Number of divisors of a number - in NP?
$begingroup$
I'm trying to show that the language ${(m,n) | m space text{has exactly} space n space text{divisors}}$ is in NP.
The input $(m,n)$ is in binary.
The non-deterministic Turing machine for the language would be:
1) Guess the prime factors of $m.$
2) Verify that $prod i(di+1) = n.$
The problem is that I can't find a way to factorize in polynomial time (in the input) the number $m.$
If stage $1$ takes m steps then it would be $m = 2^{log (m)}$ and the whole algorithm would run in exponential time.
How can I factorize a number in binary in polynomial time (polynomial in the input)?
turing-machines np-complete
edited Dec 15 '18 at 8:39
Gaby Alfonso
1,098317
asked Dec 15 '18 at 8:18
caffeincaffein
11
$endgroup$
|
show 5 more comments
$begingroup$
I'm trying to show that the language ${(m,n) | m space text{has exactly} space n space text{divisors}}$ is in NP.
The input $(m,n)$ is in binary.
The non-deterministic Turing machine for the language would be:
1) Guess the prime factors of $m.$
2) Verify that $prod i(di+1) = n.$
The problem is that I can't find a way to factorize in polynomial time (in the input) the number $m.$
If stage $1$ takes m steps then it would be $m = 2^{log (m)}$ and the whole algorithm would run in exponential time.
How can I factorize a number in binary in polynomial time (polynomial in the input)?
turing-machines np-complete
edited Dec 15 '18 at 8:39
Gaby Alfonso
1,098317
asked Dec 15 '18 at 8:18
caffeincaffein
11
$endgroup$
$begingroup$
As far as I know, it is an open problem whether integer factorization is in $P$ or not. No efficient (that is polynomial in time) method is currently known.
$endgroup$
– Peter
Dec 15 '18 at 8:24
$begingroup$
en.wikipedia.org/wiki/…
$endgroup$
– David Peterson
Dec 15 '18 at 8:25
1
$begingroup$
The whole point is that on a non-deterministic turing machine you can guess the correct solution non-deterministically in a single step... you only need to verify it in polynomial time.
$endgroup$
– spaceisdarkgreen
Dec 15 '18 at 8:29
$begingroup$
@spaceisdarkgreen But the problem is that the amount of numbers i need to guess for the correct answer might be polynomial in the size of the binary input. If i have log(m) bits and we have m factors that would be m = 2 ^ (logm) --> exponential
$endgroup$
– caffein
Dec 15 '18 at 8:31
2
$begingroup$
The number of prime factors of $m$ is logarithmic in $m.$ And in any event this has nothing to do with 'factorizing a number in polynomial time', just 'listing the (known) factors in polynomial time'. (Though if the latter were impossible, the former would be trivially impossible, rather than probably impossible, but a hard open question.)
$endgroup$
– spaceisdarkgreen
Dec 15 '18 at 8:53
|
show 5 more comments
$begingroup$
I'm trying to show that the language ${(m,n) | m space text{has exactly} space n space text{divisors}}$ is in NP.
The input $(m,n)$ is in binary.
The non-deterministic Turing machine for the language would be:
1) Guess the prime factors of $m.$
2) Verify that $prod i(di+1) = n.$
The problem is that I can't find a way to factorize in polynomial time (in the input) the number $m.$
If stage $1$ takes m steps then it would be $m = 2^{log (m)}$ and the whole algorithm would run in exponential time.
How can I factorize a number in binary in polynomial time (polynomial in the input)?
turing-machines np-complete
edited Dec 15 '18 at 8:39
Gaby Alfonso
1,098317
asked Dec 15 '18 at 8:18
caffeincaffein
11
$endgroup$
I'm trying to show that the language ${(m,n) | m space text{has exactly} space n space text{divisors}}$ is in NP.
The input $(m,n)$ is in binary.
The non-deterministic Turing machine for the language would be:
1) Guess the prime factors of $m.$
2) Verify that $prod i(di+1) = n.$
The problem is that I can't find a way to factorize in polynomial time (in the input) the number $m.$
If stage $1$ takes m steps then it would be $m = 2^{log (m)}$ and the whole algorithm would run in exponential time.
How can I factorize a number in binary in polynomial time (polynomial in the input)?
turing-machines np-complete
turing-machines np-complete
edited Dec 15 '18 at 8:39
Gaby Alfonso
1,098317
asked Dec 15 '18 at 8:18
caffeincaffein
11
edited Dec 15 '18 at 8:39
Gaby Alfonso
1,098317
asked Dec 15 '18 at 8:18
caffeincaffein
11
edited Dec 15 '18 at 8:39
Gaby Alfonso
1,098317
edited Dec 15 '18 at 8:39
Gaby Alfonso
1,098317
edited Dec 15 '18 at 8:39
Gaby Alfonso
1,098317
1,098317
asked Dec 15 '18 at 8:18
caffeincaffein
11
asked Dec 15 '18 at 8:18
caffeincaffein
11
asked Dec 15 '18 at 8:18
caffeincaffein
11
11
$begingroup$
As far as I know, it is an open problem whether integer factorization is in $P$ or not. No efficient (that is polynomial in time) method is currently known.
$endgroup$
– Peter
Dec 15 '18 at 8:24
$begingroup$
en.wikipedia.org/wiki/…
$endgroup$
– David Peterson
Dec 15 '18 at 8:25
1
$begingroup$
The whole point is that on a non-deterministic turing machine you can guess the correct solution non-deterministically in a single step... you only need to verify it in polynomial time.
