Convex-round Graphs Are Circular Perfect - Semantic Scholar
Convex-round Graphs Are Circular Perfect J rgen Bang-Jensen and Jing Huangy October 6, 2000
Abstract For 1 ? , dk denotes the graph with vertices 0 1 ? 1, in which is adjacent to if and only if j ? j ? . A graph is circular perfect if, for every induced subgraph of , the in mum for which admits a homomorphism to d is equal to the supremum for which dk admits a homomorphism to . In this k paper, we show that all convex-round graphs are circular perfect. To do so, we show that a convex-round graph is a core if and only if it is isomorphic to dk for some with ( ) = 1. Our proof relies on several structural properties of convex-round graphs. We also completely characterize maximal r -free convex-round graphs. Our characterization shows that a convex-round graph is a maximal r -free graph if and only if it is obtained from trt?1 for some 1 by substituting an independent set for each vertex of trt?1 . d
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1 Introduction Let G; G0 be graphs. A homomorphism from G to G0 is a mapping f : V (G) ! V (G0 ) which preserves adjacencies, that is, f (u)f (v) 2 E (G0 ) whenever uv 2 E (G). If there is a homomorphism from G to G0, we write G ! G0 and say that G is homomorphic to G0 If G ! G0 and G0 ! G, then G and G0 are homomorphically equivalent. For 1 d k ? d, let Gdk be the graph with vertices 0; 1; : : : ; k ? 1,0 in which i is adjacent to j if and only if d ji ? j j k ? d. It is easy to see that Gdk ! Gdk when d0 d and that Gdk and G0 dlkl are homomorphically equivalent. In general, Bondy and Hell [2] proved that Gdk ! Gdk0 if and only if k=d k0 =d0. When gcd(d; k) = 1, Gdk is called a circular clique. The circular chromatic number c(G) of a graph G (originally called the star-chromatic number, [7]) is the in mum of those k=d for which G ! Gdk . It follows from a result of Bondy Department of Mathematics and Computer Science, University of Southern Denmark, Odense, DK5230, Denmark (email: [email protected]) Research supported by NSERC and the Danish National Science Research Council y Department Math and Stats, University of Victoria, P.O. Box 3045, Victoria, B.C., V8W 3P4, Canada (email: [email protected]) Research supported by NSERC
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and Hell [2] that c(G) = minfk=d : G ! Gdk and k jV (G)jg. Hence, c(G) is always a rational number. The circular clique number !c(G) of a graph G is the supremum of those k=d for which Gdk ! G. It was proved by Zhu [9] that !c(G) is a rational number and in fact is equal to the maximum of those k=d such that G contains Gdk as an induced subgraph. 0 By the remark above, if Gdk ! Gdk0 then k=d k0=d0. This implies that c(G) !c(G). The concept of the circular chromatic number (resp. circular clique number), generalizes the `classical' chromatic number (resp. clique number). When d = 1, the circular chromatic number (resp. circular clique number) coincides with the chromatic number (G) (resp. clique number !(G)) of G. It was proved [7, 9] that, for every graph G, (G) ? 1 < c (G) (G) and !(G) !c(G) < !(G) + 1. A graph is perfect if (H ) = !(H ) for every induced subgraph H of G. A graph G is circular perfect if c(H ) = !c(H ) for every induced subgraph H of G. Circular perfect graphs were recently introduced by Zhu [9]. It follows from the above inequalities that every perfect graph is circular perfect. There are many circular perfect graphs which are not perfect. For instance, for the odd cycle C2r+1 , c(C2r+1) = (2r + 1)=r = !c(C2r+1 ) (since C2r+1 = Gr2r+1) and further, every proper induced subgraph of C2r+1 is bipartite and hence perfect. More generally, Zhu [9] showed that every Gdk is circular perfect. In order to understand more about those graphs which are circular perfect but not perfect, it is of interest to nd large classes of graphs which are circular perfect but not in general perfect. The purpose of this paper is to present a family of circular perfect graphs, which forms a superclass of fGdk : 1 d k ? dg. A graph G is called convex-round if the vertices of G can be circularly ordered L = v1 ; v2; : : : ; vn such that the neighbourhood N (vi ) of each vertex vi forms an `interval', i.e., is of the form fvl ; vl +1; : : : ; vr g where additions are modulo n. We shall refer to the circular ordering L as a convex-round enumeration of G. By an interval [vi; vj ] (w.r.t. a convexround enumeration L), we mean the set of vertices vi ; vi+1; : : : ; vj where additions are modulo n. Similarly, we de ne intervals (vi; vj ) = [vi; vj ] ? fvi; vj g, [vi ; vj ) = [vi ; vj ] ? fvj g, and (vi; vj ] = [vi; vj ] ? fvi g. In [1], the authors together with Yeo showed that convex-round graphs have many nice structural and algorithmic properties. In particular, it was shown that convex-round graphs can be recognized and a convex-round enumeration can be found in time O(m + n) (here and in the remainder of the paper, m and n denote the numbers of the edges and vertices of the graph). Furthermore, several optimization problems, e.g., the chromatic number, the maximum clique, the maximum matching, can be solved very eciently for this class of graphs. It is easy to see that every induced graph of Gdk is convex-round. We shall show that there are convex-round graphs which are not induced subgraphs of any Gdk . Furthermore, we shall show that convex-round graphs are all circular perfect. Thus this implies Zhu's result that every Gdk is circular perfect. The proof of our theorem uses several structural properties of convex-round graphs developed in [1]. i
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2 Cores of convex-round graphs A subgraph H of G is called a core of G if there is a homomorphism G ! H but no homomorphism G ! H 0 for any proper subgraph H 0 of H . A graph which is its own core is called a core. The core of a graph is unique up to isomorphism, and it is NP-complete to decide if a given graph is a core [4]. Here we show that the cores of convex-round graphs are precisely the circular cliques and, furthermore, the core of a convex-round graph can be found in O(n2).
Lemma 2.1 [1] Let G be a connected convex-round graph with a convex-round enumeration L = v1 ; v2 ; : : : ; vn. If vi vi+k 62 E (G) for some i, then either G is bipartite, or at least one of [vi ; vi+k ], [vi+k ; vi ] is independent. Moreover it can be decided in time O(n + m) which of the above properties holds for G. Corollary 2.2 [1] Let G = (V; E ) be convex-round with convex-round enumeration
L = v1 ; v2; : : : ; vn. Suppose there are indices i; j; k; l such that the vertices vi ; vj ; vk ; vl; vi occur in that order in L and vivj ; vk vl 2 E but vi vk 62 E . Then G is bipartite. The following lemma is an easy consequence of Corollary 2.2.
