube Γn(1s) for Lucas cube is the family
Λn(1
s) of subgraphs of Qn induced by strings without 1s as a substring in a circular man-
ner where s ≥ 2 is a given integer. More formally [10] for any binary strings b1b2 . . . bn and
each 1 ≤ i ≤ n, call bibi+1 . . . bnb1 . . . bi−1 the i-th circulation of b1b2 . . . bn. The gener-
alized Lucas cube Λn(1s) is the subgraph of Qn induced by strings without a circulation
containing 1s as a substring.
In [16] the existence of a perfect code in Γn(1s) is proved for n = 2p−1 and s ≥ 3.2p−2
for any integer p ≥ 2.
The strategy used in this construction is to build a perfect code C in Qn such that no
vertex of C contains 1s as substring. The set C is also a perfect code in Γn(1s) since each
vertex of Γn(1s) belongs to the unique closed neighbourhood in Qn thus in Γn(1s) of a
vertex in C. Because of the following proposition we cannot use the same idea for Λn(1s)
and s ≤ n− 1.
Proposition 3.1. Let n be an integer and 2 ≤ s ≤ n− 1. If C is a perfect code in Qn then
some word of C contains a circulation of 1s as a substring.
Proof. Let C be a such a perfect code in Qn then 1n /∈ C. Thus 1n must be neighbour
of a vertex c in C. Since c = 1i01n−1−i for some integer i the i+ 1th-circulation of c is
1n−10.
We can supplement this proposition by the two following results.
Proposition 3.2. Let p ≥ 2 and n = 2p − 1 then there exists a perfect code in Λn(1n) of
order |C| = 2
n
n+1 .
Proof. Let D be a Hamming code of length n and C = {d + (0n−11); d ∈ D}. Since
1n ∈ D, the set C is a perfect code of Qn such that 1n /∈ C. Since Λn(1n) is obtained
from Qn by the deletion of 1n every vertex of Λn(1n) is in the closed neighbourhood of
exactly one vertex of C.
Proposition 3.3. Let p ≥ 2 and n = 2p − 1 then there exists a perfect code in Λn(1n−1)
and in Λn(1n−2) of order |C| = 2
n
n+1 − 1.
Proof. Let D be a Hamming code of length n. Then D is a perfect code of Qn such that
1n ∈ D. Since Λn(1n−1) is obtained from Qn by the deletion of the closed neighbourhood
of 1n every vertex of Λn(1n−1) is in the closed neighbourhood of exactly one vertex of
C = D − {1n}. Furthermore since 1n ∈ D there is no vertex of weight n− 2 in D. Let u
be a vertex of Λn(1n−2) and f(u) be the vertex in D such that u ∈ NQn [u]. Since there is
no vertex in D with weight n− 1 or n− 2 there is no circulation of f(u) containing 1n−2
as a substring. Therefore f(u) is a vertex of Λn(1n−2) and u ∈ NΛn(1n−2)[f(u)]. Since a
code in Qn is a code in each of its subgraph C is a perfect code of Λn(1n−2).
ORCID iDs
Michel Mollard https://orcid.org/0000-0002-3602-8019
524 Ars Math. Contemp. 22 (2022) #P3.10 / 519–525
References
[1] A. R. Ashrafi, J. Azarija, A. Babai, K. Fathalikhani and S. Klavžar, The (non-) existence of
perfect codes in Fibonacci cubes, Inf. Process. Lett. 116 (2016), 387–390, doi:10.1016/j.ipl.
2016.01.010.
[2] N. Biggs, Perfect codes in graphs, J. Comb. Theory, Ser. B 15 (1973), 289–296, doi:10.1016/
0095-8956(73)90042-7.
[3] A. Castro, S. Klavžar, M. Mollard and Y. Rho, On the domination number and the 2-packing
number of Fibonacci cubes and Lucas cubes, Comput. Math. Appl. 61 (2011), 2655–2660,
doi:10.1016/j.camwa.2011.03.012.
[4] A. Castro and M. Mollard, The eccentricity sequences of Fibonacci and Lucas cubes, Discrete
Math. 312 (2012), 1025–1037, doi:10.1016/j.disc.2011.11.006.
[5] G. Cohen, I. Honkala, S. Litsyn and A. Lobstein, Covering Codes, volume 54 of North-Holland
Math. Libr., Elsevier, Amsterdam, 1997, doi:10.1016/s0924-6509(97)x8001-8.
[6] E. Dedó, D. Torri and N. Zagaglia Salvi, The observability of the Fibonacci and the Lucas
cubes, Discrete Math. 255 (2002), 55–63, doi:10.1016/s0012-365x(01)00387-9.
[7] R. W. Hamming, Error detecting and error correcting codes, Bell Syst. Tech. J. 29 (1950), 147–
160, doi:10.1002/j.1538-7305.1950.tb00463.x.
[8] W.-J. Hsu, Fibonacci cubes-a new interconnection topology, IEEE Transactions on Parallel
and Distributed Systems 4 (1993), 3–12, doi:10.1109/71.205649.
[9] A. Ilić, S. Klavžar and Y. Rho, Generalized Fibonacci cubes, Discrete Math. 312 (2012), 2–11,
doi:10.1016/j.disc.2011.02.015.
[10] A. Ilić, S. Klavžar and Y. Rho, Generalized Lucas cubes, Appl. Anal. Discrete Math. 6 (2012),
82–94, doi:10.2298/aadm120108002i.
[11] A. Ilić and M. Milošević, The parameters of Fibonacci and Lucas cubes, Ars Math. Contemp.
12 (2017), 25–29, doi:10.26493/1855-3974.915.f48.
