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 2008-06-23, 19:32 #23 Raman Noodles     "Mr. Tuch" Dec 2007 Chennai, India 3×419 Posts Sieving was done rapidly on Core 2 Duos at my university (NIT, Trichy) which helped me to sieve rapidly at that time. When vacation started on 29 Apr 2008, the sieving was 86% done on this number. After that for 20 days, I was without any resources, so sieving was suspended On 20 May 2008, we bought a new Core 2 Quad @ 2.4 GHz at home which helped to finish the sieving rapidly. Around June 9th the sieving was sufficient enough with about 78 million special-q sieved. Five days ago, the linear algebra was started on my Core 2 Duo laptop @ 1.7 GHz. Since there wasn't enough virtual memory available in normal mode, the post processing went in safe mode with the /3GB switch. Regarding square root, each dependency takes about two hours to solve it up, the first dependency failed. Cleverly simultaneously I picked up the 4th dependency on the other core of my laptop. The dependency was a good choice to give me away with the factors! I have chosen up with the fourth dependency in the square root stage because 2,1039- gave away the factors at the 4th dependency! Notice that 6,305- took 8 months to complete. But 7,295- which is twice as harder took only 6 months, eventhough I was idle for sometime between. Sieving was rushed through with those Core 2 Duos at my college. 10,312+ is half-way through sieved. It will take a couple of weeks if 30 million special-q suffice. Last fiddled with by Raman on 2008-06-23 at 20:01 Reason: M1039 gave up factors in 4th dependency
2008-06-23, 19:58   #24
xilman
Bamboozled!

"𒉺𒌌𒇷𒆷𒀭"
May 2003
Down not across

11,423 Posts

Quote:
 Originally Posted by Raman 7,295- Code: Tue Jun 24 00:30:50 2008 prp81 factor: 204239004182680605398190478754212368873366912490836010105265524712426411236134031 Tue Jun 24 00:30:51 2008 prp111 factor: 393263672474017252292660491631044385409360056708958704520879019006886885032467377758314801669636946200575798561 One minute... Let me mail Prof. Sam Wagstaff before posting further information about it...
Nice one!

I'm glad it worked out in the end. Good luck with the next.

Paul

2008-06-23, 20:33   #25
Raman
Noodles

"Mr. Tuch"
Dec 2007
Chennai, India

3×419 Posts

Quote:
 Originally Posted by xilman Nice one! I'm glad it worked out in the end. Good luck with the next.
What is the best polynomial that I can use so for 10,375-

Since 3 and 5 both divide 375,

So, the polynomial that I currently think so of, is
x10+x5+1 divided by x2+x+1
which is,

$x^8-x^7+x^5-x^4+x^3-x+1$

which has SNFS difficulty of 200 digits

Last fiddled with by Raman on 2008-06-23 at 20:38

2008-06-23, 21:15   #26
frmky

Jul 2003
So Cal

243610 Posts

Quote:
 Originally Posted by Raman $x^8-x^7+x^5-x^4+x^3-x+1$ which has SNFS difficulty of 200 digits
Yep, but make it degree 4. Not great, but the best you can do.

$x^4-x^3-4x^2+4x+1$

$10^{25}x-(10^{50}+1)$

Greg

2008-06-25, 13:27   #27
Raman
Noodles

"Mr. Tuch"
Dec 2007
Chennai, India

3·419 Posts

Quote:
 Originally Posted by frmky Yep, but make it degree 4. Not great, but the best you can do. $x^4-x^3-4x^2+4x+1$ $10^{25}x-(10^{50}+1)$
Sure? Is biquadratic (aka quartic) the best polynomial that I can use so
for 10,375-? No quintics or sextics are available for it, of
course with difficulty 200?

And eighth degree is not feasible? I think that it makes the
algebraic coefficients too larger, right?

Code:
Similarly I think that for a multiple of 11, say 7,319-
you will certainly not be using
$\sum_{i=0}^{10} x^i$ and $x-7^{29}$
You would be reducing it to degree 5, right?

And for a multiple of 13, for example 6,299-
$\sum_{i=0}^{12} x^i$ should be reduced to degree 6.

Although both of these are reduced to degree 5 and 6,
a multiple of 17 or higher cannot be reduced this way to degree 8 or higher
and should be treated up as a prime exponent, right?

For example, for 2,799- Dr. Kleinjung et al. would certainly not have
used $\sum_{i=0}^{16} x^i$ and $x-2^{47}$ or of course, the one
reduced up to degree 8 for it.

