Is the language { | p and n are natural numbers and there's no prime number in [p,p+n]} belongs to NP...

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Is the language {


| p and n are natural numbers and there's no prime number in [p,p+n]} belongs to NP class?


Is determining if there is a prime in an interval known to be in P or NP-complete?If the language of a TM is TMs which cannot self recognize, can the original TM?Why is $A_{TM}$ reducible to $HALT_{TM}$?Proving a language is not Turing-recognizable by reduction from $D = {langle Mrangle mid M text{ rejects input }langle Mrangle}$Show that the set of all TMs that move only to the right and loop for some input is decidableProving that a set of grammars for a given finite language is decidableDecidability of the TM's computing a non-empty subset of total functionssmn-theorem: Application by instantiating s<m, n> with other functionUnion of R.E. and Non R.E. languageShow: “Checking no solution for system of linear equations with integer variables and coefficients” $in mathbf{NP}$Rice's Theorem - usage on $DFA$ or $LBA$













1












$begingroup$


I was wondering if the following language belongs to NP class and if its complimentary belongs to NP class:



begin{align}
C=left{langle p,nranglemidright.& left. p text{ and $n$ are natural numbers}right.\
&left.text{ and there's no prime number in the range}left[p,p+nright]right}
end{align}



could you please check if my reasoning is okay to deduce NP?
(I am not sure, but here's what I think):
for each word $langle p,nrangle in C$ we know that the word belongs to C because there exists a primal certificate - an nontrivial divisor to any of the numbers between $[p,p+n]$, though I am not really sure it is in NP.



regarding the complement: I think it is in NP because the compliment compositeness can be decided by guessing a factor nondeterministically. But again I am not so sure about it and I don't know how to correctly prove and show it.



Would really appreciate your input on that as I am quite unsure and also checked textbooks and internet (and this site) about it.



Edit: for the sake of solving the problem, due to xskxzr's comment, let's assume p and n are represented by binary, as there's a difference according to his comment between p and n being represented in unary and binary(this is also quite interesting).










share|cite|improve this question









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  • $begingroup$
    It may depend on how $p$ and $n$ are represented (by unary or binary).
    $endgroup$
    – xskxzr
    16 hours ago






  • 1




    $begingroup$
    Possible duplicate of Is determining if there is a prime in an interval known to be in P or NP-complete?
    $endgroup$
    – xskxzr
    15 hours ago










  • $begingroup$
    My question is about: 1) is my reasoning is enough/correct to deduce C being NP class? 2)regarding its complimentary(which is not written in the aforementioned link - they don't deal with it). those are 2 different questions and i would appreciate learning from my problem in order to understand my mistake and additionally if the compliment of C is in NP
    $endgroup$
    – hps13
    15 hours ago






  • 1




    $begingroup$
    @xskxzr That's not the same question. The OP is asking whether it is in NP, not NP-complete. That question is also the complement of this one.
    $endgroup$
    – orlp
    15 hours ago






  • 1




    $begingroup$
    @hps13 Your certificate does not work. To prove that (p,n) is in C, you need to show that every number in that interval is not a prime, so you will need n divisors. But then your certificate will be of size O(n *log(n+p)), which is exponential compared to the size log(p)+log(n) of the word.
    $endgroup$
    – mlk
    12 hours ago
















1












$begingroup$


I was wondering if the following language belongs to NP class and if its complimentary belongs to NP class:



begin{align}
C=left{langle p,nranglemidright.& left. p text{ and $n$ are natural numbers}right.\
&left.text{ and there's no prime number in the range}left[p,p+nright]right}
end{align}



could you please check if my reasoning is okay to deduce NP?
(I am not sure, but here's what I think):
for each word $langle p,nrangle in C$ we know that the word belongs to C because there exists a primal certificate - an nontrivial divisor to any of the numbers between $[p,p+n]$, though I am not really sure it is in NP.



regarding the complement: I think it is in NP because the compliment compositeness can be decided by guessing a factor nondeterministically. But again I am not so sure about it and I don't know how to correctly prove and show it.



