Valiant converts $SAT$ formula to a $0/1$ matrix such that $Permanent$ of the matrix is $4^m\#SAT$.

We know $Permanent$ can be approximated to $1+\epsilon$ factor with probability $1-\frac1\delta$ in time $poly(m|\epsilon|^{-1}\log\frac1\delta)$.

Why does not these two help me tell when number of witnesses to $SAT$ formula is $0$ in randomized polynomial time?

What do I miss?

  • $\begingroup$ You can determine whether a non-negative matrix has permanent zero in polynomial time, using bipartite perfect matching. So I doubt your description of Valiant’s reduction. $\endgroup$ May 26 '19 at 18:58

As you state, Valiant shows how to construct, given a CNF $\varphi$, a polynomial size matrix whose permanent is a (known) multiple of the number of satisfying assignments of $\varphi$. Unfortunately, this matrix is not non-negative.

Given a non-negative matrix, you can determine whether its permanent is zero or not using an algorithm for bipartite perfect matching; and you can efficiently estimate its value. The same doesn't hold for general matrices.

Valiant further shows how to construct, given a CNF $\varphi$, a polynomial size 0/1 matrix, from whose permanent you can deduce the number of satisfying assignments of $\varphi$. However, the formula is not as simple as before, and in particular, neither bipartite perfect matching nor the efficient permanent estimation algorithms can determine whether $\varphi$ is satisfiable or not.

For details of a later version of this last reduction, see Section 5.1 of Ben-Dor and Halevi, Zero-One Permanent is #P-Complete, A Simpler Proof.

  • $\begingroup$ As I understand we get a $\mathbb Z$ matrix with permanent $4^{3m}\#\phi$ where $m$ is number of clauses in $n$ variable formula $\phi$ and then a $\{0,1\}$ matrix with $Perm\equiv4^{3m}\#\phi\bmod 2^{m'+1}$ at some $m'\gg m$? $\endgroup$
    – Turbo
    May 27 '19 at 11:51
  • $\begingroup$ Right, something of this sort. $\endgroup$ May 27 '19 at 12:58
  • $\begingroup$ I am not understanding the non-triviality in formula for second reduction. If $\#\mbox{perfect-matchings}=2^{cm}\#\phi$ can't we decide no matchings and hence no witnesses? $\endgroup$
    – Turbo
    Dec 14 '20 at 21:48
  • $\begingroup$ The formula for the second reduction is not what you state. It is more complicated. $\endgroup$ Dec 14 '20 at 21:50
  • 1
    $\begingroup$ You can find the 0/1 matrix in Section 5.3. Also, according to Ben-Dor and Halevi, their construction combines the work of Valiant with that of Zanko (1991). $\endgroup$ Dec 15 '20 at 8:49

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