Here's a solution based on for/fold that hasn't been suggested
yet. Runs in O(n) time.
(let-values ([(results last-sum)
(for/fold ([sofar '()]
[value 1])
([i (in-range 2 22)])
(values (cons value sofar)
(+ value i)))])
(reverse results))
2 hours ago, Joe Gilray wrote:
> Hi,
>
> I'm trying to come up with the most idiomatic way to generate triangle
> numbers that is also fast:
>
> (for/list ([i (in-range 2 22)]) (for/sum ([j (in-range i)]) j))
>
> works but is slow because it regenerates the sums each time
>
> (define s 0) (for/list ([i (in-range 1 21)]) (set! s (+ s i)) s)
>
> is fast, but ugly.
>
> How do I keep the sum in an idiomatic fashion?
>
> Thanks,
> -Joe
--
Take care,
_mike
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