20070917, 18:08  #1 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
2^{3}·5·11·13 Posts 
factoring programs
what is the best factoring program to use for different ranges of digits
say: 150 51 to 70 71 to 90 91 to 100 101 to 130 130 to 150 150+ it is probably simpler that this but i just want several different programs i am particularly interested in finding a program that will factor a 160 digit number in maybe 30ish hours i need windows binarys really unless they are easy to compile 
20070917, 18:32  #2  
"Ben"
Feb 2007
2·5·7·47 Posts 
Quote:
Otherwise, see http://www.loria.fr/~zimmerma/records/rsa160 Granted, this was 4 years ago, but the algorithm used has not changed, and neither has the hardware much. QS, (the Quadratic Sieve) is best for general numbers with number of digits N > 20 and < 100. Msieve is your best option there. There is no 'plug n' play' program for anything bigger, that I know of. You should spend some time here: http://en.wikipedia.org/wiki/Integer_factorization 

20070918, 06:34  #3 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
2^{3}×5×11×13 Posts 
from what i have heard ggnfs should be faster than QS
i know that msieve's ggnfs is supposed to be slower than its qs but that is for other reasons is there a ggnfs program that is working as fast as the ggnfs algorithm should the number is a factor of M1470 it have been reduced as much as it can be with ECM would it be faster to start again on M1470 with snfs also could u point out to me a snfs program thanks 
20070918, 20:51  #4  
Tribal Bullet
Oct 2004
3,529 Posts 
Quote:
The GGNFS package is your only practical choice for large GNFS jobs. It can handle both SNFS and GNFS problems. Go to the sourceforge page, download and install it, and try out some smaller jobs (i.e. RSA100) while you do some more background reading. If more ECM is not going to help, then nothing else will work. 

20070918, 22:20  #5  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
2·3^{2}·569 Posts 
Quote:
M1470 = 2^14701 = (2^735+1)(2^7351) Each of those two factors have themselves been factored completely as part of the work of the Cunningham project. There is absolutely no point in factoring them again except, perhaps, as a learning exercise. Even in the latter case there are enough unfactored numbers of comparable difficulty and of interest to at least one other person that refactoring M1470 is of questionable utility. Paul 

20070919, 13:38  #6 
Aug 2002
Buenos Aires, Argentina
3^{3}×7^{2} Posts 
As I told you in another thread, you should have used my factoring applet in order to find the factorization of 2^1470  1:
32665 824620 457402 195434 633605 443740 257550 045736 442160 628806 611546 120572 696033 233966 359012 654595 979733 069567 388664 407206 139801 581335 691697 666580 825036 395993 873374 496107 296177 432394 402764 349675 687529 782004 971186 285327 519716 941701 567289 776745 614889 041189 723697 073526 161103 983641 763554 076064 188135 517654 549958 476754 101195 985309 251236 506489 109949 465205 140536 348059 366804 345787 541934 500497 452294 145254 373550 578070 204266 401827 739721 665519 024733 159423 = 3 ^ 2 x 7 ^ 3 x 11 x 31 x 43 x 71 x 127 x 151 x 211 x 281 x 331 x 337 x 491 x 1471 x 5419 x 5881 x 29191 x 41161 x 86171 x 106681 x 122921 x 152041 x 664441 x 748819 x 1 564921 x 4 163041 x 20147 473081 x 2 340389 488711 x 4 363953 127297 x 4 432676 798593 x 27653 710336 343911 x 15 162868 758218 274451 x 26 032885 845392 093851 x 2 741672 362528 725535 068727 x 50647 282035 796125 885000 330641 x 252359 902034 571016 856214 298851 708529 738525 821631 x 631 430922 992211 190033 830999 202698 905758 039480 236241 x 23 618256 244840 618857 212522 155851 714598 259422 753496 906641 681177 748710 460515 038403 366198 473773 770441 
20070919, 13:47  #7  
Nov 2003
7460_{10} Posts 
Quote:


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