What is the fastest integer factorization to break RSA?
$begingroup$
I read on Wikipedia, the fastest Algorithm for breaking RSA is GNFS.
And in one IEEE paper (MVFactor: A method to decrease processing time for factorization algorithm), I read the fastest algorithms are TDM, FFM and VFactor.
Which of these is actually right?
factoring
edited 54 mins ago
kelalaka
8,60522351
asked 2 hours ago
user56036user56036
61
New contributor
user56036 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
I read on Wikipedia, the fastest Algorithm for breaking RSA is GNFS.
And in one IEEE paper (MVFactor: A method to decrease processing time for factorization algorithm), I read the fastest algorithms are TDM, FFM and VFactor.
Which of these is actually right?
factoring
edited 54 mins ago
kelalaka
8,60522351
asked 2 hours ago
user56036user56036
61
New contributor
user56036 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I read on Wikipedia, the fastest Algorithm for breaking RSA is GNFS.
And in one IEEE paper (MVFactor: A method to decrease processing time for factorization algorithm), I read the fastest algorithms are TDM, FFM and VFactor.
Which of these is actually right?
factoring
edited 54 mins ago
kelalaka
8,60522351
asked 2 hours ago
user56036user56036
61
New contributor
user56036 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I read on Wikipedia, the fastest Algorithm for breaking RSA is GNFS.
And in one IEEE paper (MVFactor: A method to decrease processing time for factorization algorithm), I read the fastest algorithms are TDM, FFM and VFactor.
Which of these is actually right?
factoring
factoring
edited 54 mins ago
kelalaka
8,60522351
asked 2 hours ago
user56036user56036
61
New contributor
user56036 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 54 mins ago
kelalaka
8,60522351
asked 2 hours ago
user56036user56036
61
New contributor
user56036 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 54 mins ago
kelalaka
8,60522351
edited 54 mins ago
kelalaka
8,60522351
edited 54 mins ago
kelalaka
8,60522351
8,60522351
asked 2 hours ago
user56036user56036
61
New contributor
user56036 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
user56036user56036
61
asked 2 hours ago
user56036user56036
61
61
New contributor
user56036 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user56036 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
user56036 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The IEEE paper is silly.
The factorization method they give is quite slow, except for rare cases. For example, in their table 1, where they proudly show that their improved algorithm takes 653.14 seconds to factor a 67 bit number; well, I just tried it using a more conventional algorithm, and it took 6msec; yes, that's 100,000 times as fast...
answered 1 hour ago
ponchoponcho
93.5k2146243
$endgroup$
1
$begingroup$
Well I think the point of the paper is to improve upon Fermat-Factoring class algorithms, so it is expected that the given algorithm(s) get beaten by the more standard ones for small sizes, but excel on large inputs with (relatively small) prime differences?
$endgroup$
– SEJPM♦
1 hour ago
$begingroup$
@SEJPM: if that's the case, then they probably shouldn't go on so much about RSA (where the probability of having a sufficiently small difference is tiny)
$endgroup$
– poncho
1 hour ago
add a comment |
$begingroup$
Which of these is actually right?
Both. From reading the abstract it appears the papper doesn't claim that "VFactor" or Fermat Factorization ("FFM") or Trial Division ("TDM") are the best methods in general. However, if the difference between primes $p,q$ with $n=pq$ is really small, like $ll2^{100}$$;dagger$, then FFM (and probably the VFactor variants as well) will be a lot faster.
Though in general the difference between two same-length random primes is about $sqrt{n}/2$ which is about $2^{1024}$ for realistically sized moduli, so these attacks don't work there. Even with 400-bit moduli, which are somewhat easily crackable using a home desktop using the GNFS, this difference is still about $2^{200}$ and thus way too large.
Of course the implementation of the key generation may be faulty and emit primes in a too small interval and it's in these cases where these specialized algorithms really shine.
$dagger$: "$ll$" meaning "a lot less" here
answered 1 hour ago
SEJPM♦SEJPM
29.3k659139
$endgroup$
add a comment |
$begingroup$
Quantum algorithms
There is of course Shor's algorithm, but as this algorithm only runs on quantum computers with a lot of qubits it's not capable to factor larger numbers than $21$ (reference).
