I have a question in complexities that I could not do.

There will be D, E, F, three languages belonging to NPH. Suppose that the reductions exist $D \leq _P E$ and $E \leq _P F$. Determine which of the following statements is correct:

  1. Inevitably there is a reduction $F \leq _P D$
  2. If there was an $F \leq _P D$ reduction then F, E and D are all in the NPC class.
  3. Even if there was an $F \leq _P D$ reduction, it is still possible that $D \notin NP$.
  4. if $F \in CoNP$ then $NP \neq CoNP$
  5. None of the above claims are true.

I think the answer is probably 2, because there are reductions between all the problems. So everyone should be in $NPH$, and if a problem is in $NPH$ then it is also in $NPC$. I can not understand why 4 is not true, it also seems logical.


1 Answer 1


if a problem is in $NPH$ then it is also in $NPC$

This statement is incorrect. A language is in $NPC$ if it is in $NPH$ and it is in $NP$.

Answer 4 is incorrect as well since that would imply that any language $L\in NP$ can be reduced to $F$, hence $L\le_p F$ and since $F\in coNP$ then $L\in coNP$. Hence, $NP\subseteq coNP$. Its not hard to show the other way around, and conclude that this means $NP=coNP$.

Answer 3 is correct. There are $NPH$ problems that is not in $NP$ (for example, take any $NEXP$-complete problem). If you choose $D=E=F\in NPH\setminus NP$ then the reductions are trivial, but they are still not in $NP$.


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