CIRCULAR CHROMATIC NUMBER AND MYCIELSKI GRAPHS*

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Acta Mat hematica Scientia 2006,26B(2):314-320

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CIRCULAR CHROMATIC NUMBER AND MYCIELSKI GRAPHS* Liu Hongmei (

)

College of Science, Three Gorges University, Yzchung 443002, China

Abstract For a general graph G , M ( G ) denotes its Mycielski graph. This article gives a number of new sufficient conditions for G to have the circular chromatic number x c ( M ( G ) ) equals to the chromatic number x ( M ( G ) ) ,which have improved some best sufficient conditions published up to date.

Key words Circular chromatic number, Mycielski graphs, chromatic number 2000 MR Subject Classification

05C15

1 Introduction All graphs in this article are finite and simple. Suppose G = (V,E ) is a graph with vertex set V and edge set E . Vince [l]defined the ( k , d)-coloring as follows: Suppose k 2 2d are positive integers. An (k,d)-coloring of G is a mapping c : V + { 0 , 1 , . .. , k - 1) such that for any edge zy of G , d 5 lc(z) - c(y) I 5 k - d. Vince [l]introduced the star chromatic number of a graph G , which is defined t o be the infimum of the ratio k / d where G has a ( k , d)-coloring. It was shown [l]that the infimum can be replaced by minimum. Zhu [2] equivalently considered an ( k ,d)-coloring of G as a mapping c from V ( G ) to open arcs of unit length in a circle of length k / d such that for each edge zy E E(G), c(z) n c(y) = 0. Instead of star chromatic number, Zhu used circular chromatic number and denoted it by xc(G) [3], that is, xc(G)= inf{k/d : there is a ( k , d)-coloring of G}. This article follows Zhu’s definition of circular chromatic number. In [4] Fan proved that a ( k ,d)-coloring of a graph G is also equivalent to a partition ( X O X, I ,. - . ,X k - 1 ) of V ( G )such that for each j , 0 2 j 5 k - 1, X j U Xj+l U . . . U Xj+d-l is a n independent set in G, where the addition of indices is taken mod k . Fan [4] defined the partition as a (k,d)-partition of graph G. The circular chromatic numbers of different kinds of graphs were studied extensively in the past decades [2]-[14]. It is well-known [l]that x(G) - 1 < xc(G)2 x ( G ) for any graph G , and there are graphs G with xc(G) = x ( G ) as well as graphs G with x c ( G )arbitrarily close t o x(G) - 1. A natural question is that which graphs have x c ( G )= x(G)? This question has *Received May 25, 2005; revised March 30, 2005. Supported by National Science Foundation of China (10371048)and the Science Foundation of Three Gorges University.

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attracted considerable attention [3,4,9,10,11,12,13].Unfortunately it is NP-hard to determine whether a given graph G has xc(G) = x(G) [ll]. In spite of this difficulty, some sufficient conditions for graph G to have xc(G) = x(G) have been found [2,3,4,10,11,12,13]. Mycielski graphs have been considered probably having their circular chromatic number to be the same as their chromatic number under certain conditions, and therefore become useful objects for circular chromatic number study. Let G be a graph with V(G) = {xi : 1 5 i 5 n}. The Mycielski graph of G, denoted by M(G), is the graph obtained from G by adding n 1 new vertices xi,xi,.. . , u, and then, for 1 5 i 5 n, joining x!,to the neighbors of xi and to u. The vertex xi is called the twin of x2 (xi is also the twin of xi). The vertex u is called the root of M(G). We use the triple (V, V’,u) for the vertex set of M(G), where V is the vertex set of G, V’ is the set of twins of V, and u is the root. For t 2 2, define Mt(G) = M(Mt-l(G)). It is well known that for any nonempty graph G, x(M(G)) = x(G) 1. However, there are infinitely many graphs G for which xc(M(G)) # xc(G) 1. For instance, let G be a cycle But, by a result in [5], xc(M(G)) = 4. of length 2m 1. It is known [l]that xc(G) = 2 Research has focused on finding sufficient conditions of G under which xc(M(G)) = x(M(G)). then xc(M(G)) = x(M(G)). Let S be the set of vertices Fan [4] proved that if x(G) 2 of degree n - 1 in G, Fan [4]also showed that if JSJ2 3, then xc(M(G)) = x(M(G)); and if IS( 2 5, then xc(M2(G))= x(M2(G)). D D-F Liu [12] also proved that if G has a universal vertex x, G is not a star and xc(G - x) > x(G - z)- 1/2, then xc(M(G)) = x(M(G)) [14] showed that when n is big enough, then x c ( M t ( k n ) = ) x(Mt(kn)). This article pushes the progress in the following directions. For (C,d)-partitions of a general graph G, a new concept of bad pair is defined. Then by introducing some special kinds of (C,d)-partitions, Fan’s results in [4]are improved and it is proved that, if IS( 2 4,then xc(M2(G))= x(M2(G)). Some better sufficient conditions of xc(M(G)) = x(M(G)) are also obtained.

