Mersenne Digest Thursday, October 28 1999 Volume 01 : Number 653
----------------------------------------------------------------------
Date: Tue, 26 Oct 1999 12:18:46 -0400 (EDT)
From: Darxus <[EMAIL PROTECTED]>
Subject: Re: Mersenne: mprime startup at boot-time
On Thu, 21 Oct 1999, Lars Lindley wrote:
> Hi everyone.
> I have request for help.
> I run mprime on Linux.
> What I want to do is to have mprime automatically start when I boot
> and send the -m output to a tty or something like that. Perhaps tty8?
> Is this possible??
I created an account called "gimps" and installed mprime in it's home
directory. In my startup scripts I have:
/bin/su - gimps -c "/home/gimps/mprime -d -w/mnt/c/gimps" > /dev/tty11 2>&1 &
I actually cheated with my init scripts. I created a file
/etc/init.d/local - and put this line in it, and then ln -s'ed it to
/etc/rc2.d/S20local through /etc/rc5.d/S20local.
On Sun, 24 Oct 1999, poke wrote:
> Sorry, -d, not -m. I'm not that familiar with the command line arguments
> of mprime. My expertise lies in the symantics of Linux.
>
> One way to pipe to a virtual terminal AND to a file is by using a
> combination of a tee and a redirect. Something like:
>
> /usr/src/mprime -d | tee -a /home/GIMPS/tracking.txt > /dev/tty6
There is no reason that starting mprime from /etc/inittab should be any
slower. It's still the same program, being run on the same machine, just
executed by a different program.
I might switch to the inittab method, but the only good reason to do so
would be to handle the gimps client crashing, which I've never seen it do.
And the only reason not to would be the rare occasions when I want to stop
it -- I'd have to remark out the line in /etc/inittab & do "telnit q"
instead of just "killall mprime".
I also do my mail & system log similarly.
I have a line in my /etc/syslog.conf
*.* /dev/tty12
This sends everything that goes through the system log daemon to the 12th
virtual terminal, which I find very convienient. Do a "killall -HUP
syslogd" to get it to reread the config file.
I have a ppp connection. In my /etc/ppp/ip-up script (which is executed
whenever ppp connects), I have the following:
/bin/su - darxus -c /home/darxus/.ip-up &
Which executes the script /home/darxus/.ip-up & as user "darxus" -- this
script contains:
#!/bin/bash
open -v -c 10 ~/pine.open
~/pine.open contains:
#!/bin/bash
export TERM=vt100
exec ssh -t monet 'pine -i'
I'd forgotten this was so complicated... there were reasons for the way I
did it, but I've long forgotten them :)
Anyway, the end result is, whenever I bring up my ppp connection, I
automatically get pine (the mailreader I use) in my 10th virtual terminal,
over a ssh connection.
I've also tried this with BitchX (IRC client), but it think's it's not
attached to a real terminal and runs in dumb terminal mode.
BTW, I've done
chown darxus.darxus /dev/tty10
chown gimps.gims /dev/tty11
I also have 20 virtual terminals (plus the 3 things above + X) & my
console's text resolution set at 132x60 using lilo. I don't use X much.
__________________________________________________________________
PGP fingerprint = 03 5B 9B A0 16 33 91 2F A5 77 BC EE 43 71 98 D4
[EMAIL PROTECTED] / http://www.op.net/~darxus
Find the next largest prime, be famous:
http://www.mersenne.org/prime.htm
_________________________________________________________________
Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
------------------------------
Date: Tue, 26 Oct 1999 19:50:33 +0200
From: Alexander Kruppa <[EMAIL PROTECTED]>
Subject: Mersenne: (2^p-1)/p == 0 (mod 3) ?
Hello all,
a simple Number Theory question. Is always
(2^p-1) / p ,odd prime p, divisible by 3 ?
Then 2^p == 1 (mod 3p) would also hold, can this be used to improve the
efficiency of a Rabin-Miller probable prime test?
Ciao,
Alex.
_________________________________________________________________
Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
------------------------------
Date: Tue, 26 Oct 1999 15:04:30 -0400 (EDT)
From: "Vincent J. Mooney Jr." <[EMAIL PROTECTED]>
Subject: Re: Mersenne: (2^p-1)/p == 0 (mod 3) ?
