THEOREMS AND COMPUTATIONS IN CIRCULAR COLOURINGS OF GRAPHS by Mohammad Ghebleh B.Sc., Sharif University of Technology, 1997 M.Sc., Sharif University of Technology, 1999 a thesis submitted in partial fulfillment of the requirements for the degree of Doctor of Philosophy in the Department of Mathematics c Mohammad Ghebleh 2007 SIMON FRASER UNIVERSITY Fall 2007 All rights reserved. This work may not be reproduced in whole or in part, by photocopy or other means, without the permission of the author.
The concept of circular chromatic number of graphs was first introduced by Vince [50] under
the name star chromatic number. Circular colouring has received much attention since it
provides a refinement of the (ordinary) chromatic number of a graph. In this chapter we
present several equivalent formulations of circular colouring and some preliminary results.
We assume the reader is familiar with the general notions of graph colouring which can be
found in standard graph theory textbooks such as [14] and [54].
1.1 Circular Colouring
For a positive real number r, we let Cr denote the additive group R/rZ. We identify Cr
with the interval [0, r). Intuitively, Cr can be thought of as a circle of perimeter r. For any
x, y ∈ Cr, the r–circular distance between x and y is defined by
|x− y|r = min{|x− y|, r − |x− y|}.
In other words, |x− y|r is the shortest distance between the two points corresponding to x
and y on a circle of perimeter r. When it is clear from the context, we may refer to |x− y|ras the distance between x and y. For every x, y ∈ [0, r), the r–circular interval [x, y]r, is
1
Chapter 1. An Introduction to Circular Colouring 2
defined by
[x, y]r =
[x, y] if x 6 y,
[x, r) ∪ [0, y] if x > y.
Sometimes it is convenient to use alternative representations of Cr. For x, y ∈ R with
x 6 y, we define the r–circular interval [x, y]r to be the image of the real interval [x, y]
under the map which reduces, modulo r, each real number to its unique representative
in [0, r). Note that if 0 6 x 6 y < r, this definition agrees with the above definition of
[x, y]r, and if y−x > r then [x, y]r = [0, r). It is apparent from the definition that if x 6 y,
then [x, y]r = [x− kr, y − kr]r for any integer k.
Definition 1.1. Let G be a graph and r a positive real number. An r–circular colouring
of G is a map c : V (G) → Cr such that |c(v) − c(w)|r > 1 for all edges uv of G. If such
a map exists, G is said to be r–circular colourable. The circular chromatic number of G is
defined by
χc(G) = inf {r : G is r–circular colourable} .
The terms r–circular colouring and r–circular colourable are often shortened to r–colouring
and r–colourable respectively. If r is an integer, it is known [50] that the existence of an
r–circular colouring of a graph is equivalent to the existence of an ordinary r–colouring.
Thus, using the shortened notation as described above does not cause any confusion with
ordinary colourability of graphs.
Example 1.2. It is immediate from the above definition that χc(K1) = 1 and χc(K2) = 2.
Remark 1.3. The condition |c(v) − c(w)|r > 1 in the above definition is equivalent to the
condition 1 6 |c(v) − c(w)| 6 r − 1.
By the above definition, an r–colouring of a graph G is also an r′–colouring of G for
every r′ > r. On the other hand, every k–colouring of G with colours from the set
{0, 1, . . . , k− 1} is a k–circular colouring of G. Therefore χc(G) 6 χ(G) for every graph G.
Not only the chromatic number of a graph G is an upper bound for its circular chromatic
number, it indeed is an approximation of χc(G) as stated in the following theorem.
Theorem 1.4. [50] For every graph G we have χ(G) − 1 < χc(G) 6 χ(G).
By restriction, any r–colouring of a graph G gives an r–colouring of every subgraph of G.
This proves the following.
Chapter 1. An Introduction to Circular Colouring 3
Lemma 1.5. Let G be a graph and H ⊆ G. Then χc(H) 6 χc(G).
Corollary 1.6. χc(G) = 1 if and only if G is a trivial graph.
Corollary 1.7. χc(G) = 2 if and only if G is a nonempty bipartite graph.
In particular, every even cycle has circular chromatic number 2. For a nontrivial example,
let C2k+1 = v1v2 · · · v2k+1v1 be the cycle of length 2k + 1 and r = 2 + 1/k. Then the map
c : V (C2k+1) → Cr defined by c(vi) = i + rZ is an r–colouring of C2k+1. Later we prove
that this is indeed best possible and we have the following.
Proposition 1.8. χc(C2k+1) = 2 + 1k for all k > 1.
The infimum in the definition of the circular chromatic number is attained for every finite
graph. Vince [50] proved that the circular chromatic number of every finite graph is rational.
A combinatorial proof of this result was found by Bondy and Hell [4].
1.1.1 (p, q)–Colouring
The original definition of the star chromatic number of a graph given by Vince [50] is
essentially the following.
Definition 1.9. Given positive integers p and q and a graph G, a (p, q)–colouring of G is
any map c : V (G) → {0, 1, . . . , p− 1} such that q 6 |c(v)− c(w)| 6 p− q for every edge vw
in G.
By Remark 1.3, if c is a (p, q)–colouring of a graph G, then the map c′ : v → c(v)/q is
a p/q–colouring of G. Indeed the converse of this observation is also true. Namely, the
existence of a p/q–colouring for a graph G implies (p, q)–colourability of G. A proof of this
can be found in [59]. Thus we have the following.
Theorem 1.10. For all G,
χc(G) = min
{
p
q: G has a (p, q)–colouring
}
.
This equivalent definition of the circular chromatic number is particularly useful in pre-
senting circular colourings of graphs since it only involves integers. This definition is also
more suitable for computer programming since the number of colours is finite.
Chapter 1. An Introduction to Circular Colouring 4
1.2 Graph Homomorphisms
Definition 1.11. Given graphs G and H, a homomorphism from G to H is any map
f : V (G) → V (H) such that for every edge vw of G, f(v)f(w) is an edge of H. We say G
maps to H, or G→ H for short, if such a homomorphism exists.
For positive integers p and q, let Kp/q be the graph defined by
n(p − q) + λ(x) − λ(y) 6 n(p − q) − 1. Now since pq > 2, we have p−1
q > 2. Thus
n(p− q) − (n− 1) > nq. So c′ is an (np− 1, nq)–colouring of G.
The proof of the above lemma provides an algorithm to obtain an improved colouring in
time O(|V (G)| + |E(G)|). Applying this algorithm to an acyclic (p, q)–colouring c of a
graph G, the improved colouring c′ is an (np− 1, nq)–colouring for some positive integer n.
If np−1, nq and all the colours c′(v) have a common factor, we can divide c′ by that factor
to obtain a colouring with smaller parameters. If c′ is acyclic, we can apply the algorithm
again to further improve the colouring. Repeatedly applying the algorithm results in a
sequence {ci} of (pi, qi)–colourings of G. Typically, after enough repeated applications of
the algorithm, pi becomes larger than |V (G)|, which usually implies a tight cycle cannot
exist. More precisely, if pi
gcd(pi,qi)> |V (G)|, then ci is acyclic. Lemma 2.4 now implies
that none of the numbers pi/qi equals χc(G). We let k/d be the largest valid candidate
(with respect to Lemma 2.5 and the upper bound pi/qi) for χc(G). The existence of a
(pi, qi)–colouring now proves that χc(G) 6 k/d. To obtain a proper (k, d)–colouring of G
we may proceed as follows. For each v ∈ V (G) we define ψi(v) by rounding kpici(v) to
the nearest integer. If ψi happens to be a proper (k, d)–colouring of G then we are done.
Otherwise, we apply the algorithm of Lemma 2.6 to ci in order to obtain ci+1 which is a
finer approximation to k/d, and repeat this process for ci+1. There is no guarantee that this
process will succeed, but for typical graphs one expects that to be the case. We illustrate
this discussion in the following example.
Example 2.7. Let G be the spindle graph shown in Figure 2.1(a). Since G is 4–chromatic
and it has order 7, by Lemma 2.5 we have χc(G) ∈ {7/2, 4}. We let c be the 4–colouring of
Chapter 2. Computational Aspects of Circular Colouring 11
3
1
0
2
0
3
1 λ = 0
λ = 1
λ = 1
λ = 1
λ = 2λ = 1
λ = 2
(a) (b)
9
4
1
7
2
10
5 27
13
4
22
8
31
17
(c) (d)
Figure 2.1: Application of the proof of Lemma 2.6 to the spindle graph
G given in Figure 2.1(a). The digraph Dc(G) and the corresponding values of λ are shown
in Figure 2.1(b). Note that since Dc(G) is acyclic, by Lemma 2.4 we have χc(G) = 7/2.
To obtain a (7, 2)–colouring of G we may apply the algorithm inherent in the proof of
Lemma 2.6 to c. Since the maximum value of λ is 2, we let n = 3. This gives an (11, 3)–
colouring c1 of G illustrated in Figure 2.1(c). It can easily be seen that ψ1 defined by
rounding the colouring 711c1 to the nearest integer, is not a proper (7, 2)–colouring of G.
We thus apply the algorithm of Lemma 2.6 to c1. This gives a (32, 9)–colouring c2 of G
illustrated in Figure 2.1(d). It is now observed that rounding 732c2 to nearest integers gives
a (7, 2)–colouring ψ2 of G.
Remark 2.8. In the above example, if c is a (4, 1)–colouring of G with exactly one vertex
coloured 3, then one application of Lemma 2.6 gives a (7, 2)–colouring since n can be
selected to equal 2. The less efficient colouring of Figure 2.1(a) illustrates the process of
repeated applications of the algorithm of Lemma 2.6 and “rounding”.
Chapter 2. Computational Aspects of Circular Colouring 12
2.4 Greedy Circular Colouring and Metaheuristics
Greedy colouring is the simplest graph colouring algorithm when dealing with ordinary
colourings of graphs. It starts with an ordering (permutation) of the vertices of a graph
G and iteratively assigns to the next vertex in the list of vertices the smallest available
positive integer as its colour. More precisely, if x1, x2, . . . , xn is an ordering of the vertices
of G, the greedy colouring algorithm assigns c(x1) = 1 in the first step, and for i > 2, in
the ith step it assigns
c(xi) = min (N \ {c(xj) : j < i and xjxi ∈ E(G)}) .
It is easy to make up examples in which this greedy algorithm requires many more than
χ(G) colours. For example if M = {aibi : 1 6 i 6 n} is a perfect matching in the
graph Kn,n and G = Kn,n \ M then the greedy algorithm applied to the permutation
a1, b1, a2, b2, . . . , an, bn uses n colours while G is bipartite.
On the other hand it is easily observed that there always exists a permutation of the
vertices of a given graph G, for which the greedy algorithm uses only χ(G) colours. Given
any χ(G)–colouring c of G with the colours 1, 2, . . . , χ(G), one such permutation is any
sequence v1, v2, . . . , vn such that
c(v1) 6 c(v2) 6 · · · 6 c(vn),
where n = |V (G)|. Therefore to find the minimum number of colours needed to properly
colour the vertices of a graph, is equivalent to find a permutation of the vertices of the
graph which minimizes the number of colours used by the greedy algorithm applied to
that permutation. This observation provides the opportunity of using randomized search
methods, including metaheuristics, for graph colouring. Culberson [10] has implemented
several variations of such algorithms including a tabu search. We use a similar approach
for circular colouring.
In the following we develop a greedy circular colouring algorithm and show how that can
be used in a tabu search algorithm for testing circular colourability. The main tool in our
algorithm is the fact that the imbalance of an orientation of a graph G can be computed
efficiently. This was first observed by Barbosa and Gafni [2]. They prove that imbal(−→G) can
be computed in O(|V (G)|6) for any acyclic orientation−→G of a graph G. Yeh and Zhu [56]
improved this result by introducing a new algorithm.
Chapter 2. Computational Aspects of Circular Colouring 13
Lemma 2.9. [56] Given an acyclic orientation−→G of a graph G, imbal(
−→G) can be computed
in O(|V (G)| · |E(G)|).
The algorithm of Yeh and Zhu is based on the following process: Let D0 =−→G and let
Di+1 be obtained from Di by reversing the orientation of all arcs of Di which are incident
to some sink. Since G has only finitely many orientations, there exist i and p such that
Di+p = Di. For every x ∈ V (G), let
t(x) = |{j : i 6 j < i+ p and x is a sink of Dj}|.
Then t(x) is independent of the choice of x. Moreover, if q = t(x) for some x ∈ V (G), then
imbal(−→G) = p
q .
Let π = (π1, π2, . . . , πn) be a permutation of the vertices of a graph G with |V (G)| = n.
Then G can be oriented according to π as follows: orient an edge πiπj of G from πi to πj,
if i < j, and for πj to πi otherwise. It is easy to observe that the orientation−→Gπ defined
above is acyclic. On the other hand, for any acyclic orientation−→G of G, there exists a (not
necessarily unique) permutation π such that−→G =
−→Gπ. One such permutation can be found
using a topological sort procedure on−→G . More precisely, we let πn be a sink of
−→G , and
then define (π1, . . . , πn−1) recursively to be a topological sort for−→G − πn.
We define the imbalance of a permutation of the vertices of G to be the imbalance of its
corresponding acyclic orientation. Then χc(G) is the smallest imbalance among permuta-
tions of the vertices of G. Taking advantage of the fact that the imbalance of a permutation
can be computed efficiently, one may investigate many permutations (possibly randomly
generated), to find one whose imbalance lies in a desired range.
2.4.1 Tabu search
Tabu search is a metaheuristic mathematical optimization method, which enhances the
performance of a local search method by using memory structures. Tabu search uses a
local or neighbourhood search procedure to iteratively move from a solution x to a solution
x′ in the neighbourhood of x, until some stopping criterion has been satisfied. Perhaps the
most important type of short-term memory to determine the solutions, also the one that
gives its name to tabu search, is the use of a tabu list. In its simplest form, a tabu list
contains the solutions that have been visited in the recent past.
Chapter 2. Computational Aspects of Circular Colouring 14
Given a graph G, we use tabu search for minimizing the imbalance function in the space
of all acyclic orientations of G. Since every acyclic orientation corresponds via topological
sort to a permutation of the vertices of G, we may equivalently search in the space of all
orderings of V (G). We need a neighbourhood structure in this space. The simplest choice
would be to say two permutations are adjacent if they differ by a transposition. So at each
step of the algorithm, we randomly generate a fixed number of permutations obtained from
the current permutation by applying a transposition, which are not in the tabu list (list of
k most recent permutations). Then we calculate the imbalance of each of these neighbours,
and select the one with the smallest imbalance. The algorithm stops as soon as it finds a
permutation with imbalance less than or equal to a given upper bound, or if some other
termination criterion, e.g. total number of iterations has reached a limit, is satisfied.
