I have been going through the excellent CIS 194 course when I got stuck on Part 5 of Homework 6. It revolves around implementing the ruler function without any divisibility testing.
I found that it is possible to build the ruler function by continuously interspersing an accumulator with values from an infinite list.
nats = [0,1,2,3,..]
[3]
[2,3,2]
[1,2,1,3,1,2,1]
[0,1,0,2,0,1,0,3,0,1,0,2,0]
Then I tried implementing this algorithm for Stream
datatype which is a list without nil
data Stream a = Cons a (Stream a)
streamToList :: Stream a -> [a]
streamToList (Cons x xs) = x : streamToList xs
instance Show a => Show (Stream a) where
show = show . take 20 . streamToList
streamFromSeed :: (a -> a) -> a -> Stream a
streamFromSeed f x = Cons x (streamFromSeed f (f x))
nats :: Stream Integer
nats = streamFromSeed succ 0
interleave x (Cons y ys) = Cons x (Cons y (interleave x ys))
foldStream f (Cons x xs) = f x (foldStream f xs)
ruler = foldStream interleave nats
As expected, I got stackoverflow error since I was trying to fold from the right. However, I was surprised to see the same algorithm work for normal infinite lists.
import Data.List
interleave x list = [x] ++ (intersperse x list) ++ [x]
ruler = take 20 (foldr interleave [] [0..])
What am I missing? Why one implementation works while the other doesn't?
Your interleave
is insufficiently lazy. The magic thing that right folds must do to work on infinite structures is to not inspect the result of the folded value too closely before they do the first bit of computation. So:
interleave x stream = Cons x $ case stream of
Cons y ys -> Cons y (interleave x ys)
This produces Cons x _
before inspecting stream
; in contrast, your version requires stream
to be evaluated a bit before it can pass to the right hand side of the equation, which essentially forces the entire fold to happen before any constructor gets produced.
You can also see this in your list version of interleave
:
interleave x list = [x] ++ intersperse x list ++ [x]
The first element of the returned list (x
) is known before intersperse
starts pattern matching on list
.