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Non-cover generalized Mycielski, Kneser, and Schrijver graphs
Discrete Mathematics 308 (2008) 4653 – 4659 www.elsevier.com/locate/disc
Non-cover generalized Mycielski, Kneser, and Schrijver graphs Ko-Wei Liha , Chen-Ying Linb , Li-Da Tongc,1 a Institute of Mathematics, Academia Sinica, Nankang, Taipei 115, Taiwan b Department of Computer Science and Information Engineering, Shu-Te University, Kaohsiung 824, Taiwan c Department of Applied Mathematics, National Sun Yat-sen University, Kaohsiung 804, Taiwan
Received 30 December 2005; received in revised form 19 August 2007; accepted 21 August 2007 Available online 29 September 2007
Abstract A graph is said to be a cover graph if it is the underlying graph of the Hasse diagram of a finite partially ordered set. We prove that the generalized Mycielski graphs Mm (C2t+1 ) of an odd cycle, Kneser graphs KG(n, k), and Schrijver graphs SG(n, k) are not cover graphs when m 0, t 1, k 1, and n 2k + 2. These results have consequences in circular chromatic number. © 2007 Elsevier B.V. All rights reserved. Keywords: Cover graph; Mycielski graph; Kneser graph; Schrijver graph; Circular chromatic number
1. Introduction For a simple graph G, let |G| and G denote its number of vertices and number of edges, respectively. An edge, or a directed arc, joining the vertices u and v is denoted by uv, or u → v. The Hasse diagram of a finite partially ordered set is a pictorial representation of the covering relation of elements. The reader is referred to Rival [13] for general information on diagrams. A graph is said to be a cover graph if it is the underlying graph of a Hasse diagram. The recognition problem of cover graphs is known to be NP-complete [1,10]. Let the graph G be endowed with an acyclic orientation D, i.e., there are no directed cycles with respect to this orientation. Edelman (as quoted in West [18]) defined an arc of D to be dependent if its reversal creates a directed cycle, or equivalently a closed directed walk, in D. Let dmax (G) (dmin (G)) be the maximum (minimum) number of dependent arcs over all acyclic orientations of G. Edelman showed that dmax (G) = G − |G| + 1 if G is connected. This is generalized by Fisher et al. [2] to dmax (G) = G − |G| + c if G has c components. Pretzel [11] proved that having an orientation with no dependent arcs, i.e., dmin (G) = 0, is equivalent to being a cover graph. A sufficient condition for dmin (G) = 0 is (G) < g(G), where (G) and g(G) are, respectively, the chromatic number and the girth of G (See [2]). Let D be an acyclic orientation of the graph G. A vertex is said to be a sink (source) of D if its out-degree (in-degree) is zero. The operation of reversing all arcs going into a sink is called a push-down by Pretzel [12]. This operation of turning a sink into a source preserves the acyclicity of an orientation. It is also true that a push-down transforms an orientation without dependent arcs into another such orientation. The following was established in [8]. E-mail addresses: [email protected] (K.-W. Lih), [email protected] (C.-Y. Lin), [email protected] (L.-D. Tong). 1 Supported in part by the National Science Council under Grant NSC92-2115-M-110-004.
0012-365X/$ - see front matter © 2007 Elsevier B.V. All rights reserved. doi:10.1016/j.disc.2007.08.082
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Mosesian’s result. Let D1 be an acyclic orientation of the connected graph G. Then for any vertex v of G there exists a sequence of push-downs from D1 to an orientation D2 so that v is the unique sink in D2 . In the subsequent sections, we will establish the results stated in the Abstract and apply them to determine some circular chromatic numbers. 2. Generalized Mycielski graphs Let G = (V0 , E0 ) be a graph with vertex set V0 = {0, 0, 0, 1, . . . , 0, n − 1} and edge set E0 . The generalized Mycielski graph Mm (G) of G has vertex set V =V0 ∪( m i=1 Vi )∪{u}, where Vi ={i, j |0 j n−1} for 1 i m, and E ) ∪ {m, j u|0 j n − 1}, where Ei = {i − 1, j i, k|0, j 0, k ∈ E0 } for 1 i m. In edge set E = E0 ∪ ( m i=1 i the sequel, the second coordinates of vertices in Mm (G) are always taken modulo |G|. We note that M0 (G) is obtained from G by adding a new vertex adjacent to all vertices of G, and M1 (G) is commonly known as the Mycielskian M(G) of G [9]. In Tardif [17], a generalized Mycielski graph is defined as the categorical product of a graph and a path with a loop at one end, and then identifying all the vertices whose second coordinate is the other end of that path. It has been shown ([2,12]) that the Mycielskian M(C5 ) of a 5-cycle, also known as the Grötzsch graph, is not a cover graph. We can prove the following generalization. Theorem 1. Let integers m 0 and k 1. Then the generalized Mycielski graph Mm (C2k+1 ) of an odd cycle is not a cover graph. Proof. Since an acyclic orientation of a 3-cycle must have a dependent arc, Mm (C3 ) is clearly not a cover graph. For k 2, let G = Mm (C2k+1 ) and C2k+1 = (V0 , E0 ) with V0 = {0, 0, 0, 1, . . . , 0, 2k} and E0 = {0, 00, 1, 0, 10, 2, . . . , 0, 2k − 10, 2k, 0, 2k0, 0}. Suppose on the contrary that G is a cover graph. By Mosesian’s result, we may have an acyclic orientation D of G such that D has no dependent arcs and u is its unique sink. Since the subgraph induced by V0 in D is an acyclic (2k + 1)-cycle, without loss of generality, we may assume that four particular consecutive vertices are oriented as follows: 0, 0 → 0, 1 → 0, 2 ← 0, 3. Claim. For 1t m − 1, the arcs t − 1, t + 1 → t, t + 2, t − 1, t + 3 → t, t + 2, t + 1, t + 1 → t, t + 2, and t, t + 2 → t + 1, t + 3 are in D. We prove this Claim by induction on t with the assistance of the following. Observation. Let an acyclic orientation of a 4-cycle xyzwx have no dependent arcs. It follows that if the path xyz is oriented as x → y → z, then the path xwz must be oriented as x → w → z. When t = 1, this observation implies that the path 0, 0 → 1, 1 → 0, 2 is in D. Since u is the unique sink in D, the arc 0, 2 → 1, 3 is also in D. Then the existence of the path 1, 1 → 0, 2 → 1, 3 implies that of the path 1, 1 → 2, 2 → 1, 3. Similarly, the existence of the path 0, 3 → 0, 2 → 1, 3 implies that of the path 0, 3 → 0, 4 → 1, 3. It follows from the uniqueness of the sink u that the arc 1, 3 → 2, 4 is in D. Therefore the induction basis holds. Suppose that the induction hypothesis is true for 1t = k m − 2, i.e., the four arcs k − 1, k + 1 → k, k + 2, k − 1, k + 3 → k, k + 2, k + 1, k + 1 → k, k + 2, and k, k + 2 → k + 1, k + 3 are in D. The existence of the path k + 1, k + 1 → k, k + 2 → k + 1, k + 3 in D implies that of the path k + 1, k + 1 → k + 2, k + 2 → k + 1, k + 3. The existence of the path k − 1, k + 3 → k, k + 2 → k + 1, k + 3 implies that of the path k − 1, k + 3 → k, k + 4 → k + 1, k + 3. It follows again from the uniqueness of the sink u that the arc k + 1, k + 3 → k + 2, k + 4 is in D. Hence the case for t = k + 1 is true and our Claim is established. It follows from the Claim that the path m, m → m − 1, m + 1 → m, m + 2 is in D. Our observation implies that we should have the path m, m → u → m, m + 2. This contradicts to the fact that u is a sink. We conclude that G = Mm (C2k+1 ) is not a cover graph. Theorem 2. Let m > 0 and n 3. Then Mm (Cn ) is a cover graph if and only if n is even.
