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I would like to show that Quadratic Programming is NP-hard.

I am currently reading a couple of papers which state that QP is NP-Hard and prove it by transforming SAT to QP, however I am finding the diction quite tough since I am just a beginner in the field. Would anyone happen to know the answer to this question who can maybe explain it to me in simpler terms?

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    $\begingroup$ Where should we start? Do you know anything about NP-completeness? Have you seen any examples of NP-completeness proof? $\endgroup$ – Yuval Filmus Nov 12 '13 at 9:22
  • $\begingroup$ Yes I have a good background on NP-Completeness and how a problem is proven to be NP-complete. My question is more specifically about this particular reduction. I have a feeling that I have not understood QP 100% and I am missing some small thing somewhere. $\endgroup$ – lvella Nov 12 '13 at 9:30
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    $\begingroup$ Can you give a reference to what Quadratic Programming is? $\endgroup$ – Raphael Nov 12 '13 at 16:53
  • $\begingroup$ Raphael, by Quadratic Programming I was referring to optimising a quadratic function which is subject to several linear constraints. $\endgroup$ – lvella Nov 12 '13 at 22:04
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The reduction of SAT to quadratic programming is pretty simple. The idea is that we can use the quadratic objective to "force" the variables to be Boolean, and so we can implement an integer linear program whose feasibility is equivalent to the SAT instance.

Given a SAT instance on a set of variables $x_1,\ldots,x_n$, we define the following quadratic program: $$ \begin{align*} & \min \sum_{i=1}^n x_i - \sum_{i=1}^n x_i^2 \\ \text{s.t.} & \\ & 0 \leq x_i \leq 1 & 1 \leq i \leq n \\ & x_{i_1} + \cdots + x_{i_j} + (1 - x_{k_1}) + \cdots + (1 - x_{k_\ell}) \geq 1 & \forall \text{ cl. } x_{i_1} \lor \cdots \lor x_{i_j} \lor \bar{x}_{k_1} \lor \cdots \lor \bar{x}_{k_l} \end{align*} $$ The linear constraints have a $0,1$ solution if and only if the SAT instance is satisfiable. Since $0 \leq x_i \leq 1$, a solution is $0,1$ if and only if $\sum_{i=1}^n x_i - \sum_{i=1}^n x_i^2 = 0$, and for any other solution $\sum_{i=1}^n x_i - \sum_{i=1}^n x_i^2 > 0$. Therefore the SAT instance is satisfiable if and only if the minimum is $0$ (or at most $0$).

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    $\begingroup$ They represent a clause. For example, the clause $x_7 \lor x_{18} \lor \bar{x}_9$ will result in the inequality $x_7 + x_{18} + (1-x_9) \geq 1$. $\endgroup$ – Yuval Filmus Nov 12 '13 at 21:37
  • $\begingroup$ Will it work to convert $3SAT$ to MAXCUT or INDSET preserving number of solutions (standard text reductions look like they will preserve solution count) and utilize resulting quadratic objective? $\endgroup$ – T.... May 16 '18 at 10:49

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