Mersenne: Re: Timing(?) errors

1999-09-21 Thread Steinar H. Gunderson

On Mon, Sep 20, 1999 at 05:45:17PM -0500, Willmore, David wrote:
Since it's a cache reading problem there's no real way to 'flush' it.
Normally, that means to write back dirty data to whatever backing store
exists, not 'invalidate everything'.  Even if you did, it would't solve the
problem.

How does swap space come into this? Linux isn't forced to swap the data
in exactly where it used to be, is it?

[removed excellent explanation on what's going on]

Thanks. :-)

Yes, it would probably be easier in Linux, but it might not do you any good.

Perhaps I could do a manual restart if it was a problem? (I can thing of
several crazy ways to do this... Perhaps fill the all the buffers with 
some random number, and find them in /proc/kcore? ;-) )

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Re: Mersenne: Iteration Times (was: GIMPS client output)

1999-09-21 Thread James Escamilla

Wouldn't the run time at 4.231 be about 10 years?

--- Eric Hahn [EMAIL PROTECTED] wrote:
 
 Okay, okay... obviously a lot of people were awake
 sigh
 (you can stop flooding me with emails!!)
 
 In a previous message I wrote:
 
 P.S. At the 79.3M range, you'll probably not want
 to set it
 at 100 iterations...  Per iteration time on 266MHz
 PII with
 64MB RAM is 58.781 seconds!!!  (Yes, it's true, but
 I'm also
 just checking to see if anybody's awake :))
 
 I went back to the exponent in question and ran
 another test.
 
 There are a couple of notes here:
   1) This originally was done for a particular test
 in QA.
   2) George didn't have the new timings up at the
 time.
   3) I thought it was high myself, but what did I
 know?
 
 What I found was:
   1) I obviously had something running in the
 background
  I was not aware of.
   2) The actual time dropped to 4.231 sec/iter
   3) Amazingly, there didn't appear to be much HDD
 paging
  happening except went you hit 'STOP'!
 
 BTW, for those of you who don't know (or actually
 asked),
 these exponents use 4096K FFT runlengths, and 16M
 save
 files...
 
 Eric Hahn
 
 

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RE: Mersenne: Re: Timing(?) errors

1999-09-21 Thread Willmore, David

 On Mon, Sep 20, 1999 at 05:45:17PM -0500, Willmore, David wrote:
 Since it's a cache reading problem there's no real way to 'flush' it.
 Normally, that means to write back dirty data to whatever backing store
 exists, not 'invalidate everything'.  Even if you did, it would't solve
 the
 problem.
 
 How does swap space come into this? Linux isn't forced to swap the data
 in exactly where it used to be, is it?
 
Correct, it does not.  Normally, though, when you're swapping, proper L2
cache coloring is the least of your performance problems.

 Yes, it would probably be easier in Linux, but it might not do you any
 good.
 
 Perhaps I could do a manual restart if it was a problem? (I can thing of
 several crazy ways to do this... Perhaps fill the all the buffers with 
 some random number, and find them in /proc/kcore? ;-) )
 
A syscall that walked the page tables to find the translation for an address
would probably the easiest.

Cheers,
David
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Mersenne: FDIV Pentium error

1999-09-21 Thread Floris Looyesteyn

I was wondering if Prime95 is affected by the Pentium
FDIV bug. (or some name like that).
I ask this because now i'm also using it on my laptop
(great work george!) and when i installed linux some time
ago it said the processor had this bug.

Floris Looyesteyn


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Re: Mersenne: FDIV Pentium error

1999-09-21 Thread Lars Lindley

 I was wondering if Prime95 is affected by the Pentium
 FDIV bug. (or some name like that).
 I ask this because now i'm also using it on my laptop
 (great work george!) and when i installed linux some time
 ago it said the processor had this bug.

It should not be a problem because Linux recognizes the bug and uses
a workaround.

Regards
/Lars

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Mersenne: Re: Re: Timing(?) errors

1999-09-21 Thread Steinar H. Gunderson

On Tue, Sep 21, 1999 at 02:03:54PM -0500, Willmore, David wrote:
Correct, it does not.  Normally, though, when you're swapping, proper L2
cache coloring is the least of your performance problems.

Yes, but if you _force_ swap-out-swap-in, like ReCache does?

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Mersenne: Re: FDIV Pentium error

1999-09-21 Thread Steinar H. Gunderson

On Tue, Sep 21, 1999 at 09:25:02PM +0200, Floris Looyesteyn wrote:
I was wondering if Prime95 is affected by the Pentium
FDIV bug. (or some name like that).

I've run it with on a P60 (with the FDIV bug) for 2-3 years now
(at least pre-PrimeNet), and it has never been a problem. Remember
that the bug influences FDIV only (which is very slow -- and thus
George doesn't use it that much ;-) ), and not to a very great
extent either.