Lemma 2.3 Let G = (V; E ) be a convex-round graph with convex-round enumeration
L = v1; v2; : : : ; vn. Suppose there are indices i and i + 1 such that the vertex vr +1 belongs to the interval (vi+1 ; vr ). Then G is bipartite. i
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Lemma 2.4 [1] If a convex-round graph is not bipartite, then it is connected.
Lemma 2.5 Let G = (V; E ) be a convex-round graph with convex-round enumeration
L = v1 ; v2; : : : ; vn. Then one of the following holds: (i) G is bipartite.
(ii) G is isomorphic to Gdk for some integers k; d such that 1 d k ? d.
(iii) There exist vertices vi ; vi+1 such that either N (vi ) N (vi+1 ) or N (vi+1 ) N (vi ).
Proof: We may assume that G is not bipartite since otherwise (i) holds. By Lemmas 2.4 and 2.3, G is connected and vr +1 2 [vr ; vi] for i = 1; 2; : : : ; n. If vl 2 [vl +1 ; vr ], then N (vi ) N (vi+1) and (iii) holds. So we may assume that vl +1 2 (vl ; vr +1 ]. Now we either have N (vi+1) N (vi ) or vr +1 2 (vr ; vi]. In the rst case (iii) holds and we are done. So assume that vr +1 2 (vr ; vi]. We shall show that vr +1 = vr +1. i
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Suppose that vr +1 6= vr +1, i.e., vr +1 2 (vr +1; vi]. To simplify notation, we denote vk = vr +1?1 and thus vk+1 = vr +1 . Consider the vertex vk . Clearly, vk is not adjacent to vi . We claim that vk is adjacent to vi+1 . By Lemma 2.1 and the fact that [vi ; vk ] is not independent, we conclude that [vk ; vi] is independent. Since G is connected, vk has at least one neighbour. Now we see that vk must be adjacent to vi+1 as otherwise none of the intervals [vk ; vi+1] and [vi+1; vk ] are independent, contradicting Lemma 2.1. By our observations above N (vk ) is contained in the interval [vi+1 ; vk ). This implies that either N (vk ) N (vk+1 ) or N (vk+1 ) N (vk ) and hence (iii) holds. Using the above argument, we can show that either (iii) holds or we have vl +1 = vl +1. Therefore, either (iii) holds or vl +1 = vl +1 and vr +1 = vr +1 for each i = 1; 2; : : : ; n. When vl +1 = vl +1 and vr +1 = vr +1 for each i = 1; 2; : : : ; n, G is isomorphic to Gdn where d = l1 ? 1. i
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Lemma 2.6 The graph Gdk is a core if and only if gcd(k; d) = 1. Proof: Since Gidik is homomorphic to Gdk , Gdk is not a core if gcd(k; d) >0 1. Suppose
that gcd(k; d) = 1 and Gdk is not a core. By Lemma 2.5, Gdk contains some Gdk0 as a proper0 induced subgraph. This implies that k=d k0=d0. On the other hand, since Gdk ! Gdk0 , k=d k0 =d0. Hence k=d = k0=d0, contradicting the fact that gcd(k; d) = 1 and k0 < k.
Theorem 2.7 A convex-round graph is a core if and only if G is isomorphic to Gdk for
some integers k; d with gcd(k; d) = 1. Furthermore, given any convex-round graph we can nd its core in time O(n2 ).
Proof: By Lemma 2.4 and the fact that a bipartite graph is a core if and only if it is just an edge, we see that a bipartite G is a core if and only G is isomorphic to G12 and any edge is the core of G. Furthermore, we can decide whether G is bipartite in time O(n + m). Hence we may assume that G is not bipartite. It follows from Lemma 2.5 that if G is a core then G is isomorphic to Gdk for some for some integers k; d. Now the rst claim follows from Lemma 2.6. The complexity claim follows from the fact that we can nd the next pair of vertices to identify in linear time.
3 Convex-round graphs are circular perfect Theorem 3.1 Every convex-round graph G is circular perfect. Furthermore, a circular
clique Gdk with k=d = ! c(G) and a homomorphism from G to Gdk can both be found in time O(n2).
Proof: To verify the rst claim, it suces to show that every convex-round graph G has
c(G) = !c(G) as every induced subgraph of a convex-round graph is again convex-round 4
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4 Figure 1: The net N and the unique convex-round enumeration of N (one can use the same enumeration). Since G is homomorphic to H if and only if the core of G is homomorphic to H and the core of every convex-round graph is convex-round (it is an induced subgraph) it suces to consider those convex-round graphs which are cores. By Theorem 2.7 these are exactly the circular cliques Gdk (gcd(k; d) = 1) and for these we have c(Gdk ) = !c(Gdk ). Now if G is an arbitrary convex-round graph and H is its core we have c(G) = c(H ) = !c(H ) !c(G) c(G) implying that c(G) = !c(G). The second claim follows from the observation above and Theorem 2.7.
Corollary 3.2 [9] The graphs Gdk are circular perfect.
It is worth noting that Theorem 3.1 does not follow directly from Corollary 3.2. As we indicate below, there are in nitely many convex-round graphs which are not induced subgraphs of any Gdk . The net N is the graph shown in the left part of Figure 1. It is easy to show that the convex-round enumeration shown in the right part is unique up to permuting the labels cyclically.
Proposition 3.3 The net N is not an induced subgraph of any regular convex-round
graph and hence it is not an induced subgraph of any Gdk .
Proof: Suppose there is some k such that the net is an induced subgraph of a k-regular
convex-round graph G. It is easy to see that in every convex-round enumeration of G the vertices of the net must appear in the order 1; 2; 3; 4; 5; 6 where we use the numbering from Figure 1. Let L be a convex-round enumeration of G and ai denote the number of vertices of G that occur between the vertex i and i + 1 of N in the ordering L. Using the fact that N is an induced subgraph of G and that L is a convex-round enumeration of G, we see that the following inequalities must hold (considering the neighbours in G of the vertices numbered 5,6,1,2,3,4 in N in this order): 5
a1 + a2 + 3 a2 + a3 + 1 a3 + a4 + 3 a4 + a5 + 1 a5 + a6 + 3 a6 + a1 + 1
k k k k k k:
From this we easily derive a contradiction and it follows that G cannot exist.
