20220606, 02:12  #1 
May 2022
25_{8} Posts 
GMPECM and ECM Frustrations [Was: About M1277]
Wow, I didn't expect this many responses.
Thank you all for your time! EDIT: I am currently running ECM on this, how long would it take to finish t65? (or whatever its called) Last fiddled with by BigNumberGuy on 20220606 at 02:18 
20220606, 05:44  #2  
"Curtis"
Feb 2005
Riverside, CA
17^{2}×19 Posts 
Quote:
Since Ryan has already done a t70 or the majority of one, all ECM curves should be run at T75 level (or larger) which is around B1=7e9. It's not important what exact B1 you choose, if it's in that vicinity (so 6e9 or 8e9 are fine, but 1e9 is far too small and wasted effort). It would take a while to complete the T75 level, but we don't have to run the whole level to contribute! I've run about 800 curves with B1 around 7e9 myself, at about a CPUday per curve on a very old Core2Xeon (Dell 6100). It's in your best interest to use the maxmem option, unless you have 30+GB available for GMPECM. 

20220606, 06:11  #3 
Aug 2020
79*6581e4;3*2539e3
601 Posts 
edit: Took to long pondering over the post...
What do you mean by "running ECM"? Make sure you don't replicate all the lower B1 values, chances of finding a factor are miniscule once a lot of curves were run already. You'd need to start at B1=850M, preferrably higher. Try and see how long a curve takes... :) To get an idea about the probabilities, this page is very helpful. Enter the number of curves for the three highest B1 values that were run (smaller ones won't have any significant effect). You can use this table to see how many curves are usually run at each B1. And for M1277 there were likely run much more curves at each B1 value already. I also have a question about ECM. Is there exactly _one_ sigma value that will produce the factor or are there several ones? Is it guaranteed there is a "winning" sigma? If so, within which bounds? I guess not, because then ECM would be deterministic by testing all possible sigmas. And are the probabilities depending on the composite candidate? Are there composites for which more sigmas will produce a factor? Last fiddled with by bur on 20220606 at 06:12 
20220606, 15:03  #4  
Random Account
Aug 2009
Not U. + S.A.
2×3×389 Posts 
Quote:
Quote:


20220606, 16:27  #5 
"Curtis"
Feb 2005
Riverside, CA
1010101110011_{2} Posts 
B1 is a single value, not a range. B2 is *much* larger than B1. Your example makes no sense.
Really, I mean it consider all curves at B1 below 2e9 useless for M1277. 
20220606, 19:14  #6  
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
6876_{10} Posts 
Quote:
A sigma corresponds to a curve which corresponds to a set of points in a field. How many possible sigmas and curves are there? (Many; enough that it is expected that relatively few will be duplicated by selecting hundreds, to hundreds of thousands, of sigmas randomly for given bounds.) Are the curves guaranteed to not intersect at points corresponding to factors, so that factors are associated with unique sigmas? (Seems unlikely, given the large number of curves, and a guarantee of multiple intersections for each pair of curves; mn = 3x3 in Bezout's Law.) https://math.mit.edu/research/highsc...72%20Rhee.pdf And a response would be very welcome from someone who understands it well. If multiple sigmas will find the same factors for some small exponent, an existence proof by finding at least two sigmas yielding the same small factor(s) ought not be difficult. And it turns out to be easy to find such. (Probably by poorly chosen bounds for the small task, but prime95 enforces a minimum B1.) As a fast experiment, M79, with two small factors 2687 and 202029703, which are also easily found with trial factoring. Code:
prime95 v30.8b14 (minimum B1=50k) M79 s B1 B2 Factor found 4011674131322832 50000 2,000,000 542853811961 = 2687 x 202029703 4273748159069600 50000 2,000,000 2687 5375224073660968 50000 2,000,000 542853811961 = 2687 x 202029703 2914336596096327 50000 2,000,000 2687 7353150075666063 50000 2,000,000 202020703 6507411175456731 50000 2,000,000 2687 5176102690618798 50000 2,000,000 202020703 5758632202560064 50000 2,000,000 2687 7719672998703375 50000 2,000,000 202020703 Code:
Sigma=17500281932646, B1=50000, B2=4000000. M1009 has a factor: 3454817 Sigma=17499395095926, B1=50000, B2=4000000. M1009 has a factor: 3454817 Sigma=908088127246459, B1=50000, B2=4000000. M1009 has a factor: 686066834105492663 = 3454817 x 198582684439 Sigma=2080153454287642, B1=50000, B2=4000000. M1009 has a factor: 3454817 Sigma=2080154903510001, B1=50000, B2=4000000. M1009 has a factor: 3454817 Sigma=2830004924066461, B1=50000, B2=4000000. M1009 has a factor: 686066834105492663 Sigma=3720592991825075, B1=50000, B2=4000000. M1009 has a factor: 198582684439 Sigma=3720595492188816, B1=50000, B2=4000000. M1009 has a factor: 686066834105492663 Sigma=4892655365568389, B1=50000, B2=4000000. M1009 has a factor: 3454817 Sigma=5642512694970646, B1=50000, B2=4000000. M1009 has a factor: 21624641697047 

20220606, 20:22  #7  
If I May
"Chris Halsall"
Sep 2002
Barbados
2^{3}×3×5×89 Posts 
Quote:
What do you advise based on this research? What are the next experiments based on yours? Perspiring minds want to know. Ken et al... These are serious questions. 