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RE: Mersenne: Re: Re: Timing(?) errors

1999-09-21 Thread Willmore, David

Random chance.  I wouldn't count on it.

 -Original Message-
 From: Steinar H. Gunderson [SMTP:[EMAIL PROTECTED]]
 Sent: Tuesday, September 21, 1999 4:12 PM
 To:   [EMAIL PROTECTED]
 Subject:  Mersenne: Re: Re: Timing(?) errors
 
 On Tue, Sep 21, 1999 at 02:03:54PM -0500, Willmore, David wrote:
 Correct, it does not.  Normally, though, when you're swapping, proper L2
 cache coloring is the least of your performance problems.
 
 Yes, but if you _force_ swap-out-swap-in, like ReCache does?
 
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Re: Mersenne: Prime95 v19 oops...

1999-09-21 Thread George Woltman

Hi,

At 10:09 PM 9/20/99 +0300, Jukka Santala wrote:
Something I forgot from earlier playing, the manual factoring savefiles
on Prime95 v19 at least don't work out too well especially on dual-CPU
machines... Since these savefiles will always be named "p000"
regardless of the -A parameter and exponent to test ;)

This is a v18 bug that I declined to fix.  Maybe I should delete that
Advanced/Factor menu choice :)

To work around this create one directory for each CPU.

This is the only known flaw in dual-CPU environments.

Regards,
George

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Mersenne: Factoring

1999-09-21 Thread Eric Hahn


  Does anybody know if there is an exponent where the
factor is, or know whether there is a proof on whether
a factor can (or can't) be, a root??  A square??

To clarify this: 
We know that any factor of 2^p-1 is in the form 2kp+1.
Letting x =2, 
  Can (2kp+1)^x = 2^p-1 ??
  Can (2kp+1)^x * (2kp+1) ... = 2^p-1 ??

Eric Hahn


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Re: Mersenne: v19 manual workdodo.ini error.

1999-09-21 Thread George Woltman

Hi,

At 02:42 PM 9/18/99 +0200, Lars Lindley wrote:
I discovered a lost exponent in the team-report and thought I would
reassign that exponent for myself.
I manually edited the worktodo.ini by adding the row
DoubleCheck=3393469,61 on the first line.
I thought that prime95 would put the exponent I was working on on
hold to do the doublecheck first as v18.1 would have.
To my surprise it continued with the old exponent.

I've tracked down and fixed this bug that Lars and Rick reported.
Look for a new v19 beta in the next day or two (I'll announce it to
the mailing list).

Regards,
George

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Mersenne Digest V1 #629

1999-09-21 Thread Mersenne Digest


Mersenne Digest  Tuesday, September 21 1999  Volume 01 : Number 629




--

Date: Sun, 19 Sep 1999 21:36:01 -0400
From: Jeff Woods [EMAIL PROTECTED]
Subject: Re: Mersenne: M(M(127)) and other M(M(p))

At 08:51 PM 9/19/99 -0400, you wrote:

prime, unless we find a factor.  Interestingly enough, when we find the next
Mersenne prime, searching for a factor of M(M(p)) might allow us to find an
even bigger prime.  If for example, 6*M(p)+1 divides M(M(p)), then it must
be prime!

Which one must be prime?   6*M(p)+1, or M(M(p))?

And why?   Enquiring minds, and all   Thanks!

Wait, that might just be the reason to search!  Will only searched up to
k=4 for M(M(6972593)), but if 2*k*M(p)+1 divides M(M(p)), then you've just
beaten the world record!  Non-Mersenne's might once again grace the top
10 list!

An interesting concept -- what sort of time factor would it take to prove 
such a thing with an average computer?   
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Date: Sun, 19 Sep 1999 21:33:53 -0400
From: "Chris Nash" [EMAIL PROTECTED]
Subject: Re: Mersenne: M(M(127)) and other M(M(p))

 even bigger prime.  If for example, 6*M(p)+1 divides M(M(p)), then it must
 be prime!

Before anybody gets overexcited at the last posting...

It is TRUE that if 2k.M(p)+1 divides M(M(p)), M(p) is prime, and k2M(p)+2,
then 2k.M(p)+1 is prime.

However, unless I'm mistaken, non-divisibility does not prove compositeness.
You could walk past a prime (in fact, you'd expect to walk past several) and
you'd never know...

Chris


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Date: Sun, 19 Sep 1999 22:27:59 -0400 (EDT)
From: Darxus [EMAIL PROTECTED]
Subject: Mersenne: GIMPS client output

Iteration: 164000 / 8410531 [1%].  Clocks: 115665753 = 0.496 sec.

