20210603, 13:34  #34  
Apr 2020
5×73 Posts 
Quote:
Last fiddled with by charybdis on 20210603 at 13:35 

20210603, 13:45  #35 
"Curtis"
Feb 2005
Riverside, CA
1001100000000_{2} Posts 
For GNFS, Qmax to Qmin ratio around 8 is best after Charybdis did his testing on a large job, I repeated the tests for C110C140 and confirmed his findings. I now use Qmin = 150k for the C120 file.
For SNFS, a wider ratio seems to work okay, but more testing is needed. I don't think it's a terrible idea to use 80k1M, but if you expanded down to the 10k range you'll find that you need more raw relations because the duplicate ratio is higher (as Charybdis explained). I'd aim for Qmax to Qmin of 10x for SNFS jobs, and I wouldn't worry about it rising to around 12x on the occasions where a poly has lowerthanexpected yield. 
20210604, 19:04  #36 
Aug 2020
2^{2}×3×5^{2} Posts 
Ok, I suspected something something like those relations being not as useful, but I thought maybe it was different for SNFS.
I ran a better comparison on 1281979*2^547+1 using the poly and parameters listed below and got CPU times of 123936 s for the selfconstructued poly and 132316 s for the yafu poly and parameters. Yafu was going with the algebraic side again, also the parameters were slightly different from VBCurtis' ones. But the difference in total runtime is by no means as large as I had before, so disregard that. I commented the lambdas for yafu parameters as suggested by charybdis. And I just realized I forgot to switch lambda0 and lambda1 for the nonyafu run. The rels_wanted for yafu were calculated with the "fudge"formula but turned out to be too low, the matrix was build with 17e6 relations. c125.poly Code:
n: 64114249202003782174714518503455061682048018480541440980450169650489022687410445023173449432639279431900621347038011616801357412684045342625257617782577139 skew: 0.12599 c0: 8 c5: 1281979 Y0: 1298074214633706907132624082305024 Y1: 1 # f(x) = 1281979*x^5+8 # g(x) = x1298074214633706907132624082305024 Code:
n: 64114249202003782174714518503455061682048018480541440980450169650489022687410445023173449432639279431900621347038011616801357412684045342625257617782577139 # 1281979*2^547+1, difficulty: 170.77, anorm: 9.45e+38, rnorm: 3.53e+40 # scaled difficulty: 170.77, suggest sieving algebraic side # size = 1.652e12, alpha = 0.931, combined = 1.178e10, rroots = 1 type: snfs #size: 170 skew: 0.0455 c0: 1 c5: 5127916 Y0: 649037107316853453566312041152512 Y1: 1 # f(x) = 5127916*x^5+1 # g(x) = x+649037107316853453566312041152512 Code:
tasks.lim0 = 5500000 tasks.lim1 = 5500000 tasks.lpb0 = 28 tasks.lpb1 = 28 tasks.sieve.mfb0 = 54 tasks.sieve.mfb1 = 52 tasks.sieve.ncurves0 = 19 tasks.sieve.ncurves1 = 15 tasks.sieve.lambda0 = 1.83 tasks.sieve.lambda1 = 1.91 tasks.I = 13 tasks.sieve.adjust_strategy = 2 tasks.sieve.rels_wanted = 22900000 tasks.qmin = 50000 tasks.sieve.qrange = 10000 tasks.sieve.sqside = 0 Code:
tasks.lim0 = 7000000 tasks.lim1 = 7000000 tasks.lpb0 = 27 tasks.lpb1 = 27 tasks.sieve.mfb0 = 54 tasks.sieve.mfb1 = 54 tasks.sieve.ncurves0 = 15 tasks.sieve.ncurves1 = 19 #tasks.sieve.lambda0 = 1.83 #tasks.sieve.lambda1 = 1.91 tasks.I = 13 tasks.sieve.adjust_strategy = 2 tasks.sieve.rels_wanted = 10000000 tasks.qmin = 50000 tasks.sieve.qrange = 10000 tasks.sieve.sqside = 1 Last fiddled with by bur on 20210604 at 19:07 
20210604, 23:49  #37 
"Ben"
Feb 2007
DBC_{16} Posts 
This change has been checked in. I left the existing test sieving the same  top three polys  then if the best has norms within 5 orders of magnitude of each other it checks the opposite side.

