2
$\begingroup$

Assume we are given a graph of $N$ vertices $V_1, \ldots, V_N$. For each vertex $V_i$, there is a corresponding weight $v_i$. For a graph of $N$ vertices, we want to find the minimum weight subgraph consisting of disjoint cliques which includes all vertices (that is, all vertices are present in the subgraph and each vertex is exactly in one clique) under two following conditions:

  1. For each clique (complete subgraph) of the original graph, the weight is defined as the maximum of the weights of the vertices of the clique. More specifically, if vertices $X_1, \ldots, X_k$ form a clique where $x_i$ is the weight of vertex $X_i$, then the weight of this clique is $\max\{x_1,\ldots,x_k \}$.
  2. The total weight of a subgraph with multiple cliques is the sum of the weights of the cliques.

I want to know whether this problem is NP-complete? Is there any approximation algorithm for this problem which runs in polynomial time of $N$.

1 Answers 1

2

The special case when each vertex weight $v_i$ is equal to $1$ sums up to solving the following problem: partition the vertices into cliques and minimize the number of cliques.

This is known as the Vertex Clique Cover problem (see here). [Edit: it is straightforward to see that Vertex Clique Cover can be reduced to your problem, and so it's NP-hard.]

Unfortunately, as stated on the Wikipedia page, the problem cannot approximated within a $n^{1 - \varepsilon}$ factor, and so there is polytime approximation algorithm achieving a reasonable factor.

If you have structural knowledge of the graph you are studying, that might help (e.g. I think if the degree is bounded, you may be able to approximate something).

  • 0
    Thanks for your response. So, the Vertex Clique Cover problem can be reduced to our problem where all vertices have weight 1. From, this we can conclude that our problem is NP-hard (and since there is a polynomial time verification for it, it is NP-complete). Am I right? – 2017-01-08
  • 1
    Yes, you're right! Sorry I forgot to mention it's NP-hard (and, as you say, NP-complete). – 2017-01-08
  • 0
    Thanks. Regarding whether the graph of this question has a specific structure, yes it has. The structure has been explained here: http://math.stackexchange.com/questions/1833883/a-graph-problem/2082662#2082662 I was guessing that because that question was long, it is better to ask it in a more general way again because it is no easy to read that question. – 2017-01-08
  • 0
    By the way, to show that problem A is NP-hard, knowing that problem B is NP-hard, if I am right we should reduce B to A, then it means if we can solve A, we can actually solve B, but we know that we cannot solve B (in polynomial time) unless P=NP. So, I guess your edit should be: "it is straightforward to see that Vertex Clique Cover can be reduced to your problem, and so your problem is NP-hard." Am I right? – 2017-01-08
  • 0
    Yes, the direction of the reduction is an extremely important point, and that was a dumb Saturday evening mistake from me! About your other post, the NP-hardness of the specific graphs you construct is certainly not implied by Vertex Clique Cover. In fact, I can't see an easy way of proving it's hard. My hunch is that it's in P if all weights are equal, but NP-hard if arbitrary weights are allowed. But that's just wild conjectuing. – 2017-01-08
  • 0
    One question, the Vertex Clique Cover problem is not NP-complete for all graphs? If it is, we can begin from a graph similar to my other question. Then, it can be reduced to a special case of my problem (all vertices with the same weights). It mean a special case of my problem is NP-hard, so the general case is definitely harder and again is NP-hard. Is this wrong? – 2017-01-08
  • 1
    No we can't assume it's NP-complete for all classes of graphs. For instance, Vertex Clique Cover is in P for all bipartite graphs (because it's the problem of finding a maximum matching). The graphs that are "hard" are those created in proof of hardness of Vertex Clique Cover. And also those graph classes for which a specific proof of hardness exists (for instance, I think that VCC on 3-regular graphs is NP-hard, but that required its own proof). – 2017-01-08
  • 0
    I just figured out that the graphs of my other post are chordal graphs and hence, they are perfect. We know that minimum Vertex Clique cover on perfect graphs can be solved in polynomial time. Do you think this can help to solve the question of my other post is polynomial time? – 2017-01-09
  • 0
    That's interesting! Well, at least you have an algorithm for the case when all weights are $1$. I don't know if this can help with the weighted version...maybe if you're really lucky, but it depends on the algorithm. Do you have a reference for the VCC algorithm on perfect graphs? – 2017-01-09
  • 0
    No, I found some references, but that's not something that I can understand. – 2017-01-10
  • 1
    It does seem that this paper could help you: https://www.zib.de/groetschel/pubnew/paper/groetschellovaszschrijver1981a.pdf. From what I understand, at the end of page 194, they find in polytime a minimum weight VCC on perfect graphs, where the weight is simply given for each clique (what I say here has to be verified though). But like you, I don't have time to understand it fully. Maybe it's something you can ask as another question. – 2017-01-10
  • 0
    I found the results in a book. Can you take a look here?http://math.stackexchange.com/questions/1833883/a-graph-problem – 2017-01-10
  • 0
    Hmm sorry I don't know how that particular method would extend to your problem. Also, you have taken out your graph description right? It's too bad - I would've liked to see it again. I don't see how they're perfect, but maybe I got the wrong def in memory. If you,re willing to someday you could put it back :) – 2017-01-13
  • 0
    Yes, I thought the long description makes it hard to understand. I have put it back. The graphs are perfect because they are chordal graphs. You can see it from the graphs' structures. I think if I show that the minimum weighted clique cover has a solution (among all possible solutions) where the cliques are disjoint, then the minimum weighted clique cover defined in the literature becomes exactly the problem I have. I asked this question here http://mathoverflow.net/questions/259460/minimum-disjoint-weighted-clique-covering Take a look if you can. thanks – 2017-01-13