I am wondering if the following problem has a name, or any results related to it.

Let $G = (V,w)$ be a weighted graph where $w(u,v)$ denotes the weight of the edge between $u$ and $v$, and for all $u,v \in V$, $w(u,v) \in [-1,1]$. The problem is to find a subset of vertices that maximizes the sum of the weights of the edges adjacent to them: $$\max_{S \subseteq V} \sum_{(u,v) : u \in S\ \textrm{or}\ v\in S} w(u,v)$$ Note that I am counting edges both that are inside the subset and that are outside the subset, which is what distinguishes this problem from max-cut. However, even if both $u$ and $v$ are in $S$, I only want to count the edge $(u,v)$ once (rather than twice), which is what distinguishes the objective from merely being the sum of the degrees.

Note that the problem is trivial if all edge weights are non-negative -- simply take the whole graph!

  • $\begingroup$ Your definition doesn't match your note later on about not counting duplicate edges. Are you summing over ordered pairs or 2-element subsets ? (the latter would give you the property you need, I think) $\endgroup$ Apr 28, 2011 at 21:28
  • 1
    $\begingroup$ Another note: the only edge weights NOT counted are those inside V \ S. Are you interested in hardness results or approximations, because in the former case, minimizing the sum of edge weights inside S' = V \ S might be the more natural problem. $\endgroup$ Apr 28, 2011 at 21:31
  • $\begingroup$ @Suresh: The formal definition in the question is correct as long as approximation ratio is concerned. It just gives exactly the twice of what one expects from the words. $\endgroup$ Apr 28, 2011 at 22:08
  • $\begingroup$ @TsuyoshiIto: oh I see, because edges across the cut are also counted twice. $\endgroup$ Apr 28, 2011 at 22:09
  • 1
    $\begingroup$ The exact problem is NP-hard because, as Suresh wrote in his comment, the problem is equivalent to the unrestricted {0,1} quadratic programming, which is NP-hard. $\endgroup$ May 1, 2011 at 13:49

2 Answers 2


Not really a solution but some observations.

This is a special case of the following problem: given a universe $U = \{1, \ldots, m\}$, and a collection of sets $S_1, \ldots, S_n \subseteq U$, and a weight function $w:U \rightarrow [-1, 1]$, find the set $I \subseteq [n]$ such that $w(\bigcup_{i \in I}{S_i})$ is maximized (the weight of a set is the total weight of its elements). Your problem corresponds to the case where each element of $U$ appears in exactly two sets (but I am not sure how to exploit this restriction, although it might help).

This is a coverage problem: similar to Max-k-Set-Cover, but without the restriction to use $k$ sets and with negative weights allowed. The greedy approximation of Max-k-Set-Cover (at each step output the set that has the largest weight of uncovered elements so far) outputs a sequence of sets such that the first $k$ sets are a $1 + 1/e$ approximation to the optimum (so this is a simultaneous approximation for all $k$). Unfortunately, as usual, there is a problem with analyzing it when weights could be negative. The basic observation of the analysis of the greedy algorithm is that if $S_1$ is the first set that is output, then $w(S_1) \geq OPT_k/k$ ($OPT_k$ being the maximum weight covered by $k$ sets), because $OPT_k$ is less than the sum of the weights of the $k$ sets in the optimal solution, and each of them has weight less than $w(S_1)$. However, with negative weights it's no longer true that $OPT_k$ is less than the sum of the weights in the optimal solution. In general, a union bound is no longer true.


FWIW, your problem is hard to approximate within a multiplicative factor of $n^{1-\epsilon}$ for any $\epsilon>0$.

We show that below by giving an approximation-preserving reduction from Independent Set, for which the hardness of approximation is known.

Reduction from Independent Set

Let undirected graph $G=(V,E)$ be an instance of Independent Set. Let $d_v$ denote the degree of vertex $v$ in $G$. Let $n$ be the number of vertices in $G$.

Construct edge-weighted graph $G'=(V',E')$ from $G$ as follows. Give each edge in $E$ weight 1. For each non-isolated vertex $v\in V$, add $d_v-1$ new edges, each with weight $-1$, to $d_v-1$ new vertices. For each isolated vertex $v\in V$, add one new edge of weight 1 to a new vertex.

(Note: each new vertex (in $G'$ but not $G$) has exactly one neighbor, which is in $G$.)

Lemma. $G$ has an independent set of size $k$ iff $G'$ (as an instance of your problem) has a solution of value at least $k$.

Proof. Let $S$ be any independent set in $G$. Then, since the vertices in $S$ are independent in $G'$, the value of $S$ in $G'$ (by your objective) is $$\sum_{v\in S} d_v - (d_v-1) ~=~ |S|.$$

Conversely, let $S$ be a solution for $G'$ of value at least $k$. Without loss of generality, assume $S$ contains no new vertices. (Each new vertex $v'$ is on a single edge $(v',v)$. If $v$ was not isolated in $G$, then the weight of the edge is $-1$, so removing $v'$ from $S$ increases the value of $S$. If $v$ was isolated, then the weight of the edge is 1, so removing $v'$ from $S$ and adding $v$ maintains the value of $S$.)

Without loss of generality, assume that $S$ is an independent set in $G$. (Otherwise, let $(u,v)$ be an edge such that $u$ and $v$ are in $S$. The total weight of $v$'s incident edges in $G'$ is $d_v - (d_v-1) = 1$, so the total weight of $v$'s incident edges other than $(u,v)$ is at most zero. Thus, removing $v$ from $S$ would not increase the value of $S$.)

Now, by the same calculation as at the start of the proof, the value of $S$ is $|S|$. It follows that $|S| \ge k$. QED

As an aside, you might ask instead for an additive approximation, of, say, $O(n)$ or $\epsilon m$.

It seems possible to me that for your problem even deciding whether there is a positive-value solution could be NP-hard.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.