The example above can be generalized to an in nite family of convex-round graphs, none of which is an induced subgraph of any Gdk . For instance, the graph Np obtained from the complete graph Kp (p 3) with vertices u1; u2; : : : ; up by adding p new vertices w1; w2; : : : ; wp so that each wi adjacent to all vertices in Np except ui, ui+1, wi?1, and wi+1. Observe that u1; w1; u2; w2; : : : ; up; wp is a convex-round enumeration of Np. A similar proof as for N shows that Np is not an induced subgraph of any Gdk . Also observe that Np is not an induced subgraph of Nq if p 6= q.
4 Convex-round graphs containing no Kr In this section we always assume that r 3. It is well-known that there are triangle-free graphs of an arbitrarily high chromatic number. For a perfect graph, the chromatic number is equal to the maximum size of a clique contained in the graph. Similarly, for a circular perfect graph, the circular chromatic number is equal to the maximum k=d for which Gdk is contained in the graph. Thus, if a circular perfect graph does not contain Kr (i.e., Kr -free), then its circular chromatic number is less than r.
Proposition 4.1 If G is a Kr -free circular perfect graph, then G ! Gtrt?1 for some
t 1.
Proof: Suppose that c(G) = !c(G) = k=d. Since G is Kr -free, k=d < r. Let t 1 be an integer with r ? 1=t k=d. Since k=d r ? 1=t = (rt ? 1)=t, Gdk ! Gtrt?1. Since c(G) = k=d, G ! Gdk . Hence G ! Gtrt?1 . A Kr -free graph is maximal if adding any edge between two non-adjacent vertices creates a subgraph Kr . Observe that a graph with n < r vertices is always Kr -free and, it is maximal if and only if it is isomorphic to Kn. It is also easy to see that every Kr -free graph is a spanning subgraph of a maximal Kr -free graph.
Theorem 4.2 Every Kr -free convex-round graph is a spanning subgraph of a maximal
Kr -free convex-round graph.
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Proof: Let G be a Kr -free convex-round graph. Suppose that G is bipartite. Partition V (G) into r ? 1 independent sets and form an complete (r ? 1)-partite graph with these sets as partite sets. Clearly the resulting graph is a maximal Kr -free convex-round graph which contains G as a spanning subgraph. Hence we may assume below that G is not bipartite. It suces to show that if G is not maximal, then we can add a new edge to G so that the resulting graph is convex-round and still Kr -free. So assume that G is not maximal. Let L = v1 ; v2; : : : ; vn be a convex-round enumeration of G. Since G is not maximal, there exists vivj 2= E (G) such that G + vivj is Kr -free. Since G is not bipartite, by Lemma 2.1 either N (vi ) [ N (vj ) (vi; vj ) or N (vi ) [ N (vj ) (vj ; vi). We may assume that N (vi ) [ N (vj ) (vj ; vi). (A similar discussion applies if N (vi ) [ N (vj ) (vj ; vi).) Let v ; v 2 [vj ; vi] satisfy the following properties:
v v 2= E (G), [v ; v ] [vj ; vi], N (v ) [ N (v ) (v ; v ), G + v v is Kr -free, and j[v ; v ]j is as small as possible.
We claim that G + v v is convex-round. Let N (v ) = [va ; vb]. Since G + v v is Kr -free, G + v va?1 is Kr -free. This is because any vertex adjacent to both v and va?1 must be adjacent to v . Hence the choice of v ; v implies that v = va?1 . A similar argument shows that if N (v ) = [vc; vd], then v = vd+1 . Therefore, G + v v is convex-round. We say a graph G satis es property Ir if, for every independent set I , there is a complete subgraph H with r ? 2 vertices such that I N (x) for all x 2 V (H ). The following lemma is due to Stephan Brandt [3].
Lemma 4.3 Every maximal triangle-free convex-round graph satis es I3 .
The following theorem is a generalization of the above lemma.
Theorem 4.4 Every maximal Kr -free convex-round graph with at least r ? 1 vertices satis es Ir . Proof: Let G be a maximal Kr -free convex-round graph. Clearly if G has exactly r ? 1 vertices, then G is isomorphic to Kr?1 and hence G satis es Ir . So assume that G has at least r vertices. Since G is Kr -free and has at least r vertices, there are two non-adjacent vertices u; v contained in G. Since G is maximal, G + uv contains a complete subgraph G0 with r vertices. As G0 must contain both u and v, H = G0 ? fu; vg is a complete subgraph of G on r ? 2 vertices with the property that fu; vg NG(x) for every x 2 V (H ). 7
Furthermore, G0 ? v is a complete subgraph of G with r ? 1 vertices. These observations imply in particular that every vertex of G is contained in a complete subgraph with r ? 1 vertices. Let I be an independent set of G. We show by induction on jI j = k that there is a complete subgraph H with r ? 2 vertices such that each vertex of I is adjacent to all vertices of H . The cases when k = 1 or 2 follow from our argument above. So assume k > 2 and that the statement is true for every independent set with fewer than k vertices. By Lemma 4.3, we further assume that r 4. Observe that G can not be bipartite. Let L = v1 ; v2; : : : ; vn be a convex-round enumeration of G and let I = fvi1 ; vi2 ; : : : ; vi g where i1 < i2 < : : : < ik . By the inductive hypothesis, there is a complete subgraph H 0 with r ? 2 vertices such that I ? fvi g N (x) for all x 2 V (H 0). If vi is adjacent to all vertices of H 0, then we are done. So assume that vi is not adjacent to at least one vertex in V (H 0). Since G is not bipartite, V (H 0) (vi ; vi +1 ) for some j = 1; 2; : : : ; k (additions are modulo k). Since vi is not adjacent to at least one vertex in V (H 0), V (H 0) (vi ?1 ; vi ) or V (H 0) (vi ; vi1 ). Without loss of generality, assume that V (H 0) (vi ?1 ; vi ). Similarly, there is a complete subgraph H 00 with r ? 2 vertices such that I ? fvi1 g N (x) for all x 2 V (H 00). Again if vi1 is adjacent to all vertices of H 00 then we are done; otherwise a similar argument as for H 0 shows that either V (H 00) (vi ; vi1 ) or V (H 00) (vi1 ; vi2 ). Since vi2 is adjacent to at least one vertex from each of H 0 and H 00 and G is convex-round, vi2 is adjacent to vi , a contradiction to the assumption that vi2 and vi are not adjacent. k
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Let G; S be graphs and v a vertex of G. To substitute S for v in G means to form a new graph G0 from G by replacing v with S so that in G0 every vertex of S is adjacent to every vertex of G that is adjacent to v.
