Quasi-parity and perfect graphs Irena Rusu L.R.I., U.R.A. 410 du C.N.R.S., bât. 490, 91405 Orsay-cedex, France Abstract In order to prove the Strong Perfect Graph Conjecture, the existence of a simple property P holding for any minimal non-quasi-parity Berge graph G would really reduce the difficulty of the problem. We prove here that this property cannot be of type G is F-free, where F is any fixed family of Berge graphs. Keywords: combinatorial problems, algorithms, graph theory. 1 Introduction. The chromatic number χ(g) of a graph G = (V, E) is the minimum number of colours necessary to colour G in such a way that any two adjacent vertices have different colours. The clique number ω(g) of G is the maximum number of pairwise adjacent vertices. It is quite easy to deduce that for any graph G, the chromatic number is not smaller than the clique number. Claude Berge [1] called perfect the graphs for which the equality χ = ω holds not only for the graph itself, but also for every induced subgraph of it. Also, he noticed that the induced odd cycles of length at least five (also called holes) and their complements (also called antiholes) are not perfect graphs, although all their subgraphs are. As a consequence, Berge formulated the conjecture below, known as the Strong Perfect Graph Conjecture (abbreviated SPGC): Conjecture 1 (SPGC) A graph G = (V, E) is perfect if and only if it contains no hole and no antihole. Many classes of graphs have been proved to be perfect or to satisfy the SPGC since Berge introduced the notion of perfection. Also, many properties of minimal imperfect graphs (i.e. imperfect graphs such that all their proper subgraphs are perfect) have been discovered. In spite of these efforts, the Strong Perfect Graph Conjecture is still open. However, now, that we have this variety of results, we may suspect that an appropriate combination of some known properties would be enough to deduce the conjecture. Nevertheless, any such preceding attempt has fallen. Our aim here is to examine a recent attempt and to bring another argument against this type of conjectures. Any definition used along this paper and not given below may be found in [2]. 2 Preliminary considerations. A graph is said to be a Berge graph if it contains no hole and no antihole. It is called an F-free graph, where F is a family of graphs, if it contains no induced subgraph isomorphic to a graph in the family F. An even pair of the graph G is a pair (x, y) of vertices such that every chordless path joining x and y in G has an even length. A graph G such that, for every subgraph H of G, either H or H has an even pair is called a quasi-parity graph. Meyniel [4] proved that a minimal imperfect graph cannot contain an even pair and, using the fact that a graph is perfect if and only if the same holds for its complement (Lovász [6]), deduced that quasi-parity graphs are perfect. 1
A star-cutset C of G = (V, E) is a disconnecting subset of V which contains a vertex adjacent to all the other vertices in C. It was proved by Chvátal [3] that no minimal imperfect graph contains a star-cutset. Developing this idea of forbidden disconnecting sets, Tucker [11] also proved that no minimal imperfect graph contains a stable cutset, i.e. a disconnecting set inducing no edge. Taking into account these results, Reed [9] formulated the conjecture below: Conjecture 2 Let G be a Berge graph such that: 1. neither G nor Ḡ contains an even pair; 2. neither G nor Ḡ contains a star-cutset. Then G or Ḡ is the line graph of a bipartite graph. Stefan Hougardy [5] disproved this conjecture, so that Hoàng, Maffray and Reed, independently, modified it to the following (the diamond is the graph obtained from a complete graph on four vertices by deleting an edge): Conjecture 3 Let G be a Berge graph such that: 1. neither G nor Ḡ contains an even pair; 2. neither G nor Ḡ contains a star-cutset; 3. neither G nor Ḡ contains a stable-cutset. Then G or Ḡ is diamond-free. In [10], we have invalidated this conjecture too by providing a method of building, starting with a particular class of graphs, a large class of non-conventional perfect graphs, that is, which satisfy all the hypothesis of the conjecture above. Using the same technique of combining properties in order to cover the entire class of perfect graphs, we shall now analyse the problem of finding a simple property P such that all minimal non-quasi-parity graphs satisfy P. Equivalently, we are interested to identify a simple class of graphs C such that the proposition below is true: Proposition 1 If G is a Berge graph satisfying the odd-connectivity condition (OC) neither G nor Ḡ contains an even pair then G is in C. This problem arrises, on the one hand, from the preceding invalidated conjectures and, on the other hand, from the necessity of finding the limits of the class of quasi-parity graphs, which is one of the largest classes known. The requirement for C to be simple is not precise, but means in practice easy enough to be proved perfect. The first results obtained in the theory of perfect graphs show that simple classes are often (but not always) the classes of F-free graphs, where F is a class of Berge graphs. Our aim here is to show that in fact C cannot be a class of F-free graphs. This means that some other properties, and probably global ones, must be added to obtain a satisfactory result. 