Circular chromatic numbers of Mycielski’s graphs
Gerard J. Chang
a;1, Lingling Huang
a;1, Xuding Zhu
b;∗;2 aDepartment of Applied Mathematics, National Chiao Tung University, Hsinchu 30050, Taiwan bDepartment of Applied Mathematics, National Sun Yat-sen University, Kaohsiung 28608, TaiwanReceived 30 May 1996; revised 23 June 1998; accepted 23 November 1998
Abstract
In a search for triangle-free graphs with arbitrarily large chromatic numbers, Mycielski devel-oped a graph transformation that transforms a graph G into a new graph (G), we now call the Mycielskian of G, which has the same clique number as G and whose chromatic number equals (G) + 1. Let n(G) = (n−1(G)) for n¿2. This paper investigates the circular chromatic
num-bers of Mycielski’s graphs. In particular, the following results are proved in this paper: (1) for any graph G of chromatic number n; c(n−1(G))6 (n−1(G)) −12; (2) if a graph G satis es
c(G)6 (G) −1d with d = 2 or 3, then c(2(G))6 (2(G)) − 1d; (3) for any graph G of
chromatic number 3, c((G)) = ((G)) = 4; (4) c((Kn)) = ((Kn)) = n + 1 for n¿3 and
c(2(Kn)) = (2(Kn)) = n + 2 for n¿4. c 1999 Elsevier Science B.V. All rights reserved.
Keywords: Circular chromatic number; Mycielski’s graphs; Girth; Homomorphism; Connectivity; Critical graph
1. Introduction
All graphs in this paper are simple, i.e., nite, undirected, loopless, and without multiple edges.
In a search for triangle-free graphs with arbitrarily large chromatic numbers, Mycielski [15] developed an interesting graph transformation as follows. For a graph G with vertex set V (G) = V and edge set E(G) = E, the Mycielskian of G is the graph (G) with vertex set V ∪ V0∪ {u}, where V0= {x0: x ∈ V }, and edge set E ∪
{xy0: xy ∈ E} ∪ {y0u: y0∈ V0}. The vertex x0 is called the twin of the vertex x (and x is
∗Corresponding author.
E-mail address: [email protected] (X. Zhu)
1Spported in part by National Science Council under grant NSC85-2121-M009-24 2Supported in part by the National Science Council under grant NSC87-2115-M-11–004 0012-365X/99/$ - see front matter c 1999 Elsevier Science B.V. All rights reserved. PII: S0012-365X(99)00033-3
also called the twin of x0); and the vertex u is called the root of (G). If there is no
ambiguity we shall always use u as the root of (G). For n¿2, let n(G)=(n−1(G)).
Mycielski [15] showed that ((G))= (G)+1 for any graph G and !((G))=!(G) for any graph G with at least one edge. Hence n(K2) is a triangle-free graph of
chromatic number n + 2. Besides such interesting properties involving clique num-bers and chromatic numnum-bers, Mycielski’s graphs also have some other parameters that behave in a predictable way. For example, it was shown by Larsen et al. [14] that f((G)) = f(G) + f(G)1 for any graph G, where f(G) is the fractional chromatic
number of G. Mycielski’s graphs were also used by Fisher [6] as examples of optimal fractional colorings that have large denominators.
The purpose of this paper is to investigate the circular chromatic numbers of Mycielski’s graphs. The circular chromatic number c(G) of a graph G is a variation
of the chromatic number of G, introduced by Vince [17] in 1988, as the ‘star chromatic number’ of a graph. Let k and d be integers such that 0 ¡ d6k. A (k; d)-coloring of G is a coloring c of vertices of G with k colors {0; 1; : : : ; k − 1} such that for any edge xy, d6|c(x) − c(y)|6k − d. The circular chromatic number c(G) of G is the
minimum ratio k
d for which there exists a (k; d)-coloring of G. (To be precise, the
minimum in the de nition should be in mum. However, it was shown in [17] that the in mum is attained.) Observe that a (k; 1)-coloring of a graph G is just an ordinary k-coloring of G. It follows that c(G)6 (G). On the other hand, it is also not dicult
to see [3,17,18] that (G) − 1 ¡ c(G). Therefore, (G) = d c(G)e. In some sense the
circular chromatic number is a re nement of the chromatic number of a graph, and it contains more information about the graph. Readers are referred to [1–5,7–13,16–23] for more information on circular chromatic numbers of graphs.
In this paper, we show that circular chromatic numbers of Mycielski’s graphs ex-hibit interesting patterns. The problem of determining if c(G) = (G) or c(G) is
‘close to’ (G) − 1 is hard and has been extensively studied for general graphs. This paper reports some progress for Mycielski’s graphs in this direction. In Section 3, we prove that c(n−1(G))6 (n−1(G)) − 12 for any graph G of chromatic number
n, and c(2(G))6 (2(G)) −1d for any graph G with c(G)6 (G) − 1d; d = 2 or
3. Section 4 establishes that c((G)) = ((G)) = 4 for any graph G of chromatic
number 3, c((Kn)) = ((Kn)) = n + 1 for n¿3, and c(2(Kn)) = (2(Kn)) = n + 2
for n¿4.
