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 2003-08-20, 12:59 #1 dsouza123     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 ?
 2003-08-20, 17:21 #2 Xyzzy     Aug 2002 22×11×191 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 64-67, 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...
2003-08-20, 18:06   #3
eepiccolo

Dec 2002
Frederick County, MD

17216 Posts

Quote:
 Originally Posted by Xyzzy 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...
The problem with this is that once you get past 64 bits, things slow down. so for you 65 bits might be something more like 7,000 s, but I'm not sure exactly. The SSE2 in the P4 keeps it from slowing down as much, but it still might end up at something like 6,000 s for 65 bits, if 64 bits took 2,400 s.

2003-08-20, 18:49   #4
sdbardwick

Aug 2002
North San Diego County

19·37 Posts

Quote:
 Originally Posted by eepiccolo The problem with this is that once you get past 64 bits, things slow down. so for you 65 bits might be something more like 7,000 s, but I'm not sure exactly. The SSE2 in the P4 keeps it from slowing down as much, but it still might end up at something like 6,000 s for 65 bits, if 64 bits took 2,400 s.
The slowdown over 64 bits is significant; my Athlon XP 1900+(1600Mhz) takes over a day just to go from 66 bit to 67 bit factoring on 23M exponents.

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.

 2003-08-20, 20:51 #5 Xyzzy     Aug 2002 20D416 Posts We know the work being done is doubled for each additional bit we go deeper... Why exactly does the time increase so much?
 2003-08-20, 20:54 #6 Prime95 P90 years forever!     Aug 2002 Yeehaw, FL 11110010101012 Posts 64 bits fits in one floating point or two integer resisters. To do 65-bit factors you must use two floats or three integer registers, resulting in a big jump in the number of multiplies and adds.
 2003-08-21, 00:12 #7 dsouza123     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 )
 2003-08-21, 01:07 #8 garo     Aug 2002 Termonfeckin, IE ACC16 Posts Well if you notice the time quadruples from the 2^62 to 2^63 step as well. This is because non-SSE2 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]
 2003-08-21, 07:35 #9 bayanne     "Tony Gott" Aug 2002 Yell, Shetland, UK 5148 Posts I think that this topic has important considerations for Lone Mersenne Hunters. All work on slower boxen (which is what this sub-project seems to attract) must be attempting to t-f 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 .....
2003-08-21, 12:15   #10

"Richard B. Woods"
Aug 2002
Wisconsin USA

11110000011002 Posts

Quote:
 Originally Posted by garo Well if you notice the time quadruples from the 2^62 to 2^63 step as well. This is because non-SSE2 machines have very efficient code upto that limit. Some of those optimizations are not applicable for 2^63 and 2^64.
I thought the increase in iteration times between 2^62 and 2^63 was due to the switchover from integer arithmetic (used up to 2^62) to floating-point arithmetic (for the 2^63 and 2^64 ranges) on some CPUs.

Classic Pentia, compared to the 486s, have far more efficient floating-point 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 FP-heavy 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:
 Originally Posted by bayanne All work on slower boxen (which is what this sub-project seems to attract) must be attempting to t-f all exponents up to 62 bit, rather than trying to move on upwards to 64 bit.
It would make sense for classic Pentia to concentrate on 2^62 -> 2^64 rather than any other range.

2003-08-21, 16:38   #11
patrik

"Patrik Johansson"
Aug 2002
Uppsala, Sweden

52·17 Posts
Re: Shortest time to complete a 2^67 trial factor (no factor

Quote:
 Originally Posted by dsouza123 What is the shortest amount of time to complete a 2^67 TF ( without finding a factor ) ? Time in days::hours:minutes .
0::20:42 2^67 M23052089,60
P4, 3141 MHz, L1 8 L2 512, RedHat Linux
Quote:
 Originally Posted by dsouza123 Are there any systems that can complete a 2^67 ( Mxxxxxxxx,57 without finding a factor) in 1 day ?
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