5 is an odd prime. 2^5 = 32 and minus one, is 31. 31 is not divisible by 3.
At 07:50 PM 10/26/99 +0200, you wrote:
>Hello all,
>
>a simple Number Theory question. Is always
>
>(2^p-1) / p ,odd prime p, divisible by 3 ?
>
>Then 2^p == 1 (mod 3p) would also hold, can this be used to improve the
>efficiency of a Rabin-Miller probable prime test?
>
>Ciao,
> Alex.
>_________________________________________________________________
>Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
>Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
>
_________________________________________________________________
Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
------------------------------
Date: Tue, 26 Oct 1999 15:41:09 -0400
From: Pierre Abbat <[EMAIL PROTECTED]>
Subject: Re: Mersenne: (2^p-1)/p == 0 (mod 3) ?
On Tue, 26 Oct 1999, Alexander Kruppa wrote:
>Hello all,
>
>a simple Number Theory question. Is always
>
>(2^p-1) / p ,odd prime p, divisible by 3 ?
I assume you mean truncated integer division, as real division makes no sense
here.
It isn't if p=3 (7/3=2), but 2^p-2 is divisible by 3. (2^n) mod 3 is 1 if n is
even, 2 if n is odd.
If p=5, 2^p-1=31, and 31/5=6, so it is true for 5.
phma
_________________________________________________________________
Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
------------------------------
Date: Tue, 26 Oct 1999 14:09:45 -0600
From: "Blosser, Jeremy" <[EMAIL PROTECTED]>
Subject: RE: Mersenne: (2^p-1)/p == 0 (mod 3) ?
> -----Original Message-----
> From: Vincent J. Mooney Jr. [mailto:[EMAIL PROTECTED]]
> Sent: Tuesday, October 26, 1999 2:05 PM
> To: [EMAIL PROTECTED]
> Subject: Re: Mersenne: (2^p-1)/p == 0 (mod 3) ?
>
>
> 5 is an odd prime. 2^5 = 32 and minus one, is 31. 31 is not
> divisible by 3.
>
Umm... 31 == 1 (mod 3) which holds to 2^p == 1 (mod 3p) below...
>
> At 07:50 PM 10/26/99 +0200, you wrote:
> >Hello all,
> >
> >a simple Number Theory question. Is always
> >
> >(2^p-1) / p ,odd prime p, divisible by 3 ?
> >
> >Then 2^p == 1 (mod 3p) would also hold, can this be used to
> improve the
> >efficiency of a Rabin-Miller probable prime test?
> >
> >Ciao,
> > Alex.
> >_________________________________________________________________
> >Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
> >Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
> >
>
> _________________________________________________________________
> Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
> Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
>
_________________________________________________________________
Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
------------------------------
Date: Tue, 26 Oct 1999 21:55:23 +0200
From: "Steinar H. Gunderson" <[EMAIL PROTECTED]>
Subject: Mersenne: Re: mprime startup at boot-time
On Tue, Oct 26, 1999 at 12:18:46PM -0400, Darxus wrote:
>I also do my mail & system log similarly.
>
>I have a line in my /etc/syslog.conf
>
> *.* /dev/tty12
But then, any cracker with even three seconds of physical access to your
machine can easily find out what is `allowed' before the system warns.
Danger, danger... (OK, paranoia.)
/* Steinar */
- --
Homepage: http://members.xoom.com/sneeze/
_________________________________________________________________
Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
------------------------------
Date: Tue, 26 Oct 1999 18:13:02 -0400 (EDT)
From: [EMAIL PROTECTED]
Subject: Re: Mersenne: (2^p-1)/p == 0 (mod 3) ?
Hi folks
> a simple Number Theory question. Is always
> (2^p-1) / p ,odd prime p, divisible by 3 ?
Those of you who've already answered this question know what Alex meant to ask was
is (M(p)-1)/p divisible by 3, for odd prime p>3.
ie (2^p-2)/p. The answer is yes, since 2^2=1 mod 3, 2^p=2 mod 3, (2^p-2) is divisible
by 3 and hence the result follows.
> Then 2^p == 1 (mod 3p) would also hold, can this be
> used to improve the efficiency of a Rabin-Miller
> probable prime test?
2^p==2.
