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 $LBA$
$begingroup$
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).
complexity-theory turing-machines computability np decision-problem
New contributor
$endgroup$
|
show 2 more comments
$begingroup$
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).
complexity-theory turing-machines computability np decision-problem
New contributor
$endgroup$
$begingroup$
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
|
show 2 more comments
$begingroup$
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).
complexity-theory turing-machines computability np decision-problem
New contributor
$endgroup$
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
complexity-theory turing-machines computability np decision-problem
New contributor
New contributor
edited 15 hours ago
hps13
New contributor
asked 16 hours ago
hps13hps13
205
205
New contributor
New contributor
$begingroup$
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
|
show 2 more comments
$begingroup$
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
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
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.
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
add a comment |
$begingroup$
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],
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– hps13
15 hours ago
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"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!
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– Wojowu
9 hours ago
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"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!
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– Wojowu
9 hours ago
add a comment |
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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.
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add a comment |
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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.
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add a comment |
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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.
$endgroup$
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.
answered 6 hours ago
gnasher729gnasher729
11.6k1217
11.6k1217
add a comment |
add a comment |
hps13 is a new contributor. Be nice, and check out our Code of Conduct.
hps13 is a new contributor. Be nice, and check out our Code of Conduct.
hps13 is a new contributor. Be nice, and check out our Code of Conduct.
hps13 is a new contributor. Be nice, and check out our Code of Conduct.
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It may depend on how $p$ and $n$ are represented (by unary or binary).
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– xskxzr
16 hours ago
1
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Possible duplicate of Is determining if there is a prime in an interval known to be in P or NP-complete?
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– xskxzr
15 hours ago
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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
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– hps13
15 hours ago
1
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@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.
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– orlp
15 hours ago
1
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@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.
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– mlk
12 hours ago