There are multiple apparent new records using adiabatic quantum computation, although some are apparently stunts: See fgrieu's answer on a related question.
Classical algorithms
The general number field sieve is the fastest known classical algorithm for factoring numbers over $10^{100}$.
The Quadratic sieve algorithm is the fastest known classical algorithm for factoring numbers under $10^{100}$.
answered 1 hour ago
AleksanderRasAleksanderRas
2,9101935
$endgroup$
2
$begingroup$
Actually, the factorization of 56153 was a stunt; the factors were deliberately chosen to have a special relation (differed in only 2 bits) and it's easy to factor when the factors have a known relation. AFAIK, the largest number that has been factored to date using a generic quantum factorization algorithm is 21.
$endgroup$
– poncho
1 hour ago
$begingroup$
I've always wondered why QS is (at least, consensually said to be) faster than GNFS below a certain thresold (not so consensual), and how much of that is due to lack of work on optimizing GNFS for smaller values.
$endgroup$
– fgrieu
1 hour ago
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The IEEE paper is silly.
The factorization method they give is quite slow, except for rare cases. For example, in their table 1, where they proudly show that their improved algorithm takes 653.14 seconds to factor a 67 bit number; well, I just tried it using a more conventional algorithm, and it took 6msec; yes, that's 100,000 times as fast...
answered 1 hour ago
ponchoponcho
93.5k2146243
$endgroup$
1
$begingroup$
Well I think the point of the paper is to improve upon Fermat-Factoring class algorithms, so it is expected that the given algorithm(s) get beaten by the more standard ones for small sizes, but excel on large inputs with (relatively small) prime differences?
$endgroup$
– SEJPM♦
1 hour ago
$begingroup$
@SEJPM: if that's the case, then they probably shouldn't go on so much about RSA (where the probability of having a sufficiently small difference is tiny)
$endgroup$
– poncho
1 hour ago
add a comment |
$begingroup$
The IEEE paper is silly.
The factorization method they give is quite slow, except for rare cases. For example, in their table 1, where they proudly show that their improved algorithm takes 653.14 seconds to factor a 67 bit number; well, I just tried it using a more conventional algorithm, and it took 6msec; yes, that's 100,000 times as fast...
answered 1 hour ago
ponchoponcho
93.5k2146243
$endgroup$
1
$begingroup$
Well I think the point of the paper is to improve upon Fermat-Factoring class algorithms, so it is expected that the given algorithm(s) get beaten by the more standard ones for small sizes, but excel on large inputs with (relatively small) prime differences?
$endgroup$
– SEJPM♦
1 hour ago
$begingroup$
@SEJPM: if that's the case, then they probably shouldn't go on so much about RSA (where the probability of having a sufficiently small difference is tiny)
$endgroup$
– poncho
1 hour ago
add a comment |
$begingroup$
The IEEE paper is silly.
The factorization method they give is quite slow, except for rare cases. For example, in their table 1, where they proudly show that their improved algorithm takes 653.14 seconds to factor a 67 bit number; well, I just tried it using a more conventional algorithm, and it took 6msec; yes, that's 100,000 times as fast...
answered 1 hour ago
ponchoponcho
93.5k2146243
$endgroup$
The IEEE paper is silly.
The factorization method they give is quite slow, except for rare cases. For example, in their table 1, where they proudly show that their improved algorithm takes 653.14 seconds to factor a 67 bit number; well, I just tried it using a more conventional algorithm, and it took 6msec; yes, that's 100,000 times as fast...
answered 1 hour ago
ponchoponcho
93.5k2146243
answered 1 hour ago
ponchoponcho
93.5k2146243
answered 1 hour ago
ponchoponcho
93.5k2146243
answered 1 hour ago
ponchoponcho
93.5k2146243
93.5k2146243
1
$begingroup$
Well I think the point of the paper is to improve upon Fermat-Factoring class algorithms, so it is expected that the given algorithm(s) get beaten by the more standard ones for small sizes, but excel on large inputs with (relatively small) prime differences?