It wouldn't make a great deal of difference... you might save a few bits in the length
of the exponent, but that isn't a great saving. As far as I can see though the problem
with this reasoning is that the extra factor 3 in the known factor of M(p)-1 doesn't
occur because p is prime, it occurs because p is odd. So it's not actually giving you
any additional information that could improve your test's effectiveness.
Chris Nash
Lexington KY
UNITED STATES
_________________________________________________________________
Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
------------------------------
Date: Tue, 26 Oct 1999 21:19:24 -0400 (EDT)
From: Darxus <[EMAIL PROTECTED]>
Subject: Re: Mersenne: Re: mprime startup at boot-time
On Tue, 26 Oct 1999, Steinar H. Gunderson wrote:
> On Tue, Oct 26, 1999 at 12:18:46PM -0400, Darxus wrote:
> >I also do my mail & system log similarly.
> >
> >I have a line in my /etc/syslog.conf
> >
> > *.* /dev/tty12
>
> But then, any cracker with even three seconds of physical access to your
> machine can easily find out what is `allowed' before the system warns.
> Danger, danger... (OK, paranoia.)
If such a person can gain physical access to your machine, he owns it
already. There is no defense against physical access.
And if that *were* useful information, it'd be almost as easy to find it
in /var/log.
I'm afraid I just don't agree with you :)
ipchains & tripwire are *much* more effective than that type of paranoia
:)
(don't get me wrong, I incourage paranoia.. but it can be channeled more
usefully)
__________________________________________________________________
PGP fingerprint = 03 5B 9B A0 16 33 91 2F A5 77 BC EE 43 71 98 D4
[EMAIL PROTECTED] / http://www.op.net/~darxus
Find the next largest prime, be famous:
http://www.mersenne.org/prime.htm
_________________________________________________________________
Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
------------------------------
Date: Wed, 27 Oct 1999 10:20:10 +0000
From: "Steinar H . Gunderson" <[EMAIL PROTECTED]>
Subject: Re: Mersenne: Re: mprime startup at boot-time
On Tue, Oct 26, 1999 at 09:19:24PM -0400, Darxus wrote:
>If such a person can gain physical access to your machine, he owns it
>already. There is no defense against physical access.
There's a BIG difference here:
1. Unlimited physical access.
2. Physical access for three seconds or so...
>And if that *were* useful information, it'd be almost as easy to find it
>in /var/log.
If that was public.
>I'm afraid I just don't agree with you :)
As I said, it it paranoia ;-)
/* Steinar */
_________________________________________________________________
Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
------------------------------
Date: Wed, 27 Oct 1999 02:07:54 PDT
From: "Alan Simpson" <[EMAIL PROTECTED]>
Subject: RE: Mersenne: More Schlag
hi,
recently Joth Tupper posted a message with the following
"I seem to recall a non-intuititive theorem about rational approximations to
numbers (this is from c. 1968). If you can approximate a number too closely,
then it is transcendental. S.Lang wrote a book on trancendental numbers and
degrees around 1973 and a precise statement might be there.
Does anyone recall this?"
Yes!
This is the result that Liouville proved to show that Liouville's number
must be transcendental. The result is known, funnily enough as Liouville's
theorem.
It says that if \alpha is an algebraic number of degree n \geq 2, then there
exists a constant c(\alpha)>0 such that
| \alpha - p/q | > c(\alpha)/|q|^{n}
always holds.
Or if you prefer, for any m>n,
| \alpha - p/q | < |q|^{-m} ---- call this (*)
has only finitely many solutions.
At the root of the proof is the extremely "profound" fact that there are no
integers between 0 and 1 (or rather that the integers are discrete):
Let f(x)=x^n+a_{n-1}x^{n-1}+...+a_{1}x+a_{0} be the "minimal polynomial" for
\alpha over the rationals. Let p/q be a rational number.
Then
q^{n}f(p/q) is an integer, it cannot be zero because f(x) is the minimal
polynomial for \alpha, therefore
|q^{n}f(p/q)| \geq 1.
f(x) =(x-\alpha)g(x) where g(x) is some other polynomial, so
|q^{n}(p/q-\alpha)g(p/q)| \geq 1 or
|p/q-\alpha| \geq |1/(g(p/q)q^{n})|.