Chapter 3
Bounding Circular Chromatic
Number
In this chapter we present some upper and lower bounds for the circular chromatic number
of a graph.
3.1 Upper Bounds
Obtaining upper bounds on the circular chromatic number is often considered to be easier
than lower bounds. Nevertheless, there are several open problems concerning upper bounds.
The circular colourability of planar graphs with large girth (Conjecture 3.15) and the
circular colourability of cubic graphs of large girth (Problem 3.13) are two such problems.
In all the alternate definitions for the circular chromatic number of a graph given in Chap-
ter 1, χc is defined to be the minimum of a function, taken over some combinatorial objects.
Therefore given a graph G, each such object provides an upper bound for χc(G). For ex-
ample, finding a (p, q)–colouring of G proves that χc(G) 6pq .
Alternatively, any acyclic orientation−→G of G provides the upper bound χc(G) 6 imbal(
−→G).
For some graphs, for example graphs embedded in a surface with all faces having even
length, one could use the special structure of the graph to obtain “good” orientations
which have optimum or near-optimum imbalance. We use this approach when dealing with
15
Chapter 3. Bounding Circular Chromatic Number 16
even-faced projective planar graphs, and also in Theorem 3.7.
One other approach for finding upper bounds on the circular chromatic number is to use
the transitivity of graph homomorphism. More specifically, if G maps to H and H maps
to K then G maps to K. The special case K = Kp/q for some p and q implies that if
G → H and H is (p, q)–colourable, then G is (p, q)–colourable. In particular, if G → H,
then χc(G) 6 χc(H).
3.1.1 Even-Faced projective planar graphs
Embedded graphs have a more controlled behaviour with respect to colourings. For example
the 4–colour theorem states that every planar graph has chromatic number at most 4. It
is well-known that the chromatic number of graphs embedded in an orientable surface
are bounded by a function of the genus of the surface. Even-faced embeddings provide
even more structure for colouring. For example, every even-faced plane graph is bipartite.
Although there exist non-bipartite even-faced projective plane graphs, the following lemma
provides us with a powerful tool in dealing with their orientations.
Lemma 3.1. [21] Let G be a projective plane graph such that there exists an orientation−→G of G with respect to which all faces of G are perfectly balanced. Then G is bipartite.
Since odd cycles cannot be perfectly balanced in any orientation, in order for all faces to
be perfectly balanced in the above lemma, G needs to be an even-faced projective plane
graph. Goddyn and Verdian [21] proved that every even-faced projective plane graph G
has an orientation−→G such that the discrepancy of each cycle of G with respect to
−→G is
at most 2. This bounds the circular chromatic number of G away from 2. Indeed we can
determine χc(G) exactly for such graphs.
Given a graph G with a 2–cell embedding π in a surface X, a subgraph H of G is a
surface subgraph, if the embedding of H in X induced from π is a 2–cell embedding. For
an embedded graph G, we let maxfl(G) be the greatest length of a face of G. The following
result of Goddyn and Verdian-Rizi is one of our principal tools.
Theorem 3.2. [21] Let G be an even-faced projective plane graph and let 2ℓ = minH maxfl(H),
where H ranges over all surface subgraphs H of G. Then
χc(G) =2ℓ
ℓ− 1.
Chapter 3. Bounding Circular Chromatic Number 17
Figure 3.1: An even-faced embedding of P − x in the projective plane
Given a projective plane graph G, every contractible cycle C is the boundary of a face of
the surface subgraph obtained by deleting all the vertices inside C. On the other hand, if
C is a non-contractible cycle in G, C itself is a surface subgraph with exactly one face of
length 2|C|.
It is worth mentioning that Theorem 3.2 is a generalization of a result of Youngs [57] which
proves every non-bipartite quadrangulation of the projective plane is 4–chromatic.
Example 3.3. Let P be the Petersen graph and x ∈ V (P ). Then χc(P − x) = 3.
Proof. Consider the unique embedding of K3,3 in the projective plane and subdivide the
edges which go “through the cross-cap”. This gives an embedding of P −x in the projective
plane (Figure 3.1) with four faces of length 6. Since P − x has girth 5, we have 2ℓ = 6 and
Theorem 3.2 gives χc(P − x) = 3.
Note that the circular chromatic number of every cubic graph which contains P − x as
a subgraph equals 3. Thus we may combine P − x with cubic graphs of girth at least 5
to obtain girth 5 cubic graphs of arbitrary large order whose circular chromatic number
equals 3.
The proof of Example 3.3 can be applied to other subdivisions of K3,3, provided every
cycle has the same parity as its corresponding cycle in P − x. These graphs turn out as
subgraphs of some of the interesting graphs we discuss in this thesis.
Chapter 3. Bounding Circular Chromatic Number 18
3.2 Lower Bounds
The following lemma proves that a well-known lower bound for the ordinary chromatic
number also holds for the circular chromatic number.
Lemma 3.4. For every graph G, χc(G) >|V (G)|α(G) .
This lower bound is weak in the sense that it is not tight for many “interesting” graphs.
Moreover, the independence number of a graph is hard to compute. In the following we
prove a more general bound. For a graph G and a positive integer k, αk(G) denotes the
maximum size of a union of k independent sets in G. In other words, αk(G) is the maximum
order of an induced k–partite subgraph of G. The following lemma is also implied by
Proposition 1.22 of [30, Page 14].
Lemma 3.5. [58] Let G be a graph and let k 6 χc(G) be a positive integer. Then
χc(G) >k|V (G)|αk(G)
.
Proof. Suppose χc(G) = p/q and let c be a (p, q)–colouring of G. In the following all colours
are reduced modulo p. For all 0 6 t 6 p − 1, we let It = c−1({t, t + 1, . . . , t + q − 1}).Then It is an independent set in G. Since k 6 p/q, for all t, the sets It, It+1, . . . , It+q−1 are
pairwise disjoint. We let Jt = It ∪ It+1 ∪ . . . ∪ It+q−1. Then
|Jt| = |It| + · · · + |It+q−1| 6 αk(G). (3.1)
On the other hand a vertex v ∈ V (G) belongs to Jt if and only if c(v) − kq + 1 6 t 6 c(v).
Thereforep−1∑
t=0
|Jt| = kq|V (G)|. (3.2)
The result follows from (3.1) and (3.2).
Lemma 1.5 gives a lower bound for χc(G) in terms of its subgraphs. This method could be
useful when a graph is regularly structured, or when it is sparse. In particular, χc(G) >
ωc(G) where
ωc(G) = min
{
p
q: Kp/q ⊆ G
}
Chapter 3. Bounding Circular Chromatic Number 19
is the circular clique number of G.
Similarly, one may consider odd cycles in G to obtain a lower bound on χc(G). Note that
since K(2k+1)/k is isomorphic to C2k+1, this lower bound is weaker than the one obtained
via the circular clique number. On the other hand, finding short odd cycles in a graph is
much easier than finding general Kp/q subgraphs.
3.2.1 The odd-girth bound
The odd-girth of a graph G is the length of the shortest odd cycle in G. By Lemma 1.5,
if G has odd-girth 2k + 1, then χc(G) > 2 + 1k . Most of the time, e.g. when G is not
3–colourable, this bound is not tight. If this bound is tight, then G is said to be odd-girth
colourable. A result of Gerards [17] gives a forbidden subgraph characterization of graphs
which have an orientation in which each cycle has discrepancy at most 1.
Theorem 3.6. [17] If a non-bipartite graph G is not odd-girth colourable, then G contains
an odd K4 or an odd K23 as a subgraph.
In the above theorem, an odd K4 is any subdivision G of K4 in which every cycle has the
same parity as its corresponding cycle in K4. Equivalently, in a planar drawing of G, all
faces must be odd. An odd K23 is a graph consisting of three odd cycles C1, C2 and C3
and three vertex-disjoint paths P1, P2, P3, possibly of length 0, such that Pi joins a vertex
of Ci to a vertex of Ci+1. Here the indices are reduced modulo 3. See Figure 3.2 for an
illustration.
Theorem 3.6 motivates the study of the circular chromatic number of odd K4 and odd K23
graphs. In the following we establish the exact values of the circular chromatic number of
these graphs. A thread in a graph G is a maximal path which is internally disjoint from the
rest of G. Namely, a thread is a path ax1 · · · xkb such that in G, d(xi) = 2 for 1 6 i 6 k,
and d(a) 6= 2 and d(b) 6= 2.
Theorem 3.7. Let G be an odd K4 or an odd K23 with odd-girth g. Then G is odd-girth
colourable if and only if it has an even cycle of length at least 2g. Moreover, if the longest
even cycle in G has length 2ℓ < 2g, then
χc(G) =2ℓ
ℓ− 1.
Chapter 3. Bounding Circular Chromatic Number 20
Proof. Consider a drawing of K4 in the plane consisting of a convex 4–gon and its two
diagonals. By routing the two diagonals through a cross-cap, we obtain an embedding of
K4 in the projective plane, with three faces of length 4. Subdividing this embedded graph
according to an odd K4 graph G, we obtain an even-faced embedding of G in the projective
plane. The result now follows from Theorem 3.2.
Let G be an odd K23 consisting of odd cycles C1, C2, C3 and paths P1, P2, P3 such that Pi
joins yi ∈ V (Ci) to xi+1 ∈ V (Ci+1). Let L be the combined length of the paths P1, P2, P3,
and for i ∈ {1, 2, 3} let ai and bi be the lengths of the two xiyi–paths on Ci. Since ai and bi
have opposite parity, we may assume ai has the same parity as L. Given an orientation−→G
of G, we claim that at least one even cycle of G is unbalanced with respect to−→G . Consider
the planar drawing of G given in Figure 3.2, and fix the clockwise direction as the positive
direction. Let a+i (resp. a−i ) denote the number of edges of the xiyi–path of length ai on
Ci whose direction agrees (resp. disagrees) with the positive direction (from xi to yi). We
similarly define b+i , b−i , L+, and L−. Since the even cycles of G have lengths L+a1+a2+a3
and L+ ai + bj + bk where {i, j, k} = {1, 2, 3}, if all the even cycles of G are balanced with
in G (See Figure 4.3 for an illustration). We label the vertices of G by v1, v2, . . . , v2n
according to C. For 1 6 i 6 2n, we define c(vi) to equal iq reduced modulo p. Obviously
c is a (p, q)–colouring of C. On the other hand, for each vivj ∈ E(G) \ E(C) we have
|j − i| ∈ {4, 7, 10} by the choice of C. Therefore c(vi) − c(vj) equals one of 4q = 4k,
7q = 5k − 1 and 10q = 6k − 2 modulo p. The first two distances are always valid while
the last one is valid only when k 6 5. Therefore every edge of G is valid with respect to c
which gives χc(G) 6 p/q = 8n3n−2 .
Let k > 5 and let c0 be the (44, 16)–colouring of P (22, 2) obtained above. Since for all
vertices v, c0(v) is a multiple of 16 reduced modulo 44, and since gcd(44, 16) = 4, we
Chapter 4. Circular Chromatic Number of Some Special Graphs 31
A
B
C
A
B
C
Figure 4.3: A Hamiltonian cycle of P (14, 2) used in the proof of Theorem 4.7. Semiedgeswrap around according to their labels.
5
9
2
6 10
3
7
0
Figure 4.4: A block of the (11, 4)–colouring of P (22, 2) used in the proof of Theorem 4.7.
see c0(v) is a multiple of 4 and hence all colours can be divided by 4 to obtain an (11, 4)–
colouring c1 of P (22, 2). In this colouring, the vertices xi and yi of P (22, 2) with 19 6 i 6 22
are coloured as shown in Figure 4.4. Since c1(x19) and c1(c22) have distance 6, the colouring
of this block can be repeated t times to obtain an (11, 4)–colouring of P (22 + 4t, 2).
For n > 22, the lower bound 11/4 is already proved in 4.5. For n = 6 the equality holds
since P (6, 2) has triangles. For n = 10, 14, 18 (also for n = 22) the lower bound is implied
by Lemma 3.5 as follows. We let
I = {x2, x4, y1, y5} ∪ {x2i+5 : 1 6 i 6 2k − 2} ∪ {y4i+2 : 1 6 i 6 k},
and
I ′ = {y2} ∪ {x2i+4 : 1 6 i 6 2k − 1} ∪ {y4i−1 : 1 6 i 6 k}.
Then I and I ′ are disjoint independent sets in G and |I ∪ I ′| = |I| + |I ′| = (3k + 2) + 3k.
Thus α2(G) > 6k + 2 and by Lemma 4.6 we have α2(G) = 6k + 2. Lemma 3.5 now gives
χc(G) >2(4k + 2)
6k + 2=
8n
3n− 2.
Corollary 4.8. The Dodecahedron graph P (10, 2) has circular chromatic number 20/7.
Chapter 4. Circular Chromatic Number of Some Special Graphs 32
It remains to determine χc(P (n, 2)) when n is odd. As we mentioned before, P (5, 2) is the
Petersen graph which has circular chromatic number 3. In the following we prove that the
same holds for P (7, 2).
Theorem 4.9. χc(P (7, 2)) = 3.
Proof. By an easy case analysis we show that every 3–colouring of P (7, 2) has a tight
6–cycle. Note that permuting the colours in a 3–colouring does not affect tight cycles.
This significantly reduces the number of cases to be considered. Let c be a 3–colouring of
P (7, 2). Since in every 3–colouring of the cycle C = x1x2 · · · x7x1, at least one colour class
has size 3, we may assume that c(x2) = c(x4) = c(x6) = 0, c(x1) = 1, and c(x7) = 2. This
forces c(y2) = c(y6) = α and c(y4) = β with {α, β} = {1, 2}.
Case 1. One other colour, say 1, is used three times on C. Then c(x1) = c(x3) = c(x5) = 1.
Hence c(y1) = c(y5) = α′ and c(y3) = β′ with {α′, β′} = {0, 2}. Since y1 is adjacent to y6
we have α 6= α′ and since the neighbours of y7 are coloured 2, α, and α′, one of α and α′
equals 2. The two subcases are shown in Figure 4.5 (a) and (b). In each case a tight cycle
is highlighted.
Case 2. c(x3) = 1 and c(x5) = 2. Then one of y1 and y3 is coloured 0 and similarly one
of y5 and y7 is coloured 0. By symmetry, we may assume that c(y3) = 0. This forces the
colours of the remaining vertices. This case is shown in Figure 4.5 (c).
Case 3. c(x3) = 2 and c(x5) = 1. Based on the value of α we have two subcases shown in
Figure 4.5 (d) and (e).
Computationally, by finding a (p, q)–colouring of the graph at hand and verifying that all
such colourings have a tight cycle, we proved
χc(P (n, 2)) =
17/6 for n ∈ {9, 11},
14/5 for n ∈ {13, 15, 17, 19}.
