Here's my awful terrible code which is a direct translation from a
java version I wrote and needs a fast functional sieve: I prefer
Cristophe Grande's.
http://clj-me.cgrand.net/index.php?s=Everybody%20loves%20the%20Sieve%20of%20Eratosthenes
The one in contrib is pretty good as well.
(letfn [(n
I looked through some of my Project Euler solutions and found this
version. It is essentially the same as Uncle Bob's, but to my eye it
is a bit easier to read.
(defn least-nontrivial-divisor [n];; integer n > 1
(loop [k 2]
(cond
(zero? (rem n k)) k ;; k divides n,
I focused more on idiom than correctness (obviously). Took a risk with not
much effort except to see if I could do any better with a toy problem. The
answer was no, Uncle Bob's is "idiomatic" modulo some minor things.
On Sat, Jun 12, 2010 at 12:59 PM, Steve Purcell wrote:
> On 12 Jun 2010, at 16
On 12 Jun 2010, at 16:18, Russell Christopher wrote:
> You're right. Hope I haven't offended with the fail, I thought I had tested
> it - by iterating over a range and comparing it to Uncle Bob's but obviously
> I didn't do that right and then realized that factorization is likely not
> O(n) an
You're right. Hope I haven't offended with the fail, I thought I had tested
it - by iterating over a range and comparing it to Uncle Bob's but obviously
I didn't do that right and then realized that factorization is likely not
O(n) anyway. I'll probably take more time next time.
Regards,
Russell
On 11 Jun 2010, at 20:35, Russell Christopher wrote:
> didn't need the assoc in my previous try
>
> (defn of [n]
> (letfn [(f [res k]
> (if (= 0 (rem (:n res) k))
>{:n (/ (:n res) k) :fs (conj (:fs res) k)}
>res))]
> (:fs (reduce f {:n n :fs
didn't need the assoc in my previous try
(defn of [n]
(letfn [(f [res k]
(if (= 0 (rem (:n res) k))
{:n (/ (:n res) k) :fs (conj (:fs res) k)}
res))]
(:fs (reduce f {:n n :fs []} (range 2 n)
On Fri, Jun 11, 2010 at 3:15 PM, russellc wrote:
Not sure it's better than Uncle Bobs version but it seems a little
more idiomatic?
(defn of [n]
(letfn [(f [res k]
(if (= 0 (rem (:n res) k))
(assoc (assoc res :n (quot (:n res) k)) :fs (conj (:fs
res) k))
res))]
(:fs (reduce f {:n n :fs []} (r