Theorem 4.5 Let G be a convex-round graph with at least r ? 1 vertices. The following
statements are equivalent.
1. G is a maximal Kr -free graph; 2. G is Kr -free and satis es Ir ; 3. G is obtained from Gtrt?1 for some t 1 by substituting an independent set for each vertex of Gtrt?1 .
Proof: The implication (1) ) (2) was treated in Theorem 4.4. The implication (3) ) (1) can also be easily veri ed. So we only need to prove (2) ) (3). If suces to show that if G is reduced, that is, no two vertices have the same neighbourhood, then G is isomorphic to Gtrt?1 for some t 1. Thus we assume that G is reduced. If G has exactly r ? 1 vertices, then G must be complete as G satis es Ir . Hence G is isomorphic to Gtrt?1 with t = 1. So we may assume that G contains at least r vertices. This implies that G can not be bipartite. By Lemma 2.5, either G is isomorphic to Gdk for some integers k; d, or G contains two vertices x; y with N (x) N (y). Suppose that the latter case occurs. Since G is reduced, N (x) 6= N (y). Hence N (y) ? N (x) 6= ; and there is a vertex z adjacent to y but not to x. Since G satis es Ir , G has a complete subgraph H with r ? 2 vertices such 8
that each vertex of H is adjacent to both x and z. Now we see that V (H ) [ fy; zg induces a complete subgraph isomorphic to Kr , contradicting the assumption that G is Kr -free. Therefore G must be isomorphic to Gdk for some integers k; d. Since G is Kr -free, k rd ? 1 and, since G satis es Ir , k rd ? 1. Therefore k = rd ? 1 and G is isomorphic to Gdrd?1. Note that when G is not convex-round, the statements in Theorem 4.5 may not be equivalent. For instance, the Petersen graph is a maximal triangle-free graph which fails I3 and which is not isomorphic to Gt3t?1 for any t 1. However, Pach [6] showed that a triangle-free graph G satis es I3 if and only if it is obtained from Gt3t?1 for some t 1 by substituting an independent set for each vertex of Gt3t?1 . Combining Theorems 4.2 and 4.5, we have the following:
Corollary 4.6 Every Kr -free convex-round graph is a spanning subgraph of the graph obtained from Gtrt?1 for some t 1 by substituting an independent set for each vertex of Gtrt?1 .
5 Questions and remarks We say that a graph G is interval enumerable if there exists a linear enumeration I = v1 ; v2; : : : ; vn of V (G) such that, for each vertex vi, there exists numbers 1 `i ; hi i ? 1 such that N (vi ) \ [v1 ; vi?1] = [v` ; vh ], where `i > hi is used to indicate that vi has no neighbour vi with an index less than i. That is, we require that for each vi the neighbours with indices less than i form an interval (note that there is no restriction on the neighbours of vi with index higher than i). We shall refer to the enumeration I as an interval enumeration of G. It is easy to see that every convex-round enumeration of a graph G is also an interval enumeration of G. However the latter class is much larger and in fact contains graphs which are not circular perfect. One such example is the graph H in Figure 2 due to Zhu [9], for which we have c(H ) = 8=3 and !c(H ) = 5=2. i
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Problem 5.1 Characterize circular perfect interval enumerable graphs. It was shown in [1] that one can nd a maximum clique in any interval enumerable graph in linear time. By Theorem 2.7 we can nd a maximum circular clique as well as the circular chromatic number of any convex-round graph in polynomial time. Hence it may be of interest to design polynomial algorithms for nding a maximum circular clique and for determining the circular chromatic number of a given interval enumerable graph. As we saw above the graph Np (p 3) is convex-round but not an induced subgraph of any Gdk . In view of Theorem 4.5, the following problem might be interesting.
Problem 5.2 Characterize those convex-round graphs that are induced subgraphs of a
circular clique.
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Figure 2: An interval enumerable graph H with c(H ) > !c(H ). The labelling shown is an interval enumeration of H . The complement of a convex-round graph is called concave-round, [1]. Hence Problem 5.2 is equivalent to characterizing concave-round graphs which are induced subgraphs of a power of a cycle. Since no Np is an induced subgraph of any Gdk , it follows that its complement (which is concave-round) is not an induced subgraph of any power of a cycle. Concave-round graphs may have (G) > !(G) + 1 (which implies that not all concaveround graphs are circular perfect). In fact for C3i?1 i?1 , the (i ? 1)st power of a (3i ? 1)-cycle, i ?1 i ?1 i?1 3 i ?1 3 i ?1 3 we have (C3i?1 ) = d 2 e and !(C3i?1) = d 3 e implying that (C3i?1 i?1 ) = d 2 ! (C3i?1 )e 3 and we always have (G) d 2 !(G)e for every concave-round graph G, [1].
Problem 5.3 Characterize circular perfect concave-round graphs. It is well-known that the complement of a perfect graph is perfect, [5]. We have seen that all convex-round graphs are circular perfect but not all concave-round graphs are circular perfect. Thus the following problem should be interesting.
Problem 5.4 For which graphs G are both G and its complement circular perfect. Note that all odd cycles belong to this class since both C2r+1 and and its complement are convex-round graphs for every integer r 1.
Acknowledgement We wish to thank Stephan Brandt for stimulating discussions on triangle-free convex-round graphs.
References [1] J. Bang-Jensen, J. Huang, and A. Yeo, Convex-round graphs and concave-round graphs, SIAM J. Discrete Math. 13 (2000) 179 - 193. 10
[2] J.A. Bondy and P. Hell, A note on the star chromatic number, J. Graph Theory 14 (1990) 479 - 482. [3] S. Brandt, Structure of dense triangle-free graphs, Combinatorics, Probability and Computing 8 (1999) 237-245. [4] P. Hell and J. Nesetril, The core of a graph, Discrete Math. 109 (1992) 117 - 126. [5] L. Lovasz, Normal hypergraphs and the perfect graph conjecture, Discrete Math. 2 (1972) 253 - 267. [6] J. Pach, Graphs whose every independent set has a common neighbour, Discrete Mathematics 37 (1981) 217-228. [7] A. Vince, Star chromatic number, J. Graph Theory 12 (1988) 551 - 559. [8] X. Zhu, The circular chromatic number, a survey, Combinatorics, Graph Theory, Algorithms and Applications, M. Fiedler, J. Kratochvil, J. Nesetril, eds., North Holland, Amsterdam (2000), to appear. [9] X. Zhu, Circular perfect graphs (2000), manuscript.
