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#23 | |
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Jun 2003
2·59 Posts |
Quote:
Thomas
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#24 |
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Aug 2002
865810 Posts |
We could send him a note every time we (as a group) complete 1000 curves... Or maybe email him once we've completed this bit depth...
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#25 |
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Aug 2002
207228 Posts |
126 curves done... (B1=44000000 & B2=184367799127)
BTW, if you post your curves, make sure that you are not posting your cumulative total... For example, every time I post, I clear my log file... |
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#26 |
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Mar 2003
New Zealand
13·89 Posts |
Has anyone done a curve at the 55 digit (B1=110 million) level to see how long it will take and how much memory is used? I don't have a machine with enough memory to do stage two without increasing the -k parameter.
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#27 | |
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Aug 2002
2×32×13×37 Posts |
Quote:
Code:
PID USER PR NI VIRT RES SHR S %CPU %MEM TIME+ COMMAND 9809 mv 39 19 227m 222m 5252 R 93.9 45.0 46:10.69 ecm -c 0 -k 32 11e7 |
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#28 | |
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Jun 2003
2·59 Posts |
Quote:
BTW: [Mon Oct 18 16:48:25 2004] M1061 completed 100 ECM curves, B1=44000000, B2=4290000000 Who should keep the group curve count and e-mail George ? Thomas
Last fiddled with by thomasn on 2004-10-19 at 08:59 |
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#29 | |
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Bamboozled!
"๐บ๐๐ท๐ท๐ญ"
May 2003
Down not across
2·17·347 Posts |
Quote:
Alex? Paul |
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#30 |
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"Nancy"
Aug 2002
Alexandria
2,467 Posts |
Memory use is, iirc, dF*(log_2(dF)+6), not sure about the 6, residues.
B2=B2min + d*dF*k, and d is approximately (well, not quite) proportional to dF. Multiplying k by 4 therefore approx. halves dF, and so approx. halves memory use, for a given B2min-B2 interval. For a first approximation, memory use is a function of 1/sqrt(k). Alex Last fiddled with by akruppa on 2004-10-19 at 15:08 |
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#31 |
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Aug 2002
207228 Posts |
I just upgraded from a 3200+ to a 3400+... AMD's naming scheme is a bit convoluted, so I'll describe the differences... The 3200+ I had was a 2GHz part with a 1024K L2... The 3400+ I have now is a 2.4GHz part with 512K L2...
Assuming the L2 cache size is irrelevant, this is a 20% jump in clock speed... 2000MHz/1024K Code:
GMP-ECM 5.0.3 [powered by GMP 4.1.4] [ECM] Input number is 24707306311927565716857342128774085333197833223161879682238935306082805123046306993647507776054336486228891340858985829027076261887914242781617846672453431386903982455635542158748401823985988322905245077938567513252198179128990807936780194781391547404884040101606295111368825026273254703636026307207764436438929167613951 (320 digits) Using B1=44000000, B2=184367799127, polynomial Dickson(30), sigma=3552224530 Step 1 took 973900ms Step 2 took 1134027ms Code:
GMP-ECM 5.0.3 [powered by GMP 4.1.4] [ECM] Input number is 24707306311927565716857342128774085333197833223161879682238935306082805123046306993647507776054336486228891340858985829027076261887914242781617846672453431386903982455635542158748401823985988322905245077938567513252198179128990807936780194781391547404884040101606295111368825026273254703636026307207764436438929167613951 (320 digits) Using B1=44000000, B2=184367799127, polynomial Dickson(30), sigma=2461182436 Step 1 took 812757ms Step 2 took 965094ms Stage 1 = 19.8% faster... Stage 2 = 17.5% faster... I'm sure the fact I used different sigma values makes this less accurate... Anyways, in a nutshell, for ECM on the K8, a super large L2 cache isn't too important... |
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#32 | |
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Mar 2003
New Zealand
13·89 Posts |
Quote:
I noticed that the Celeron is comparable to the P4 for numbers up to M8192, but much slower for numbers above M32768, so somewhere in between is probably the point where the number no longer fits in a 128k cache. |
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#33 |
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Aug 2002
2×32×13×37 Posts |
Are we even dealing with FFTs here, like we are when we do GIMPS work? If so that would make a lot of sense... Hmm...
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