Re: [Haskell-cafe] Re: OT: Literature on translation of lambda calculus to combinators

Cool, Thanks :D also quickcheck says the two algorithms are equivalent :) On Fri, Jan 29, 2010 at 4:33 AM, Nick Smallbone wrote:

Job Vranish writes:

...

Ideally we'd like the type of convert to be something like: convert :: LambdaExpr -> SKIExpr but this breaks in several places, such as the nested converts in the RHS of the rule: convert (Lambda x (Lambda y e)) | occursFree x e = convert (Lambda x (convert (Lambda y e)))

A while ago I tried modifying the algorithm to be pure top-down so that it wouldn't have this problem, but I didn't have much luck.

Anybody know of a way to fix this?

The way to do it is, when you see an expression Lambda x e, first convert e to a combinatory expression (which will have x as a free variable, and will obviously have no lambdas). Then you don't need nested converts at all.

Not-really-tested code follows.

Nick

data Lambda = Var String | Apply Lambda Lambda | Lambda String Lambda deriving Show

data Combinatory = VarC String | ApplyC Combinatory Combinatory | S | K | I deriving Show

compile :: Lambda -> Combinatory compile (Var x) = VarC x compile (Apply t u) = ApplyC (compile t) (compile u) compile (Lambda x t) = lambda x (compile t)

lambda :: String -> Combinatory -> Combinatory lambda x t | x `notElem` vars t = ApplyC K t lambda x (VarC y) | x == y = I lambda x (ApplyC t u) = ApplyC (ApplyC S (lambda x t)) (lambda x u)

vars :: Combinatory -> [String] vars (VarC x) = [x] vars (ApplyC t u) = vars t ++ vars u vars _ = []

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