Would really appreciate your input on that as I am quite unsure and also checked textbooks and internet (and this site) about it.



Edit: for the sake of solving the problem, due to xskxzr's comment, let's assume p and n are represented by binary, as there's a difference according to his comment between p and n being represented in unary and binary(this is also quite interesting).










share|cite|improve this question









New contributor




hps13 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    It may depend on how $p$ and $n$ are represented (by unary or binary).
    $endgroup$
    – xskxzr
    16 hours ago






  • 1




    $begingroup$
    Possible duplicate of Is determining if there is a prime in an interval known to be in P or NP-complete?
    $endgroup$
    – xskxzr
    15 hours ago










  • $begingroup$
    My question is about: 1) is my reasoning is enough/correct to deduce C being NP class? 2)regarding its complimentary(which is not written in the aforementioned link - they don't deal with it). those are 2 different questions and i would appreciate learning from my problem in order to understand my mistake and additionally if the compliment of C is in NP
    $endgroup$
    – hps13
    15 hours ago






  • 1




    $begingroup$
    @xskxzr That's not the same question. The OP is asking whether it is in NP, not NP-complete. That question is also the complement of this one.
    $endgroup$
    – orlp
    15 hours ago






  • 1




    $begingroup$
    @hps13 Your certificate does not work. To prove that (p,n) is in C, you need to show that every number in that interval is not a prime, so you will need n divisors. But then your certificate will be of size O(n *log(n+p)), which is exponential compared to the size log(p)+log(n) of the word.
    $endgroup$
    – mlk
    12 hours ago














1












1








1





$begingroup$


I was wondering if the following language belongs to NP class and if its complimentary belongs to NP class:



begin{align}
C=left{langle p,nranglemidright.& left. p text{ and $n$ are natural numbers}right.\
&left.text{ and there's no prime number in the range}left[p,p+nright]right}
end{align}



could you please check if my reasoning is okay to deduce NP?
(I am not sure, but here's what I think):
for each word $langle p,nrangle in C$ we know that the word belongs to C because there exists a primal certificate - an nontrivial divisor to any of the numbers between $[p,p+n]$, though I am not really sure it is in NP.



regarding the complement: I think it is in NP because the compliment compositeness can be decided by guessing a factor nondeterministically. But again I am not so sure about it and I don't know how to correctly prove and show it.



Would really appreciate your input on that as I am quite unsure and also checked textbooks and internet (and this site) about it.



Edit: for the sake of solving the problem, due to xskxzr's comment, let's assume p and n are represented by binary, as there's a difference according to his comment between p and n being represented in unary and binary(this is also quite interesting).










share|cite|improve this question









New contributor




hps13 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I was wondering if the following language belongs to NP class and if its complimentary belongs to NP class:



begin{align}
C=left{langle p,nranglemidright.& left. p text{ and $n$ are natural numbers}right.\
&left.text{ and there's no prime number in the range}left[p,p+nright]right}
end{align}



could you please check if my reasoning is okay to deduce NP?
(I am not sure, but here's what I think):
for each word $langle p,nrangle in C$ we know that the word belongs to C because there exists a primal certificate - an nontrivial divisor to any of the numbers between $[p,p+n]$, though I am not really sure it is in NP.



regarding the complement: I think it is in NP because the compliment compositeness can be decided by guessing a factor nondeterministically. But again I am not so sure about it and I don't know how to correctly prove and show it.



Would really appreciate your input on that as I am quite unsure and also checked textbooks and internet (and this site) about it.



Edit: for the sake of solving the problem, due to xskxzr's comment, let's assume p and n are represented by binary, as there's a difference according to his comment between p and n being represented in unary and binary(this is also quite interesting).







complexity-theory turing-machines computability np decision-problem






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New contributor




hps13 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









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share|cite|improve this question




share|cite|improve this question








edited 15 hours ago







hps13













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asked 16 hours ago









hps13hps13

205




205




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New contributor





hps13 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






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Check out our Code of Conduct.












