Gorgeous! Thanks Justin.<div><br></div><div>To speed it up, change the second "101" to "first"</div><div><br></div><div>To speed it up even a little more, you could do this, but it not quite as pretty:</div> <div><br></div><div><div>(define (euler4e)</div><div> (for*/fold ([greatest 0])</div><div> ([first (in-range 101 1000)]</div><div> [second (in-range first 1000)])</div><div> (let ([prod (* first second)])</div> <div> (if (palindromic? prod)</div><div> (max greatest prod)</div><div> greatest))))</div></div><div><br></div><div>Just out of curiosity, have you used for/fold with >1 accum? It seems very powerful.</div> <div><br></div><div>thanks again,</div><div>-Joe<br><br><div class="gmail_quote">On Sun, Mar 11, 2012 at 1:33 PM, Justin Zamora <span dir="ltr"><<a href="mailto:justin@zamora.com">justin@zamora.com</a>></span> wrote:<br> <blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">This is a great situation to use Racket's advanced list<br> comprehensions. There is no need to use set!. You can keep track of<br> the maximum as you loop. Here is my solution.<br> <br> ; Problem 4<br> ; Find the largest palindrome made from the product of two 3-digit numbers.<br> ; 3-digit numbers are the numbers 100-999<br> (define (pe-004)<br> (for*/fold ([greatest 0])<br> ([first (in-range 101 1000)]<br> [second (in-range 101 1000)]<br> #:when (palindrome? (* first second)))<br> (max greatest (* first second))))<br> <br> Justin<br> <div><div class="h5"><br> On Sun, Mar 11, 2012 at 4:12 PM, Joe Gilray <<a href="mailto:jgilray@gmail.com">jgilray@gmail.com</a>> wrote:<br> > Hi, I'm redoing the ProjectEuler problems to learn Racket (great fun!).<br> > Below are three solutions to PE#4. Which is most in the spirit of Racket?<br> > Of course, I'd love to see solutions that are more idiomatic too. Is there<br> > a better way that doesn't use "for" at all?<br> ><br> > Thanks!<br> > -Joe<br> ><br> > ; function that turns a positive integer into a list of digits (in reverse<br> > order)<br> > ; invoke as (listize 234511) or (set-count (list->set (listize 112342)))<br> > (define (listize num)<br> > (if (= num 0)<br> > '()<br> > (cons (remainder num 10) (listize (quotient num 10)))))<br> ><br> > ; function that returns #t if the passed number is palindromic<br> > ; invoke as (palindromic? n)<br> > (define (palindromic? n)<br> > (if (equal? (listize n) (reverse (listize n))) #t #f))<br> > )<br> ><br> > ; ProjectEuler problem #4<br> > (define (euler4a)<br> > (let ([max 100000])<br> > (for ([i (build-list 899 (lambda (x) (+ 100 x)))])<br> > (for ([j (build-list (- 999 i) (lambda (y) (+ i y)))])<br> > (let ([prod (* i j)])<br> > (when (palindromic? prod)<br> > (begin<br> > (when (> prod max) (set! max prod))<br> > (printf "~a * ~a = ~a~n" i j prod))))))<br> > (printf "Max palindromic product is ~a~n" max)))<br> ><br> > ; ProjectEuler problem #4 using for*<br> > (define (euler4b)<br> > (let ([max 100000])<br> > (for* ([i (build-list 899 (lambda (x) (+ 100 x)))]<br> > [j (build-list 899 (lambda (y) (+ 100 y)))]<br> > #:when (and (>= j i) (palindromic? (* i j))))<br> > (let ([prod (* i j)])<br> > (when (> prod max) (set! max prod))<br> > (printf "~a * ~a = ~a~n" i j prod)))<br> > (printf "Max palindromic product is ~a~n" max)))<br> ><br> > ; ProjectEuler problem #4 - a mix of 4a and 4b<br> > (define (euler4c)<br> > (let ([max 100000])<br> > (for* ([i (build-list 899 (lambda (x) (+ 100 x)))]<br> > [j (build-list (- 999 i) (lambda (y) (+ i y)))]<br> > #:when (palindromic? (* i j)))<br> > (let ([prod (* i j)])<br> > (when (> prod max) (set! max prod))<br> > (printf "~a * ~a = ~a~n" i j prod)))<br> > (printf "Max palindromic product is ~a~n" max)))<br> ><br> </div></div>> ____________________<br> > Racket Users list:<br> > <a href="http://lists.racket-lang.org/users" target="_blank">http://lists.racket-lang.org/users</a><br> ><br> </blockquote></div><br></div>