My Problem
Whether there's an efficient algorithm to find a max-weight (or min-weight) k-clique in a complete k-partite graph (a graph in which vertices are adjacent if and only if they belong to different partite sets according to wikipedia)?
More Details about the Terms
Max-weight Clique: Every edge in the graph has a weight. The weight of a clique is the sum of the weights of all edges in the clique. The goal is to find a clique with the maximum weight.
Note that the size of the clique is k which is the largest possible clique size in a complete k-partite graph.
What I have tried
I met this problem during a project. Since I am not a CS person, I am not sure about the complexity etc.
I have googled several related papers but none of them deals with the same problem. I have also programmed a greedy algorithm + simulated annealing to deal with it (the result seems not good). I have also tried something like Dynamic Programming (but it does not seem efficient). So I wonder whether the exact optimal can be computed efficiently. Thanks in advance.
EDIT Since my input can be really large (e.g. the number of vertices in each clique is 2^k), I hope to find a really fast algorithm (e.g. polynomial of k in time) that works out the optimal result. If it's not possible, can we prove some lower bound of the complexity?
The maximum clique problem in a weighted graph in general is intractable. In your case, if the graph contains N nodes, you can enumerate through all possible k-cliques in N ** k time. If k is fixed (don't know if it is), your problem is trivially polynomially solvable, as this is a polynomial in N. I don't believe the problem to be tractable if k is a free parameter because I can't see how the assumption of a k-partite graph would make the problem significantly simpler from the general one.
How hard your problem is in practice depends also on how the weights are distributed. If all the weights are very near to each others, i.e. the difference between "best" and "good" is relatively small, the problem is very hard. If you have wildly different weights on the edges, the problem can be easier, because a greedy algorithm can give you a good "initial" solution, and you can use that and subsequent good solutions to limit your combinatorial search using the well-known branch-and-bound method.