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Proof. We have already known from Theorem 1 that Mm (Cn ) is not a cover graph when n is odd. Now let n be an even integer. Since (Mm (Cn )) = 3 < 4 = g(Mm (Cn )), Mm (Cn ) is a cover graph. Remark 1. For n = 2k, an acyclic orientation D of Mm (Cn ) without dependent arcs can be explicitly defined as follows. Let Cn = (V0 , E0 ) with V0 = {0, 0, 0, 1, . . . , 0, 2k − 1} and E0 = {0, 00, 1, 0, 10, 2, . . . , 0, 2k − 20, 2k − 1, 0, 2k − 10, 0}. The edges of Mm (Cn ) are oriented as follows. The cycle Cn is alternately oriented by letting 0, 2i → 0, 2i + 1 and 0, 2i + 2 → 0, 2i + 1 for 0 i k − 1. Orient all other edges from a vertex with a smaller first coordinate to a vertex with a larger first coordinate or to the top vertex u. 3. Tower graphs A homomorphism f from a graph G1 (V1 , E1 ) to a graph G2 (V2 , E2 ) is a mapping from V1 to V2 such that uv ∈ E1 implies f (u)f (v) ∈ E2 . Lemma 3. Suppose that there is a homomorphism f from a graph G1 to a graph G2 . If G2 is a cover graph, so is G1 . Proof. Since G2 is a cover graph, there exists an acyclic orientation D2 of G2 having no dependent arcs. Define an orientation D1 of G1 as follow. If u and v are adjacent in G1 and f (u) → f (v) in D2 , then let u → v in D1 . The existence of a directed cycle C in D1 would imply that of a directed closed walk f (C) in D2 , contradicting the acyclicity of D2 . Therefore D1 is acyclic. Suppose on the contrary that G1 is not a cover graph. Then there exists a dependent arc in D1 . Let the reversal of the arc v1 → vt of D1 create a directed cycle C : v1 → v2 → · · · → vt → v1 for some t 3. Consider the directed path P : v1 → v2 → · · · → vt in D1 . Since D2 is acyclic, f (P ) must be a directed path in D2 . This forces f (v1 ) → f (vt ) to be a dependent arc in D2 , contradicting the assumption on D2 . We conclude that G1 is a cover graph. Now we are going to define the tower graph TCn of a cycle Cn . For integers n 3 and k = n/2 , let the graph G = Mk−1 (Cn ) have the vertex set V (G) = ( k−1 i=0 Vi ) ∪ {u} with Vi = {i, j |0 j n − 1} for 0 i k − 1. The tower graph TCn is the subgraph of G induced by the set W = {i, j |0 i k − 1, i j n − 1 − i} ∪ {u} together with new added edges i, i + 1i + 1, n − i − 2 and i, n − i − 2i + 1, i + 1 for 0 i k − 2. We may write the vertex u as k, k. Theorem 4. For k 1, the tower graph TC2k+1 is not a cover graph. Proof. Let G = Mk−1 (C2k+1 ) and H = TC2k+1 . Define a function f from V (G) to V (H ) as follows: Let f (u) = u and i, j if 0 i k − 1 and i j 2k − i, f (i, j ) = i, i if (j < i and i + j is even) or (j > 2k − i and i + j is odd), i, 2k − i if (j < i and i + j is odd) or (j > 2k − i and i + j is even). We are going to verify that f is a homomorphism from G to H. Let xy be an edge of G. If x, y ∈ V (H ), then f (x)f (y) = xy ∈ E(H ). If x = i, j satisfies j < i or j > 2k − i and y ∈ Vi−1 , then y = i − 1, j − 1 or i − 1, j + 1. Case 1: j < i. If i + j is even, then f (x) = i, i and f (y) = i − 1, i − 1. If i + j is odd, then f (x) = i, 2k − i and f (y) = i − 1, 2k − i + 1 or i − 1, i. We always have f (x)f (y) ∈ E(H ). Case 2: j > 2k − i. If i + j is even, then f (x) = i, 2k − i and f (y) = i − 1, 2k − i + 1. If i + j is odd, then f (x) = i, i and f (y) = i − 1, i − 1 or i − 1, 2k − i. Again we always have f (x)f (y) ∈ E(H ). Consequently, f is a homomorphism form G to H. Hence by Lemma 3 and Theorem 1, H is not a cover graph. Remark 2. It is easy to color TC2k+1 with four colors. The fact that Mm (C2k+1 ) is 4-chromatic was first proved in Stiebitz [16]. Tardif [17] gave a different proof. Therefore, the existence of a homomorphism from Mk−1 (C2k+1 ) to TC2k+1 shows that the latter is also 4-chromatic.
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Theorem 5. Suppose n3 is an integer. Then TCn is a cover graph if and only if n is even. Proof. By Theorem 4, T C n is not a cover graph when n is odd. When n 4 is even, it is again easy to see that (TCn ) = 3. Hence the inequality (TCn ) = 3 < 4 = g(TCn ) implies that TCn is a cover graph. Remark 3. An acyclic orientation of TCn having no dependent arcs can be explicitly defined by a construction similar to the one used in Remark 1. 4. Kneser graphs and Schrijver graphs Let [n] denote the set {0, 1, . . . , n − 1}. For n > 2k. The Kneser graph KG(n, k) has the set of all k-subsets of [n] as its vertex set, and two vertices A and B are adjacent if and only if A ∩ B = ∅. One of the well-known facts about Kneser graphs is (KG(n, k)) = n − 2k + 2 proved in Lovász [7]. We note that (KG(2k + 1, k)) = 3 < g(KG(2k + 1, k)) when k > 1, hence KG(2k + 1, k) is a cover graph. From now on, we focus on the case n = 2k + 2 and k 1 unless otherwise stated. We define the following sets for 0 j 2k. {0, 2, . . . , j − 2, j + 1, . . . , 2k − 3, 2k − 1} if j is even, X0,j = {1, 3, . . . , j − 2, j + 1, . . . , 2k − 2, 2k} if j is odd. Then we define the following sets for 1 i k. {1, 3, . . . , i − 1, 2k − (i − 1), . . . , 2k − 3, 2k − 1} if i is even, Zi = {0, 2, . . . , i − 1, 2k − (i − 1), . . . , 2k − 2, 2k} if i is odd. We also let Z−1 = Z0 = ∅ for convenience. When i > 0, we can partition Zi into two equal-sized parts ZiL and ZiR such that ZiL = Zi ∩ [i]. Since i k, we observe that i − 1 < 2k − i + 1, and hence |Zi | = |ZiL | + |ZiR | = 2 i/2. For 1 i k and i j 2k − i, we observe that L Zi if the parities of i and j are different, (1) X0,j ∩ Zi = ZiR if the parities of i and j are the same. Consequently, |X0,j ∩ Zi | = i/2. For 1i k and i j 2k − i, we further define the following sets. (X0,j − Zi ) ∪ Zi−1 if i is even, Xi,j = (X0,j − Zi ) ∪ Zi−1 ∪ {2k + 1} if i is odd. We note that i j 2k−i Xi,j = ∅ for any i, 1 i k and X0,j = (X0,j − Z0 ) ∪ Z−1 . Lemma 6. Let 0 i k and i j 2k − i. Then we have the following: (a) Each |Xi,j | = k. (b) The Xi,j ’s are all distinct. Proof. (a) This is obviously true when i = 0. Suppose that i 1. Observe that Zi−1 ∩ Zi = ∅. If i is even, then |Xi,j | = |X0,j ∪ Zi−1 | − |X0,j ∩ Zi | = |X0,j | + |Zi−1 | − |X0,j ∩ Zi−1 | − |X0,j ∩ Zi | = k + 2 (i − 1)/2 − (i − 1)/2 − i/2 = k. If i is odd, then |Xi,j | = |X0,j ∪ Zi−1 ∪ {2k + 1}| − |X0,j ∩ Zi | = |X0,j | + |Zi−1 | − |X0,j ∩ Zi−1 | + 1 − |X0,j ∩ Zi | = k + (i − 1) − (i − 1)/2 + 1 − i/2 = k.
