Efficiently Recognizing the P 4-Structure of Trees and of Bipartite Graphs Without Short Cycles

A 4-uniform hypergraph represents the P4-structure of a graph G if its hyperedges are the vertex sets of the P4's in G. By using the weighted 2-section graph of the hypergraph we propose a simple e1⁄2cient algorithm to decide whether a given 4-uniform hypergraph represents the P4-structure of a bipartite graph without 4-cycle and 6-cycle. For trees, our algorithm is di ̈erent from that given by G. Ding and has a better running time namely O…n2† where n is the number of vertices. The P4-structure of a graph G ˆ …V ;E† is the set of all vertex sets of induced paths with four vertices (denoted by P4) in G. The importance of the P4-structure of graphs is well-known from investigations of ChvaÂtal [3] and Reed [5] showing that if two graphs have the same P4-structure and one of them is perfect then the other is perfect as well. Recently, Ding in [4] gave an incremental polynomial-time algorithm proceeding leaf by leaf which for a given 4-uniform hypergraph H ˆ …V ;E† decides whether there is a tree T such that H represents the P4-structure of T. We propose here a conceptually simpler and more direct algorithm to decide this question by investigating weighted 2-section graphs. Using an appropriate implementation it leads to a better time bound O…jV j†. Note that this time bound is optimal in the case that jEj is quadratic in jV j. Our concept works even for the larger class of bipartite graphs without cycles of length 4 and 6 (denoted by C4, respectively, C6). Furthermore our approach can be extended to another class containing trees, namely the class of block graphs [2], but this case is technically much more involved. Let now H ˆ …V ;E† be a hypergraph and let jV j ˆ n and jEj ˆ m. The 2section graph G…H† has the same vertex-set as H; two vertices are adjacent in * The work of the ®rst author was supported, in part, by DIMACS and RUTCOR, the third author was supported by NSF grants CCR-9407180 and CCR-9522093 and by ONR grant 00014-95-1-0779 G…H† if there is a hyperedge in H containing both of them. For vertices u and v in V, let the multiplicity mH…u; v† denote the number of hyperedges of H containing both. Throughout this note we assume that the edges uv of G…H† are weighted by mH…u; v†. We call a hyperedge e ˆ fx1; x2; x3; x4g of type i …i A f0; 1; . . . ; 6g† if in e exactly i pairs xkxl with k 0 l and k; l A f1; . . . ; 4g have multiplicity 1. If the hyperedge e ˆ fx1; x2; x3; x4g is of type 1 with mH…x1; x4† ˆ 1 then x2x3 is called the mid-edge of e. In a graph G, the vertices of degree 1 are called endvertices; if G is a tree, its endvertices are also called leaves. Let Pk…Ck† denote an induced path (cycle) with k vertices. The graph P is obtained from a C4 by adding one pendant vertex. If Q ˆ x1x2x3x4 is an induced P4, then x1; x4 are endpoints, and x2x3 is the mid-edge of Q. Lemma 1. Suppose that H ˆ …V ;E† is the P4-structure of the …P;C6†-free connected bipartite graph G which is not a P4. Then for every hyperedge e A E exactly one of the following conditions hold: (i) e is of type 1. Then the mid-edge of e is an edge in G. (ii) e is of type 2. The pairs of multiplicity 1 have a common vertex which is an endvertex in G and is adjacent, in G, to the vertex of e not contained in the two pairs of multiplicity 1. (iii) e is of type 3. The pairs of multiplicity 1 have a common vertex which is an endvertex of G. In particular, if e contains no endvertex then e is of type 1. Proof. Since G is a …P;C6†-free bipartite graph, it is obvious that for each of its P4's; the pair of its endpoints is contained in exactly one hyperedge of H: …1† Thus, no hyperedge is of type 0. Next we claim that in each of its P4's; there is one of the endpoints contained in all of the pairs of multiplicity 1: …2† In particular, every hyperedge is of type 1, 2, or 3. To see (2), consider a P4 Q ˆ x1x2x3x4 in G. If x1 (or x4) has further neighbor x, then xx1x2x3 (or x2x3x4x) is an induced P4 (else G would have a P), hence all pairs of multiplicity 1 in Q must have x4 (or x1) in common. If x2 (or x3) has further neighbor y, then yx2x3x4 (or yx3x2x1) is an induced P4 (else G would have a P), hence all pairs of multiplicity 1 in Q must have x1 (or x4) in common. Since G is not a P4, we get (2). Now, let e ˆ fx1; x2; x3; x4g be a hyperedge and let Q be the P4 induced by e. First, if x1; x4 is the only pair with multiplicity 1, then by (1), x1 and x4 must be the endpoints of Q, hence x2x3 is an edge of G. Second, assume that e is of type 2. By (2) we may assume that mH…x1; x3† ˆ mH…x1; x4† ˆ 1, say. Thus, x1 is an endpoint of Q. By (1) we may assume further, that x4 is the other endpoint, say. Then x1 and x3 are nonadjacent in G; otherwise 382 A. BrandstaÈdt et al.