20200805, 05:31  #23  
Jun 2003
2^{2}×5×79 Posts 
Quote:
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. 

20200829, 02:57  #24 
Romulan Interpreter
Jun 2011
Thailand
10010101001011_{2} 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^p2) that you compute for PRP, haha.

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