One can estimate g(p/q) quite easily and use this estimate to get the value
for c(\alpha) mentioned above.
You prove that the Liouville number (or any similar type number) is
transcendental from this result is by noting that the rational
approximations are getting more and more accurate at each step (you are
basically adding lots more zeros each time). But Liouville's theorem forbids
this, if the number is algebraic. Thus the number is transcendental.
It can be formulated somewhat different yet again (in terms of "heights" of
numbers) and it is an extremely important result that is used almost
everywhere in diophantine analysis and transcendental number theory (so
ultimately most proofs in these areas depend upon cleverly showing that some
"deep" result depends on the fact that there are no integers between 0 and 1
:-)).
Liouville proved his theorem around 1840. It has since been improved.
Thue (1909) showed that you can weaken the condition on m in (*) above to
m>n/2
Siegel (1922) shows that m>2\sqrt{n} suffices
Dyson and Gelfond independently showed in the 40's that m>\sqrt{2n} suffices
and finally (sort of) Roth in 1954 showed that m>2 suffices.
Roth won a Fields medal for this (and some other) work.
Does that answer your questions?
______________________________________________________
Get Your Private, Free Email at http://www.hotmail.com
_________________________________________________________________
Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
------------------------------
Date: Wed, 27 Oct 1999 12:56:05 +0200
From: Alexander Kruppa <[EMAIL PROTECTED]>
Subject: Re: Mersenne: (2^p-1)/p == 0 (mod 3) ?
"Vincent J. Mooney Jr." wrote:
>
> 5 is an odd prime. 2^5 = 32 and minus one, is 31. 31 is not divisible by 3.
>
>
> At 07:50 PM 10/26/99 +0200, you wrote:
> >Hello all,
> >
> >a simple Number Theory question. Is always
> >
> >(2^p-1) / p ,odd prime p, divisible by 3 ?
> >
Well, I managed to mess up the equation so badly that noone could
possibly make any sense of it. It should have been: (2^(p-1) -1) /p, and
by now I figured out why these numbers are divisible by 3 for odd (and
not just for prime) p>3 (if you kick out the /p, its for p>2).
a^k == 1 (mod a^k -1)
a^n == a^(n-k) (mod a^k -1), n>=k
a^(nk) == 1 (mod a^k -1)
(or a^(nk)-1 == 0 (mod a^k-1) ,which shows that Mersenne primes must
have prime exponents)
Since 2^2-1=3, 2^(2k)-1 == 0 (mod 3) for all integer k and, in the above
formula,
2^(p-1)-1 == 0 (mod 3) for odd p.
Thats what happens when you ask a question before thinking it to the
end, sorry for the confusion.
Ciao,
Alex.
_________________________________________________________________
Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
------------------------------
Date: Wed, 27 Oct 1999 01:12:37 +0200
From: Remigiusz Urbaniak <[EMAIL PROTECTED]>
Subject: Mersenne: Prime Theory
Hi all,
I am quite new for the prime numbers subject, and when go deeper in
theory of them I encounter a serious problem - all docs and faqs are in
"mathematical english", but "mathematical english" is not my favorite
one :). Please let me know, if there is somewhere in the "netland" such
a documentation in pure mathematic form ( I mean math equations instead
english poetry ) in TeX, pdf, ps or similar "graphic" format, which I
can understand...( I hope :))) ).
Thanks in advance/Remi.
_________________________________________________________________
Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
------------------------------
Date: Wed, 27 Oct 1999 07:03:27 -0700
From: "Joth Tupper" <[EMAIL PROTECTED]>
Subject: Re: Mersenne: More Schlag
Thanks! Good Stuff!
Joth
PS: There is no Nobel Prize in math. This is perhaps unfortunate, as the
winner of a Nobel Prize
seems to get over $500,000 (US) in cash and generally a new building at the
university of choice.
The closest math counterpart is the Fields Medal which does include a prize:
$15,000 Canadian
(or about $10,000 US) and sometimes a corner office.
Ah, the peaceful uses of dynamite.
- ----- Original Message -----
From: Alan Simpson <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
Sent: Wednesday, October 27, 1999 2:07 AM
Subject: RE: Mersenne: More Schlag
>
> hi,
>
> recently Joth Tupper posted a message with the following
>
> "I seem to recall a non-intuititive theorem about rational approximations
to
> numbers (this is from c. 1968). If you can approximate a number too
closely,
> then it is transcendental. S.Lang wrote a book on trancendental numbers
and
> degrees around 1973 and a precise statement might be there.
