PERFECT TRIANGLE FAMILIES ZSOLT TUZA

THEOREM 3. IfG is planar andT c T(G), then a collection o/v(T) mutually edgedisjoint triangles ofT and a z(T)-element edge set containing at least one edge of each Te T can be found in polynomial time.
d(W) = T(T').
Since a(H') = #(H') holds for every H' < H if and only if the chromatic number = of each H' is equal to its maximum clique size [2], the above equalities imply that Theorem 2 is valid for T if and only if H is a perfect graph. 3. Observe next that K4 — e (an edge deleted from the complete graph of order 4) is not an induced subgraph of H. Indeed, suppose for a contradiction that four vertices a, b, c, d induce K4 — e in H, c and d being non-adjacent. The sets {a,b, c) and {a, b,d] induce triangles in H, and therefore 1 implies that there exist edges uv = a n b D c and xy = a 0 b n d in G. Since cd is not an edge of H, the cyclic triangle corresponding to c in G does not contain xy, and the cyclic triangle belonging to d does not contain uv. Hence xy # uv. On the other hand, uv and xy are contained in a D b, implying that those two cyclic triangles have the same vertex set in G. Hence the contradiction a = b follows. 4. It is known (see [4, 5]) that the Strong Perfect Graph Conjecture is true for the class of (K4 — e)-free graphs; consequently, 3 yields that H is perfect if and only if neither H nor its complement contains any odd cycle Ck of length k ^ 5 as an induced subgraph. Since the complement of Cfc contains an induced K 4 - e whenever k ^ 7 (and C5 is self-complementary), it suffices to show that every induced cycle of length ^ 4 in H is even. 5. Fix a planar embedding of G. By the region of a triangle TeT(G) we mean the finite region surrounded by the edges of Tin the plane. By the planarity of G, no two of its triangles T, T can overlap; that is, if their regions share an internal point, then one of T and T' is entirely contained in the other. We denote by T < V the situation where the region of 7" contains that of T, and write T\\ T' if the two regions are internally disjoint. Since the boundary of T is a closed curve, it separates the edges of all triangles T' <T from the edges of all r"eT(G) for which T" < T does not hold. Here ' separation' is meant in a weak sense, namely if T' n T" # 0 , then T' n T" is an edge (or a vertex) of T. The triangles reT(G) may have clockwise or counterclockwise orientations (determined by the embedding of G). Suppose that Tand 7'eT(G) share an edge. If r|| T', then precisely one of T and T has a clockwise orientation, and if T < T', then the orientations of T and T' are the same. 6. Recalling the conclusion of 4, let 7^,..., Tk be a sequence of cyclic triangles in G, corresponding to an induced cycle in H; that is, Tt shares an edge with 7J (/
ZSOLT TUZA
2. Define the 'intersection graph' H of T in the following way. Assign a distinct vertex vT to each TeJ, and join vT to vT. {T # T'eT) if and only if T and 7" share an edge. The subcollections T' c: T are in one-to-one correspondence with the induced subgraphs H' c H. Denote by a(H') the independence number of H' (the maximum number of pairwise non-adjacent vertices) and by #(H') the clique covering number (the minimum number of complete subgraphs whose union contains all vertices of H'). The definitions immediately imply a(H') = v(T). Moreover, 1 above yields that the edges contained in more than one triangle of T' are in one-to-one correspondence with those maximal complete subgraphs of H' which are not isolated vertices in H'. Thus
There are several related problems which remain open; they are discussed in the concluding section. 2. The Proof Let T be a collection of cyclic triangles in a planar oriented graph G. We prove that T(T) = v(T). 1. We begin with the observation that v(T') = 1 implies T(T') = 1 for every T' cz T (that is, the cyclic triangles of an oriented graph always satisfy the Helly property). Indeed, let T, T'eT be two triangles sharing an oriented edge e = uv. Then in the subgraph formed by the four edges of (T\j T') — e, the out-degree of u as well as the in-degree of v is zero, so that every triangle T"eT' has to contain e (for otherwise T" would be edge-disjoint from T or T).