20210630, 15:11  #38 
"Max"
Jun 2016
Toronto
1575_{8} Posts 
factoring Lucas numbers by GCDs
Could anybody please point me in the direction of the analog of the two formulas below, with the same/similar structure but for the Lucas numbers?
I think I saw them before somewhere, they would be important for factoring to rule out the GCDs. 
20210630, 15:52  #39 
Apr 2020
5×73 Posts 
Change all the Fs to Ls and the 1 at the end to 25.

20210630, 16:18  #40  
Feb 2017
Nowhere
5·7^{2}·19 Posts 
Quote:
Then combining with F_{n+2}F_{n2}  F_{n}*F_{n} = /+1 (don't ask me to prove it off the cuff, I'd screw it up) gives the first identity. For Lucas numbers, [scribble scribble] the corresponding identities appear to be L_{n+1}L_{n1}  L_{n}*L_{n} = +/5 and L_{n+2}L_{n2}  L_{n}*L_{n} = /+5 so L_{n2}L_{n1}L_{n+1}L_{n+2} = L_{n}^{4}  25. Vaguely similar identities for sums and differences are in this thread. 

20210630, 17:13  #41  
"Max"
Jun 2016
Toronto
1575_{8} Posts 
Quote:
Exactly what I kinda remember and was looking for! Any idea on how to approach the conversion of the second formula to Lucas universe? 

20210710, 05:21  #42 
Aug 2020
2^{2}×3×5^{2} Posts 
I wanted to factor the c119 cofactor of 1281979*2^559+1 and after 5% of sieving got an ETA of about 1:30 hours. A cado GNFS run had an ETA of 0:40 hours.
Code:
n: 12698277665702007683889275815276693326593280837281482940452961346094193789732001920277335420071981852156946378079178463 skew: 0.0958 c5: = 1281979 c0: = 2 y1: = 1 y0: = 5192296858534827628530496329220096 edit: I used the c120 parameters because the cofactor was small and it was stated here every 30 bits I should switch to 5 digits more/less of params file. Now that I think about it, was that meant for the original Proth number? I.e. regardless of the cofactor's size if the Proth number has an exponent around 550 I should use the c125 parameters? edit2: That also wasn't the case. I did the same with the c125 params I used to factor a c155 cofactor of 1281979*2^557+1 and it had an ETA for the c119 of more than 2 hours. Is there something wrong with the polynomial? I did the "multiply by 2^n" before, it was usually slower than "pull 2^x" but not by that much and and especially not when it was 2^1, in that case it's usually faster. Last fiddled with by bur on 20210710 at 05:43 
20210710, 05:48  #43 
Jun 2003
2^{3}·5·127 Posts 
This is a case where GNFS _is_ the faster option. Although probably not by a factor of 2.
Rather than SNFS on 2*(1281979*2^559+1), you could try (1281979*2^4)(2^111)^5+1 with appropriate skew which might be a shade faster. You have to chose parameters appropriate for SNFS175 (including sieving on rational side and sieve region size), not for the composite c119. Last fiddled with by axn on 20210710 at 05:49 
20210710, 05:59  #44  
Aug 2020
2^{2}·3·5^{2} Posts 
Interesting, is there a way to know if GNFS is faster or does it just turn out by experiment?
Quote:
Quote:


Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
A new tool to identify aliquot sequence margins and acquisitions  garambois  Aliquot Sequences  24  20210225 23:31 
Comparison of GNFS/SNFS With Quartic (Why not to use SNFS with a Quartic)  EdH  EdH  14  20200420 16:21 
new candidates for M...46 and M48  cochet  Miscellaneous Math  4  20081024 14:33 
Please identify!  BrianE  Lounge  24  20080801 14:13 
Easily identify composites using multiplication  Nightgamer360  Miscellaneous Math  9  20070709 17:38 