I want to commit to some `g = ZK(f)` and prove that I know `x` where `f(x) == True` without revealing `x` or `f`. Is that possible?
Discussion
Nope
Chad GPT for the case where I'm ok with revealing x and f is a Schnorr signature verification function.
Have no idea what you're talking about and if this works but this is what my self hosted Llama 2 gave so feel free to let me know is it valid answer or just a hallucination
Could you paste the entire code snipped here? 🙏
You need to build a polynomial that's 1 iff f(x) is 1, 0 otherwise. But the tricky part is to know that f(x) == f'(x), since the verifier don't know f.
Ask Riemann