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(b) First we show that Xi,j = Xi,t when j = t. This is again obviously true when i = 0. Suppose that i 1. It suffices to show that (X0,j − Zi ) ∪ Zi−1 are all distinct for distinct j’s. Since (X0,j − Zi ) ∪ Zi−1 = (X0,j − (X0,j ∩ Zi )) ∪ Zi−1 , it suffices to show that X0,j − ZiL if the parities of i and j are different, X0,j − (X0,j ∩ Zi ) = (2) X0,j − ZiR if the parities of i and j are the same are all distinct for distinct j’s. It follows from the definition of X0,j and (2) that X0,j − (X0,j ∩ Zi ) = X0,t − (X0,t ∩ Zi ) when the parities of j and t are the same. Suppose that j and t have different parities. We may assume that j is even and t is odd. If i is even, then 0 ∈ (X0,j − ZiR ) − (X0,t − ZiL ) and 2k ∈ (X0,t − ZiL ) − (X0,j − ZiR ). If i is odd, then 1 ∈ (X0,t − ZiR ) − (X0,j − ZiL ) and 2k − 1 ∈ (X0,j − ZiL ) − (X0,t − ZiR ). We conclude that Xi,j = Xi,t . Next, we want to show that Xi,j = Xs,t when i = s. If the parities of i and s are different, then we may assume that i is odd and s is even. In this case, Xi,j = Xs,t since 2k + 1 ∈ Xi,j − Xs,t . Now assume that 0 i < s both have the same parity. Then {s − 2, 2k − (s − 2)} ⊆ Zs−1 − Zi−1 . Since Xi,j − {2k + 1} = (X0,j − Zi ) ∪ Zi−1 and Xs,t − {2k + 1} = (X0,t − Zs ) ∪ Zs−1 , the equality Xi,j = Xs,t would imply {s − 2, 2k − (s − 2)} ⊆ X0,j . Therefore, {s − 2, 2k − (s − 2)} ⊆ X0,j ∩ Zs−1 . However, X0,j ∩ Zs−1 is equal to either L or Z R according to (1). Neither of the two sets can include {s − 2, 2k − (s − 2)} as a subset. We have arrived Zs−1 s−1 at a contradiction, and hence Xi,j = Xs,t . Theorem 7. For k 1, the tower graph TC2k+1 is isomorphic to a subgraph of the Kneser graph KG(2k + 2, k). Proof. Let G denote the graph KG(2k + 2, k). By Lemma 6, the Xi,j ’s are k-subsets of the set [2k + 2] for 0 i k and i j 2k − i. If 0 j 2k − 1, we have X0,j ∩ X0,j +1 = ∅ and hence X0,j X0,j +1 ∈ E(G). Moreover, X0,0 consists of odd integers and X0,2k consists of even integers, and hence X0,0 X0,2k ∈ E(G). This shows that X0,0 , . . . , X0,2k form a cycle of length 2k + 1. Next we want to show that Xi,j Xi−1,j −1 and Xi,j Xi−1,j +1 are edges of G for 1 i k and i j 2k − i. We know that X0,t ∩ X0,t+1 and Zt ∩ Zt+1 are both empty for all t. By definition, we have the following equalities. (X0,j − Zi ) ∪ Zi−1 , i is even, Xi,j = (X0,j − Zi ) ∪ Zi−1 ∪ {2k + 1}, i is odd. (X0,j −1 − Zi−1 ) ∪ Zi−2 ∪ {2k + 1}, i is even, Xi−1,j −1 = (X0,j −1 − Zi−1 ) ∪ Zi−2 , i is odd. (X0,j +1 − Zi−1 ) ∪ Zi−2 ∪ {2k + 1}, i is even, Xi−1,j +1 = i is odd. (X0,j +1 − Zi−1 ) ∪ Zi−2 , Hence, Xi,j ∩ Xi−1,j −1 = Xi,j ∩ Xi−1,j +1 = ∅. In the third step, we want to show that Xi+1,i+1 Xi,2k−i−1 and Xi+1,2k−i−1 Xi,i+1 are edges of G for 0 i k − 1. Again by definition, we have the following equalities. i is odd, (X0,i+1 − Zi+1 ) ∪ Zi , Xi+1,i+1 = (X0,i+1 − Zi+1 ) ∪ Zi ∪ {2k + 1}, i is even. (X0,2k−i−1 − Zi ) ∪ Zi−1 ∪ {2k + 1}, i is odd, Xi,2k−i−1 = (X0,2k−i−1 − Zi ) ∪ Zi−1 , i is even. (X0,i+1 − Zi ) ∪ Zi−1 ∪ {2k + 1}, i is odd, Xi,i+1 = (X0,i+1 − Zi ) ∪ Zi−1 , i is even. (X0,2k−i−1 − Zi+1 ) ∪ Zi , i is odd, Xi+1,2k−i−1 = (X0,2k−i−1 − Zi+1 ) ∪ Zi ∪ {2k + 1}, i is even. Since it is straightforward to show X0,i+1 ∩X0,2k−i−1 ⊆ Zi ∪Zi+1 , we then have Xi+1,i+1 ∩Xi,2k−i−1 =Xi+1,2k−i−1 ∩ Xi,i+1 = ∅.
