20150222, 17:11  #1 
Sep 2011
3×19 Posts 
NFS success probability in practice
Hello, I figured out a while back the problem with my python implementation. It turns out I wasn't checking that the (a,b) pairs were coprime. Anyway, it works now it and factors numbers up to 30 digits in reasonable time. However, it seems like my implementation factors with a chance of 1/4 per dependency versus 1/2 from literature. Is 1/2 the actual probability in practice?

20150222, 19:16  #2 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{2}×7×337 Posts 
It is 1/2 in practice (if there are exactly two factors).
Maybe your probability is because there were more than two factors? What was your test set  random numbers? 
20150223, 03:55  #3 
Sep 2011
111001_{2} Posts 
No, somehow, after computing the sqrt using CRT, the numbers will sometimes fail to be congruent. I will github the code soon.

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