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#23 | |
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May 2007
Kansas; USA
101×103 Posts |
Quote:
Last fiddled with by gd_barnes on 2010-02-16 at 19:44 |
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#24 | |
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Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
17·251 Posts |
Quote:
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#25 | |||
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Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
17·251 Posts |
Environment: i5 on XP 32-bit with the other 3 cores working on other LLR 3.8.0 work
Program: LLR 3.8.0 Quote:
Raw results: Code:
3*2^33219-1 is not prime. LLR Res64: A376F1DAC8711F6A Time : 627.730 ms. 7*2^33219-1 is not prime. LLR Res64: 9E8805FFADF7975A Time : 628.117 ms. 9*2^33219-1 is not prime. LLR Res64: 79561A8C7B4ECD75 Time : 626.235 ms. 2*3^20959-1 is not prime. RES64: D0BBDC84AD2E4430. OLD64: F0BD1921158D3FE3 Time : 774.398 ms. 4*3^20959-1 is not prime. RES64: CAE19F16E1DE1038. OLD64: 60A4DD44A59A30A5 Time : 773.138 ms. 6*3^20959-1 is not prime. RES64: 9855F56DADEBEFB5. OLD64: C901E04909C3CF1C Time : 771.612 ms. 3*2^33219-1 is not prime. LLR Res64: A376F1DAC8711F6A Time : 628.925 ms. 7*2^33219-1 is not prime. LLR Res64: 9E8805FFADF7975A Time : 625.828 ms. 9*2^33219-1 is not prime. LLR Res64: 79561A8C7B4ECD75 Time : 627.717 ms. 2*3^20959-1 is not prime. RES64: D0BBDC84AD2E4430. OLD64: F0BD1921158D3FE3 Time : 774.332 ms. 4*3^20959-1 is not prime. RES64: CAE19F16E1DE1038. OLD64: 60A4DD44A59A30A5 Time : 771.940 ms. 6*3^20959-1 is not prime. RES64: 9855F56DADEBEFB5. OLD64: C901E04909C3CF1C Time : 771.234 ms. Average base 2 time: 627.425 ms Average base 3 time: 772.776 ms At 10,000 digits with very small k, base 2 is 18.81% faster than base 3. Quote:
Raw results: Code:
301*2^332191-1 is not prime. LLR Res64: 838E673AE827270E Time : 90.516 sec. 303*2^332191-1 is not prime. LLR Res64: 8490D6606D386526 Time : 90.488 sec. 307*2^332191-1 is not prime. LLR Res64: 77653C95F8BAB70C Time : 90.503 sec. 302*3^209591-1 is not prime. RES64: 6F4E51BFDCF6E592. OLD64: 4DEAF53F96E4B0B3 Time : 116.848 sec. 304*3^209591-1 is not prime. RES64: 060AB7310C91F6EE. OLD64: 7B286806ECDF83B8 Time : 117.237 sec. 310*3^209591-1 is not prime. RES64: 7750769DF8641EEA. OLD64: 1B27494C9B00E8EA Time : 116.754 sec. Average base 2 time: 90.502 sec Average base 3 time: 116.946 sec At 100,000 digits with k about 300, base 2 is 22.61% faster than base 3. Quote:
Raw results: Code:
3*2^332191-1 is not prime. LLR Res64: F7D86C663D06F365 Time : 90.509 sec. 7*2^332191-1 is not prime. LLR Res64: F9F68E1E93C80AD4 Time : 90.423 sec. 9*2^332191-1 is not prime. LLR Res64: 647E0ADBF224F9BB Time : 90.590 sec. 3*5^143067-1 is not prime. RES64: CAF5E343491784FF. OLD64: 60E1A9C9DB468EFA Time : 97.941 sec. 4*5^143067-1 is not prime. RES64: 13F277C23E6E6CDE. OLD64: 2CE26F06449181B1 Time : 97.896 sec. 6*5^143067-1 is not prime. RES64: 57F729330E0D6FE8. OLD64: 07E57B992A284FB5 Time : 97.920 sec. Average base 2 time: 90.507 sec Average base 5 time: 97.919 sec At 100,000 digits with very small k, base 2 is 7.57% faster than base 5. Last fiddled with by Mini-Geek on 2010-02-16 at 14:34 |
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#26 |
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Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
17×251 Posts |
I have a feeling that the granularity coming from the FFT size choosing is, in a way, throwing the data off. That might explain the large difference between the tests where base 2 is about 7% faster and where it's about 20% faster than other bases.
