Red-Blue Clique Partitions and (1-1)-Transversals

Motivated by a problem of Gallai on (1-1)-transversals of 2-intervals, it was proved by the authors in 1969 that if the edges of a complete graph K are colored with red and blue (both colors can appear on an edge) so that there is no monochromatic induced C4 and C5 then the vertices of K can be partitioned into a red and a blue clique. Aharoni, Berger, Chudnovsky and Ziani recently strengthened this by showing that it is enough to assume that there is no induced monochromatic C4 and there is no induced C5 in one of the colors. Here we strengthen this result further, showing that it is enough to assume that there is no monochromatic induced C4 and there is no K5 on which both color classes induce a C5. We also answer a question of Kaiser and Rabinovich, giving an example of six 2-convex sets in the plane such that any three intersect but there is no (1-1)-transversal for them. 1 Red-blue clique partitions of complete graphs In 1968, thinking on a problem about piercing cycles in digraphs, Gallai arrived at the problem of piercing 2-intervals. He defined 2-intervals as subsets of the real line R having two interval components, one in (−∞, 0) and one in (0,∞) and asked: how many points are needed to pierce a family of pairwise intersecting 2-intervals? His question generated the paper [7] in which (as a special case of a general upper bound) we proved that two points always pierce pairwise intersecting 2-intervals and one of them can be selected from (−∞, 0) and the other from (0,∞). Let’s call such a pair of points a (1-1)-transversal. This result can be extended to 2-trees, where a 2-tree is the union of two subtrees, one a subtree of T1 the other a subtree of T2, where T1 and T2 are vertex-disjoint trees. In [7] we proved a stronger result (see Theorem 1 below) using only properties of the intersection graph of subtrees of a tree. Consider 2-colored complete graphs, where edges are colored with red, blue, or both colors. Edges of one color only are called pure edges and can be ∗Research supported in part by the OTKA Grant No. K104343. the electronic journal of combinatorics 23(3) (2016), #P3.40 1 pure red or pure blue. Another view is to consider a complete graph (clique) as the union of a red and a blue graph on the same vertex set. Theorem 1. (Gyárfás, Lehel [7], 1970) Assume that G is a 2-colored complete graph containing no monochromatic induced C4 and C5. Then V (G) can be partitioned into a red and a blue clique. Given a set of n pairwise intersecting 2-subtrees, one can represent their intersections by a 2-colored complete graph Kn. Then both colors determine chordal graphs i.e. graphs in which every cycle of length at least four has a chord. In particular, there is no monochromatic induced C4 and C5. Applying Theorem 1, the vertices of Kn can be partitioned into a red and a blue clique (empty sets or single vertices are accepted as cliques) and by the Helly-property of subtrees we have a (1-1)-transversal for the 2-subtrees. Thus Theorem 1 implies the following. Corollary 2. (Gyárfás, Lehel [7], 1970) Pairwise intersecting 2-subtrees have a (1-1)transversal. Since 2-colorings of complete graphs with pure edges only can be considered as a graph and its complement, furthermore, the complement of C4 is 2K2, we get another consequence of Theorem 1. Corollary 3. (Földes, Hammer [6], 1977) Assume that a graph G does not contain C4, 2K2, C5 as an induced subgraph. Then G is a split graph, i.e. its vertices can be partitioned into a clique and an independent set. The seminal paper of Tardos [11] (1995) introduced topological methods, he proved that 2-intervals without k+1 pairwise disjoint members have (k−k)-transversals. Further results were obtained by Kaiser [8] (1997), Alon [2, 3] (1998, 2002), Matousek [10] (2001), Berger [5] (2005), and this list of references is very far from being complete. In this note we only consider the graph coloring approach. Very recently Theorem 1 was generalized as follows. Theorem 4. (Aharoni, Berger, Chudnovsky, Ziani [1], 2015) Assume that G is a 2-colored complete graph such that there is no monochromatic induced C4 and there is no red induced C5. Then V (G) can be partitioned into a red and a blue clique. We show that the proof of Theorem 1 in [7] yields an even stronger result. Let K∗ 5 denote the 2-colored K5 where every edge is pure and both colors span a C5. Theorem 5. Assume that G is a 2-colored complete graph such that there is no monochromatic induced C4 and there is no K ∗ 5 . Then V (G) can be partitioned into a red and a blue clique. Proof. We prove the result by induction on |V (G)|. For 1 6 |V (G)| 6 3 the theorem is obvious. Fixing any p ∈ V (G), by the inductive hypothesis we have V (G − p) = R ∪ B where R and B are disjoint vertex sets spanning a red and a blue clique. the electronic journal of combinatorics 23(3) (2016), #P3.40 2