5

I have a problem $\Pi_1$ that I want to show that is NP-hard. I know that I must find an NP-hard problem $\Pi_2$ and a polynomial time reduction $f()$ from instances of $\Pi_2$ to $\Pi_1$ such that $I_2$ is an Yes-instance of $\Pi_2$ iff $I_1=f(I_2)$ is an Yes-instance of $\Pi_1$.

What if I find a (constant sized) family of reductions $f_i()$ such that $I_2$ is an Yes-instance of $\Pi_2$ iff at least one $f_i(I_2)$ is an Yes-instance of $\Pi_1$? Is this enough? Is there a way of translating this one in the "classical" definition? How to formalize this?

I know that in the second situation I can say that I can't solve $\Pi_1$ in polynomial time unless P=NP, but I'm no sure that is equivalent of saying that $\Pi_1$ is NP-hard.

Raphael
  • 73,212
  • 30
  • 182
  • 400
Vinicius dos Santos
  • 155
  • 7

1 Answers1

6

NP-hardness is usually defined using many-to-one reductions, for various reasons, for example to separate NP from coNP (there was some question about it either here or in cstheory). What you're describing is a specific case of more general Turing reductions, which is more common in classical recursion theory, and it is natural since it has the consequences you mention. Unfortunately, the definition of NP-hardness requires a many-to-one reduction.

Yuval Filmus
  • 280,205
  • 27
  • 317
  • 514