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Abstract For 1 ? , dk denotes the graph with vertices 0 1 ? 1, in which is adjacent to if and only if j ? j ? . A graph is circular perfect if, for every induced subgraph of , the in mum for which admits a homomorphism to d is equal to the supremum for which dk admits a homomorphism to . In this k paper, we show that all convex-round graphs are circular perfect. To do so, we show that a convex-round graph is a core if and only if it is isomorphic to dk for some with ( ) = 1. Our proof relies on several structural properties of convex-round graphs. We also completely characterize maximal r -free convex-round graphs. Our characterization shows that a convex-round graph is a maximal r -free graph if and only if it is obtained from trt?1 for some 1 by substituting an independent set for each vertex of trt?1 . d
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1 Introduction Let G; G0 be graphs. A homomorphism from G to G0 is a mapping f : V (G) ! V (G0 ) which preserves adjacencies, that is, f (u)f (v) 2 E (G0 ) whenever uv 2 E (G). If there is a homomorphism from G to G0, we write G ! G0 and say that G is homomorphic to G0 If G ! G0 and G0 ! G, then G and G0 are homomorphically equivalent. For 1 d k ? d, let Gdk be the graph with vertices 0; 1; : : : ; k ? 1,0 in which i is adjacent to j if and only if d ji ? j j k ? d. It is easy to see that Gdk ! Gdk when d0 d and that Gdk and G0 dlkl are homomorphically equivalent. In general, Bondy and Hell [2] proved that Gdk ! Gdk0 if and only if k=d k0 =d0. When gcd(d; k) = 1, Gdk is called a circular clique. The circular chromatic number c(G) of a graph G (originally called the star-chromatic number, [7]) is the in mum of those k=d for which G ! Gdk . It follows from a result of Bondy Department of Mathematics and Computer Science, University of Southern Denmark, Odense, DK5230, Denmark (email: [email protected]) Research supported by NSERC and the Danish National Science Research Council y Department Math and Stats, University of Victoria, P.O. Box 3045, Victoria, B.C., V8W 3P4, Canada (email: [email protected]) Research supported by NSERC
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and Hell [2] that c(G) = minfk=d : G ! Gdk and k jV (G)jg. Hence, c(G) is always a rational number. The circular clique number !c(G) of a graph G is the supremum of those k=d for which Gdk ! G. It was proved by Zhu [9] that !c(G) is a rational number and in fact is equal to the maximum of those k=d such that G contains Gdk as an induced subgraph. 0 By the remark above, if Gdk ! Gdk0 then k=d k0=d0. This implies that c(G) !c(G). The concept of the circular chromatic number (resp. circular clique number), generalizes the `classical' chromatic number (resp. clique number). When d = 1, the circular chromatic number (resp. circular clique number) coincides with the chromatic number (G) (resp. clique number !(G)) of G. It was proved [7, 9] that, for every graph G, (G) ? 1 < c (G) (G) and !(G) !c(G) < !(G) + 1. A graph is perfect if (H ) = !(H ) for every induced subgraph H of G. A graph G is circular perfect if c(H ) = !c(H ) for every induced subgraph H of G. Circular perfect graphs were recently introduced by Zhu [9]. It follows from the above inequalities that every perfect graph is circular perfect. There are many circular perfect graphs which are not perfect. For instance, for the odd cycle C2r+1 , c(C2r+1) = (2r + 1)=r = !c(C2r+1 ) (since C2r+1 = Gr2r+1) and further, every proper induced subgraph of C2r+1 is bipartite and hence perfect. More generally, Zhu [9] showed that every Gdk is circular perfect. In order to understand more about those graphs which are circular perfect but not perfect, it is of interest to nd large classes of graphs which are circular perfect but not in general perfect. The purpose of this paper is to present a family of circular perfect graphs, which forms a superclass of fGdk : 1 d k ? dg. A graph G is called convex-round if the vertices of G can be circularly ordered L = v1 ; v2; : : : ; vn such that the neighbourhood N (vi ) of each vertex vi forms an `interval', i.e., is of the form fvl ; vl +1; : : : ; vr g where additions are modulo n. We shall refer to the circular ordering L as a convex-round enumeration of G. By an interval [vi; vj ] (w.r.t. a convexround enumeration L), we mean the set of vertices vi ; vi+1; : : : ; vj where additions are modulo n. Similarly, we de ne intervals (vi; vj ) = [vi; vj ] ? fvi; vj g, [vi ; vj ) = [vi ; vj ] ? fvj g, and (vi; vj ] = [vi; vj ] ? fvi g. In [1], the authors together with Yeo showed that convex-round graphs have many nice structural and algorithmic properties. In particular, it was shown that convex-round graphs can be recognized and a convex-round enumeration can be found in time O(m + n) (here and in the remainder of the paper, m and n denote the numbers of the edges and vertices of the graph). Furthermore, several optimization problems, e.g., the chromatic number, the maximum clique, the maximum matching, can be solved very eciently for this class of graphs. It is easy to see that every induced graph of Gdk is convex-round. We shall show that there are convex-round graphs which are not induced subgraphs of any Gdk . Furthermore, we shall show that convex-round graphs are all circular perfect. Thus this implies Zhu's result that every Gdk is circular perfect. The proof of our theorem uses several structural properties of convex-round graphs developed in [1]. i
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2 Cores of convex-round graphs A subgraph H of G is called a core of G if there is a homomorphism G ! H but no homomorphism G ! H 0 for any proper subgraph H 0 of H . A graph which is its own core is called a core. The core of a graph is unique up to isomorphism, and it is NP-complete to decide if a given graph is a core [4]. Here we show that the cores of convex-round graphs are precisely the circular cliques and, furthermore, the core of a convex-round graph can be found in O(n2).
Lemma 2.1 [1] Let G be a connected convex-round graph with a convex-round enumeration L = v1 ; v2 ; : : : ; vn. If vi vi+k 62 E (G) for some i, then either G is bipartite, or at least one of [vi ; vi+k ], [vi+k ; vi ] is independent. Moreover it can be decided in time O(n + m) which of the above properties holds for G. Corollary 2.2 [1] Let G = (V; E ) be convex-round with convex-round enumeration
L = v1 ; v2; : : : ; vn. Suppose there are indices i; j; k; l such that the vertices vi ; vj ; vk ; vl; vi occur in that order in L and vivj ; vk vl 2 E but vi vk 62 E . Then G is bipartite. The following lemma is an easy consequence of Corollary 2.2.