We conjecture the following.
Conjecture 4.10. For all odd n > 13, χc(P (n, 2)) = 14/5.
The upper bound is proved in the following proposition. We leave the lower bound open.
Chapter 4. Circular Chromatic Number of Some Special Graphs 33
(a)
A
B
C A
B
C
1 1 1 2
0 0 0
0 2 0 1
2 1 2
(b)
A
B
C A
B
C
1 1 1 2
0 0 0
2 0 2 0
1 2 1
(c)
A
B
C A
B
C
1 1 2 2
0 0 0
2 0 1 0
1 2 1
(d)
A
B
C A
B
C
1 2 1 2
0 0 0
2 0
1 2 1
(e)
A
B
C A
B
C
1 2 1 2
0 0 0
0 1
2 1 2
Figure 4.5: 3–Colourings of P (7, 2). Semiedges wrap around according to their labels.
Chapter 4. Circular Chromatic Number of Some Special Graphs 34
A
B
C A
B
C
10
1
7
2
8
13
4
10
5
12
3
9
4
5 12 3 9 0 8 13
6 11 6 1 7 0
A
B
C A
B
C
10
1
9
4
13
7
12
6
1
10
5
12
3
9
4
5 0 8 3 9 0 8 13
6 11 2 11 2 7 0
Figure 4.6: (14, 5)–colourings of P (13, 2) and P (15, 2). Semiedges wrap around accordingto their labels
Proposition 4.11. Given odd n > 13, we have χc(P (n, 2)) 6 14/5.
Proof. In Figure 4.6 we present (14, 5)–colourings of P (13, 2) and P (15, 2). In each colour-
ing, the block contained in the shaded box can be repeated t times to obtain (14, 5)–
colourings of the graphs P (13 + 4t, 2) and P (15 + 4t, 2).
4.1.2 The graphs P (n, 3)
By Proposition 4.2, when d is odd the study of the circular chromatic number of P (n, d) is
non-trivial only when n is also odd. For d = 3, by the definition we must have n > 6. It is
not difficult to see that P (7, 3) is isomorphic to P (7, 2) and hence it has circular chromatic
number 3. We begin by proving an upper bound.
Proposition 4.12. For all odd n > 9,
χc(P (n, 3)) 62n
n− 3.
Proof. If 3|n then n = 6k + 3 for some k > 1 and 2nn−3 = 2k+1
k . A (2k + 1, k)–colouring of
P (6k + 3, 3) is given by c(x1) = c(y2) = c(x3) = 0, c(y1) = c(x2) = c(y3) = k, c(xi+3) =
c(xi) + k, and c(yi+3) = c(yi) + k.
Chapter 4. Circular Chromatic Number of Some Special Graphs 35
If n is not a multiple of 3, let (p, q) =(
n, n−32
)
. For 1 6 i 6 n we define c(y3i) = iq where
3i and iq are both reduced modulo n = p. Then c(yi+1) = c(yi) ± q. Therefore if we let
c(xi) = c(yi+1), c is a (p, q)–colouring of P (n, 3).
Theorem 4.13. For odd n > 7 the following are true.
• If 3|n, then χc(P (n, 3)) = 2 + 6n−3 .
• If n ≡ −1 (mod 6), then 2 + 6n+1 6 χc(P (n, 3)) 6 2 + 6
n−3 .
• If n ≡ 1 (mod 6), then 2 + 6n+2 6 χc(P (n, 3)) 6 2 + 6
n−3 .
Proof. All upper bounds are from Proposition 4.12. When n = 6k + 3 is a multiple of 3,
the subgraph of P (n, 3) induced by the vertices yi consists of three cycles of length 2k+ 1.
Thus the upper bound is tight.
For n = 6k − 1, the graph P (n, 3) − {x3i, y3i : 1 6 i 6 2k − 1} has an embedding in the
projective plane with one face of length 6, 2k − 1 faces of length 8, and one face of length
4k + 2. Using Theorem 3.2 one gets the lower bound. The case k = 3 is illustrated in
Figure 4.7.
For n = 6k + 1, the graph P (n, 3) − {x3i, y3i : 1 6 i 6 2k} has an embedding in the
projective plane with 2k faces of length 8 and one face of length 4k + 4. Again applying
Theorem 3.2 gives the lower bound. The case k = 3 is illustrated in Figure 4.7.
The lower and upper bounds of Theorem 4.13 for χc(P (n, 3)) are asymptotically equal.
On the other hand, for small values of n these bounds are far apart. In particular since
P (7, 3) ∼= P (7, 2), the above theorem gives 3 = χc(P (7, 2)) ∈[
83 ,
72
]
.
We conjecture that the upper bound of Proposition 4.12 is tight when n is large enough.
Conjecture 4.14. For odd n > 9,
χc(P (n, 3)) =2n
n− 3.
By Theorem 4.13, this conjecture is true when n is a multiple of 3. We verified this
conjecture computationally for n 6 30.
Chapter 4. Circular Chromatic Number of Some Special Graphs 36
A
B
A
B
x1
x2
x4
x5
x7
x8
y1
y2
y4
y5
y7
y8
B
A
B
Ax1
x2
x4
x5
x7
x8
y1
y2
y4
y5
y7
y8
Figure 4.7: Even-faced projective plane embeddings of subgraphs of P (11, 3) and P (13, 3).Semiedges are routed through a cross-cap, according to their labels.
4.2 Squares of Graphs
Colourings of powers of graphs are of importance in coding theory. For example, the
chromatic number of second and third powers of hypercube graphs is of interest in the
study of scalability of optical networks [51].
Given a graph G and an integer k > 1, the kth power of G, denoted G(k) is the graph
with the same vertex set as G in which two vertices are adjacent if and only if they are at
distance at most k in G. The graph G(2) is called the square of G.
An obvious lower bound for χc
(
G(2))
is ∆(G)+1. This is because the closed neighbourhood
of each vertex of G induces a clique in G(2). In this section we study the circular chromatic
numbers of squares for some families of graphs. Before that, as an example we compute
the circular chromatic number of squares of cycles.
Example 4.15. For all n > 3,
χc
(
C(2)n
)
=n
⌊n/3⌋ .
Proof. Let G = C(2)n . Since every two non-adjacent vertices of G have distance at least
3 in Cn, we have α(G) 6 ⌊n/3⌋. Thus χc(G) >n
⌊n/3⌋ by Lemma 3.4. On the other
hand, if v1, v2, . . . , vn are the vertices of Gn in the cyclic order, the colouring c defined by
Chapter 4. Circular Chromatic Number of Some Special Graphs 37
c(vi) = i ⌊n/3⌋ (modulo n) is a proper (n, ⌊n/3⌋)–colouring of G.
4.2.1 Squares of hypercube graphs
It is proved in [51] that for all positive d,
d+ 1 6 χ(
Q(2)d
)
6 2⌊log2d⌋+1. (4.1)
It is conjectured in [51] that the above upper bound is tight. Our computations provide
evidence that even the circular chromatic number of Q(2)d equals the upper bound of (4.1).
Note that the value of the upper bound in (4.1) is constant when d lies between consecutive
powers of 2. On the other hand Qd contains Qd−1 as a subgraph. Thus it suffices to prove
the above conjecture when d is a power of 2. The cases d = 1, 2 are trivial since Q(2)1 is
isomorphic to K2 and Q(2)2 is isomorphic to K4.
Proposition 4.16. χc
(
Q(2)4
)
= 8.
Proof. Let G = Q(2)4 . For each x ∈ V (G), the vertices at distance 1 or 2 from x in Q4 are
adjacent to x in G, and the vertices at distance 3 or 4 from x in Q4 form a clique in G.
Therefore α(G) 6 2, which gives χc(G) > 8. Indeed every set {x, x}, where x ∈ V (G) and
x is the unique vertex at distance 4 from x in Q4, is an independent set in G. Assigning
the same colour to any such pair gives an 8–colouring of G.
By the above proposition and by (4.1) we have χc
(
Q(2)d
)
= 8 for d = 4, 5, 6, 7.
Conjecture 4.17. For all positive d, χc
(
Q(2)d
)
= 2⌊log2d⌋+1.
The first open case of Conjecture 4.17 is when d = 8. An independent set in Q(2)8 is in
fact a binary code with codewords of length 8 and with minimum distance at least 3. It is
known (cf. [8]) that the maximum size of such code is 20. Therefore Q(2)8 has independence
number 20, and the bound of Lemma 3.4 gives χc
(
Q(2)8
)
> 12.8 while the conjectured
value is 16.
Chapter 4. Circular Chromatic Number of Some Special Graphs 38
4.2.2 Squares of prisms
We consider prisms of cycles, namely the graphs K2 � Cn, where � denotes the Cartesian
product. The study of the circular chromatic number of squares of prisms is partly moti-
vated by the fact that the hypercube graph Q3 is isomorphic to K2 �C4. Throughout this
section we denote Gn = K2 � Cn.
The graph G3 has diameter 2 thus G(2)3 is a 6–clique and χc
(
G(2)3
)
= 6. Since C(2)5
is a 5–clique, we see that χc
(
G(2)5
)
> 5. The results of this section prove that indeed
χc
(
G(2)5
)
= 5. The aim of this section is to prove the following.
Theorem 4.18. For all n > 3,
χc
(
G(2)n
)
=
4 if 4|n,4n
n−1 if n is odd,
4nn−2 if n ≡ 2 (mod 4).
It turns out that these values are all equal to the lower bound of Lemma 3.4. Thus we need
to prove the upper bounds and calculate the independence numbers of the graphs G(2)n for
the above result. The upper bounds are less trivial, and we deal with them first in the
following lemmas.
Lemma 4.19. If n is a multiple of 4, then χc
(
G(2)n
)
6 4.
Proof. Let V (Cn) = {1, 2, . . . , n} in the cyclic order and V (K2) = {1, 2}. Then V (G) =
{(i, j) : 1 6 i 6 2 and 1 6 j 6 n}. We define c(1, j) = j (mod 4) and c(2, j) = j + 2
(mod 4). It is easy to see that c is a 4–colouring of G(2)n .
Lemma 4.20. For odd n > 3 we have χc
(
G(2)n
)
64n
n−1 .
Proof. Similarly to the previous proof we have V (G) = {(i, j) : 1 6 i 6 2 and 1 6 j 6
2k + 1}. For convenience, we write xj = (1, j) and yj = (2, j).
If n = 4q + 1 for some integer q, then a (4q + 1, q)–colouring of G(2)n can be defined by
c(i, j) = (2i+ j)q (mod n). Here the end-vertices of each edge of G(2) receive colours with
distance q, 2q, or 3q which are all valid distances for a (4q + 1, q)–colouring.
Chapter 4. Circular Chromatic Number of Some Special Graphs 39
Figure 4.8: A Hamiltonian cycle in G(2)7 used in the proof of Lemma 4.20
If n ≡ −1 (mod 4), then let (p, q) = (2n, n−12 ). We define
and define c(xi) = c(yi+2) = iq (mod p). Similarly to the previous case, one sees that c is
a proper (p, q)–colouring of V(2)2n .
Chapter 4. Circular Chromatic Number of Some Special Graphs 42
Proposition 4.26. If n > 4 is a multiple of 4, then χc
(
V(2)2n
)
64n
n−2 .
Proof. Let (p, q) =(
2n, n−22
)
and define c(vi) = iq (mod p). Then
|c(v2) − c(v1)| = |c(v2n) − c(v1)| = q,
|c(v3) − v(v1)| = |c(v2n−1 − c(v1)| = 2q,
|c(vn+1) − c(v1)| = nq = 2q + 2 (mod p),
|c(vn) − v(v1)| = q + 2 (mod p), and
|c(vn+2 − c(v1)| = 3q + 2 (mod p).
Therefore c is a proper (p, q)–colouring of V(2)2n .
Lemma 4.27. For all n > 3 we have
α(V(2)2n ) =
n2 if n ≡ 2 (mod 4),
n−12 if n ≡ ±1 (mod 4),
n−22 if n ≡ 0 (mod 4).
Proof. Note that the colourings given in Propositions 4.24, 4.25, and 4.26 give independent
sets of the desired size in the graphs V(2)2n . So we only need to prove the upper bounds on
the independence number.
Let the vertices of V2n be v1, v2, . . . , v2n where for 1 6 i 6 2n, the vertex vi is adjacent to
vi−1, vi+1, and vi+n (all subscripts are reduced modulo n). Let I be an independent set in
V(2)2n . If vi ∈ I, then none of the vertices vi+1, vi+n, and vi+n+1 can be in I. Therefore I can
have at most one vertex from any two consecutive diagonals vivi+n and vi+1vi+n+1. This
proves the upper bound when n is odd and when n ≡ 2 (mod 4). On the other hand, if n is
a multiple of 4, the above analysis shows that an independent set of size n/2, if one exists,
has one vertex from every second diagonal. Namely, one vertex from the diagonals v2iv2i+n.
Moreover, since vi is adjacent to vi+2, such independent set has to alternate between the
vertices v2i and v2i+n where 1 6 i 6 n/2. Therefore, up to isomorphism, such independent
set equals {v2, v4+n, v6, v8+n . . . , vn+n}. But v2n is adjacent to v2. This contradiction proves
that α(V(2)2n ) 6 n/2 − 1.
Chapter 4. Circular Chromatic Number of Some Special Graphs 43
4.2.4 Squares of flower snarks
Flower snarks were first introduced by Isaacs [31] as the first infinite family of well-connected
cubic graphs which are not 3–edge colourable. We study circular edge colourings of these
graphs in Section 7.2. Unlike prisms and Mobius ladders, for squares of flower snarks the
lower bound of Lemma 3.4 is not tight. This makes the study of the circular chromatic
number of squares of these graphs more interesting.
Given an odd integer n > 3, the flower snark Jn consists of two cycles x1x2 · · · xnx1 and
y1y2 · · · y2ny1 and n claws {zixi, ziyi, ziyn+i}, where 1 6 i 6 n. Some flower snarks are illus-
trated in Figure 4.9. In this figure, the vertices zi are on the top row of each picture while
the top horizontal line represents the cycle x1x2 · · · xnx1 and the bottom two horizontal
lines represent the cycle y1y2 · · · y2ny1.
Proposition 4.28. For all odd n > 3, we have α(J(2)n ) = n.
Proof. Each claw {zixi, ziyi, ziyi+n} of Jn induces a 4–clique in J(2)n , thus α(J
(2)n ) 6 n. On
the other hand, the set {z1, z2, . . . , zn} is an independent set in J(2)n .