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Our proof implies that the mapping taking the vertex i, j of TC2k+1 to the vertex Xi,j of G is an injective homomorphism. The images of all the edges of TC2k+1 form a subgraph of G that is isomorphic to TC2k+1 . Corollary 8. The Kneser graphs KG(2k + 2, k) are not cover graphs for k 1. This is a consequence of Lemma 3 and Theorems 4, and 7. Furthermore, since KG(2k + 2, k) is a subgraph of KG(2k + i, k) for i 2, hence we have the following. Theorem 9. The Kneser graphs KG(n, k) are not cover graphs for n 2k + 2 and k 1. A k-subset X of [n] is said to be 2-stable if min{|u − v|, n − |u − v|}2 for any two distinct elements u and v of X, or equivalently, X contains no two consecutive numbers in the cyclic order of [n]. The Schrijver graph SG(n, k) is the subgraph of KG(n, k) induced by all vertices that are 2-stable subsets. Schrijver graphs were first introduced in [14] where it was established that (SG(n, k)) = (KG(n, k)) and SG(n, k) is vertex color critical. Let the 2-stable k-subset X of [n] contain a fixed number u. Starting from u, we enumerate the n-circular distances between consecutive elements of X in the cyclic order. The sequence of “gaps” xi 2, 1 i n, satisfy the equation x1 + x2 + · · · + xk = n. Conversely, any such sequence of integers determine a 2-stable k-subset of [n] containing u. 2 Therefore, the vertex set of the Schrijver graph SG(n, k) has size nk ( n−k−1 k−1 ). In particular, |SG(2k + 2, k)| = (k + 1) . Lemma 10. The Xi,j ’s are 2-stable subsets of [2k + 2] for 0 i k and i j 2k − i. Proof. It is obvious that the X0,j ’s are 2-stable sets for all 0 j 2k. Now assume 1 i k and i j 2k − i. When L , hence X R the parities of i − 1 and j are different, X0,j ∩ Zi−1 = Zi−1 0,j ∪ Zi−1 = X0,j ∪ Zi−1 . When the parities of R L i − 1 and j are the same, X0,j ∩ Zi−1 = Zi−1 , hence X0,j ∪ Zi−1 = X0,j ∪ Zi−1 . In both cases, there are no consecutive integers in (X0,j − Zi ) ∪ Zi−1 . If i is odd, then{0, 2k} ⊆ Zi . It follows from the definition of Xi,j that {0, 2k} ∩ Xi,j = ∅. If i is even, then 2k + 1 ∈ / Xi,j . We now can conclude that the Xi,j ’s are 2-stable sets. Theorem 11. The Schrijver graphs SG(2k + 2, k) are not cover graphs for k 1. Proof. The number of vertices of TC2k+1 is equal to 1 + 3 + 5 + · · · + (2k + 1) = (k + 1)2 . Thus TC2k+1 is a spanning subgraph of SG(2k + 2, k) by Theorem 7 and Lemma 10. Then Lemma 3 implies that SG(2k + 2, k) is not a cover graph for k 1. Since SG(2k + 2, k) is a subgraph of SG(2k + i, k) for i 2, hence we have the following. Theorem 12. The Schrijver graphs SG(n, k) are not cover graphs for n2k + 2 and k 1. 5. Applications to circular chromatic number Let D be an acyclic orientation of the graph G. For an undirected cycle C of G, we choose one of the two traversals of C as the positive direction. An arc is said to be forward if its orientation under D is along the positive direction of C, otherwise it is said to be backward. We use (C, D)+ (or (C, D)− ) to denote the set of all forward (or backward) arcs of C with respect to D. The imbalance Imb(D) of D is defined to be |C| |C| C is a cycle of G . max max , |(C, D)+ | |(C, D)− | By convention, Imb(D) = 2 if G has no cycles. The circular chromatic number c (G) of a graph is a refinement of the ordinary chromatic number. The reader is referred to Zhu [19] for a comprehensive survey of this concept. A result of Goddyn et al. [3] implies that c (G) = min{Imb(D)|D is an acyclic orientationof G}. Lemma 13. Let G be a graph with (G) = g(G). If G is not a cover graph, then c (G) = (G).
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Proof. Since G is not a cover graph, there exists a dependent arc in any acyclic orientation D of G. The reversal of a dependent arc creates a directed cycle in D. Let C be the underlying cycle of that directed cycle. Then max{|C|/|(C, D)+ |, |C|/|(C, D)− |} g(G) = (G). It follows that c (G)(G). On the other hand, obviously c (G) (G). All the graphs referred to in the following Corollaries are 4-chromatic and of girth 4. Since they are not cover graphs, the results are immediate consequences of Lemma 13. Corollary 14 (Lin et al. [6]). If m 1 and k 2, then c (Mm (C2k+1 )) = 4. Corollary 15. If k 3, then c (TC2k+1 ) = 4. Corollary 16 (Johnson et al. [4]). If k 3, then c (KG(2k + 2, k)) = 4. Corollary 17 (Simonyi and Tardas [15]). If k 3, then c (SG(2k + 2, k)) = 4. We note that Corollary 17 was also obtained independently by Li and Lih [5]. The stronger result c (SG(n, k)) = (SG(n, k)) was established for even n in Simonyi and Tardos [15]. References [1] G. Birghtwell, On the complexity of diagram testing, Order 10 (1993) 297–303. [2] D.C. Fisher, K. Fraughnaugh, L. Langley, D.B. West, The number of dependent arcs in an acyclic orientation, J. Combin. Theory Ser. B 71 (1997) 73–78. [3] L.A. Goddyn, M. Tarsi, C.Q. Zhang, On (k, d)-colorings and fractional nowhere-zero flows, J. Graph Theory 28 (1998) 155–161. [4] A. Johnson, F.C. Holroyd, S. Stahl, Multichromatic numbers, star chromatic numbers and Kneser graphs, J. Graph Theory 26 (1997) 137–145. [5] W.-T. Li, K.-W. Lih, The circular extremality of some reduced Kneser graphs, preprint, 2003. [6] W. Lin, P.C.B. Lam, W.C. Shiu, Circular chromatic numbers of the generalized Mycielskians of cycles, Nanjing Daxue Xuebao Shuxue Bannian Kan 23 (2006) 232–241. [7] L. Lovász, Kneser’s conjecture, chromatic number and homotopy, J. Combin. Theory Ser. A 25 (1978) 319–324. [8] K.M. Mosesian, Some theorems on strongly basable graphs, Akad. Nauk. Armyan. SSR. Dokl. 54 (1972) 241–245 (in Russian). [9] J. Mycielski, Sur le coloriage des graphes, Colloq. Math. 3 (1955) 161–162. [10] J. Nešetˇril, V. Rödl, More on complexity of the diagrams, Comment. Math. Univ. Carolin. 2 (1995) 269–278. [11] O. Pretzel, On graphs that can be oriented as diagrams of ordered sets, Order 2 (1985) 25–40. [12] O. Pretzel, On reorienting graphs by pushing down maximal vertices, Order 3 (1986) 135–153. [13] I. Rival, The diagram, in: I. Rival (Ed.), Graphs and Order, Reidel, Dordrecht, 1985, pp. 103–133. [14] A. Schrijver, Vertex-critical subgraphs of Kneser graphs, Nieuw Arch. Wisk. 26 (3) (1978) 454–461. [15] G. Simonyi, G. Tardos, Local chromatic number, Ky Fan’s theorem, and circular colorings, Combinatorica 26 (2006) 587–626. [16] M. Stiebitz, Beiträge zur Theorie def farbungskritschen Graphen, Abilitation Thesis, Technical University Ilmenau, 1985. [17] C. Tardif, Fractional chromatic numbers of cones over graphs, J. Graph Theory 38 (2001) 87–94. [18] D.B. West, Acyclic orientations of complete bipartite graphs, Discrete Math. 138 (1995) 393–396. [19] X. Zhu, Circular chromatic number: a survey, Discrete Math. 229 (2001) 371–410.
