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” $in mathbfNP$Rice's Theorem - usage on $DFA$ or $LBA$
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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” $in mathbfNP$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:
beginalign
C=leftlangle p,nranglemidright.& left. p text and $n$ are natural numbersright.\
&left.text and there's no prime number in the rangeleft[p,p+nright]right
endalign
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
asked 7 hours ago
hps13hps13
205
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.
$endgroup$
|
show 1 more comment
$begingroup$
I was wondering if the following language belongs to NP class and if its complimentary belongs to NP class:
beginalign
C=leftlangle p,nranglemidright.& left. p text and $n$ are natural numbersright.\
&left.text and there's no prime number in the rangeleft[p,p+nright]right
endalign
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
asked 7 hours ago
hps13hps13
205
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
7 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
6 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
6 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
6 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
3 hours ago
|
show 1 more comment
$begingroup$
I was wondering if the following language belongs to NP class and if its complimentary belongs to NP class:
beginalign
C=leftlangle p,nranglemidright.& left. p text and $n$ are natural numbersright.\
&left.text and there's no prime number in the rangeleft[p,p+nright]right
endalign
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
asked 7 hours ago
hps13hps13
205
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:
beginalign
C=leftlangle p,nranglemidright.& left. p text and $n$ are natural numbersright.\
&left.text and there's no prime number in the rangeleft[p,p+nright]right
endalign
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
asked 7 hours ago
hps13hps13
205
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.
asked 7 hours ago
hps13hps13
205
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.
edited 6 hours ago
hps13
asked 7 hours ago
hps13hps13
205
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.
asked 7 hours ago
hps13hps13
205
asked 7 hours ago
hps13hps13
205
205
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.
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.
hps13 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
It may depend on how $p$ and $n$ are represented (by unary or binary).
$endgroup$
– xskxzr
7 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
6 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
6 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
6 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
3 hours ago
|
show 1 more comment
$begingroup$
It may depend on how $p$ and $n$ are represented (by unary or binary).
$endgroup$
– xskxzr
7 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
6 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
6 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
6 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
3 hours ago
$begingroup$
It may depend on how $p$ and $n$ are represented (by unary or binary).
$endgroup$
– xskxzr
7 hours ago
$begingroup$
It may depend on how $p$ and $n$ are represented (by unary or binary).
$endgroup$
– xskxzr
7 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
6 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
6 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
6 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
6 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
6 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
6 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
3 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
3 hours ago
|
show 1 more comment
1 Answer
1
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.
answered 6 hours ago
orlporlp
5,9551826
$endgroup$
$begingroup$
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
6 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
6 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
49 mins ago
add a comment |
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1 Answer
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1 Answer
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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.
answered 6 hours ago
orlporlp
5,9551826
$endgroup$
$begingroup$
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
6 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
6 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
49 mins 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.
answered 6 hours ago
orlporlp
5,9551826
$endgroup$
$begingroup$
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
6 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
6 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
49 mins 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.
answered 6 hours ago
orlporlp
5,9551826
$endgroup$
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 6 hours ago
orlporlp
5,9551826
answered 6 hours ago
orlporlp
5,9551826
answered 6 hours ago
orlporlp
5,9551826
answered 6 hours ago
orlporlp
5,9551826
5,9551826
$begingroup$
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
6 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
6 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
49 mins 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
6 hours ago
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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
6 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
49 mins ago
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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
6 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
6 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
6 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
6 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
49 mins 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
49 mins ago
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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$begingroup$
It may depend on how $p$ and $n$ are represented (by unary or binary).
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– xskxzr
7 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
6 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
6 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.
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– orlp
6 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.
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– mlk
3 hours ago