Why was primality test thought to be NP?












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To check if $n$ is prime, one only need to try dividing $n$ by numbers up to $sqrt{n}$, meaning that the complexity would be $O(sqrt{n})$. In my opinion, $O(sqrt{n}) < O(n)$ so this simple algorithm is already P. But why did people think that primality test is NP, and were surprised by the AKS primality test?






complexity-theory time-complexity np







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$endgroup$








  • 2




    $begingroup$
    Possible duplicate of Primality testing: Why is dividing a number $n$ by every integer between 2 and $sqrt{n}$ an inefficient test?
    $endgroup$
    – David Richerby
    1 hour ago






  • 4




    $begingroup$
    If something is in P then it is automatically in NP. You seem to be using "NP" as shorthand for either "NP - P" or "NP-complete".
    $endgroup$
    – John Coleman
    1 hour ago
















3












$begingroup$


To check if $n$ is prime, one only need to try dividing $n$ by numbers up to $sqrt{n}$, meaning that the complexity would be $O(sqrt{n})$. In my opinion, $O(sqrt{n}) < O(n)$ so this simple algorithm is already P. But why did people think that primality test is NP, and were surprised by the AKS primality test?






complexity-theory time-complexity np







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Possible duplicate of Primality testing: Why is dividing a number $n$ by every integer between 2 and $sqrt{n}$ an inefficient test?
    $endgroup$
    – David Richerby
    1 hour ago






  • 4




    $begingroup$
    If something is in P then it is automatically in NP. You seem to be using "NP" as shorthand for either "NP - P" or "NP-complete".
    $endgroup$
    – John Coleman
    1 hour ago














3












3








3





$begingroup$


To check if $n$ is prime, one only need to try dividing $n$ by numbers up to $sqrt{n}$, meaning that the complexity would be $O(sqrt{n})$. In my opinion, $O(sqrt{n}) < O(n)$ so this simple algorithm is already P. But why did people think that primality test is NP, and were surprised by the AKS primality test?






complexity-theory time-complexity np







share|cite|improve this question











$endgroup$




To check if $n$ is prime, one only need to try dividing $n$ by numbers up to $sqrt{n}$, meaning that the complexity would be $O(sqrt{n})$. In my opinion, $O(sqrt{n}) < O(n)$ so this simple algorithm is already P. But why did people think that primality test is NP, and were surprised by the AKS primality test?






complexity-theory time-complexity np




complexity-theory time-complexity np






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 1 hour ago









David Richerby

71k16109199




71k16109199










asked 5 hours ago









DimenDimen

364




364








  • 2




    $begingroup$
    Possible duplicate of Primality testing: Why is dividing a number $n$ by every integer between 2 and $sqrt{n}$ an inefficient test?
    $endgroup$
    – David Richerby
    1 hour ago






  • 4




    $begingroup$
    If something is in P then it is automatically in NP. You seem to be using "NP" as shorthand for either "NP - P" or "NP-complete".
    $endgroup$
    – John Coleman
    1 hour ago














  • 2




    $begingroup$
    Possible duplicate of Primality testing: Why is dividing a number $n$ by every integer between 2 and $sqrt{n}$ an inefficient test?
    $endgroup$
    – David Richerby
    1 hour ago






  • 4




    $begingroup$
    If something is in P then it is automatically in NP. You seem to be using "NP" as shorthand for either "NP - P" or "NP-complete".
    $endgroup$
    – John Coleman
    1 hour ago








2




2




$begingroup$
Possible duplicate of Primality testing: Why is dividing a number $n$ by every integer between 2 and $sqrt{n}$ an inefficient test?
$endgroup$
– David Richerby
1 hour ago




$begingroup$
Possible duplicate of Primality testing: Why is dividing a number $n$ by every integer between 2 and $sqrt{n}$ an inefficient test?
$endgroup$
– David Richerby
1 hour ago




4




4




$begingroup$
If something is in P then it is automatically in NP. You seem to be using "NP" as shorthand for either "NP - P" or "NP-complete".
$endgroup$
– John Coleman
1 hour ago




$begingroup$
If something is in P then it is automatically in NP. You seem to be using "NP" as shorthand for either "NP - P" or "NP-complete".
$endgroup$
– John Coleman
1 hour ago










1 Answer
1






active

oldest

votes


















3












$begingroup$

To start, any decision problem with a witness verifiable for a "yes" answer in polynomial time is in NP.



