Sign up ×
Computer Science Stack Exchange is a question and answer site for students, researchers and practitioners of computer science. It's 100% free, no registration required.

I am interesting in proving that there is no search problem that is polynomial bounded and self-reducible, as long as ${\sf P} \neq {\sf NP} \cap {\sf coNP}$.

The problem is I don't know how to approach the proof, below I wrote few ideas with open questions.

We can start by denoting the search problem in set ${\sf NP} \cap {\sf coNP}$ in terms of search problem relations $R_1$ and $R_2$ such that $S = \left \{ x:R_1(x) \neq \emptyset \right \} = \left \{ x:R_2(x) = \emptyset \right \} $. But how to present that the decision problem $S$ is not in ${\sf P}$. I don't know (but it seems to be crucial to show that $S$ is not in ${\sf P}$).

Having defined $S$ the next step would be to show that there is a relation $R$ that is self-reducible to $S$, but is not polynomial bounded.

In short, the question is how to define relation $R$ that is self-reducible to $S$. How to prove that $R$ is not polynomial bounded. Actually proving that $R$ is polynomial bounded may be redundant because $S$ is in ${\sf NP} \cap {\sf coNP}$ and it's given that ${\sf NP} \cap {\sf coNP} \neq {\sf P} $.

Addendum: I was given a hint

$R = \left \{ (x,1y):(x,y) \in R_1 \right \} \cup \left \{ (x,0y):(x,y) \in R_2 \right \}$

If I will be able to show that search problem relation R is self-reducible to S, than I think the problem is solved.

share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.