20030307, 23:44  #1 
Aug 2002
6F_{16} Posts 
About trial factoring
Hi all,
It has been a hard week and now is too late in Europe, so forgive me if I write something wrong. Here you are ... Is well know that for a prime p, if the Mersenne Number Mp = 2^p  1 is not prime and thus it has factors, they are in the form F = k * q + 1 where q = 2 * p I'm wonder whether we can take any advantage of this special form of the Mersenne number factors in Trial factoring task, specially when candidate factors are longer than 64 bits. For trial factoring, once we have a candidate for factor passing the filters: 1) F=+1 or 1 (mod 8 ) 2) F have no small factors We have to make modular exponentiation (mod F). And see whether 2^p = 1 (mod F) The most expensive task in the modular exponentiation is the modular square, i. e. to compute y = x * x (mod F) If we write x < F in the form x = a1 * q + a0, a0 < q, a1 < k (1) it looks like a semi representation in base q . Note than a1 can be high enough (a1 >> q). We can write x * x = (a1 * ((a1 * q) + 2 * a0 ))* q + a0 * a0; (2) or x * x = (a1 * a1 * q + 2 * a0 * a1 ) * q + a0 * a0; (3) OTOH if we want to compute (A * q) mod F = A * q mod ( k * q + 1) = = r * q  c where r = A mod k c = A / k (integer division) i. e. A = c * k + r Usually, c can be computed fast in a FPU. even more, we can see A * q^2 mod F = (A * q) * q mod F = = ((A * q) (mod F)) * q) (mod F) = (B * q) (mod F) = = ((r0 * q  c0) * q) (mod F) = = (r1 * q  c1) (4) where A = c0 * k + r0 B = r0 * q  c0 = c1 * k + r1 Then to compute (3) we would need to perform three kmodulo and a qmodulo plus some more work. Let's see and example. Suposse candidate F having 70 bits, 25 bits for q and the remaining 45 to k. What we need to do: I) C = a1 * a1 * q^2 mod F This requires i ) 45 * 45 bits square x = (a1 * a1) ii) compute c0 = int (x / k) iii) compute r0 = x mod k iv) compute B = r0 * q  c0 (70 bits arithmetic) v) compute c1 = int (B / k) vi) compute r1 = B mod k II) D = 2 * a0 * a1 * q mod F i) a 26 * 45 bits mul y = 2 * a0 * a1 ii) compute c2 = int (y / k) iii) compute r2 = y mod k III) a0 * a0 i) compute c3 = (a0 * a0  c1  c2) mod q ii) compute r3 = int (a0 * a0  c1  c2) / q IV) final i) C = (r1 + r2 + r3) * q + c3 ii) normalization Is a long list of things to do, but I don't know if is a lot of work compared with multiprecision arithmetic we would make other way The proposed scheme only uses modular arithmetic for lengths where there fast and optimized algorithms. Is this a cracy or wrong idea?. May be. Guillermo Ballester Valor 
20030308, 01:34  #2  
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}×3×641 Posts 
Re: About trial factoring
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Going to mod 840 congruence classes would enable also easily skipping all multiples of 7, but I presume George analyzed that long ago and found it not worth the extra complication compared to the cost of eliminating multiples of 7 in other steps. 

20030308, 07:42  #3 
Aug 2002
3·37 Posts 
Hi,
Yes, I know this previous sieve work that all trial factoring code has to do. And in fact, is a great advantage the form of the factors. But what I meant, sorry if I not explained well, is try to use this form in the modular exponentiation we have to do with the few candidates we still have after sieve work, when the size is over 64 bits . Guillermo. 
20030522, 01:37  #4  
Aug 2002
Portland, OR USA
2·137 Posts 
Quote:
When the candidate factor F is over 64 bits, you want to use the method described in your first post to break it into several operations of < 64 bits. I think the question you want is: Is your method faster than [sp] karisuba multiplication (or other), plus a size*2 bit mod. You'll want to compare your method for factor sizes of 64, <128, <192 bits, ... to see if it beats the others at that size range. If you could code a function gbv(m, f) that uses your method to calculate 2^m  1 (mod f), and someone codes it using several other big number methods, then we could run a time trial and see. Any takers? Bruce 

20030522, 02:04  #5  
Aug 2002
Portland, OR USA
274_{10} Posts 
Re: About trial factoring
Quote:
I finally reduced my goal to writing a simple one that works, and wrote it as an html form using javascript. Now, someone can easily convert my source to c, mesh it with one of the libraries, and we'll be in business. After reading the info on Trial Factoring, I suspect that this won't be usefull with the methods Prime95 uses to store and sieve candidate factors  Since each candidate is separated by at least 2*p. But if anyone is interested, I can email it; or post it here as an attachment; or upload it to someones server and post a link. Bruce 

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