[12] S. Klavžar, Structure of Fibonacci cubes: a survey, J. Comb. Optim. 25 (2013), 505–522, doi:
10.1007/s10878-011-9433-z.
[13] S. Klavžar and M. Mollard, Cube polynomial of Fibonacci and Lucas cubes, Acta Appl. Math.
117 (2012), 93–105, doi:10.1007/s10440-011-9652-4.
[14] S. Klavžar, M. Mollard and M. Petkovšek, The degree sequence of Fibonacci and Lucas cubes,
Discrete Math. 311 (2011), 1310–1322, doi:10.1016/j.disc.2011.03.019.
[15] J. Liu and W.-J. Hsu, Distributed algorithms for shortest-path, deadlock-free routing and broad-
casting in a class of interconnection topologies, in: Proceedings Sixth International Parallel
Processing Symposium, 1992 pp. 589–596, doi:10.1109/ipps.1992.222963.
[16] M. Mollard, The existence of perfect codes in a family of generalized Fibonacci cubes, Inf.
Process. Lett. 140 (2018), 1–3, doi:10.1016/j.ipl.2018.07.010.
[17] E. Munarini, C. Perelli Cippo and N. Salvi, On the Lucas cubes, Fibonacci Q. 39 (2001),
https://www.fq.math.ca/39-1.html.
[18] M. Ramras, Congestion-free routing of linear permutations on Fibonacci and Lucas cubes,
Australas. J. Comb. 60 (2014), 1–10, https://ajc.maths.uq.edu.au/?page=get_
volumes&volume=60.
[19] N. Z. Salvi, On the existence of cycles of every even length on generalized Fibonacci
cubes, Matematiche 51 (1996), 241–251, https://lematematiche.dmi.unict.it/
index.php/lematematiche/article/view/475.
M. Mollard: The (non-)existence of perfect codes in Lucas cubes 525
[20] E. Saygi and Ö. Eğecioğlu, q-counting hypercubes in Lucas cubes, Turk. J. Math. 42 (2018),
190–203, doi:10.3906/mat-1605-2.
[21] F. I. Solov’eva, On perfect binary codes, in: General theory of information transfer and com-
binatorics, Elsevier, Amsterdam, pp. 371–372, 2005, doi:10.1016/j.endm.2005.07.059.
[22] Y. Vasil’ev, On nongroup close-packed codes, Probl. Kibern. 8 (1962), 337–339.
[23] X. Wang, X. Zhao and H. Yao, Structure and enumeration results of matchable Lucas cubes,
Discrete Appl. Math. 277 (2020), 263–279, doi:10.1016/j.dam.2019.09.011.

Author Guidelines
Before submission
Papers should be written in English, prepared in LATEX, and must be submitted as a PDF
file. The title page of the submissions must contain:
• Title. The title must be concise and informative.
• Author names and affiliations. For each author add his/her affiliation which should
include the full postal address and the country name. If avilable, specify the e-mail
address of each author. Clearly indicate who is the corresponding author of the paper.
• Abstract. A concise abstract is required. The abstract should state the problem stud-
ied and the principal results proven.
• Keywords. Please specify 2 to 6 keywords separated by commas.
• Mathematics Subject Classification. Include one or more Math. Subj. Class. (2020)
codes – see https://mathscinet.ams.org/mathscinet/msc/msc2020.html.
After acceptance
Articles which are accepted for publication must be prepared in LATEX using class file
amcjoucc.cls and the bst file amcjoucc.bst (if you use BibTEX). If you don’t use BibTEX,
please make sure that all your references are carefully formatted following the examples
provided in the sample file. All files can be found on-line at:
https://amc-journal.eu/index.php/amc/about/submissions/#authorGuidelines
Abstracts: Be concise. As much as possible, please use plain text in your abstract and
avoid complicated formulas. Do not include citations in your abstract. All abstracts will be
posted on the website in fairly basic HTML, and HTML can’t handle complicated formulas.
It can barely handle subscripts and greek letters.
Cross-referencing: All numbering of theorems, sections, figures etc. that are referenced
later in the paper should be generated using standard LATEX \label{. . . } and \ref{. . . }
commands. See the sample file for examples.
Theorems and proofs: The class file has pre-defined environments for theorem-like state-
ments; please use them rather than coding your own. Please use the standard \begin{proof}
. . . \end{proof} environment for your proofs.
Spacing and page formatting: Please do not modify the page formatting and do not use
\medbreak, \bigbreak, \pagebreak etc. commands to force spacing. In general, please
let LATEX do all of the space formatting via the class file. The layout editors will modify the
formatting and spacing as needed for publication.
Figures: Any illustrations included in the paper must be provided in PDF format, or
via LATEX packages which produce embedded graphics, such as TikZ, that compile with
PdfLATEX. (Note, however, that PSTricks is problematic.) Make sure that you use uniform
lettering and sizing of the text. If you use other methods to generate your graphics, please
provide .pdf versions of the images (or negotiate with the layout editor assigned to your
article).
xxi
Subscription
Yearly subscription: 150 EUR
Any author or editor that subscribes to the printed edition will receive a complimentary
copy of Ars Mathematica Contemporanea.
Subscription Order Form
Name: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
E-mail: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Postal Address: . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
I would like to subscribe to receive . . . . . . copies of each issue of
Ars Mathematica Contemporanea in the year 2022.
I want to renew the order for each subsequent year if not cancelled by e-mail:
□ Yes □ No
Signature: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Please send the order by mail, by fax or by e-mail.
By mail: Ars Mathematica Contemporanea
UP FAMNIT
Glagoljaška 8
SI-6000 Koper
Slovenia
By fax: +386 5 611 75 71
By e-mail: info@famnit.upr.si
xxii

Printed in Slovenia by IME TISKARNE