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2021-06-23, 23:00   #23
JuanTutors

"Juan Tutors"
Mar 2004

24×5×7 Posts

Quote:
 Originally Posted by axn Current P-1 stage 2 uses 1 multiplication to handle two q's. In your new scheme, it will take 4 multiplications to handled 1 q (Hq, M*Hq, (Hq-1) * (M*Hq-1), and Π).
Ah I missed the step of multiplying by MH^q-1, making it 4 multiplications, not 3.

Can you lead me to an explainer that shows how one multiplication is used to handle two q's? I was under the impression that two multiplications are used to handle one q. One multiplication was used to calculate H^q and then another multiplication was used to calculate Π(H^q-1).

Quote:
 Sadly, most of the 13.45 factors are completely useless. A factor q will divide a p-1 with probability 1/q (well, technically 1/(q-1)) ...
I was trying to find this theorem! Which one is it?

Quote:
 ... Which means, for example, a 10-digit q is only 1% as good as an 8-digit q. A 20-digit q is 1/10 billion as good as a 10-digit q. So, while we gain a lot of factors, there is not a chance in our life time to ever see one appear in a found factor's P-1 decomposition. You're better off to just increase the B2 bound in traditional stage 2.
So that even though MH^q-1 has I believe on average ln(2) (? or 1?) factors between B2 = 10^7 and B2^2 = 10^14, the probability that any factor we happen to come upon will help us find a factor of Mp is about 10^-10.5, which is about 10^-(10.5-6) = 10^-4.5 times the probability that any factor between B1 and B2 helps. So for whatever cost we add via this method, ignoring all potential factors larger than B^2 due to diminishing benefits, we multiply the probability of finding a factor during this modified stage 2 by about 1+ln(2)*10^-4.5. Does that sound about right? Even if it's on the same order, that's pretty weak.