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Suppose I have a set of unique elements $A=\{a_1, a_2, ..., a_n\}$. Suppose I also have a metric function $f:A\times A \rightarrow R^+$. I want to choose $k$ elements from $A$ (i.e. $a_{i_1},a_{i_2}, ..., a_{i_k}$, $I=\{i_1,...,i_k\}$) such that it maximizes the sum:

$$S=\sum _ {i,.j\in I, i\neq j}^{}{f(a_i,a_j)}$$

Is there any known algorithm to do something like this? I know the Hungarian Algorithm can be used to solve the case of $k=n$. Can it be adapted for $k

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    The Hungarian algorithm cannot solve the case $k=n$. The case $k=n$ is trivial since there is only one way to select $n$ elements from $A$. – 2017-01-20

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This problem is NP-hard. The decision problem of max-clique is NP-complete. Deciding whether a clique of size $k$ exists in a graph of $N$ vertices can be translated to your problem as follows. Let $A$ be the vertex set, and let $f(a_i,a_j)=1$ if the vertices are adjacent, $f(a_i,a_j)=0$ otherwise. A clique of size $k$ exists if and only if the optimal value of $S$ is $0.5 k (k-1)$.

This doesn't mean that the problem is not easy for some values of $k$ (such as $k=1$) but that an algorithm that works for any $k$ can probably not solve it in polynomial time.

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    how you use the metric function in your translation? – 2017-01-21
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    I define the metric in my translation, as is typical in proofs where you show that an NP-complete problem is a special case of the problem at hand. Whoever downvoted me, please indicate why my solution is [an egregiously sloppy, no-effort-expended post, or is clearly and perhaps dangerously incorrect](http://stackoverflow.com/help/privileges/vote-down). – 2017-01-21