These results yield many graphs with special properties having particular circular chromatic numbers. For example, it follows that there are triangle-free 4-critical graphs whose circular chromatic numbers are 4. This disproves a conjecture in [16]. It also follows from these results that there are triangle-free and color-critical graphs G of high connectivity for which c(G)6 (G) −12.
Along the way to proving these results, we also re ne some tools used by others in the study of the relationship between the circular chromatic number and the chro-matic number of a graph. We believe that the results obtained here are just a fraction of a family of interesting properties concerning the circular chromatic numbers of Mycielski’s graphs. In Section 5, a few questions are raised.
2. Preliminary results
The connectivity Ä(G) of a graph G is the minimum non-negative integer k such that G\S is disconnected or trivial for some vertex set S of size k. A graph G is k-critical if (H) ¡ (G) = k for any proper subgraph H of G; or equivalently, G is connected and (G\e) ¡ (G)=k for any edge e in G. The following lemma is surely folkloric:
Lemma 1. If G has no isolated vertices; then Ä((G))¿Ä(G) + 1. If G is k-critical; then (G) is (k + 1)-critical.
Proof. Suppose V (G)=V and V ((G))=V ∪ V0∪ {u}. Let S be a subset of V ((G))
of size Ä(G). If |S ∩ V | ¡ Ä(G), then G\(S ∩ V ) is connected. Also, for any vertex x ∈ V; x0 is adjacent to at least Ä(G) vertices of V in (G). So, any such vertex x0
of (G)\S is adjacent to at least one vertex in G\(S ∩ V ). And u is adjacent to all such vertices x0 of (G)\S. Thus, (G)\S is connected. If |S ∩ V | = Ä(G), then S ⊆ V .
Since G has no isolated vertices, any vertex x ∈ V \S is adjacent to some vertex y0
in V0, which is in turn adjacent to u. Thus, (G)\S is also connected. Therefore,
Ä((G))¿Ä(G) + 1.
For the proof of the second half of this lemma, assume that G is k-critical. Since G is connected, so is (G). Let e be any edge of (G). We consider the following three cases.
Suppose e = ab for some a ∈ V and b ∈ V . Let c be a proper (k − 1)-coloring of G\e. Then the following c0 is a proper k-coloring of (G)\e: c0(x) = c(x) for all
x ∈ V; c0(x0) = k − 1 for all x0∈ V0, and c0(u) = 0.
Suppose e = ab0 for some a ∈ V and b0∈ V0. Let c be a proper (k − 1)-coloring of
G\ab. Then the following c0 is a proper k-coloring of (G)\e: c0(b)=k−1; c0(x)=c(x)
for all x ∈ V \{b}; c0(x0) = c(x) for all x0∈ V0, and c0(u) = k − 1.
Suppose e = a0u for some a0∈ V0. Suppose c is a proper (k − 1)-coloring of G\a.
Then the following c0 is a proper k-coloring of (G)\e: c0(x) = c0(x0) = c(x) for all
x ∈ V \{a} and c0(a) = c0(a0) = c0(u) = k − 1.
For an n-coloring c : V (G) 7→ {0; 1; : : : ; n − 1} of G, we denote by Dc(G) the
directed graph with vertex set V (G) in which there is an arc from x to y if and only if xy ∈ E(G) and c(x) + 1 ≡ c(y) (mod n). It was shown in [10], in the corollary of Theorem 1, that an n-chromatic graph G satis es c(G) ¡ n if and only if G has an
n-coloring c for which Dc(G) is acyclic. For our purposes in this paper, we re ne this
result in two respects.
Lemma 2. If x0 is a vertex of an n-chromatic graph G for which c(G) ¡ n; then
there is an n-coloring c of G such that Dc(G) is acyclic; c(x0) = 1; and c(x) 6∈ {0; 1}
Proof. Suppose c(G)=kd¡ n and d ¿ 1. Then G has a (k; d)-coloring h with h(x0)=
d − 1. De ne c: V (G) 7→ {0; 1; : : : ; n − 1} by c(v) = bh(v)+1d c for each v ∈ V (G). It is straightforward to check that c is a proper coloring, Dc(G) is acyclic, c(x0) = 1, and
c(x) 6∈ {0; 1} for all vertices x adjacent to x0.
Corollary 3. If (G) with root u satisÿes c((G)) ¡ ((G)) = n; then there is an
n-coloring c of (G) such that Dc((G)) is acyclic; c(u) = 1; and c(x0) 6∈ {0; 1} for
all x0∈ V0. Moreover; for any such coloring c; there is an edge ab ∈ E(G) such that
c(a) = 0; c(b) = 1; and c(a0) = c(b0).
Proof. Applying Lemma 2 to (G) with x0= u, we obtain an n-coloring c such that
Dc((G)) is acyclic, c(u)=1, and c(x0) 6∈ {0; 1} for all x0∈ V0. To prove the ‘moreover’
part, we assume to the contrary that c(a0) 6= c(b0) for all edges ab ∈ E(G) with c(a)=0
and c(b) = 1. Let c0 be the coloring de ned by c0(x) = c(x) if c(x) 6∈ {0; 1} and c0(x) =
c(x0) if c(x) ∈ {0; 1}. It is straightforward to verify that c0 is an (n − 2)-coloring of G,
contrary to the assumption that ((G)) = n.
