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Old 2015-07-26, 21:00   #56
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Quote:
Originally Posted by Madpoo View Post
and a verified residue of 0x7AFB854059F08BEF (well, technically the last 64 bits of it, but you know what I mean).
There lies the problem. You only have the residue mod 2^64. Therefore after performing the loop, you know s = (res + k * 2^64) mod f for some unknown k. The problem is that you can always find a value of k that works for any residue. For your M2000143 example, the decimal residue following the loop using f is 1838775420032325409. From this and the verified residue, you can find k = 815412537616906440. For the invalid residue 0x6AFB854059F08BEF, you can find k = 1201294030677328637.
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Old 2015-07-26, 21:20   #57
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I'll take all of seven on the second list, 34769731 to 45951173 inclusive.
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Old 2015-07-26, 22:18   #58
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Quote:
Originally Posted by frmky View Post
There lies the problem. You only have the residue mod 2^64. Therefore after performing the loop, you know s = (res + k * 2^64) mod f for some unknown k. The problem is that you can always find a value of k that works for any residue. For your M2000143 example, the decimal residue following the loop using f is 1838775420032325409. From this and the verified residue, you can find k = 815412537616906440. For the invalid residue 0x6AFB854059F08BEF, you can find k = 1201294030677328637.
Well, I guess I'm still confused on how exactly I could verify the residue given the factor, since all we have are the last 64 bits of the residue.

I won't worry about it terribly though.

There's around 26,000-27,000 exponents where the residue is verified, but has a factor. Another nearly 18K exponents where a factor was found after only 1 LL test was done.

I thought it'd be cool to figure out, for those 18K tests, whether the residue was good or not. It'd be a pretty impressive and large set of data to figure out the reliability of the LL tester.
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Old 2015-07-26, 22:44   #59
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Originally Posted by Madpoo View Post
I thought it'd be cool to figure out, for those 18K tests, whether the residue was good or not. It'd be a pretty impressive and large set of data to figure out the reliability of the LL tester.
At the very least, I've found 104 instances where a verifying LL test exists for an exponent, but there is also a bad/mismatched residue from some yahoo. That's 104 more "bad" results I can glean out of the data right away.

Plus the tens of thousands of "good" results I can pull out, where those verifying residues exist but it was factored.

Yeah, I think that kind of info is useful, and maybe I can get George to approve adding new "result types" instead of the single "factored" code that exists now. Like:
Factored - Unverified
Factored - Verified
Factored - Bad

I don't know if we'd need a "Factored - Suspect" to indicate if the residue was originally marked "suspect" or not. Don't know if that matters as much.
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Old 2015-07-26, 22:54   #60
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Quote:
Originally Posted by frmky View Post
There lies the problem. You only have the residue mod 2^64. Therefore after performing the loop, you know s = (res + k * 2^64) mod f for some unknown k. The problem is that you can always find a value of k that works for any residue. For your M2000143 example, the decimal residue following the loop using f is 1838775420032325409. From this and the verified residue, you can find k = 815412537616906440. For the invalid residue 0x6AFB854059F08BEF, you can find k = 1201294030677328637.
does this change fail the test ?

Quote:
(19:52) gp > f=6174103888966755151;s=4;b=8861823203214920687;for(x=3,2000143,s=Mod(Mod(s,f)^2-2,1<<64));if(lift(s)==lift(Mod(Mod(b,f),1<<64)),print("factor found:"f))
factor found:6174103888966755151
edit:never mind it says it finds the factor either way with the hex to decimal converter I used to convert it to decimal for testing.

Last fiddled with by science_man_88 on 2015-07-26 at 23:11
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Old 2015-07-26, 23:17   #61
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Huh?

Code:
(16:16) gp > f=6174103888966755151;s=4;b=8861823203214920687;for(x=3,2000143,s=Mod(Mod(s,f)^2-2,1<<64));if(lift(s)==lift(Mod(Mod(b,f),1<<64)),print("factor found:"f))
(16:16) gp > lift(s)
%8 = 1838775420032325409
(16:16) gp > lift(Mod(Mod(b,f),1<<64))
%9 = 2687719314248165536
Edit: OK, you found the problem. :-)

Last fiddled with by frmky on 2015-07-26 at 23:19
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Old 2015-07-27, 00:21   #62
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Quote:
Originally Posted by Madpoo View Post
There are 47 in the list.
I just took the last 23 in this list. As in, I now "own" all of the candidates from the list of 47. Better build another list Aaron....