  • $begingroup$
    It may depend on how $p$ and $n$ are represented (by unary or binary).
    $endgroup$
    – xskxzr
    16 hours ago






  • 1




    $begingroup$
    Possible duplicate of Is determining if there is a prime in an interval known to be in P or NP-complete?
    $endgroup$
    – xskxzr
    15 hours ago










  • $begingroup$
    My question is about: 1) is my reasoning is enough/correct to deduce C being NP class? 2)regarding its complimentary(which is not written in the aforementioned link - they don't deal with it). those are 2 different questions and i would appreciate learning from my problem in order to understand my mistake and additionally if the compliment of C is in NP
    $endgroup$
    – hps13
    15 hours ago






  • 1




    $begingroup$
    @xskxzr That's not the same question. The OP is asking whether it is in NP, not NP-complete. That question is also the complement of this one.
    $endgroup$
    – orlp
    15 hours ago






  • 1




    $begingroup$
    @hps13 Your certificate does not work. To prove that (p,n) is in C, you need to show that every number in that interval is not a prime, so you will need n divisors. But then your certificate will be of size O(n *log(n+p)), which is exponential compared to the size log(p)+log(n) of the word.
    $endgroup$
    – mlk
    12 hours ago


















  • $begingroup$
    It may depend on how $p$ and $n$ are represented (by unary or binary).
    $endgroup$
    – xskxzr
    16 hours ago






  • 1




    $begingroup$
    Possible duplicate of Is determining if there is a prime in an interval known to be in P or NP-complete?
    $endgroup$
    – xskxzr
    15 hours ago










  • $begingroup$
    My question is about: 1) is my reasoning is enough/correct to deduce C being NP class? 2)regarding its complimentary(which is not written in the aforementioned link - they don't deal with it). those are 2 different questions and i would appreciate learning from my problem in order to understand my mistake and additionally if the compliment of C is in NP
    $endgroup$
    – hps13
    15 hours ago






  • 1




    $begingroup$
    @xskxzr That's not the same question. The OP is asking whether it is in NP, not NP-complete. That question is also the complement of this one.
    $endgroup$
    – orlp
    15 hours ago






  • 1




    $begingroup$
    @hps13 Your certificate does not work. To prove that (p,n) is in C, you need to show that every number in that interval is not a prime, so you will need n divisors. But then your certificate will be of size O(n *log(n+p)), which is exponential compared to the size log(p)+log(n) of the word.
    $endgroup$
    – mlk
    12 hours ago
















$begingroup$
It may depend on how $p$ and $n$ are represented (by unary or binary).
$endgroup$
– xskxzr
16 hours ago




$begingroup$
It may depend on how $p$ and $n$ are represented (by unary or binary).
$endgroup$
– xskxzr
16 hours ago




1




1




$begingroup$
Possible duplicate of Is determining if there is a prime in an interval known to be in P or NP-complete?
$endgroup$
– xskxzr
15 hours ago




$begingroup$
Possible duplicate of Is determining if there is a prime in an interval known to be in P or NP-complete?
$endgroup$
– xskxzr
15 hours ago












$begingroup$
My question is about: 1) is my reasoning is enough/correct to deduce C being NP class? 2)regarding its complimentary(which is not written in the aforementioned link - they don't deal with it). those are 2 different questions and i would appreciate learning from my problem in order to understand my mistake and additionally if the compliment of C is in NP
$endgroup$
– hps13
15 hours ago




$begingroup$
My question is about: 1) is my reasoning is enough/correct to deduce C being NP class? 2)regarding its complimentary(which is not written in the aforementioned link - they don't deal with it). those are 2 different questions and i would appreciate learning from my problem in order to understand my mistake and additionally if the compliment of C is in NP
$endgroup$
– hps13
15 hours ago




1




1




$begingroup$
@xskxzr That's not the same question. The OP is asking whether it is in NP, not NP-complete. That question is also the complement of this one.
$endgroup$
– orlp
15 hours ago