Since J(2)n has 4n vertices, Lemma 3.4 gives χc
(
J(2)n
)
> 4 which is the trivial lower bound
for any graph with a vertex of degree 3 or more. The actual value of χc
(
J(2)n
)
on the other
hand, turns out to be always greater than 4. In the following we first calculate the circular
chromatic number of J(2)3 and J
(2)5 . For the latter graph our proof is computer-assisted.
Then we establish all other cases in Theorem 4.31.
Proposition 4.29. χc
(
J(2)3
)
= 7.
Proof. The subgraph of J(2)3 induced by the set {y1, y2, . . . , y6, z1} is a 7–clique. Therefore
we have χc
(
J(2)3
)
> 7. On the other hand, a 7–colouring of this clique can be easily
extended to J(2)3 .
Proposition 4.30. χc
(
J(2)5
)
= 6.
Proof. The subgraph of J(2)5 induced by the xi is a 5–clique. Therefore we need at least 5
colours to colour this graph. Suppose J(2)5 has a 5–colouring c. We may assume c(xi) = i.
Then c(z1) equals one of c(x3) and c(x4). Without loss of generality we assume c(z1) = 3.
Chapter 4. Circular Chromatic Number of Some Special Graphs 44
Now c(z2) ∈ {4, 5} and c(y2), c(y7) ∈ {1, 4, 5}. Therefore one of c(y2) and c(y7) equals 1.
This implies that c(z3) = 4. Similarly, since c(y4), c(y9) ∈ {1, 2, 3} and c(z4) ∈ {1, 2}, one
of c(y4) and c(y9) must be 3, which implies c(z5) = 2. Finally we have c(y1), c(y6) ∈ {3}.This is a contradiction since y1 and y6 are adjacent. This proves χ
(
J(2)5
)
= 6.
To show χc
(
J(2)5
)
= 6, it suffices to show that every 6–colouring of J(2)5 has a tight cycle.
We verified this using a computer program.
Before the next result, we need the following definition. Let r > 4 and let (α, β, γ) be
a triple of colours in Cr. We say this triple has positive sign if β ∈ [α, γ]r and we say
this triple has negative sign otherwise. For example, the triples (1, 2, 3) and (3, 1, 2) have
positive sign while (1, 3, 2) has negative sign.
Theorem 4.31. For all odd n > 7, we have χc
(
J(2)n
)
= 5.
Proof. Since Jn is 3–regular, χc
(
J(2)n
)
> 4. Suppose χc
(
J(2)n
)
= r = 4 + ε < 5 and let c
be an r–circular colouring of J(2)n .
Without loss of generality we may assume that c(z1) = 0. Since the vertices {z1, x1, y1, yn+1}induce a 4–clique, each of the colours c(x1), c(y1), and c(yn+1) must be in one of the in-
tervals Ij = [j, j + ε] with j = 1, 2, 3, and no two of these can be selected from the same
interval. Similarly, since the vertices {z1, x2, y2, yn+2} induce a 4–clique, c(x2), c(y2), and
c(yn+2) must satisfy the same conditions. On the other hand, the colours of the end-
vertices of the edges x1x2, y1y2, and yn+1yn+2 must be selected from different intervals.
Note that since ε < 1, the intervals Ij are mutually disjoint and therefore the triples of
colours (c(x1), c(y1), c(yn+1)) and (c(x2), c(y2), c(yn+2)) have the same sign.
A similar argument proves that for all 1 6 i 6 n, the triples (c(xi), c(yi), c(yn+i)) and
(c(xi+1), c(yi+1), c(yn+i+1)) have the same sign. Here xn+1 = x1 and y2n+1 = y1. In par-
ticular, the triple (c(x1), c(y1), c(yn+1)) has the same sign as (c(xn+1), c(yn+1), c(y2n+1)) =
(c(x1), c(yn+1), c(y1)). This is a contradiction. Therefore χc
(
J(2)n
)
> 5.
It remains to give a 5–colouring of J(2)n for n > 7. In Figure 4.9 we present 5–colourings of
J(2)7 , J
(2)9 , and J
(2)11 . In each picture the shaded block can be repeated t > 0 times to obtain
5–colourings for J(2)7+3t, J
(2)9+3t, and J
(2)11+3t.
Chapter 4. Circular Chromatic Number of Some Special Graphs 45
1 2 3 1 4 2 3
2 3 1 2 3 1 2
3 1 2 4 1 3 4
1 2 3 1 2 3 1 2 3
2 3 1 2 3 4 2 1 4
3 1 2 3 1 2 4 3 1
1 2 3 1 2 3 1 4 3 2 4
2 3 1 2 3 1 2 3 1 4 2
3 1 2 3 1 2 3 1 2 3 1
Figure 4.9: 5–Colourings of squares of flower snarks. Vertices with no label are coloured 0.Semiedges wrap around.
Chapter 4. Circular Chromatic Number of Some Special Graphs 46
a b
x
y
Figure 4.10: The unit distance graph Ha,b
Remark 4.32. The flower snarks can be similarly defined for even n > 4. The resulting
graphs are all bipartite and thus trivial for vertex or edge colouring. The squares of these
graphs on the other hand, all have circular chromatic number 5. This is because the proof
of Theorem 4.31 does not depend on the parity of n.
4.3 The Plane Unit-Distance Graph
The results presented in this section are published in [13].
The plane unit distance graph R is defined to be the graph with vertex set R2 in which two
vertices (points in the plane) are adjacent if and only if they are at Euclidean distance 1.
Every subgraph of R is also said to be a unit distance graph. It is known that (cf. [27, 42])
4 6 χ(R) 6 7.
It is known (cf. [48, pp. 59–65]) via a lower bound on the fractional chromatic number of
R, that χc (R) > 32/9. We improve this lower bound to 4.
Let a and b be two points in the plane, and let d(a, b) denote the Euclidean distance
between a and b. If d(a, b) =√
3, then we may find points x and y in the plane such that
the subgraph of R induced on the set {a, b, x, y} is isomorphic to the graph H obtained by
deleting one edge from K4 (see Figure 4.10). We denote this unit distance graph by Ha,b.
On the other hand, it is easy to see that, in any embedding of H as a unit distance graph
in the plane, the Euclidean distance between the two vertices of degree 2 in H is√
3.
Lemma 4.33. Let 0 < ε < 1, r = 3 + ε, and let a, b ∈ R2 with d(a, b) =
√3. Let c be an
r–circular colouring of Ha,b. Then |c(a) − c(b)|r 6 ε.
Chapter 4. Circular Chromatic Number of Some Special Graphs 47
Proof. We label the vertices of H as in Figure 4.10. Without loss of generality, we may
assume c(a) = 0. Since a, x, y form a triangle in Ha,b, we have c(x) ∈ [1, 1 + ε]r and
c(y) ∈ [2, 2 + ε]r or vice versa. On the other hand, b is adjacent to both x and y. Thus
Since this is true for every µ′ < µ, we have µ > 1. This is a contradiction since µ 6 ε <
1.
We do not know of a finite subgraph of R with circular chromatic number 4. In the
remainder of this section. we construct finite subgraphs of R with circular chromatic
number arbitrarily close to 4.
Chapter 4. Circular Chromatic Number of Some Special Graphs 48
Gn
Gn
Gn
Gn
x
y
y′
z
Figure 4.11: Construction of Gn+1 from Gn.
Figure 4.12: The Moser (spindle) graph.
The graph G0 = K2 is obviously a unit distance graph. In our construction of graphs
Gn (n > 0) we distinguish two vertices in each of them. To emphasize the distinguished
vertices x and y of Gn, we write Gx,yn . We identify each subgraph of R with its geometric
representation given by its vertex set.
For n > 0, the graph Gn+1 is constructed recursively from four copies of Gn. Let S =
V (Gx,yn ) ⊆ R
2. Let us rotate the set S in the plane about the point x, so that the image
y′ of y under this rotation is at distance 1 from y. Let S′ be the image of S under this
rotation. Let T be the set of all points in S ∪ S′ and their reflections about the line yy′.
In particular let z ∈ T be the reflection of x about the line yy′. We define Gx,zn+1 to be the
subgraph of R induced on T . This construction is depicted in Figure 4.11.
Note that G1 is the graph Ha,b of Figure 4.10 and G2 contains the Moser graph shown in
Figure 4.12 as a subgraph. The Moser graph, also known as the spindle graph, was the
first 4–chromatic unit distance graph discovered [42].
Lemma 4.35. For every n > 1, we have χc(Gn) > 4 − 21−n. Moreover, for every r =
4 − 21−n + ε with 0 6 ε < 21−n, and for every r–circular colouring c of Gx,zn , we have
|c(x) − c(z)|r 6 2n−1ε.
Chapter 4. Circular Chromatic Number of Some Special Graphs 49
Proof. We use induction on n. The nontrivial part of the case n = 1 is proved in
Lemma 4.33. Let n > 1 and Gx,zn+1 be as shown in Figure 4.11. Let r = 4 − 21−n + ε
for some ε > 0, and let c be a r–circular colouring of Gx,zn+1. Without loss of generality
we may assume that c(x) = 0. By the induction hypothesis, |0 − c(y)|r and |0 − c(y′)|rare both at most 2n−1ε. Hence |c(y) − c(y′)|r 6 2nε. On the other hand, since y and
y′ are adjacent in Gx,zn+1, we have |c(y) − c(y′)|r > 1. Therefore ε > 2−n, and we have
χc(Gn+1) > 4 − 21−n + 2−n = 4 − 2−n.
Now let r = 4 − 2−n + ε for some 0 6 ε < 2−n, and let c be a r–circular colouring of
Gn+1 with c(x) = 0. Note that r = 4 − 21−n + ε′, with ε′ = 2−n + ε < 21−n. By the
induction hypothesis, |0 − c(y)|r, |0 − c(y′)|r, |c(z) − c(y)|r, and |c(z) − c(y′)|r are all at
most 2n−1ε′ < 1. Therefore we have
c(y), c(y′) ∈[
−2n−1ε′, 2n−1ε′]
r
and
c(z) ∈[
c(y) − 2n−1ε′, c(y) + 2n−1ε′]
r∩
[
c(y′) − 2n−1ε′, c(y′) + 2n−1ε′]
r.
Since |c(y)−c(y′)|r > 1, one of the colours c(y) and c(y′), say c(y), is in the circular interval[
−2n−1ε′, 2n−1ε′ − 1]
r, and c(y′) ∈
[
−2n−1ε′ + 1, 2n−1ε′]
r. Therefore
[
c(y) − 2n−1ε′, c(y) + 2n−1ε′]
r⊆
[
−2nε′, 2nε′ − 1]
r=
[
−2nε′, 2nε]
r
and[
c(y′) − 2n−1ε′, c(y′) + 2n−1ε′]
r⊆
[
−2nε′ + 1, 2nε′]
r=
[
−2nε, 2nε′]
r.
Finally, since ε′ < 21−n, we have 2nε′ < r − 2nε′. Hence
c(z) ∈[
−2nε′, 2nε]
r∩
[
−2nε, 2nε′]
r= [−2nε, 2nε]r .
This completes the induction step.
Let us observe that, when constructing Gn+1 from four copies of Gn, it may happen that
vertices in distinct copies of Gn correspond to the same points in the plane. Additionally, it
may happen that some new edges between vertices in distinct copies of Gn are introduced
because their ends have distance 1. Let H1,H2, . . . be the sequence of abstract graphs
defined exactly as with G1, G2, . . . except we do not allow vertex identifications nor new
edges to be introduced in this way. Clearly χc(Gn) > χc(Hn), but we cannot argue equality
in general. The proof of Lemma 4.35 applied to the graphsHn gives slightly more, as follows.
Chapter 4. Circular Chromatic Number of Some Special Graphs 50
Theorem 4.36. For every n > 0, χc(Hn) = 4 − 21−n.
Proof. The cases n = 0, 1 are trivial. Let n > 1, and let Hn+1 be as in Figure 4.11. Let
r = 4−2−n = 4−21−n+2n. By the proof of Lemma 4.35, Hx,yn admits an r–circular colouring
c1, with c1(x) = 0 and c1(y) = 1/2. Similarly the graphs Hx,y′
n , Hy,zn , and Hy′,z
n admit r–
circular colourings c2, c3, and c4, respectively, with c2(x) = 0, c2(y′) = c4(y
′) = −1/2,
c3(y) = 1/2, and c3(z) = c4(z) = 0. Now an r–circular colouring c of Hn+1 can be obtained
by combining the partial colourings c1, c2, c3, c4.
4.4 The Projective Plane Orthogonality Graph
In this section we study the circular chromatic number of another geometrically defined
infinite graph. Let RP 2 be the set of all lines through the origin in R3. We define a graph O
with V (O) = RP 2 in which two vertices are adjacent if and only if they are perpendicular
as lines in R3. Certain properties of O (e.g. independent sets which meet every triangle)
are of interest in quantum physics [46]. If O is 3–colourable, then every colour class of a 3–
colouring of O is an independent set which meets every triangle. Kochen and Specker [37]
proved that O has no such independent set. Indeed O has a finite subgraph in which
independent set meets every triangle. This proves χ(O) > 4. A 4–colouring of O was
discovered by Godsil and Zaks [23], thus proving the following.
Theorem 4.37. χ(O) = 4.
It remains open whether O has circular chromatic number 4. By investigating finite sub-
graphs of O, we give lower bounds on χc (O).
Given n > 1, let Xn be the set of all non-zero vectors (a, b, c) such that a, b, c are relatively
prime integers with absolute value at most n. Moreover, we assume that the leading non-
zero component of each (a, b, c) ∈ Xn is positive. Therefore elements of Xn correspond to
unique vertices of O. Let Gn be the subgraph of O induced by Xn.
the remaining vertices of G1 all have degree 3, and each of them has a neighbourhood of
them form {ui, vj , wk}. Moreover, no two of the vertices of degree 3 have more than one
common neighbour. For i ∈ {1, 2}, let ti be the common neighbour of u1 and vi. Then
c(t1) ∈ {9, 0, 1}. Therefore t1 is not adjacent to w2 and we have N(ti) = {u1, vi, wi}. Now
let t′i be the common neighbour of w2 and vi. Then since t′1 = t1 is adjacent to u1, we have
that t′2 is adjacent to u2. This is a contradiction since any choice of c(t′2) contradicts with
one of c(u2), c(v2), and c(w2).
The graphs Gn for n > 2 are much more complicated in structure than G1. For example
Chapter 4. Circular Chromatic Number of Some Special Graphs 52
G2 has order 49 and size 138. With a computer, we were able to prove that the graph
H2 = G2 − {111, 211, 211, 211, 211} with order 44 and size 117, has circular chromatic
number 11/3. We first found an (11, 3)–colouring of H2 to prove the upper bound, then
investigated all (11, 3)–colourings of H2 and verified that every such colouring of H2 has a
tight cycle. Thus by Lemma 2.4 we have χc (H2) = 11/3. This proves the following.