Lemma 2.3 Let G = (V; E ) be a convex-round graph with convex-round enumeration
L = v1; v2; : : : ; vn. Suppose there are indices i and i + 1 such that the vertex vr +1 belongs to the interval (vi+1 ; vr ). Then G is bipartite. i
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Lemma 2.4 [1] If a convex-round graph is not bipartite, then it is connected.
Lemma 2.5 Let G = (V; E ) be a convex-round graph with convex-round enumeration
L = v1 ; v2; : : : ; vn. Then one of the following holds: (i) G is bipartite.
(ii) G is isomorphic to Gdk for some integers k; d such that 1 d k ? d.
(iii) There exist vertices vi ; vi+1 such that either N (vi ) N (vi+1 ) or N (vi+1 ) N (vi ).
Proof: We may assume that G is not bipartite since otherwise (i) holds. By Lemmas 2.4 and 2.3, G is connected and vr +1 2 [vr ; vi] for i = 1; 2; : : : ; n. If vl 2 [vl +1 ; vr ], then N (vi ) N (vi+1) and (iii) holds. So we may assume that vl +1 2 (vl ; vr +1 ]. Now we either have N (vi+1) N (vi ) or vr +1 2 (vr ; vi]. In the rst case (iii) holds and we are done. So assume that vr +1 2 (vr ; vi]. We shall show that vr +1 = vr +1. i
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Suppose that vr +1 6= vr +1, i.e., vr +1 2 (vr +1; vi]. To simplify notation, we denote vk = vr +1?1 and thus vk+1 = vr +1 . Consider the vertex vk . Clearly, vk is not adjacent to vi . We claim that vk is adjacent to vi+1 . By Lemma 2.1 and the fact that [vi ; vk ] is not independent, we conclude that [vk ; vi] is independent. Since G is connected, vk has at least one neighbour. Now we see that vk must be adjacent to vi+1 as otherwise none of the intervals [vk ; vi+1] and [vi+1; vk ] are independent, contradicting Lemma 2.1. By our observations above N (vk ) is contained in the interval [vi+1 ; vk ). This implies that either N (vk ) N (vk+1 ) or N (vk+1 ) N (vk ) and hence (iii) holds. Using the above argument, we can show that either (iii) holds or we have vl +1 = vl +1. Therefore, either (iii) holds or vl +1 = vl +1 and vr +1 = vr +1 for each i = 1; 2; : : : ; n. When vl +1 = vl +1 and vr +1 = vr +1 for each i = 1; 2; : : : ; n, G is isomorphic to Gdn where d = l1 ? 1. i
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Lemma 2.6 The graph Gdk is a core if and only if gcd(k; d) = 1. Proof: Since Gidik is homomorphic to Gdk , Gdk is not a core if gcd(k; d) >0 1. Suppose
that gcd(k; d) = 1 and Gdk is not a core. By Lemma 2.5, Gdk contains some Gdk0 as a proper0 induced subgraph. This implies that k=d k0=d0. On the other hand, since Gdk ! Gdk0 , k=d k0 =d0. Hence k=d = k0=d0, contradicting the fact that gcd(k; d) = 1 and k0 < k.
Theorem 2.7 A convex-round graph is a core if and only if G is isomorphic to Gdk for
some integers k; d with gcd(k; d) = 1. Furthermore, given any convex-round graph we can nd its core in time O(n2 ).
Proof: By Lemma 2.4 and the fact that a bipartite graph is a core if and only if it is just an edge, we see that a bipartite G is a core if and only G is isomorphic to G12 and any edge is the core of G. Furthermore, we can decide whether G is bipartite in time O(n + m). Hence we may assume that G is not bipartite. It follows from Lemma 2.5 that if G is a core then G is isomorphic to Gdk for some for some integers k; d. Now the rst claim follows from Lemma 2.6. The complexity claim follows from the fact that we can nd the next pair of vertices to identify in linear time.
3 Convex-round graphs are circular perfect Theorem 3.1 Every convex-round graph G is circular perfect. Furthermore, a circular
clique Gdk with k=d = ! c(G) and a homomorphism from G to Gdk can both be found in time O(n2).
Proof: To verify the rst claim, it suces to show that every convex-round graph G has
c(G) = !c(G) as every induced subgraph of a convex-round graph is again convex-round 4
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4 Figure 1: The net N and the unique convex-round enumeration of N (one can use the same enumeration). Since G is homomorphic to H if and only if the core of G is homomorphic to H and the core of every convex-round graph is convex-round (it is an induced subgraph) it suces to consider those convex-round graphs which are cores. By Theorem 2.7 these are exactly the circular cliques Gdk (gcd(k; d) = 1) and for these we have c(Gdk ) = !c(Gdk ). Now if G is an arbitrary convex-round graph and H is its core we have c(G) = c(H ) = !c(H ) !c(G) c(G) implying that c(G) = !c(G). The second claim follows from the observation above and Theorem 2.7.
Corollary 3.2 [9] The graphs Gdk are circular perfect.
It is worth noting that Theorem 3.1 does not follow directly from Corollary 3.2. As we indicate below, there are in nitely many convex-round graphs which are not induced subgraphs of any Gdk . The net N is the graph shown in the left part of Figure 1. It is easy to show that the convex-round enumeration shown in the right part is unique up to permuting the labels cyclically.
Proposition 3.3 The net N is not an induced subgraph of any regular convex-round
graph and hence it is not an induced subgraph of any Gdk .
Proof: Suppose there is some k such that the net is an induced subgraph of a k-regular
convex-round graph G. It is easy to see that in every convex-round enumeration of G the vertices of the net must appear in the order 1; 2; 3; 4; 5; 6 where we use the numbering from Figure 1. Let L be a convex-round enumeration of G and ai denote the number of vertices of G that occur between the vertex i and i + 1 of N in the ordering L. Using the fact that N is an induced subgraph of G and that L is a convex-round enumeration of G, we see that the following inequalities must hold (considering the neighbours in G of the vertices numbered 5,6,1,2,3,4 in N in this order): 5
a1 + a2 + 3 a2 + a3 + 1 a3 + a4 + 3 a4 + a5 + 1 a5 + a6 + 3 a6 + a1 + 1
k k k k k k:
From this we easily derive a contradiction and it follows that G cannot exist.
