If graph isomorphism is in P, is then P = NP?
$begingroup$
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?
complexity-theory np-complete p-vs-np graph-isomorphism
edited Jan 30 at 17:02
Discrete lizard♦
4,24411536
asked Jan 30 at 16:17
PascalIvPascalIv
495
$endgroup$
add a comment |
$begingroup$
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?
complexity-theory np-complete p-vs-np graph-isomorphism
edited Jan 30 at 17:02
Discrete lizard♦
4,24411536
asked Jan 30 at 16:17
PascalIvPascalIv
495
$endgroup$
2
$begingroup$
If P=NP then every nontrivial problem in NP is NP-complete.
$endgroup$
– Yuval Filmus
Jan 30 at 16:30
add a comment |
$begingroup$
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?
complexity-theory np-complete p-vs-np graph-isomorphism
edited Jan 30 at 17:02
Discrete lizard♦
4,24411536
asked Jan 30 at 16:17
PascalIvPascalIv
495
$endgroup$
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
complexity-theory np-complete p-vs-np graph-isomorphism
edited Jan 30 at 17:02
Discrete lizard♦
4,24411536
asked Jan 30 at 16:17
PascalIvPascalIv
495
edited Jan 30 at 17:02
Discrete lizard♦
4,24411536
asked Jan 30 at 16:17
PascalIvPascalIv
495
edited Jan 30 at 17:02
Discrete lizard♦
4,24411536
edited Jan 30 at 17:02
Discrete lizard♦
4,24411536
edited Jan 30 at 17:02
Discrete lizard♦
4,24411536
4,24411536
asked Jan 30 at 16:17
PascalIvPascalIv
495
asked Jan 30 at 16:17
PascalIvPascalIv
495
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Jan 30 at 16:40
dkaeaedkaeae
2,0201721
$endgroup$
add a comment |
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
$begingroup$
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.
answered Jan 30 at 16:40
dkaeaedkaeae
2,0201721
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 30 at 16:40
dkaeaedkaeae
2,0201721
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 30 at 16:40
dkaeaedkaeae
2,0201721
$endgroup$
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.
answered Jan 30 at 16:40
dkaeaedkaeae
2,0201721
edited Jan 31 at 8:14
answered Jan 30 at 16:40
dkaeaedkaeae
2,0201721
answered Jan 30 at 16:40
dkaeaedkaeae
2,0201721
answered Jan 30 at 16:40
dkaeaedkaeae
2,0201721
2,0201721
add a comment |
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
2
$begingroup$
If P=NP then every nontrivial problem in NP is NP-complete.
$endgroup$
– Yuval Filmus
Jan 30 at 16:30