Saturday, 8 August 2009

Lambda Calculus Normal Order Reducer

// This implementation support evaluation of anonymous recursion through Y Combinator (ex. Y Factorial)

type exp = | Var of string
| Lambda of string * exp
| Apply of exp * exp

let rec subst x v a =
match a with
| Var y ->
x = y then v else a
| Lambda(y, a') ->
x = y then a else Lambda(y, subst x v a')
| Apply(a', a'') ->
Apply(subst x v a', subst x v a'')

let rec reduce e =
let rec reduce' e =
match e with
| Var _ -> e
| Lambda (s, e') -> Lambda(s, reduce' e')
| Apply(e1, e2) ->
e1 with
| Lambda(s, e3) -> subst s e2 e3
| _ -> Apply(reduce' e1, reduce' e2)
reduce' e

let rec loop f x =
let x' = f x
if x = x' then x' else loop f x'

let normalOrderReducer = loop reduce


devdevowicz said...

Now how can I reduce the lambda expression using this feature in F# Interactvie?

holoed said...

1) Paste the reducer code in F# Interactive
2) Type the following (This example shows that: (\x.x) (\y.y) reduces to (\y.y))

> normalOrderReducer (Apply(Lambda("x", Var "x"), Lambda("y", Var "y")));;
val it : exp = Lambda ("y",Var "y")

devdevowicy said...

thanks man.

When I want to reduce expression with number f.e. ( (/x . x) 3 )
How do this with F#?

holoed said...

Numbers in the lambda calculus can be encoded using Church Encoding (http://
3 ≡ \f.\x. f (f (f x))

f# beginner said...

hi, i'm trying to bulid applicative order reducer. I'm thinking i've to change a Apply part in reduce function, right?