Research Rutcor Report
Even-hole-free and Balanced Circulants
Diogo Andrade a Endre Borosb Vladimir Gurvichc
RRR 8, February 2005
RUTCOR Rutgers Center for Operations Research Rutgers University 640 Bartholomew Road Piscataway, New Jersey 08854-8003 Telephone: Telefax: Email: 732-445-3804 732-445-5472 [e-mail]
http://rutcor.rutgers.edu/rrr
Rutgers University, 640 Bartholomew Road, Piscataway NJ 08854-8003; email: [e-mail] b RUTCOR, Rutgers University, 640 Bartholomew Road, Piscataway NJ 08854-8003; email: [e-mail] c RUTCOR, Rutgers University, 640 Bartholomew Road, Piscataway NJ 08854-8003; email: [e-mail]
a RUTCOR,
Rutcor Research Report
RRR 8, February 2005
Even-hole-free and Balanced Circulants
Diogo Andrade
Endre Boros
Vladimir Gurvich
Abstract. In this paper some well-known conjectures about the even-hole-free graphs and balanced graphs are veried under the additional assumption of circular symmetry. Keywords: balanced graph, circulant, circular symmetry, even hole, even-hole-free graph
Acknowledgements: The rst author was supported in part by DIMACS, the National Science Foundations Center for Discrete Mathematics and Theoretical Computer Science, research of the second author was partially supported by the National Science Foundation (Grant IIS-0118635), the third author was partially supported by both and by American Institute of Mathematics.
Page 2
RRR 8
1
Introduction
A circulant graph (or circulant) is a graph on n vertices where the ith vertex is connected to the j th vertex if |i j| belongs to the list of edge lengths L. We denote circulants by {l1 , . . . , lm }n , where n is the number of vertices and {l1 , . . . , lm } is the list of edge lengths. Such graphs have circular symmetry. Figure 1 shows the circulant {1, 2}6: 0 5 1
4 3
2
Figure 1: Circulant {1, 2}6. Many standard properties can be reformulated for circulants in arithmetic terms. For example, the circulant {l1 , . . . , lm }n is connected i GCD(n, l1 , . . . , lm ) = 1 and it is bipartite i n is even, while all l1 , . . . , lm are odd. Such simplications are naturally used in computer experiments allowing to study significantly larger graphs. Given a conjecture C in the form: all graphs from a family F have a property P , it may be dicult to analyze C in general (for the whole family F ), but much easier for the circulants of F . For example, it was conjectured in [8] that each connected kernel-less digraph, except for odd dicycles, contains an edge which can be deleted and the remaining graph is still kernel-less (see also [3]). Yet, in [1] it was shown that the directed circulant {1, 7, 8}47 is a counterexample. It can be also much easier to prove P for the circulants from F than for the whole family F ; see e.g. [2, 4, 10, 7, 9, 11], where C = SPGC, Berges Strong Perfect Graph Conjecture. Let us remark that two dierent approaches were applied: in [10] F is the family of graphs without induced odd holes and antiholes (the so-called Berge graphs) and P means that a graph is perfect, while in [2, 4, 7, 9, 11] F is the family of partitionable graphs and P means that a graph is not Berge. Respectively, the Berge and partitionable circulants are studied in [2, 4, 7, 10, 9, 11]. In this paper we consider balanced and even-hole-free circulants. We will always assume that GCD(n, l1 , . . . , lm ) = 1; in other words, we restrict ourselves to connected circulants. It is easy to see that the circulant (tl1 , . . . , tlm )tn is just t vertex disjoint copies of the circulant (l1 , . . . , lm )n .
