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












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












$begingroup$


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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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






active

oldest

votes


















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











$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "419"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: false,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: null,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcs.stackexchange.com%2fquestions%2f103618%2fif-graph-isomorphism-is-in-p-is-then-p-np%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    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






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Computer Science Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcs.stackexchange.com%2fquestions%2f103618%2fif-graph-isomorphism-is-in-p-is-then-p-np%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Human spaceflight

            Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

            張江高科駅