I think that they would only have used up so with
$2x^6-1$ and $x-2^{133}$ in the Bonn University.
What about reducing the degree 14 for 10,375- (since it is a multiple of 15)
this way up to degree 7 directly?

$\sum_{i=0}^{14} x^i$ and $x-10^{25}$

Last fiddled with by Raman on 2008-06-25 at 14:24 Reason: Please remove this reason feature. One always edits so to add up with more points in the post.

2008-06-25, 15:09   #28
jasonp
Tribal Bullet

Oct 2004

354510 Posts

Quote:
 Originally Posted by Raman Sure? Is biquadratic (aka quartic) the best polynomial that I can use so for 10,375-? No quintics or sextics are available for it, of course with difficulty 200? And eighth degree is not feasible? I think that it makes the algebraic coefficients too larger, right?
Correct, the algebraic sieve values grow too large too quickly, so the number of algebraic sieve values that are smooth enough drops too fast. The asymptotic estimates for NFS indicate that a degree-7 polynomial is feasible only for inputs that have many hundreds, if not thousands, of digits.

Most of the smallest cunningham numbers that are left have similar difficulty; if an available cunningham number is unusually small, it's probably because the NFS polynomials involved are unusually bad :)

Last fiddled with by jasonp on 2008-06-25 at 15:09

2008-06-26, 15:02   #29
R.D. Silverman

"Bob Silverman"
Nov 2003
North of Boston

23·3·311 Posts

Quote:
 Originally Posted by jasonp Correct, the algebraic sieve values grow too large too quickly, so the number of algebraic sieve values that are smooth enough drops too fast. The asymptotic estimates for NFS indicate that a degree-7 polynomial is feasible only for inputs that have many hundreds, if not thousands, of digits. Most of the smallest cunningham numbers that are left have similar difficulty; if an available cunningham number is unusually small, it's probably because the NFS polynomials involved are unusually bad :)

Actually, there are a fair number of composites left under 230 digits that
do not require a quartic.

10,312+ Raman; in progress
2,2106L quartic; yech
10,378+
7,384+
5,341- reserved
2,1694M
3,517+ I will do shortly
7,393+
2,1104+ in progress; LA 75%
10,259+
10,339-
2,1119+
2,1128+
2,1149-
2,1161+
2,1161-
10,339+
7,396+

 2008-06-26, 18:08 #30 fivemack (loop (#_fork))     Feb 2006 Cambridge, England 2×7×461 Posts I'm about to start 10,259+ if nobody else is interested in it.
 2008-06-26, 19:09 #31 bsquared     "Ben" Feb 2007 5·727 Posts I'm going after 10,339-
2008-06-26, 19:25   #32
R.D. Silverman

"Bob Silverman"
Nov 2003
North of Boston

164508 Posts

Quote:
 Originally Posted by R.D. Silverman Actually, there are a fair number of composites left under 230 digits that do not require a quartic. 10,312+ Raman; in progress 2,2106L quartic; yech 10,378+ 7,384+ 5,341- reserved 2,1694M 3,517+ I will do shortly 7,393+ 2,1104+ in progress; LA 75% 10,259+ 10,339- 2,1119+ 2,1128+ 2,1149- 2,1161+ 2,1161- 10,339+ 7,396+
And there are also lots of them that do require a quartic:

3,565-, 580+

6,335-
6,370+

5,370+, 400+, 410+ 430+

7,335- 320+, 340+

2,860+, 865+, 925+.....

etc. etc. etc.

7,320+, 340+

3,580+

2008-07-20, 13:58   #33
Raman
Noodles

"Mr. Tuch"
Dec 2007
Chennai, India

125710 Posts

Quote:
 Originally Posted by frmky Yep, but make it degree 4. Not great, but the best you can do. $x^4-x^3-4x^2+4x+1$ $10^{25}x-(10^{50}+1)$ Greg
So, can you please explain to me up how you derived the 4th degree
polynomial from the 8th degree one for $10,375-$
$x^8-x^7+x^5-x^4+x^3-x+1$
$x-10^{25}$
I am starting to sieve for 10,375- now.
10,312+ is in Linear Algebra and will finish up
within about 12 hours or so
(Matrix has less than 20 million rows!)

EMERGENCY
Also that I can't enter the value of m in the GGNFS
poly file too, because of the fact that
$\division_{10^{25}}^{(10^{50}+1)}$
is again not an integer at all

Last fiddled with by Raman on 2008-07-20 at 14:46

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