>
> Does anyone recall this?"
>
> Yes!
>
> This is the result that Liouville proved to show that Liouville's number
> must be transcendental. The result is known, funnily enough as Liouville's
> theorem.
>
> It says that if \alpha is an algebraic number of degree n \geq 2, then
there
> exists a constant c(\alpha)>0 such that
>
> | \alpha - p/q | > c(\alpha)/|q|^{n}
>
> always holds.
>
> Or if you prefer, for any m>n,
> | \alpha - p/q | < |q|^{-m} ---- call this (*)
>
> has only finitely many solutions.
>
> At the root of the proof is the extremely "profound" fact that there are
no
> integers between 0 and 1 (or rather that the integers are discrete):
>
> Let f(x)=x^n+a_{n-1}x^{n-1}+...+a_{1}x+a_{0} be the "minimal polynomial"
for
> \alpha over the rationals. Let p/q be a rational number.
> Then
> q^{n}f(p/q) is an integer, it cannot be zero because f(x) is the minimal
> polynomial for \alpha, therefore
> |q^{n}f(p/q)| \geq 1.
>
> f(x) =(x-\alpha)g(x) where g(x) is some other polynomial, so
> |q^{n}(p/q-\alpha)g(p/q)| \geq 1 or
> |p/q-\alpha| \geq |1/(g(p/q)q^{n})|.
> One can estimate g(p/q) quite easily and use this estimate to get the
value
> for c(\alpha) mentioned above.
>
> You prove that the Liouville number (or any similar type number) is
> transcendental from this result is by noting that the rational
> approximations are getting more and more accurate at each step (you are
> basically adding lots more zeros each time). But Liouville's theorem
forbids
> this, if the number is algebraic. Thus the number is transcendental.
>
> It can be formulated somewhat different yet again (in terms of "heights"
of
> numbers) and it is an extremely important result that is used almost
> everywhere in diophantine analysis and transcendental number theory (so
> ultimately most proofs in these areas depend upon cleverly showing that
some
> "deep" result depends on the fact that there are no integers between 0 and
1
> :-)).
>
> Liouville proved his theorem around 1840. It has since been improved.
>
> Thue (1909) showed that you can weaken the condition on m in (*) above to
> m>n/2
> Siegel (1922) shows that m>2\sqrt{n} suffices
> Dyson and Gelfond independently showed in the 40's that m>\sqrt{2n}
suffices
> and finally (sort of) Roth in 1954 showed that m>2 suffices.
>
> Roth won a Fields medal for this (and some other) work.
>
> Does that answer your questions?
>
> ______________________________________________________
> Get Your Private, Free Email at http://www.hotmail.com
> _________________________________________________________________
> Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
> Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
>
_________________________________________________________________
Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
------------------------------
Date: Wed, 27 Oct 1999 17:23:11 -0700
From: Eric Hahn <[EMAIL PROTECTED]>
Subject: Mersenne: Trial-factorers
I'm looking for program(s) capable of trial-factoring
prime exponent Mersenne numbers (using 2kp+1) meeting
the following requirements:
1) Capable of trial-factoring any exponent > 1
(at least to some considerably large number,
say 1 trillion?)
As I recall, Brian [Beesley] mentioned something once
about having a program that could test an exponent
of an arbitrary size... Brian??
2) Capable of testing a factor of any size.
(even over the 2^76 limit of Prime95).
I just know somebody is going to have to mention the
time involved in testing factors of such a large size.
Let me just say, I realize *exactly* how much time
would be required...
3) Capable of trial-factoring a range of k's.
(example: from k=1000 to k=2500)
It would be best if all three of the requirements
could be fulfilled by a single program...
Can anybody be of some help???
Eric Hahn
_________________________________________________________________
Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
------------------------------
Date: Wed, 27 Oct 1999 21:13:44 -0400
From: Albert Garrido <[EMAIL PROTECTED]>
Subject: Mersenne: Questions about prime.ini syntax, and hardware advice.
- -----BEGIN PGP SIGNED MESSAGE-----
Hash: SHA1
Hi.