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Some Sufficient Conditions for x c ( M ( G ) )= x ( M ( G ) )

As a starting point of investigating the properties of (Ic, d)-partition of Mycielski graphs, Fan [4] proved some basic properties of (k,d)-partition, which can be summarized into the following two lemmas: Lemma 2.1 ([4]) Let (Xo,X I , . . . ,Xk-1)be an (k,d)-partition of G, where Ic and d are integers such that gcd(k, d ) = 1. If X t = 0 for;some t , then xc(G) < f . Lemma2.2 ([4]) Let Gbeagraph, V(G) ={x1,...,xn},andV(M(G)) = (x1,.-.,xn;xi, ...,xh;u}. If there is a (Ic,d)-partition of M(G), then there is a (k,d)-partition ( X o , X l , . . . , Xk-1) of M(G) such that: 1) U € X o 2) Define the set

then for any x E V(G)\A, {x,x’) c Xj for some j , d 5 j 5 Ic - d.

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Lemma 2.2 reveals an important class of (k,d)-partition of Mycielski graphs. To investigate this kind of partition, it is convenient to introduce the following definition [14]. Definition 2.3 Suppose ( X O , . . ,X k - 1 ) is an ( k ,d)-partition of a graph G . For each 2 E V ( G ) ,there is j , 0 5 j I k - 1, such that 2 E X j . The set

is called the d-field of z in the (k,d)-partition. X j is called the partition-class of 2 in the (Ic, d)-partition and denoted by C(z). Let 2 be a vertex of G , A neighbor of z is a vertices that is adjacent to z. Denote by N ( z ) the set of neighbors of z. If H is a subset of V ( G ) ,define N ( H ) = U z E ~ N ( z ) . From Definition 2.3, any vertex in 6(z) is not a neighbor of z in G , that is, b ( z ) n N ( x )= 8. Definition 2.4 Let M ( G ) be a Mycielski graph with vertex set (V,V’,u).Suppose x c ( M ( G ) )= where gcd(k,d) = 1. Let P = ( X o , X l , . . - , X k - l )be a (k,d)-partition of M ( G ) . A bad pair of P is two consecutive X i and Xi+l such that ( X i U X i + l ) n V = 8. The following two lemmas can be proved using Definition 2.4. Lemma 2.5 Let M ( G ) be a Mycielski graph with vertex set (V,V’,u). Suppose x c ( M ( G ) )= where gcd(k,d) = 1. Let P = ( X o , X l , . . . , X k - l )be an (k,d)-partition of M ( G ) . If d = 2, then P has at most one bad pair. Proof Suppose, on the contrary, that {X,, X p + l } and { X q ,X q + l } are two bad pairs of P, where p < q. Since d = 2, k can be written in the form k = 2t 1, where t 2 2 . If q = p 1, consider P\{Xp, X p + l ,Xp+2} whose elements can be paired into t - 1 independent sets: Bi = Xp+2,+l U Xp+2i+2,1 5 i 5 t - 1, where the addition of indices is taken mod k. Let Fi = Bi 3 V, 1 5 i 5 t - 1. Then { F l ,F 2 ; . . , F t - 1 } is a proper ( t - 1)-coloring of G . Therefore, x ( G ) 5 t - 1, and so x ( M ( G ) ) = x ( G ) 1 5 t = This means that x c ( M ( G ) )5 x ( M ( G ) )5 contradicts the fact that x c ( M ( G ) )= If q > p 1, let

