Mersenne Digest        Thursday, March 29 2001        Volume 01 : Number 834




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Date: Tue, 27 Mar 2001 13:23:36 +0100
From: Steve <[EMAIL PROTECTED]>
Subject: Re: Mersenne: atomic clock

On Sun, Mar 25, 2001 at 10:13:40PM -0800, John R Pierce wrote:
 
> which atomic clock would that be?  The US Naval Observatory master clock?
> http://tycho.usno.navy.mil/time.html   probably has contact info.  my local
> server time is referenced to a time server that is a few stratums removed
> from USNO time, my time seems to stay within about 50mS of USNO at all
> times.

Strange that this subject should be brought up here, I noticed something
in my logs a day or so ago where my machine was 4-5 minutes different from
a server that I was using, can't remember weather it was mail, news or 
mersenne.  Every time I connect to the net (probably five to six times per
day), my machine synchronises its self with a time server in Manchester,
North West England.  So I know that my machine isn't wrong. 

- -- 
Cheers
Steve              email mailto:[EMAIL PROTECTED]

%HAV-A-NICEDAY Error not enough coffee  0 pps. 

web http://www.zeropps.uklinux.net/

or  http://start.at/zero-pps

  1:07pm  up 53 days, 13:50,  2 users,  load average: 1.00, 1.00, 1.00
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Date: Tue, 27 Mar 2001 23:09:31 -0800
From: Spike Jones <[EMAIL PROTECTED]>
Subject: Mersenne: LL question

OK I understand how you start with 4, then square and
subtract 2, then take the mod M, then repeat P times.
If the remainder is 0, then M is prime.

But what if the mod M comes out to 1 on one of
the intermediate steps?  Then 1^2 - 2 = -1
Then what?  spike

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Date: Wed, 28 Mar 2001 09:41:22 +0200
From: "Hoogendoorn, Sander" <[EMAIL PROTECTED]>
Subject: RE: Mersenne: LL question

> But what if the mod M comes out to 1 on one of
> the intermediate steps?  Then 1^2 - 2 = -1
> Then what?  spike

Then you will be stuck in a loop, same thing happens when the outcome is
- -2, -1, 0 and 2
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Date: Wed, 28 Mar 2001 10:13:22 +0200 (MET DST)
From: [EMAIL PROTECTED]
Subject: RE: Mersenne: LL question

- - From: "Hoogendoorn, Sander" <[EMAIL PROTECTED]>

- -> But what if the mod M comes out to 1 on one of
- -> the intermediate steps?  Then 1^2 - 2 = -1
- -> Then what?  spike

- -Then you will be stuck in a loop, same thing happens when the outcome is
- - -2, -1, 0 and 2

       This case never occurs.  If M = 2^p - 1 where p is prime and p > 2,
then M == 7 (mod 12).  By quadratic reciprocity, the 
Jacobi symbol (3 over M) is -1, so 3 is not a square modulo M.
Therefore x^2 - 2 == 1 (mod M) has no solution.



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Date: Wed, 28 Mar 2001 02:54:45 -0600
From: "Steve" <[EMAIL PROTECTED]>
Subject: Re: Mersenne: LL question

I don't believe that can ever happen, but if it did then the next
step would just use mod(-1) which is p-1. The mod function
never returns a negative number.

Steve Harris

- -----Original Message-----
From: Spike Jones <[EMAIL PROTECTED]>
To: [EMAIL PROTECTED] <[EMAIL PROTECTED]>
Date: Wednesday, March 28, 2001 1:47 AM
Subject: Mersenne: LL question


>OK I understand how you start with 4, then square and
>subtract 2, then take the mod M, then repeat P times.
>If the remainder is 0, then M is prime.
>
>But what if the mod M comes out to 1 on one of
>the intermediate steps?  Then 1^2 - 2 = -1
>Then what?  spike
>
>_________________________________________________________________________
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------------------------------

Date: Wed, 28 Mar 2001 03:26:55 -0600
From: "Steve" <[EMAIL PROTECTED]>
Subject: Re: Mersenne: LL question

Oops, I meant mod M(-1) which is M-1...

