20220607, 00:01  #12  
Random Account
Aug 2009
Not U. + S.A.
7·11·31 Posts 
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My example was simply that, just an example. If I had used "e9" instead of "e6". It should not be hard to devise an experiment to see what works and what does not using a small exponent with a known small factor. 

20220607, 00:23  #13  
Random Account
Aug 2009
Not U. + S.A.
7×11×31 Posts 
This did not take long to devise. Somebody tell me why this works when, in theory, it should not...
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20220607, 01:00  #14 
May 2022
15_{16} Posts 
Can somebody please tell me why tf does my computer say "segmentation fault" whenever I only set B1 and it completes the first curve? I am so confused
(I'm using B1=110000000) (Does it have anything to do with my RAM? I have 8 GB) EDIT: I know that using low B1 is useless, but still, why does my computer kill the process when it goes to stage 2? (MacOS) Last fiddled with by BigNumberGuy on 20220607 at 01:23 
20220607, 01:16  #15 
May 2022
3×7 Posts 
Also this "GMPECM" thing you guys are talking about, how do you use it?
(I'm not an expert please don't judge) 
20220607, 01:57  #16  
Apr 2020
1101101011_{2} Posts 
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One would have to be an idiot to want to only multiply by large primes, as that drastically reduces the chance that the group order divides our product of primes. GMPECM therefore has inbuilt idiotproofing and will refuse to run if you specify B1 as a range without a saved residue. Of course noone would ever be so perverse as to specify a backwards range for B1... or so I'd have thought until storm5510 just did it. The authors of GMPECM apparently didn't foresee this possibility  frankly, I don't blame them  and the result is that it runs a curve with no stage 1 at all. You'll see the exact same results if you use B1=0. 

20220607, 10:58  #17 
"Oliver"
Sep 2017
Porta Westfalica, DE
10011001110_{2} Posts 
There is at least some truth behind this. Yes, you will set a range for B1 if you resume a save file. In this case, you will set it to {B1 it was run at}{B1 it was run at}. This enables GMPECM to select a good B2, since it would not know the B1 otherwise.
A second possibility to use ranges is to split stage 2. If you have completed stage 1 and saved the residue, you could select a sensible B2, choose in how many parts you want to split it, and would run GMPECM while resuming the stage 1 file like this (step size is (selected B2  original B1) / step count): run B1={original B1}{original B1}, B2={original B1}{original B1 + step size} run B1={original B1}{original B1}, B2={original B1 + step size}{original B1 + 2 * step size} run B1={original B1}{original B1}, B2={original B1 + 2 * step size}{original B1 + 3 * step size} run B1={original B1}{original B1}, B2={original B1 + 3 * step size}{original B1 + 4 * step size} \(\hspace{20em}\)⋮ run B1={original B1}{original B1}, B2={original B1 + (step count  2) * step size}{original B1 + (step count  1) * step size} run B1={original B1}{original B1}, B2={original B1 + (step count  1) * step size}{selected B2} 
20220607, 11:09  #18  
Aug 2020
79*6581e4;3*2539e3
601 Posts 
edit: Sorry, I thought my edit was fast enough. It wasn't, so retina's reply is to the original version.
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The winning sigma just happened to have factors that math the current B1/B2 settings. And the probability to find a factor with a given B1/B2 just corresponds to the probability that a sigma will be found that is B1/B2 smooth. The question that remains, is there an upper bound for the group order of a sigma/factor combination? If so, then there'd be a B1 that would be guarenteed to find that factor, i.e. B1 = (max_GroupOrder)^1/2. Last fiddled with by bur on 20220607 at 11:28 Reason: complete rewriting 

20220607, 11:17  #19  
Undefined
"The unspeakable one"
Jun 2006
My evil lair
6641_{10} Posts 
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Plus the software we currently have only uses a small subset of all possible sigmas. You only get 53 bits to choose from. So that will hinder your ability to find factors even if you could run through all 2^53 available values. Last fiddled with by retina on 20220607 at 11:18 

20220607, 14:33  #20 
Apr 2020
5^{3}×7 Posts 
Yes: for a prime p, the group order is always within 2sqrt(p) of p+1, by Hasse's theorem. Of course the resulting B1/B2 that are guaranteed to find the factor are absurdly high for factors that we care about. As retina says, at that point you might as well be running TF.

20220607, 14:36  #21  
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
2^{3}·7·127 Posts 
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I guess you meant match there not math. (I don't think you mean the sigma value's factors matter. Among the least smooth winning sigmas in this post are 7974 985130 155996 = 2^{2} × 1993 746282 538999, and 5912 327664 438779 = 47 × 125 794205 626357; in a previous post, at least sigma value 4892 655365 568389 is prime.) Perhaps you meant something like, the winning sigma generates a curve containing points corresponding to some factors of the number of interest, that are findable with the current B1 and B2 settings. Quote:
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I ran a single curve at t65 level for M1619, with elapsed time ~7.3 hours on i3370M (after searching several machines to find one with gmpecm installed). Step1 was 4+ hours of that. Running 1 M1277 t65 curve now. Last fiddled with by kriesel on 20220607 at 15:07 

20220607, 14:56  #22 
Aug 2020
79*6581e4;3*2539e3
601 Posts 
As R.D. Silverman and kriesel kindly pointed out I made some annoying errors in my post above, of course not sigma has to be smooth but the group order.
Thanks for the explanations, I finally understood, I think, what the probability for ECM to find a factor actually means. 
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