3 The results. As before, let F be a family of Berge graphs, and consider C the class of F-free graphs. In order to prove that Proposition 1 cannot be valid we have to construct a Berge graph G which has no even pair (neither in G nor in Ḡ), but contains at least one graph H F. To this end we have to consecutively distroy all the even pairs of H and H, by adding an appropriate graph on the edges of H and H. This is possible, as shown by the following theorem. To formulate it, let us say that a graph G is hole-equivalent to its induced subgraph G if all the holes and antiholes of G are contained in G. Also, G will be called odd-connected if neither G nor Ḡ has an even pair. 2
Theorem 1 For every graph H, there exist an odd-connected graph G H such that: 1. H is an induced subgraph of G H ; 2. G H is hole-equivalent to H. We may suppose, without loss of generality, that H and H are 2-connected; otherwise we can add vertices to H in order to obtain a graph H with these properties and which contains H as an induced subgraph. Then the graph G H obtained for H is also hole-equivalent to H. We denote by F ab the graph in Figure 1 which, as it can easily be verified, is odd-connected. Lemma 1 Let G be a graph and e an edge of G. The graph G obtained from G and F ab by identifying the edges e and ab is hole-equivalent to G. Proof. We firstly notice that F ab is a Berge graph. Indeed, since it has eight vertices, any induced holes or antiholes must be of size five or seven. Moreover, the vertices 1, 2, 3, 4, 5, 6 induce a C 6, so every hole or antihole must contain at least one of the vertices a and b (in fact, both of them). It is easy to verify that this is not possible. Consequently, an odd chordless cycle induced in G but not included in G should have vertices in both G \ {a, b} and F ab \ {a, b}, but that is impossible since ab would be a chord. Also, an antihole should have vertices in both graphs, but then ab would be a star-cutset of the antihole, and this would contradict Chvátal s result. We deduce that G is hole-equivalent to G. Starting with our initial graph H, consider now the following prolonging algorithm (where m = E(H), m = E( H) ): Prolonging Algorithm Input: H, F aib i (i = 1,...,m), F cjd j (j = 1,...,m ); Output: G H for i = 1, m do identify e i E(H) and a i b i E(F aib i ) end; for j = 1, m do identify e j E( H) and c j d j E(F cjd j ) end. By lemma 1, the graph H obtained by performing the prolonging algorithm is hole-equivalent to H. To prove that it is in fact the graph G H we are looking for it is sufficient to show that in H and H every two non-adjacent vertices are joined by at least one chordless path of odd length. To do this, notice first 1 2 3 4 5 6 a b Figure 1: The graph F ab. 3
the way in which the different subgraphs of G are joined: - for every i {1,..., m}, every proper vertex of F aib i is joined to every proper vertex in F cjd j, j {1,...,m }, to no proper vertex in F ai b i (i i), and to no vertex in H (except for a i, b i ); - for every j {1,...,m }, every proper vertex of F cjd j is adjacent to every vertex in H F cjd j. (A proper vertex of F ab is any vertex of F ab except for a and b; a basic vertex is either a or b.) Lemma 2 The graph H obtained from H using the prolonging algorithm is odd-connected. Proof. As we already noticed, the graph F ab in Figure 1 is odd-connected. Indeed, any two nonadjacent vertices in F ab or in F ab are joined by a chordless path on four vertices. Thus, in H and H, these pairs are not even. Further, it is easily verified that, in F ab, if x is a proper vertex non-adjacent to the basic vertex u, then x and u can be joined both by odd and even chordless paths not containing the remaining basic vertex. Finally, it is easy to see that in F ab every proper vertex is non-adjacent to at least one basic vertex. To prove that in H and H every two non-adjacent vertices x, y are joined by an odd chordless path, we have to consider the six cases below. The notation we use is F ab for any subgraph F aib i in H and F cd for any subgraph F cjd j in H. Moreover, x F ab or x F cd means that x is a proper vertex of the corresponding graph. Case 1. x H, y H. If x, y are non-adjacent in H, then F xy guarantees the existence of a P 4 joining x and y in H ; otherwise, they are adjacent and F xy insures that x