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Old 2021-06-23, 02:53   #22
axn
 
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Jun 2003

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Quote:
Originally Posted by JuanTutors View Post
Modification 1 of Stage 2 calculates the following:
1. Calculate M. (First 1% of the PRP test.)
2. Calculate Hq. (First multiplication, from an array, and part of the original Stage 2 algorithm)
3. Multiply M by Hq and subtract 1. (Second multiplication, and the only new multiplication. The subtraction is essentially free.)
4. Multiply by Hq-1. (Third multiplication, and part of the original stage 2 algorithm)
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).

Quote:
Originally Posted by JuanTutors View Post
The time cost of this Modification 1 of Stage 2 algorithm are 50% larger, plus the cost of arriving at iteration m = 10^6 based on this procedure, which according to my computer's speed adds about 25% more time depending on the length of Mp. Thus, in total, including Stage 1, this Pm1 test is about 38% more time expensive:
That makes it roughly 8x costlier stage 2. Since stage 2 is already longer than stage 1, overall this will be about 5x more time than regular P-1.

Quote:
Originally Posted by JuanTutors View Post
The gain is all the unknown factors of each 2^m+q, which have on average 13.45 distinct prime factors each, a huge step from the single prime factor q. Note that 2^m+q may have common factors with E. However, since 2^m+q has over 3*105 digits, that does not prevent us from gaining many prime factors.
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)). 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.
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