\# \mathrm{SAT} \in \mathrm{IP}
Contents
\(\# \mathrm{SAT} \in \mathrm{IP}\)#
#SAT problem#
(#SAT)
Let \(\phi\) be any Boolean formula on \(n\) variables of size \(S=\operatorname{poly}(n)\) The goal of #SAT is to compute the number of satisfying assignments of \(\phi\) :
The interactive proof for #SAT#
(arithmetization)
The process of replacing the Boolean formula \(\phi\) with an arithmetic circuit \(\psi\) computing an extension polynomial \(g\) of \(\phi\).
We can turn \(\phi\) into an arithmetic circuit \(\psi\) over \(\mathbb{F}\) that computes the desired extension \(g\) of \(\phi\).
For any gate in \(\phi\) computing the AND of two inputs \(y, z, \psi\) replaces \(\operatorname{AND}(y, z)\) with multiplication of \(y\) and \(z\) over \(\mathbb{F}\). It is easy to check that the bivariate polynomial \(y \cdot z\) extends the Boolean function \(\operatorname{AND}(y, z)\), i.e., \(\operatorname{AND}(y, z)=y \cdot z\) for all \(y, z \in\{0,1\}\).
Likewise, \(\psi\) replaces any gate computing \(\operatorname{OR}(y, z)\) by \(y+z-y \cdot z\).
Any formula leaf of the form \(\bar{y}\) (i.e., the negation of variable \(y\) ) is replaced by \(1-y\)
It is easy to check that \(\psi(x)=\phi(x)\) for all \(x \in\{0,1\}^{n}\), and that the number of gates in the arithmetic circuit \(\psi\) is at most \(3 S\).
Then apply the sum-check protocol to compute \(\sum_{x \in\{0,1\}^{n}} g(x)\)