I'm currently trying to configure the Time command, as listed in the
docs, to
get the prime95 client to function as follows.
User ID=XYZABC
Time=1-5/18:00-0:00,1-5/0:00-08:00,6-7/0:00-24:00
(reset of Prime.ini)
Anyhow, it doesn't seem to work, and I tried a few other syntax forms,
but
what would usually happen is that either the app runs at 6:00PM and then
goes
to sleep at 6:01. I inserted the line Priority=1 after the Time line,
and it
doesn't do much. I couldn't find an explanation of the time variables
on the
FAQ, I'm hoping someone's been down this road before.
I don't know if the undocumented commands are being supported, or it's
been
discussed already. I'm using the
prime.ini=yourfilename and local.ini=yourfilename.
Anyhow. If I try to start the Prime95 Program, with the renamed.ini it
will
give me the user information dialog box. And will prompt to download
new
exponents (after doing a self test). if I leave the prime.ini and
local.ini
files without changing. It seems to work, except for an odd
inconsistency
that might be due to me adding all of these commands to my .ini file
LockUserInfo=1 (Unsure if this is an Undocumented as it appears on the
FAQ
now)
PercentPrecision=4
prime.ini=newfilename
local.ini=newfilename
worktodo.ini=newfilename
prime.log=newfileaname
prime.spl=newfilename
resultst.txt=newfilename
workingdir=C:\winnt\blahblah\blah
I know it's slightly off-topic, but I am in need of advice as far as how
well
Prime will run on this machine. I currently have an order in place for
an
Kryotech Cool Athlon� 900. For anyone who hasn't seen it.
http://www.kryotech.com/maintext900.asp
Anyhow, it's an Athlon that's running at 900Mhz, besides the novelty
value of
that. How fast would something like this get through an average LL
test?
Something in the 10202624 range? I won't know until it it arrives, and
this
is somewhat assuming that Kryotech doesn't announce the Super G product
until
then, and/or they ship on schedule.
Anyhow, any comments on the above are welcome.
- -----BEGIN PGP SIGNATURE-----
Version: PGPfreeware 6.5.1 for non-commercial use <http://www.pgp.com>
iQA/AwUBOBejG6svR6n7TjeiEQL7OwCfc3tPSrChhK6A8KHGjJ8To+nXFAwAnRBh
E9arZvjP3zwwyoPtdb9XDUZz
=sEYB
- -----END PGP SIGNATURE-----
_________________________________________________________________
Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
------------------------------
Date: Wed, 27 Oct 1999 22:29:59 -0700
From: Eric Hahn <[EMAIL PROTECTED]>
Subject: Re: Mersenne: Questions about prime.ini syntax, and hardware advice.
Albert Garrido wrote:
>I'm currently trying to configure the Time command, as
>listed in the docs, to get the prime95 client to function
>as follows.
>
>User ID=XYZABC
>Time=1-5/18:00-0:00,1-5/0:00-08:00,6-7/0:00-24:00
>(reset of Prime.ini)
If you're trying to run from Midnight to 6AM and 6PM to
Midnight on Mon-Fri, the time line should read as follows:
Time=1-5/0:00-08:00,1-5/18:00-24:00,6-7/0:00-24:00
Note Midnight is 0:00 when starting the day and 24:00 when
ending it. It's exactly like what you have for Sat-Sun.
It's also best if you put them in order by time (you'll
notice I switched the two Mon-Fri times around). There's
not usually any problems, but glitches do happen.
>I don't know if the undocumented commands are being
>supported, or it's been discussed already. I'm using the
>prime.ini=yourfilename and local.ini=yourfilename.
Perhaps somebody else can help you here. They look okay,
but I have never used them myself (no need!)
>I know it's slightly off-topic, but I am in need of advice
>as far as how well Prime will run on this machine. I
>currently have an order in place for an Kryotech Cool
>Athlon� 900.
>[...]
>How fast would something like this get through an average LL
>test? Something in the 10202624 range?
>[...]
Based on the known information about LL-testing times,
Prime95 should be able to LL-test a number such as
10,202,623 in 2 to 2.5 weeks. A 10M digit exponent
could be tested in about 7-8 months....
