why does $ A≤_p \#SAT$ if $A \in BPP$

Question

hello and thank you for helping me understand the following:

I really don't understand this, why if language $A \in BPP$ then $A≤_P\#SAT$?

language A is in BPP class, if for a probabilistic turing machine M, M outputs 1 for all $x \in A$ in a probability $\geq 2/3$, and for all $x \not \in A$ M outputs 1 in a probability of $\leq 1/3$ and of course M must run in a polynomial time on all inputs.

so if $A \in BPP$, why does it mean that $A≤_P\#SAT$? if M is reduced to boolean F, it means that we'll get output of 1 in a probability of $\geq 2/3$ for every $x \in A$? why?

can someone please show me algorithmically or mathmatically why if language $A \in BPP$ then $A≤_P\#SAT$? quite lost

thank you, would appreciate if you could explain along.

Answer 1

Using the proof of the Cook–Levin theorem, for every input $x$ you can construct in polynomial time a SAT instance $\phi(r,z)$ which encodes "$M$ accepts when run on input $x$ and randomness $r$". Here $r$ is a vector of $m = \mathit{poly}(n)$ bits, representing the random bits of $M$, and $z$ is an auxiliary vector, with the following property: in any accepting run of $M$, there is exactly one setting of $z$ which satisfies $\phi$.

By definition, if $x \in L$ then $\phi$ has at least $(2/3)2^m$ satisfying assignments, and if $x \notin L$ then $\phi$ has at most $(1/3)2^m$ satisfying assignments. You can distinguish between the two cases using a $\mathsf{\#SAT}$ oracle.