RRR 8
Page 3
2
Even-Hole-Free Graphs
A hole is a chordless cycle of length at least 4. A graph is even-hole-free if it contains no hole of even length. Conjecture 2.1. (Ho`ng). Every even-hole-free graph is 3-divisible. a Conjecture 2.2. (Ho`ng). If G is an even-hole-free graph then (G) 2(G) 1. a Conjecture 2.3. (Hayward and Reed). An even-hole-free graph contains a vertex whose neighborhood can be partitioned into two cliques. These conjecture are presented in [12] together with a short proof that Conjecture 2.3 implies Conjecture 2.2 which implies Conjecture 2.1. Indeed, let G be an even-hole-free graph, Conjecture 2.3 implies that each induced subgraph H of G has a vertex of degree at (F ) most 2(H) 2, and therefore (G) 2(G) 1. Since any graph F is (F )1 -divisible, G is 3-divisible. Denition 2.1. Given integers k 1 and m 0, let us introduce a graph GE (k, m) = (V, E) with circular symmetry as follows: V = {0, ..., n 1}, where n = k(2m + 1), and (i, j) E i j = {1, 0, +1} (mod (2m + 1)). For convenience, the loops i = j are included. For example if k = 1, the circulant is an odd cycle of size 2m + 1. If m = 0 (or 1), the circulant is a complete graph of size k (or 3k). Figure 2 illustrates the example where k = 2, m = 2 then n = 10 and (i, j) E i j (mod 5) {9, 0, 1; 4, 5, 6}. 9 8 7 6 5 4 0 1 2 3
Figure 2: Circulant {1, 4, 5}10. We now prove that graphs GE (k, m) are even-hole-free and then verify for them the three conjectures above. Denition 2.2. Let , 0 , + be edge types of GE (k, m) corresponding to edges (i, j) where i j = {1, 0, +1} + (2m + 1)t (t N), respectively. Lemma 2.1. Given a graph GE (k, m), any cycle of size greater than 3 that contains two consecutive edges of dierent types or two consecutive edges of type 0 has a chord.
Page 4
RRR 8
Proof. Let l1 , l2 be the two consecutive edges of dierent types (or two consecutive edges of type 0 ) in the cycle. Summing the lengths of these edges we have: + (2m + 1) ( N), where is the contribution of the {1, 0, +1} term, and (2m + 1) is the contribution of the term (2m + 1)t. Notice that {1, 0, +1}, because the edges are of dierent type (or = 0 if both edges are of type 0 ). Then, l3 = l1 + l2 = + (2m + 1) is a chord creating a chordless cycle of size 3. Theorem 2.1. The graph GE (k, m) has no even holes. Moreover, every chordless cycle in GE (k, m) is of size 3 or 2m + 1. Proof. Assume by contradiction that there is an even-hole. By Lemma 2.1 we know that such hole cannot contain 2 consecutive edges of dierent type, or 2 consecutive edges of type 0 , because such hole would have a chord and form a triangle. Thus, if there is an even-hole it must consist only of edge lengths or only + . So assume there is an even-hole consisting of only lengths of type + , let H = {h1 , , h2 , . . . , hp } (p even) be such hole, then we have:
p
sum of lengths of H =
i=1
hi = p + (2m + 1), ( N)
since H is a hole, p + (2m + 1) must be a multiple of 2m + 1, therefore p must be a multiple of 2m + 1. Moreover, since p is even, p 2(2m + 1). Now, consider the rst subgraph of H containing the rst 2m edges:
2m
sum of rst 2m lengths =
i=1
hi = 2m + (2m + 1) = (2m + 1) 1, (, N)
and then, by the denition of GE , there is a chord of type + that forms an odd-hole of size 2m + 1. The proof is analogous for a hole consisting of edges of type . We now present a dierent way of dening the graphs GE (k, m), as well as an alternate proof for Theorem 2.1. Let G be a graph with vertex set V (G) = {v1 , v2 , . . . , vn }, let Gk be a graph constructed from G by the following operations:
1 1 2 2 k k 1. Take k copies of each vertex in V (G), i.e. V (Gk ) = {v1 , . . . , vn , v1 , . . . , vn , . . . , v1 , . . . , vn };
2. For each edge (u, v) E(G), connect {u1 , . . . , uk , v 1, . . . , v k } in a 2k clique; Figure 3 illustrates an example of this operation applied to a cycle of length 5 (C5 ). The resulting graph is the circulant {1, 4, 5}10. Theorem 2.2. A graph G is even-hole-free if and only if Gk is even-hole-free.
