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

On 02/27/2014 07:55 AM, Jens Axel Søgaard wrote:
> 2014-02-27 15:23 GMT+01:00 Cristian Baboi <cristian.baboi at gmail.com>:
>> Hello,
>>
>> I recently used racket for a math assignment in a crypto class because of
>> big numbers support. Others used python, java, haskell and bragged with
>> short execution times. Is there anything I can do to speed up the following
>> code or is my computer too old ?
>
> First make sure you use the same algorithm. Post it!
>
> 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.
>
> Since all your calculations are modulo p, you can use with-modulus and mod*.
>
> Third: When benchmarking turn off display and use racket (not DrRacket).
>
> Fourth: Use Racket hash tables rather than rnrs ones. (I haven't looked, so
> I am unsure how they are implemented - maybe they are slower - maybe they
> are the ok)/
>
> Without changing the algorithm, I get this:
>
> #lang racket
> (require math)
> (require rnrs/hashtables-6)
> (define p 13407807929942597099574024998205846127479365820592393377723561443721764030073546976801874298166903427690031858186486050853753882811946569946433649006084171)
> (define g 11717829880366207009516117596335367088558084999998952205599979459063929499736583746670572176471460312928594829675428279466566527115212748467589894601965568)
> (define h 3239475104050450443565264378728065788649097520952449527834792452971981976143292558073856937958553180532878928001494706097394108577585732452307673444020333)
> (define B (expt 2 20))
>
> (with-modulus p
>    (define gB  (modexpt g B))
>    (define gm1 (modular-inverse g p))
>    (define (mx x0 x1) (+ (* B x0) x1))
>
>    (define (hash n ht l x1)
>      (cond
>        [(> n 0)  (hashtable-set! ht l x1)
>                  (when (< x1 10) (displayln (list x1 l)))
>                  (hash (- n 1) ht  (mod* l gm1) (+ x1 1))]
>        [else     (hashtable-set! ht l x1)
>                  ht]))
>
>    (define (htbl) (hash (- B 1) (make-eqv-hashtable B) h 0))
>    (define (gol)  (make-eqv-hashtable B))
>
>    (define (caut n ht l x0)
>      (cond
>        [(> n 0) (define x1 (hashtable-ref ht l -1))
>                 (when (< x0 10) (displayln (list x0 l)))
>                 (if (eqv? x1 -1)
>                     (caut (- n 1) ht  (mod* l gB) (+ x0 1))
>                     (cons x0 x1))]
>        [else (define x1 (hashtable-ref ht l -1))
>              (if (eqv? x1 -1)
>                  (cons -1 -1)
>                  (cons x0 x1))]))
>
>    (define run (caut (- B 1) (htbl) 1 0))
>    (define x (mx (car run) (cdr run)))
>    (displayln x))

I'm not sure what the hash tables are for, but they're apparently the 
culprit. When I removed them, I got 1048575 in a few seconds. (I had 
`hash' return `x1', which I passed to `caut' instead of a hash table.)

For this algorithm, is it necessary to keep all the intermediate values 
computed by `hash' in a hash table?

Neil ⊥