$begingroup$
@xskxzr That's not the same question. The OP is asking whether it is in NP, not NP-complete. That question is also the complement of this one.
$endgroup$
– orlp
15 hours ago




1




1




$begingroup$
@hps13 Your certificate does not work. To prove that (p,n) is in C, you need to show that every number in that interval is not a prime, so you will need n divisors. But then your certificate will be of size O(n *log(n+p)), which is exponential compared to the size log(p)+log(n) of the word.
$endgroup$
– mlk
12 hours ago




$begingroup$
@hps13 Your certificate does not work. To prove that (p,n) is in C, you need to show that every number in that interval is not a prime, so you will need n divisors. But then your certificate will be of size O(n *log(n+p)), which is exponential compared to the size log(p)+log(n) of the word.
$endgroup$
– mlk
12 hours ago










2 Answers
2






active

oldest

votes


















5












$begingroup$

Note that there's always a prime between $p$ and $2p$ for any $p$ thus we can eliminate $n$ as a complexity parameter, as for any non-trivial instance we have $n < p$.



The complement (there exists a prime in range $[p, p+n]$) is quite easy (due to the shoulders of giants). There are various well-known polynomial size certificates of primality of $q$ you could return along with some $q in [p, p+n]$ if that prime $q$ exists. Thus your language is in co-NP.



However the question of whether it is in NP or not seems very hard to me. Our input size is $b$ such that $2^b approx p$. You need the existence of a $O(b^c)$ sized certificate that in the worst case asserts the compositeness of an exponentially large ($|[p, p+n]| approx |[p, 2p]| approx p approx 2^b$) series of consecutive integers. From my intuition about number theory, finding such a certificate would be a major result, but disproving its existence would be as well. But perhaps someone with more number theoretic knowledge than me can pitch in if either of those turn out to be easy.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you very much for your answer. it really helps me understand the background behind the problem, but as you've written, i am trying to check whether it is an NP problem and it is indeed a difficult yet interesting thing. thank you so much for what you've written, it really helps
    $endgroup$
    – hps13
    15 hours ago










  • $begingroup$
    I am wondering - did the explanation i tried to give is wrong? I am not sure, but here's what I think: for each word ⟨p,n⟩∈C we know that the word belongs to C because there exists a primal certificate - an nontrivial divisor to any of the numbers between [p,p+n],
    $endgroup$
    – hps13
    15 hours ago










  • $begingroup$
    "disproving its existence would be [a major result] as well" Well, proving this problem is not in NP would prove $NPneq co-NP$, which has a corollary $Pneq NP$. So yes, it would be quite a result!
    $endgroup$
    – Wojowu
    9 hours ago



















1












$begingroup$

As orlp says, it's trivial to give a certificate if the answer is "YES" (just give a non-trivial factor of each of the numbers in question), the problem is whether this certificate is polynomial sized.



There is a conjecture, backed by heuristics and some computational evidence, that there is a constant c > 0 such that for any k-bit prime p, there is a prime between p and p + c * k^2. If this conjecture is true, then the certificate is actually of size O (k^3) whenever the answer is YES, so the problem is in NP. But, it's just a conjecture.



But of course if we might be able to prove the much weaker conjecture "for some c, there is always a prime between p and p + c * k^3" or the even weaker conjecture "there is a polynomial pol (k) such that there is always a prime between p and p + pol(k)", then this problem would also be in NP. So we can prove that is in NP by proving something about prime gaps.



There is also the possibility that a less obvious and shorter certificate exists.






share|cite|improve this answer









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    2 Answers
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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    Note that there's always a prime between $p$ and $2p$ for any $p$ thus we can eliminate $n$ as a complexity parameter, as for any non-trivial instance we have $n < p$.



    The complement (there exists a prime in range $[p, p+n]$) is quite easy (due to the shoulders of giants). There are various well-known polynomial size certificates of primality of $q$ you could return along with some $q in [p, p+n]$ if that prime $q$ exists. Thus your language is in co-NP.