$endgroup$
– SEJPM♦
1 hour ago
$begingroup$
@SEJPM: if that's the case, then they probably shouldn't go on so much about RSA (where the probability of having a sufficiently small difference is tiny)
$endgroup$
– poncho
1 hour ago
add a comment |
1
$begingroup$
Well I think the point of the paper is to improve upon Fermat-Factoring class algorithms, so it is expected that the given algorithm(s) get beaten by the more standard ones for small sizes, but excel on large inputs with (relatively small) prime differences?
$endgroup$
– SEJPM♦
1 hour ago
$begingroup$
@SEJPM: if that's the case, then they probably shouldn't go on so much about RSA (where the probability of having a sufficiently small difference is tiny)
$endgroup$
– poncho
1 hour ago
1
1
$begingroup$
Well I think the point of the paper is to improve upon Fermat-Factoring class algorithms, so it is expected that the given algorithm(s) get beaten by the more standard ones for small sizes, but excel on large inputs with (relatively small) prime differences?
$endgroup$
– SEJPM♦
1 hour ago
$begingroup$
Well I think the point of the paper is to improve upon Fermat-Factoring class algorithms, so it is expected that the given algorithm(s) get beaten by the more standard ones for small sizes, but excel on large inputs with (relatively small) prime differences?
$endgroup$
– SEJPM♦
1 hour ago
$begingroup$
@SEJPM: if that's the case, then they probably shouldn't go on so much about RSA (where the probability of having a sufficiently small difference is tiny)
$endgroup$
– poncho
1 hour ago
$begingroup$
@SEJPM: if that's the case, then they probably shouldn't go on so much about RSA (where the probability of having a sufficiently small difference is tiny)
$endgroup$
– poncho
1 hour ago
add a comment |
$begingroup$
Which of these is actually right?
Both. From reading the abstract it appears the papper doesn't claim that "VFactor" or Fermat Factorization ("FFM") or Trial Division ("TDM") are the best methods in general. However, if the difference between primes $p,q$ with $n=pq$ is really small, like $ll2^{100}$$;dagger$, then FFM (and probably the VFactor variants as well) will be a lot faster.
Though in general the difference between two same-length random primes is about $sqrt{n}/2$ which is about $2^{1024}$ for realistically sized moduli, so these attacks don't work there. Even with 400-bit moduli, which are somewhat easily crackable using a home desktop using the GNFS, this difference is still about $2^{200}$ and thus way too large.
Of course the implementation of the key generation may be faulty and emit primes in a too small interval and it's in these cases where these specialized algorithms really shine.
$dagger$: "$ll$" meaning "a lot less" here
answered 1 hour ago
SEJPM♦SEJPM
29.3k659139
$endgroup$
add a comment |
$begingroup$
Which of these is actually right?
Both. From reading the abstract it appears the papper doesn't claim that "VFactor" or Fermat Factorization ("FFM") or Trial Division ("TDM") are the best methods in general. However, if the difference between primes $p,q$ with $n=pq$ is really small, like $ll2^{100}$$;dagger$, then FFM (and probably the VFactor variants as well) will be a lot faster.
Though in general the difference between two same-length random primes is about $sqrt{n}/2$ which is about $2^{1024}$ for realistically sized moduli, so these attacks don't work there. Even with 400-bit moduli, which are somewhat easily crackable using a home desktop using the GNFS, this difference is still about $2^{200}$ and thus way too large.
Of course the implementation of the key generation may be faulty and emit primes in a too small interval and it's in these cases where these specialized algorithms really shine.
$dagger$: "$ll$" meaning "a lot less" here
answered 1 hour ago
SEJPM♦SEJPM
29.3k659139
$endgroup$
add a comment |
$begingroup$
Which of these is actually right?
Both. From reading the abstract it appears the papper doesn't claim that "VFactor" or Fermat Factorization ("FFM") or Trial Division ("TDM") are the best methods in general. However, if the difference between primes $p,q$ with $n=pq$ is really small, like $ll2^{100}$$;dagger$, then FFM (and probably the VFactor variants as well) will be a lot faster.
