R J wrote:
2. I believe that the reverse implementation--namely, implementing
foldr in terms of foldl--is impossible. What's the proof of that?
As others have said, foldr in terms of foldl is impossible when infinite
lists are taken into account. For finite lists it's easy though:
(\f z
On Wed, 11 Mar 2009, R J wrote:
foldl and foldr are defined as follows:
foldr :: (a - b - b) - b - [a] - b
foldr f e [] = e
foldr f e (x : xs) = f x (foldr f e xs)
foldl :: (b - a - b) - b - [a] - b
foldl f e [] = e
foldl f e (x
foldl and foldr are defined as follows:
foldr:: (a - b - b) - b - [a] - b
foldr f e [] = e
foldr f e (x : xs) = f x (foldr f e xs)
foldl:: (b - a - b) - b - [a] - b
foldl f e [] = e
foldl f e (x : xs) = foldl f (f e x) xs
1.
Read this excellent paper:
http://www.cs.nott.ac.uk/~gmh/fold.pdf
Am 11.03.2009 um 19:24 schrieb R J:
foldl and foldr are defined as follows:
foldr:: (a - b - b) - b - [a] - b
foldr f e [] = e
foldr f e (x : xs) = f x (foldr f e xs)
foldl
2009/3/11 R J rj248...@hotmail.com:
2. I believe that the reverse implementation--namely, implementing foldr in
terms of foldl--is impossible. What's the proof of that?
That's correct. Consider their behaviour on infinite lists.
--Max
___
On Wednesday 11 March 2009 2:24:55 pm R J wrote:
foldl and foldr are defined as follows:
foldr:: (a - b - b) - b - [a] - b
foldr f e [] = e
foldr f e (x : xs) = f x (foldr f e xs)
foldl:: (b - a - b) - b - [a] - b
foldl f e [] =
Am Mittwoch, 11. März 2009 19:24 schrieb R J:
foldl and foldr are defined as follows:
foldr:: (a - b - b) - b - [a] - b
foldr f e [] = e
foldr f e (x : xs) = f x (foldr f e xs)
foldl:: (b - a - b) - b - [a] - b
foldl f e [] = e
2009/3/11 R J rj248...@hotmail.com:
3. Any advice on how, aside from tons of practice, to develop the intuition
for rapidly seeing solutions to questions like these would be much
appreciated. The difficulty a newbie faces in answering seemingly simple
questions like these is quite
