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2020-08-05, 05:31   #23
Citrix

Jun 2003

157310 Posts

Quote:
 Originally Posted by Citrix 1) Why do we have to use all the primes in B1. A smaller subset set of primes in B1 might have an lower hammer weight N. We then take this smaller subset and the remaining B1 primes (or move them to the B2 stage). Have you tried this?
Thinking about this further... all primes in B1 are not necessary and it does not need to be an absolute multiple of powersmooth(1000000). Some primes can be twice, some three times etc.

Turning the problem around - instead of fixing a number powersmooth (x) and then searching we leave both input and output variable.

If we are flexible with above --- then we could generate a 2^n+c that has a lot of factors (smooth) and c is relatively small and we can control number of bits in n. This should have a low hammer weight.

OR generate a number with low hammer weight that has a lot of factors.

 2020-08-29, 02:57 #24 LaurV Romulan Interpreter     Jun 2011 Thailand 2×3×31×47 Posts Mihai, there is a way to reduce that number of bits, but the multiple becomes such big, that would be impossible to use, as you will need soooOOO many more squarings, you gain nothing. For example, instead of computing b^e (with odd e), you can compute b^(2^ord(2,e))/b. What's in parenthesis is a power of 2, so it has no additional "pops", and the resulted power (after division) is a multiple of e (by definition of ord). But what you get there is HUUGE, way beyond b^(2^p-2) that you compute for PRP, haha.

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