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#27 | |
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May 2007
Kansas; USA
28A316 Posts |
Quote:
I'm wondering if my I7 would have the same difference for those various small tests. I'll make time to check it later today. |
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#28 |
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Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
10AB16 Posts |
Since the FFT length increases in discrete jumps, the time it takes for tests does not increase linearly, (smoothly) but in big jumps, (granularly...whether I just invented that word or not
) and that the effects of those jumps might easily mean the difference between 0%, 7%, or 20% improvement (rather than only the factors we're trying to vary: digit size, base, and k).To sum the whole statement up: take our results with a grain of salt (no pun intended), since they don't account for FFT changes. Last fiddled with by Mini-Geek on 2010-02-16 at 20:07 |
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#29 | |
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May 2007
Kansas; USA
101·103 Posts |
Quote:
To sum it up: It seems like our results here would be useful since the FFTlen changes are non-existent due to same size tests and k-value. Or maybe I'm not understanding something. I had assumed that FFTlen jumps would be solely related to k-value and size of the test. Could those be impacted by base also even though the other 2 variables are almost the same? If so, that would explain the differences and what we'd need to do is test across a variety of sizes and average all of the speed differences to get a better of idea of what the "real" speed difference is. Assuming that the above is the case, then REALLY what determines the speed difference for a particular test is where it falls within the range of the FFTlen for both the base 2 and non-base 2 tests. So what we'd need to do is find somewhat similar sizes and k's and verify where that size falls within the particular FFTlen for the bases in question to get a real apples to apples comparison. As an example, if the base 2 test falls on the low end of the FFTlen (i.e. right after an increase in FFTlen), and the non-base 2 test falls on the high end of the FFTlen for that base, then there might only be a 1-2% difference or no difference at all. Whereas in the reverse scenario, we might get as much as a 25-30% difference. This may be more complicated to test than I had imagined. Last fiddled with by gd_barnes on 2010-02-16 at 20:30 |
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#30 | |
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Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
16F916 Posts |
Quote:
Or at what size number the FFT length changes for different bases. One problem with this is that AFAIK pfgw doesn't output FFT length(that has always bugged me if your reading rouge) |
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#31 |
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Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
10AB16 Posts |
3*2^332191-1, Used fftlen = 20480
2*3^209591-1, Using FFT length 20K, a = 3 (Gary's scenario, same FFT length, 7% difference) 3*2^33219-1, Used fftlen = 1792 2*3^20959-1, Using FFT length 1792, a = 3 (scenario 1: digit length ~10,000, same FFT length, 18.81% difference) 301*2^332191-1, Used fftlen = 20480 302*3^209591-1, Using FFT length 24K, a = 3 (scenario 3: k~=300, different FFT lengths, 22.61% difference) 3*2^332191-1, Used fftlen = 20480 3*5^143067-1, Using FFT length 20K, a = 3 (scenario 4: b=5, same FFT length, 7.57% difference) It is definitely possible to have different FFT lengths at similar k and sizes, but we'd need more data (be it experimental or by the math or viewing the code) to have any good idea of how often that happens. I know that there's a tool called fft_len that tells you what FFT length will be used for the given k in the given n range for base 2, but I don't know about other bases. |
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#32 | |
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"Mark"
Apr 2003
Between here and the
24·397 Posts |
Quote:
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#33 |
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A Sunny Moo
Aug 2007
USA (GMT-5)
3·2,083 Posts |
I think LLR always tells you the FFT length; for its new N-1/N+1 tests you have to wait a bit for some initialization stuff to finish, but that shouldn't be more than 10-15 seconds even for very large tests.
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