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#1 |
Sep 2008
Krefeld, Germany
3468 Posts |
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Hi,
I'm currently sieving on 8*10^230-1, parameters are: c6: 25 c0: -2 skew: 0.66 type: snfs rlim: 50000000 alim: 50000000 lpbr: 30 lpba: 30 mfbr: 59 mfba: 59 rlambda: 2.7 alambda: 2.7 sieving on rational side only. I started from Q=1M to 15M: 14e siever, ~23M relations Q=15M to 25M: 15e siever, ~40M relations Currently working at Q=26M. about 131 CPU days so far, still far from a usable matrix. The problem is the duplicate rate, last 8M relations had 6M duplicates in it, at a total of just 10M duplicates! I'm quite sure the sieving ranges dont overlap. At this rate it will take forever to complete the job. Any hints/suggestions? Thanks in advance Last fiddled with by Syd on 2009-08-12 at 18:33 Reason: 10M total of cause |
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#2 |
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
5·1,171 Posts |
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if the algebraic side isnt too slow you might try that
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#3 |
Sep 2008
Krefeld, Germany
2×5×23 Posts |
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#4 |
Tribal Bullet
Oct 2004
24·13·17 Posts |
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If you are sieving special-q far below the factor base limit, then I would expect a huge number of duplicates. How about special-q above the factor base limit?
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#5 |
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
41·229 Posts |
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You will need at least 92-95M unique relations (a bit more is going to make for a better matrix).
I wouldn't recommend sieving on algebraic side; this will give you even more duplicates. Most of these existing duplicates came from sieving from a very low starting point. (Better would have started from 10-15M.) The parameters look fine. Hopefully the future relations will not be as redundant. Use 15e. For a job of this size you should not use the vanilla scripts. If you are using them (and the script has MINRELS.txt spells in it), then add a file MINRELS.txt in the project directory with 90000000 in it. You will save a lot of time by not filtering until there's at least a chance of convergence. When you will have 90M raw relations, filter and have a look at redundancy, then revise the next time to filter by putting a larger number in MINRELS.txt This is quite a big number for home computing, but not impossible. Good luck! |
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#6 |
Jul 2003
So Cal
2,083 Posts |
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You started sieving very low. You need to move to q above the factor base limits, and once you do for this number the algebraic side will be faster than the rational side. Sieve on the algebraic side starting at q=50M. This should significantly reduce your rate of duplicates and give you plenty of relations to finish the factorization. Also for future factorizations, for 30 bit large primes mfbr/a of 60 or 61 would be better than 59.
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#7 |
Sep 2008
Krefeld, Germany
2×5×23 Posts |
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Thank you!
92M unique, thats quite a lot more than I expected. Anyway, too late to stop now. Just started sieving at Q=50M on algebraic side, only about 20% less relations than with Q=15M on rational side. Hope that will give enough unique relations! I always started low with Q=1M or even lower because it yields more relations per second. On small jobs that gave about 20% duplicates. Is this also a bad idea? |
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#8 |
Jul 2003
So Cal
2,083 Posts |
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Serge was referring to 92M total including duplicates when you start sieving at q of half the factor base limit. You should be able to build a matrix at about 78M unique, but a few more will improve it. I usually get at least 83M - 85M unique before starting the LA.
For small numbers, if what you're doing works, keep doing it. ![]() |
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#9 |
Sep 2008
Krefeld, Germany
23010 Posts |
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Thanks again, I was finally able to finish it.
75M unique (92M total relations) resulted in this 3-split: prp70 factor: 7194989070351007241001770794481202899232920377811344337193524816903113 prp74 factor: 28405869825449471447004672858393144480378760112284964510299428948566224457 prp80 factor: 22777672993897316831397692267111749997397458530603944165484429112280289213344407 |
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#10 |
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
938910 Posts |
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Nice job, and let me be the first to welcome you to the top of that list!
Triple-splits all around the house. |
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