Let $G = (V,E)$ be an undirected graph, with $e \in E$ has positive weight $w_e$.

Given a set of integers $I = \{i_1,\dots,i_n\}$ such that $\sum_{k=1}^n i_k = |V|$. I want to find a partition $P$ of $G$ of cardinality $n$ such that each element of the partition has cardinality corresponding exactly to one of the elements of $I$.

The maximization function sums all the edges weights $w_{(v_1,v_2)}$ if both $v_1$ and $v_2$ belong the same subset within the solution - i.e. $v_1, v_2 \in P_j$, for any $P_j \in P$.

Similarly, the objective function might be minimizing all the weights $w_{(v_1,v_2)}$, where $v_1$ and $v_2$ belong different element of the partition.

My question is whether such a problem has already an official formulation in literature. Or if there is an interesting equivalent problem.

  • 1
    $\begingroup$ The question confuses me. I'll take a guess. Are you looking for the max-k-cut problem? Goldschmidt, O., & Hochbaum, D. S. (1994). A polynomial algorithm for the k-cut problem for fixed k. Mathematics of operations research, 19(1), 24-37. Closely related is correlation clustering: Demaine, E. D., Emanuel, D., Fiat, A., & Immorlica, N. (2006). Correlation clustering in general weighted graphs. Theoretical Computer Science, 361(2-3), 172-187. $\endgroup$
    – ktnr
    Commented May 18, 2021 at 17:47
  • $\begingroup$ Thank you. It is actually almost equivalent to a the capacitated max-k-cut problem. $\endgroup$ Commented May 18, 2021 at 21:17

2 Answers 2


When the objective function relies on the weights of edges within the subgraphs (as opposed to edges connecting subgraphs), I believe your problem is equivalent to a quadratic multiple knapsack problem. There seems to be a fair bit of literature on that problem (of all of which I am blissfully ignorant).

  • $\begingroup$ May I extend my question to the case where knapsacks have dependencies among each other. E.g. a couple of knapsacks which together are worth more than another couple. Does this problem exists? $\endgroup$ Commented May 24, 2021 at 10:11
  • $\begingroup$ I'm not sure how to interpret that. In what I had in mind, the $k$-th knapsack would have capacity $i_k$, and $\sum_k i_k = \vert V \vert$ would imply that all knapsacks must be used (and filled to capacity). So I do not understand what you have in mind for knapsack dependencies. $\endgroup$
    – prubin
    Commented May 24, 2021 at 22:14
  • $\begingroup$ Indeed my reasoning was originally with partition of a graph. It is more intuitive my question by thinking that formulation. Hence, an edge can have a weight varying with the considered solution; namely, it can worth more when connecting two elements of the partition, instead of other twos. $\endgroup$ Commented May 25, 2021 at 9:26

The capacitated max-k-cut formulation models the problem I describe.

Besides, it is less constraining, since the equality $\sum_{k=1}^n i_k = |V|$ becomes $\sum_{k=1}^n i_k \geq |V|$, where elements $i_k$ are referred as capacity.


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