Theorem 4.40. χc (O) > 11/3.
The graph G3 has order 145 and size 546. We were able to find a 4–colouring of this
graph which has no tight cycles. Applying Lemma 2.6, this gives a (27, 7)–colouring of G3.
Therefore we have 4 − 13 = χc (H2) 6 χc (G3) 6 4 − 1
7 . This colouring cannot be improved
further since it has a tight cycle. Our study of colourings of the graphs Gn suggests the
following conjecture.
Conjecture 4.41. χc (O) = 4.
Chapter 5
Circular Colourability vs. Girth
The pentagon colouring problem (Problem 3.13) asks whether every cubic graph with large
enough girth maps to C5 (i.e. has circular chromatic number at most 5/2). In this chapter
we present some work related to this problem.
5.1 How About Odd-Girth?
For planar graphs, the folding lemma of [36] asserts that short even facial cycles are not
obstructions to colourability. Specifically, every planar graph G with odd-girth g admits
a homomorphism to a plane graph G′ with all faces having length g. The fact that one
can get rid of some of the short even cycles in a planar graph without affecting circular
colourability inspires the following two problems.
Problem 5.1. Does every cubic graph with sufficiently large odd-girth map to C5?
Problem 5.2. Does every cubic graph with sufficiently large odd-girth have circular chro-
matic number strictly less than 3?
In this section we prove both these problems have negative answers. By Theorem 3.2 it
suffices to find projective plane graphs with all faces having length 4 or 6 which have no
short odd cycles. The construction of such graphs follows.
By Euler’s formula, a cubic projective plane graph every face of which has length 4 or 6,
must have exactly three faces of length 4. We construct a sequence of projective plane
53
Chapter 5. Circular Colourability vs. Girth 54
Figure 5.1: The spiderweb graphs S1 and S2. Semiedges are routed through a cross-cap.
graphs with this property, called spiderweb graphs. The spiderweb graph Sn consists of
n + 1 “layers”. The graph S0 is the complete graph K4 and the graphs S1 and S2 are
illustrated in Figure 5.1. To further clarify the construction of the spiderweb graphs we
show a projective plane embedding of S3 in Figure 5.2. Since in an even-faced projective
plane graph odd cycles are precisely the non-contractible cycles, adding more layers is
expected to increase the odd-girth of the spiderweb graphs.
Obviously the graphs Sn are all cubic, and embedded in the projective plane with three
faces of length 4 and all other faces having length 6. Therefore by Theorem 3.2 except for
S0, which does not have any hexagonal faces, they all have circular chromatic number 3.
The proof of the following proposition is straight-forward, but technical and we omit it
here.
Proposition 5.3. Given n > 0, the graph Sn has odd-girth 2 ⌈3n/2⌉ + 3.
The results of this section are summed up in the following.
Theorem 5.4. There exist cubic graphs with sufficiently large odd-girth, whose circular
chromatic number equals 3.
Chapter 5. Circular Colourability vs. Girth 55
Figure 5.2: The spiderweb graph S3. Semiedges are routed through a cross-cap.
5.2 An Alternate View
The pentagon colouring problem asks whether for cubic graphs, large girth implies that
the circular chromatic number is at most 5/2. Alternately, one might ask if the circular
chromatic number of cubic graphs with a certain girth is bounded below 3. More specifically,
for all g > 3, let
f(g) = sup {χc(G) : G is a cubic graph with girth > g} .
The pentagon colouring problem asks whether f(g) 6 5/2 for large enough g. By definition,
f is a non-increasing function of g, so equivalently, the pentagon colouring problem asks
whether
limg→∞
f(g) 6 5/2.
In this section we show that f(9) > 5/2 and thus the minimum girth required in the
pentagon colouring problem is at least 10. Since every subcubic graph G is a subgraph of
a cubic graph having the same girth as G, the pentagon colouring problem is equivalent
to the assertion that every subcubic graph with large enough girth has circular chromatic
number at most 5/2. By extending the domain of the pentagon colouring problem we gain
Chapter 5. Circular Colourability vs. Girth 56
BC
A
B C
A
Figure 5.3: A subcubic graph with girth 6 and circular chromatic number 3
more generality which is useful for inductive proofs. The best advantage of this extension
for us is that we may construct subcubic graphs of large girth as subdivisions of cubic
graphs. In particular this process can be carried out on a surface such as the projective
plane. While cubic graphs embedded in the plane or the projective plane have bounded
girth, subcubic embedded graphs can have arbitrary large girth.
Since χc (K4) = 4 and χc (P ) = 3 where P is the Petersen graph, we have f(3) = 4
and f(4) = f(5) = 3. The graph shown in Figure 5.3 is a subcubic projective plane
hexangulation with girth 6. Therefore by Theorem 3.2 it has circular chromatic number 3.
This proves f(6) = 3.
We do not know any (sub)cubic graph with girth at least 7 and circular chromatic number 3.
We determined the circular chromatic number of small (order at most 34) cubic graphs with
girth 7, looking for such graph. The results of these computations are presented in Table 5.1.
From this table one can see that there exists a cubic graph of order 26 and girth 7 with
circular chromatic number 14/5. An illustration of this graph is presented in Figure 5.4.
Indeed this is the best we know for girth 7. It provides us with the bound f(7) > 14/5.
In Proposition 5.6 we prove that f(g) >2g
g−2 for all even g. In particular f(8) > 8/3 and
f(10) > 5/2. These are the best known bounds for g = 8, 10.
We also computed the circular chromatic number of small cubic graphs with girth 9. Since
the smallest of these graphs have order 58, this computation and even generating such
Chapter 5. Circular Colourability vs. Girth 57
A
A
A
B
B
B
Figure 5.4: A cubic graph with girth 7 and circular chromatic number 14/5. The semiedgeslabelled A are incident with a new vertex A and the semiedges labelled B are incident witha new vertex B.
graphs could not be completed for orders greater than 60. The (partial) results of these
computations are presented in Table 5.2. One can see in this table that f(9) >2911 , thus
proving the following.
Proposition 5.5. The girth requirement in the pentagon colouring problem is at least 10.
The results of further computations of circular chromatic number of small graphs are pre-
sented in Section 5.4.
The following is yet another application of Theorem 3.2.
Proposition 5.6. For all even g > 4, we have f(g) >2g
g−2 .
Proof. By Theorem 3.2, it suffices to find a non-bipartite subcubic projective plane graph
with girth g whose faces all have length g. For g = 4 the complete graph K4, and for
g = 6 the graph of Figure 5.3 have the desired properties. In the following construction we
assume g = 2n > 6.
We start with the Mobius ladder Vg embedded in the projective plane with one face of
length g and every other face having length 4 (the diagonals of the main cycle go through a
cross-cap). We label the vertices of Vg so that it consists of the cycle v1v2 · · · vgvg+1, where
vg+1 = v1, and the diagonals vivn+i for 1 6 i 6 n. For each 1 6 i 6 g, we subdivide the
Chapter
5.
Cir
cular
Colourabil
ity
vs.
Gir
th
58
nnumber of graphs with circular chromatic number ...
Table 5.1: Circular chromatic number of small cubic graphs of girth 7.Here a tilde in a cell means that the corresponding upper bound is verified for the graphs counted in that cell, but theequality is not verified. We have verified that every graph counted in a column to the right of the 5/2 column is not(5, 2)–colourable. The same rules apply to all tildes in all the tables of this thesis.
nnumber of graphs with circular chromatic number ...
125
177
3213
5221
52
3313
2811
239
187
3112
135
218
2911
58 5 1∼ 1∼ 9∼ 1∼ 1∼
60 2∼ 1∼ 1∼ 1∼ 210 1∼ 6∼ 12∼ 136∼ 11∼ 93∼
Table 5.2: Circular chromatic number of small cubic graphs of girth 9
Chapter 5. Circular Colourability vs. Girth 59
Figure 5.5: A subcubic projective plane graph with girth 10 and all faces of length 10
edge vivi+1 to a path Pi of length n − 1 and denote the vertex of Pi adjacent to vi by xi.
Let Gg be the disjoint union of this graph with a cycle y1y2 · · · ygy1 in which xi is joined to
yi by an edge if i is odd, and by a path of length n− 1 if i is even. It is easy to see that Gg
is embedded in the projective plane with all faces having length g, and that Gg has girth g.
The graph G6 is shown in Figure 5.3, and the graph G10 is shown in Figure 5.5.
As it can be seen from the construction of the graph Gg in the above proposition, these
graphs are very sparse. This makes it possible to obtain new graphs with the same girth,
starting from several copies of the graph Gg and joining pairs of vertices of degree 2 with
new edges. Intuitively, one expects to be able to get graphs with larger circular chromatic
number. We tried this approach for g = 10 but we failed to find any subcubic graphs with
girth 10 and circular chromatic number strictly greater than 5/2. We investigated hundreds
of thousands of cubic and subcubic randomly generated graphs containing one or several
copies of G10. Since this was a failed attempt, we omit the details here.
5.3 Circular Critical Subcubic Graphs
We observed in Section 5.2 that f(4) = f(5) = f(6) = 3. Namely there exist cubic graphs
with girth 4, 5, and 6 whose circular chromatic numbers are 3. This suggests investigating
Chapter 5. Circular Colourability vs. Girth 60
Order Number of graphs Girth 4 critical Girth 5 critical
4 1
5 2
6 5
7 8
8 23
9 48 1
10 148
11 399 1
12 1339 2 1
13 4395 1
14 16183 2 2
15 60717 2
16 242087 1 3
17 991019 1 1
18 4210628 3 2
19 18345575 3 2
20 82111548 5 6
Table 5.3: Summary of small triangle-free subcubic 3–circular critical graphs
cubic graphs of girth 7 with the same property. As mentioned in Section 5.2, we do not
know of any such graph. In this section we study minimal graphs with χc = 3. More
precisely, a graph G is said to be r–circular critical, if χc(G) = r and χc(H) < r for all
proper subgraphsH of G. It is an easy observation that every graph with circular chromatic
number r has an r–circular critical subgraph. Thus the existence of a cubic graph with girth
7 and circular chromatic number 3 is equivalent to the existence of a subcubic 3–circular
critical graph with girth 7.
One can easily see that a 3–circular critical graph is triangle-free, unless it is the triangle K3.
Also, every such graph is 2–connected. We investigated all 2–connected triangle-free sub-
cubic graphs of order up to 20 looking for 3–circular critical graphs. Table 5.3 summarizes
the results.
It can be observed in Table 5.3 that all 3–circular critical graphs of order at most 20
Chapter 5. Circular Colourability vs. Girth 61
have girth at most 5. The existence of cubic graphs with girth 6 and circular chromatic
number 3 proves the existence of 3–circular critical subcubic graphs with girth 6. The
graph of Figure 5.3 is an example of such a graph with order 21.
The graph obtained by deleting one vertex from the Petersen graph, the generalized Pe-
tersen graph P (7, 2), and the Triplex graph are examples of 3–circular critical graphs. A
complete catalogue of subcubic 3–circular critical graphs with order at most 20 is given in
Appendix A.
5.4 Further Computational Results
In Tables 5.1 and 5.2 we presented the results of our computation of cubic graphs with
girth at least 7 and cubic graphs with girth at least 9. In this section we present the results
of similar computations for girths 5, 6, and 8. These results are shown in Tables 5.4, 5.5
and 5.6.
nnumber of graphs with circular chromatic number ...
3–colouring of G which is a contradiction. Thus c(z′1) = 0. We define five extensions of
c to G, which result in five directed paths in Dc(H) as before. Two of these extensions
have c(x) = c(y) = 0, c(x1) = c(x2) = 2, c(y1) = c(y2) = 1, and c(z), c(z1) ∈ {1, 2}. As
above, Dc(H) has a directed y′2z′1–path and a directed z′1x
′2–path. The third extension
has c(x2) = 0, c(z) = c(x1) = c(y1) = c(y2) = 1, and c(x) = c(y) = c(z) = 2. This
gives a directed x′2x′1–path in Dc(H). The fourth extension has c(z) = c(x2) = c(y2) = 0,
c(x) = c(y1) = c(z1) = 1, and c(y) = c(x1) = 2. Thus there exists a directed x′1y′1–path
in Dc(H). The fifth extension is an obvious modification of the third, giving a directed
y′1y′2–path in Dc(H). These five directed paths combine to a nontrivial directed closed walk
in Dc(H), a contradiction.
Chapter 6. Subcubic Graphs with Circular Chromatic Number 3 69
v2v1
v4 v3
x′2x′1
x′4 x′3
x2x1
x4 x3
(a) (b) (c)
Figure 6.2: Further forbidden configurations for a 3–circular critical subcubic graph G.White vertices have degree 2 in G.
Lemma 6.8. A 3–circular critical graph G does not contain any of the configurations in
Figure 6.2.
Proof. Suppose G contains the configuration (c) of Figure 6.2, labelled as shown. Let
H = G− {v1, v2, v3, v4, x1, x2, x3, x4}, and let c be an acyclic 3–colouring of H. Let α be a
colour not in {c(x′1), c(x′3)}. We define c(x1) = c(x3) = c(v2) = c(v4) = α and arbitrarily
extend to a (proper) 3–colouring of G. Then v1 and v3 are terminal vertices in Dc(G),
proving Dc(G) is acyclic, a contradiction.
For configurations (a) and (b) the proof is similar, but it involves considering many cases.
If G contains either of these two configurations, we let H be the graph obtained by deleting
the internal vertices of that configuration. Then H has an acyclic 3–colouring. We analyze
different combinations of colours appearing on the connecting vertices of the deleted con-
figuration. For each case we show that either c can be extended to an acyclic 3–colouring of
G, or that Dc(H) contains a nontrivial directed closed walk. Both of these outcomes con-
tradict our assumptions, thus proving the respective configuration cannot occur in G. We
present the details of this case analysis for configurations (a) and (b) in Appendix B.
We are now ready to present the main result of this section.
Theorem 6.9. Let G be a 3–circular critical subcubic graph and for i = 1, 2 let Vi(G) be
the set of all x ∈ V (G) such that dG(x) = i. Then |V2(G)| 6710 |V3(G)|.
Proof. Let G′ be the cubic graph obtained by suppressing all degree 2 vertices in G. Let
Chapter 6. Subcubic Graphs with Circular Chromatic Number 3 70
H = G− V2(G). Then
|V2(G)| = |E(G′)| − |E(H)| =3
2|V3(G)| − |E(H)|. (6.1)
On the other hand, by Lemmas 6.5 – 6.8 every connected component of H has at least 5
vertices. Thus c 615 |V3(G)|, where c is the number of connected components of H, and we
have
|E(H)| > |V (H)| − c >4
5|V3(G)|. (6.2)
Finally by (6.1) and (6.2) we have
|V2(G)| 63
2|V3(G)| − 4
5|V3(G)| =
7
10|V3(G)|.