    However the question of whether it is in NP or not seems very hard to me. Our input size is $b$ such that $2^b approx p$. You need the existence of a $O(b^c)$ sized certificate that in the worst case asserts the compositeness of an exponentially large ($|[p, p+n]| approx |[p, 2p]| approx p approx 2^b$) series of consecutive integers. From my intuition about number theory, finding such a certificate would be a major result, but disproving its existence would be as well. But perhaps someone with more number theoretic knowledge than me can pitch in if either of those turn out to be easy.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      thank you very much for your answer. it really helps me understand the background behind the problem, but as you've written, i am trying to check whether it is an NP problem and it is indeed a difficult yet interesting thing. thank you so much for what you've written, it really helps
      $endgroup$
      – hps13
      15 hours ago










    • $begingroup$
      I am wondering - did the explanation i tried to give is wrong? I am not sure, but here's what I think: for each word ⟨p,n⟩∈C we know that the word belongs to C because there exists a primal certificate - an nontrivial divisor to any of the numbers between [p,p+n],
      $endgroup$
      – hps13
      15 hours ago










    • $begingroup$
      "disproving its existence would be [a major result] as well" Well, proving this problem is not in NP would prove $NPneq co-NP$, which has a corollary $Pneq NP$. So yes, it would be quite a result!
      $endgroup$
      – Wojowu
      9 hours ago
















    5












    $begingroup$

    Note that there's always a prime between $p$ and $2p$ for any $p$ thus we can eliminate $n$ as a complexity parameter, as for any non-trivial instance we have $n < p$.



    The complement (there exists a prime in range $[p, p+n]$) is quite easy (due to the shoulders of giants). There are various well-known polynomial size certificates of primality of $q$ you could return along with some $q in [p, p+n]$ if that prime $q$ exists. Thus your language is in co-NP.



    However the question of whether it is in NP or not seems very hard to me. Our input size is $b$ such that $2^b approx p$. You need the existence of a $O(b^c)$ sized certificate that in the worst case asserts the compositeness of an exponentially large ($|[p, p+n]| approx |[p, 2p]| approx p approx 2^b$) series of consecutive integers. From my intuition about number theory, finding such a certificate would be a major result, but disproving its existence would be as well. But perhaps someone with more number theoretic knowledge than me can pitch in if either of those turn out to be easy.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      thank you very much for your answer. it really helps me understand the background behind the problem, but as you've written, i am trying to check whether it is an NP problem and it is indeed a difficult yet interesting thing. thank you so much for what you've written, it really helps
      $endgroup$
      – hps13
      15 hours ago










    • $begingroup$
      I am wondering - did the explanation i tried to give is wrong? I am not sure, but here's what I think: for each word ⟨p,n⟩∈C we know that the word belongs to C because there exists a primal certificate - an nontrivial divisor to any of the numbers between [p,p+n],
      $endgroup$
      – hps13
      15 hours ago










    • $begingroup$
      "disproving its existence would be [a major result] as well" Well, proving this problem is not in NP would prove $NPneq co-NP$, which has a corollary $Pneq NP$. So yes, it would be quite a result!
      $endgroup$
      – Wojowu
      9 hours ago














    5












    5








    5





    $begingroup$

    Note that there's always a prime between $p$ and $2p$ for any $p$ thus we can eliminate $n$ as a complexity parameter, as for any non-trivial instance we have $n < p$.



    The complement (there exists a prime in range $[p, p+n]$) is quite easy (due to the shoulders of giants). There are various well-known polynomial size certificates of primality of $q$ you could return along with some $q in [p, p+n]$ if that prime $q$ exists. Thus your language is in co-NP.



    However the question of whether it is in NP or not seems very hard to me. Our input size is $b$ such that $2^b approx p$. You need the existence of a $O(b^c)$ sized certificate that in the worst case asserts the compositeness of an exponentially large ($|[p, p+n]| approx |[p, 2p]| approx p approx 2^b$) series of consecutive integers. From my intuition about number theory, finding such a certificate would be a major result, but disproving its existence would be as well. But perhaps someone with more number theoretic knowledge than me can pitch in if either of those turn out to be easy.






    share|cite|improve this answer









    $endgroup$



    Note that there's always a prime between $p$ and $2p$ for any $p$ thus we can eliminate $n$ as a complexity parameter, as for any non-trivial instance we have $n < p$.