Lemma 4. Suppose G is an n-chromatic graph and that there is an n-coloring c: V (G) 7→ {0; 1; : : : ; n − 1} of G such that Dc(G) is acyclic. Let P be the set of all directed
paths of Dc(G). For any P ∈ P; let z(P) be the number of vertices of P which are
colored 0 and let d = max{z(P) + 1: P ∈ P}. If n¿3; then c(G)6n − 1d.
Proof. For each vertex x of G, let Px be the set of all directed paths of Dc(G) that
end at x and let ‘(x) = max {z(P):P ∈ Px}. De ne an (nd − 1; d)-coloring h of G
by h(x) = (c(x)d + ‘(x)) mod (nd − 1). Since 06c(x)6n − 1 and ‘(x)6d − 1, it follows that 06c(x)d + ‘(x)6nd − 1 and then h(x) = c(x)d + ‘(x), except h(x) = 0 for c(x) = n − 1 and ‘(x) = d − 1. We show that h is indeed an (nd − 1; d)-coloring of G.
Suppose xy is an edge of G. Assume that c(x) ¡ c(y). First consider the case that 26c(y) − c(x)6n − 2. If c(y)6n − 2 or c(y) = n − 1 but ‘(y) ¡ d − 1, then c(y)d6h(y)6c(y)d + d − 1 and c(x)d6h(x)6c(x)d + d − 1. Hence,
d6c(y)d−(c(x)d+d−1)6h(y)−h(x)6c(y)d+d−1−c(x)d6nd−1−d: If c(y) = n − 1 and ‘(y) = d − 1, then h(y) = 0. Since 16c(x)6n − 3, it follows that d6h(x)6(n − 2)d − 1. Hence, d6h(x) − h(y)6nd − 1 − d.
Next, we assume that c(y)−c(x)=1. In this case, xy is an arc of Dc(G). Therefore,
‘(y)¿‘(x). If c(y)6n − 2 or c(y) = n − 1 but ‘(y) ¡ d − 1, then h(y) = c(y)d + ‘(y) and h(x) = c(x)d + ‘(x). Hence, d6h(y) − h(x)62d − 16nd − 1 − d. If c(y) = n − 1 and ‘(y) = d − 1, then h(y) = 0. Since c(x) = n − 2, it follows that (n − 2)d6h(x)6 (n − 2)d + d − 1. Hence, d6h(x) − h(y)6nd − 1 − d.
Finally, we assume that c(y)=n−1 and c(x)=0. In this case, yx is an arc of Dc(G)
h(x) = ‘(x)¿‘(y) + 1. It follows that d6h(y) − h(x)6nd − 1 − d. This completes the proof of the lemma.
Corollary 5. Suppose n¿3 and that G is an n-chromatic graph having an n-coloring c such that Dc(G) is acyclic. Let P be the set of all directed paths of Dc(G). For
each P ∈ P; let s(P) be the number of arcs in P and let s = max {s(P): P ∈ P}. If d = bs
nc + 2; then c(G)6n −d1.
Proof. Since each directed path of Dc(G) has at most s arcs, it follows that the
path contains at most bs
nc + 1 vertices with color 0. The result then follows from
Lemma 4.
3. Graphs G with c((G)) ¡ ((G))
Note that (K2) is a pentagon that has circular chromatic number 52. Indeed, it is not
dicult to see that for any bipartite graph G; (G) has circular chromatic number 5 2.
The purpose of this section is to study G and m for which c(m(G))6 (m(G)) −d1
for some d.
To work with such graphs, we need to take special care with the names of the vertices. We now introduce a system for naming the vertices of m(G). It turns out
that using this naming system provides an easy method for determining the adjacency of vertices and for telling which vertex is the twin of another vertex at a certain level. For any two non-negative integers i and j, let i&j denote the integer whose binary representation is the logical ‘and’ of the binary representations of i and j. For instance, 14&25 = 011102&110012= 010002= 8 and 10&17 = 010102&100012= 000002= 0.
If i ¿ 0, let f(i) denote the maximum factor of i that is a power of 2. For instance, f(1)=f(3)=1; f(6)=f(18)=2 and f(12)=4. For any graph G and any non-negative integer m, let Gm be the graph whose
vertex set V (Gm) = {xi: x ∈ V (G) and 06i ¡ 2m} ∪ {ui: 16i ¡ 2m};
edge set E(Gm) = {xiyj: xy ∈ E(G) and i&j = 0} ∪ {xiuj: i&j = f(j)}
∪ {uiuj: i&j = max {f(i); f(j)}andf(i) 6= f(j)}:
Note that G∼=G0.
Lemma 6. For any graph G and any non-negative integer m; Gm+1 is isomorphic to
(Gm). Consequently; m(G) ∼= Gm for any m¿0.
Proof. Consider the function h: V (Gm+1) 7→ V ((Gm)) de ned by
h(xi) = xi and h(xi+2m
) = (xi)0 for x ∈ V (G) and 06i ¡ 2m;
h(ui) = ui and h(ui+2m
) = (ui)0 for 16i ¡ 2m;
h(u2m
Fig. 1. 2(G).
It is straightforward to check that h is an isomorphism between Gm+1 and (Gm) by
using the following facts:
(i) If 06i; j ¡ 2m, then (i + 2m)&j = i&(j + 2m) = i&j and (i + 2m)& (j + 2m) =
(i&j) + 2m.
