Coloring 2-Intersecting Hypergraphs

A hypergraph is 2-intersecting if any two edges intersect in at least two vertices. Blais, Weinstein and Yoshida asked (as a first step to a more general problem) whether every 2-intersecting hypergraph has a vertex coloring with a constant number of colors so that each hyperedge has at least min{|e|, 3} colors. We show that there is such a coloring with at most 5 colors (which is best possible). A proper coloring of a hypergraph is a coloring of its vertices so that no edge is monochromatic, i.e. contains at least two vertices with distinct colors. It is well-known that intersecting hypergraphs without singleton edges have proper colorings with at most three colors. This statement is from the seminal paper of Erdős and Lovász [3]. Recently Blais, Weinstein and Yoshida suggested a generalization in [1]. They consider t-intersecting hypergraphs, in which any two edges intersect in at least t vertices and they call a coloring of the vertices c-strong if every edge e is colored with at least min{|e|, c} distinct colors. One of the problems they consider is the following. Problem 1. ([1]) Suppose that H is a t-intersecting hypergraph. Is there a (t + 1)strong vertex coloring of H where the number of colors is bounded by a function of t? In particular, is there a t + 1-strong vertex coloring with at most 2t + 1 colors? If true, it would be best possible, as the 2t-element sets of a 3t element set demonstrate. Notice that for t = 1 the answer to Problem 1 is affirmative (for both parts) according to the starting remark but open for t > 2 [1]. Our aim is to give an affirmative answer to both parts of the problem in case of t = 2. Notice that intersecting hypergraphs do not always have 3-strong colorings with any fixed number of colors: if every edge of a ∗Research coordinator of the junior author at the Elective Undergraduate Research Program of Budapest Semesters in Mathematics, 2013 Summer program the electronic journal of combinatorics 16 (2009), #R00 1 (k + 1)-chromatic graph is extended by the same new vertex, the resulting intersecting hypergraph has no 3-strong coloring with k colors. Thus the 2-intersecting condition is important in the following theorem. Theorem 2. Every 2-intersecting hypergraph G has a 3-strong coloring with at most five colors. We learned from a referee that a weaker form of Theorem 2 (with 21 colors instead of 5) is proved recently in [2]. We also prove a lemma that will be used in the proof of Theorem 2 but has independent interest. A hypergraph has property Pt for some integer t > 2 if any i edges intersect in at least t+ 1− i vertices, for all i, 2 6 i 6 t. Lemma 3. Suppose that H is a hypergraph with property Pt. Then H has a t-strong coloring with at most t+ 1 colors. Proof. Let H be a hypergraph with property Pt for t > 2. Select an edge e of H which is minimal for containment. Let F be the hypergraph defined on the vertex set of e with edge set {h ∩ e : h ∈ E(H)}. Color each vertex not in e with color t + 1. If t = 2, color the vertices of e arbitrarily using colors 1,2 (or just color 1 if e has just one vertex). If |e| = t − 1, color vertices of e by 1, 2, . . . , t − 1. Otherwise, since F has property Pt−1, we can find by induction a (t− 1)-strong coloring C on F with colors 1, 2, . . . , t. We may suppose that C uses all colors 1, 2, . . . , t on e, otherwise we may change some repeated colors to the missing colors maintaining the (t− 1)-strong coloring. Thus C colors e with at least t colors and, since for any other edge h ∈ H, |h ∩ e| > t− 1, C uses at least t− 1 colors on h ∩ e and h also has at least one vertex of color t + 1. Therefore we have a t-strong coloring of H with t+ 1 colors. It is worth noting that Lemma 3 does not hold if we require a t-strong coloring with at most t colors. Indeed, all t-sets of t+1 elements have property Pt but a t-strong coloring must use t+ 1 colors. Proof of Theorem 2. By the condition, there are no singleton edges. Also, if some edge e has just two vertices, coloring them with colors 1, 2 and all other vertices by 3, we obviously have a 3-strong coloring. Thus we may assume that every edge has at least three vertices, therefore a 3-strong coloring