Importance: High ✭✭✭
Author(s): Alon, Noga
Recomm. for undergrads: no
Posted by: fhavet
on: March 1st, 2013
Problem   Is there a minimum integer $ f(d) $ such that the vertices of any digraph with minimum outdegree $ d $ can be partitioned into two classes so that the minimum outdegree of the subgraph induced by each class is at least $ d $?

Thomassen [T] proved the conjecture when $ d=1 $ and showed $ f(1)=3 $. In fact, this case is equivalent to the case $ k=2 $ of the Bermond-Thomassen Conjecture.

The existence of the corresponding function $ f $ for the undirected analogue is easy and has been observed by many authors. Stiebitz [S] even proved the following tight result: if the minimum degree of an undirected graph $ G $ is $ d_1+d_2+ \cdots + d_k $, where each $ d_i $ is a non-negative integer, then the vertex set of $ G $ can be partitioned into $ k $ pairwise disjoint sets $ V_1,\dots , V_k $, so that for all $ i $, the induced subgraph on $ V_i $ has minimum degree at least $ d_i $. This is clearly tight, as shown by an appropriate complete graph.


*[A] Noga Alon, Disjoint Directed Cycles, Journal of Combinatorial Theory, Series B, 68 (1996), no. 2, 167-178.

[S] M. Stiebitz, Decomposing graphs under degree constraints, J. Graph Theory 23 (1996), 31-324.

[T] C. Thomassen, Disjoint cycles in digraphs, Combinatorica 3 (1983), 393 - 396.

* indicates original appearance(s) of problem.


Comments are limited to a maximum of 1000 characters.
More information about formatting options