(ii) If 16i ¡ 2m, then f(2m) = 2m¿ f(i) = f(i + 2m).
An induction with the basis G∼=G0 proves that m(G)∼=Gm for any m¿0.
It follows from Lemma 6 that for any graph G, we may simply take the de nition of Gm as a naming system for the vertices of m(G). For the remainder of this paper,
we use V (Gm) and E(Gm) to denote the vertex set and the edge set of m(G),
respec-tively. Fig. 1 shows 2(G). Note that a link between the two sets {xi: x ∈ V (G)} and
{yj: y ∈ V (G)} means that i&j = 0, i.e., xiyj∈ E(2(G)) if and only if xy ∈ E(G);
and a link between {xi: x ∈ V (G)} and uj means i&j = f(j), i.e., xiuj∈ E(2(G)) for
all x ∈ V (G).
Theorem 7. If G is a graph for which c(G)6 (G)−d1 with d=2 or 3; then c(2(G))
6 (2(G)) − 1 d.
Proof. Suppose (G) = k and that c(G)6k −d1. Let c : V (G) 7→ {0; 1; : : : ; dk − 2} be
a (dk − 1; d)-coloring of G. De ne c0: V (2(G)) 7→ {0; 1; : : : ; dk + 2d − 2} as c0(xi) = dk + d if i = 0 and c(x) = dk − d; c(x) + dk − 1 if i = 2 and c(x)6d − 2; dk − 1 if i = 3 and c(x)6d − 2; c(x) otherwise; c0(ui) = dk + d − 1 if i = 1; dk + 2d − 2 if i = 2; dk + d − 2 if i = 3:
It is straightforward to verify that d6|c0(a) − c0(b)|6(dk + 2d − 1) − d for each
Corollary 8. If G is a graph for which c(G)6 (G) − 1d; d = 2 or 3; and k is a
non-negative integer; then c(2k(G))6 (2k(G)) −1d.
For any integer n¿4, in order to nd an n-chromatic graph G for which c((G))6
((G)) −1
2, we may take any graph H such that (H) = n − 1 and c(H)6n −32 (see
[17] for a proof of the existence of such graphs) and let G = (H). It follows from Theorem 7 that c((G)) = c(2(H))6 (2(H)) −12= ((G)) −12. Therefore there
are many graphs G whose Mycielskians have circular chromatic numbers strictly less than their chromatic numbers. Our next result concerns graphs obtained by repeatedly taking Mycielski transformations of a graph.
Theorem 9. If G is a graph of chromatic number n; then c(n−1(G))6
(n−1(G)) −1 2.
Proof. First of all, we construct a (2n − 1)-coloring c of n−1(K
n) such that
Dc(n−1(Kn)) is acyclic. For the sake of clarity, we rst color the vertices of n−2(Kn).
Let the vertices of Kn be x1; x2; : : : ; xn. Then by our naming system, the vertex
set of n−2(K
n) is {xji: 16j6n and 06i ¡ 2n−2} ∪ {ui: 16i ¡ 2n−2}. Let I ={i: 06
i ¡ 2n−2}. We partition I into subsets I
t for 16t6n−1, where It={i ∈ I: 2n−2−2t−16
i ¡ 2n−2− 2t−2}. Let c be the (2n − 1)-coloring of n−2(K
n) de ned as follows: c(xi j) = 2j − 1 if 26j6n − 1; 0 if j = n and i ∈ I1= {2n−2− 1}; 2 if j = 1 and i ∈ I1∪ In−1; 2t if j ∈ {1; n} and i ∈ It for 26t6n − 2; 2n − 2 if j = n and i ∈ In−1; c(ui) = ( 17 if i = 2n−3;
2t + 4 otherwise; where f(i) = 2t:
We rst verify that c is a proper coloring of n−2(K
n). The graph n−2(Kn) has three
types of edges: xi jxi
0
j0; uiuj, and xijuk. If xijxi 0
j0∈ E(n−2(Kn)), then j 6= j0 and i&i0= 0.
It follows that c(xi
j) 6= c(xi
0
j0), since i&i0 6= 0 when i and i0 are both in It for some
26t6n−2. If uiuj∈ E(n−2(K
n)), then f(i) 6= f(j), which implies that c(ui) 6= c(uj).
Suppose xi
juk∈ E(n−2(Kn)). Note that u2n−3 is the only vertex of color 1. Thus, we
may assume that k 6= 2n−3. Then c(uk) is an even integer and 46c(uk)62n − 4.
Suppose to the contrary that both end vertices of the edge xi
juk are colored with color
c(xi
j) = c(uk) = 2s + 4, where 26s + 26n − 2. It then follows from the de nition
that f(k) = 2s, j ∈ {1; n}, and i ∈ I
s+2 = {2n−2− 2s+1; : : : ; 2n−2− 2s− 1}. However
this implies that i&k 6= f(k) and hence, xi
juk6∈ E(n−2(Kn)). Therefore, c is indeed a
proper coloring of n−2(K n).