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L1 = {Xp+2, X p + 3 , * . * , X q - l } , L2 = {Xq+2, .. . , Xk-1,Xo, X1, - * . , X p - l } . Since k is odd, 1L1)and )L2l have different parity. Suppose that, without loss of generality, lLll is even and lLzl is odd. Then, as seen above, the elements of L1 can be paired into independent sets. The first 1,521 - 1 elements of L2 can be paired into i(lL21 - 1) independent sets and the last elements form an additional independent set. In totality, we have 1 k-3 -(IL11+ lLZl+ 1) = 2 = t - 1

2

independent sets, which yield, as seen above, a proper ( t - 1)-coloring of G, and therefore x ( G ) 5 t - 1. As above, x c ( M ( G ) )5 a contradiction again. Lemma 2.6 Let M ( G ) be a Mycielski graph with vertex set (V,V’,u). Suppose that x c ( M ( G ) )= where gcd(k,d) = 1. Let ( X O , X ~ , . - . , Xbe ~ -an ~ )(k,d)-partition of M ( G ) with u E X O .If d = 2 and 1 x01= 1, then there is no odd i such that { X i , X i + l } is a bad pair. Proof Suppose, on the contrary, that { X p , X p + l }is a bad pairs with an odd number p . Since d = 2, we have that k is odd. Remove Xo, X , , Xp+l from the partition, and let

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L1 = {X1,Xz,...XP-1} and Lz = {Xp+Z,Xp+3,...,Xk-1). Since p is odd, both lLll and 1,521 are even. As in the proof of Lemma 2.5, we can pair the elements of each Li into independent sets. Totally we have + ( l L ~ l [ L Z = ~ )b(k - 3) independent sets, which give, since Xo = { u } , a proper coloring of G. Therefore x ( G ) 5 ;(k - 3), and so x c ( M ( G ) )5 a contradiction. Employing these two lemmas, the following sufficient condition for G having x c ( M ( G ) )= x ( M ( G ) )can be obtained. Theorem 2.7 Let G be a graph with n vertices. If G contains an K4 such that N(K4) # V ( G )and each vertex of K4 has degree at least n - 3 in G , then x c ( M ( G ) )= x ( M ( G ) ) . Proof Suppose, on the contrary, that x c ( M ( G ) )< x ( M ( G ) ) .Then x c ( M ( G ) )= where d 2 2 and gcd(k, d) = 1. Let (V,V’,u ) be the vertex set of M(G). With Lemma 2.2, we can assume that ( X O X, I , . . ,X k - 1 ) is an ( k ,d)-partition of M ( G ) with u E XO.Set

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Since u E XO and u is adjacent to each vertex of V’, we have that A n V’ = 0. For x E V u V’, denote by x‘ the twin of x ; for X C V U V’,X’ = {x‘ : x E X}. Let x E V\A, say 5 E X j (d 5 j 5 k - d). If x’ 4 X j , say x’ E Xi with 1 # j, we may move x’ to X j to obtain a new (k,d)-partition(Yo,Yl,...,Yk-l), whereY, =Xi i f i ${j,Z},q =Xju{x’},andYi =Xl\{x’}. Now, x and x’ lie in the same y3 in the new (k,d)-partition. In view of this fact, we are able to assume that ( X o ,X I , . . . ,X k - 1 ) has been chosen such that for any x E V\A,

{x,x‘} C Xj for some j , d 5 j 5 k - d.

