20030820, 12:59  #1 
Sep 2002
2×331 Posts 
Shortest time to complete a 2^67 trial factor (no factor)
What is the shortest amount of time to complete a 2^67 TF ( without finding a factor ) ?
Time in days::hours:minutes . What setup ( CPU, speed, cache size, OS ) ? For example using my Athlon 1200 the shortest elapsed time is 3::2:22 . 3::02:22 2^67 M23128319,57 Athlon, 1200, L1 128 L2 256, Win98SE They do start with different amounts of factoring done: 57,58 etc. which does affect the amount of time so if you know the starting point include it. Most M23 on my PC are M23xxxxxx,57 with one ,58. Are there any systems that can complete a 2^67 ( Mxxxxxxxx,57 without finding a factor) in 1 day ? 
20030820, 17:21  #2 
"Mike"
Aug 2002
2×11×19^{2} Posts 
I bet the fastest will be a P4... Even though the AMD wins by a big margin up to 64 bits, the vast majority of the overall work will be from 6467, which is where the SSE2 crap kicks in...
I can run 23128319 from 58 to 59 bits in 75 or so seconds on an overclocked (2100MHz) 2500+ Barton, so... 59 = 75 60 = 150 61 = 300 62 = 600 63 = 1,200 64 = 2,400 65 = 4,800 66 = 9,600 67 = 19,200 If you add all that up you get 38,325 seconds or 10 hours 38 minutes 45 seconds... 
20030820, 18:06  #3  
Dec 2002
Frederick County, MD
2×5×37 Posts 
Quote:


20030820, 18:49  #4  
Aug 2002
North San Diego County
682_{10} Posts 
Quote:
OTOH, my P4 2.5 (OC'd 2.4) generally takes about 25 hours total to TF a 23M exponent to 67 bits (starting bits 58, 59, 60). I'll check my work 2.66 and see if it is less than a day. 

20030820, 20:51  #5 
"Mike"
Aug 2002
2×11×19^{2} Posts 
We know the work being done is doubled for each additional bit we go deeper... Why exactly does the time increase so much?

20030820, 20:54  #6 
P90 years forever!
Aug 2002
Yeehaw, FL
7^{2}·149 Posts 
64 bits fits in one floating point or two integer resisters. To do 65bit factors you must use two floats or three integer registers, resulting in a big jump in the number of multiplies and adds.

20030821, 00:12  #7 
Sep 2002
2·331 Posts 
Time in hours to do a 2^67 on an Athlon 1200
skipping smaller steps 2^60 1/12 = 5 minutes 2^61 1/6 2^62 1/3 2^63 1 1/4 2^64 2 1/2 2^65 10 2^66 20 2^67 40 74 1/3 hours Up to and including 2^64 the time doubles There is a big jump at 2^65 the time quadruples. At 2^66 it resumes time doubling. This is using the x87 ( no SSE2 on the Athlon ) 
20030821, 01:07  #8 
Aug 2002
Termonfeckin, IE
2757_{10} Posts 
Well if you notice the time quadruples from the 2^62 to 2^63 step as well. This is because nonSSE2 machines have very efficient code upto that limit. Some of those optimizations are not applicable for 2^63 and 2^64. I believe this code was contributed by someone other than George. If you mosey around in the forum you should find George's post about this.
So to sum it all up, 2^65 is about 32 times as slow as 2^62 as opposed to 8 times which is what we would have expected if tha scaling were regular. This proportion is very different for P4s and is actually closer to 8 than to 32. In fact here are some timing comparisons I made about 7 months ago. [code:1]Time taken to complete 14366959 to 0.98% To Bit On PII 450 On P4 2533 Improv On 1333TB Improv 59 6 2.5 2.4 1.85 3.25 60 12 5 2.4 3.7 3.25 61 24 10 2.4 7.4 3.25 62 48 20 2.4 14.8 3.25 63 170 37.5 4.5 51 3.33 64 340 75 4.5 102 3.33 65 1540 180 8.5 480 3.21 [/code:1] 
20030821, 07:35  #9 
"Tony Gott"
Aug 2002
Yell, Shetland, UK
2×3×53 Posts 
I think that this topic has important considerations for Lone Mersenne Hunters. All work on slower boxen (which is what this subproject seems to attract) must be attempting to tf all exponents up to 62 bit, rather than trying to move on upwards to 64 bit.
Thoughts anyone, or is this beginning to move OT ..... 
20030821, 12:15  #10  
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}×3×641 Posts 
Quote:
Classic Pentia, compared to the 486s, have far more efficient floatingpoint because AFAIK that's when Intel introduced FP pipelining. But their integer performance is not as dramatically better per clock cycle than 486s as the FP ratio is. As a result, classic Pentia get their best throughput per clock cycle, relative to other types, on tasks that are FPheavy but below the SSE2 range. TF M67108913 on a P75: 2^61 to 2^62: 0.279 sec./iteration 2^62 to 2^63: 0.320 sec./iteration 2^63 to 2^64: 0.320 sec./iteration 2^64 to 2^65: 1.314 sec./iteration Note the small step from 2^62 iteration time to 2^63/2^64 iteration time. Its relative total times per bit range are about: 2^61 to 2^62: 1.0 2^62 to 2^63: 2.3 2^63 to 2^64: 4.6 2^64 to 2^65: 37.7 Quote:


20030821, 16:38  #11  
"Patrik Johansson"
Aug 2002
Uppsala, Sweden
5^{2}·17 Posts 
Re: Shortest time to complete a 2^67 trial factor (no factor
Quote:
P4, 3141 MHz, L1 8 L2 512, RedHat Linux Quote:


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