Next we color the remaining vertices of n−1(Kn) = (n−2(Kn)). All these vertices
will be colored the same color as their twins in n−2(K
n), except that c(x2nn−1−1) =
c(u2n−2+2n−3
) = 2n − 2 and c(u2n−2
To verify that c is a proper coloring of n−1(K
n), it suces to consider the three
exceptionally colored vertices x2n−1−1
n ; u2
n−2+2n−3
, and u2n−2
. The only other vertex with color 1 is u2n−3
, which is not adjacent to u2n−2
. The only other vertices of color 2n−2 are those xi
n with 06i ¡ 2n−3, which are not adjacent to x2
n−1−1
n and u2
n−2+2n−3
. Therefore, c is indeed a proper coloring of n−1(K
n).
Next we show that Dc(n−1(Kn)) is acyclic. Assume to the contrary that Dc(n−1(Kn))
contains a directed cycle. Note that x2n−2−1
n is the only vertex of color 0. We conclude
that this cycle has a length of 2n − 1. It starts with x2n−2−1
n ; and then u2
n−3
, which is the unique vertex of color 1 adjacent to x2n−2−1
n ; and ends with u2
n−2+2n−3
, which is the unique vertex of color 2n − 2 adjacent to x2n−2−1
n .
Let us call the vertices of this cycle Y =(y0; y1; : : : ; y2n−2), where c(yi)=i. Therefore,
y0= x2nn−2−1; y1= u2n−3, and y2n−2= u2n−2+2n−3.
De ne J∗
t = {xij: 06i ¡ 2t−1 or 06i − 2n−2¡ 2t−1} and It∗ = {xji : i ∈ It or i −
2n−2∈ I
t} for 16t6n − 2. It is easy to verify that for any 16t6n − 2 and xijxi
0
j0∈
E(n−1(K
n)), if xij∈ It∗, then i&i0= 0 and so xji00∈ Jt∗. Moreover, any vertex xij∈ Jt∗
is not adjacent to any vertex uk colored by 2t + 2, since 06i (or i − 2n−2) ¡ 2t−1
and f(k) = 2t−1. Since y
1= u2n−3, we may conclude that y2∈ I1∗. It then follows
that y3∈ J1∗; y4∈ I2∗; y5∈ J2∗; : : : ; y2n−4∈ In−2∗ , and y2n−3∈ Jn−2∗ . Thus, y2n−3 is not
ad-jacent to y2n−2= u2n−2+2n−3, contrary to the assumption that Y is a cycle. Therefore,
Dc(n−1(Kn)) is indeed acyclic. Since there is only one vertex colored with color 0, it
follows from Lemma 4 that c(n−1(Kn))62n − 1 −21 = (n−1(Kn)) −12.
If G is an arbitrary n-chromatic graph, then there is a homomorphism from G to Kn. It follows that there is a homomorphism from n−1(G) to n−1(Kn). Therefore,
c(n−1(G))6 (n−1(Kn)) −12= (n−1(G)) −12.
The following corollary follows easily from Theorems 7 and 9:
Corollary 10. If G is an n-chromatic graph and t is a non-negative integer; then c(n−1+2t(G))6 (n−1+2t(G)) −12.
By Lemma 1, if G is color-critical, then so is m(G). It also follows from Lemma 1
that m(G) has high connectivity. If G is triangle-free, then so is m(G). Thus, it
follows from Corollary 10 that there are triangle-free and color-critical graphs G of high connectivity for which c(G)6 (G)−12 (for example, k−3(C5) is such a k-critical
graph). This gives another proof of Theorem 4 in [2], which asserts that there exist k-critical (k − 1)-connected triangle-free graphs G for which c(G)6k −12.
4. Graphs G with c((G)) = ((G))
This section investigates graphs G for which c((G)) = ((G)). We rst prove
Theorem 11. If (G) = 3; then c((G)) = ((G)) = 4.
Proof. Suppose that, to the contrary, there is a 3-chromatic graph G for which c((G)) ¡ 4. By Corollary 3, there is a 4-coloring c of (G) such that Dc((G))
is acyclic, c(u) = 1, c(x0) 6∈ {0; 1} for all x0∈ V0; and there is an edge xy ∈ E(G) such
that c(x) = 0; c(y) = 1, and c(x0) = c(y0). Assume that c is such a coloring with a least
number of 0–1 edges (i.e., edges with two end vertices colored 0 and 1, respectively). Assume c(x0) = c(y0) = 2 (the case in which c(x0) = c(y0) = 3 is symmetric). Then
c(z) = 1 for each z ∈ NG(x), otherwise xyx0z is a directed 4-cycle in Dc((G)). For
each z ∈ NG(x), if c(z0) = 3, then c(w) = 0 for each w ∈ NG(z)\{x}, otherwise xzwz0 is
a directed 4-cycle in Dc((G)). We re-color z0 with color 2 for each z ∈ NG(x), and
re-color x and x0 with color 3. It is straightforward to verify that this new coloring c0
is still a proper 4-coloring of (G) and that Dc0(G) is acyclic. However c0 has fewer
0–1 edges than c, contrary to the choice of c.
It was conjectured in [16] that triangle-free n-critical graphs have circular chromatic numbers strictly less than n. However, it follows from Lemma 1 and Theorem 11 that for k¿2, (C2k+1) is a triangle-free 4-critical graph that has the circular chromatic
number 4. Therefore the conjecture fails for n = 4. We do not know whether the conjecture fails for any other integer n.
