20060613, 23:51  #1 
Aug 2005
Brazil
2·181 Posts 
SNFS trivial factorization
Hi guys! I'm factoring a number using SNFS, and for all dependencies, it returned the result:
From dependence x, sqrt obtained: r1=1 Is this possible or does it reveal some bugs in the ggnfs software I'm using? Can I complete the factorization without having to start sieving all over again? 
20060614, 01:13  #2 
Jun 2003
The Texas Hill Country
3^{2}·11^{2} Posts 
Assuming that the "problem" is actually that you found only trivial factorizations (as opposed to some "bug" in either your hardware, software, or parameters), you can always do some additional sieving and add the new relations to the previous set.
However, before I would be willing to say anything more, I would need a lot more details in order to make an educated guess as to what went wrong. 
20060614, 01:30  #3  
Aug 2005
Brazil
552_{8} Posts 
Quote:
Last fiddled with by fetofs on 20060614 at 01:30 

20060614, 06:09  #4 
"Nancy"
Aug 2002
Alexandria
9A3_{16} Posts 
How many different dependencies were there? The odds of getting only trivial factorisatons in n dependencies is 2^{[I]n[/I]} (assuming 2 prime factors in your number). If you only had maybe five or six dependencies, it might have been simple bad luck. If you had twenty and they all failed to find a proper factor, I'd invesitagte for possible software problems before spending more time sieving.
Alex Last fiddled with by akruppa on 20060614 at 07:43 
20060614, 06:53  #5  
Mar 2004
Belgium
839 Posts 
Quote:
A time ago, when I did a C130+ digit SNFS factorization, I also had the problem that the GGNFS Suite (0.77.1 snapshot 200509...) returned a trivial factor of r1=1 for each of the 32 dependencies. After that, I factored other number successfully with SNFS & GNFS, all be it with number much smaller (C87 => C115). BTW, I use the factLat.pl script Regards Cedric Last fiddled with by ValerieVonck on 20060614 at 06:55 

20060614, 09:37  #6  
Nov 2003
1D24_{16} Posts 
Quote:
(1) Duplicate relations (2) The number being factored was actually prime. 

20060614, 11:46  #7  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
3·3,529 Posts 
Quote:
(2) The number being factored was actually a prime power. Paul 

20060614, 13:06  #8  
Aug 2005
Brazil
2·181 Posts 
Quote:
I thought that might be a bit strange, that's why I posted this... Is there any way to check if my number is a prime power? 

20060614, 13:23  #9 
"Nancy"
Aug 2002
Alexandria
4643_{8} Posts 
First of all test if it is a prime. Technically, NFS will fail for prime powers as well, but the chance that a large number is a prime power but no prime and without small factors is really slim. If you want to test for prime powers as well, you could try if the square root, cube root, fifth root, generally: the pth root (p prime) are integers, until N^{1/p} is below your trial division limit.
Alex 
20060614, 14:00  #10  
Aug 2005
Brazil
362_{10} Posts 
Quote:


20060614, 14:25  #11 
"Sander"
Oct 2002
52.345322,5.52471
29·41 Posts 
Dump some relations and sieve a bit further before rebuilding the matrix again.

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