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Given two problems $P_1$ and $P_2$. $P_1$ is NP-complete in the strong sense and we want to prove that $P_2$ is also NP-complete but the reduction from $P_1$ to $P_2$ is not polynomial. Can we say that $P_2$ is NP-complete?

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No. As a counterexample pick any non-trivial problem $P_2$, i.e., a problem with at least one yes instance $I_{\text{yes}}$ and at least one no instance $I_{\text{no}}$.

To reduce an instance $I_1$ of $P_1$ to an instance $I_2$ of $P_2$ first solve $I_1$ (e.g., by brute force). If the answer to $I_1$ is yes, then let $I_2 = I_{\text{yes}}$, otherwise $I_2 = I_{\text{no}}$.

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  • $\begingroup$ what de you mean please by yes instance and no instance? $\endgroup$ – Farah Mind Apr 7 at 23:39
  • $\begingroup$ and what is the relation between the fact that $I_2=I_{yes}$ and the non polynomial reduction? $\endgroup$ – Farah Mind Apr 7 at 23:45
  • $\begingroup$ The argument in my answer shows that, if you don't require reductions to have a polynomial running time, you can reduce any decision problem $P_1$ to any non-trivial decision problem $P_2$ (regardless of whether $P_2$ is NP-complete). This disproves the fact that such a reduction from a NP-complete problem $P_1$ implies that $P_2$ must also be NP-complete. $\endgroup$ – Steven Apr 7 at 23:56
  • $\begingroup$ A decision problem is a triple $P=\langle\mathcal{I},S,\pi\rangle$, where $\mathcal{I} \subseteq\{0,1\}^*,S:\mathcal{I}\to\{0,1\}^*$, and $\pi:\mathcal{I}\times S\to\{ \text{true},\text{false}\}$. See, e.g., the definition in Bovet, Crescenzi "Introduction to the theory of complexity" (freely available). Solving $P$ means deciding whether the set $\mathcal{S}(I)=\{\sigma\in S(I)\wedge\pi(I,\sigma) = \text{true}\}$ is non-empty. By yes-instance of $P$ I mean an instance $I\in \mathcal{I}$ such that $\mathcal{S}(I)\neq\emptyset$. $\endgroup$ – Steven Apr 8 at 0:00

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