On 6 May 2009, at 19:27, Adrian Neumann wrote:
Hello,
I'm trying to prove the unfold fusion law, as given in the chapter "Origami Programming" in "The Fun of Programming". unfold is defined like this:
unfold p f g b = if p b then [] else (f b):unfold p f g (g b)
And the law states:
unfold p f g . h = unfold p' f' g' with p' = p.h f' = f.h h.g' = g.h
Foremost I don't really see why one would want to fuse h into the unfold. h is executed once, at the beginning and is never needed again. Can someone give me an example?
So, this is what I got so far:
unfold p f g.h = (\b -> if p b then [] else (f b): unfold p f g (g b).h = if p (h b) then [] else (f (h b)) : unfold p f g (g (h b)) = if p' b then [] else f' b: unfold p f g (h (g' b))
= if p' b then [] else f' b : (unfold p f g . h) g' b = if p' b then [] else f' b : unfold p' f' g' (g' b) -- NB! = unfoldr p' f' g' b This "proof" is actually biting itself on the tail; however, you can make it work, for example, like this: (A_n) take n ((unfold p f g . h) b) = take n (unfold p' f' g' b) Now, A_0 is obvious (take 0 whatever = []), and A_n follows from A_{n-1} by the previous argument. By induction, A_n holds for all n.