# algorithms – Reduction between Parity-SAT and approximate counting

Consider two problems as defined here.

Approximate counting: Given a Boolean function $$f(x)$$, for $$x in {0, 1}^{n}$$, distinguish between the two cases:

1. The number of satisfying assignments for $$f(x)$$ is $$geq 2^{k+1}$$.
2. The number of satisfying assignments for $$f(x)$$ is $$leq 2^{k}$$.

Parity-SAT: Given a Boolean function $$f(x)$$, for $$x in {0, 1}^{n}$$, output $$1$$ if the number of satisfying assignments to $$f(x)$$ is even.

Is there a way to reduce Parity-SAT to approximate counting (or vice versa)?