Non-cover generalized Mycielski, Kneser, and Schrijver graphs Ko-Wei Liha , Chen-Ying Linb , Li-Da Tongc,1 a Institute of Mathematics, Academia Sinica, Nankang, Taipei 115, Taiwan b Department of Computer Science and Information Engineering, Shu-Te University, Kaohsiung 824, Taiwan c Department of Applied Mathematics, National Sun Yat-sen University, Kaohsiung 804, Taiwan
Received 30 December 2005; received in revised form 19 August 2007; accepted 21 August 2007 Available online 29 September 2007
Abstract A graph is said to be a cover graph if it is the underlying graph of the Hasse diagram of a finite partially ordered set. We prove that the generalized Mycielski graphs Mm (C2t+1 ) of an odd cycle, Kneser graphs KG(n, k), and Schrijver graphs SG(n, k) are not cover graphs when m 0, t 1, k 1, and n 2k + 2. These results have consequences in circular chromatic number. © 2007 Elsevier B.V. All rights reserved. Keywords: Cover graph; Mycielski graph; Kneser graph; Schrijver graph; Circular chromatic number
1. Introduction For a simple graph G, let |G| and G denote its number of vertices and number of edges, respectively. An edge, or a directed arc, joining the vertices u and v is denoted by uv, or u → v. The Hasse diagram of a finite partially ordered set is a pictorial representation of the covering relation of elements. The reader is referred to Rival [13] for general information on diagrams. A graph is said to be a cover graph if it is the underlying graph of a Hasse diagram. The recognition problem of cover graphs is known to be NP-complete [1,10]. Let the graph G be endowed with an acyclic orientation D, i.e., there are no directed cycles with respect to this orientation. Edelman (as quoted in West [18]) defined an arc of D to be dependent if its reversal creates a directed cycle, or equivalently a closed directed walk, in D. Let dmax (G) (dmin (G)) be the maximum (minimum) number of dependent arcs over all acyclic orientations of G. Edelman showed that dmax (G) = G − |G| + 1 if G is connected. This is generalized by Fisher et al. [2] to dmax (G) = G − |G| + c if G has c components. Pretzel [11] proved that having an orientation with no dependent arcs, i.e., dmin (G) = 0, is equivalent to being a cover graph. A sufficient condition for dmin (G) = 0 is (G) < g(G), where (G) and g(G) are, respectively, the chromatic number and the girth of G (See [2]). Let D be an acyclic orientation of the graph G. A vertex is said to be a sink (source) of D if its out-degree (in-degree) is zero. The operation of reversing all arcs going into a sink is called a push-down by Pretzel [12]. This operation of turning a sink into a source preserves the acyclicity of an orientation. It is also true that a push-down transforms an orientation without dependent arcs into another such orientation. The following was established in [8]. E-mail addresses: [email protected] (K.-W. Lih), [email protected] (C.-Y. Lin), [email protected] (L.-D. Tong). 1 Supported in part by the National Science Council under Grant NSC92-2115-M-110-004.
0012-365X/$ - see front matter © 2007 Elsevier B.V. All rights reserved. doi:10.1016/j.disc.2007.08.082
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Mosesian’s result. Let D1 be an acyclic orientation of the connected graph G. Then for any vertex v of G there exists a sequence of push-downs from D1 to an orientation D2 so that v is the unique sink in D2 . In the subsequent sections, we will establish the results stated in the Abstract and apply them to determine some circular chromatic numbers. 2. Generalized Mycielski graphs Let G = (V0 , E0 ) be a graph with vertex set V0 = {0, 0, 0, 1, . . . , 0, n − 1} and edge set E0 . The generalized Mycielski graph Mm (G) of G has vertex set V =V0 ∪( m i=1 Vi )∪{u}, where Vi ={i, j |0 j n−1} for 1 i m, and E ) ∪ {m, j u|0 j n − 1}, where Ei = {i − 1, j i, k|0, j 0, k ∈ E0 } for 1 i m. In edge set E = E0 ∪ ( m i=1 i the sequel, the second coordinates of vertices in Mm (G) are always taken modulo |G|. We note that M0 (G) is obtained from G by adding a new vertex adjacent to all vertices of G, and M1 (G) is commonly known as the Mycielskian M(G) of G [9]. In Tardif [17], a generalized Mycielski graph is defined as the categorical product of a graph and a path with a loop at one end, and then identifying all the vertices whose second coordinate is the other end of that path. It has been shown ([2,12]) that the Mycielskian M(C5 ) of a 5-cycle, also known as the Grötzsch graph, is not a cover graph. We can prove the following generalization. Theorem 1. Let integers m 0 and k 1. Then the generalized Mycielski graph Mm (C2k+1 ) of an odd cycle is not a cover graph. Proof. Since an acyclic orientation of a 3-cycle must have a dependent arc, Mm (C3 ) is clearly not a cover graph. For k 2, let G = Mm (C2k+1 ) and C2k+1 = (V0 , E0 ) with V0 = {0, 0, 0, 1, . . . , 0, 2k} and E0 = {0, 00, 1, 0, 10, 2, . . . , 0, 2k − 10, 2k, 0, 2k0, 0}. Suppose on the contrary that G is a cover graph. By Mosesian’s result, we may have an acyclic orientation D of G such that D has no dependent arcs and u is its unique sink. Since the subgraph induced by V0 in D is an acyclic (2k + 1)-cycle, without loss of generality, we may assume that four particular consecutive vertices are oriented as follows: 0, 0 → 0, 1 → 0, 2 ← 0, 3. Claim. For 1t m − 1, the arcs t − 1, t + 1 → t, t + 2, t − 1, t + 3 → t, t + 2, t + 1, t + 1 → t, t + 2, and t, t + 2 → t + 1, t + 3 are in D. We prove this Claim by induction on t with the assistance of the following. Observation. Let an acyclic orientation of a 4-cycle xyzwx have no dependent arcs. It follows that if the path xyz is oriented as x → y → z, then the path xwz must be oriented as x → w → z. When t = 1, this observation implies that the path 0, 0 → 1, 1 → 0, 2 is in D. Since u is the unique sink in D, the arc 0, 2 → 1, 3 is also in D. Then the existence of the path 1, 1 → 0, 2 → 1, 3 implies that of the path 1, 1 → 2, 2 → 1, 3. Similarly, the existence of the path 0, 3 → 0, 2 → 1, 3 implies that of the path 0, 3 → 0, 4 → 1, 3. It follows from the uniqueness of the sink u that the arc 1, 3 → 2, 4 is in D. Therefore the induction basis holds. Suppose that the induction hypothesis is true for 1t = k m − 2, i.e., the four arcs k − 1, k + 1 → k, k + 2, k − 1, k + 3 → k, k + 2, k + 1, k + 1 → k, k + 2, and k, k + 2 → k + 1, k + 3 are in D. The existence of the path k + 1, k + 1 → k, k + 2 → k + 1, k + 3 in D implies that of the path k + 1, k + 1 → k + 2, k + 2 → k + 1, k + 3. The existence of the path k − 1, k + 3 → k, k + 2 → k + 1, k + 3 implies that of the path k − 1, k + 3 → k, k + 4 → k + 1, k + 3. It follows again from the uniqueness of the sink u that the arc k + 1, k + 3 → k + 2, k + 4 is in D. Hence the case for t = k + 1 is true and our Claim is established. It follows from the Claim that the path m, m → m − 1, m + 1 → m, m + 2 is in D. Our observation implies that we should have the path m, m → u → m, m + 2. This contradicts to the fact that u is a sink. We conclude that G = Mm (C2k+1 ) is not a cover graph. Theorem 2. Let m > 0 and n 3. Then Mm (Cn ) is a cover graph if and only if n is even.