Though in general the difference between two same-length random primes is about $sqrt{n}/2$ which is about $2^{1024}$ for realistically sized moduli, so these attacks don't work there. Even with 400-bit moduli, which are somewhat easily crackable using a home desktop using the GNFS, this difference is still about $2^{200}$ and thus way too large.
Of course the implementation of the key generation may be faulty and emit primes in a too small interval and it's in these cases where these specialized algorithms really shine.
$dagger$: "$ll$" meaning "a lot less" here
answered 1 hour ago
SEJPM♦SEJPM
29.3k659139
$endgroup$
Which of these is actually right?
Both. From reading the abstract it appears the papper doesn't claim that "VFactor" or Fermat Factorization ("FFM") or Trial Division ("TDM") are the best methods in general. However, if the difference between primes $p,q$ with $n=pq$ is really small, like $ll2^{100}$$;dagger$, then FFM (and probably the VFactor variants as well) will be a lot faster.
Though in general the difference between two same-length random primes is about $sqrt{n}/2$ which is about $2^{1024}$ for realistically sized moduli, so these attacks don't work there. Even with 400-bit moduli, which are somewhat easily crackable using a home desktop using the GNFS, this difference is still about $2^{200}$ and thus way too large.
Of course the implementation of the key generation may be faulty and emit primes in a too small interval and it's in these cases where these specialized algorithms really shine.
$dagger$: "$ll$" meaning "a lot less" here
answered 1 hour ago
SEJPM♦SEJPM
29.3k659139
answered 1 hour ago
SEJPM♦SEJPM
29.3k659139
answered 1 hour ago
SEJPM♦SEJPM
29.3k659139
answered 1 hour ago
SEJPM♦SEJPM
29.3k659139
29.3k659139
add a comment |
add a comment |
$begingroup$
Quantum algorithms
There is of course Shor's algorithm, but as this algorithm only runs on quantum computers with a lot of qubits it's not capable to factor larger numbers than $21$ (reference).
There are multiple apparent new records using adiabatic quantum computation, although some are apparently stunts: See fgrieu's answer on a related question.
Classical algorithms
The general number field sieve is the fastest known classical algorithm for factoring numbers over $10^{100}$.
The Quadratic sieve algorithm is the fastest known classical algorithm for factoring numbers under $10^{100}$.
answered 1 hour ago
AleksanderRasAleksanderRas
2,9101935
$endgroup$
2
$begingroup$
Actually, the factorization of 56153 was a stunt; the factors were deliberately chosen to have a special relation (differed in only 2 bits) and it's easy to factor when the factors have a known relation. AFAIK, the largest number that has been factored to date using a generic quantum factorization algorithm is 21.
$endgroup$
– poncho
1 hour ago
$begingroup$
I've always wondered why QS is (at least, consensually said to be) faster than GNFS below a certain thresold (not so consensual), and how much of that is due to lack of work on optimizing GNFS for smaller values.
$endgroup$
– fgrieu
1 hour ago
add a comment |
$begingroup$
Quantum algorithms
There is of course Shor's algorithm, but as this algorithm only runs on quantum computers with a lot of qubits it's not capable to factor larger numbers than $21$ (reference).
There are multiple apparent new records using adiabatic quantum computation, although some are apparently stunts: See fgrieu's answer on a related question.
Classical algorithms
The general number field sieve is the fastest known classical algorithm for factoring numbers over $10^{100}$.
The Quadratic sieve algorithm is the fastest known classical algorithm for factoring numbers under $10^{100}$.
answered 1 hour ago
AleksanderRasAleksanderRas
2,9101935
$endgroup$
2
$begingroup$
Actually, the factorization of 56153 was a stunt; the factors were deliberately chosen to have a special relation (differed in only 2 bits) and it's easy to factor when the factors have a known relation. AFAIK, the largest number that has been factored to date using a generic quantum factorization algorithm is 21.
$endgroup$
– poncho
1 hour ago
$begingroup$
I've always wondered why QS is (at least, consensually said to be) faster than GNFS below a certain thresold (not so consensual), and how much of that is due to lack of work on optimizing GNFS for smaller values.