_________________________________________________________________
Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
------------------------------
Date: Thu, 28 Oct 1999 11:23:30 +0200
From: Alexander Kruppa <[EMAIL PROTECTED]>
Subject: Re: Mersenne: Trial-factorers
Eric Hahn wrote:
>
> I'm looking for program(s) capable of trial-factoring
> prime exponent Mersenne numbers (using 2kp+1) meeting
> the following requirements:
>
> [factors and exponents of arbitrary size]
>
mersfacgmp from the Will Edgingtons mers package uses gmp's mpz type for
factors (thus the size of the factor is only limited by your available
memory) and long ints for the exponent. If you have a 64 bit cpu, that
should allow exponentes up to 18446744 trillion.
> I just know somebody is going to have to mention the
> time involved in testing factors of such a large size.
> Let me just say, I realize *exactly* how much time
> would be required...
Ok, but know that gmp is not particulary efficient on "small" number of
only a few dozend decimals, the time required might be what you expect
and then some.
> 3) Capable of trial-factoring a range of k's.
> (example: from k=1000 to k=2500)
Hmmm, I dont think mersfacgmp can process this directly, but it would be
easy to compute the proper factoring limits with a shell script that
calls bc.
The mers package can be found on Will's home page,
http://www.garlic.com/~wedgingt/mersenne.html
Ciao,
Alex.
_________________________________________________________________
Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
------------------------------
Date: Thu, 28 Oct 1999 11:37:12 +0200
From: Alexander Kruppa <[EMAIL PROTECTED]>
Subject: Mersenne: LL and Pollar-rho in one?
Hi,
the Lucas-Lehmer iteration
L_0 = 4
L_{n+1} = L_n ^2 -2
looks suspiciously like an iteration used in Pollard-Rho:
x_0 = a
x_{n+1} = x_n ^2 +b
Can this be used to do a LL test and Pollard-rho factoring attempt at
once?
I remember faintly having read something that b=-2 is not a good choice
for Pollard-rho, but I don't remember any explanation why.. would it
work here?
Ciao,
Alex.
_________________________________________________________________
Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
------------------------------
Date: Thu, 28 Oct 1999 08:08:53 -0700
From: Will Edgington <[EMAIL PROTECTED]>
Subject: Mersenne: Trial-factorers
Eric Hahn writes:
I'm looking for program(s) capable of trial-factoring
prime exponent Mersenne numbers (using 2kp+1) meeting
the following requirements:
1) Capable of trial-factoring any exponent > 1
(at least to some considerably large number,
say 1 trillion?)
If your C compiler supports a 64 bit integer type, the mers package
factorers mersfacgmp and mersfaclip can use that for the exponent,
allowing exponents to about 2^64/24 (about 768 quadrillion, ~7e17).
The '/24' can be reduced to '/2' by using a lower number of sieve
passes. If you don't have a 64 bit integer type, then you're limited
to 2^32/24 or so (about 178 million). See setup.h for the #define's
of BIG_LONG.
As I recall, Brian [Beesley] mentioned something once
about having a program that could test an exponent
of an arbitrary size... Brian??
The mersfac* code could be modified to allow arbitrary exponents
without a lot of work, but it wouldn't be trivial to do either. I do
intend on doing it someday, however, to support the factoring of the
M(M(p))'s, as the sieving of trial factors in mmtrial is not as good
as that in mersfac*.
2) Capable of testing a factor of any size.
(even over the 2^76 limit of Prime95).
Mersfacgmp and mersfaclip use the FSF's libgmp and A. Lenstra's
freeLIP libraries, respectively, for the trial factors.
3) Capable of trial-factoring a range of k's.
(example: from k=1000 to k=2500)
As Alexander Kruppa noted, this is not supported directly, but could
be done fairly easily with bc. And it could be added to mersfac*
directly without a lot of effort; see rw.c where it reads the command
line and the input file(s), especially the -m flag. The -a flag is
probably also of interest; without it, mersfac* will stop when it
finds a factor rather than looking for all the factors in a range.
The simplest way to do it with bc would be to create 'G' (gap) lines
like:
M( <exp> )G: <start> <stop>
where '<exp>' is the exponent and '<start>' and '<stop>' are the first
and last trial factors to test.
In any case, part of the output will be 'J' (join) lines, that show
exactly what range of trial factors were actually tested; please
include them in any output of mersfac* that you send to me.