$endgroup$
– spaceisdarkgreen
Dec 15 '18 at 8:29
$begingroup$
@spaceisdarkgreen But the problem is that the amount of numbers i need to guess for the correct answer might be polynomial in the size of the binary input. If i have log(m) bits and we have m factors that would be m = 2 ^ (logm) --> exponential
$endgroup$
– caffein
Dec 15 '18 at 8:31
2
$begingroup$
The number of prime factors of $m$ is logarithmic in $m.$ And in any event this has nothing to do with 'factorizing a number in polynomial time', just 'listing the (known) factors in polynomial time'. (Though if the latter were impossible, the former would be trivially impossible, rather than probably impossible, but a hard open question.)
$endgroup$
– spaceisdarkgreen
Dec 15 '18 at 8:53
|
show 5 more comments
$begingroup$
As far as I know, it is an open problem whether integer factorization is in $P$ or not. No efficient (that is polynomial in time) method is currently known.
$endgroup$
– Peter
Dec 15 '18 at 8:24
$begingroup$
en.wikipedia.org/wiki/…
$endgroup$
– David Peterson
Dec 15 '18 at 8:25
1
$begingroup$
The whole point is that on a non-deterministic turing machine you can guess the correct solution non-deterministically in a single step... you only need to verify it in polynomial time.
$endgroup$
– spaceisdarkgreen
Dec 15 '18 at 8:29
$begingroup$
@spaceisdarkgreen But the problem is that the amount of numbers i need to guess for the correct answer might be polynomial in the size of the binary input. If i have log(m) bits and we have m factors that would be m = 2 ^ (logm) --> exponential
$endgroup$
– caffein
Dec 15 '18 at 8:31
2
$begingroup$
The number of prime factors of $m$ is logarithmic in $m.$ And in any event this has nothing to do with 'factorizing a number in polynomial time', just 'listing the (known) factors in polynomial time'. (Though if the latter were impossible, the former would be trivially impossible, rather than probably impossible, but a hard open question.)
$endgroup$
– spaceisdarkgreen
Dec 15 '18 at 8:53
$begingroup$
As far as I know, it is an open problem whether integer factorization is in $P$ or not. No efficient (that is polynomial in time) method is currently known.
$endgroup$
– Peter
Dec 15 '18 at 8:24
$begingroup$
As far as I know, it is an open problem whether integer factorization is in $P$ or not. No efficient (that is polynomial in time) method is currently known.
$endgroup$
– Peter
Dec 15 '18 at 8:24
$begingroup$
en.wikipedia.org/wiki/…
$endgroup$
– David Peterson
Dec 15 '18 at 8:25
$begingroup$
en.wikipedia.org/wiki/…
$endgroup$
– David Peterson
Dec 15 '18 at 8:25
1
1
$begingroup$
The whole point is that on a non-deterministic turing machine you can guess the correct solution non-deterministically in a single step... you only need to verify it in polynomial time.
$endgroup$
– spaceisdarkgreen
Dec 15 '18 at 8:29
$begingroup$
The whole point is that on a non-deterministic turing machine you can guess the correct solution non-deterministically in a single step... you only need to verify it in polynomial time.
$endgroup$
– spaceisdarkgreen
Dec 15 '18 at 8:29
$begingroup$
@spaceisdarkgreen But the problem is that the amount of numbers i need to guess for the correct answer might be polynomial in the size of the binary input. If i have log(m) bits and we have m factors that would be m = 2 ^ (logm) --> exponential
$endgroup$
– caffein
Dec 15 '18 at 8:31
$begingroup$
@spaceisdarkgreen But the problem is that the amount of numbers i need to guess for the correct answer might be polynomial in the size of the binary input. If i have log(m) bits and we have m factors that would be m = 2 ^ (logm) --> exponential
$endgroup$
– caffein
Dec 15 '18 at 8:31
2
2
$begingroup$
The number of prime factors of $m$ is logarithmic in $m.$ And in any event this has nothing to do with 'factorizing a number in polynomial time', just 'listing the (known) factors in polynomial time'. (Though if the latter were impossible, the former would be trivially impossible, rather than probably impossible, but a hard open question.)
$endgroup$
– spaceisdarkgreen
Dec 15 '18 at 8:53
$begingroup$
The number of prime factors of $m$ is logarithmic in $m.$ And in any event this has nothing to do with 'factorizing a number in polynomial time', just 'listing the (known) factors in polynomial time'. (Though if the latter were impossible, the former would be trivially impossible, rather than probably impossible, but a hard open question.)
$endgroup$
– spaceisdarkgreen
Dec 15 '18 at 8:53
|
show 5 more comments
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$begingroup$
As far as I know, it is an open problem whether integer factorization is in $P$ or not. No efficient (that is polynomial in time) method is currently known.
$endgroup$
– Peter
Dec 15 '18 at 8:24
$begingroup$
en.wikipedia.org/wiki/…
$endgroup$
– David Peterson
Dec 15 '18 at 8:25
1
$begingroup$
The whole point is that on a non-deterministic turing machine you can guess the correct solution non-deterministically in a single step... you only need to verify it in polynomial time.
$endgroup$
– spaceisdarkgreen
Dec 15 '18 at 8:29
$begingroup$
@spaceisdarkgreen But the problem is that the amount of numbers i need to guess for the correct answer might be polynomial in the size of the binary input. If i have log(m) bits and we have m factors that would be m = 2 ^ (logm) --> exponential
$endgroup$
– caffein
Dec 15 '18 at 8:31
2
$begingroup$
The number of prime factors of $m$ is logarithmic in $m.$ And in any event this has nothing to do with 'factorizing a number in polynomial time', just 'listing the (known) factors in polynomial time'. (Though if the latter were impossible, the former would be trivially impossible, rather than probably impossible, but a hard open question.)
$endgroup$
– spaceisdarkgreen
Dec 15 '18 at 8:53