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
http://thomashartman-learning.googlecode.com/svn/trunk/haskell/lazy-n-strict...
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"
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. --- This e-mail may contain confidential and/or privileged information. If you are not the intended recipient (or have received this e-mail in error) please notify the sender immediately and destroy this e-mail. Any unauthorized copying, disclosure or distribution of the material in this e-mail is strictly forbidden.