(1)

Furthermore, among all such (k, d)-partitions, we assume that (XO,XI,. . . , Xk-1) has been chosen such that IAl is as small as possible, (2) where, J A Jis the number of elements in A. By Lemma 2.1, X i # 0,O 5 i 5 k - 1. Since each pair of the four vertices of K4 is adjacent, let { x q , x r , x s , x t }be the four vertices of K4 such that xi E Xi,i E { q , T , s , t } , and 0 5 q < T < s < t 5 k - 1. By the given condition N ( K 4 ) # V, there is

u

z E V\(

“Xi)),

iE{q,r,s,t)

and hence d(xi) 5 n - 2 in G for each i E { q , T , s,t } . Claim For any i E { q , r , s , t } ,if d 1 5 i 5 k - d - 1, then d(xi) = n - 3 in G , Xi= { x i , x i } ,z’ E Xi-l u Xi+l, and [Xi-, n V’I = IXi+l n V’(= 1. Proof of the claim: Since d 1 5 i 5 k - d - 1, we have (Xi-1 U Xi+l) n A = 0, and thus by (l),no vertex of Xi-1 can be the twin of a vertex of Xi+l. Therefore, if let

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then IRnVl 2 2. Note that xi E V, and in M ( G ) ,xi is not adjacent to any vertex of Xi-luXi+l and thus not to any vertex of R. Also, xi is not adjacent to any vertex of X i UX;. It follows that d(xi) 5 n - 3 in G with equality only if z E R, X i = Xi) = { x i ,x:}, and IXi-1 n V’(= (Xi+l n V’I = 1. By the given condition, equality d(zi) = n - 3 holds and therefore z E R, Xi = { x i ,xi},

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and IXi-1 n V’I = IXi+l n V’I = 1. By (l),z E R implies that z’ E Xi-1 U X i + l . This proves the claim. Since each pair of the four vertices is adjacent, we have that q + d < r , r + d l s , and s l t - d l k - d - 1 . By the claim, d(z,) = n - 3 in G, X , = {x,,zb}, z’ E X,-1 U X s + l , and IX,-1 n V’(= IXs+l n V’(= 1. Now, if T L d 1, then we also have that d(x,) = n - 3 in G, X , = {z,-,xk}, z’ E X,-1 U X,+l, and IX,-1 n V’I = IX,+, n V’I = 1. It follows that

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+ 2, X , = {x,,z;}, X , = {z,,xi}, x , + ~n v’= {z’}.

(3)

By (l),either z E Xr+l or z E A. If z E A , by (3) we may move z to X,+1 and obtain a contradiction to (2). Therefore, z E XT+l and so