It was shown in [16] that n-critical graphs of ‘large girth’ have circular chromatic numbers ‘close to’ n − 1. However, it is unknown how large the girth of an n-critical graph G must be to guarantee that c(G) ¡ n. For each integer n, let g(n) be the
minimum integer such that any n-critical graph of girth greater than g(n) has c(G) ¡ n.
It follows from the corollary of Theorem 1 in [10] that g(n)6n. The above argument shows that g(4)¿4 and hence g(4) = 4. It is easy to show that g(3) = 3. The value of g(n) is unknown for n¿5.
The next four results concern the Mycielskian of the complete graphs Kn. When n=2,
(K2) is the pentagon, and hence has circular chromatic number 52. It follows from
Theorem 11 that c(2(K2))=4, and follows from Corollary 8 that c(2k+1(K2))62k+
2 + 1
2. In the following we consider the case of n¿3.
Theorem 12. If n¿3; then c((Kn)) = ((Kn)) = n + 1.
Theorem 13. If n¿4; then c(2(Kn)) = (2(Kn)) = n + 2.
To prove Theorems 12 and 13, it suces to show that for any (n + 1)-coloring c of (Kn) and any (n + 2)-coloring c0 of 2(Kn), the directed graphs Dc((Kn)) and
Dc0(2(Kn)) contain directed cycles (see Lemma 2). However, we prove two stronger
results that seem to be potentially useful for more general graphs.
First, we introduce notation. Suppose G is a graph and that c : V (G) 7→ {0; 1; : : : ; k − 1} is a proper coloring of G. Let C =(x1; x2; : : : ; xm) be a cycle of G. We say that C is
c(x1) ¡ · · · ¡ c(xi−1). We may view the colors as cyclically ordered, such that i
pre-cedes i + 1 and k − 1 prepre-cedes 0. Then a cycle C is consistently colored only if the colors of the vertices of C are in the same cyclic order as C. To be precise, for two colors i and j we let [i; j]k denote the set {i; i+1; i+2; : : : ; j}, where addition is carried out modulo k. For example, [2; 5]8= {2; 3; 4; 5} and [5; 2]8= {5; 6; 7; 0; 1; 2}. We let (i; j)k= [i; j]k − {i; j} and [i; j)k = [i; j]k− {j}. Then a cycle C = (x1; x2; : : : ; xm) is
consistently colored if for any index q 6∈ {p; p + 1}, c(xq) 6∈ [c(xp); c(xp+1)]k.
It is trivial that for any proper coloring of Kn there is a consistently colored n-cycle.
In the next two theorems, we show that for n¿3, every proper coloring of (Kn) has
a consistently colored (n+1)-cycle; and for n¿4, every proper coloring of 2(K n) has
a consistently colored (n + 2)-cycle.
Theorem 14. If n¿3 and c : V ((Kn)) 7→ {0; 1; : : : ; k − 1} is a proper coloring of
(Kn); then there is a consistently colored (n + 1)-cycle.
Proof. The restriction of c to V = V (Kn) has a consistently colored n-cycle which
we assume is C = (x0
1; x02; : : : ; x0n). The colors (c(x01); c(x02); : : : ; c(x0n)) form a cycle with
respect to the cyclic order of the colors. For each j, we consider the color c(x1
j) (recall
that x1
j is the twin of x0j). If c(xj1) ∈ [c(x0i); c(x0i+1)]k for some i 6∈ {j − 1; j}, then we
obtain an (n+1)-cycle (x0
1; x20; : : : ; xi0; x1j; x0i+1; : : : ; x0n) which is consistently colored (note
that x1
j is adjacent to both x0i+1 and x0i, hence c(xj1) ∈ (c(xi0); c(x0i+1))).
Assume now that for each j, we have c(x1
j) ∈ (c(x0j−1); c(x0j+1))k. Let i be the
index such that c(u1) ∈ [c(x0
i); c(x0i+1))k. We now consider the relative positions of
the colors c(x1
i), c(xi+11 ), and c(u1). If c(u1) ∈ (c(x1i); c(x1i+1))k⊆(c(xi−10 ); c(x0i+2))k,
then (x0
1; x02; : : : ; xi−10 ; xi1; u1; xi+11 ; x0i+2; : : : ; x0n) is a consistently colored (n + 1)-cycle. If
c(u1) ∈ (c(x1
i+1); c(x0i+1))k⊆(c(xi+11 ); c(x1i+2))k, then (x01; x02; : : : ; xi0; x1i+1; u1; x1i+2; x0i+3; : : : ;
x0
n) is a consistently colored (n + 1)-cycle. Otherwise c(u1) ∈ [c(x0i); c(x1i))k⊆(c(xi−11 );
c(x1
i))k, and in this case (x10; x02; : : : ; x0i−2; x1i−1; u1; x1i; xi+10 ; x0i+2; : : : ; x0n) is a consistently
colored (n + 1)-cycle.
We note from the proof of Theorem 14 that there are two types of consistently colored (n + 1)-cycles in (Kn) (see Fig. 2). Type I is an (n + 1)-cycle CI(i; j)=
(x0
1; : : : ; xi0; x1j; x0i+1; : : : ; xn0) obtained from the n-cycle C =(x01; x20; : : : ; x0n) in Kn by adding
a vertex x1
j between x0i and x0i+1. Type II is an (n + 1)-cycle CII(i) = (x01; : : : ; x0i; xi+11 ; u1;
x1
i+2; x0i+3; : : : ; xn0) obtained from the n-cycle in Kn by replacing xi+10 ; x0i+2 with
x1
i+1; u1; x1i+2.
