Adrian Neumann schrieb:
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?
Maybe you should read the euqation from the other direction. Then the h
becomes "fused out" and is called only once instead of many times. But
you can only do this if you can factor out h from p', f' and g'.
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))
not very much. I kinda see why it works after I "unfold" some more, but
I can't really prove it. I suspect I need some technique I haven't
learned yet. I've heard about fixpoint induction, that looks promising,
but Google knows very little about it.
I hope somebody can give me some hints.
Regards,
Adrian
Regards,
Martin.
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