Now consider different cases in which q takes different values: Case 1 q = 0. Since xq E X O ,z E X , + l , z # X I U Xk-1, and I(X1 U Xk-1) n VI 2 2, it follows that dc(xq) 5 n - 4, a contradiction. Case 2 q 2 1. Then, T 2 q d 2 d 1, and so equations (3) and (4) hold. We note that (3) implies that d = 2. If q 2 3 (= d l ) ,then by the claim, z’ E Xq-l U X q + l , which contradicts (4).Therefore, q I 2. If q = 2, then xq is not adjacent to any vertex in X I U X3 U X : U XA. Moreover, xq is not adjacent to z E X,+1. It follows that X3 = X i and 1x11 = 1. By moving the only vertex of X1 to X Z ,we obtain a new (k, d)-partition that is contrary to (2). Therefore, q = 1, that is, xq E X I . By similar argument, we have xt E X k - 1 . Now consider the twin x: of x q . Suppose that x: E X j . Since x: is adjacent to x,, x,,xt in M ( G ) , we have 1 < j < T - 1 or s + 1 < j < k - 2. As before, let J = Xj-1 U X j U X j + l , and we see that xqv # E ( G ) for any w E V n J. Note that z # J and d ( x q ) n - 3 in G. If j > 2, then IV n JI 2 1. But if V n J = 0, then { X j - l , X j } and { X j , X j + l } are two bad pairs, which contradicts Lemma 2.5. Thus, IV n JI = 1. Let w E V n J. Then, d(zq) = n - 3 which implies that X O = { u } . Furthermore, since and xqw E E ( G ) for every ‘u E V\{xq,w,z}, zq’u # E(G)for any w E V n (X2 u X i ) , we have that V n ( X ZU X i ) G {w,z } , which implies that V n (X2 U X i ) = {w}.Since w # A, by (l),w E X Z . But w E J, and so j = 3. Since IV n JJ= 1, we have (X3 U X 4 ) n V = 0. That is, { X 3 , X 4 } is a bad pair, contrary to Lemma 2.6. This shows that j = 2, that is, xb E X2. Similarly, xi E X k - 2 . Since xb E X Z , x q has a neighbor a E V such that a # X3 U X Z , a’ E X3 U X Z . Otherwise we can move xq to X Z and obtain an (k,d)-partition that contradicts (2). By (l),a E A and thus a E X k - 1 . Since xi E X k - 2 , we see that a # xt. Using the same argument for the proof of x: E Xk-2, we have some b E V such that b E X1 and b # x q . Since xq E X1 and z : E X Z , we have that xqu # E ( G ) for any w E V n ( X 2 u X3). If V n ( X ZU X 3 ) # 0, then, since x q z , z q b # E(G), we have d ( x q ) I n - 4, a contradiction. , s } is a bad pair. Similarly, ( X k - 2 , Xk-3) is a bad Therefore, V n (X2 U X 3 ) = 0, and so { X Z X pair as well. So there are two bad pairs, contrary to Lemma 2.5. This completes the proof of the theorem.

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Theorem 2.7 specifies a sufficient condition for G to have x c ( M ( G ) )= x ( M ( G ) ) It . is also a useful tool to obtain further results. As an application, the following important corollary can be derived. Corollary 2.8 Let G be a graph with n vertices and K the set of vertices of degree n - 1. If IKI 2 4, then x c ( M 2 ( G ) = ) x(M2(G)). Proof Let H = M ( G ) . Then H is a graph with 2 n 1 vertices, and any four vertices of K induce an K4 in H , such that each vertex of K4 has degree 2 ( n - 1) = IV(H)I - 3 in H . Since the root of M ( G ) ( = H ) is not in N ( K 4 ) , applying Theorem 2.7 to H , we have that x c ( M ( H ) )= x ( M ( H ) ) ,which is equivalent to x c ( M 2 ( G ) = ) x(M2(G)). It is easy to see that Corollary 2.8 generalizes Fan’s sufficient condition in [4] described above. An immediate consequence is the following result: Corollary 2.9 If n 2 4, then x c ( M 2 ( K n )= ) x ( M 2 ( K n )= ) n 2. This is exactly the result of [5] stated in Introduction. The sufficient condition in Theorem 2.7 requires G containing K4. This can be relaxed to K3, but with a trade-off of more rigorous requirement on degrees of the vertices of K3 : Theorem 2.10 Let G be a graph with n vertices, if G contains an K3, N ( K 3 ) # V ( G ) , and each vertex of K3 has degree n - 2 in G , then x , ( M ( G ) ) = x ( M ( G ) ) . Proof Suppose, on the contrary, that x c ( M ( G ) )< x ( M ( G ) ) .Then x c ( M ( G ) )= fiere d 2 2 , and gcd(k,d) = 1. Let V ( M ( G ) )= { 2 1 , 2 2 , . . . , 2 n ; x ~ , x ~ , . ~ ~ where ,x~;u}7 V ( G )= { x 1 , 2 2 , .. ,x,} is the set of vertices of G , V ’ ( G )= {xi,xh,... , x;} is the set of twins of V(G), and u is the root of M ( G ) .For any z E V U V’, denote by x‘ the twin of z. According to Lemma 2.2, M ( G ) has an ( k ,d)-partition ( X o ,X I , .. Xk-1) such that u E X O ,and if denote