We call {x0
i; x0i+1} the base of a Type I cycle CI(i; j) and {x0i; x0i+1; xi+20 ; x0i+3} the
base of a Type II cycle CII(i).
Theorem 15. If n¿4 and c : V (2(K
n)) 7→ {0; 1; : : : ; k − 1} is a proper coloring of
2(Kn); then there is a consistently colored (n + 2)-cycle.
Proof. Recall that the vertex set of 2(K
n) is (S3i=0Vi) ∪ {u1; u2; u3}, where Vi =
{xi: x ∈ V (Kn)} for 06i63 (see the naming system introduced in the previous section
and Fig. 1). Then we have ve copies of (Kn) in 2(Kn) that are induced by the
following vertex sets: V0∪ V1∪ {u1}, V0∪ V1∪ {u3}, V0∪ V2∪ {u2}, V0∪ V3∪ {u1},
V0∪ V3∪ {u2}. By Theorem 14, the copy of (Kn) with vertex set V0∪ Vr∪ {us} has
a consistently colored (n + 1)-cycle Cr;s, which is either of Type I or of Type II.
Assume to the contrary that 2(K
n) has no consistently colored (n + 2)-cycles. We
consider the relative positions of the above ve (n+1)-cycles Cr;s. First of all, the bases
of any two (n + 1)-cycles have at least two common vertices, otherwise their union would induce a consistently colored (n + 2)-cycle. Therefore, for any two consistently colored (n + 1)-cycles Cr;s and Cr0;s0, one of the following relative positions holds.
(i) Cr;s= CI(i; j) and Cr0;s0= CI(i; j0) (see Fig. 3).
(ii) Cr;s= CII(i) and Cr0;s0 = CI(i; j) or CII(i) or CI(i + 2; j) or CII(i + 2) (see
Fig. 4), or vice versa.
(iii) Cr;s= CII(i) and Cr0;s0= CI(i + 1; j) or CII(i + 1) (see Fig. 5), or vice versa.
Claim. If us is adjacent to all vertices of V
r0 and us0 is adjacent to all vertices of Vr;
then (i) or (ii) holds.
Proof. Suppose to the contrary that (iii) holds. For the case in which Cr;s= CII(i) and
Cr0;s0= CI(i + 1; j), since xrj0 is adjacent to us (i.e., the dashed line in Fig. 5 is an edge
in 2(K
n)), we have
(x0
1; : : : ; x0i+1; xr
0
j ; us; xi+2r ; xi+30 ; : : : ; x0n) or (x10; : : : ; x0i; xri+1; us; xr
0
j; x0i+2; : : : ; xn0)
Fig. 4. Relative position (ii).
Fig. 5. Relative position (iii).
is a consistently colored (n + 2)-cycle. Similarly, for the case in which Cr;s= CII(i)
and Cr0;s0= CII(i + 1), xri+20 is adjacent to us and so,
(x0
1; : : : ; x0i+1; xr
0
i+2; us; xri+2; x0i+3; : : : ; x0n)
or (x0 1; : : : ; xi0; xi + 1r; us; xr 0 i+2; us 0 ; xr0 i+3; x0i+4; : : : ; x0n)
is a consistently ordered (n + 2)-cycle.
Note that if Cr;sand Cr0;s0 have relative position (i) or (ii), and that Cr0;s0 and Cr00;s00
have relative position (i) or (ii), then Cr;s and Cr00;s00 also have relative position (i) or
(ii). In other words, two (n + 1)-cycles having relative positions (i) or (ii) form an equivalence relation among the ve (n + 1)-cycles.
This fact along with an application of the above claim to the (n + 1)-cycle pairs (C2;2; C3; 2), (C3; 2; C3;1), (C3;1; C1;1), (C1;1; C1;3) leads to the conclusion that C2; 2 and
Suppose C2;2 and C1;3 have relative position (ii), say C2;2= CII(i) and C1;3= CI(i; j)
or CII(i) or CI(i + 2; j) or CII(i + 2). If C2;2= CII(i) and C1;3= CI(i; j), then x1j is
adjacent to x2
i+1 and so
(x0
1; : : : ; x0i; xj1; x2i+1; u2; x2i+2; x0i+3; : : : ; x0n) or (x01; : : : ; x0i; xi+12 ; x1j; x0i+1; : : : ; x0n)
is a consistently colored (n + 2)-cycle. If C2;2= CII(i) and C1;3= CII(i), then u3 is
adjacent to u2 and so
(x0
1; : : : ; x0i; xi+11 ; u3; u2; xi+22 ; x0i+3; : : : ; x0n)
or (x0
1; : : : ; xi0; xi+12 ; u2; u3; xi+21 ; x0i+3; : : : ; x0n)
is a consistently colored (n + 2)-cycle. If C2;2= CII(i) and C1;3= CI(i + 2; j), then x1j
is adjacent to x2
i+2 and so
(x0
1; : : : ; x0i+2; xj1; x2i+2; x0i+3; : : : ; x0n) or (x01; : : : ; x0i; xi+12 ; u2; xi+22 ; x1j; x0i+3; : : : ; x0n)
is a consistently colored (n + 2)-cycle. If C2;2= CII(i) and C1;3= CII(i + 2), then x1i+3
is adjacent to x2
i+2 and so
(x0
1; : : : ; x0i+2; xi+31 ; x2i+2; x0i+3; : : : ; x0n)
or (x0
1; : : : ; xi0; xi+12 ; u2; x2i+2; x1i+3; u3; x1i+4; xi+50 ; : : : ; x0n)
is a consistently colored (n + 2)-cycle.
