On Mon, Mar 19, 2012 at 04:35, Joe Gilray <jgil...@gmail.com> wrote:
> Thanks Rodolfo and Eli for the education, very elegant solutions.
>
> I really like the clever use of the "(and (right-triangle? a b c) (list a
> b c))))" idiom.
>
> I had to look up in-value... unfortunately the manual is a bit sparse
> there, but I got the gift by running some examples... thanks.
>
> After going "D'oh" about the infinite loop, here is the code I ended up
> with:
>
> (define (pythagorean-triple n)
> (let loop-ab ([a 1] [b 2])
> (define c (- n a b))
> (cond [(>= a n) '()]
> [(<= c b) (loop-ab (add1 a) (+ a 2))]
> [(right-triangle? a b c) (list a b c)]
> [else (loop-ab a (add1 b))])))
>
> I noticed that the sequence-based solutions are quite a bit slower than
> the code above probably because they don't short-cut on (<= c b), is there
> an elegant way to speed them up?
>
>
Before you asked I wrote this:
; by Rodolfo Carvalho
(define (pythagorean-triple/alt n)
(for*/first ([a (in-range 1 (ceiling (/ n 3)))]
[b (in-range (add1 a) (ceiling (/ (- n a) 2)))]
[c (in-value (- n a b))]
#:when (and (< b c)
(right-triangle? a b c)))
(list a b c)))
; by Eli on the mailing list, modified by Rodolfo
(define (pythagorean-triple/alt2 n)
(for*/or ([a (in-range 1 n)]
[b (in-range (add1 a) n)]) ; start from `a+1' instead of `a'.
(define c (- n a b))
(and (< b c) (right-triangle? a b c) (list a b c)))) ; added `(< b c)'
check.
And counted how many times they call right-triangle -- the same number of
times.
Indeed your (first) solution with let loops seems slightly faster, but not
by a significant margin in my experiments.
I didn't take the time to analyze it much, but looking at the code
expansion using the Macro Stepper suggested that the for macros generate a
lot more code to be executed than the nested lets.
[]'s
Rodolfo Carvalho
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