Mon May 19 17:55:17 CEST 2003
On Monday, May 19, 2003, at 04:23 PM, Vladimir Sekissov wrote:
> Good day,
> sean.hinde> > 2^BigNumber mod ReasonableNumber, then there's no need to
> sean.hinde> > calculate 2^BigNumber.
> sean.hinde> Aha! Any pointers on how to do this?
> Volume 2 of Donald Knuth's "The Art of Computer Programming" is now
> your best friend.
> Or search google with "arbitrary precision arithmetic".
Thanks. Here's what turned up. Translated from C:
% Calculate M ^ P mod Q
modexp(M, P, Q) ->
modexp(M, P, Q, 1).
modexp(M, 0, Q, Z) ->
modexp(M, P, Q, Z) ->
Z1 = if ((P band 1) == 1) ->
Z * M rem Q;
M1 = M * M rem Q,
P1 = P bsr 1,
modexp(M1, P1, Q, Z1).
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