If graph isomorphism is in P, is then P = NP?












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I think that, since graph isomorphism is not known to be $textbf{NP}$-complete, we can not reduce all problems in $textbf{NP}$ to it, and therefore the implication does not hold.



Additionally, in the accepted answer to this question it is stated, that a proof that graph isomorphism is not $textbf{NP}$-complete would prove $textbf{P} neq textbf{NP}$. Why?










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$endgroup$








  • 2




    $begingroup$
    If P=NP then every nontrivial problem in NP is NP-complete.
    $endgroup$
    – Yuval Filmus
    Jan 30 at 16:30
















5












$begingroup$


I think that, since graph isomorphism is not known to be $textbf{NP}$-complete, we can not reduce all problems in $textbf{NP}$ to it, and therefore the implication does not hold.



Additionally, in the accepted answer to this question it is stated, that a proof that graph isomorphism is not $textbf{NP}$-complete would prove $textbf{P} neq textbf{NP}$. Why?










share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    If P=NP then every nontrivial problem in NP is NP-complete.
    $endgroup$
    – Yuval Filmus
    Jan 30 at 16:30














5












5








5


1



$begingroup$


I think that, since graph isomorphism is not known to be $textbf{NP}$-complete, we can not reduce all problems in $textbf{NP}$ to it, and therefore the implication does not hold.



Additionally, in the accepted answer to this question it is stated, that a proof that graph isomorphism is not $textbf{NP}$-complete would prove $textbf{P} neq textbf{NP}$. Why?










share|cite|improve this question











$endgroup$




I think that, since graph isomorphism is not known to be $textbf{NP}$-complete, we can not reduce all problems in $textbf{NP}$ to it, and therefore the implication does not hold.



Additionally, in the accepted answer to this question it is stated, that a proof that graph isomorphism is not $textbf{NP}$-complete would prove $textbf{P} neq textbf{NP}$. Why?







complexity-theory np-complete p-vs-np graph-isomorphism






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edited Jan 30 at 17:02









Discrete lizard

4,24411536




4,24411536










asked Jan 30 at 16:17









PascalIvPascalIv

495




495








  • 2




    $begingroup$
    If P=NP then every nontrivial problem in NP is NP-complete.
    $endgroup$
    – Yuval Filmus
    Jan 30 at 16:30














  • 2




    $begingroup$
    If P=NP then every nontrivial problem in NP is NP-complete.
    $endgroup$
    – Yuval Filmus
    Jan 30 at 16:30








2




2




$begingroup$
If P=NP then every nontrivial problem in NP is NP-complete.
$endgroup$
– Yuval Filmus
Jan 30 at 16:30




$begingroup$
If P=NP then every nontrivial problem in NP is NP-complete.
$endgroup$
– Yuval Filmus
Jan 30 at 16:30










1 Answer
1






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20












$begingroup$

We don't know.



We do know that $textbf{P} = textbf{NP}$ implies graph isomorphism is in $textbf{P}$, but the other implication has not been proven (to the best of my knowledge). It is suspected graph isomorphism is $textbf{NP}$-intermediate (i.e., it is in $textbf{NP} setminus textbf{P}$ and not $textbf{NP}$-complete). This question as well as this other one list evidences supporting said suspicions.





Regarding your second question: If $textbf{P} = textbf{NP}$, then any (nontrivial) problem in $textbf{NP}$ is trivially $textbf{NP}$-complete because any (nontrivial) problem in $textbf{P}$ is $textbf{P}$-complete (and we have assumed $textbf{P} = textbf{NP}$). Hence, by contraposition, if the conclusion of this implication is false (i.e., there is a problem in $textbf{NP}$ which is not $textbf{NP}$-complete), then the premise (i.e., $textbf{P} = textbf{NP}$) must also be false.






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    1 Answer
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    1 Answer
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    active

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    20












    $begingroup$

    We don't know.



