Michael Mossey wrote:
In S. Thompson's book, problem 9.13 asks us to define 'init' in terms of foldr. I was baffled at first because I didn't see a natural way to do this. It would look something like
init xs = foldr f initialValue xs
where f would cons on each character except the rightmost.
f <when passed rightmost char> b = [] f <when passed any other char a> b = a : b
How does f "know" when it is passed the first character? initialValue has to signal this somehow. On #haskell, one person suggested doing it with some post-processing:
init xs = snd $ foldr f (True,[]) xs where f _ (True,_) = (False,[]) f a (False,b) = (False,a:b)
I had an idea. If the initial value is the entire list, then its length can function as the "signal" that we are dealing with the rightmost char. This requires no post-processing:
init xs = foldr f xs xs where f a b | length b == length xs = [] | otherwise = a:b
These seem contrived. I wonder if there is a more natural solution that Thompson had in mind. Any comments?
It is best to see foldr f b as an operation that takes a list x0 : x1 : x2 : ... : [] and replaces every (:) with f and the [] with b : x0 `f` x1 `f` x2 `f` ... `f` b See also http://en.wikipedia.org/wiki/Fold_(higher-order_function) for Cale's nice pictures. It is then clear that we have to choose b to signal the end of the list. Furthermore, b should be the same as init [] . Unfortunately, this expression is a run-time error, but this is a fault of the type signature init :: [a] -> [a] which should really be init' :: [a] -> Maybe [a] to make it clear that some lists like the empty one simply don't have an initial segment. And this version has a natural implementation in terms of foldr : init' = foldr f Nothing where f _ Nothing = Just [] f x (Just xs) = Just (x:xs) Of course, we need some post-processing to obtain the original init from this, but I think that it's very natural. Regards, apfelmus -- http://apfelmus.nfshost.com