3
$\begingroup$

This may be a naive question but I am a bit clueless. I was under the opinion that given two NP complete problems, $P1$ and $P2$, if $P2 \in APX$ and $P1 \notin APX$, then we cannot have a direct reduction $P2\leq P1$ but we can have a reduction $P1\leq P2$.

In this paper on page 165 http://www.lancaster.ac.uk/staff/letchfoa/articles/cut-projection.pdf, there is a reduction $MAXCUT \leq STABLESET$. That is, given an instance of STABLESET, I can find an instance of MAXCUT. Why cant I use the approximate solution for MAXCUT to find an approximate solution to STABLESET?

Improve this question
$\endgroup$
6
$\begingroup$

In short, a reduction need not preserve approximations. For more details, see my answer on cstheory to a similar question.

Improve this answer
$\endgroup$
1
  • $\begingroup$ ah, i wanted to give the vertex-cover VS independent-set example, but Yuval is just too quick. $\endgroup$
    – Igor Shinkar
    Oct 2 '13 at 20:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.