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Proof. We have already known from Theorem 1 that Mm (Cn ) is not a cover graph when n is odd. Now let n be an even integer. Since (Mm (Cn )) = 3 < 4 = g(Mm (Cn )), Mm (Cn ) is a cover graph. Remark 1. For n = 2k, an acyclic orientation D of Mm (Cn ) without dependent arcs can be explicitly defined as follows. Let Cn = (V0 , E0 ) with V0 = {0, 0, 0, 1, . . . , 0, 2k − 1} and E0 = {0, 00, 1, 0, 10, 2, . . . , 0, 2k − 20, 2k − 1, 0, 2k − 10, 0}. The edges of Mm (Cn ) are oriented as follows. The cycle Cn is alternately oriented by letting 0, 2i → 0, 2i + 1 and 0, 2i + 2 → 0, 2i + 1 for 0 i k − 1. Orient all other edges from a vertex with a smaller first coordinate to a vertex with a larger first coordinate or to the top vertex u. 3. Tower graphs A homomorphism f from a graph G1 (V1 , E1 ) to a graph G2 (V2 , E2 ) is a mapping from V1 to V2 such that uv ∈ E1 implies f (u)f (v) ∈ E2 . Lemma 3. Suppose that there is a homomorphism f from a graph G1 to a graph G2 . If G2 is a cover graph, so is G1 . Proof. Since G2 is a cover graph, there exists an acyclic orientation D2 of G2 having no dependent arcs. Define an orientation D1 of G1 as follow. If u and v are adjacent in G1 and f (u) → f (v) in D2 , then let u → v in D1 . The existence of a directed cycle C in D1 would imply that of a directed closed walk f (C) in D2 , contradicting the acyclicity of D2 . Therefore D1 is acyclic. Suppose on the contrary that G1 is not a cover graph. Then there exists a dependent arc in D1 . Let the reversal of the arc v1 → vt of D1 create a directed cycle C : v1 → v2 → · · · → vt → v1 for some t 3. Consider the directed path P : v1 → v2 → · · · → vt in D1 . Since D2 is acyclic, f (P ) must be a directed path in D2 . This forces f (v1 ) → f (vt ) to be a dependent arc in D2 , contradicting the assumption on D2 . We conclude that G1 is a cover graph. Now we are going to define the tower graph TCn of a cycle Cn . For integers n 3 and k = n/2 , let the graph G = Mk−1 (Cn ) have the vertex set V (G) = ( k−1 i=0 Vi ) ∪ {u} with Vi = {i, j |0 j n − 1} for 0 i k − 1. The tower graph TCn is the subgraph of G induced by the set W = {i, j |0 i k − 1, i j n − 1 − i} ∪ {u} together with new added edges i, i + 1i + 1, n − i − 2 and i, n − i − 2i + 1, i + 1 for 0 i k − 2. We may write the vertex u as k, k. Theorem 4. For k 1, the tower graph TC2k+1 is not a cover graph. Proof. Let G = Mk−1 (C2k+1 ) and H = TC2k+1 . Define a function f from V (G) to V (H ) as follows: Let f (u) = u and i, j if 0 i k − 1 and i j 2k − i, f (i, j ) = i, i if (j < i and i + j is even) or (j > 2k − i and i + j is odd), i, 2k − i if (j < i and i + j is odd) or (j > 2k − i and i + j is even). We are going to verify that f is a homomorphism from G to H. Let xy be an edge of G. If x, y ∈ V (H ), then f (x)f (y) = xy ∈ E(H ). If x = i, j satisfies j < i or j > 2k − i and y ∈ Vi−1 , then y = i − 1, j − 1 or i − 1, j + 1. Case 1: j < i. If i + j is even, then f (x) = i, i and f (y) = i − 1, i − 1. If i + j is odd, then f (x) = i, 2k − i and f (y) = i − 1, 2k − i + 1 or i − 1, i. We always have f (x)f (y) ∈ E(H ). Case 2: j > 2k − i. If i + j is even, then f (x) = i, 2k − i and f (y) = i − 1, 2k − i + 1. If i + j is odd, then f (x) = i, i and f (y) = i − 1, i − 1 or i − 1, 2k − i. Again we always have f (x)f (y) ∈ E(H ). Consequently, f is a homomorphism form G to H. Hence by Lemma 3 and Theorem 1, H is not a cover graph. Remark 2. It is easy to color TC2k+1 with four colors. The fact that Mm (C2k+1 ) is 4-chromatic was first proved in Stiebitz [16]. Tardif [17] gave a different proof. Therefore, the existence of a homomorphism from Mk−1 (C2k+1 ) to TC2k+1 shows that the latter is also 4-chromatic.
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Theorem 5. Suppose n3 is an integer. Then TCn is a cover graph if and only if n is even. Proof. By Theorem 4, T C n is not a cover graph when n is odd. When n 4 is even, it is again easy to see that (TCn ) = 3. Hence the inequality (TCn ) = 3 < 4 = g(TCn ) implies that TCn is a cover graph. Remark 3. An acyclic orientation of TCn having no dependent arcs can be explicitly defined by a construction similar to the one used in Remark 1. 4. Kneser graphs and Schrijver graphs Let [n] denote the set {0, 1, . . . , n − 1}. For n > 2k. The Kneser graph KG(n, k) has the set of all k-subsets of [n] as its vertex set, and two vertices A and B are adjacent if and only if A ∩ B = ∅. One of the well-known facts about Kneser graphs is (KG(n, k)) = n − 2k + 2 proved in Lovász [7]. We note that (KG(2k + 1, k)) = 3 < g(KG(2k + 1, k)) when k > 1, hence KG(2k + 1, k) is a cover graph. From now on, we focus on the case n = 2k + 2 and k 1 unless otherwise stated. We define the following sets for 0 j 2k. {0, 2, . . . , j − 2, j + 1, . . . , 2k − 3, 2k − 1} if j is even, X0,j = {1, 3, . . . , j − 2, j + 1, . . . , 2k − 2, 2k} if j is odd. Then we define the following sets for 1 i k. {1, 3, . . . , i − 1, 2k − (i − 1), . . . , 2k − 3, 2k − 1} if i is even, Zi = {0, 2, . . . , i − 1, 2k − (i − 1), . . . , 2k − 2, 2k} if i is odd. We also let Z−1 = Z0 = ∅ for convenience. When i > 0, we can partition Zi into two equal-sized parts ZiL and ZiR such that ZiL = Zi ∩ [i]. Since i k, we observe that i − 1 < 2k − i + 1, and hence |Zi | = |ZiL | + |ZiR | = 2 i/2. For 1 i k and i j 2k − i, we observe that L Zi if the parities of i and j are different, (1) X0,j ∩ Zi = ZiR if the parities of i and j are the same. Consequently, |X0,j ∩ Zi | = i/2. For 1i k and i j 2k − i, we further define the following sets. (X0,j − Zi ) ∪ Zi−1 if i is even, Xi,j = (X0,j − Zi ) ∪ Zi−1 ∪ {2k + 1} if i is odd. We note that i j 2k−i Xi,j = ∅ for any i, 1 i k and X0,j = (X0,j − Z0 ) ∪ Z−1 . Lemma 6. Let 0 i k and i j 2k − i. Then we have the following: (a) Each |Xi,j | = k. (b) The Xi,j ’s are all distinct. Proof. (a) This is obviously true when i = 0. Suppose that i 1. Observe that Zi−1 ∩ Zi = ∅. If i is even, then |Xi,j | = |X0,j ∪ Zi−1 | − |X0,j ∩ Zi | = |X0,j | + |Zi−1 | − |X0,j ∩ Zi−1 | − |X0,j ∩ Zi | = k + 2 (i − 1)/2 − (i − 1)/2 − i/2 = k. If i is odd, then |Xi,j | = |X0,j ∪ Zi−1 ∪ {2k + 1}| − |X0,j ∩ Zi | = |X0,j | + |Zi−1 | − |X0,j ∩ Zi−1 | + 1 − |X0,j ∩ Zi | = k + (i − 1) − (i − 1)/2 + 1 − i/2 = k.
