Algorithm to Maximize Degree Centrality of Subgraph

Question

Say I have some graph with nodes and undirected edges (the edges may have a weight associated to them).

I want to find all (or at least one) connected subgraphs that maximize the sum of the degree centrality of all nodes in the subgraph (the degree centrality is based on the original graph) under the constraint that the sum of the weighted edges is < X.

Is there an algorithm that will do this?

Answer 1

A quick search took me to this description of degree centrality. It turns out that the "degree centrality" of a vertex is simply its degree (neighbour count).

Unfortunately your problem is NP-hard, so it's very unlikely that any algorithm exists that can solve every instance quickly. First notice that, assuming edge weights are positive, the edges in any optimal solution necessarily form a tree, since in any non-tree you can delete at least 1 edge without destroying connectivity, and doing so will decrease the total edge weight of the subgraph. (So, as a positive spinoff: If you compute the minimum spanning tree of your input graph and find that it happens to have total weight < X, then you can simply include every vertex in the graph in your solution.)

Let's formulate a decision version of your problem. Given a graph G = (V, E) with positive (I'll assume) weights on the edges, a number X and a number Y, we want to know: Does there exist a connected subgraph G' = (V', E') of G such that the sum of the edge weights in E' is at most X, and the sum of the degrees of V' (w.r.t. G) is at least Y? (Clearly this is no harder than your original problem: If you had an algorithm to solve your problem, then you could just run it, add up the degrees of the vertices in the subgraph it found and compare this to Y to answer "my" problem.)

Here's a reduction from the NP-hard Steiner Tree in Graphs problem, where we are given a graph G = (V, E) with positive weights on the edges, a subset S of its vertices, and a number k, and the task is to determine whether it's possible to connect the vertices in S using a subset of edges with total weight at most k. (As I showed above, the solution will necessarily be a tree.) If the sum of all degrees in G is d, then all we need to do to transform G into an input for your problem is the following: For each vertex s_i in S we add enough new "ballast" vertices that are each connected only to s_i, via edges with weight X+1, to bring the degree of s_i up to d+1. We set X to k, and set Y to |S|(d+1).

Now suppose that the solution to the Steiner Tree problem is YES -- that is, there exists a subset of edges having total weight <= k that does connect all the vertices in S. In that case, it's clear that the same subgraph in the instance of your problem constructed above connects (possibly among others) all the vertices in S, and since each vertex in S has degree d+1, the total degree is at least |S|(d+1), so the answer to your decision problem is also YES.

In the other direction, suppose that the answer to your decision problem is YES -- that is, there exists a subset of edges having total weight <= X ( = k) that connects a set of vertices having total degree at least |S|(d+1). We need to show that this implies a YES answer to the original Steiner Tree problem. Clearly it suffices to show that the vertex set V' of any subgraph satisfying the conditions above (i.e. edges have total weight <= k and vertices have total degree >= |S|(d+1)) contains S (possibly among other vertices). So let V' be the vertex set of such a solution, and suppose to the contrary that there is some vertex u in S that is not in V'. But then the largest sum of degrees that we could possibly make would be to include all other non-ballast vertices in the graph in V', which would give a degree total of at most (|S|-1)(d+1) + d (the first term is the degree sum for the other vertices in S; the second is an upper bound on the degree sum of all non-S vertices in G; note that none of the ballast vertices we added in could be in the subgraph, because the only way to include any of them is to use an edge of weight X+1, which we obviously can't do). But clearly (|S|-1)(d+1) + d = |S|(d+1) - 1, which is strictly less than |S|(d+1), contradicting our assumption that V' has a degree total at least |S|(d+1). So it follows that S is a subset of V', and thus that it is possible to use the same subset of edges to connect the vertices in S for a total weight of at most k, i.e. that the answer to the Steiner Tree problem is also YES.

So a YES answer to either problem implies a YES answer to the other one, in turn implying that a NO answer to either implies a NO answer to the other. Thus if it were possible to solve the decision version of your problem in polynomial time, it would imply a polynomial-time solution to the NP-hard Steiner Tree in Graphs problem. This means the decision version of your problem is itself NP-hard, and so is the optimisation version (which as I said above is at least as hard). (The decision form is also NP-complete, since a YES answer can be easily verified in polynomial time.)

---

Sidenote: At first I thought I had a very straightforward reduction from the NP-hard Knapsack problem: Given a list of n weights w_1, ..., w_n and a list of n profits p_1, ..., p_n, make a single central vertex c, and n other vertices v_1, ..., v_n. For each v_i, attach it to c with an edge of weight w_i, and add p_i other leaf vertices, each attached only to v_i with an edge of weight X+1. However this reduction doesn't actually work, because the profits can be exponential in the input size n, meaning that the constructed instance of your problem might need to have an exponential number of vertices, which isn't allowed for a polynomial-time reduction.