and y is not an even pair in H. Case 2. x H, y F ab. We may suppose that x a, b, so that x and y are non-adjacent. By a preceding remark, y is non-adjacent to at least one basic vertex in F ab, say a. Since H is a 2-connected graph, there exists a chordless path P in H joining x and a which does not contain b. Moreover, y and a are joined in F ab both by odd and even chordless paths not containing b, so that we can find a path P whose parity guarantees that P P is an odd path from x to y. This path is obviously chordless. Case 3. x H, y F cd. Again, we may suppose that x c, d, so x and y are adjacent. By translating the reasoning in H, we have that x H, y F cd and x, y are non-adjacent. As in case 2, we deduce the existence of the suitable path. Case 4. x F ab, y F a b. Let a be the basic vertex non-adjacent to x, and a the basic vertex adjacent to y (if it exists) or an arbitrary basic vertex of F a b (in the other case). In both cases, b y E. Then from a to a we can find a chordless path P which does not contain b (but possibly contains b ). From y to a we can find a chordless path P in F a b not containing a neighbour of b, consequently P P is a chordless path joining a to y. As before, we can choose in F ab a suitable path P joining x to a such that P P P be an odd chordless path. Case 5. x F ab, y F cd. Then x and y are adjacent, so we must continue the reasoning in H. Here we have x F ab, y F cd and x, y are non-adjacent. Moreover, the proper vertices of F ab are joined to all the vertices in H \ {a, b}. If c is the basic vertex non-adjacent to y, then either xc E( H ) (and we can find a suitable path in F cd joining c to y) or c = a (so we can suitable join x to a in F ab and y to c in F cd ). Case 6. x F cd, y F cd. Then x, y are adjacent and, while translating the reasoning in H, we obtain the situation in case 4. Consequently, for every pair of non-adjacent vertices in H or H there is an odd chordless path joining them. Lemma 2 is proved. 4
Proof of Theorem 1. The lemmas 1 and 2 guarantee that the graph G H obtained by the prolonging algorithm contains H, is odd-connected and hole-equivalent to H. Corollary 1 In Proposition 1, the class C cannot be a class of F-free graphs. Proof. The graph G H obtained for a graph H F is Berge (since H itself is a Berge graph), oddconnected and not contained in C, thus it contradicts Proposition 1 with C = {F free Berge graphs}. Remark. The prolonging algorithm adds a graph isomorphic to F ab on every edge of H and on every edge of H, in order to introduce odd chordless paths when necessary. In fact, there could be some pairs of vertices which do not need such an operation (since they are already joined by an odd chordless path), but it seems very difficult to identify them algorithmically. That s why the version of the prolonging algorithm we presented is preferable in our case. Consider now the two decision problems below, concerning the recognition of Berge graphs: Problem 1. Instance: A graph H. Question: Is H a Berge graph? Problem 2. Instance: An odd-connected graph H. Question: Is H a Berge graph? From Theorem 1 we can also deduce: Corollary 2 Problem 1 and Problem 2 are polynomially equivalent. Proof. Obviously, Problem 2 is nothing else but a particular case of Problem 1, so it reduces polynomial to Problem 1. To reduce Problem 1 to Problem 2, it is sufficient to build, starting with the instance H of Problem 1, the graph G H given by Theorem 1. Since G H has exactly the same holes and antiholes as G, a positive answer for H in Problem 1 induces a positive answer for G H in Problem 2 and vice-versa. References [1] C. Berge - Farbun von Graphen, deren samtliche bzw, deren ungerade Kreise starr sind, Wiss. Z. Martin Luther Univ., Halle Wittenberg (1961). [2] C. Berge - Graphes, Paris, Dunod, 1970. [3] V. Chvátal - Star-cutsets and perfect graphs, J. Combin. Theory Ser. B 39 (1985), 189 199. [4] J. Fonlupt, J. P. Uhry - Transformations which preserve perfectness and h-perfectness of graphs, in Annals of Discrete Mathematics 16 (1982), 83 95. [5] S. Hougardy - Counterexamples to three conjectures concerning perfect graphs, Discrete Math., to appear. [6] L. Lovász - A characterization of perfect graphs, J. Combin. Theory Ser. B 13 (1972), 95 98. [7] H. Meyniel - A new property of critical imperfect graphs and some consequences, European J. Combin. 8 (1987), 313 316. 5
[8] F. Maffray - Une étude structurelle des graphes parfaits, Thèse d Etat, Université Joseph Fourier, Grenoble, France (1992). [9] B. Reed - A semi-strong perfect graph theorem, Ph.D. Thesis, Dept. of Comp. Sci, McGill University, Montréal, Québec, Canada (1985). [10] I. Rusu - Building counterexamples, 1st Workshop on Perfect Graphs, Princeton, New Jersey, (1993). [11] A. Tucker - Coloring graphs with stable cutsets, J. Combin. Theory Ser. B 34 (1983), 258 267. 6