    The complement (there exists a prime in range $[p, p+n]$) is quite easy (due to the shoulders of giants). There are various well-known polynomial size certificates of primality of $q$ you could return along with some $q in [p, p+n]$ if that prime $q$ exists. Thus your language is in co-NP.



    However the question of whether it is in NP or not seems very hard to me. Our input size is $b$ such that $2^b approx p$. You need the existence of a $O(b^c)$ sized certificate that in the worst case asserts the compositeness of an exponentially large ($|[p, p+n]| approx |[p, 2p]| approx p approx 2^b$) series of consecutive integers. From my intuition about number theory, finding such a certificate would be a major result, but disproving its existence would be as well. But perhaps someone with more number theoretic knowledge than me can pitch in if either of those turn out to be easy.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 15 hours ago









    orlporlp

    5,9651826




    5,9651826












    • $begingroup$
      thank you very much for your answer. it really helps me understand the background behind the problem, but as you've written, i am trying to check whether it is an NP problem and it is indeed a difficult yet interesting thing. thank you so much for what you've written, it really helps
      $endgroup$
      – hps13
      15 hours ago










    • $begingroup$
      I am wondering - did the explanation i tried to give is wrong? I am not sure, but here's what I think: for each word ⟨p,n⟩∈C we know that the word belongs to C because there exists a primal certificate - an nontrivial divisor to any of the numbers between [p,p+n],
      $endgroup$
      – hps13
      15 hours ago










    • $begingroup$
      "disproving its existence would be [a major result] as well" Well, proving this problem is not in NP would prove $NPneq co-NP$, which has a corollary $Pneq NP$. So yes, it would be quite a result!
      $endgroup$
      – Wojowu
      9 hours ago


















    • $begingroup$
      thank you very much for your answer. it really helps me understand the background behind the problem, but as you've written, i am trying to check whether it is an NP problem and it is indeed a difficult yet interesting thing. thank you so much for what you've written, it really helps
      $endgroup$
      – hps13
      15 hours ago










    • $begingroup$
      I am wondering - did the explanation i tried to give is wrong? I am not sure, but here's what I think: for each word ⟨p,n⟩∈C we know that the word belongs to C because there exists a primal certificate - an nontrivial divisor to any of the numbers between [p,p+n],
      $endgroup$
      – hps13
      15 hours ago










    • $begingroup$
      "disproving its existence would be [a major result] as well" Well, proving this problem is not in NP would prove $NPneq co-NP$, which has a corollary $Pneq NP$. So yes, it would be quite a result!
      $endgroup$
      – Wojowu
      9 hours ago
















    $begingroup$
    thank you very much for your answer. it really helps me understand the background behind the problem, but as you've written, i am trying to check whether it is an NP problem and it is indeed a difficult yet interesting thing. thank you so much for what you've written, it really helps
    $endgroup$
    – hps13
    15 hours ago




    $begingroup$
    thank you very much for your answer. it really helps me understand the background behind the problem, but as you've written, i am trying to check whether it is an NP problem and it is indeed a difficult yet interesting thing. thank you so much for what you've written, it really helps
    $endgroup$
    – hps13
    15 hours ago












    $begingroup$
    I am wondering - did the explanation i tried to give is wrong? I am not sure, but here's what I think: for each word ⟨p,n⟩∈C we know that the word belongs to C because there exists a primal certificate - an nontrivial divisor to any of the numbers between [p,p+n],
    $endgroup$
    – hps13
    15 hours ago




    $begingroup$
    I am wondering - did the explanation i tried to give is wrong? I am not sure, but here's what I think: for each word ⟨p,n⟩∈C we know that the word belongs to C because there exists a primal certificate - an nontrivial divisor to any of the numbers between [p,p+n],
    $endgroup$
    – hps13
    15 hours ago












