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R.D. Silverman 2004-04-23 12:54

[QUOTE=aaronl]What about relatively small numbers? How difficult is it to factor a 384 bit number with GNFS?[/QUOTE]

"About" a month on my 3.2Ghz PIV to do the sieving and another few days
and several hundred Mbytes of memory to do the linear algebra.

xilman 2004-04-23 14:44

[QUOTE=Bob Silverman]"About" a month on my 3.2Ghz PIV to do the sieving and another few days
and several hundred Mbytes of memory to do the linear algebra.[/QUOTE]
That's easy by my measure :smile:

Cheap consumer PCs here in the UK are being shipped with half a gig of RAM. Upgrading them to a gigabyte is not very expensive.

Lots of people on this forum seem to be entirely unfazed by the idea of spending a month or two running a LL test. If they are equally enthusiastic about factoring, they won't mind spending the same resources on a factorization.

Paul

R.D. Silverman 2004-04-23 15:08

[QUOTE=xilman]That's easy by my measure :smile:

Cheap consumer PCs here in the UK are being shipped with half a gig of RAM. Upgrading them to a gigabyte is not very expensive.

Lots of people on this forum seem to be entirely unfazed by the idea of spending a month or two running a LL test. If they are equally enthusiastic about factoring, they won't mind spending the same resources on a factorization.

Paul[/QUOTE]

I agree; you are preaching to the choir.... :smile:

Note that part of the time should be devoted to searching for a good
polynomial. I never implemented software to do this (i.e. Brian Murphy's
ideas). If you just use a naiive 'base m' representation, the computation
probably will take 3-4 times as long. The exact total time will depend on
the quality of the polynomial and the time you spend searching for a good
one. This is why I put 'about' in quotes.

BTW, I think it would be useful if you or someone could add (daily?/weekly?)
info to the NFSNET home page about the progress of the linear algebra for
M811 (and 10,223+), as well as the info about the current sieve effort.

What was the weight of the final matrix?

JHansen 2004-04-23 21:44

[QUOTE=Bob Silverman]If you just use a naiive 'base m' representation, the computation
probably will take 3-4 times as long.[/QUOTE]


Surely you wouldn't do that anymore? The Franke/Kleinjung polynomial selection tool (+ script) is extremely easy to use, and it produces polynomials where the coef's are nowhere near the size of the common root. For example: I'm currently working on a C133 GNFS factorization, and I'm using polynomials f(X):=2077740*x^5 - 868387138833*x^4 + 644699090935711*X^3 - 909508722977946890078*X^2 - 153100527692752699232273159*X + 14919416027434566785965893689715 and g(X):=28864624875229*X - 20703253390555537351847126, which share the root M:=762763124104274928687715937066818931526538706071874614433284596644025798338619371964842522017015\\ 6068238067416396308603682486719481305

This means that there is more than a factor 10^100 between any of the coefficients and M.

If you use the (tweaked) base-m method some of the coef's can be of order m/2, right?

Jes


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