Author Topic: New RSA Algorithm discussion  (Read 63198 times)

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Offline ruler501

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Re: New RSA Algorithm discussion
« Reply #90 on: February 27, 2011, 04:21:57 pm »
Could you generate a range reducing the numbers that have to be searched. Aren't the two numbers supposed to be pretty close in bit size? We could then use an optimized algorithm that would just have to go through a much smaller amount of numbers.

If wahat I said is stupid I'm sorry I know almost nothing about this.
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Offline fb39ca4

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Re: New RSA Algorithm discussion
« Reply #91 on: February 27, 2011, 04:41:37 pm »
I think there may be a simple method we have overlooked.  This isn't neccissarily the best resource, but this is a list of the top 500 known prime numbers.  Perhaps we can multiply them to each other until we find n?  I know it's brute force, but hey, it could work.
Th prime numbers used in the key are around 2^512. The ones in that list are much, much larger. Even if we found all the primes near 2^512, we would have a lot of work to do because by the prime number theorem, there is about 2^503 of them.

Offline jnesselr

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Re: New RSA Algorithm discussion
« Reply #92 on: February 27, 2011, 04:52:19 pm »
I think there may be a simple method we have overlooked.  This isn't neccissarily the best resource, but this is a list of the top 500 known prime numbers.  Perhaps we can multiply them to each other until we find n?  I know it's brute force, but hey, it could work.
Quite possibly, but unlikely.  It would work if it only had one factor, but if it has both, it won't work.  The best thing dealing with that I've seen, is start off with some prime number S1.  Then, multiply 1 S1 with 2 S2. S2 is the next prime lower than S1.  This would take huge initial amounts of memory and would output the file that had this number for some bound B.  So, for example:
B=11
S1=11
S2=7
S3=5
S4=3
S5=2
11*(7*7)*(5*5*5)*(3*3*3*3)*(2*2*2*2*2)= 174636000

So, for numbers like 35, we get:
174636000 mod 35 = 0, so we do 174636000 / 35 = 4,989,600.
4989600 mod 35 = 0, so we do 4989600 / 35 =142,560
142,560 mod 35 = 5, hey look a factor.

It gets slightly strange on numbers like 33. (Pretty much any number that doesn't have it's other prime next to it.  I'm just going to use // to mean divide it out as far as it will go out completely. e.g., no fraction or decimal points after dividing.
174,636,000 // 33 =5,292,000
5,292,000 mod 33 = 21. Strange I know, but you only have to do gcd(21,33) to get 3, which is a factor.

So the problem involves setting your maximum bound, and figuring out how many times each number should be multiplied, and sending that task out to whatever...  hmmm... Me has an idea.  I could figure that out, by just saying the maximum bound B minus whatever prime number you are on S, plus 1.  So say the upper bound is 11 (B), and we need to know how many times to multiply 5 in.  So 11-5+1=7.  So that adds slight redundancy but it's easier to split on many machines.

I couldl set up a site that allows you to download data file(s), and a program (Most likely java for easy portability), and gives rewards based on how many you have completed.  Simply download the file, run the program, upload the output, and you get a point or something.  Sounds good to me!  What do you think?

SirCmpwn

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Re: New RSA Algorithm discussion
« Reply #93 on: February 27, 2011, 04:53:50 pm »
Go for it.  I just started my own program as well, and I encourage others to do the same.  Let's get this damn thing done.

Offline ruler501

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Re: New RSA Algorithm discussion
« Reply #94 on: February 27, 2011, 05:00:11 pm »
I agree. I hope we can get the range we need lowered down. 2^503 is better than 2^512. I think we could narrow it down using simple algorithms couldn't we? does anyone know where a list of primes around 5^12 are? We just need everything from 2^511 to 2^513 shouldn't we. then an algorithm that sorts through those for likely matches. after that another algorithm to sort for the one match we need.
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-----BEGIN GEEK CODE BLOCK-----
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GCM/CS/M/S d- s++: a---- C++ UL++ P+ L++ E---- W++ N o? K- w-- o? !M V?
PS+ PE+ Y+ PGP++ t 5? X R tv-- b+++ DI+ D+ G++ e- h! !r y

