Unidentified Combinatorial Problem - MathOverflow

Unidentified Combinatorial Problem

2011-02-28T12:28:51Z

Given a (finite, simple, undirected) graph $\mathcal{G} = (V, E)$, an edge binning associates each $e_{ij} \in E$ with one or the other of its vertices $v_i, v_j \in V$. Let $c_i$ be the number of edges associated with vertex $v_i$ in a given edge binning. Find an edge binning such that $\max_{v_i \in V}(c_i)$ is minimized.

Is this (or its dual) a well-known problem, or reducible to a well-known problem?

Edit:

The proper formal problem statement follows (derived from Asahiro 2009), with $d^+(u)$ denoting the outdegree of vertex $u$.

Minimum Maximum Outdegree: Given a finite, simple, undirected graph $\mathcal{G}= (V, E)$, find an orientation $\Lambda$ of $\mathcal{G}$ that minimizes $\max_{u \in V}[d^+_\Lambda(u)]$.

This can equivalently be stated in terms of indegree.

Note that Asahiro et al. primarily study the problem involving a weighted graph and weighted outdegree, which is generally NP-hard.

Answer by Gjergji Zaimi for Unidentified Combinatorial Problem

2011-02-28T13:14:36Z

This problem is equivalent to the graph orientation problem also known as the graph balancing problem. One is given an undirected graph and has to give an orientation of the edges which minimizes the maximum out-degree. If this value is $k$, then the graph is called $k$-orientable. Here are some articles on the topic "Graph Orientation Algorithms to Minimize the Maximum Outdegree", and "A note on graph balancing problems with restrictions".

Answer by Fedor Petrov for Unidentified Combinatorial Problem

2011-02-28T14:01:12Z

Let me also state an explicit criterion of the existence of orientation with out-degrees at most $d$: any induced subgraph on some, say, $k$ vertices contains at most $dk$ edges. This is clearly necessary, and the proof that it is sufficient is not hard: orient edges arbitrarily and consider the following procedure.

If out-degree of some vertex $a$ is at least $d+1$, then consider the set of vertices $x$, for which there exist oriented path from $a$. If all out-degrees of such vertices are at least $d$, then the set of them contradicts to our assumption. If deree of $x$ is less then $d$, then invert all edges on the path from $a$ to $x$.

Repeating this stuff we kill all high (more then $d$) out-degree after a fnite number of steps.