20200625, 22:21  #177  
Nov 2003
2·3·17·73 Posts 
Quote:
12^3191 has an algebraic factor. We are not factoring 12^3191, we are factoring (12^3191)/(12^291) ~12^290 ~ C313. The resulting polynomial is reciprocal, so we can do this number with a quintic. However, quintic polynomials for numbers this size result in matrices that are significantly larger than those for numbers of similar size done with sextics. Greg is LA constrained right now, so he skipped 12^3191 for the time being. He did C314, C315, C316 C317 and is now working on C318's via 3^6671 etc. Greg may indeed do R323 before he does 12^3191. I think he will. R323 might well be done by a reciprocal octic to take advantage of the algebraic factor 10^191. Whether the octic would be easier than the obvious sextic might be an interesting experiment. It might also be interesting to see if a septic would be any better. I think a septic will be slightly better in general for numbers of this size. Let's do a "back of the envelope" look at the norms. Take (10^6, 10^6) == (a,b) as a 'typical lattice point'. For a sextic, an algebraic norm is ~ a^6 ~ 10^36 and a linear norm is ~ b * (10^324/6) ~ 10^60. For a septic an anorm is ~a^7 ~ 10^42 and a linear norm is b *(10^322/7) ~ 10^52. The norms are closer for the septic and their product is slightly smaller. A septic seems slightly superior. For the reciprocal octic an anorm is a^8 ~ 10^48 and a linear norm is b * (10^38) ~ 10^44 which seems even better still. Note that one also needs to adjust these estimates by the specialq. The estimates also ignore the effect of variance on the norms. Since we want smooth numbers we are more concerned with the tails of the distributions of the norms rather than the means. However, it does give a quick comparison. NFS works best when the norms are as nearly equal as possible, other things being equal. This very rough estimate is based on the assumption that (10^6, 10^6) is a typical lattice point. Adjust the analysis if this assumption is not a good enough estimate. I do now know what sieve areas the lasievef siever uses. Noone has been calling for him to do 12^3191. It is possible that Greg missed the reciprocal octic for R323. He will get to it. Doing R323 seems to be a compulsion with you. 

20200627, 07:33  #178 
Jul 2003
So Cal
3750_{8} Posts 

20200627, 07:48  #179  
Jul 2003
So Cal
2024_{10} Posts 
Quote:


20200628, 19:57  #180  
Nov 2003
7446_{10} Posts 
Quote:
a degree 7 polynomial would be better (than degree 6) for Greg to use moving forward for numbers that NFS@Home is about to undertake. 

20200629, 10:29  #181  
"Bo Chen"
Oct 2005
Wuhan,China
A0_{16} Posts 
Quote:
deg 6 need 102 CPU years to collect 1200M raw relations, while deg 7 need 182 CPU years on an i3 CPU. I attach the poly and test files. 

20200629, 15:39  #182  
Nov 2003
2·3·17·73 Posts 
Quote:
algebraic and decrease the linear. 

20200629, 20:50  #183 
Nov 2003
1D16_{16} Posts 

20200702, 00:04  #184 
"Bo Chen"
Oct 2005
Wuhan,China
2^{5}×5 Posts 
I guess you mean increase alim and decrease rlim, use option a.
But I test again with these changes, the situation is the same. When I use alim=800M rlim =200M a, and binary lasieve5_f compiled, it need 100 CPU years to collect 1200M raw relations on an i7 CPU; While use alim=rlim =400M,r with the same binary and processor,it need 40 CPU years to collect 1200M raw relations. Though I don't know why,it is a little strange. 
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