Dan Weston westondan at imageworks.com
Wed Dec 12 16:01:52 EST 2007

```Dan Weston wrote:
> scanl above is not strict in its second argument. The data dependencies
> cause the strictness. Cf:
>
> 1

The first claim is of course false, nore would the example show it anyway.

scanl is not strict in its third argument (I forgot about the initial
value as the second argument):

Prelude Data.List> let z = [1,4,undefined,8,9] in scanl (\x y -> 5) 8 z
[8,5,5,5,5,5]

It is the data dependence in the first argument of scanl that would make
the above strict:

Prelude Data.List> let z = [1,4,undefined,8,9] in scanl (+) 8 z
[8,9,13,*** Exception: Prelude.undefined

Also note that it is better not to introduce the / operator in your
test, as it fails with large numbers. Multiply both sides by the
denominator before the comparison and leave everything as Num a instead
of Floating a. You can do the division at the end.

> Thomas Hartman wrote:
>>
>>
>>  >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.
>
> Not true:
>
> Prelude Data.List> let f a = (\(a,b,c)->c) . head . dropWhile (\(s,n,_)
> -> s <=n*a) . scanl (\(s,n,_) x ->(s+x,n+1,x)) (0,0,0) in f (10^5) [1,3..]
> 200001
>
>
>> 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.
>
> scanl above is not strict in its second argument. The data dependencies
> cause the strictness. Cf:
>
> 1
>
>> 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
>>
>
>
> You have
>
> Prelude Control.Arrow Data.List>
>   let avg5 = uncurry (/) . foldl' (\(s,n) x -> (s + x,n + 1)) (0,0)
>    in avg5 [1..10000000]
> *** Exception: stack overflow
> -- This fails in 100 sec
>
> Try this. It is not foldl' that needs to be strict, but the function
> folded:
>
> Prelude Data.List> let avg5 = uncurry (/) . foldl' (\(!s,!n) x -> (s +
> x,n + 1)) (0,0) in avg5 [1..10000000]
>
> You will need -fbang-patterns for this (there are other ways to do this
>
>>
>>
>> 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
>>     Re: [Haskell-cafe] eager/strict eval katas
>>
>>
>>
>>
>>
>>
>>
>>
>> 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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