20220606, 21:40  #8 
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
2^{2}·3^{2}·191 Posts 
Selecting M199 for a test run that would require stage 2 in P1; 10 separate 1curve B1=50000, B2=4000000 runs, succeeded in 7 of 10:
Code:
ECM found a factor in curve #1, stage #1 Sigma=5162481759413312, M199 has a factor: 164504919713 Cofactor is a probable prime! ECM found a factor in curve #1, stage #1 Sigma=5912327664438779, M199 has a factor: 164504919713 Cofactor is a probable prime! M199 completed 1 ECM curve ECM found a factor in curve #1, stage #2 Sigma=7974985130155996, M199 has a factor: 164504919713 ECM found a factor in curve #1, stage #2 Sigma=8865572054271244, M199 has a factor: 164504919713 ECM found a factor in curve #1, stage #1 Sigma=608224716041860, M199 has a factor: 164504919713 Cofactor is a probable prime! ECM found a factor in curve #1, stage #2 Sigma=1498813620767224, M199 has a factor: 164504919713 M199 completed 1 ECM curve M199 completed 1 ECM curve ECM found a factor in curve #1, stage #2 Sigma=4311317535291977, M199 has a factor: 164504919713 Or M257, B1=120000, B2=4000000, 10 separate curves run individually, of which two found the 49 bit factor that would not be found by recommended level TF (to 4044 bits): Code:
ECM found a factor in curve #1, stage #2 Sigma=5976369241598935, M257 has a factor: 535006138814359 M257 completed 1 ECM curve ECM found a factor in curve #1, stage #2 Sigma=8788874036099382, M257 has a factor: 535006138814359 M257 completed 1 ECM curve M257 completed 1 ECM curve M257 completed 1 ECM curve M257 completed 1 ECM curve M257 completed 1 ECM curve M257 completed 1 ECM curve M257 completed 1 ECM curve Last fiddled with by kriesel on 20220606 at 21:41 
20220606, 21:49  #9  
If I May
"Chris Halsall"
Sep 2002
Barbados
2^{3}·3·5·89 Posts 
Quote:
Care to "wheel, and come again? 

20220606, 22:50  #10 
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
15334_{8} Posts 
(Edit time slot expired on my previous post)
A little experimentation with prime95 or whatever is handy can be enlightening. A modest laptop is sufficient. On M257, B1=120000, B2=4000000, the 10 separate curves run individually found the 49 bit factor that would not have been found with P1 and the same bounds (requiring much bigger P1 B2), but did not find the 80 bit factor that would have been found with P1 and the same bounds as used here. A set of another 10 singlecurve runs gave the factor found just once: ECM found a factor in curve #1, stage #2 Sigma=70969448337440, B1=120000, B2=4000000. M257 has a factor: 535006138814359 (ECM curve 1, B1=120000, B2=4000000) A quick look at prime95 v30.7b9 source code seems to confirm what the seed values listed above imply, that sigma ranges up to ~2^{53}. Max sigma seen in my samples ~ 8788874036099382 ~ 2^{52.965} (nearly 10^{16}) From ecm.cpp (// comments are mine present here only): Code:
double sigma; /* Sigma for the current curve */ Code:
/* Choose curve with order divisible by 16 and choose a point (x/z) on said curve. */ do { uint32_t hi, lo; ecmdata.sigma = (rand () & 0x1F) * 65536.0 * 65536.0 * 65536.0; // highest five bits randomized ecmdata.sigma += (rand () & 0xFFFF) * 65536.0 * 65536.0; //next 16 bits if (CPU_FLAGS & CPU_RDTSC) rdtsc (&hi, &lo); ecmdata.sigma += lo ^ hi ^ ((unsigned long) rand () << 16); // the other 32 bits of 53? } while (ecmdata.sigma <= 5.0); if (w>curve > 5.0 && w>curve < 9007199254740992.0) { // 5 < curve <2^{53} ecmdata.sigma = w>curve; w>curves_to_do = 1; } Last fiddled with by kriesel on 20220606 at 22:52 
20220606, 23:09  #11 
If I May
"Chris Halsall"
Sep 2002
Barbados
24670_{8} Posts 

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