![]() |
Different bases = different times?
I'm wondering how the amount of time to do primality tests changes with the base that is used with k.b^n+-1. I know that an odd base is not as quick as a base that is a power of two, but don't know if an even base has different times to complete primality tests than a power of two base.
Eg: k*2^n+1 = 10000 digits. k*62^n+1 = 10000 digits. Does the test time change? Thanks, Roger |
[QUOTE=roger;104496]I'm wondering how the amount of time to do primality tests changes with the base that is used with k.b^n+-1. I know that an odd base is not as quick as a base that is a power of two, but don't know if an even base has different times to complete primality tests than a power of two base.
Eg: k*2^n+1 = 10000 digits. k*62^n+1 = 10000 digits. Does the test time change? [/QUOTE] Yes. The number of modular squarings is the same. Since k*b^n + 1, b!= 2 will have higher Hamming weight, the latter will have more multiplications by b. [but note that multiplication by a single precision number does not take much time.] The biggest gain comes from the fact that it is MUCH faster to do modular multiplication modulo k*2^n + 1 than for k*b^n + 1, b != 2. |
| All times are UTC. The time now is 17:49. |
Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2021, Jelsoft Enterprises Ltd.