On Dec 12, 2007 2:31 PM, Thomas Hartman
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 + ยทยทยท + 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.