We prove in Section 6.3 that the above inequality does not hold for embedded subcubic
graphs with girth at least 9 and nonnegative Euler characteristic, thus proving such graphs
all have circular chromatic number strictly less than 3.
6.3 Embedded Graphs with Girth 9
In this section we apply the results of the previous section to embedded graphs. Given
a surface X, an X–graph is any graph which admits an embedding in X (faces need not
be 2–cells). We limit our attention to subcubic graphs with girth at least 9. For a fixed
surface X, suppose there exists a subcubic X–graph G with χc(G) = 3. Then G contains a
3–circular critical subgraph. Obviously any subgraph of G is subcubic, it has girth at least
9, and it embeds in X. Therefore to prove non-existence of subcubic X–graphs with girth
at least 9 and with circular chromatic number 3, it suffices to prove non-existence of such
3–circular critical graphs. As in previous section, for a graph G and i > 0, we let Vi(G) be
the set of all vertices of degree i in G.
Lemma 6.10. Let X be a surface with Euler characteristic c, and G be a subcubic 3–
circular critical X–graph with girth at least 9. Then |V3(G)| 6 −90c.
Proof. For i = 2, 3, let ni = |Vi(G)|. Then |V (G)| = n2 + n3 and 2|E(G)| = 2n2 + 3n3.
Applying Euler’s formula to an embedding of G in X, we get
(n2 + n3) −2n2 + 3n3
2+ f > c,
Chapter 6. Subcubic Graphs with Circular Chromatic Number 3 71
where f is number of faces of G (we have equality if and only if each face is a 2–cell). On
the other hand, since G is connected, the boundary of each face is a closed walk in G, thus
it has length at least 9. Thus
9f 6 2|E(G)| = 2n2 + 3n3.
Therefore,
9(c+1
2n3) = 9f 6 2n2 + 3n3,
which gives
n2 >9
2c+
3
4n3.
On the other hand, by Theorem 6.9 we have n2 6 710n3. Therefore,
9
2c+
3
4n3 6
7
10n3,
which after simplification gives the desired result.
Since the plane, the projective plane, the torus and the Klein bottle all have nonnegative
Euler characteristic, the following is an immediate corollary of the above lemma.
Theorem 6.11. Every subcubic graph with girth at least 9 which is embeddable in either
the plane, the projective plane, the torus, or the Klein bottle, has circular chromatic number
strictly less than 3.
Remark 6.12. For surfaces with negative Euler characteristic, Lemma 6.10 bounds the
order of a 3–circular critical subcubic embedded graph with girth at least 9. Thus one can
extend Theorem 6.11 to include more surfaces, or exclude a surface, by examining only a
finite number of subcubic embedded graphs with girth at least 9. Despite this, we were
unable to perform this computation for the surface N3 of Euler characteristic −1, due to
the enormous number of subcubic N3–graphs with girth at least 9 and order at most 153.
Chapter 7
The Circular Chromatic Index
7.1 Introduction
Edge colouring is of fundamental importance in graph theory. One of the main motivations
in the study of edge colourings of graphs, is P.G. Tait’s [49] observation that the four colour
problem is equivalent to the statement: every 2–connected cubic planar graph is 3–edge
colourable. Eventually the four colour theorem was proved via this approach. Tait’s result
motivated the study of non-Tait-colourable graphs, namely 2–connected cubic graphs which
are not 3–edge colourable.
Until 1975, only four “non-trivial” non-Tait-colourable graphs were known. The difficulty
in finding such graphs led Gardner [16] to propose calling such graphs snarks. The am-
biguous term “non-trivial” in Gardner’s definition has attracted considerable attention in
the literature. It is known that a non-Tait-colourable graph can be reduced to a smaller
one, if it has multiple edges, a triangle or a C4, or if it has a 2–edge cut or a 3–edge cut
consisting of non-adjacent edges. It is generally accepted by graph theorists that a snark
is a cyclically 4–edge connected cubic graph which is not 3–edge colourable. A graph G is
said to be cyclically k–edge connected, if the removal of any set of fewer than k edges from
G results in at most one non-tree connected component. The cyclic edge-connectivity of G
is the largest k such that G is cyclically k–edge connected.
Cameron, Chetwynd, and Watkins [9] prove that every snark with cyclic edge connectivity 4
either “contains” a smaller snark, or is the dot product of two smaller snarks. We delay the
72
Chapter 7. The Circular Chromatic Index 73
definition of the dot product to Section 7.4. It is also proved in [9] that every snark, other
than the Petersen graph, with cyclic edge connectivity 5 can be reduced to smaller snarks.
One might interpret these reducibility results as “triviality” and thus define a snark to be
either the Petersen graph, or a cyclically 6–edge connected cubic graph which is not 3–edge
colourable.
For circular edge colourings, we know of no reductions other than the trivial reduction of
bridges. One of the easy reductions for 3–edge colourability of cubic graphs is contracting
a triangle to a single vertex. We see in Section 7.2 that this reduction does not work
for circular edge colouring, since the Petersen graph which is obtained by contracting a
triangle in J3 is not (7, 2)–edge colourable, while J3 is (7, 2)–edge colourable. We define a
(3–circular) snark to be a 2–connected cubic graph which is not 3–edge colourable.
In this chapter we study circular edge colourings of graphs. Vizing’s theorem partitions
the set of simple graphs into two classes according to their circular chromatic indices; those
with χ′ = ∆ and those with χ′ = ∆ + 1. So in the context of ordinary edge colouring, all
snarks are the same since they all have chromatic index 4. Circular edge colouring provides
a finer scale for measuring how close is a (sub)cubic graph to being 3–edge colourable. For
example, Theorem 7.1 proves that the Petersen graph is the “most difficult” cubic graph
with respect to circular edge colouring.
The circular chromatic index of a graph G, is defined by χ′c (G) = χc (L(G)), where L(G)
is the line graph of G. By Vizing’s theorem, and by Theorem 1.4, for every graph G we
have
∆(G) 6 χ′c (G) 6 ∆(G) + 1.
Moreover, χ′c (G) = ∆(G) if and only if G is class 1. In particular, if G is a class 2 subcubic
graph, then 3 < χ′c (G) 6 4. In Figure 7.1 we present two subcubic graphs whose circular
chromatic indices equal 4. Theorem 7.1 states that these two graphs are the only bridgeless
subcubic graphs with this property.
Theorem 7.1. [1] Every bridgeless subcubic graph, other than the graphs of Figure 7.1,
has circular chromatic index at most 11/3.
It is known [59] that the Petersen graph P has circular chromatic index 11/3. Thus the
above theorem proves that P has the largest circular chromatic index among all bridgeless
Chapter 7. The Circular Chromatic Index 74
Figure 7.1: The two subcubic 2–connected graphs with circular chromatic index 4
cubic graphs. This behaviour is expected because of Jaeger’s Petersen colouring conjec-
ture [33] which asserts that for all bridgeless cubic graphs G, L(G) admits a homomorphism
to L(P ).
One other consequence of Theorem 7.1 is that no number in the interval(
113 , 4
)
is the
circular chromatic index of a cubic graph. Therefore(
113 , 4
)
is a gap in the set
S = {χ′c (G) : G is a cubic graph}.
Note any cubic graph which contains one of the graphs illustrated in Figure 7.1 as a
subgraph, has circular chromatic index 4. It is worth mentioning here that indeed S =
{χ′c (G) : G is a subcubic graph}. This is because there exist subcubic graphs with exactly
one vertex of degree 2 whose circular chromatic index is arbitrarily close to 3. Given k > 1,
the necklace graph Nk is obtained by doubling the edges v2iv2i+1 (1 6 i 6 k) in a cycle
C2k+1 = v1v2 · · · v2k+1v1. Then every vertex in Nk other than v1 is of degree 3. It is proved
in [43] that χ′c (Nk) = 3+1/k. Given a subcubic graph G, we let k be large enough so that
3 + 1/k 6 χ′c (G). For each v ∈ V (G), we attach 3 − dG(v) copies of the necklace Nk to v
via 3−dG(v) bridges. The resulting graph is cubic and has circular chromatic index χ′c (G).
By Theorem 7.1, 4 is not a member of the set
Sb = {χ′c (G) : G is a 2–connected cubic graph}.
Thus Sb 6= S. Is is not known whether Sb = S \ {4}. It is also open whether replacing
“cubic” with “subcubic” in the definition of Sb results in the same set.
One might ask about the distribution of the sets S and Sb in the interval [3, 4]. Kaiser, Kra ’l,
and Skrekovski [35] prove that class 2 cubic graphs with large enough girth have circular
chromatic index arbitrarily close to 3. On the other hand, Kochol [38] proved that there
exist snarks with arbitrary large girth. Therefore Sb is infinite and 3 is an accumulation
point of Sb. Until recently, only a finite number of values in S were known. Mazak [40]
Chapter 7. The Circular Chromatic Index 75
proves that all numbers of the form 3 + 2/3k belong to Sb. We show in Section 7.4 that all
numbers 3 + 1/k belong to Sb. All the known values of circular chromatic index of cubic
graphs, are of the form 3 + 2/k. This motivates the following problem.
Problem 7.2. Let k > 2 be an integer. Does there exist a cubic graph whose circular
chromatic index equals 3 + 2/k? Does there exist a snark whose circular chromatic index
is not of the form 3 + 2/k for some k > 2?
The following is folklore.
Conjecture 7.3. Every bridgeless cubic graph, other than the Petersen graph, has circular
chromatic index at most 7/2.
It is proved in [35] that every bridgeless cubic graph with girth at least 14 has circular
chromatic index at most 7/2. If the above conjecture is true, then(
72 ,
113
)
is a gap in Sb.
This author believes that S = Sb ∪{4}, and that 3 is the only accumulation point of either
of these sets.
In this chapter we determine the circular chromatic index of some snarks. For convenience,
when dealing with edge colourings we allow graphs to have semiedges, i.e. edges with only
one end-vertex. This notion helps to study edge colourings of building blocks of our graphs,
and then paste the edge colourings together to obtain an edge colouring of the graph being
considered.
7.2 Flower Snarks
The results of this section are published in [19].
Before flower snarks were introduced by Isaacs [31], only four snarks were known. As well
as introducing this first infinite family of snarks, Isaacs gave a construction in [31], known
as the dot product, which constructs from two snarks a new snark. We discuss the dot
product in Section 7.4 where we study the circular chromatic index of an infinite family of
graphs obtained via dot product from several copies of the Petersen graph.
We recall from Section 4.2.4, that given an odd integer n > 3, the flower snark Jn consists of
two cycles x1x2 · · · xnx1 and y1y2 · · · y2ny1 and a claw {zixi, ziyi, ziyn+i} for each 1 6 i 6 n.
Chapter 7. The Circular Chromatic Index 76
The following is the main result of this section.
Theorem 7.4. For all odd n > 3,
χ′c (Jn) =
7/2 if n = 3,
17/5 if n = 5,
10/3 if n > 7.
We first present a proof of the fact that flower snarks are class 2. This proof is the key
idea in the proof of Theorem 7.4, which just mimics this proof for (3 + ε)–circular edge
colourings of flower snarks. The following lemma, known as the parity lemma is a powerful
tool in edge colourings of cubic graphs.
Lemma 7.5. [31] Let H be a cubic graph with n vertices. Given any 3–edge colouring of
H and 0 6 i 6 2, let ni denote the number of semiedges of H coloured i. Then ni ≡ n
modulo 2.
Proposition 7.6. For all odd n > 3, the flower snark Jn is class 2.
Proof. Let H be the graph shown in Figure 7.2. Let c be an 3–edge colouring of H. The
labels αi and βi in the picture, indicate the colours assigned to the semiedges of H by c.
Since H has four vertices, by the parity lemma, each of the colours 0, 1, 2 appears an even
number of times. Since αi 6= βi for i = 0, 1, 2, each colour appears at most three times, thus
exactly twice, as the colour of a semiedge of H. Therefore, if the αi are all distinct, then
βi are also all distinct. Moreover, since the edges incident with z receive different colours,
the permutations α0α1α2 and β0β1β2 have the same sign. In this case we say the triple
(α0, α1, α2) has type A. Otherwise, the sets {α0, α1, α2} and {β0, β1, β2} both have size 2,
and the colour in the intersection of these two sets appears exactly once on each side. In
this case we say the triple (α0, α1, α2) has type B.
Note that Jn consists of n copies of H. Suppose Jn is 3–edge colourable and let c be one
such edge colouring. For each 1 6 i 6 n, let σi = (c(xixi+1), c(yiyi+1), c(yn+iyn+i+1)),
where xn+1 = x1 and y2n+1 = y1. By the above observation, σi are either all of type A or
all of type B. In the former case, all σi must have the same sign as permutations of {0, 1, 2}.This is a contradiction since σ1 has the same sign as (c(xnx1), c(y2ny1), c(ynyn+1)) which
differs from σn by a single transposition.
Chapter 7. The Circular Chromatic Index 77
z
x
y
y′
α0
α1
α2
β0
β1
β2
Figure 7.2: The building block of the flower snarks
Thus all σi are of type B. Let αi be the colour with majority in σi. Then by the above
observation, αi alternate between two colours. This contradicts the fact that n is odd.
As we mentioned before, our proof of Theorem 7.4 mimics the above proof of Proposi-
tion 7.6. For that, we need to first modify the definitions of type A and type B triples of
colours and their signs.
Let 3 6 r < 10/3. We aim to prove that Jn is not r–circular edge colourable. Suppose
0 6 α 6 β < r are colours in Cr. If β − α < r + α − β, then we say β follows α, and
otherwise, we say α follows β. Two colours α, β ∈ Cr are said to be close, if |α−β|r < 2/3,
and otherwise, they are said to be far apart. A sequence (α0, α1, α2) of colours is of
type A, if α0, α1 and α2 are pairwise far apart, and it is of type B if two of the colours
are close and the remaining one is far apart from both of the other two. A sequence
σ = (α0, α1, α2) of type A has positive sign if 0 6 αi 6 αi+1 6 αi+2 < r for some 0 6 i 6 2,
where the indices are reduced modulo 3. The sequence σ is said to have negative sign
otherwise. Let σ = (α0, α1, α2) be a sequence of colours of type B, and let i, j, k be such
that {i, j, k} = {0, 1, 2}, and αi and αj are close. The sequence σ has positive sign, if αk
follows both αi and αj, and it has negative sign, if αi and αj both follow αk. For example, if
r = 13/4, the sequences (0, 1, 2) and (0, 2, 1) are both of type A, and the sequences (0, 1, 0)
and (2, 1.8, 1) are both of type B. In each case, the first sequence has positive sign, and the
second has negative sign. Note that the sign of a sequence of type B is not always defined,
and that when the sign is defined, it does not depend on the order of the elements in the
sequence. For example for r = 13/4, the sequence (0, 1.5, 2) is of type B. The sign of this
sequence is undefined since 1.5 follows 0 and 0 follows 2. Another example of a type B
sequence whose sign is undefined is (0, 0, r/2).
