# Complexity of the min edge-colored cut problem

Given an undirected graph $G=(V,E)$ with a color on each edge, the problem is to find a 2-partition $(V_1,V_2)$ of $V$ s.t. the number of colors used by the edges $uv, u \in V_1, v \in V_2$ is minimized.

For example, a cut with 1 blue edge, 1 red edge, and 1 yellow edge between $V_1$ and $V_2$ is worst than a cut with 10 blue edges and 15 red edges between $V_1$ and $V_2$.

Of course, the uncolored cut is well-known to be polynomial-time solvable. The complexity of this colored version is reported as an open-problem in http://hal.archives-ouvertes.fr/hal-00371100/en. It surprise me. Do I miss some known result?

• What is exactly the function that is used to calculate the cost of the cut? (is it equal to the number of colors involved in the cut?) – Marzio De Biasi Jan 12 '12 at 16:14
• @Vor : exactly. – richard Jan 12 '12 at 16:23
• is the number of colours fixed? If it is, then the question is poly-time solvable, but the most obvious approach has a factor of $k!$ where $k$ is the number of colours. – Artem Kaznatcheev Jan 12 '12 at 17:38
• I am not sure what the approach that has a factor of $k!$ is, but it's easy to get an algorithm that depends on $k$ as $2^k$: guess the colors in the optimal solution, set the edges with the guessed colors to have weight 0 and the edges with the remaining colors to have infinite weight, and run a min cut algorithm. – Sasho Nikolov Jan 12 '12 at 17:53
• What is the constraint on the cut? Is it just neither V1 nor V2 can be empty, or does it have to be an (s,t)-cut for some given vertices s and t? (Needless to say, if there is no constraint, setting V1=∅ and V2=V is trivially optimal.) – Tsuyoshi Ito Jan 13 '12 at 11:54