$bullet$ The problem 'Is $p$ composite?' has a witness $k$, and it can be tested in polynomial time whether $k$ divides $p$, therefore is in NP.



$bullet$ The problem 'Is $p$ prime?' has a polynomial certificate (which can be found here), therefore is also in NP.



Now, your complexity analysis is not correct. If your input is a number $p$, then the input size is $log (p)$, since you need $log(p)$ bits to represent it. Therefore, $n=log(p)$ and $sqrt{p}=frac{1}{2} log(p)$ bits. So the time required (in respect to $n$) checking all the numbers up to $sqrt{p}$ will be $2^{frac{1}{2}n}$, and is in fact exponential in terms on input size $n$.




The question "Where exactly does the Primality problem lie" has been answered in the link here






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    "Is p composite" is obviously in NP, because you can bring a witness easily. "Is p prime" is in NP, but not obviously at all - there is a quite deep theorem that every prime p has a short prove of being prime, which I think involves complete factorisations of p+1 and/or p-1.
    $endgroup$
    – gnasher729
    2 hours ago










  • $begingroup$
    @gnasher729 thank you for spotting it! I edited it in.
    $endgroup$
    – lox
    1 hour ago












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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









3












$begingroup$

To start, any decision problem with a witness verifiable for a "yes" answer in polynomial time is in NP.



$bullet$ The problem 'Is $p$ composite?' has a witness $k$, and it can be tested in polynomial time whether $k$ divides $p$, therefore is in NP.



$bullet$ The problem 'Is $p$ prime?' has a polynomial certificate (which can be found here), therefore is also in NP.



Now, your complexity analysis is not correct. If your input is a number $p$, then the input size is $log (p)$, since you need $log(p)$ bits to represent it. Therefore, $n=log(p)$ and $sqrt{p}=frac{1}{2} log(p)$ bits. So the time required (in respect to $n$) checking all the numbers up to $sqrt{p}$ will be $2^{frac{1}{2}n}$, and is in fact exponential in terms on input size $n$.




The question "Where exactly does the Primality problem lie" has been answered in the link here






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    "Is p composite" is obviously in NP, because you can bring a witness easily. "Is p prime" is in NP, but not obviously at all - there is a quite deep theorem that every prime p has a short prove of being prime, which I think involves complete factorisations of p+1 and/or p-1.
    $endgroup$
    – gnasher729
    2 hours ago










  • $begingroup$
    @gnasher729 thank you for spotting it! I edited it in.
    $endgroup$
    – lox
    1 hour ago
















3












$begingroup$

To start, any decision problem with a witness verifiable for a "yes" answer in polynomial time is in NP.



$bullet$ The problem 'Is $p$ composite?' has a witness $k$, and it can be tested in polynomial time whether $k$ divides $p$, therefore is in NP.



$bullet$ The problem 'Is $p$ prime?' has a polynomial certificate (which can be found here), therefore is also in NP.



Now, your complexity analysis is not correct. If your input is a number $p$, then the input size is $log (p)$, since you need $log(p)$ bits to represent it. Therefore, $n=log(p)$ and $sqrt{p}=frac{1}{2} log(p)$ bits. So the time required (in respect to $n$) checking all the numbers up to $sqrt{p}$ will be $2^{frac{1}{2}n}$, and is in fact exponential in terms on input size $n$.




