P vs. NP,algorithm

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Is there a known, explicit example of an algortihm with the property such that if $Pneq NP$ then this algorithm doesn't run in polynomial time and if $P=NP$ then it does run in polynomial time?






ds.algorithms np p-vs-np







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    Sort of. If P = NP, Levin’s universal search algorithm runs in polynomial time on accepting instances en.wikipedia.org/wiki/…
    – Emil Jeřábek
    2 hours ago














up vote
4
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favorite












Is there a known, explicit example of an algortihm with the property such that if $Pneq NP$ then this algorithm doesn't run in polynomial time and if $P=NP$ then it does run in polynomial time?






ds.algorithms np p-vs-np







share|cite|improve this question

















  • 1




    Sort of. If P = NP, Levin’s universal search algorithm runs in polynomial time on accepting instances en.wikipedia.org/wiki/…
    – Emil Jeřábek
    2 hours ago












up vote
4
down vote

favorite









up vote
4
down vote

favorite











Is there a known, explicit example of an algortihm with the property such that if $Pneq NP$ then this algorithm doesn't run in polynomial time and if $P=NP$ then it does run in polynomial time?






ds.algorithms np p-vs-np







share|cite|improve this question













Is there a known, explicit example of an algortihm with the property such that if $Pneq NP$ then this algorithm doesn't run in polynomial time and if $P=NP$ then it does run in polynomial time?






ds.algorithms np p-vs-np




ds.algorithms np p-vs-np






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asked 3 hours ago









user2925716

1375




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  • 1




    Sort of. If P = NP, Levin’s universal search algorithm runs in polynomial time on accepting instances en.wikipedia.org/wiki/…
    – Emil Jeřábek
    2 hours ago












  • 1




    Sort of. If P = NP, Levin’s universal search algorithm runs in polynomial time on accepting instances en.wikipedia.org/wiki/…
    – Emil Jeřábek
    2 hours ago







1




1




Sort of. If P = NP, Levin’s universal search algorithm runs in polynomial time on accepting instances en.wikipedia.org/wiki/…
– Emil Jeřábek
2 hours ago




Sort of. If P = NP, Levin’s universal search algorithm runs in polynomial time on accepting instances en.wikipedia.org/wiki/…
– Emil Jeřábek
2 hours ago










1 Answer
1






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3
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If you assume that $P=^?NP$ is provable in PA (or ZFC), a trivial example is the following:



Input: N (integer in binary format)
For I = 1 to N do
begin
 if I is a valid encoding of a proof of P = NP in PA (or ZFC)
 then halt and accept
End
Reject





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  • How do I quicly decide if "I is a valid encoding of a proof of P = NP in PA (or ZFC)" ?
    – user2925716
    2 hours ago










  • @user2925716 You can do it in polynomial time (imagine that $I$ is a string that represents the full proof in any reasonable deduction system).
    – Marzio De Biasi
    1 hour ago










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote



accepted










If you assume that $P=^?NP$ is provable in PA (or ZFC), a trivial example is the following:



Input: N (integer in binary format)
For I = 1 to N do
begin
 if I is a valid encoding of a proof of P = NP in PA (or ZFC)
 then halt and accept
End
Reject





share|cite|improve this answer






















  • How do I quicly decide if "I is a valid encoding of a proof of P = NP in PA (or ZFC)" ?
    – user2925716
    2 hours ago










  • @user2925716 You can do it in polynomial time (imagine that $I$ is a string that represents the full proof in any reasonable deduction system).
    – Marzio De Biasi
    1 hour ago














up vote
3
down vote



accepted










If you assume that $P=^?NP$ is provable in PA (or ZFC), a trivial example is the following:



Input: N (integer in binary format)
For I = 1 to N do
begin
 if I is a valid encoding of a proof of P = NP in PA (or ZFC)
 then halt and accept
End
Reject





share|cite|improve this answer






















  • How do I quicly decide if "I is a valid encoding of a proof of P = NP in PA (or ZFC)" ?
    – user2925716
    2 hours ago










  • @user2925716 You can do it in polynomial time (imagine that $I$ is a string that represents the full proof in any reasonable deduction system).
    – Marzio De Biasi
    1 hour ago












up vote
3
down vote



accepted







up vote
3
down vote



accepted






If you assume that $P=^?NP$ is provable in PA (or ZFC), a trivial example is the following:



Input: N (integer in binary format)
For I = 1 to N do
begin
 if I is a valid encoding of a proof of P = NP in PA (or ZFC)
 then halt and accept
End
Reject





share|cite|improve this answer














If you assume that $P=^?NP$ is provable in PA (or ZFC), a trivial example is the following:



Input: N (integer in binary format)
For I = 1 to N do
begin
 if I is a valid encoding of a proof of P = NP in PA (or ZFC)
 then halt and accept
End
Reject






share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 8 mins ago

























answered 2 hours ago









Marzio De Biasi

18k240108




18k240108











  • How do I quicly decide if "I is a valid encoding of a proof of P = NP in PA (or ZFC)" ?
    – user2925716
    2 hours ago










  • @user2925716 You can do it in polynomial time (imagine that $I$ is a string that represents the full proof in any reasonable deduction system).
    – Marzio De Biasi
    1 hour ago
















  • How do I quicly decide if "I is a valid encoding of a proof of P = NP in PA (or ZFC)" ?
    – user2925716
    2 hours ago










  • @user2925716 You can do it in polynomial time (imagine that $I$ is a string that represents the full proof in any reasonable deduction system).
    – Marzio De Biasi
    1 hour ago















How do I quicly decide if "I is a valid encoding of a proof of P = NP in PA (or ZFC)" ?
– user2925716
2 hours ago




How do I quicly decide if "I is a valid encoding of a proof of P = NP in PA (or ZFC)" ?
– user2925716
2 hours ago












@user2925716 You can do it in polynomial time (imagine that $I$ is a string that represents the full proof in any reasonable deduction system).
– Marzio De Biasi
1 hour ago




@user2925716 You can do it in polynomial time (imagine that $I$ is a string that represents the full proof in any reasonable deduction system).
– Marzio De Biasi
1 hour ago

















 

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