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I have a question from a test that I could not pass, I could not answer the question and I am looking for help with this question

This is the question

Will be $A\in NP$

Suppose that $A\notin CoNP$. determine which of the claims is correct:

  1. $P=CoNP$
  2. $P=NP$
  3. $P\neq NP$
  4. $NP \cap CoNP = \phi $

According to the data in the question, you need to choose the correct answer from the 4 possible answers.

I do not understand if there is any connection at all between the language A, and the answers themselves.

What I think is that A must only be in NP and it cannot be in P, because it is not in CoNP, and CoNP itself is in P.

But I do not find an answer that can fit it, maybe 3 is correct, but it has nothing to do with the question at all, it is always true.

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  • $\begingroup$ it is incorrect to say that $coNP\subseteq P$. If this were true then $P=NP$, and we don't know that. $\endgroup$
    – nir shahar
    Jul 21 '21 at 8:24
  • $\begingroup$ Thanks, if a problem is in $coNP$, it does not have to be in $P$ too? Or it can be but not necessarily. $\endgroup$
    – dust
    Jul 21 '21 at 8:32
  • $\begingroup$ Not necessarily. Its an open problem. Knowing whether $coNP\subseteq P$ or $coNP \neq P$ is equivalent to knowing whether $P=NP$ or $P\neq NP$. $\endgroup$
    – nir shahar
    Jul 21 '21 at 8:37
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The correct answer is no. 3.

Suppose $A \in \text{NP}$ and $A \notin \text{co-NP}$. Clearly this shows $\text{NP} \neq \text{coNP}$, but that's not a possible choice for this question.

Observe that complement of the machine output can be trivially implemented in a deterministic polynomial machine. That is to say, $\text{P} = \text{co-P}$. Thus, if $\text{P} = \text{NP}$, then $P = \text{co-NP}$, which together imply $\text{NP} = \text{co-NP}$. This cannot be the case by assumption, so $\text{P} \neq \text{NP}$.

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$P$ is closed under complement. The rest is up to you.

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  • $\begingroup$ Thanks, if I understood you correctly, actually, P = COP, and by the question $A\in NP $ and $A\notin CoNP $, therefore $CoNP \neq NP$ and get that $CoP \neq P$ , but I do not see an answer to it $\endgroup$
    – dust
    Jul 21 '21 at 9:04
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1+2 are not proven true and false, considered to be extremely hard to be proven either way, and most people assume they are false. That obviously answers 3.

4: P is a subset of both NP and co-NP, and since P is not empty, 4 is false.

Language A is indeed irrelevant. If the question had said "assume A in co-NP" that would have shown much more trivially that 4 is false, but it's false anyway.

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  • $\begingroup$ The language $A$ is not irrelevant. If there is a language $A\in NP$ but $A\not\in coNP$, as the question assumes, then it follows that $P\ne NP$, so 3 is the only correct statement of the four. $\endgroup$
    – Lieuwe Vinkhuijzen
    Aug 20 '21 at 18:52

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