On Tue, Apr 9, 2013 at 2:51 PM, Laurent <laurent.ors...@gmail.com> wrote:
> On Tue, Apr 9, 2013 at 2:42 PM, Matthew Flatt <mfl...@cs.utah.edu> wrote:
>
>> (for/product ([p (in-range (+ m 1) (+ n 1))]) p)
>
>
>
> But I don't fully understand why a simple (for/product ([p (in-range 1
> (add1 n))]) p) is as fast as the iota variants.
>
I don't understand either.
FWIW here's the kind of multiplications performed by the doctored version
attached below:
> (fact 1000000)
#hash(((fíx fíx fíx) . 48577) ((big big) . 209717) ((fíx big) . 46338)
((fíx fíx) . 646789))
while the for version should be almost all big x big.
====
(define x*-counter 'not-initialized)
(define (x* . args)
(hash-update! x*-counter (map (λ (x)
(if (fixnum? x)
'fíx
'big))
args)
add1
0)
(apply * args))
;; from included; to excluded.
(define (mult from to)
(case (- to from)
((0) 1)
((1) from)
((2) (x* from (+ from 1)))
((3) (x* from (+ from 1) (+ from 2)))
;; ...
(else
(let ((middle (quotient (+ from to) 2)))
(x* (mult from middle) (mult middle to))))))
(define (fact n)
(set! x*-counter (make-hash))
(define f (if (zero? n)
1
(mult 2 (add1 n))))
(values ;f
x*-counter))
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