Thomas Hartman thomas.hartman at db.com
Wed Dec 12 15:09:57 EST 2007

```>Note that 1 + ··· + n = n * (n+1) / 2, so the average of [1..n] is (n+1)
/ 2

fair enough.

But I believe  if I restate the problem  so that you need to find the
average of an arbitrary list, your clever trick doesn't work and we need
eager eval or we blow the stack.

Also... on second thought, I actually solved a slightly different problem
than what I originally said:  the problem of detecting when the moving
average of an increasing list is greater than 10^6; but my solution
doesn't give the index of the list element that bumped the list over the
average. However I suspect my code could be tweaked to do that (still
playing around with it):

Also I actually used a strict scan not a strict fold and... ach, oh well.

As you see I wrote a customized version of foldl' that is strict on the
tuple for this to work. I don't think this is necessarily faster than what
you did  (haven't quite grokked your use of unfold), but it does have the
nice property of doing everything in one one fold step (or one scan step I
guess, but isn't a scan

t.

t1 = average_greater_than (10^7) [1..]

average_greater_than max xs = find (>max) \$ averages xs

averages = map fst . myscanl' lAccumAvg (0,0)
average = fst . myfoldl' lAccumAvg (0,0)
lAccumAvg (!avg,!n) r = ( (avg*n/n1) + (r/n1),(n1))
where n1 = n+1

myfoldl' f (!l,!r) [] = (l,r)
myfoldl' f (!l,!r) (x:xs) = ( myfoldl' f q xs )
where q = (l,r) `f` x

myscanl f z []  = z : []
myscanl f z (x:xs) =  z : myscanl f (f z x) xs

myscanl' f (!l,!r) []  = (l,r) : []
myscanl' f (!l,!r) (x:xs) =  (l,r) : myscanl' f q xs
where q = (l,r) `f` x

"Felipe Lessa" <felipe.lessa at gmail.com>
12/12/2007 02:24 PM

To
Thomas Hartman/ext/dbcom at DBAmericas
cc
Subject

On Dec 12, 2007 2:31 PM, Thomas Hartman <thomas.hartman at db.com> wrote:
> exercise 2) find the first integer such that average of [1..n] is >
[10^6]
>   (solution involves building an accum list of (average,listLength)
tuples.
> again you can't do a naive fold due to stack overflow, but in this case
even
> strict foldl' from data.list isn't "strict enough", I had to define my
own
> custom fold to be strict on the tuples.)

What is wrong with

Prelude> snd . head \$ dropWhile ((< 10^6) . fst) [((n+1) / 2, n) | n <-
[1..]]
1999999.0

Note that 1 + ··· + n = n * (n+1) / 2, so the average of [1..n] is
(n+1) / 2. The naive

Prelude Data.List> let avg xs = foldl' (+) 0 xs / (fromIntegral \$ length
xs)
Prelude Data.List> snd . head \$ dropWhile ((< 10^6) . fst) [(avg
[1..n], n) | n <- [1..]]

works for me as well, only terribly slower (of course). Note that I
used foldl' for sum assuming the exercise 1 was already done =). How
did you solve this problem with a fold? I see you can use unfoldr:

Prelude Data.List> last \$ unfoldr (\(x,s,k) -> if s >= k then Nothing
else Just (x, (x+1,s+x,k+10^6)))  (2,1,10^6)

I'm thinking about a way of folding [1..], but this can't be a left
fold (or else it would never stop), nor can it be a right fold (or
else we wouldn't get the sums already done). What am I missing?

Cheers,

--
Felipe.

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