Armando, thanks for a plausible explanation. Here is what happened
after I made the suggested change:
user=> (prime-factors 600851475143)
n: 600851475143 , f: 71
n: 8462696833 , f: 839
n: 10086647 , f: 1471
n: 6857 , f: 6857
#<CompilerException java.lang.ClassCastException: java.lang.Integer
cannot be cast to clojure.lang.IFn (REPL:91)>
I guess this something subtle with unordered vs. sorted-set as
implicit factors accumulator. May be that is why the author of the
solution had to use unordered set followed by (apply max ...).
On Jan 13, 11:23 am, Armando Blancas <armando_blan...@yahoo.com>
wrote:
> 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.
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