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Old 2019-05-22, 22:32   #12
AG5BPilot
 
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Quote:
Originally Posted by ATH View Post
You can probably add:
CpuSupportsAVX512F=0

to LLR.ini, or use the -oCpuSupportsAVX512F=0 in the command line.

I cannot test this since I do not have an AVX512 cpu, but for me using -oCpuSupportsFMA3=0 turns off AVX2 and uses AVX instead.
That was the first thing we tried. (See my message right before yours.)

Then we looked in the LLR source code. It's not there.
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Old 2019-05-23, 05:41   #13
Darkclown
 
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That flag does not work w/ the 3.8.23 release of LLR.
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Old 2019-07-08, 13:39   #14
ATH
Einyen
 
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Error with ForcePRP=1 in the ini file:

https://mersenneforum.org/showthread.php?t=22981
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Old 2019-09-20, 06:56   #15
stream
 
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What happens if LLR cannot factorize 'b' ? It may happen with big GFN numbers, having very high 'b'. Probably aprcl.exe should help with factorization, but we don't include it in Boinc setups. Without aprcl, internal factorization fails and a warning is printed:


>cllr64.3.8.23.exe -d -q"1814570322897374^65536+1"
Error 2 while trying to create new process
Error 2 while trying to create new process
Base factorized as : 2*7*67*1934509939123*1
Base cofactor : 1934509939123, 1 (Must be proven prime or factorized externally)


In the example above, 1934509939123 is prime. But what happens in theoretical case if LLR accepts composite cofactor? Will the prime test itself work correctly?
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Old 2020-05-19, 02:37   #16
Batalov
 
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Phi(4,2^7658614+1)/2

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I am posting a patch to LLR that allows to run the "Proth.exe" -like test for k*b^n+1 dividing some large cyclotomic Phi(b^m,2).

The input file should have the following magic header:
Code:
ABC DivPhi($a*$b^$c+1)
2 3 822
2 3 897
2 3 1252
2 3 1454
2 3 4217
The output will be:
Code:
2*3^822+1 does not divide 2^3^822-1  Time : 263.342 ms.
2*3^897+1 Divides 2^3^897-1  Time : 116.718 ms.
2*3^897+1 Divides 2^3^896-1  Time : 103.216 ms.
2*3^897+1 does not divide 2^3^895-1  Time : 101.769 ms.
Conclusion: 2*3^897+1 Divides Phi(3^896,2)
2*3^1252+1 does not divide 2^3^1252-1  Time : 97.786 ms.
2*3^1454+1 does not divide 2^3^1454-1  Time : 109.374 ms.
2*3^4217+1 Divides 2^3^4217-1  Time : 37.989 ms.
2*3^4217+1 does not divide 2^3^4216-1  Time : 50.651 ms.
Conclusion: 2*3^4217+1 Divides Phi(3^4217,2)
Attached Files
File Type: zip Llr_DivPhi.zip (2.2 KB, 31 views)
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Old 2020-05-21, 13:53   #17
Jean Penné
 
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May 2004
FRANCE

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Default Thank you Serge!

Quote:
Originally Posted by Batalov View Post
I am posting a patch to LLR that allows to run the "Proth.exe" -like test for k*b^n+1 dividing some large cyclotomic Phi(b^m,2).

The input file should have the following magic header:
Code:
ABC DivPhi($a*$b^$c+1)
2 3 822
2 3 897
2 3 1252
2 3 1454
2 3 4217
The output will be:
Code:
2*3^822+1 does not divide 2^3^822-1  Time : 263.342 ms.
2*3^897+1 Divides 2^3^897-1  Time : 116.718 ms.
2*3^897+1 Divides 2^3^896-1  Time : 103.216 ms.
2*3^897+1 does not divide 2^3^895-1  Time : 101.769 ms.
Conclusion: 2*3^897+1 Divides Phi(3^896,2)
2*3^1252+1 does not divide 2^3^1252-1  Time : 97.786 ms.
2*3^1454+1 does not divide 2^3^1454-1  Time : 109.374 ms.
2*3^4217+1 Divides 2^3^4217-1  Time : 37.989 ms.
2*3^4217+1 does not divide 2^3^4216-1  Time : 50.651 ms.
Conclusion: 2*3^4217+1 Divides Phi(3^4217,2)
Thank you, Serge, for this code ; in included it in my new Llr.c file and tested it locally. So, it will be included in the future LLR 3.8.24 release I hope to upload soon. Also, this release will include the Gerbicz tests for base two numbers, and also the -oCpuSupportsAVX512F=0 option that was required for a while...
I hope you are well and take care of you by these times...
Best Regards,
Jean
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Old 2020-05-21, 16:06   #18
Batalov
 
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Thank you Jean!

This code is 5-6 years old so it passed a lot of internal testing (though this time I restored it from memory from scratch and of course, with modern LLR, one can now use m/t).
I have extensively used it previously and I have already tested all known "Divides Phi numbers" (not just visible top 20; all such numbers including tiny ones can be found by UTM search, using option "all"). All past Proth.exe-based results are reproduced.

There are additional checks that I can provide later -- so that we can quit this test very early; if by Euler's criterion we can immediately check that 2^b^n-1 will not be zero without computation.
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