A literal set is a unordered hash-set. To get the factors in order
change #{f} for (sorted-set f).On Jan 13, 7:09 am, Vitaly Peressada <vit...@ufairsoft.com> wrote: > The following solution by <b>mtgred</b> for <a href="http://clojure- > euler.wikispaces.com/">Project Euler Clojure</a> problem 003 uses > implicit recursion. > > <pre> > (use '[clojure.contrib.lazy-seqs :only (primes)]) > (defn prime-factors [n] > (let [f (some #(if (= 0 (rem n %)) %) primes)] > (if (= f n) #{f} (conj (prime-factors (/ n f)) f)))) > (apply max (prime-factors 600851475143)) > </pre> > > Here is above with added println > > (defn prime-factors [n] > (let [f (some #(if (= 0 (rem n %)) %) primes)] > (println "n:" n ", f:" f) > (if (= f n) > #{f} > (conj (prime-factors (/ n f)) f)))) > > Which produces > > n: 600851475143 , f: 71 > n: 8462696833 , f: 839 > n: 10086647 , f: 1471 > n: 6857 , f: 6857 > #{71 839 6857 1471} > > Can anybody explain why 6857 comes 3rd? I would expect to be the last. -- You received this message because you are subscribed to the Google Groups "Clojure" group. To post to this group, send email to clojure@googlegroups.com Note that posts from new members are moderated - please be patient with your first post. To unsubscribe from this group, send email to clojure+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/clojure?hl=en
