Stefan O'Rear wrote:
On Sat, Oct 20, 2007 at 08:05:37PM +0100, Andrew Coppin wrote:
I want to construct a program that prints out something like this:
[\fx -> f(fx)] [\f -> [\x -> f(fx)]] [\f -> S[\x -> f][\x -> fx]] [\f -> S(Kf)[\x -> fx]] [\f -> S(Kf)f] S[\f -> S(Kf)][\f -> f] S(S[\f -> S][\f -> Kf])[\f -> f] S(S(KS)[\f -> Kf])[\f -> f] S(S(KS)K)[\f -> f] S(S(KS)K)I
I can quite happily construct a program which, given the first line, yields the last line. But getting it to print all the intermediate steps is harder. And, like I said, when something is "hard" in Haskell, it usually means you're doing it the wrong way... ;-)
Thought it sounded fun, so I did it:
data Term = Lam String Term | Term :$ Term | Var String
paren act = if act then \ a -> ('(':) . a . (')':) else id
ppr i (Lam s t) = paren (i > 0) $ (:) '\\' . (++) s . (++) ". " . ppr 0 t ppr i (a :$ b) = paren (i > 1) $ ppr 1 a . (:) ' ' . ppr 2 b ppr i (Var s) = paren (i > 2) $ (++) s
reduce (Lam nm bd :$ obj) = Just (subst bd) where subst (Lam nm' bd') = Lam nm' (if nm == nm' then bd' else subst bd') subst (a :$ b) = subst a :$ subst b subst (Var nm') = if nm == nm' then obj else Var nm' reduce (left :$ right) = fmap (:$ right) (reduce left) reduce other = Nothing
trail' ob = ppr 0 ob "\n" : maybe [] trail' (reduce ob) trail = concat . trail'
The important part is the co-recursive trail, which produces a value using (:) before calling itself. No monads necessary.
Unbelievable... I spend an entire day coding something, and somebody else manages to write a complete working solution in under 20 minutes. Heh. 8^) I feel it might take me another 20 minutes just to figure out how this works...