RRR 8 v1 v5 v2
2 v4 2 v3 2 v5 1 v1 1 v2 1 v3 1 v4 2 v2 1 v5
Page 5
v4
v3
2 v1
2 Figure 3: C5 transformed into C5 (which is the circulant {1, 4, 5}10).
Proof. (=) If Gk is even-hole-free, then all its induced subgraphs are even-hole-free, and G is clearly an induced subgraph of Gk . (=) Assume that G is even-hole-free but Gk is not. Then, there is an even-hole H = k k km {vi11 , vi22 , . . . , vim }. Clearly i1 = i2 . . . = im , otherwise there is a chord in H. But since k km i1 = i2 . . . = im the induced subgraph on {vi11 , . . . , vim } in Gk is the same as the induced subgraph on {vi1 , . . . , vim } in G. Therefore, there is no even-hole on Gk . Lemma 2.2. Let G be an odd cycle of length 2m + 1, then Gk is GE (k, m). Proof. Immediate from the denition of GE (k, m). It is easy to see that Theorem 2.1 follows from Theorem 2.2 and Lemma 2.2 Proposition 2.1. If m > 1 the maximum cliques of GE (k, m) are of size 2k, otherwise if m = 0 or 1 the maximum clique is n = k(2m + 1). Proof. If m = 0 (or 1), GE (k, m) is a complete graph on n vertices, therefore the maximum clique is n. Now assume m > 1. Since the graph has circular symmetry, every maximum clique has not only the same size, but also the same structure. So consider an initial vertex u: it is connected to 3k1 vertices. Lets divide those vertices into 3 types: v , v0 , v+ , corresponding to vertices connected to u by an , 0 , + edge, respectively (notice also that u is a v0 type vertex). And by the denition of GE (k, m), the 2k vertices v0 , v+ form a clique (and by symmetry, so do the 2k vertices v , v0 ). Moreover, no vertex v is connected to any vertex v+ , and the vertices v0 are not connected to any other vertices on the graph. So, 2k is the maximum. Proposition 2.2. If m 1, the maximum independent sets of GE (k, m) are of size m. Proof. Notice that to nd the maximum independent set, it suces to consider an interval of 2m + 1 consecutive vertices (say, from 0 to 2m), because every vertex outside this interval is equivalent 1 to some vertex within this interval.
Equivalent in the sense that it is connected to the same set of vertices. More specically, a vertex v (2m + 1 v < n) is equivalent to mod(v, 2m + 1).
1
Page 6
RRR 8
Now, let H be the induced subgraph of an interval of 2m + 1 consecutive vertices of GE . It is easy to see that H is a cycle on 2m + 1 vertices, and therefore its maximum independent set is m. Proposition 2.3. The chromatic number of GE (k, m) is at most 2k + k/m + 2. Proof. Follows from Proposition 2.2. For every group of 2m consecutive vertices there are two disjoint independent sets of size m, so we assign one color to each. To the next group of 2m vertices we assign two new colors, and so on. The graph has a total of n = 2km + k vertices. Assigning the coloring proposed above, the chromatic number (GE ) 2 2km+k = 2k + k/m. If mod(k, 2m) = 0, (GE ) = 2k + k/m, 2m if mod(k, 2m) = 1, (GE ) = 2k + k/m + 1, otherwise ) (GE = 2k + k/m + 2. Now from Propositions 2.1 and 2.2, Conjectures 2.1 and 2.2 follow immediately. Yet, to prove all 3 conjectures, is suces to prove Conjecture 2.3. This is done by the following theorem. Theorem 2.3. An even-hole-free circulant GE (k, m) contains a vertex whose neighborhood can be partitioned into two cliques. Proof. Since a circulant has circular symmetry, it does not matter which vertex we pick. The theorem shall work for every vertex v GE (k, m). Let v , v0 , v+ be dened as above, and assign type v0 to the chosen vertex v. The neighborhood of v consists of k vertices of type v+ , k vertices of type v and k 1 vertices of type v0 . By denition, all vertices v+ are connected among themselves, and so are all vertices v . The v0 vertices are connected to both types. Then, one possible partition into two cliques is: one clique contains all neighbors of type v+ and v0 , and the other clique contains all neighbors of type v . Conjecture 2.4. Every non-empty even-hole-free circulant is isomorphic to a GE (k, m). This conjecture was tested by computer for circulants up to 60 vertices.
3
Balanced Graphs
A graph is balanced if every induced cycle has length 0 (mod 4). Obviously, balanced graphs are bipartite. Conjecture 3.1. (Conforti, Rao) In every balanced graph, there exists an edge that can be deleted, so that the resulting graph remains balanced. Conjecture 3.2. (Conforti, Cornuejls, Rao) Every balanced graph is either basic or has a o 2-join or a skew-partition.
