[racket] making Racket code more idiomatic
Blindly refactoring the code, I'd use `for/fold' and add a `lst'
accumulator to `loop':
(define (euler29c)
; calculate 99^2 - duplicates
(- (sqr 99)
(for/sum ([d '(2 3 5 6 7 10)])
(let loop ([lst '()] [exp 1])
(if (> (expt d exp) 100)
(- (length lst) (length (remove-duplicates lst)))
(loop (for/fold ([lst lst]) ([i (in-range 2 101)])
(cons (* i exp) lst))
(add1 exp)))))))
At Tue, 17 Apr 2012 09:45:50 -0700, Joe Gilray wrote:
> Hi,
>
> To continue our conversation about creating idiomatic Racket code, here is
> some code I wrote last night to solve projecteuler.net problem #29:
>
> (define (euler29a)
> ; calculate 99^2 - duplicates
> (- (sqr 99)
> (for/sum ([d '(2 3 5 6 7 10)])
> (let ([lst '()])
> (let loop ([exp 1])
> (if (> (expt d exp) 100) (- (length lst) (length
> (remove-duplicates lst)))
> (begin
> (for ([i (in-range 2 101)]) (set! lst (cons (* i
> exp) lst)))
> (loop (add1 exp)))))))))
>
> It's fast (it avoids calculating a bunch of huge numbers), it gives the
> correct answer, so what's not to love?!
>
> Well, it starts off OK, but my eye stumbles over the following:
>
> 1) predeclaring lst and accessing it twice, related to each other
> 2) ugly single parameter named-let loop
> 3) ugly "begin" - not a big deal, but I just dislike when having to use
> begin
> 4) use of set!
>
> Here is a quick rewrite:
>
> (define (euler29b)
> ; calculate 99^2 - duplicates
> (- (sqr 99)
> (for/sum ([d '(2 3 5 6 7 10)])
> (let ([lst '()])
> (do ([exp 1 (add1 exp)])
> ((> (expt d exp) 100) (- (length lst) (length
> (remove-duplicates lst))))
> (for ([i (in-range 2 101)]) (set! lst (cons (* i exp)
> lst))))))))
>
> It solves #2 and #3 above, but it is still fundamentally clunky.
>
> Can someone help and teach us all some tricks? My instincts say it should
> be possible to use append-map, for/list and/or foldl to build a list of the
> duplicates then simply count them in the for/sum loop, but am still unable
> to do it.
>
> Thanks,
> -Joe
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