If we prove that:

The existance of a $(2-\epsilon)$-approximation algorithm for Problem P1 implies $P = NP$,

can we conclude:

There exists no PTAS for Problem P1, and so P1 is APX-hard?


You can conclude There exists no PTAS for Problem P1 if $P \neq NP$, but you can NOT conclude P1 is APX-hard. Precisely:

  1. If someone proofs $P = NP$ you get a trivial PTAS;
  2. Assumung $P \neq NP$, it still could be APX-intermediate (see https://en.wikipedia.org/wiki/APX#APX-intermediate).
  • $\begingroup$ Thanks. So, based on the Wiki page, we cannot expect to find a PTAS for Problem P1, but it doesn't mean it is APS-hard. Am I right? $\endgroup$ – Mostafa May 6 '20 at 6:50
  • $\begingroup$ yes that is correct $\endgroup$ – user3680510 May 6 '20 at 6:54
  • $\begingroup$ So, I edit your answer based on that. $\endgroup$ – Mostafa May 6 '20 at 7:29

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