I think I would like to make another note: when we talk about the complexity of a function, we are talking about the time taken to completely evaluate the result. Otherwise any expression in haskell will be O(1), since it just creates a thunk.
And then the user of the program is to be blamed for running the program, since that is what caused evaluation of those thunks :)


Abhay

2008/5/31 Martin Geisler <mg@daimi.au.dk>:
Tillmann Rendel <rendel@daimi.au.dk> 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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