RRR 8
Page 7
Recall that a balanced graph G is called basic if all its vertices on one side of the bipartition have degree at most 2 or G contains a hole H such that the vertices of G \ H induce a complete bipartite graph. The denitions of 2-join and skew-partition can be found in literature, see e.g. [5]. In fact, they are not needed here. We will only make use of the concept of star-cutset which is a special case of skew-partition. This notion was introduced by Chvtal in [6] as follows. a A graph is a star if it has a vertex incident to all other vertices. Given a graph G = (V, E), a vertices-set S V is a star-cutset if the induced subgraph G[S] is a star and G[V \ S] is not connected. Denition 3.1. Given integer k 1 and m 1, let us introduce a graph GB (k, m) = (V, E) with circular symmetry as follows: V = Zn = {0, 1, ..., n 1}, where n = 4km, and (i, j) E i j = {1, +1} (mod 4m), for some integer t. For example, if k = 1, then the circulant is a cycle of length 4m. Figure 4 illustrates the case where k = 2, m = 1; then n = 8 and (i, j) E i j (mod 8) {1, 3, 5, 7}. 0 7 6 5 4 Figure 4: Circulant {1, 3}8. First, we will prove that graphs GB (k, m) are balanced and then verify for them the above 2 conjectures. Denition 3.2. Let , + be edge types of GB (k, m) corresponding to edges (i, j) where i j = {1, +1} + (4m)t (t N), respectively. Lemma 3.1. Any cycle of size greater than 4 in a given graph GB (k, m) that contains two consecutive edges of dierent types has a chord. Proof. Let l1 , l2 be the two consecutive edges of dierent type in the cycle. Summing the lengths of this edges we have: +1 1 + 4m ( N), where 4m is the contribution of the term (4m)t. Let l3 be an edge of any type following l2 , summing up l1 + l2 + l3 we have + 4m ( N). Notice that {1, +1}, therefore l4 = l1 + l2 + l3 is a chord, closing a hole of length 4 in the cycle. 3 1 2
Page 8
RRR 8
Theorem 3.1. The graph GB (k, m) is balanced. Moreover, every hole in GB (k, m) has size 4 or 4m. Proof. Assume for contradiction that the graph is not balanced. By Lemma 3.1 we know that such hole cannot contain edges of dierent types, since there will be a chord forming a hole of size 4. Thus, if there is a cycle of length dierent than 0 (mod 4), it contains only edges of type , or only of type + . Assume it contains only + edges, let H = {h1 , . . . , hp } (p = 0 (mod 4)) be the hole, then we have:
p
sum of lengths of H =
i=1
hi = p + 4m, ( N)
but since H is a hole, p + 4m must be a multiple of 4km, which implies that p must be 0 (mod 4m). Now, taking the sum of the lengths of 4m 1 edges of type + , we have: 4m 1 + 4m = 1 + 4m ( N), then, by denition of GB , there is and + edge e such that e + 1 + 4m = t(4km) (t N), which implies that p = 4m. The proof is analogous for a hole consisting of edges of type . Theorem 3.2. Every balanced circulant GB (k, m) is either basic or has a star-cutset. Proof. If k = 1, then GB is a cycle of length 4m which is basic. If k > 1, GB is not basic, however it has a star-cutset. Indeed, pick any vertex u in GB , and partition all other vertices in three sets V , V0 , V+ , respectively, v V , V0 , V+ if u v = {1, 0, +1} (mod 4m). Then S = {u} V V+ is a star-cutset. Indeed, clearly S is a star with the center u and S is a cutset, since G[V0 ] consists of isolated vertices. Clearly, this theorem implies Conjecture 3.2, since star-cutset is a special case of skewpartition. u v v+
v+ v0
v
Figure 5: Illustration of Theorem 3.2 on Circulant {1, 3}8. Figure 5 illustrates Theorem 3.2 on Circulant {1, 3}8 . The star cutset consists of u and the four vertices of type v and v+ . The vertex v0 is disconnected after the removal of the cutset.
RRR 8 Lemma 3.2. Given a graph GB (k, m), we have: (i) Every cycle of length 4i + 2 has at least two chords.
Page 9
(ii) A (4i + 2)-cycle has exactly two chords if and only if m > 1, i = 1, that is the cycle is of length 6, and the edge type sequence is + , + , + , , , . Proof. Notice that a (4i + 2)-cycle must contain at least one edge of each type, otherwise, the sum of the edge lengths will not be a multiple of n = 4km (in other words, it would not be a cycle). Let l1 , l2 be the consecutive edges of dierent type, l0 be the edge preceding l1 , and l3 be the edge following l2 in the cycle. Then, by Lemma 3.1, c1 = l0 + l1 + l2 is a chord, and c2 = l1 + l2 + l3 is a chord. This proves (i). To prove (ii), we start by proving that no (4i + 2)-cycle of length greater than 6 can have 2 chords. Let us split into 2 cases: 1. There is 1 edge of one type and 4i + 1 edges of the other. Assume the 1 edge is of type and the remaining of type + . Let l0 , l1 , l2 , l3 , l4 be consecutive edges of type + , + , , + , + , respectively. By Lemma 3.1, c1 = l0 + l1 + l2 , c2 = l1 + l2 + l3 and c3 = l2 + l3 + l4 are chords. The same reason...