on the minimal edges of G is also a 3-strong coloring on G. Thus we may assume that G is an antichain. If any three edges of G have non-empty intersection, we can apply Lemma 3 and get a 3-strong coloring with at most 4 colors. Thus, we may suppose that G contains three edges with empty intersection, select them with the smallest possible union, let these edges be e1, e2, e3 and set X = e1 ∪ e2 ∪ e3. A vertex v ∈ X is called a private part of ei (i = 1, 2, 3) if v ∈ ei but v is not covered by any of the other two ej-s. We color the vertices in X as follows. The private parts of e1, e2, e3 (if they exist) are colored with 1, 2, 3 respectively. Notice that each intersection has at least two vertices, color e1 ∩ e3 with colors 1, 3 so that color 1 is used only once, color e1 ∩ e2 with colors 2, 4 so that color 2 is used only once. Vertices in e2 ∩ e3 are all colored with color 5. The coloring outside X varies according to the number of private parts of ei-s. the electronic journal of combinatorics 16 (2009), #R00 2 Case 1. Each ei has private parts, i = 1, 2, 3. Here we color vertices not covered by X one by one with 1 or 2 by the following greedy type algorithm: if an uncolored vertex w / ∈ X completes an edge f such that all vertices of f − {w} are colored with colors 2, 3 only (both present otherwise |f ∩ e1| 6 1 or |f ∩ e2| 6 1) then color w with color 1, otherwise color it with color 2. We claim that a 3-strong coloring is obtained. Suppose there is an edge fij with colors i, j only, 1 6 i < j 6 5. Edges f12, f14, f24 would intersect e3 in at most one vertex, edge f25 would intersect e1 in at most one vertex and f13 would not intersect e2 at all. Edges f35, f45 would form a proper subset of e3, e2, respectively, contradicting the antichain property. Edge f34 cannot exist because the triple f34, e2, e3 has no intersection and Y = f34 ∪ e2∪e3 is a proper subset of X because e1 has a private vertex. Thus we get a contradiction with the definition of e1, e2, e3. The same argument can be applied to exclude f15, f23 ⊂ X (with Y = f15 ∪ e1 ∪ e2, Y = f23 ∪ e2 ∪ e3 and using that e3, e1 have private vertices). Thus the only possibility is that there is an edge f15 or f23 with some vertex w / ∈ X. However, no such f15 exists since w / ∈ X is colored with 1 only if there exists edge f of G such that f − {w} is colored with colors 2, 3 only thus |f ∩ f15| = 1 contradiction. Moreover, no such f23 can exist either, because its vertex in V −X colored last got color 1 according to the rule governing Case 1. Case 2. Two of e1, e2, e3 have private parts, by suitable relabeling we may suppose that the private part of e2 is empty. In this case vertices not covered by X are colored with color 2 and claim that we have a 3-strong coloring. The nonexistence of f12, f13, f14, f24, f25 follow as in Case 1 and here f23 can be excluded the same way since |f23 ∩ e2| 6 1. The exclusion of f34, f35, f45 and f15 ⊂ X is also exactly the same as in Case 1. Thus here we have to exclude only the existence of an edge f15 containing some vertices w / ∈ X. However, this cannot happen since here every vertex outside X is colored with color 2. Case 3. Exactly one of e1, e2, e3 has a private part, by suitable relabeling we may suppose that it is e2. Here all vertices not covered by X are colored with 1. Edges f12, f13, f14, f15, f24, f25 are all excluded since there is some ei intersecting them in at most one vertex. The edges f34, f35, f45 are excluded since they are proper subsets of some ei. The only possible edge is f23 but in this case we can replace the triple e1, e2, e3 by the non-intersecting triple f23, e2, e3 which has the same union but they have two private parts: the vertices of color 4 in e2 and the vertex of color 1 in e3. This reduces Case 3 to Case 2. Case 4. None of the edges e1, e2, e3 have private parts. Vertices uncovered by X are colored with 1. Here f12, f13, f14, f15, f23, f24, f25 are all excluded since there is some ei intersecting them in at most one vertex. The other three edges f34, f35, f45 are excluded since they are proper subsets of some ei. In all cases we found a 3-strong coloring with at most five colors. the electronic journal of combinatorics 16 (2009), #R00 3