20070629, 16:42  #1 
(loop (#_fork))
Feb 2006
Cambridge, England
2^{3}×797 Posts 
Better measures of SNFS difficulty
A recent thread pointed out to me that, given N = 15*R, you can construct a quartic SNFS polynomial for the part of a^N1 that is explained by the degree8 factor of x^151 by taking advantage of the symmetry of that factor and writing it in terms of x+1/x; by the usual SNFSdifficulty measurement, this has difficulty (8/15) N log a.
So I tried setting this up for 10^3451, where the difficulty estimate is 184, and get Code:
n: 28687509643492892075221128006449237350451146619508500482915984762059137296478553404775990467913950190401092476853539226200030840255611742267377635357890631 type: snfs skew: 1 c4: 1 c3: 1 c2: 4 c1: 4 c0: 1 Y1: 100000000000000000000000 Y0: 10000000000000000000000000000000000000000000001 I assumed that this was the situation in which you tune the 'skew' parameter, but skew=100 and skew=0.01 were both very significantly slower than skew=1. To get relations at a reasonable rate is possible with sieve parameters something like Code:
skew: 1 alim: 9000000 rlim: 72000000 lpbr: 29 lpba: 29 mfbr: 53 mfba: 53 
20070629, 17:16  #2  
Nov 2003
2^{2}×5×373 Posts 
Quote:
The reason is that for optimal performance you want the norms on the rational side and on the algebraic side to be approximately equal. When doing numbers in the 180+ digit range, a quartic yields norms on the rational side that are significantly larger than those on the algebraic. The correct thing to do in this case is to use a far bigger factor base (and bound on the large primes) for the rational side than the algebraic. BTW, this number is better done with the lattice sieve using rational side special q. Are you reserving this number??? It is one of two that I had planned to do after I finish 7,348+. The other is 7,553L. For another example, consider 11^2351. The "difficulty" with (x^5  1)/(x1) and x = 11^47 is about 196. However, it uses a quartic polynomial and will be more difficult than would (say) 11^191 + 1 [already factored] with a quintic. 

20070630, 00:10  #3 
(loop (#_fork))
Feb 2006
Cambridge, England
2^{3}×797 Posts 
Is there a central spot that handles reservations for the Cunningham project? 11^251+1 (by GNFS on the C151) is so tempting (prime exponent and lots of ECM factors) that I'd assume it's already spoken for.
If you're already set up to do 10^3451, go for it, and I'll try to do Fibonacci(1009) instead, where I'm a little more confident I've got the right parameters for the sieving. That'll take me most of July. 
20070630, 01:13  #4  
Jun 2003
The Texas Hill Country
3^{2}×11^{2} Posts 
Quote:
If you do not have his email address, PM me and I will provide it to you. 

20070630, 09:20  #5 
"Sander"
Oct 2002
52.345322,5.52471
4A5_{16} Posts 

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