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Given
$$\mathrm{\#3SAT} = \{ (w, y) \mid w\text{ is a $\mathrm{3SAT}$ instance with at least $y$ satisfying assignments}\}\,,$$ prove that $\mathrm{\#3SAT}$ is NP-Hard.

I am currently stuck with this one.

Basically $\mathrm{\#3SAT}$ is the counting version of $\mathrm{3SAT}$ and to me it is seems clear that establishing membership in the language is more complex than establish the classic $\mathrm{3SAT}$ membership.

I'm trying to reason through the prover-verifier paradigm: let's say you have a $\mathrm{3SAT}$ instance, now suppose the certificate for that instance is $n$ bits. It is obvious that the certificate for the same instance but translated into $\mathrm{\#3SAT}$ with let's say $y=10$ would be at least $10n$ bits. So if the first can be examined in time $t$ then the second will take at least $10t$ and so on. However I am not satisfied with this reasoning, it seems incomplete and superficial. I would greatly appreciate your advice on how to proceed with this proof. Thanks in advance.

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It seems that you're trying to prove that $\#\mathrm{3SAT}$ is in $\mathrm{NP}$. Since $\#\mathrm{3SAT}$ is $\mathrm{\#P}$-complete, and $\mathrm{\#P}$ seems to be harder than anything in the polynomial hierarchy, it's very unlikely that $\#\mathrm{3SAT}\in\mathrm{NP}$.

The error in your attempted proof is that your certificate doesn't have polynomial length. A satisfying assignment has length $\Theta(|x|)$ (the formula can't have more variables than its length, but it could be of the form $(x_1\lor x_2\lor x_3) \land (x_4\lor x_5\lor x_6) \land\dots$). The number $y$ could have any value between zero and $2^{|y|}-1$, so your certificate has length about $|x|2^{|y|}$, which is exponential in the input size.

However, the question asks you to prove that it's $\mathrm{NP}$-hard, not that it's $\mathrm{NP}$-complete. You just need to prove that there's a polynomial-time reduction from some $\mathrm{NP}$-complete problem to $\#\mathrm{3SAT}$.

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  • $\begingroup$ thanks for the answer. What about this sketch: a NTM can decide membership to 3SAT in polynomial time. Then to decide membership in #3SAT it will take a polynomial expansion on the time implied to decide 3SAT. $\endgroup$ – Yamar69 Jun 17 '19 at 10:02
  • $\begingroup$ @Yamar69 You're supposed to be reducing problems to $\mathrm{\#3SAT}$, not designing algorithms for $\mathrm{\#3SAT}$. $\endgroup$ – David Richerby Jun 17 '19 at 10:07
  • $\begingroup$ David Richerby i know but that is the hard part for me :( can you give me a direction? $\endgroup$ – Yamar69 Jun 17 '19 at 10:11
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    $\begingroup$ Not being rude, but I'm not sure you do know. You started off trying to prove $\mathrm{\#3SAT}\in\mathrm{NP}$. I said don't do that; reduce an $\mathrm{NP}$-complete problem to it. Then you tried to produce an algorithm for $\mathrm{\#3SAT}$ and I said don't do that; reduce an $\mathrm{NP}$-complete problem to it. It doesn't seem to me that you've spent any time trying to actually reduce an $\mathrm{NP}$-complete problem to $\mathrm{\#3SAT}$. It's only eight minutes since I last told you to do that and you showed no evidence of having tried to do it before then. $\endgroup$ – David Richerby Jun 17 '19 at 10:17
  • $\begingroup$ One last question, I noticed that you have edited your answer by adding a comment on the length of the certificate. I understand what you wrote but it seems to me that it is correct only if you take into consideration the space of all possible certificates. Basically what I was saying was this: if you have an instance of # 3SAT with y = 10, The certificate that must be given as a meal to the verifier must contain only 10 different distributions of variables that make the input belonging to # 3SAT(10) and therefore it will be only 10 times larger than the 3SAT's one. $\endgroup$ – Yamar69 Jun 17 '19 at 10:36
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The subset for $y = 1$ is 3SAT, which is NP-complete (thus NP-hard); the full set can't be easier...

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