$endgroup$
– fgrieu
1 hour ago
add a comment |
$begingroup$
Quantum algorithms
There is of course Shor's algorithm, but as this algorithm only runs on quantum computers with a lot of qubits it's not capable to factor larger numbers than $21$ (reference).
There are multiple apparent new records using adiabatic quantum computation, although some are apparently stunts: See fgrieu's answer on a related question.
Classical algorithms
The general number field sieve is the fastest known classical algorithm for factoring numbers over $10^{100}$.
The Quadratic sieve algorithm is the fastest known classical algorithm for factoring numbers under $10^{100}$.
answered 1 hour ago
AleksanderRasAleksanderRas
2,9101935
$endgroup$
Quantum algorithms
There is of course Shor's algorithm, but as this algorithm only runs on quantum computers with a lot of qubits it's not capable to factor larger numbers than $21$ (reference).
There are multiple apparent new records using adiabatic quantum computation, although some are apparently stunts: See fgrieu's answer on a related question.
Classical algorithms
The general number field sieve is the fastest known classical algorithm for factoring numbers over $10^{100}$.
The Quadratic sieve algorithm is the fastest known classical algorithm for factoring numbers under $10^{100}$.
answered 1 hour ago
AleksanderRasAleksanderRas
2,9101935
edited 1 hour ago
answered 1 hour ago
AleksanderRasAleksanderRas
2,9101935
answered 1 hour ago
AleksanderRasAleksanderRas
2,9101935
answered 1 hour ago
AleksanderRasAleksanderRas
2,9101935
2,9101935
2
$begingroup$
Actually, the factorization of 56153 was a stunt; the factors were deliberately chosen to have a special relation (differed in only 2 bits) and it's easy to factor when the factors have a known relation. AFAIK, the largest number that has been factored to date using a generic quantum factorization algorithm is 21.
$endgroup$
– poncho
1 hour ago
$begingroup$
I've always wondered why QS is (at least, consensually said to be) faster than GNFS below a certain thresold (not so consensual), and how much of that is due to lack of work on optimizing GNFS for smaller values.
$endgroup$
– fgrieu
1 hour ago
add a comment |
2
$begingroup$
Actually, the factorization of 56153 was a stunt; the factors were deliberately chosen to have a special relation (differed in only 2 bits) and it's easy to factor when the factors have a known relation. AFAIK, the largest number that has been factored to date using a generic quantum factorization algorithm is 21.
$endgroup$
– poncho
1 hour ago
$begingroup$
I've always wondered why QS is (at least, consensually said to be) faster than GNFS below a certain thresold (not so consensual), and how much of that is due to lack of work on optimizing GNFS for smaller values.
$endgroup$
– fgrieu
1 hour ago
2
2
$begingroup$
Actually, the factorization of 56153 was a stunt; the factors were deliberately chosen to have a special relation (differed in only 2 bits) and it's easy to factor when the factors have a known relation. AFAIK, the largest number that has been factored to date using a generic quantum factorization algorithm is 21.
$endgroup$
– poncho
1 hour ago
$begingroup$
Actually, the factorization of 56153 was a stunt; the factors were deliberately chosen to have a special relation (differed in only 2 bits) and it's easy to factor when the factors have a known relation. AFAIK, the largest number that has been factored to date using a generic quantum factorization algorithm is 21.
$endgroup$
– poncho
1 hour ago
$begingroup$
I've always wondered why QS is (at least, consensually said to be) faster than GNFS below a certain thresold (not so consensual), and how much of that is due to lack of work on optimizing GNFS for smaller values.
$endgroup$
– fgrieu
1 hour ago
$begingroup$
I've always wondered why QS is (at least, consensually said to be) faster than GNFS below a certain thresold (not so consensual), and how much of that is due to lack of work on optimizing GNFS for smaller values.
$endgroup$
– fgrieu
1 hour ago
add a comment |
user56036 is a new contributor. Be nice, and check out our Code of Conduct.
user56036 is a new contributor. Be nice, and check out our Code of Conduct.
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