Tillmann Rendel <[EMAIL PROTECTED]> writes: Hi! (Cool, another guy from DAIMI on haskell-cafe!)
> Another (n - 1) reduction steps for the second ++ to go away.
>
> last ("o" ++ "l")
> A) ~> last ('o' : "" ++ "l"))
> L) ~> last ("" ++ "l")
> A) ~> last ("l")
> L) ~> 'l'
>
> And the third and fourth ++ go away with (n - 2) and (n - 3) reduction
> steps. Counting together, we had to use
>
> n + (n - 1) + (n - 2) + ... = n!
>
> reduction steps to get rid of the n calls to ++, which lies in O(n^2).
> Thats what we expected of course, since we know that each of the ++
> would need O(n) steps.
I really liked the excellent and very clear analysis! But the last
formula should be:
n + (n - 1) + (n - 2) + ... = n * (n+1) / 2
which is still of order n^2.
--
Martin Geisler
VIFF (Virtual Ideal Functionality Framework) brings easy and efficient
SMPC (Secure Multi-Party Computation) to Python. See: http://viff.dk/.
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