Suppose C2;2and C1;3have relative position (i), say C2;2=CI(i; j) and C1;3=CI(i; j0).
Then j = j0, otherwise, x2
j is adjacent to x1j0 and so
(x0
1; : : : ; x0i; xj2; xj10; x0i+1; : : : ; x0n) or (x01; : : : ; xi0; xj10; x2j; xi+10 ; : : : ; x0n)
is a consistently colored (n + 2)-cycle. We may assume that both V0∪ V2∪ {u2} and
V0∪ V1∪ {u3} have only one Type I (n+1)-cycle, for otherwise the union of two cycles
with di erent bases is a consistently colored (n+2)-cycle, or we may choose j; j0so that
j 6= j0, contrary to the conclusion above. Therefore c(x1
p), c(xp2) ∈ (c(xp−10 ); c(x0p+1))k
for each p 6= j. We may assume that none of V0∪ V2∪ {u2} and V0∪ V1∪ {u3}
has Type II cycles, for otherwise we may choose C2; 2 and C1; 3 so that they have
relative position (ii), which has been discussed in the previous paragraph. It follows that c(u2) ∈ (c(x2
j−1); c(x2j+1))kand c(u3) ∈ (c(x1j−1); c(x1j+1))k, (cf. the proof of Theorem
14). Note that i 6= j − 1; j, without loss of generality, we may assume that 16i6j − 2 ¡ n and c(u2) ∈ (c(u3); c(x0
j+2))k. If i ¡ j − 2, then since u3 is adjacent to u2,
(x0
1; : : : ; x0i; xj1; x0i+1; : : : ; x0j−2; xj−11 ; u3; u2; xj+12 ; x0j+2; : : : ; x0n)
is a consistently colored (n + 2)-cycle. If i = j − 2, then either (x0
1; : : : ; x0j−2; x1j−1; x2j; xj−10 ; : : : ; xj+20 ; : : : ; x0n)
or (x0
1; : : : ; xj−20 ; xj2; xj−11 ; u3; u2; x2j+1; xj+20 ; : : : ; x0n)
is a consistently colored (n + 2)-cycle.
We have omitted some details of the proof, which can be added easily. For example, we did not explicitly use the condition n¿4, which is a necessary condition. Indeed,
when n = 4, the fourth picture in Fig. 4 looks di erent. The vertices x0
i+4; x0i+5 are
equal to x0
i; x0i+1, respectively. A dashed line should be added between xri+1 and xi+4r .
However, a consistently colored (n + 2)-cycle can still be found in the corresponding cases.
It follows from Theorems 14 and 15 that for any (n+1)-coloring c of (Kn) and any
(n + 2)-coloring c0 of 2(K
n), the directed graphs Dc((Kn)) and Dc0(2(Kn)) contain
directed cycles. Therefore, Theorems 12 and 13 have been proven. We close this section with the following conjecture:
Conjecture 1. If n¿m + 2, then c(m(Kn)) = (m(Kn)) = n + m.
5. Further research
We have established some results for circular chromatic numbers on Mycielski’s graphs. However, many questions remain open. We list below some related questions. Question 1. Given a graph G, what can we say about the sequence ( (m(G)) −
c(m(G)): m=1; 2; : : :)? Does it approach a limit? What are the possible accumulating
points of such a sequence?
It follows from Corollary 10 that there are in nitely many integers m for which (m(G))−
c(m(G))¿12. We do not know whether there are in nitely many integers
m for which (m(G)) − c(m(G)) ¡1 2.
Question 2. What is c(n(Kn))?
We know that c(2(K2))= (2(K2))=4, but we do not know the value c(n(Kn))
for any other n.
Question 3. What determines whether c((G)) = ((G))?
We have many examples G for which c((G))= ((G)), and also many examples
G for which c((G)) ¡ ((G)). However, it seems dicult to characterize those
graphs G for which c((G)) = ((G)). For two integers k and d such that k ¿ 2d,
Gd
k is the graph with vertex set {0; 1; : : : ; k − 1} in which ij is an edge if and only if
d6|i − j|6k − d. Vince [17] showed that c(Gkd) = kd. It is easy to prove (see [3])
that a graph G is (k; d)-colorable if and only if there exists a homomorphism from G to Gd
k. Therefore, in the study of circular chromatic numbers, graphs Gkd play the role
of complete graphs in the study of chromatic numbers. Theorem 12 says that for n¿3, c((Kn)) = ((Kn)). An interesting question is:
Remark. Question 4 has now been answered in the armative in [13]. Acknowledgements
The authors thank the referees for many constructive suggestions on the revision of this paper.
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