Chapter 7. The Circular Chromatic Index 78
Lemma 7.7. Let H be the graph of Figure 7.2, let 3 6 r < 10/3, and let the labels αi
and βi in the picture denote the colours assigned to the semiedges in an r–circular edge
colouring of H. Then either (α0, α1, α2) and (β0, β1, β2) are both of type A and have the
same sign, or they are both of type B and have opposite signs.
Proof. Let r = 3 + ε for some 0 < ε < 1/3, and let c be an r–circular edge colouring of H.
For each x ∈ Cr, we define I(x) = [x+ 1, x+ 1 + ε]r and J(x) = [x+ 2, x+ 2 + ε]r. Let
γ0 = c(zx), γ1 = c(zy), and γ2 = c(zy′). Then (γ0, γ1, γ2) is of type A, and by symmetry we
may assume that it has positive sign. Namely, for all 0 6 i 6 2, then γi ∈ I(γi−1)∩J(γi+1),
where the indices are reduced modulo 3. Note that since each circular interval I(x) or J(x)
has length ε < 1/3, when two such intervals intersect, every two colours in their union are
close. This holds for I(γi−1) ∪ J(γi+1), for all 0 6 i 6 2. Moreover, every y ∈ I(γi) ∪ J(γi)
is far apart from every x ∈ I(γi−1) ∪ J(γi+1).
For each 0 6 i 6 2, one of αi and βi is in I(γi) and the other is in J(γi). By symmetry we
may assume that α0 ∈ I(γ0) and β0 ∈ J(γ0). Therefore, we have the following four cases
for the values of α1, α2, β1, β2.
Case 1. α1 ∈ I(γ1) and α2 ∈ I(γ2). By the above discussion, the colours αi are pairwise
far apart, and αi+1 follows αi for all 0 6 i 6 2. The same conclusion holds for the colours
βi similarly. Thus (α0, α1, α2) and (β0, β1, β2) are both of type A and they have the same
sign as (γ0, γ1, γ2).
Case 2. α1 ∈ I(γ1) and α2 ∈ J(γ2). In this case, α0, α2 ∈ I(γ0)∪J(γ2) are close, while they
are both far apart from α1 ∈ I(γ1). Therefore, (α0, α1, α2) has type B. Since α1 follows
both α0 and α2, this triple has positive sign. Similarly, β1, β2 ∈ J(γ1)∪ I(γ2) are close and
both far apart from β0 ∈ I(γ0). Thus (β0, β1, β2) is of type B, and since β1 and β2 both
follow β0, this triple has negative sign.
Case 3. α1 ∈ J(γ1) and α2 ∈ I(γ2). This case is similar to Case 2.
Case 4. α1 ∈ J(γ1) and α2 ∈ J(γ2). This case is similar to Case 2.
Similarly to the proof of Proposition 7.6, the above lemma gives the following.
Corollary 7.8. For all odd n > 3, we have χ′c (Jn) > 10/3.
To complete the proof of Theorem 7.4 for n > 7, it remains to prove the upper bound.
Chapter 7. The Circular Chromatic Index 79
A
B
C
A
C
B
3 9 2 5 1 4 0 3 0 3
3 6 2 8 1 7 0 6 0 6
6 0 7 4 7 3 6 3 6 3
0 9 5 4 4 3 3 3 3
6 6 8 8 7 7 6 6 6
3 3 1 1 0 0 0 0 0
Figure 7.3: A (10, 3)–edge colouring of J9. Semiedges wrap-around according to theirlabels.
In Figure 7.3, we give an (10, 3)–edge colouring of J9. The colours of the two blocks in
the shaded area can be repeated t times to obtain an (10, 3)–edge colouring of J9+2t. An
(10, 3)–edge colouring of J7 can be obtained by deleting the shaded blocks in Figure 7.3.
In the remainder of this section we prove Theorem 7.4 for n = 3, 5. Suppose that χ′c (G) =
p/q for some graph G and relatively prime integers p and q. By Lemma 2.4, L(G) has a
tight cycle with respect to any edge (p, q)–colouring of G. In particular, all the colours
0, 1, . . . , p−1 are used in any (p, q)–edge colouring of G. On the other hand by Lemma 2.5,
we have q 6 α′(G).
Proposition 7.9. χ′c (J3) = 7/2.
Proof. A (7, 2)–edge colouring of J3 is given in Figure 7.4. Therefore 10/3 6 χ′c (J3) 6 7/2.
On the other hand, if χ′c (J3) = p/q, where p and q are relatively prime positive integers,
then q 6 α′(J3) = 6. Hence χ′c (J3) ∈
{
103 ,
175 ,
72
}
.
Suppose χ′c (J3) = 10/3 and let c be an (10, 3)–edge colouring of J3. Since J3 has 18
edges, at least one colour, say 9, is used exactly once by c. For i ∈ {0, 3, 6}, let Mi =
c−1({i, i+ 1, i+ 2}). Then each Mi is a matching in J3 and |M0 ∪M3 ∪M6| = 17. Thus at
least two of the Mi are perfect matchings of J3. But J3 does not have two disjoint perfect
matchings since it is class 2.
Suppose χ′c (J3) = 17/5 and let c be an (17, 5)–edge colouring of J3. Then all but one of
the colours are used exactly once by c. Assume that the exceptional colour is 1. Then
c−1({0, 1, 2, 3, 4}) and c−1({1, 2, 3, 4, 5}) are two perfect matchings of J3 which differ in
exactly one edge. This contradicts the fact that J3 is a simple graph.
Chapter 7. The Circular Chromatic Index 80
A
B
C
A
C
B
0 2 4 0
1 5 0 2
2 5 3 1
3 3 4
5 6 2
0 1 6
Figure 7.4: A (7, 2)–edge colouring of J3
A
B
C
A
C
B
0 5 10 0 11
2 14 4 11 6
13 8 15 3 8
7 9 16 1 1
12 15 5 6 6
1 3 10 13 13
Figure 7.5: A (17, 5)–edge colouring of J5
For the graph J5, a (17, 5)–edge colouring is given in Figure 7.5 which along with the
lower bound of Corollary 7.8, proves χ′c (J5) ∈
{
103 ,
278 ,
175
}
. We were not able to prove
χ′c (J5) > 17/5 without the assistance of a computer. A brute force algorithm for checking
(27, 8)–edge colourability of J5 would be too slow. We designed a faster algorithm for
verifying the existence of (p, q)–edge colouring of J5 based on the following idea: first, we
construct an auxiliary graph G with the vertex set {0, 1, . . . , p − 1}3. Two such vertices
(α0, α1, α2) and (β0, β1, β2) are joined by an edge if there exists an (p, q)–edge colouring
of the graph in Figure 7.2 which agrees with these two triples on the values of the αi and
the βi. It is easy to see that J5 is (p, q)–edge colourable, if and only if G contains a walk
from a triple (α0, α1, α2) to the triple (α0, α2, α1). For (p, q) = (10, 3) and (p, q) = (27, 8),
the construction of the auxiliary graph and searching for the desired walk can be completed
quickly. In this way, we verified that J5 is neither (10, 3)–edge colourable nor (27, 8)–edge
colourable, thus proving the following.
Proposition 7.10. χ′c (J5) = 17/5.
This completes the proof of Theorem 7.4.
Chapter 7. The Circular Chromatic Index 81
a
b
a′
b′
c
Figure 7.6: The Loupekine construction
7.3 Goldberg Snarks
The results of this section are published in [18].
In this section we study the circular chromatic index of the Goldberg snarks and of a closely
related family of snarks, the twisted Goldberg snarks. Goldberg snarks were introduced by
Goldberg [24] as counterexamples to the critical graph conjecture [3, 34] which asserts that
every critical graph (a graph all of whose proper subgraphs have smaller chromatic index)
has odd order. Goldberg’s counterexamples to the critical graph conjecture are not cubic.
Goldberg snarks are obtained from those graphs by adding edges to make them cubic.
The building block of Goldberg snarks is obtained via Loupekine’s construction, i.e. via
deleting the vertices of a path of length 2, from the Petersen graph. This operation is
depicted in Figure 7.6. Suppose G is a snark, and let H be obtained from G by deleting the
vertices of a path of length 2. If we label the resulting semiedges similarly to Figure 7.6,
then the parity lemma (Lemma 7.5) easily implies that in every 3–edge colouring of H,
either a = b and a′ 6= b′, or a 6= b and a′ = b′. In other words, in every 3–edge colouring,
the semiedges on one side of H match, while the semiedges on the other side of H mismatch.
Given odd n > 3, the Goldberg snark Gn is constructed from n copies of this block, arranged
cyclically, by attaching the semiedges a and b in each block to the semiedges a′ and b′ in the
next block, and then attaching the semiedges c of all blocks to vertices of an n–cycle, in the
same cyclic order. More precisely, V (Gn) = {vtj | 1 6 t 6 n, 1 6 j 6 8}, and adjacencies
are defined as shown in Figure 7.7.
Since in every 3–edge colouring of the graph on the right in Figure 7.6, one of the pairs
(a, b) and (a′, b′) is a match and the other is a mismatch, we see that the Goldberg snarks
Gn are not 3–edge colourable. We mimic this argument in computation of the circular
chromatic indices of these snarks.
Chapter 7. The Circular Chromatic Index 82
vt−12 vt
1 vt
2 vt+11
vt
8vt
6 vt
7
vt−14
vt
3 vt
4
vt+13
vt−15
vt
5 vt+15
Figure 7.7: Construction of Goldberg snarks. Superscripts are reduced modulo n.
We can modify the construction of Gn as follows. We connect the semiedges a and b of a
block to the semiedges b′ and a′ of the next block respectively (instead of a′ and b′). We
refer to this alteration as a twist in Gn. Since any even number of twists cancel out, we
get only one different graph for each n, obtained by applying only one twist to Gn. We
refer to this graph by twisted Goldberg snark, and denote it by TGn. More precisely, the
twisted Goldberg snark TGk is obtained from Gn by replacing the edges v12v
21 and v1
4v23
with the edges v12v
23 and v1
4v21 . The same argument as for Gn shows that TGn is a snark
for all odd n > 3.
Let e, e′, e′′ be the edges incident with a vertex x in a cubic graph G and let c be an r–
circular edge colouring of G where r = 3+ε. Then one of |c(e)−c(e′)|r and |c(e)−c(e′′)|r is
in the interval [1, 1 + ε]r and the other is in the interval [2, 2 + ε]r. By repeatedly applying
this observation we have the following.
Lemma 7.11. Let r = 3 + ε for some 0 < ε < 1, and let c be an r–edge colouring of
a cubic graph G. If e, e′ ∈ E(G) are at distance d in the line graph L(G), then c(e′) ∈[c(e) + t, c(e) + t+ dε]r for some integer d 6 t 6 2d.
For every 1 6 k 6 n, we denote the edges vk1v
k−12 and vk
3vk−14 of the Goldberg snark Gn,
by ek and fk respectively.
Lemma 7.12. Given odd n > 3, and r and ε satisfying 3 < r = 3+ε < 13/4, and an r–edge
colouring c of Gn, for every 1 6 k 6 n we have |c(ek) − c(fk)|r ∈ [0, 2ε]r ∪ [1 − 2ε, r/2]r.
Proof. The edge fk is at distance 4 from ek in Gn. Thus by Lemma 7.11, c(fk) ∈ c(ek) +
Chapter 7. The Circular Chromatic Index 83
[t, t+ 4ε]r for some t ∈ {4, 5, . . . , 8}. On the other hand,
Hence, |c(ek)− c(fk)|r ∈ [0, 2ε]r if t = 6, and |c(ek)− c(fk)|r ∈ [1 − 2ε, r/2]r otherwise.
Note that since ε < 1/4, only one of the alternatives in the above lemma can hold. Figure 7.8
shows all possible colourings of a block of Gn. The circular intervals on the edges indicate
possible values for the colour of that edge. In light of the above lemma, given an r–edge
colouring c of Gn where 3 < r < 13/4, we say the pair (ek, fk) of edges of Gn is of type A
(with respect to c), if |c(ek) − c(fk)|r ∈ [0, 2ε]r, and it is of type B if |c(ek) − c(fk)|r ∈[1 − 2ε, r/2]r.
Lemma 7.13. Let n > 3 be an odd integer and 3 < r < 13/4. Let c be an r–edge colouring
of Gn. Then for every 1 6 k 6 n, one of (ek, fk) and (ek+1, fk+1) is of type A and the
other is of type B.
Proof. This proof is based on the cases presented in Figure 7.8. We may assume that in
each picture, the two semiedges on the left are ek and fk and the two semiedges on the
right are ek+1 and fk+1. Moreover, we may assume that c(ek) = 0.
If (ek, fk) is of type A, then one of the cases (i), (iii) or (v) in Figure 7.8 holds. In cases (i)
and (iii), (ek+1, fk+1) is of type B. In case (v), if c(ek+1) ∈ [2, 2 + ε]r the conclusion holds.
On the other hand, if c(ek+1) ∈ [2 + ε, 2 + 2ε]r, then it is easily deduced that c(fk+1) must
be in [−ε, ε]r, thus the conclusion still holds.
A similar argument shows that if (ek, fk) is of type B (cases (ii), (iv), and (vi) in Figure 7.8),
then (ek+1, fk+1) is of type A. Note that in these cases, the intervals given for c(ek+1) and
c(fk+1) contain points with distance more than 2ε. Thus one needs to check those choices
for the colours of ek+1 and fk+1 cannot occur simultaneously.
