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Let us say we have $m$ bins and $n$ balls. Every bin $i$ has capacity $c_i$ which is the number of balls that can be put into bin $i$. We have $c_i\geq1$ for all $i$. For each bin $i$, there is a collection of sets $S_i=\{X_1,X_2,\ldots,X_{k_i}\}$ for given $k_i$. Each $X_j\in S_i$ is the set of balls that can be put into bin $i$. We have $|X_j|\leq c_i$ and $\emptyset\in S_i$ for all $i$.

For example, for $m=2$ and $n=3$, with $c_1=1$ and $c_2=2$, say we have $k_1=4$ and $k_2=5$. Say we have $S_1=\{\emptyset,\{1\},\{2\},\{3\}\}$. $S_2=\{\emptyset,\{1\},\{2\},\{3\},\{2,3\}\}$. This means that ball $1$, $2$ or $3$ can be each assigned to bin $1$. Also, each ball can be assigned to bin $2$. Further, balls $2$ and $3$ can be together assigned to bin $2$. We might have an instance with $k_2=6$ and $S_2=\{\emptyset,\{1\},\{2\},\{3\},\{2,3\},\{1,2\}\}$ for example.

We want to assign the maximum number of balls into the bins. Is this easy or hard?

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  • $\begingroup$ If $S_i$ is a set of sets, have you tried to prove it NP-hard? What partner problems have you looked at? $\endgroup$ – D.W. Nov 1 at 17:59
  • $\begingroup$ @D.W. See my edits $\endgroup$ – zdm Nov 2 at 1:23
  • $\begingroup$ I'm still not clear on what the problem statement is. I'd still like to see a general specification of what assignments are legal, and how this relates to the $S_i$. Two examples are not a substitute for a general specification. Also, are you sure you mean $\{\{2,3\}\}$ and not $\{\{2\},\{3\}\}$? (Yet another reason we need a general specification.) I don't know what $c_i$-admissible or 2-admissible means; please define all non-standard terminology. Also, I still don't see any indication of what you have tried or whether you have tried to prove the problem NP-hard and if so how. $\endgroup$ – D.W. Nov 2 at 1:43
  • $\begingroup$ If $S_i=\{X_1, X_2, \ldots, X_{k_i}\}$, then each $X_j$ is a set of balls that can be put together into bin $i$. $\endgroup$ – zdm Nov 2 at 2:03
  • $\begingroup$ That doesn't match your first example. $\endgroup$ – D.W. Nov 2 at 2:04
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You can reduce the Exact Cover problem to your problem. The elements in the Exact Cover problem corresponds to the balls in your problem. For each subset $T$ in the Exact Cover problem, we construct a bin $i$ with $S_i=\{\emptyset, T\}$. Then there exists an exact cover if and only if all balls can be put into the bins. Hence, your problem is NP-hard.

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  • $\begingroup$ Thanks, very interesting! For what values of $c_i$, is it still NP-hard? If $c_i=2$? $\endgroup$ – zdm Nov 2 at 5:27
  • $\begingroup$ @zdm It is still NP-hard by a reduction from Rainbow Matching. $\endgroup$ – xskxzr Nov 2 at 6:39
  • $\begingroup$ So, can we say that exact cover problem is NP-hard even if each subset has size at most 2? $\endgroup$ – zdm Nov 4 at 14:22
  • $\begingroup$ If we add the constraint that if $X_j\in S_i$ then all subsets of $X_j$ are also in $S_i$, can we still prove NP-hardness? If balls $\{2,3\}$ can be put into bin $i$ then ball $2$ can be put alone and also ball $3$ can be put alone. $\endgroup$ – zdm Nov 4 at 14:49
  • $\begingroup$ The hardness of this problem comes from the constraint that you can only select one set from $S_i$. Exact 2-set cover has no such constraint. Comments are not for extended disccusion, you can ask a new question. $\endgroup$ – xskxzr Nov 4 at 16:40

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