[racket] slow g^x=h mod p meet in the middle code

În data de Thu, 27 Feb 2014 17:33:32 +0200, Cristian Baboi  
<cristian.baboi at gmail.com> a scris:

>>> Second:  (modulo (* l gm1) p) looks inefficient.
>>>                If l and gm1 are large, then it is more efficient
>>>                to reduce modulo p first, then multiply, then reduce  
>>> again.
>
> Well, gm1 and gB are already reduced modulo p ...
> The only one who is not is l which is lower than B which is lower than p.
>
Sorry. x0 and x1 are less than B. l is already mod p.