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in Polyhedral Study of the Cardinality Constrained Knapsack Problem the authors prove that the Cardinality Constrained Knapsack Problem is NP-Hard by reducing PARTITION to it.

Besides, it's easy to see that the KP problem is a special case of the QKP.

How should one proceed to prove the NP-Hardness of the Cardinality Constrained Quadratic Knapsack Problem?

CCQKP:

$$max\ \sum_{i=1}^{n} \sum_{j=1}^{n} x_i x_j c_{ij}$$ $s.t$ $$\sum_{j=1}^{n} a_j x_j \leq C$$ $$\sum_{j=1}^{n}x_j = 1$$ $$0 \leq x_j \leq 1,\ j = 1,...,n$$ $$\sum_{j=1}^{n}z_j = k$$ $$z_j \in \{0,1\},\ j = 1,...,n$$

I'm aware that we usually talk about the hardness of Decision Problems even tough I'm formalizing the Optimization version of it.

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  • $\begingroup$ Is there anything to do? The decision problem directly reduces to the optimization problem: if you want to know "Is there a widget with weight at least $w$?" you can just compute the maximum weight of any widget and answer "yes" if that is at least $w$. $\endgroup$ – David Richerby Oct 2 '18 at 14:57
  • $\begingroup$ I understand that the decision problem reduces to the optimization problem. What I'm rather interested in is how I can prove the hardness of the CCQKP. I'm sorry if the question is vague. $\endgroup$ – Arthur M Oct 2 '18 at 15:53
  • $\begingroup$ When we say that an optimization problem is NP-hard, we really mean that its decision version is NP-hard. For a maximization problem, the decision version gets an instance and a value $\theta$, and the goal is to decide whether there is a feasible solution of value at least $\theta$. $\endgroup$ – Yuval Filmus Oct 2 '18 at 16:26
  • $\begingroup$ @YuvalFilmus sure, I understand that. What I'm asking, though, is how I can prove that the CCQKP is NP-Hard given that both the CCKP and the QKP are NP-Hard. $\endgroup$ – Arthur M Oct 2 '18 at 22:40
  • $\begingroup$ If QKP is NP-hard then so is your problem - just don’t use any cardinality constraints. $\endgroup$ – Yuval Filmus Oct 2 '18 at 22:55
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The quadratic knapsack problem is NP-hard. Therefore, the cardinality-constrained quadratic knapsack problem is NP-hard, too (as Yuval says, just don't use any cardinality constraints).

If problem X is a special case of problem Y, and problem X is NP-hard, then so is problem Y. Now let X = CCQKP and Y = QKP.

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