The question "Where exactly does the Primality problem lie" has been answered in the link here






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    "Is p composite" is obviously in NP, because you can bring a witness easily. "Is p prime" is in NP, but not obviously at all - there is a quite deep theorem that every prime p has a short prove of being prime, which I think involves complete factorisations of p+1 and/or p-1.
    $endgroup$
    – gnasher729
    2 hours ago










  • $begingroup$
    @gnasher729 thank you for spotting it! I edited it in.
    $endgroup$
    – lox
    1 hour ago














3












3








3





$begingroup$

To start, any decision problem with a witness verifiable for a "yes" answer in polynomial time is in NP.



$bullet$ The problem 'Is $p$ composite?' has a witness $k$, and it can be tested in polynomial time whether $k$ divides $p$, therefore is in NP.



$bullet$ The problem 'Is $p$ prime?' has a polynomial certificate (which can be found here), therefore is also in NP.



Now, your complexity analysis is not correct. If your input is a number $p$, then the input size is $log (p)$, since you need $log(p)$ bits to represent it. Therefore, $n=log(p)$ and $sqrt{p}=frac{1}{2} log(p)$ bits. So the time required (in respect to $n$) checking all the numbers up to $sqrt{p}$ will be $2^{frac{1}{2}n}$, and is in fact exponential in terms on input size $n$.




The question "Where exactly does the Primality problem lie" has been answered in the link here






share|cite|improve this answer











$endgroup$



To start, any decision problem with a witness verifiable for a "yes" answer in polynomial time is in NP.



$bullet$ The problem 'Is $p$ composite?' has a witness $k$, and it can be tested in polynomial time whether $k$ divides $p$, therefore is in NP.



$bullet$ The problem 'Is $p$ prime?' has a polynomial certificate (which can be found here), therefore is also in NP.



Now, your complexity analysis is not correct. If your input is a number $p$, then the input size is $log (p)$, since you need $log(p)$ bits to represent it. Therefore, $n=log(p)$ and $sqrt{p}=frac{1}{2} log(p)$ bits. So the time required (in respect to $n$) checking all the numbers up to $sqrt{p}$ will be $2^{frac{1}{2}n}$, and is in fact exponential in terms on input size $n$.




The question "Where exactly does the Primality problem lie" has been answered in the link here







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 17 mins ago

























answered 4 hours ago









loxlox

611110




611110








  • 1




    $begingroup$
    "Is p composite" is obviously in NP, because you can bring a witness easily. "Is p prime" is in NP, but not obviously at all - there is a quite deep theorem that every prime p has a short prove of being prime, which I think involves complete factorisations of p+1 and/or p-1.
    $endgroup$
    – gnasher729
    2 hours ago










  • $begingroup$
    @gnasher729 thank you for spotting it! I edited it in.
    $endgroup$
    – lox
    1 hour ago














  • 1




    $begingroup$
    "Is p composite" is obviously in NP, because you can bring a witness easily. "Is p prime" is in NP, but not obviously at all - there is a quite deep theorem that every prime p has a short prove of being prime, which I think involves complete factorisations of p+1 and/or p-1.
    $endgroup$
    – gnasher729
    2 hours ago










  • $begingroup$
    @gnasher729 thank you for spotting it! I edited it in.
    $endgroup$
    – lox
    1 hour ago








1




1




$begingroup$
"Is p composite" is obviously in NP, because you can bring a witness easily. "Is p prime" is in NP, but not obviously at all - there is a quite deep theorem that every prime p has a short prove of being prime, which I think involves complete factorisations of p+1 and/or p-1.
$endgroup$
– gnasher729
2 hours ago




$begingroup$
"Is p composite" is obviously in NP, because you can bring a witness easily. "Is p prime" is in NP, but not obviously at all - there is a quite deep theorem that every prime p has a short prove of being prime, which I think involves complete factorisations of p+1 and/or p-1.
$endgroup$
– gnasher729
2 hours ago












$begingroup$
@gnasher729 thank you for spotting it! I edited it in.
$endgroup$
– lox
1 hour ago




$begingroup$
@gnasher729 thank you for spotting it! I edited it in.
$endgroup$
– lox
1 hour ago


















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