    $begingroup$
    "disproving its existence would be [a major result] as well" Well, proving this problem is not in NP would prove $NPneq co-NP$, which has a corollary $Pneq NP$. So yes, it would be quite a result!
    $endgroup$
    – Wojowu
    9 hours ago




    $begingroup$
    "disproving its existence would be [a major result] as well" Well, proving this problem is not in NP would prove $NPneq co-NP$, which has a corollary $Pneq NP$. So yes, it would be quite a result!
    $endgroup$
    – Wojowu
    9 hours ago











    1












    $begingroup$

    As orlp says, it's trivial to give a certificate if the answer is "YES" (just give a non-trivial factor of each of the numbers in question), the problem is whether this certificate is polynomial sized.



    There is a conjecture, backed by heuristics and some computational evidence, that there is a constant c > 0 such that for any k-bit prime p, there is a prime between p and p + c * k^2. If this conjecture is true, then the certificate is actually of size O (k^3) whenever the answer is YES, so the problem is in NP. But, it's just a conjecture.



    But of course if we might be able to prove the much weaker conjecture "for some c, there is always a prime between p and p + c * k^3" or the even weaker conjecture "there is a polynomial pol (k) such that there is always a prime between p and p + pol(k)", then this problem would also be in NP. So we can prove that is in NP by proving something about prime gaps.



    There is also the possibility that a less obvious and shorter certificate exists.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      As orlp says, it's trivial to give a certificate if the answer is "YES" (just give a non-trivial factor of each of the numbers in question), the problem is whether this certificate is polynomial sized.



      There is a conjecture, backed by heuristics and some computational evidence, that there is a constant c > 0 such that for any k-bit prime p, there is a prime between p and p + c * k^2. If this conjecture is true, then the certificate is actually of size O (k^3) whenever the answer is YES, so the problem is in NP. But, it's just a conjecture.



      But of course if we might be able to prove the much weaker conjecture "for some c, there is always a prime between p and p + c * k^3" or the even weaker conjecture "there is a polynomial pol (k) such that there is always a prime between p and p + pol(k)", then this problem would also be in NP. So we can prove that is in NP by proving something about prime gaps.



      There is also the possibility that a less obvious and shorter certificate exists.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        As orlp says, it's trivial to give a certificate if the answer is "YES" (just give a non-trivial factor of each of the numbers in question), the problem is whether this certificate is polynomial sized.



        There is a conjecture, backed by heuristics and some computational evidence, that there is a constant c > 0 such that for any k-bit prime p, there is a prime between p and p + c * k^2. If this conjecture is true, then the certificate is actually of size O (k^3) whenever the answer is YES, so the problem is in NP. But, it's just a conjecture.



        But of course if we might be able to prove the much weaker conjecture "for some c, there is always a prime between p and p + c * k^3" or the even weaker conjecture "there is a polynomial pol (k) such that there is always a prime between p and p + pol(k)", then this problem would also be in NP. So we can prove that is in NP by proving something about prime gaps.



        There is also the possibility that a less obvious and shorter certificate exists.






        share|cite|improve this answer









        $endgroup$



        As orlp says, it's trivial to give a certificate if the answer is "YES" (just give a non-trivial factor of each of the numbers in question), the problem is whether this certificate is polynomial sized.



        There is a conjecture, backed by heuristics and some computational evidence, that there is a constant c > 0 such that for any k-bit prime p, there is a prime between p and p + c * k^2. If this conjecture is true, then the certificate is actually of size O (k^3) whenever the answer is YES, so the problem is in NP. But, it's just a conjecture.



        But of course if we might be able to prove the much weaker conjecture "for some c, there is always a prime between p and p + c * k^3" or the even weaker conjecture "there is a polynomial pol (k) such that there is always a prime between p and p + pol(k)", then this problem would also be in NP. So we can prove that is in NP by proving something about prime gaps.



        There is also the possibility that a less obvious and shorter certificate exists.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 6 hours ago









        gnasher729gnasher729

        11.6k1217




        11.6k1217






















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