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(b) First we show that Xi,j = Xi,t when j = t. This is again obviously true when i = 0. Suppose that i 1. It suffices to show that (X0,j − Zi ) ∪ Zi−1 are all distinct for distinct j’s. Since (X0,j − Zi ) ∪ Zi−1 = (X0,j − (X0,j ∩ Zi )) ∪ Zi−1 , it suffices to show that X0,j − ZiL if the parities of i and j are different, X0,j − (X0,j ∩ Zi ) = (2) X0,j − ZiR if the parities of i and j are the same are all distinct for distinct j’s. It follows from the definition of X0,j and (2) that X0,j − (X0,j ∩ Zi ) = X0,t − (X0,t ∩ Zi ) when the parities of j and t are the same. Suppose that j and t have different parities. We may assume that j is even and t is odd. If i is even, then 0 ∈ (X0,j − ZiR ) − (X0,t − ZiL ) and 2k ∈ (X0,t − ZiL ) − (X0,j − ZiR ). If i is odd, then 1 ∈ (X0,t − ZiR ) − (X0,j − ZiL ) and 2k − 1 ∈ (X0,j − ZiL ) − (X0,t − ZiR ). We conclude that Xi,j = Xi,t . Next, we want to show that Xi,j = Xs,t when i = s. If the parities of i and s are different, then we may assume that i is odd and s is even. In this case, Xi,j = Xs,t since 2k + 1 ∈ Xi,j − Xs,t . Now assume that 0 i < s both have the same parity. Then {s − 2, 2k − (s − 2)} ⊆ Zs−1 − Zi−1 . Since Xi,j − {2k + 1} = (X0,j − Zi ) ∪ Zi−1 and Xs,t − {2k + 1} = (X0,t − Zs ) ∪ Zs−1 , the equality Xi,j = Xs,t would imply {s − 2, 2k − (s − 2)} ⊆ X0,j . Therefore, {s − 2, 2k − (s − 2)} ⊆ X0,j ∩ Zs−1 . However, X0,j ∩ Zs−1 is equal to either L or Z R according to (1). Neither of the two sets can include {s − 2, 2k − (s − 2)} as a subset. We have arrived Zs−1 s−1 at a contradiction, and hence Xi,j = Xs,t . Theorem 7. For k 1, the tower graph TC2k+1 is isomorphic to a subgraph of the Kneser graph KG(2k + 2, k). Proof. Let G denote the graph KG(2k + 2, k). By Lemma 6, the Xi,j ’s are k-subsets of the set [2k + 2] for 0 i k and i j 2k − i. If 0 j 2k − 1, we have X0,j ∩ X0,j +1 = ∅ and hence X0,j X0,j +1 ∈ E(G). Moreover, X0,0 consists of odd integers and X0,2k consists of even integers, and hence X0,0 X0,2k ∈ E(G). This shows that X0,0 , . . . , X0,2k form a cycle of length 2k + 1. Next we want to show that Xi,j Xi−1,j −1 and Xi,j Xi−1,j +1 are edges of G for 1 i k and i j 2k − i. We know that X0,t ∩ X0,t+1 and Zt ∩ Zt+1 are both empty for all t. By definition, we have the following equalities. (X0,j − Zi ) ∪ Zi−1 , i is even, Xi,j = (X0,j − Zi ) ∪ Zi−1 ∪ {2k + 1}, i is odd. (X0,j −1 − Zi−1 ) ∪ Zi−2 ∪ {2k + 1}, i is even, Xi−1,j −1 = (X0,j −1 − Zi−1 ) ∪ Zi−2 , i is odd. (X0,j +1 − Zi−1 ) ∪ Zi−2 ∪ {2k + 1}, i is even, Xi−1,j +1 = i is odd. (X0,j +1 − Zi−1 ) ∪ Zi−2 , Hence, Xi,j ∩ Xi−1,j −1 = Xi,j ∩ Xi−1,j +1 = ∅. In the third step, we want to show that Xi+1,i+1 Xi,2k−i−1 and Xi+1,2k−i−1 Xi,i+1 are edges of G for 0 i k − 1. Again by definition, we have the following equalities. i is odd, (X0,i+1 − Zi+1 ) ∪ Zi , Xi+1,i+1 = (X0,i+1 − Zi+1 ) ∪ Zi ∪ {2k + 1}, i is even. (X0,2k−i−1 − Zi ) ∪ Zi−1 ∪ {2k + 1}, i is odd, Xi,2k−i−1 = (X0,2k−i−1 − Zi ) ∪ Zi−1 , i is even. (X0,i+1 − Zi ) ∪ Zi−1 ∪ {2k + 1}, i is odd, Xi,i+1 = (X0,i+1 − Zi ) ∪ Zi−1 , i is even. (X0,2k−i−1 − Zi+1 ) ∪ Zi , i is odd, Xi+1,2k−i−1 = (X0,2k−i−1 − Zi+1 ) ∪ Zi ∪ {2k + 1}, i is even. Since it is straightforward to show X0,i+1 ∩X0,2k−i−1 ⊆ Zi ∪Zi+1 , we then have Xi+1,i+1 ∩Xi,2k−i−1 =Xi+1,2k−i−1 ∩ Xi,i+1 = ∅.
