20190522, 05:25  #45  
May 2019
24_{10} Posts 
Quote:
Quote:
Code:
(note: g(x) = (a*a)x^2 + (2*a*b)x + (b*bn)) =============== Smooth #1 =============== Polynomial g(x) = 66725635853624133104491845866161 x^2 + 152328794209151404760083205350018 x + 4493699262681020585053171287604495116568 Smooth found at x = 6057 g( 6057 ) = 2046641900942309291498167743817830894505 LHS (ax+b) = 49467741495615522472 =============== Smooth #2 =============== Polynomial g(x) = 70563600195904987918333753599889 x^2 + 132234389859675357708720776316226 x + 4493699287668381330025165273238560877368 Smooth found at x = 2988 g( 2988 ) = 3864092420317814097540854938717866384640 LHS (ax+b) = 25091969418551765843 =============== Smooths #3 & 4 (same polynomial) =============== Polynomial g(x) = 51537154140545974176898100656321 x^2 + 158251595892937553938363990151170 x + 4493699228136277324690277448438158865064 Smooth found at x = 9618 g( 9618 ) = 275314899814652154417599991477095633600 LHS (ax+b) = 69058049852526267917 Smooth found at x = 11544 g( 11544 ) = 2376171639294564454200188189849665980872 LHS (ax+b) = 82884684887582914631 

20190522, 06:23  #46  
May 2019
2^{3}·3 Posts 
Quote:
I think I may have implemented my normal distribution function incorrectly, because I just had something very interesting happen—when I changed the sigma to 1 (not 1*mean), there was one polynomial that picked up 132 smooths! (Of course, the program halted because only one set of qs could satisfy such an excessively strict standard deviation, hahaha) Admittedly, my understanding of statistics is very poor (it was a requirement at my university even for humanities majors like myself, and I nearly failed). I just copied the function from Wikipedia: , where Now, the original function has as input, but only uses in the body of the function. If this isn't a typo, then you can use the function even if you just define the standard deviation, and not the variance, right? (I understand standard deviation but don't really understand intuitively what the variance is, so I'd rather use that if I can.) The other thing is that I was getting odd behavior when I scaled by . So I took that out. So when you say "use a sigma of 1", I am interpreting that to mean this: In the case of this 50digit number, my ideal q, . So . Thus primes in the factor base between fall within 1 standard deviation of the mean. Is that correct? 

20190522, 07:55  #47 
May 2019
2^{3}·3 Posts 
On second thought, I think they are all probably duplicates, because on a typical goround with these parameters, the program will cycle between the same two sets of qs, and the huge spike in smooths is not seen until the second sieving of the same polynomial. That is, they are probably all smooths formed from partials left over by the same x and polynomial. Oops :/

20190522, 14:16  #48  
"Ben"
Feb 2007
2·1,789 Posts 
Quote:
[edit] If you do run it on a bunch of B's and verify that the sieve code is finding everything it should, then at least you've eliminated that as a source of concern. Last fiddled with by bsquared on 20190522 at 14:24 

20190522, 14:45  #49  
"Tilman Neumann"
Jan 2016
Germany
1D9_{16} Posts 
Quote:
Sorry, I should have looked at my source code. I am _not_ using a normal distribution. What I am doing is: * Given a_best = the theoretically best avalue and q_count = the wanted number of q's, find the index indexCentre of the prime base element such that primeBase[indexCentre]^q_count is nearest to a_best. * Compute some "variance" as indexVariance = 0.75*sqrt(primeBaseSize) * Then for each of the first q_count1 q's, I select a prime base index randomly from the uniform distribution [indexCentre  indexVariance, indexCentre + indexVariance]. * The last q is computed deterministically to match a_best as good as possible. What I meant with "sigma" is the 0.75 factor. The procedure seems to be very similar to henryzz's. Last fiddled with by Till on 20190522 at 15:03 Reason: clarity, last q 

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