I should re-test before post. This version of the main loop actually
terminates (for smaller numbers), but will still blow up:

(loop [ p (first perms) r (rest perms) s 0 c 0 ]
  (if (> c 8)
    s
    (if (every? #(%1 p) guards)
      (do
        (println p)
        (recur (first r) (rest r) (+ s (seq-to-val p)) (inc c)))
      (recur (first r) (rest r) s c))))

Regards,
Dirk

On Mon, Mar 9, 2009 at 4:55 PM, Dirk Vleugels <dirk.vleug...@gmail.com> wrote:
> Hi,
>
> $ time clojure euler43.clj
> (1 4 0 6 3 5 7 2 8 9)
> (1 4 3 0 9 5 2 8 6 7)
> (1 4 6 0 3 5 7 2 8 9)
> java.lang.OutOfMemoryError: Java heap space
> Dumping heap to java_pid7476.hprof ...
>
> The hprof file is 330mb in size, i can't read it with jhat (even if i
> give it 1.5G ....).
>
> I assume perms is the problem, but i dont know how to write this in a
> more memory efficient way - I'm not looking for a more efficient
> algorithm though....
>
> Thanx for any hint,
> Dirk
>
> (use 'clojure.contrib.combinatorics)
>
> (def perms (permutations (range 10)))
>
> (defn seq-to-val [ s ]
>    (reduce #(+ (* %1 10) %2)  s))
>
> (defn extract [ s i ]
>   (conj [] (nth s i) (nth s (inc i)) (nth s (inc (inc i)))))
>
> (defn f24 [ s ]
>   (= (mod (seq-to-val (extract s 1)) 2) 0))
>
> (defn f35 [ s ]
>   (= (mod (seq-to-val (extract s 2)) 3) 0))
>
> (defn f46 [ s ]
>   (= (mod (seq-to-val (extract s 3)) 5) 0))
>
> (defn f57 [ s ]
>   (= (mod (seq-to-val (extract s 4)) 7) 0))
>
> (defn f68 [ s ]
>   (= (mod (seq-to-val (extract s 5)) 11) 0))
>
> (defn f79 [ s ]
>   (= (mod (seq-to-val (extract s 6)) 13) 0))
>
> (defn f810 [ s ]
>   (= (mod (seq-to-val (extract s 7)) 17) 0))
>
> (def guards [ f24 f35 f46 f57 f68 f79 f810 ])
>
> (loop [ p (first perms) r (rest perms) sum 0 count 0 ]
>    (when (= count 10)
>        sum)
>    (if (every? #(%1 p) guards)
>        (do
>           (println p)
>           (recur (first r) (rest r) (+ sum (seq-to-val p)) (inc count)))
>       (recur (first r) (rest r) sum (inc count))))
>



-- 
Dirk Vleugels
2scale GmbH

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