    We do know that $textbf{P} = textbf{NP}$ implies graph isomorphism is in $textbf{P}$, but the other implication has not been proven (to the best of my knowledge). It is suspected graph isomorphism is $textbf{NP}$-intermediate (i.e., it is in $textbf{NP} setminus textbf{P}$ and not $textbf{NP}$-complete). This question as well as this other one list evidences supporting said suspicions.





    Regarding your second question: If $textbf{P} = textbf{NP}$, then any (nontrivial) problem in $textbf{NP}$ is trivially $textbf{NP}$-complete because any (nontrivial) problem in $textbf{P}$ is $textbf{P}$-complete (and we have assumed $textbf{P} = textbf{NP}$). Hence, by contraposition, if the conclusion of this implication is false (i.e., there is a problem in $textbf{NP}$ which is not $textbf{NP}$-complete), then the premise (i.e., $textbf{P} = textbf{NP}$) must also be false.






    share|cite|improve this answer











    $endgroup$


















      20












      $begingroup$

      We don't know.



      We do know that $textbf{P} = textbf{NP}$ implies graph isomorphism is in $textbf{P}$, but the other implication has not been proven (to the best of my knowledge). It is suspected graph isomorphism is $textbf{NP}$-intermediate (i.e., it is in $textbf{NP} setminus textbf{P}$ and not $textbf{NP}$-complete). This question as well as this other one list evidences supporting said suspicions.





      Regarding your second question: If $textbf{P} = textbf{NP}$, then any (nontrivial) problem in $textbf{NP}$ is trivially $textbf{NP}$-complete because any (nontrivial) problem in $textbf{P}$ is $textbf{P}$-complete (and we have assumed $textbf{P} = textbf{NP}$). Hence, by contraposition, if the conclusion of this implication is false (i.e., there is a problem in $textbf{NP}$ which is not $textbf{NP}$-complete), then the premise (i.e., $textbf{P} = textbf{NP}$) must also be false.






      share|cite|improve this answer











      $endgroup$
















        20












        20








        20





        $begingroup$

        We don't know.



        We do know that $textbf{P} = textbf{NP}$ implies graph isomorphism is in $textbf{P}$, but the other implication has not been proven (to the best of my knowledge). It is suspected graph isomorphism is $textbf{NP}$-intermediate (i.e., it is in $textbf{NP} setminus textbf{P}$ and not $textbf{NP}$-complete). This question as well as this other one list evidences supporting said suspicions.





        Regarding your second question: If $textbf{P} = textbf{NP}$, then any (nontrivial) problem in $textbf{NP}$ is trivially $textbf{NP}$-complete because any (nontrivial) problem in $textbf{P}$ is $textbf{P}$-complete (and we have assumed $textbf{P} = textbf{NP}$). Hence, by contraposition, if the conclusion of this implication is false (i.e., there is a problem in $textbf{NP}$ which is not $textbf{NP}$-complete), then the premise (i.e., $textbf{P} = textbf{NP}$) must also be false.






        share|cite|improve this answer











        $endgroup$



        We don't know.



        We do know that $textbf{P} = textbf{NP}$ implies graph isomorphism is in $textbf{P}$, but the other implication has not been proven (to the best of my knowledge). It is suspected graph isomorphism is $textbf{NP}$-intermediate (i.e., it is in $textbf{NP} setminus textbf{P}$ and not $textbf{NP}$-complete). This question as well as this other one list evidences supporting said suspicions.





        Regarding your second question: If $textbf{P} = textbf{NP}$, then any (nontrivial) problem in $textbf{NP}$ is trivially $textbf{NP}$-complete because any (nontrivial) problem in $textbf{P}$ is $textbf{P}$-complete (and we have assumed $textbf{P} = textbf{NP}$). Hence, by contraposition, if the conclusion of this implication is false (i.e., there is a problem in $textbf{NP}$ which is not $textbf{NP}$-complete), then the premise (i.e., $textbf{P} = textbf{NP}$) must also be false.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 31 at 8:14

























        answered Jan 30 at 16:40









        dkaeaedkaeae

        2,0201721




        2,0201721






























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