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6(u) = Xk-d+l

u Xk-df2 u ’ ‘ ’ u Xk-1 u XO u x1 u ’ ‘. u Xd-1,

then for any z E V\6(u), {z,x’}c X j , for some j, d 5 j 5 k - d. Since each pair of three vertices of K3 are adjacent, assume {z,,x,,xt} are the three vertices of K3 with xi E X i , i E {r,s, t } , and 0 5 T < s < t 5 k - 1. With the given condition, there is 2 E V\ N(zi),

U

iE{T,S,t)

and hence d(zi) 5 n - 2 in G, for each i E {r,s,t}. We have that d(xi) = n - 2 , i E Since each pair of the three vertices is adjacent, we have that

{T,

s,t } .

d 5 T + d 5 S, s 5 t - d 5 k - d - 1. Because for any j such that d I j 5 k - d, there exists {x,z’} c X j , so x,, xi E X , . Note that d(xi) = n - 2 , for i E {r,s, t } in G , X,-l U X , U Xs+l is an independent set. This implies that, for any v E V\{z}, v,v’ E N ( x i ) , i E { T , s , ~ } and , v,v’ $ X,-1 UX-,UX,+1 (except x,,x/Q E X , ) . With X j # 0 for any 0 I j 5 k - 1, we have { z , z ’ } = X s - l U X s + l . Because d 1 I s 1 5 k - d, if z E X s + l , we obtain that z , z‘ E X s + l , X,-l = 0, which is impossible. So z E z‘ E X s + l . Also x,,x’, E X , , and d = 2, as otherwise Xs+z = 0. Since 0 5 r I s - d, 2,. E X O , z E X I . Also u E XO and z i is adjacent to every vertex of V ’ ( G ) . This means that Xk-1 nW’(G) = 0. On the other hand, d ( x T )= n - 2 in G , z E X I , hence Xk-1 n V ( G ) = 0. Therefore x k - 1 = 0. From Lemma 2.1, it is impossible. This prove the theorem.

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Acknowledgments The author is indebted to Professor Genghua Fan for his invaluable comments. References 1 Vince A. Star chromatic number. J Graph Theory, 1988, 12: 551-559 2 Zhu Xuding. Circular chromatic number. Discrete Mathematics, 2001, 229: 371-410 3 Zhu Xuding. Star chromatic number and products of graphs. J Graph Theory, 1992, 18: 557-569 4 Fan Genghua. Circular chromatic number and mycielski graphs. Combinatorica, 2004,24(27): 127-135 5 Chang Grrard J. Huang Lingling, Zhu Xuding. Circular chromatic numbers of Mycielski’sgraphs. Discrete Mathematics, 1999, 205: 23-37 6 Hossein Hajiabolhassan, Zhu Xuding. Mycielski’s graphs with circular chromatic number equal chromatic number. J Graph Theory, 2003, 44: 106-115 7 Bondy J A, Hell P. A note on the star chromatic number. J Graph Theory, 1990, 14: 479-482 8 Abbott H L, Zhou B. The star chromatic number. J Graph Theory, 1993,17: 349-360 9 Steffen E,Zhu X. On the star chromatic numbers of graphs. Combinatorica, 1996, 16: 439-448 10 Zhu X.Graphs whose circular chromatic number equal the chromatic number. Combinatorica, 1999,19: 139-149 11 Guichard D R. Acyclic graph coloring and the star chromatic number. J Graph Theory, 1993, 17: 63-71 12 Liu Daphne Derfen. Circular chromatic number for iterated mycielski graphs. Manuscript. 13 Liu Hongmei, Nie Xiaodong. Some results about the circular chromatic number of Mm(Kn). J of Mathematical Study, 2004,4(37): 407-416 14 Liu Hongmei, Gao Shichen, Feng Dehong. Circular chromatic number of mycielski graphs. J of Systems Science and Information, 2005,3(3): 547-552