Might be nice to display the percentage out to an accuracy that changes
every hundred iterations.  Hmm, looks like that's an integer of the
percentage, not rounded.  Guess it doesn't matter.  For the one I'm
working on it looks like 3 decimal places would be needed to see a change
every 100 iterations.

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Date: Sun, 19 Sep 1999 22:59:37 -0400
From: "Rick Pali" [EMAIL PROTECTED]
Subject: RE: Mersenne: GIMPS client output

From: Darxus

 Iteration: 164000 / 8410531 [1%].  Clocks: 115665753 = 0.496 sec.

 Might be nice to display the percentage out to an accuracy that
 changes every hundred iterations.

If you're using version 19, add "PercentPrecision=3" to the prime.ini file.
If you want more than three decimal places, change the number...I believe
it'll go up to six. I believe that four should be sufficient as even
M33219281 changes every three or four hundred iterations with three places
displayed.

Rick.
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Date: Sun, 19 Sep 1999 23:05:39 -0400 (EDT)
From: Lucas Wiman  [EMAIL PROTECTED]
Subject: Re: Mersenne: GIMPS client output

 Might be nice to display the percentage out to an accuracy that changes
 every hundred iterations.  Hmm, looks like that's an integer of the
 percentage, not rounded.  Guess it doesn't matter.  For the one I'm
 working on it looks like 3 decimal places would be needed to see a change
 every 100 iterations.

This is changed in V19 (currently in Beta).  I believe (George correct
me if I'm wrong) that you can specify it up to 6 decimal places.

- -Lucas
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Date: Sun, 19 Sep 1999 23:24:41 -0400
From: "Chris Nash" [EMAIL PROTECTED]
Subject: Re: Mersenne: M(M(127)) and other M(M(p))

Hi Jeff

 prime, unless we find a factor.  Interestingly enough, when we find the
next
 Mersenne prime, searching for a factor of 

Re: Mersenne: Factoring

1999-09-21 Thread Jud McCranie

At 04:27 PM 9/21/99 -0700, Eric Hahn wrote:
We know that any factor of 2^p-1 is in the form 2kp+1.
Letting x =2,
   Can (2kp+1)^x = 2^p-1 ??
   Can (2kp+1)^x * (2kp+1) ... = 2^p-1 ??


No known factors of Mersenne numbers have x1, but it hasn't been proven 
that it is impossible.


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| |
| Programming Achieved with Structure, Clarity, And Logic |
+-+


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Re: Mersenne: Iteration Times (was: GIMPS client output)

1999-09-21 Thread Eric Hahn

At 09:47 AM 9/21/1999 -0700, James Escamilla wrote:
Wouldn't the run time at 4.231 be about 10 years?

Yes, for that particular exponent (79,299,959), it would
take approx. 10 yrs. and 231 days to test.

That's assuming 4 items:
  1) A P2 266MHz PC was being used the entire time.
  2) The PC was being used exclusively to test the exponent 24/7.
  3) The 4.231 sec/iter is constant (which it isn't!)
  4) A factor isn't found (below 2^62 is unsuccessful at least!)

Eric Hahn
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Re: Mersenne: FDIV Pentium error

1999-09-21 Thread poke


All programs are affected by the FDIV bug. It is a bug in the design of
the microprocessor. Linus has a way around it. The brain dead morons at
Microsoft have to make everyone else wait for a solution (That has yet to
come AFIK). Linux on the other hand usually has problems solved in a
matter of hours. 

-Chuck



On Tue, 21 Sep 1999, Floris Looyesteyn wrote:

 I was wondering if Prime95 is affected by the Pentium
 FDIV bug. (or some name like that).
 I ask this because now i'm also using it on my laptop
 (great work george!) and when i installed linux some time
 ago it said the processor had this bug.
 
 Floris Looyesteyn
 
 
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Re: Mersenne: M(M(127)) and other M(M(p))

1999-09-21 Thread Chris Nash

Hi folks

 Wait, that might just be the reason to search!  Will only searched up to
 k=4 for M(M(6972593)), but if 2*k*M(p)+1 divides M(M(p)), then you've just
 beaten the world record!  Non-Mersenne's might once again grace the top
 10 list!

I really hope that neither Will Edgington (with M(M(6972593))) nor Chip
Kerchner (with M(M(1398269))) dedicated any computer time whatsoever to
search for factors 2*k*M(p)+1 up to k=4.

As Will's page

http://www.garlic.com/~wedgingt/MMPstats.txt

points out, since M(p)=1 mod 3, k cannot be 1 mod 3. Also, since M(p)=-1 mod
8 for odd p=3, k must be 0 or 1 mod 4 (otherwise 2 is not a quadratic
residue of this supposed factor, the 8x+-1 condition).

It follows then the first possible factor of M(M(p)) has k=5.

Chris Nash
Lexington KY
UNITED STATES


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