Chapter 7. The Circular Chromatic Index 84
0 [1, 1 + ε]
[2, 2 + ε]
[−ε, ε]
[−ε, ε] [1 − ε, 1 + ε]
[−2ε, 2ε] [2 − ε, 2 + 2ε] [1 − ε, 1 + 2ε]
[2,2 + 2ε]
[1−ε,
1+
2ε]
[−ε,ε]
0 [1, 1 + ε]
[2, 2 + ε]
[−ε, ε]
[1 − ε, 1 + 2ε] [−ε, ε]
[1 − ε, 1 + 2ε] [2 − ε, 2 + 2ε] [−2ε, ε]
[2,2 + 2ε]
[−ε,
2ε]
[1−ε,1
+ε]
(i) (ii)
0 [1, 1 + ε]
[2, 2 + ε]
[2, 2 + 2ε]
[2 − ε, 2 + 2ε] [1 − ε, 1 + ε]
[−2ε, 2ε] [2 − ε, 2 + 2ε] [1 − ε, 1 + 2ε]
[−ε,ε]
[1−ε,
1+
2ε]
[−ε,ε]
0 [1, 1 + ε]
[2, 2 + ε]
[2, 2 + 2ε]
[1 − ε, 1 + 2ε] [−ε, ε]
[1 − 2ε, 1 + 2ε] [−2ε, ε] [2 − ε, 2 + 2ε]
[−ε,ε]
[2−ε,
2+
2ε] [1−
ε,1+ε]
(iii) (iv)
0 [1, 1 + ε]
[2, 2 + ε]
[2, 2 + 2ε]
[2 − ε, 2 + 2ε] [−ε, ε]
[−2ε, 2ε] [2 − ε, 2 + 2ε] [−2ε, ε]
[−ε,ε]
[1−ε,
1+
2ε] [1−
ε,1+ε]
0 [1, 1 + ε]
[2, 2 + ε]
[2, 2 + 2ε]
[2 − ε, 2 + 2ε] [−ε, ε]
[2 − ε, 2 + 2ε] [−2ε, ε] [2 − ε, 2 + 2ε]
[−ε,ε]
[1−ε,
1+
2ε] [1−
ε,1+ε]
(v) (vi)
Figure 7.8: All possible r–edge colourings of a block of Gn where r < 13/4. By symmetry,the top-left edge is assumed to have colour 0.
Chapter 7. The Circular Chromatic Index 85
1 5 9 0 4 8 12 7 2 10 1 10 2 10 1
9 0 0 9 7 3 3 12 10 1 1 10 10 1
1 9 1 10 6 2 11 3 11 6 1 6 11 6 1
9 0 8 4 0 9 0 9
5 4 12 8 5 5 5
9
0
5
9
12
3
3
11
6
6
6
6
6
6
5 5 5 1 11 7 7 7 2 10 10 2 2 10
Figure 7.9: A (13, 4)–edge colouring of G7 (and TG7). Semiedges wrap around accordingto the definition of the corresponding graph.
Corollary 7.14. For odd n > 3, we have χ′c (Gn) > 13/4 and χ′
c (TGn) > 13/4.
Proof. Let c be an r–edge colouring of Gn for some r < 13/4. Applying Lemma 7.13 to
c with k = 1, . . . , n, we conclude that (e1, f1) must be both of type A and type B. This
contradicts Lemma 7.12, thus χ′c (Gn) > 3 + 1/4. This argument is valid for the twisted
Goldberg snarks TGn as well. Thus χ′c (TGk) > 13/4 for all odd n > 3.
Theorem 7.15. For odd n > 5, we have χ′c (Gn) = χ′
c (TGn) = 13/4.
Proof. By Corollary 7.14, we only need to show that for all odd n > 5 we have χ′c (Gn) 6
13/4. A (13, 4)–edge colouring of G7 is given in Figure 7.9. The two blocks in the shaded
area can be deleted to obtain a (13, 4)–edge colouring of G5, or repeated t times for t > 1,
to obtain a (13, 4)–edge colouring of G7+2t. On the other hand, for all odd n > 5, since
the colouring of Gn given above assigns the same colour to the edges e1 and f1, “twisting”
these two edges results in a proper (13, 4)–edge colouring of TGn.
It remains to find the circular chromatic index of the graphs G3 and TG3. We exhibit
10/3–edge colourings of these two graphs in Figures 7.10 and 7.11. Hence χ′c (G3) 6 10/3
and χ′c (TG3) 6 10/3. On the other hand, an exhaustive computer search shows that both
these graphs have a tight cycle with respect to any (10, 3)–edge colouring. Therefore, by
Lemma 2.4, we have χ′c (G3) = χ′
c (TG3) = 10/3.
We summarize the results of this section in the following theorem.
Chapter 7. The Circular Chromatic Index 86
5 8 1 8 5 9 5
2 9 4 7 9 6
6 3 0 4 8 2 6
0 3 6 0
6 0 3
2
5
4
1
2
2
9 6 7 1 5 9
Figure 7.10: A (10, 3)–edge colouring of G3. Semiedges wrap around according to thedefinition of G3.
6 3 0 7 1 8 5
4 7 0 7 9 2
5 8 1 4 8 2 6
3 6 0 3
0 3 6
0
7
4
4
5
2
1 4 7 1 5 9
Figure 7.11: A(10, 3)–edge colouring of TG3. Semiedges wrap around according to thedefinition of TG3.
Chapter 7. The Circular Chromatic Index 87
Figure 7.12: The second Loupekine snark
Theorem 7.16. For all odd n > 3, we have
χ′c (Gn) = χ′
c (TGn) =
10/3 if n = 3,
13/4 if n > 5.
Let G∗3 and TG∗
3 be obtained by contracting the unique triangle in G3 and TG3 respectively.
Then G∗3 and TG∗
3 are snarks of order 22. These two snarks are known as the Loupekine
snarks. An alternate drawing of TG∗3 is presented in Figure 7.12. Note that both of the
(10, 3)–edge colourings of Figures 7.10 and 7.11 remain valid for G∗3 and TG∗
3. This proves
χ′c(G
∗3) = χ′
c(TG∗3) = 10/3.
7.4 Generalized Blanusa Snarks
Having order 18, Blanusa snarks are the smallest snarks after the Petersen graph P . They
are both obtained by a dot product from two copies of P . Given two cubic class 2 graphs
G1 and G2, the dot product G1 · G2 is constructed by adding four edges to the disjoint
union of G1 − {vw, v′w′} and G2 − {x, y}, as shown in Figure 7.13, where vw and v′w′ are
non-adjacent edges in G1 and x and y are adjacent vertices in G2. This operation was first
introduced by Isaacs [31].
Note that the dot product G1 · G2 depends on the choice of the edges vw, v′w′ in G1 and
the edge xy in G2. When G1 = G2 = P , since the Petersen graph is edge-transitive, the
Chapter 7. The Circular Chromatic Index 88
G1 G2 G1 ·G2
v
w
v′
w′
v
w
v′
w′
x
y
x′
x′′
y′
y′′
x′
x′′
y′
y′′
Figure 7.13: The dot product construction
choice of xy does not matter. On the other hand, two non-adjacent edges in P can be at
distance 2 or 3 in the line graph L(P ). The two non-isomorphic snarks obtained by a dot
product P · P are called the Blanusa snarks. More precisely, if vw, v′w′ ∈ E(P ) have a
common neighbour in L(P ), the resulting graph P ·P is called the first Blanusa snark, and
otherwise, the graph P · P is called the second Blanusa snark.
This construction was generalized by Watkins [53]. Following the notation introduced
in [53], we refer to the graph obtained by cutting two edges of P which are at distance i+1,
and keeping the four resulting semiedges, as the Blanusa block Ai (i = 1, 2). The graph
obtained by deleting two adjacent vertices of P and keeping the four semiedges is called
the Blanusa block B. These blocks are shown in Figure 7.14. The labels of the semiedges
indicate connections in constructions involving these blocks. The semiedges with label a
(resp. b) are always connected to a semiedge with label a′ (resp. b′) and vice versa. By this
assumption, the first (resp. second) Blanusa snark is obtained by attaching a copy of A1
(resp. A2) to a copy of B. It can be seen in Figure 7.14 that A1 can indeed be decomposed
into a copy of B and a single edge. Thus in constructions involving A1, we may replace A1
by a B and a K2 to obtain a further decomposition. We should point out here an error in
Figure 20 of [53], where the drawings of A1 and A2 are switched.
Watkins [53], defines two families of generalized Blanusa snarks using the blocks B, A1,
and A2. The family B1 consists of the graphs B1n constructed as follows: take n−1 copies of
the block B and one copy of A1, arrange these blocks cyclically, and connect the semiedges
a and b of each block to the semiedges a′ and b′ of the next block respectively. Note that B11
is the Petersen graph, and B12 is the first Blanusa snark. The family B2 is defined similarly,
using the block A2 in place of A1. Following [53], we refer to members of B1 (resp. B2) as
Chapter 7. The Circular Chromatic Index 89
b′a
b a′
b′a
b a′
b′a
b a′
B A1 A2
Figure 7.14: The Blanusa blocks
type 1 (resp. type 2) generalized Blanusa snarks.
Unlike the flower snarks and the Goldberg snarks which take only a finite set of values for
their circular chromatic index, the graphs in B1 ∪ B2 all have distinct circular chromatic
indices. The circular chromatic index of type 1 generalized Blanusa snarks was established
by Mazak [40].
Theorem 7.17. [40] For all n > 1, χ′c
(
B1n
)
= 3 + 23n .
In this section we prove the following result for type 2 generalized Blanusa snarks.
Theorem 7.18. For all n > 1,
χ′c
(
B2n+1
)
= 3 +1
⌊1 + 3n/2⌋ =
3 + 23n+1 if n is odd,
3 + 23n+2 if n is even.
7.4.1 The upper bounds
We first prove the upper bounds of Theorem 7.18. The structure of the optimum edge
colourings of the graphs B2n helps in understanding the proof of the lower bounds, presented
in the next section.
Let G be a cubic graph which may contain semiedges. A consecutive colouring of G is any
mapping c : E(G) → Z such that for each v ∈ V (G), if e, e′, e′′ are the edges incident with
v, then the colours c(e), c(e′), c(e′′) are three consecutive integers. Obviously, reducing the
colours c(e) modulo 3, one gets a proper edge 3–colouring of G. One could also reduce
the colours c(e) modulo 3 + ε for any given 0 < ε < 1, to obtain an (3 + ε)–circular edge
Chapter 7. The Circular Chromatic Index 90
2
1
2 4
5
4
21
3
3
0
0 6
3
2
1
2 3
2
3
21
0 1
1
0
0
0 4
4
22
1
2 0
2
3
21
0 1
1
0
0
0 1
4
2
(a) (b) (c)
Figure 7.15: Consecutive colourings of Blanusa blocks B,A2, A2 used in Lemma 7.19
colouring of G. The notion of consecutive colouring helps us present circular edge colourings
of graphs by integers rather than real numbers.
Lemma 7.19. Given n > 1, let ε = 1⌊1+3n/2⌋ and r = 3 + ε. Then χ′
c
(
B2n+1
)
6 r.
Proof. If n is even, then ε = 23n+2 . Consider the consecutive colouring of the graph B
given in Figure 7.15(a). Since 3 = −ε modulo r and 6 = −2ε modulo r, we may apply the
transformation c(e) 7→ 3ε− c(e) to the second of two consecutive copies of the block B in
B2n+1, to get a colouring c for which c(a) = c(b) = 0 in the first block, and c(a′) = c(b′) = 3ε
in the second block. Since B2n+1 contains n copies of the block B, combining suitable
transformations of these colourings, we get an edge r–circular colouring c of these blocks,
for which c(a) = c(b) = 0 for the first block, and c(a′) = c(b′) = (n/2)3ε = 1 − ε for the
last block. On the other hand, since 4 = 1 − ε modulo r, the consecutive colouring of A2
given in Figure 7.15(b) can be used to extend c to an r–circular edge colouring of B2n+1.
If n is odd, then ε = 23n+1 . Similarly to the previous case, we find a partial r–circular edge
colouring c of B2n+1 which colours all the edges in all copies of B, such that for the block
A2 we have c(a) = c(b) = 0, c(a′) = 1, and c(b′) = 1 − ε. This colouring can be extended
to B2n+1 using the consecutive colouring of A2 given in Figure 7.15(c).
7.4.2 The lower bounds
To prove the lower bounds, we need to study (3 + ε)–circular edge colourings of the blocks
B and A2. The following is proved by Mazak [40].
Chapter 7. The Circular Chromatic Index 91
Lemma 7.20. [40] Let 0 < ε < 14 , r = 3 + ε, and c be an r–circular edge colouring of B.
Then |c(a) − c(a′)|r 6 2ε and |c(a) − c(a′)|r + |c(b) − c(b′)|r 6 3ε.
If c is a 3–edge colouring of A2 such that c(a) = c(a′), then by the parity lemma, c(b) = c(b′)
which is a contradiction since this gives a 3–edge colouring of the Petersen graph. Thus
by the parity lemma, in every 3–edge colouring of A2, either c(a) = c(b) 6= c(a′) = c(b′) or
c(a) = c(b′) 6= c(a′) = c(b). In our next lemma, we prove an analogue of this observation
for (3 + ε)–circular edge colourings of A2.
Lemma 7.21. Let 0 < ε < 13 , r = 3+ε, and c be an r–circular edge colouring of A2. Then
|c(a) − c(a′)|r + |c(b) − c(b′)|r > 2 − 2ε.
Proof. Let e0 be the unique edge of A2 which is at distance 3 from a, a′, b, b′. We may
assume that c(e0) = 0. For every e ∈ E(A2) let de denote the distance from e0 to e.
Then 0 6 de 6 3, and by Lemma 7.11 there exists te ∈ {de, de + 1, . . . , 2de} such that
c(e) ∈ [te, te + deε]r. Since ε < 13 , each of these intervals contains exactly one of the integers
0, 1, 2 and intervals corresponding to different integers are disjoint. Let σ(e) ∈ {0, 1, 2}be the integer corresponding to (the interval containing) c(e). Then σ is a proper 3–
edge colouring of A2 and by the parity lemma, either σ(a) = σ(b) 6= σ(a′) = σ(b′) or
σ(a) = σ(b′) 6= σ(a′) = σ(b). By symmetries of A2, we may assume that the former holds.
Now up to symmetry we either have
σ(a) = σ(b) = 0 and σ(a′) = σ(b′) = 1, or
σ(a) = σ(b) = 1 and σ(a′) = σ(b′) = 2.
(7.1)
If we reduce the colours in [−2ε, 0]r modulo r to lie in the real interval [−2ε, 0], then by (7.1)
On the other hand, since a′ and b′ are at distance 3 from b and a respectively, by Lemma 7.11
there exist integers 3 6 t′, t′′ 6 6 such that c(a′) − c(b) ∈ [t′, t′ + 3ε]r and c(b′) − c(a) ∈[t′′, t′′ + 3ε]r. Since σ(a′)−σ(b) = σ(b′)−σ(a) = 1, we have s = r = 4. Therefore c(a′)−c(b)and c(b′) − c(a) are both in the real interval [1 − ε, 1 + 2ε] and we have
By extending c to a 3–colouring of G in the following ways, we obtain directed paths in
Dc(H) which combine to a nontrivial directed closed walk in Dc(H). The implied directed
path in each extension of c is shown by an arrow. Dashed arrows in each picture indicate
directed paths in Dc(H) found similarly.
01
1
2 2
0
2 2
210
01
1
2 2
0
1 2
1
0
2
1
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