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♦
4,18411435
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♦
4,18411435
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♦
4,18411435
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♦
4,18411435
asked Jan 30 at 16:17
PascalIvPascalIv
495
edited Jan 30 at 17:02
Discrete lizard♦
4,18411435
asked Jan 30 at 16:17
PascalIvPascalIv
495
edited Jan 30 at 17:02
Discrete lizard♦
4,18411435
edited Jan 30 at 17:02
Discrete lizard♦
4,18411435
edited Jan 30 at 17:02
Discrete lizard♦
4,18411435
4,18411435
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
1,452620
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add a comment |
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1 Answer
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$begingroup$
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
1,452620
$endgroup$
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
1,452620
$endgroup$
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
1,452620
$endgroup$
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
1,452620
edited Jan 31 at 8:14
answered Jan 30 at 16:40
dkaeaedkaeae
1,452620
answered Jan 30 at 16:40
dkaeaedkaeae
1,452620
answered Jan 30 at 16:40
dkaeaedkaeae
1,452620
1,452620
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$begingroup$
If P=NP then every nontrivial problem in NP is NP-complete.
$endgroup$
– Yuval Filmus
Jan 30 at 16:30