Steve


>I don't believe that can ever happen, but if it did then the next
>step would just use mod(-1) which is p-1. The mod function
>never returns a negative number.
>
>Steve Harris
>
>-----Original Message-----
>From: Spike Jones <[EMAIL PROTECTED]>
>To: [EMAIL PROTECTED] <[EMAIL PROTECTED]>
>Date: Wednesday, March 28, 2001 1:47 AM
>Subject: Mersenne: LL question
>
>
>>OK I understand how you start with 4, then square and
>>subtract 2, then take the mod M, then repeat P times.
>>If the remainder is 0, then M is prime.
>>
>>But what if the mod M comes out to 1 on one of
>>the intermediate steps?  Then 1^2 - 2 = -1
>>Then what?  spike
>>
>>_________________________________________________________________________
>>Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
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>
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------------------------------

Date: Wed, 28 Mar 2001 09:27:06 -0500
From: vincent mooney <[EMAIL PROTECTED]>
Subject: Notification for "RE: Mersenne: LL question"

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------------------------------

Date: Thu, 29 Mar 2001 10:08:59 +1000
From: Steve Phipps <[EMAIL PROTECTED]>
Subject: Mersenne: ECM memory usage

I've been carrying out ECM for some time now on small exponents (under 
40,000) and I'm curious about the amount of memory that it uses.

According to readme.txt, the minimum memory required is 192 times the FFT
size. For the exponents I'm looking at, I suspect that the FFT size is of
order a kilobyte (BTW, can I look the FFT sizes and breakpoints up
somewhere?) and so the minimum memory required is less than 1MB.

However, I use mprime with the available memory set to 24MB and I've
noticed that ECM uses almost all of this. I'd like to know why ECM is 
using so much memory - does it enable it to run faster? And how much
memory would ECM 'like' to use if it was available?

Regards, 
Steve

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------------------------------

Date: Wed, 28 Mar 2001 21:12:06 -0500
From: George Woltman <[EMAIL PROTECTED]>
Subject: Re: Mersenne: ECM memory usage

Hi,

At 10:08 AM 3/29/2001 +1000, Steve Phipps wrote:
>For the exponents I'm looking at, I suspect that the FFT size is of
>order a kilobyte (BTW, can I look the FFT sizes and breakpoints up
>somewhere?) and so the minimum memory required is less than 1MB.

A size 1024 FFT can handle exponents up to 22599.  For rough estimating
purposes, divide the exponent by 21 and round up to the next FFT size.

>However, I use mprime with the available memory set to 24MB and I've
>noticed that ECM uses almost all of this. I'd like to know why ECM is
>using so much memory - does it enable it to run faster?

Yes, more memory can be used to save some operations.

>  And how much
>memory would ECM 'like' to use if it was available?

I've never studied this.  I suspect you quickly reach the point
of diminishing returns.  Try mprime with 8MB and I'll bet you'll
notice little difference.

Regards,
George

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------------------------------

Date: Thu, 29 Mar 2001 21:30:57 +0100
From: Gareth Randall <[EMAIL PROTECTED]>
Subject: Re: Mersenne: Getting new GIMPSers

Pierre Abbat wrote:

> I know that there's an idle process (#0), an init process (#1), and many other
> processes in a computer. Idle time accrues to the idle process. What I don't
> understand is how 2:43 of the idle time was accounted to prime95 and the other
> seventeen seconds to other processes - it should all have been accounted to > the 
>idle process.


Bit late I know, but there are two other reasons for acruing idle time:

1. My Win95 system always started with ~23 seconds of idle time accumulated. On my 
system this implied that the counter started before the video display had even been 
initialised, which makes sense. The kernel is alive and counting long before you see 
your desktop, and it isn't until later that prime95 starts.

2. When prime95 (and mprime) communicates with the server, calculations stop. In fact 
this can't be helped. If a calculation has just finished then clearly no more work can 
be done. If a calculation is in progress it's possible to envisage "update" 
connections running as a parallel thread, but that could lead to race conditions if 
the work completes just as the communication is going on. (That's one guess on why 
it's single threaded, oh and it's a lot simpler to program!)


(Note: My prime95 was replaced by mprime on Linux 6 months ago. If anyone was 
wondering about file format compatibility, I found that the prime95 file format is 
directly compatible with mprime.)

Yours,
 
======= Gareth Randall =======

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End of Mersenne Digest V1 #834
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