The example above can be generalized to an in nite family of convex-round graphs, none of which is an induced subgraph of any Gdk . For instance, the graph Np obtained from the complete graph Kp (p 3) with vertices u1; u2; : : : ; up by adding p new vertices w1; w2; : : : ; wp so that each wi adjacent to all vertices in Np except ui, ui+1, wi?1, and wi+1. Observe that u1; w1; u2; w2; : : : ; up; wp is a convex-round enumeration of Np. A similar proof as for N shows that Np is not an induced subgraph of any Gdk . Also observe that Np is not an induced subgraph of Nq if p 6= q.
4 Convex-round graphs containing no Kr In this section we always assume that r 3. It is well-known that there are triangle-free graphs of an arbitrarily high chromatic number. For a perfect graph, the chromatic number is equal to the maximum size of a clique contained in the graph. Similarly, for a circular perfect graph, the circular chromatic number is equal to the maximum k=d for which Gdk is contained in the graph. Thus, if a circular perfect graph does not contain Kr (i.e., Kr -free), then its circular chromatic number is less than r.
Proposition 4.1 If G is a Kr -free circular perfect graph, then G ! Gtrt?1 for some
t 1.
Proof: Suppose that c(G) = !c(G) = k=d. Since G is Kr -free, k=d < r. Let t 1 be an integer with r ? 1=t k=d. Since k=d r ? 1=t = (rt ? 1)=t, Gdk ! Gtrt?1. Since c(G) = k=d, G ! Gdk . Hence G ! Gtrt?1 . A Kr -free graph is maximal if adding any edge between two non-adjacent vertices creates a subgraph Kr . Observe that a graph with n < r vertices is always Kr -free and, it is maximal if and only if it is isomorphic to Kn. It is also easy to see that every Kr -free graph is a spanning subgraph of a maximal Kr -free graph.
Theorem 4.2 Every Kr -free convex-round graph is a spanning subgraph of a maximal
Kr -free convex-round graph.
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Proof: Let G be a Kr -free convex-round graph. Suppose that G is bipartite. Partition V (G) into r ? 1 independent sets and form an complete (r ? 1)-partite graph with these sets as partite sets. Clearly the resulting graph is a maximal Kr -free convex-round graph which contains G as a spanning subgraph. Hence we may assume below that G is not bipartite. It suces to show that if G is not maximal, then we can add a new edge to G so that the resulting graph is convex-round and still Kr -free. So assume that G is not maximal. Let L = v1 ; v2; : : : ; vn be a convex-round enumeration of G. Since G is not maximal, there exists vivj 2= E (G) such that G + vivj is Kr -free. Since G is not bipartite, by Lemma 2.1 either N (vi ) [ N (vj ) (vi; vj ) or N (vi ) [ N (vj ) (vj ; vi). We may assume that N (vi ) [ N (vj ) (vj ; vi). (A similar discussion applies if N (vi ) [ N (vj ) (vj ; vi).) Let v ; v 2 [vj ; vi] satisfy the following properties:
v v 2= E (G), [v ; v ] [vj ; vi], N (v ) [ N (v ) (v ; v ), G + v v is Kr -free, and j[v ; v ]j is as small as possible.
We claim that G + v v is convex-round. Let N (v ) = [va ; vb]. Since G + v v is Kr -free, G + v va?1 is Kr -free. This is because any vertex adjacent to both v and va?1 must be adjacent to v . Hence the choice of v ; v implies that v = va?1 . A similar argument shows that if N (v ) = [vc; vd], then v = vd+1 . Therefore, G + v v is convex-round. We say a graph G satis es property Ir if, for every independent set I , there is a complete subgraph H with r ? 2 vertices such that I N (x) for all x 2 V (H ). The following lemma is due to Stephan Brandt [3].
Lemma 4.3 Every maximal triangle-free convex-round graph satis es I3 .
The following theorem is a generalization of the above lemma.
Theorem 4.4 Every maximal Kr -free convex-round graph with at least r ? 1 vertices satis es Ir . Proof: Let G be a maximal Kr -free convex-round graph. Clearly if G has exactly r ? 1 vertices, then G is isomorphic to Kr?1 and hence G satis es Ir . So assume that G has at least r vertices. Since G is Kr -free and has at least r vertices, there are two non-adjacent vertices u; v contained in G. Since G is maximal, G + uv contains a complete subgraph G0 with r vertices. As G0 must contain both u and v, H = G0 ? fu; vg is a complete subgraph of G on r ? 2 vertices with the property that fu; vg NG(x) for every x 2 V (H ). 7
Furthermore, G0 ? v is a complete subgraph of G with r ? 1 vertices. These observations imply in particular that every vertex of G is contained in a complete subgraph with r ? 1 vertices. Let I be an independent set of G. We show by induction on jI j = k that there is a complete subgraph H with r ? 2 vertices such that each vertex of I is adjacent to all vertices of H . The cases when k = 1 or 2 follow from our argument above. So assume k > 2 and that the statement is true for every independent set with fewer than k vertices. By Lemma 4.3, we further assume that r 4. Observe that G can not be bipartite. Let L = v1 ; v2; : : : ; vn be a convex-round enumeration of G and let I = fvi1 ; vi2 ; : : : ; vi g where i1 < i2 < : : : < ik . By the inductive hypothesis, there is a complete subgraph H 0 with r ? 2 vertices such that I ? fvi g N (x) for all x 2 V (H 0). If vi is adjacent to all vertices of H 0, then we are done. So assume that vi is not adjacent to at least one vertex in V (H 0). Since G is not bipartite, V (H 0) (vi ; vi +1 ) for some j = 1; 2; : : : ; k (additions are modulo k). Since vi is not adjacent to at least one vertex in V (H 0), V (H 0) (vi ?1 ; vi ) or V (H 0) (vi ; vi1 ). Without loss of generality, assume that V (H 0) (vi ?1 ; vi ). Similarly, there is a complete subgraph H 00 with r ? 2 vertices such that I ? fvi1 g N (x) for all x 2 V (H 00). Again if vi1 is adjacent to all vertices of H 00 then we are done; otherwise a similar argument as for H 0 shows that either V (H 00) (vi ; vi1 ) or V (H 00) (vi1 ; vi2 ). Since vi2 is adjacent to at least one vertex from each of H 0 and H 00 and G is convex-round, vi2 is adjacent to vi , a contradiction to the assumption that vi2 and vi are not adjacent. k
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Let G; S be graphs and v a vertex of G. To substitute S for v in G means to form a new graph G0 from G by replacing v with S so that in G0 every vertex of S is adjacent to every vertex of G that is adjacent to v.