Offline AngelFish

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Re: New RSA Algorithm discussion
« Reply #95 on: February 27, 2011, 05:12:09 pm »
Here's a paper from the most successful RSA factorization thus far:

http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.156.9709&rep=rep1&type=pdf

Also, graph, your method appears to try to find the prime factorization of the keys. The fact of the matter is that the prime factors in question are 1024 bits long. That means that you're essentially brute forcing the problem. It wouldn't really reduce the complexity of the problem.
∂²Ψ    -(2m(V(x)-E)Ψ
---  = -------------
∂x²        ℏ²Ψ

Offline jnesselr

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Re: New RSA Algorithm discussion
« Reply #96 on: February 27, 2011, 05:17:18 pm »
Also, graph, your method appears to try to find the prime factorization of the keys. The fact of the matter is that the prime factors in question are 1024 bits long. That means that you're essentially brute forcing the problem. It wouldn't really reduce the complexity of the problem.
That's incorrect, actually, I only need to find factors that are 512 bits for a 1024 bit key.  And the cool part is that the more people contribute, the better chances we get to having it work.  Also, with this, we can use the same number to factor any other keys, or increase the bound.  (Although by increasing that, you would have to say larger numbers get more factors.  So, essentially, we should make the bound around 2^768, and start around the 512 bit area.  The only reason it sounds remotely reasonable, is because we can just check when someone submits something if it has it in there, or just store it otherwise for other numbers.  Sounds good enough to me to at least try it.

EDIT: Also, to you non-basic programmers, let the no ending parenthesis just bug you.  :devil:
« Last Edit: February 27, 2011, 05:17:57 pm by graphmastur »

Offline ruler501

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Re: New RSA Algorithm discussion
« Reply #97 on: February 27, 2011, 05:23:37 pm »
Must... add... ending... parentheses...
;)
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SirCmpwn

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Re: New RSA Algorithm discussion
« Reply #98 on: February 27, 2011, 05:26:50 pm »
What is the maximum 512-bit prime number?  Does anyone know?
(If not, the biggest non-prime 512-bit number works, too)
Also, what is a reasonably good distance between numbers to try?

Offline jnesselr

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Re: New RSA Algorithm discussion
« Reply #99 on: February 27, 2011, 05:33:26 pm »
What is the maximum 512-bit prime number?  Does anyone know?
(If not, the biggest non-prime 512-bit number works, too)
Also, what is a reasonably good distance between numbers to try?
The reasonable distance is essentially random.  Smallest number bellow 2^512? That would be slightly difficult.  Just start at 2^512, I guess.

Offline AngelFish

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Re: New RSA Algorithm discussion
« Reply #100 on: February 27, 2011, 05:34:35 pm »
The upper bound on the gap between two Prime numbers of the required size is approximately ln(2^513)
« Last Edit: February 27, 2011, 05:35:01 pm by Qwerty.55 »
∂²Ψ    -(2m(V(x)-E)Ψ
---  = -------------
∂x²        ℏ²Ψ

SirCmpwn

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Re: New RSA Algorithm discussion
« Reply #101 on: February 27, 2011, 05:35:06 pm »
The numbers should be close to each other, it says.  So how close is close?

Offline AngelFish

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Re: New RSA Algorithm discussion
« Reply #102 on: February 27, 2011, 05:35:35 pm »
Sir, see my post above yours  ;)
∂²Ψ    -(2m(V(x)-E)Ψ
---  = -------------
∂x²        ℏ²Ψ

Offline jnesselr

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Re: New RSA Algorithm discussion
« Reply #103 on: February 27, 2011, 05:36:40 pm »
Somewhere between two and ln(2^513)

SirCmpwn

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Re: New RSA Algorithm discussion
« Reply #104 on: February 27, 2011, 05:36:47 pm »
Could I have that number in hex? :)
Nevermind, it's about 356.
« Last Edit: February 27, 2011, 05:37:51 pm by SirCmpwn »