Hi Guys.<div> </div><div>Thanks for the education on for/fold and for*/fold.</div><div> </div><div>Cristian, your solution is very elegant, thanks, but unfortunately it does generate all the numbers so is relatively slow and won't scale too well with a and b. </div> <div> </div><div>-Joe <div class="gmail_quote">On Tue, Apr 17, 2012 at 11:01 AM, Cristian Esquivias <<a href="mailto:cristian.esquivias@gmail.com">cristian.esquivias@gmail.com</a>> wrote: <blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">I used Project Euler to try out new languages as well. Here was my attempt at Problem 29 for reference. (define (prob29 a b) (set-count (for*/fold ([nums (set)]) ([i (in-range 2 (add1 a))] [j (in-range 2 (add1 b))]) (values (set-add nums (expt i j)))))) - Cristian <div class="HOEnZb"><div class="h5"> On Tue, Apr 17, 2012 at 9:54 AM, Matthew Flatt <<a href="mailto:mflatt@cs.utah.edu">mflatt@cs.utah.edu</a>> wrote: > 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 <a href="http://projecteuler.net" target="_blank">projecteuler.net</a> 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 >> ____________________ >> Racket Users list: >> <a href="http://lists.racket-lang.org/users" target="_blank">http://lists.racket-lang.org/users</a> > ____________________ > Racket Users list: > <a href="http://lists.racket-lang.org/users" target="_blank">http://lists.racket-lang.org/users</a> </div></div></blockquote></div> </div>