PROBLEM SECTION Splitting digraphs Summary: PROBLEM SECTION Splitting digraphs Noga Alon There are several known results asserting that undirected graphs can be partitioned in a way that satisfies various imposed constrains on the degrees. The corresponding results for directed graphs, where degrees are replaced by outdegrees, often fail, and when they do hold, they are usually much harder, and lead to fascinating open problems. In this note we list three problems of this type, and mention the undirected analogs. All graphs and digraphs considered here are simple, that is, they have no loops and no multiple edges. Minimum degrees A result of Steibitz [7] asserts that if the minimum degree of an undirected graph G is d1 + d2 + . . . + dk + k - 1, where each di is a non-negative integer, then the vertex set of G can be partitioned into k pairwise disjoint sets V1, . . . , Vk, so that for all i, the induced subgraph on Vi has minimum degree at least di. This is clearly tight, as shown by an appropriate complete graph. The analogous problem for directed graphs seems more difficult. For non-negative integers d1 d2 . . . dk, let F(d1, d2, . . . , dk) denote the minimum number F (if it exists), such that if the minimum outdegree of a directed graph D is F, then the vertex set of D can be partitioned into k pairwise disjoint sets V1, . . . , Vk, so that the induced subdigraph of D on Vi has minimum outdegree at least di. If there is no such finite F, define F(d1, d2, . . . , dk) = . When d1 = d2 = . . . = dk = d, denote F(d1, d2, . . . , dk) by Fk(d). It is easy to see that for Collections: Mathematics