BTW, seven from my first batch of 14 from the original list are now completed. None matched. All but the highest two will be completed in about three hours.

BTW2... Aaron, two in the list of 47 were sub-optimally TF'ed. Anything between 40M and 50M should be TF'ed to 72. I'll TF them myself in parallel.
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Old 2015-07-27, 00:30   #63
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Quote:
Originally Posted by chalsall View Post
I just took the last 23 in this list. As in, I now "own" all of the candidates from the list of 47. Better build another list Aaron....

BTW, seven from my first batch of 14 from the original list are now completed. None matched. All but the highest two will be completed in about three hours.

BTW2... Aaron, two in the list of 47 were sub-optimally TF'ed. Anything between 40M and 50M should be TF'ed to 72. I'll TF them myself in parallel.
Here's 4 more, using the same criteria as the last shorter list. They recently became available as assignments expired, or extra TF work completed:
Code:
exponent	Bad	Good	Unk	Sus	worktodo
37952297	19	3	5	5	DoubleCheck=37952297,71,1
41573579	17	2	6	3	DoubleCheck=41573579,72,1
41626679	11	2	6	10	DoubleCheck=41626679,72,1
41678927	17	2	6	3	DoubleCheck=41678927,72,1
I'll work on a larger list in a bit.
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Old 2015-07-27, 01:23   #64
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In less than a day, I'll begin finishing the trial factoring for about 90 50M+ exponents per day. Finishing the trial factoring of the 50M+ exponents will take me approximately two weeks. That should be good for daily lists for a while. :)
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Old 2015-07-27, 01:29   #65
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Quote:
Originally Posted by Madpoo View Post
I'll work on a larger list in a bit.
Here are some more:
Code:
exponent	Bad	Good	Unk	Sus	worktodo
34865477	9	2	15	8	DoubleCheck=34865477,71,1
34949749	9	2	15	8	DoubleCheck=34949749,71,1
34996139	5	1	3	0	DoubleCheck=34996139,71,1
35143627	5	1	10	6	DoubleCheck=35143627,71,1
35476471	5	1	10	6	DoubleCheck=35476471,71,1
35578973	9	2	15	8	DoubleCheck=35578973,71,1
36131197	9	2	15	8	DoubleCheck=36131197,71,1
36337391	9	2	15	9	DoubleCheck=36337391,71,1
36453931	5	1	10	6	DoubleCheck=36453931,71,1
36529453	5	1	2	2	DoubleCheck=36529453,71,1
36808613	5	1	10	6	DoubleCheck=36808613,71,1
36815837	9	2	15	8	DoubleCheck=36815837,71,1
37260259	5	1	9	2	DoubleCheck=37260259,71,1
38501137	5	0	6	3	DoubleCheck=38501137,72,1
40882241	9	2	15	9	DoubleCheck=40882241,72,1
41471879	9	2	15	8	DoubleCheck=41471879,71,1
41501183	5	0	5	0	DoubleCheck=41501183,72,1
41620973	11	2	1	8	DoubleCheck=41620973,72,1
42019207	5	1	9	2	DoubleCheck=42019207,72,1
42022873	9	2	15	9	DoubleCheck=42022873,72,1
42100753	39	9	13	18	DoubleCheck=42100753,72,1
42174677	9	2	15	8	DoubleCheck=42174677,71,1
42240043	19	4	5	4	DoubleCheck=42240043,72,1
42791519	9	2	15	8	DoubleCheck=42791519,71,1
43263431	9	2	15	8	DoubleCheck=43263431,70,1
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Old 2015-07-27, 01:50   #66
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Quote:
Originally Posted by Madpoo View Post
Here are some more:
Code:
exponent	Bad	Good	Unk	Sus	worktodo
41471879	9	2	15	8	DoubleCheck=41471879,71,1
42174677	9	2	15	8	DoubleCheck=42174677,71,1
42791519	9	2	15	8	DoubleCheck=42791519,71,1
43263431	9	2	15	8	DoubleCheck=43263431,70,1
These four need more trial factoring. I'll start immediately.

Edit: should be finished 60 minutes from now.

Last fiddled with by Mark Rose on 2015-07-27 at 01:54
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