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If there exists a Cook reduction of a decision problem $\mathcal{P}_1$ into another decision problem $\mathcal{P}_2$ and also a Cook reduction of $\mathcal{P}_2$ into $\mathcal{P}_1$, then is there also a Karp reduction (a polynomial transformation) between $\mathcal{P}_1$ and $\mathcal{P}_2$ (in both directions)?

These are the definitions I use:

Cook reduction
$\mathcal{P}_1$ polynomially reduces to $\mathcal{P}_2$ if there is a polynomial-time oracle algorithm for $\mathcal{P}_1$ using an oracle for $\mathcal{P}_2$.

Karp reduction
$\mathcal{P}_1=(X_1,Y_1)$ polynomially transforms to $\mathcal{P}_2=(X_2,Y_2)$ if there is a function $f:X_1\rightarrow X_2$ computable in polynomial time such that for all $x\in X_1$, $x\in Y_1$ if and only if $f(x)\in X_2$.

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Héctor Andrade
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1 Answers1

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The answer depends on whether you require both of $\mathcal{P}_1,\mathcal{P}_2$ to be in NP or not.

If the problems need not be in NP

Then the answer to your question is "No". There exist problems $\mathcal{P}_1,\mathcal{P}_2$ that are Cook-reducible (in both directions) but not Karp-reducible (in either direction)

In particular, consider $\mathcal{P}_1 = $ HALT (the Halting problem) and $\mathcal{P}_2 = \overline{\mathcal{P}_1}$ (its complement). They are Cook-reducible (you just invoke the oracle and complement its answer), but not Karp-reducible.

(Similarly, if $\textsf{NP} \ne \textsf{coNP}$, then the answer is No: take $\mathcal{P}_1 = $ SAT and $\mathcal{P}_2 = $ TAUTOLOGY, so that $\mathcal{P}_1$ is NP-complete and $\mathcal{P}_2$ is coNP-complete. They're Cook-reducible, but a Karp reduction would prove that $\textsf{NP} = \textsf{coNP}$ via a standard argument.)

If the question is restricted to require both problems to be in NP

Then it's an open question; no one knows whether the answer is "Yes" or "No". See https://en.wikipedia.org/wiki/NP-completeness#Completeness_under_different_types_of_reduction and

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D.W.
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