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Our proof implies that the mapping taking the vertex i, j of TC2k+1 to the vertex Xi,j of G is an injective homomorphism. The images of all the edges of TC2k+1 form a subgraph of G that is isomorphic to TC2k+1 . Corollary 8. The Kneser graphs KG(2k + 2, k) are not cover graphs for k 1. This is a consequence of Lemma 3 and Theorems 4, and 7. Furthermore, since KG(2k + 2, k) is a subgraph of KG(2k + i, k) for i 2, hence we have the following. Theorem 9. The Kneser graphs KG(n, k) are not cover graphs for n 2k + 2 and k 1. A k-subset X of [n] is said to be 2-stable if min{|u − v|, n − |u − v|}2 for any two distinct elements u and v of X, or equivalently, X contains no two consecutive numbers in the cyclic order of [n]. The Schrijver graph SG(n, k) is the subgraph of KG(n, k) induced by all vertices that are 2-stable subsets. Schrijver graphs were first introduced in [14] where it was established that (SG(n, k)) = (KG(n, k)) and SG(n, k) is vertex color critical. Let the 2-stable k-subset X of [n] contain a fixed number u. Starting from u, we enumerate the n-circular distances between consecutive elements of X in the cyclic order. The sequence of “gaps” xi 2, 1 i n, satisfy the equation x1 + x2 + · · · + xk = n. Conversely, any such sequence of integers determine a 2-stable k-subset of [n] containing u. 2 Therefore, the vertex set of the Schrijver graph SG(n, k) has size nk ( n−k−1 k−1 ). In particular, |SG(2k + 2, k)| = (k + 1) . Lemma 10. The Xi,j ’s are 2-stable subsets of [2k + 2] for 0 i k and i j 2k − i. Proof. It is obvious that the X0,j ’s are 2-stable sets for all 0 j 2k. Now assume 1 i k and i j 2k − i. When L , hence X R the parities of i − 1 and j are different, X0,j ∩ Zi−1 = Zi−1 0,j ∪ Zi−1 = X0,j ∪ Zi−1 . When the parities of R L i − 1 and j are the same, X0,j ∩ Zi−1 = Zi−1 , hence X0,j ∪ Zi−1 = X0,j ∪ Zi−1 . In both cases, there are no consecutive integers in (X0,j − Zi ) ∪ Zi−1 . If i is odd, then{0, 2k} ⊆ Zi . It follows from the definition of Xi,j that {0, 2k} ∩ Xi,j = ∅. If i is even, then 2k + 1 ∈ / Xi,j . We now can conclude that the Xi,j ’s are 2-stable sets. Theorem 11. The Schrijver graphs SG(2k + 2, k) are not cover graphs for k 1. Proof. The number of vertices of TC2k+1 is equal to 1 + 3 + 5 + · · · + (2k + 1) = (k + 1)2 . Thus TC2k+1 is a spanning subgraph of SG(2k + 2, k) by Theorem 7 and Lemma 10. Then Lemma 3 implies that SG(2k + 2, k) is not a cover graph for k 1. Since SG(2k + 2, k) is a subgraph of SG(2k + i, k) for i 2, hence we have the following. Theorem 12. The Schrijver graphs SG(n, k) are not cover graphs for n2k + 2 and k 1. 5. Applications to circular chromatic number Let D be an acyclic orientation of the graph G. For an undirected cycle C of G, we choose one of the two traversals of C as the positive direction. An arc is said to be forward if its orientation under D is along the positive direction of C, otherwise it is said to be backward. We use (C, D)+ (or (C, D)− ) to denote the set of all forward (or backward) arcs of C with respect to D. The imbalance Imb(D) of D is defined to be |C| |C| C is a cycle of G . max max , |(C, D)+ | |(C, D)− | By convention, Imb(D) = 2 if G has no cycles. The circular chromatic number c (G) of a graph is a refinement of the ordinary chromatic number. The reader is referred to Zhu [19] for a comprehensive survey of this concept. A result of Goddyn et al. [3] implies that c (G) = min{Imb(D)|D is an acyclic orientationof G}. Lemma 13. Let G be a graph with (G) = g(G). If G is not a cover graph, then c (G) = (G).
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Proof. Since G is not a cover graph, there exists a dependent arc in any acyclic orientation D of G. The reversal of a dependent arc creates a directed cycle in D. Let C be the underlying cycle of that directed cycle. Then max{|C|/|(C, D)+ |, |C|/|(C, D)− |} g(G) = (G). It follows that c (G)(G). On the other hand, obviously c (G) (G). All the graphs referred to in the following Corollaries are 4-chromatic and of girth 4. Since they are not cover graphs, the results are immediate consequences of Lemma 13. Corollary 14 (Lin et al. [6]). If m 1 and k 2, then c (Mm (C2k+1 )) = 4. Corollary 15. If k 3, then c (TC2k+1 ) = 4. Corollary 16 (Johnson et al. [4]). If k 3, then c (KG(2k + 2, k)) = 4. Corollary 17 (Simonyi and Tardas [15]). If k 3, then c (SG(2k + 2, k)) = 4. We note that Corollary 17 was also obtained independently by Li and Lih [5]. The stronger result c (SG(n, k)) = (SG(n, k)) was established for even n in Simonyi and Tardos [15]. References [1] G. Birghtwell, On the complexity of diagram testing, Order 10 (1993) 297–303. [2] D.C. Fisher, K. Fraughnaugh, L. Langley, D.B. West, The number of dependent arcs in an acyclic orientation, J. Combin. Theory Ser. B 71 (1997) 73–78. [3] L.A. Goddyn, M. Tarsi, C.Q. Zhang, On (k, d)-colorings and fractional nowhere-zero flows, J. Graph Theory 28 (1998) 155–161. [4] A. Johnson, F.C. Holroyd, S. Stahl, Multichromatic numbers, star chromatic numbers and Kneser graphs, J. Graph Theory 26 (1997) 137–145. [5] W.-T. Li, K.-W. Lih, The circular extremality of some reduced Kneser graphs, preprint, 2003. [6] W. Lin, P.C.B. Lam, W.C. Shiu, Circular chromatic numbers of the generalized Mycielskians of cycles, Nanjing Daxue Xuebao Shuxue Bannian Kan 23 (2006) 232–241. [7] L. Lovász, Kneser’s conjecture, chromatic number and homotopy, J. Combin. Theory Ser. A 25 (1978) 319–324. [8] K.M. Mosesian, Some theorems on strongly basable graphs, Akad. Nauk. Armyan. SSR. Dokl. 54 (1972) 241–245 (in Russian). [9] J. Mycielski, Sur le coloriage des graphes, Colloq. Math. 3 (1955) 161–162. [10] J. Nešetˇril, V. Rödl, More on complexity of the diagrams, Comment. Math. Univ. Carolin. 2 (1995) 269–278. [11] O. Pretzel, On graphs that can be oriented as diagrams of ordered sets, Order 2 (1985) 25–40. [12] O. Pretzel, On reorienting graphs by pushing down maximal vertices, Order 3 (1986) 135–153. [13] I. Rival, The diagram, in: I. Rival (Ed.), Graphs and Order, Reidel, Dordrecht, 1985, pp. 103–133. [14] A. Schrijver, Vertex-critical subgraphs of Kneser graphs, Nieuw Arch. Wisk. 26 (3) (1978) 454–461. [15] G. Simonyi, G. Tardos, Local chromatic number, Ky Fan’s theorem, and circular colorings, Combinatorica 26 (2006) 587–626. [16] M. Stiebitz, Beiträge zur Theorie def farbungskritschen Graphen, Abilitation Thesis, Technical University Ilmenau, 1985. [17] C. Tardif, Fractional chromatic numbers of cones over graphs, J. Graph Theory 38 (2001) 87–94. [18] D.B. West, Acyclic orientations of complete bipartite graphs, Discrete Math. 138 (1995) 393–396. [19] X. Zhu, Circular chromatic number: a survey, Discrete Math. 229 (2001) 371–410.