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Old 2007-05-30, 17:18   #12
bdodson
 
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Quote:
Originally Posted by R.D. Silverman View Post
The remaining numbers under 768 bits are:

5,317-, 323-
6,283-
6,284+, 292+
7,263-, 269-, 271-
7,268+
The "current" info in my previous post seems to have been a few
months out-of-date, at least relative to hard-copy (or perhaps a
needed cache clearing). Reflecting post-NFSNET base-5 factors,
5, 317- is now 7th on the Most wanted list and 7, 263- is 8th.
The others are also Wanted, except that 6, 283- has been reserved ---
that would be for NFSNET's next.

All of these have had ecm pretests to 2*t50, and checking for factors
by snfs is a lot cheaper than checking ecm pretests on gnfs candidates
(three of these nine are over 200-digits, the smallest ones are
161- 168- and 173-digits; missing a p55 may be tolerable for a
difficulty 220-229 snfs, perhaps for .66*225 = 150 gnfs; but we're
accumulating a collection larger gnfs candidates. Of course, the new
state-of-the-art is 53.2% chance of missing a p70, which translates
into c. t68; or as reported elsewhere, over 3*t65 ... that's for
.66*312 = c208? Translation seems a bit off, believe I heard a
rough equivalent to 700-bit gnfs, which would be 210.7-digits;
close enough ...) OK, three of these nine give a Much cheaper
snfs model for state-of-the-art gnfs difficulty; while even the smaller
composites are measurably above current Smaller-but-Needed size,
if gnfs is the quicker choice (over snfs).

Admittedly, a rather bizarre reason; unlikely to sustain anyone through
several months of computation. Pending continued progress on getting
minimal-stats back up, tracking a larger scale sieving collaboration is
probably a more viable objective. Wonder how many of the c. 800
Cunninghams have snfs difficulty under 300, and how many have already
been sufficiently factored to have gnfs be the method of choice. I'm
still working on an updated view of the 2- list, n < 1200. -Bruce

PS - If M1061 is back on our list of un-spoken-for numbers, and the
M1039 ecm pre-test is an even moderately plausible standard, then
finishing t60 (another 33000 curves at b1=260M?) would seem to be
Very conservative. At difficulty 320, M1061 is nearly twice as hard.
Just meeting the 3*p65 standard used for M1039 would be
3*(70,000); over 200K curves with B1=850M. Still seems to me that
running t55's on the 2- numbers of difficulty above 230, say with
unfactored part above .66*234 = c156 is more likely to find a factor.
Depending upon whether one's ecm objective is finding factors,
versus ecm pre-testing near-term sieving candidates, versus ecm trophy
hunting (on other people's longer-term sieving candidates, or numbers
like Alex's F31, way-out of sieving ... err, gnfs_sieving/snfs_sieving range).
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Old 2007-06-26, 10:04   #13
VolMike
 
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Did you sieve the small factors?

Last fiddled with by VolMike on 2007-06-26 at 10:04
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Old 2007-06-26, 10:49   #14
R.D. Silverman
 
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Quote:
Originally Posted by VolMike View Post
Did you sieve the small factors?
Huh?
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Old 2007-06-26, 10:59   #15
VolMike
 
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2^772+1 has 2 small factors:17 and 43464340002838801.
Thus 2^772+1 should be divided on them and then the quotient should be tested by GNFS algorithm.So does GNFS implementation which you use eliminate by dividing these 2 small factors?
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Old 2007-06-26, 11:05   #16
R.D. Silverman
 
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Quote:
Originally Posted by VolMike View Post
2^772+1 has 2 small factors:17 and 43464340002838801.
Thus 2^772+1 should be divided on them and then the quotient should be tested by GNFS algorithm.So does GNFS implementation which you use eliminate by dividing these 2 small factors?
Your assumptions are wrong.

(1) 2^772+1 is not being done via GNFS, but by SNFS.
(2) The small prime factors of 2^772+1 are irrelevant.
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Old 2007-06-26, 11:11   #17
VolMike
 
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I see
Thus SNFS applied to 2^772+1 returns results faster then GNFS applied to quotient.
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Old 2007-06-26, 11:14   #18
Wacky
 
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Quote:
Originally Posted by VolMike View Post
2^772+1 has 2 small factors:17 and 43464340002838801.
Thus 2^772+1 should be divided on them and then the quotient should be tested by GNFS algorithm.So does GNFS implementation which you use eliminate by dividing these 2 small factors?
Yes, there are "small factors" that are removed at the appropriate time, whether doing SNFS or GNFS.

However, in a case such as this where the product of the known factors is still only a very small portion of the total number, SNFS will give sufficiently smaller polynomial coefficients that ignoring the known factors will produce relations significantly faster than the best general polynomial which takes them into account.

(It looks as if Bob was saying the same thing while I was composing my response)

Last fiddled with by Wacky on 2007-06-26 at 11:16 Reason: Clarification of source of "duplicate" responses.
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Old 2007-06-26, 11:23   #19
VolMike
 
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Quote:
Yes, there are "small factors" that are removed at the appropriate time, whether doing SNFS or GNFS.

However, in a case such as this where the product of the known factors is still only a very small portion of the total number, SNFS will give sufficiently smaller polynomial coefficients that ignoring the known factors will produce relations significantly faster than the best general polynomial which takes them into account.

(It looks as if Bob was saying the same thing while I was composing my response)
Thanks for your explanation.
As I see, factorization is not completed?
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Old 2007-06-26, 11:56   #20
Wacky
 
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On 6/14, Paul reported that he was starting the solution of the 5.6M square matrix that he was able to produce. He estimated that that phase would take 19 days. His estimates are usually quite good.
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Old 2007-06-26, 20:38   #21
xilman
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Quote:
Originally Posted by VolMike View Post
Thanks for your explanation.
As I see, factorization is not completed?
The linear algebra is progressing nicely. At the moment, it is roughly 3.38/5.54, or 61%, completed. When it has finished, a few hours more will suffice to find the factors.

As Wacky said, the linear algebra started on 14th June. Perform the extrapolation yourself.

Paul
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Old 2007-06-27, 06:11   #22
VolMike
 
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Ok. I understand. Approximately 1 week left.

Last fiddled with by VolMike on 2007-06-27 at 06:12
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