 mersenneforum.org Knuth-Schroeppel analysis
 Register FAQ Search Today's Posts Mark Forums Read 2005-04-06, 14:43 #1 akruppa   "Nancy" Aug 2002 Alexandria 2,467 Posts Knuth-Schroeppel analysis Hi, I've been wondering about how the technique of estimating root properties for NFS or Knuth-Schroeppel factors for MPQS actually works. According to various sources (i.e. TAOCP 4.5.4) they compare the average log contribution of small primes to the factorization of the numbers that are tested for smoothness with the average log contribution in uniformly, randomly chosen integers. Let f(p,S) be the average exponent of the prime p in values chosen uniformly at random from the set S. For S the natural numbers, we have f(p,\N) = 1/(p-1). If S is the set of values we test for smoothness, and s \in S chosen at random, then the log of the residual after dividing out primes p 2005-04-06, 15:06   #2
R.D. Silverman

Nov 2003

1D2416 Posts Quote:
 Originally Posted by akruppa Hi, I've been wondering about how the technique of estimating root properties for NFS or Knuth-Schroeppel factors for MPQS actually works. According to various sources (i.e. TAOCP 4.5.4) they compare the average log contribution of small primes to the factorization of the numbers that are tested for smoothness with the average log contribution in uniformly, randomly chosen integers. Let f(p,S) be the average exponent of the prime p in values chosen uniformly at random from the set S. For S the natural numbers, we have f(p,\N) = 1/(p-1). If S is the set of values we test for smoothness, and s \in S chosen at random, then the log of the residual after dividing out primes p

Yes, it is about right. Yes, it is an assumption; but "leap of faith" is
too strong. There is substantial supporting numerical evidence. I characterise something as "leap of faith" when there is no supporting
evidence. There are also some probability arguments to suggest that the
assumption is correct (or at most off by a constant factor). These
arguments [for at least quadratic extension fields] are discussed in the
"Cohen-Lenstra" heuristics.   2005-04-12, 15:21 #3 akruppa   "Nancy" Aug 2002 Alexandria 1001101000112 Posts I looked at Cohen and Lensta's paper and some follow-up papers (i.e. Williams and te Riele), but I'm afraid they require much more algebra than I know. I'm clueless about character groups, Dedekind domains and whatnot. I think a vague argument that the analysis is about right could be made supposing that the rho(alpha) function is "linear enough" for small variation of alpha that a weighted sum can be pulled inside the argument. I'll look at this idea more closely sometime later. Thanks for your reply, Alex Last fiddled with by akruppa on 2005-04-12 at 15:21   2009-12-19, 21:14 #4 ThiloHarich   Nov 2005 32·11 Posts I think the question akruppa raised is valid. I analyzed the smooth values and found out that the probability of a factor p being part of the smooth value is much higher then for a random number (which should be clear). For the SIQS (without large primes) I estimated a value of: 2.2 / (p - 1)^.72 for the expected length of a factor (of the factor base, different from 2) in a smooth decomposition. So I used this approximation for determining the multiplier. If it differs it always causes a better runtime then the approximation log (p)/p or log (p)/(p-1). Here are some examples: N = 25249581771989830594180551024377089571(38) my multiplier: 59 time : 0.24417041800000003 knuth-schroeppel: 3 time : 0.520334009 N = 24339015700034049398642312663507799431(38) my multiplier: 11 time: 0.283217839 knuth-schroeppel: 7 time: 0.45216473 N = 15028821219978351294914980921150425953(38) my multiplier: 2 time: 0.241755027 knuth-schroeppel: 1 time: 0.6213507580000001 From my point of view the p-1 approximation is better then the p since the following facts: A random number is dividable by a factor p with probability 1/p. With probability 1/p^2 it is dividable by p^2. So the resulting length of the exponent of a factor p is log (p)/p + 2*log (p)/p^2 + 3*log (p)/p^3 + ... = log (p)/p * (1 + 1/p + 1/p^2 + ..) = log (p)/p * (1/(1-1/p)) = log (p) * (1/(p-p/p)) = log (p) /(p-1) I will try running the sieve with different multipliers lets say k_1, k_2, k_3 with polynomials including the factors k_2*k_3, k_1*k_3, k_1*k_2. So the conguences are all mod k_1*k_2*k_3. Will this work?   2009-12-19, 21:54 #5 jasonp Tribal Bullet   Oct 2004 3,529 Posts Do you mean 2.2*log(p)? Msieve picks the same multipliers your code does, and I thought it uses the Knuth-Schroeppel algorithm. A factor of p contributes 2*log(p)/(p-1) if it is not part of the multiplier, and 1*log(p)/(p-1) if it is.   2009-12-19, 23:11 #6 ThiloHarich   Nov 2005 32·11 Posts No I really mean something O(p^c) where c is something around -.72. instead of 2 * log (p) / (p-1). You can see it as increasing the knuth-schröppel factors 2*log (p) / (p-1) by a factor (p-1) ^ 0.37 / log (p). I alway used 2 * log (p)/ (p-1). I did not see why a factor of 1 should work better if p divides a. This is an simple approximation of the values I observed, and it worked for my sieve.   2009-12-20, 00:38 #7 jasonp Tribal Bullet   Oct 2004 DC916 Posts Using 1 instead of 2 when p divides the multiplier is a trick to account for the factor base containing only one sieve root for p in that case, instead of the usual two roots. I think Contini's thesis first documents the idea.   2009-12-20, 10:05 #8 ThiloHarich   Nov 2005 11000112 Posts Ahh this is true. I added this correction to the approximation. But since this is only the case for some small factors this does not influence the overall result much. Since the higher factors were weighted much higher with my formula we should not stop at some factor with the summing them up, as we can do with the Knuth-Schroeppel fromula.   2010-01-08, 17:01 #9 ThiloHarich   Nov 2005 32·11 Posts I found the reason why I was getting greater and better multipliers. I took for the factor k a term - log (k), but since q(x) = (sqrt(n) + x)^2 - kn grows with O(sqrt(k*n)*x), a factor - log(k)/2 is more suitable. So even that my approximation on the resulting hit rate (or the resulting length) of the factors in the factorbase is better then in other approximations, the resulting multipliers are mostly the same then in the other approximations like the implementation of alpertron :(  Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post Dougy Math 5 2014-04-08 20:50 TObject Software 4 2013-02-05 23:53 davieddy Puzzles 9 2011-08-02 09:59 cheesehead Lounge 20 2009-08-17 03:19 troels munkner Miscellaneous Math 2 2006-07-17 03:18

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