Theorem 4.5 Let G be a convex-round graph with at least r ? 1 vertices. The following
statements are equivalent.
1. G is a maximal Kr -free graph; 2. G is Kr -free and satis es Ir ; 3. G is obtained from Gtrt?1 for some t 1 by substituting an independent set for each vertex of Gtrt?1 .
Proof: The implication (1) ) (2) was treated in Theorem 4.4. The implication (3) ) (1) can also be easily veri ed. So we only need to prove (2) ) (3). If suces to show that if G is reduced, that is, no two vertices have the same neighbourhood, then G is isomorphic to Gtrt?1 for some t 1. Thus we assume that G is reduced. If G has exactly r ? 1 vertices, then G must be complete as G satis es Ir . Hence G is isomorphic to Gtrt?1 with t = 1. So we may assume that G contains at least r vertices. This implies that G can not be bipartite. By Lemma 2.5, either G is isomorphic to Gdk for some integers k; d, or G contains two vertices x; y with N (x) N (y). Suppose that the latter case occurs. Since G is reduced, N (x) 6= N (y). Hence N (y) ? N (x) 6= ; and there is a vertex z adjacent to y but not to x. Since G satis es Ir , G has a complete subgraph H with r ? 2 vertices such 8
that each vertex of H is adjacent to both x and z. Now we see that V (H ) [ fy; zg induces a complete subgraph isomorphic to Kr , contradicting the assumption that G is Kr -free. Therefore G must be isomorphic to Gdk for some integers k; d. Since G is Kr -free, k rd ? 1 and, since G satis es Ir , k rd ? 1. Therefore k = rd ? 1 and G is isomorphic to Gdrd?1. Note that when G is not convex-round, the statements in Theorem 4.5 may not be equivalent. For instance, the Petersen graph is a maximal triangle-free graph which fails I3 and which is not isomorphic to Gt3t?1 for any t 1. However, Pach [6] showed that a triangle-free graph G satis es I3 if and only if it is obtained from Gt3t?1 for some t 1 by substituting an independent set for each vertex of Gt3t?1 . Combining Theorems 4.2 and 4.5, we have the following:
Corollary 4.6 Every Kr -free convex-round graph is a spanning subgraph of the graph obtained from Gtrt?1 for some t 1 by substituting an independent set for each vertex of Gtrt?1 .
5 Questions and remarks We say that a graph G is interval enumerable if there exists a linear enumeration I = v1 ; v2; : : : ; vn of V (G) such that, for each vertex vi, there exists numbers 1 `i ; hi i ? 1 such that N (vi ) \ [v1 ; vi?1] = [v` ; vh ], where `i > hi is used to indicate that vi has no neighbour vi with an index less than i. That is, we require that for each vi the neighbours with indices less than i form an interval (note that there is no restriction on the neighbours of vi with index higher than i). We shall refer to the enumeration I as an interval enumeration of G. It is easy to see that every convex-round enumeration of a graph G is also an interval enumeration of G. However the latter class is much larger and in fact contains graphs which are not circular perfect. One such example is the graph H in Figure 2 due to Zhu [9], for which we have c(H ) = 8=3 and !c(H ) = 5=2. i
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Problem 5.1 Characterize circular perfect interval enumerable graphs. It was shown in [1] that one can nd a maximum clique in any interval enumerable graph in linear time. By Theorem 2.7 we can nd a maximum circular clique as well as the circular chromatic number of any convex-round graph in polynomial time. Hence it may be of interest to design polynomial algorithms for nding a maximum circular clique and for determining the circular chromatic number of a given interval enumerable graph. As we saw above the graph Np (p 3) is convex-round but not an induced subgraph of any Gdk . In view of Theorem 4.5, the following problem might be interesting.
Problem 5.2 Characterize those convex-round graphs that are induced subgraphs of a
circular clique.
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Figure 2: An interval enumerable graph H with c(H ) > !c(H ). The labelling shown is an interval enumeration of H . The complement of a convex-round graph is called concave-round, [1]. Hence Problem 5.2 is equivalent to characterizing concave-round graphs which are induced subgraphs of a power of a cycle. Since no Np is an induced subgraph of any Gdk , it follows that its complement (which is concave-round) is not an induced subgraph of any power of a cycle. Concave-round graphs may have (G) > !(G) + 1 (which implies that not all concaveround graphs are circular perfect). In fact for C3i?1 i?1 , the (i ? 1)st power of a (3i ? 1)-cycle, i ?1 i ?1 i?1 3 i ?1 3 i ?1 3 we have (C3i?1 ) = d 2 e and !(C3i?1) = d 3 e implying that (C3i?1 i?1 ) = d 2 ! (C3i?1 )e 3 and we always have (G) d 2 !(G)e for every concave-round graph G, [1].
Problem 5.3 Characterize circular perfect concave-round graphs. It is well-known that the complement of a perfect graph is perfect, [5]. We have seen that all convex-round graphs are circular perfect but not all concave-round graphs are circular perfect. Thus the following problem should be interesting.
Problem 5.4 For which graphs G are both G and its complement circular perfect. Note that all odd cycles belong to this class since both C2r+1 and and its complement are convex-round graphs for every integer r 1.
Acknowledgement We wish to thank Stephan Brandt for stimulating discussions on triangle-free convex-round graphs.
References [1] J. Bang-Jensen, J. Huang, and A. Yeo, Convex-round graphs and concave-round graphs, SIAM J. Discrete Math. 13 (2000) 179 - 193. 10
[2] J.A. Bondy and P. Hell, A note on the star chromatic number, J. Graph Theory 14 (1990) 479 - 482. [3] S. Brandt, Structure of dense triangle-free graphs, Combinatorics, Probability and Computing 8 (1999) 237-245. [4] P. Hell and J. Nesetril, The core of a graph, Discrete Math. 109 (1992) 117 - 126. [5] L. Lovasz, Normal hypergraphs and the perfect graph conjecture, Discrete Math. 2 (1972) 253 - 267. [6] J. Pach, Graphs whose every independent set has a common neighbour, Discrete Mathematics 37 (1981) 217-228. [7] A. Vince, Star chromatic number, J. Graph Theory 12 (1988) 551 - 559. [8] X. Zhu, The circular chromatic number, a survey, Combinatorics, Graph Theory, Algorithms and Applications, M. Fiedler, J. Kratochvil, J. Nesetril, eds., North Holland, Amsterdam (2000), to appear. [9] X. Zhu, Circular perfect graphs (2000), manuscript.
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