If you want to do this, you can do it simpler, without the loop and temp var:
(defn twice-composite? [n]
(->> (take-while #(< % n) prime-seq)
(every #(or (divides? (* 2 %) n)
(not (divides? % n))))
but this is not what you want. See the hints I sent you off the list.
On Wed, Nov 3, 2010 at 2:34 PM, Dan Kefford <dan_keff...@hotmail.com> wrote:
> One idea that I had was to inline the factoring logic into twice-
> composite. Who cares about the factors? We just want to know if there
> are two or not:
>
> (defn twice-composite? [n]
> (loop [ps prime-seq
> tmp n
> p-count 0]
> (if (< 2 p-count)
> false
> (if (= 1 tmp)
> (= 2 p-count)
> (let [p (first ps)]
> (if (divides? tmp p)
> (recur ps (quot tmp p) (inc p-count))
> (recur (rest ps) tmp p-count))))))
> )
>
> But this lead to even slower code.
>
> On Nov 3, 2:15 pm, Mark Engelberg <mark.engelb...@gmail.com> wrote:
>> Your Clojure implementation of your particular approach is reasonable,
>> but you need a cleverer approach. I'll send you a hint offline.
>
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Cheers,
Leif
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