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#12 |
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Nov 2010
Germany
3×199 Posts |
Thanks a lot, I think now I got it ...
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#13 | |
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"Lucan"
Dec 2006
England
2·3·13·83 Posts |
Quote:
A couple of refinements: 1) The first six bits of the exponent generate the number 2^63 at most. If the bits are all 1 we get: 2^1, 2^3 , 2^7, 2^15, 2^31, 2^63, so for testing factors >2^63, the number of steps in the algorithm is (bits in exponent - 6). 2) The squaring and more importantly the division/modulo steps get more time consuming with the number of bits in the factor, 64 being a notable watershed. David Last fiddled with by davieddy on 2010-11-21 at 23:18 |
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