Is there a proof that there exists a decidable problem that is NOT NP-HARD??
Is there any decidable problem that is NOT NP-HARD?
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computational-complexity
1 Answers
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Very simple answer:
Since you need one $x \in A$, and one $x \not \in A$ for a polynomial time reduction, $A = \emptyset$ cannot be a hard language for NP.
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0haha... wonderful.. thnx – 2011-12-03
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0and under $P=NP$ only $\emptyset$ and $A^{\ast}$ are not NP-hard under polynomial time reductions. – 2011-12-03
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1@sdcvvc, not NP-hard you mean. – 2011-12-03
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0Yes, thanks for the correction. – 2011-12-03