Will
http://www.garlic.com/~wedgingt/mersenne.html
mers.tar.gz
mers.tgz
mers.zip
`
_________________________________________________________________
Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
------------------------------
Date: Thu, 28 Oct 1999 17:12:46 +0200
From: Lars Lindley <[EMAIL PROTECTED]>
Subject: Mersenne: Quiet
I think it's about time we find a new prime...
The list is so quiet now.
:)
/Lars
_________________________________________________________________
Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
------------------------------
Date: Thu, 28 Oct 1999 17:42:24 +0200
From: Alexander Kruppa <[EMAIL PROTECTED]>
Subject: Re: Mersenne: LL and Pollar-rho in one?
Jason Stratos Papadopoulos wrote:
>
> On Thu, 28 Oct 1999, Alexander Kruppa wrote:
>
> > Hi,
> >
> > the Lucas-Lehmer iteration [...]
> > looks suspiciously like an iteration used in Pollard-Rho: [...]
> > Can this be used to do a LL test and Pollard-rho factoring attempt at
> > once?
>
> Except that every once in a while you'd have to perform a GCD, and
> a GCD on million-digit size numbers takes about a day (no kidding!
> There seems to be no fast way to do one!)
Thats true (btw, I keep hearing of a n*log(n) gcd - how long would that
take then? Where can I read up on it - Knuth vol.2 ?) but doing the
Pollard-Rho along a LL test would not be particularly efficient,
anyways. For every 2 LL iterations, you'd have to do another one for the
cycle find algorithm and a multiply to store up any factor you find -
thats 9 transforms instead of 4, so your test time will more than
double. And Pollard-Rho is probably not very well suited for finding
Mersenne factors as it cannot exploit the fact that these factor are 1
mod (2p) as P-1 can.
I'm mostly asking for curiosity, whether the LL iteration really makes a
proper Pollard-Rho iteration, especially with the -2.
Ciao,
Alex.
_________________________________________________________________
Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
------------------------------
Date: Thu, 28 Oct 1999 20:54:16 +0200 (MET DST)
From: [EMAIL PROTECTED]
Subject: Re: Mersenne: LL and Pollard-rho in one?
Alexander Kruppa <[EMAIL PROTECTED]> observes:
> Hi,
>
> the Lucas-Lehmer iteration
>
> L_0 = 4
> L_{n+1} = L_n ^2 -2
>
> looks suspiciously like an iteration used in Pollard-Rho:
>
> x_0 = a
> x_{n+1} = x_n ^2 +b
>
> Can this be used to do a LL test and Pollard-rho factoring attempt at
> once?
> I remember faintly having read something that b=-2 is not a good choice
> for Pollard-rho, but I don't remember any explanation why.. would it
> work here?
If L_0 = alpha + 1/alpha (where alpha = 2 + sqrt(3) is found
by solving a quadratic) then
L_n = alpha^(2^n) + alpha^(-2^n)
L_m - L_n = alpha^(2^m) + alpha^(-2^m) - alpha^(2^n) - alpha^(-2^n)
= (alpha^(2^m) - alpha^(2^n)) + (1 - alpha^(-2^m - 2^n))
If p is a factor of the number we are trying to factor, then
the order of alpha will divide one of p+1 and p-1
(depending upon whether alpha is in the base field GF(p)
or the extension field GF(p^2)).
The size of this order will be O(p).
m and n will usually need to have size O(p) before we achieve
alpha^(2^m) == alpha^(+- 2^n) (mod p)
(which is the same as 2^m == +- 2^n (mod (order of alpha)) ).
When b <> 0, -2, we know no such formula for the iterates of Pollard
Rho. x -> x^2 + b (mod p) seems to behave like a pseudorandom function
modulo p, and Pollard Rho takes O(sqrt(p)) iterations rather than O(p).
For Lucas-Lehmer, when p = 2^q - 1 is a Mersenne prime, p+1 is
divisible by a very high :-) power of 2. The L_n values collapse to
0, -2, 2, 2, ... as alpha^(2^n) becomes sqrt(-1), -1, 1 in GF(p^2).
_________________________________________________________________
Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
------------------------------
End of Mersenne Digest V1 #653
******************************
