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Old 2005-12-29, 18:07   #1
thechickenman
 
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Default P4's a waste at 2... what?

I've been reading the old threads, saying that using a P4 below a certin bit depth is a waste... What bit depth do P4's start beign efficent?

I've been using P4's at 62 bits... assuming that's really not very efficent?

Would anyone fuss if I gave up on the ranges I took to 2^62 and changed to some more efficent for P4's? I've got 2 running things- a 2.3 Ghz and a 2.8 Ghz

Last fiddled with by thechickenman on 2005-12-29 at 18:07
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Old 2005-12-29, 19:49   #2
Xyzzy
 
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I think the P4's efficiency kicks in at 264 because of some SSE2 enhancements.
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Old 2005-12-29, 20:57   #3
thechickenman
 
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Quote:
Originally Posted by Xyzzy
I think the P4's efficiency kicks in at 264 because of some SSE2 enhancements.
So I should really be using exponents that had already been done to 2^63? Or already to 2^64?
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Old 2005-12-29, 21:21   #4
ewmayer
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Factoring an exponent in the megadigit-or-larger range to 2^64 should only take a matter of hours, so I would spin things the other way: all the Pentia are decent at factoring below 2^64, but it's above 2^64 that the P4 really begins to shine. (And that's where most of your CPU time will be spent, since each added factoring bit more than doubles the work required.)
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Old 2005-12-30, 13:54   #5
tom11784
 
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Code:
Ideally for a P4 the #s should be already to 2^64,
but the only such ranges currently available are
48.1-48.2M      64
78.0-78.026M    65
78.026-78.06M   64
79.0-79.1M      64
unless you already have another range reserved
and multiple machines running on them

My P4 is currently moving exponents from 66bit to 68bit
in the 69M range, or will be again once the spring semester
starts up in 2 weeks.
-Tom
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Old 2005-12-30, 14:00   #6
jinydu
 
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Quote:
Originally Posted by ewmayer
since each added factoring bit more than doubles the work required
Shouldn't it be almost exactly double (in fact slightly less, since the density of possible prime factors decreases as you go to higher bit lengths)?

2^1 + 2^2 + 2^3 + ... + 2^n \approx 2^{n+1}

when n >> 1

Last fiddled with by jinydu on 2005-12-30 at 14:03
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Old 2005-12-30, 14:06   #7
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but moving from 2^64 to 2^65 requires using 3 32-bit terms for each number, instead of 2, so the increase from 64 to 65 is more than dbl and then 65 to 66, 66 to 67, etc is roughly double
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