20060124, 16:37  #1 
"Bob Silverman"
Nov 2003
North of Boston
2^{3}×3×311 Posts 
Amazing!!!
In case anyone hasn't noticed, Aoki et.al just set a new record
both for NFS size/difficulty and penultimate factor size. They did a 274 digit number. Yes. 274 digits ~ 907 bits.!!!!!!!!! Holy !! So much for our doing 2,857....... They did 6,353 C274. Fantastic. 
20060124, 18:08  #2  
"Nancy"
Aug 2002
Alexandria
2,467 Posts 
This is their announcement, from Paul Zimmermann's page:
Quote:
The next record will probably be the kilobit factorisation. Not much point in putting so much work into it as will be required to beat this and not claim the grand prize. Makes me wonder what Franke et al. are up to... For the record, the estimated 17.3 years on 2GHz Opteron translates to about 40 days continuous use of our cluster. That would be very hard to justify with the other users. Alex Last fiddled with by akruppa on 20060124 at 18:16 

20060125, 19:19  #3  
"Bob Silverman"
Nov 2003
North of Boston
2^{3}·3·311 Posts 
Quote:
Alex, Why not put grid computing software on your cluster? You can now set up sieve jobs so they run only when nothing else is running (or the load is sufficiently small)? The grid software would automatically submit sieve ranges to machines as they become available. If a machine becomes busy while a job is in progress it is put to sleep. 

20060125, 19:45  #4  
Aug 2004
2^{2}×3×11 Posts 
I'm trying to understand this bit:
Quote:
I can see how that might work, but I don't see how you would be guaranteed that any combinations of the solutions of the unextended matrix would be in the null space of the extended matrix unless you have more solutions than quadratic characters. I guess I'm missing something here. Chris 

20060126, 01:59  #5  
Tribal Bullet
Oct 2004
5×709 Posts 
Quote:
I've never seen it done in code, though, so it's probably trickier than I'm making it out to be. jasonp 

20060126, 09:14  #6  
Aug 2004
10000100_{2} Posts 
Quote:
Suppose we start with an m x n matrix X (m relations, n primes, with m > p), and we find some of the null space, i.e. an s x m matrix B with BX = 0. To find the solutions we want we have to extend X with q quadratic character columns, i.e. an m x (n + q) matrix X' = (X Q). Then BX' = (0 BQ), so we want to find an r x s matrix C such that CBX' = (0 CBQ) = 0, i.e. C(BQ) = 0. The matrix BQ is s x q (a small matrix, easy for Gaussian Elimination), so to guarantee a nontrivial solution to C(BQ) = 0 we must have s > q. aoki et al don't say how many quadratic characters they use, but if their 34 "actual" solutions correspond to B, then q would have to be < 34, which seems unlikely. However, I reckon they must have done something more subtle, so can anyone explain what they meant by "actual 34 solutions" and "adjusted the unit parts" please? Chris 

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