if (++) were left associative ?

[David Feuer]

On Sun, Apr 07, 2002, Konst Sushenko wrote:
> Hello,
> 
> this is probably embarrassing, but I just realised that I do not
> understand why list concatenation operation takes quadratic time if it
> is associated from left to right (see e.g. 'The Haskell School of
> Expression' by Hudak, p. 335):
> 
>         cat1 = (++)
>         cat2 = (++)
> 
>         ( [1,2] `cat1` [3,4] ) `cat2` [5,6]
> 
> 
> My understanding is that cat2 is invoked before cat1, and as soon as the
> first element of cat1 is available, cat2 starts building its own result.
> cat2 does not need to wait until the whole list is emitted by cat1, does
> it? Where is quadratic time complexity?

Suppose we define

listseq [] x = x
listseq (a:b) x = listseq b x

l = [1..n]

inter = foldl (++) [] (map (:[]) foo)

final = listseq inter inter


Then inter will be the result of concatenating  [1],[2],[3],... from
left to right.  One meaningful question is: how long will it take to
calculate "final"?  This is (I believe) called the _shared cost_ of
inter (while inter does not actually do any significant work, it
produces a chain of suspensions that together do quite a bit.... I'm not
sure if the way they are chained still allows this to be considered the
shared cost, but I'm not sure).

1.  How long does it take to calculate final ?
Well, let's try it for n = 4:

inter = foldl (++) [] [[1],[2],[3],[4]]
      = foldl (++) ([]++[1]) [[2],[3],[4]] = foldl (++) [1] [[2],[3],[4]]
      = foldl (++) ([1]++[2]) [[3],[4]] = foldl (++) [1,2] [[3],[4]]
      = foldl (++) ([1,2]++[3]) [[4]] = foldl (++) [1,2,3] [[4]]
      = foldl (++) ([1,2,3]++[4]) [] = foldl (++) [1,2,3,4] []
      = [1,2,3,4]

Note that in successive iterations, the following are calculated:

[]++[1]
[1]++[2]
[1,2]++[3]
[1,2,3]++[4]

Since l1++l2 takes order length(l1), each concatenation takes more time
than the previous one (linearly), so the total time for all of them is
quadratic.

David Feuer