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The optimization version of TSP asks for the length of the shortest tour. Unlike the decision version of TSP, there's no obvious way to verify a proposed solution of the optimization problem in polynomial time. But is there a proof of whether or not it can be verified in polynomial time assuming P ≠ NP?

Raphael
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user541686
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1 Answers1

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If $optTSP$ is in $NP$, then $coNP = NP$. The latter is unresolved currently.

Proof: if $optTSP$ is in $NP$, then $coTSP$ is in $coNP$ (the certificate being whatever certificate was provided to verify a solution to $optTSP$ was minimal, and then comparing the value of that solution to the desired bound). And because $TSP$ is $NP$-complete then this implies $NP = coNP$.

D.W.
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Scrub
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