Subtraction of church numerals in haskell

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情歌与酒
情歌与酒 2020-12-29 06:06

I\'m attempting to implement church numerals in Haskell, but I\'ve hit a minor problem. Haskell complains of an infinite type with

Occurs check: cannot construct th

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  • 2020-12-29 06:25

    This definition of predChurch doesn't work in simply typed lambda calculus, only in the untyped version. You can find a version of predChurch that works in Haskell here.

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  • 2020-12-29 06:30

    I have encountered the same problem. And I solved it without adding type signature.

    Here's the solution, with cons car copied from SICP.

    cons x y = \m -> m x y
    car z = z (\p q -> p)
    cdr z = z (\p q -> q)
    
    next z = cons (cdr z) (succ (cdr z))
    pred n = car $ n next (cons undefined zero)
    
    sub m n = n pred m
    

    You can find full source here.

    I'm really amazed after writing sub m n = n pred m, and load it in ghci without type error!

    Haskell code is so concise! :-)

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  • 2020-12-29 06:43

    The problem is that predChurch is too polymorphic to be correctly inferred by Hindley-Milner type inference. For example, it is tempting to write:

    predChurch :: Church a -> Church a
    predChurch = \n -> \f -> \x -> n (\g -> \h -> h (g f)) (\u -> x) (\u -> u)
    

    but this type is not correct. A Church a takes as its first argument an a -> a, but you are passing n a two argument function, clearly a type error.

    The problem is that Church a does not correctly characterize a Church numeral. A Church numeral simply represents a number -- what on earth could that type parameter mean? For example:

    foo :: Church Int
    foo f x = f x `mod` 42
    

    That typechecks, but foo is most certainly not a Church numeral. We need to restrict the type. Church numerals need to work for any a, not just a specific a. The correct definition is:

    type Church = forall a. (a -> a) -> (a -> a)
    

    You need to have {-# LANGUAGE RankNTypes #-} at the top of the file to enable types like this.

    Now we can give the type signature we expect:

    predChurch :: Church -> Church
    -- same as before
    

    You must give a type signature here because higher-rank types are not inferrable by Hindley-Milner.

    However, when we go to implement subChurch another problem arises:

    Couldn't match expected type `Church'
           against inferred type `(a -> a) -> a -> a'
    

    I am not 100% sure why this happens, I think the forall is being too liberally unfolded by the typechecker. It doesn't surprise me though; higher rank types can be a bit brittle because of the difficulties they present to a compiler. Besides, we shouldn't be using a type for an abstraction, we should be using a newtype (which gives us more flexibility in definition, helps the compiler with typechecking, and marks the places where we use the implementation of the abstraction):

    newtype Church = Church { unChurch :: forall a. (a -> a) -> (a -> a) }
    

    And we have to modify predChurch to roll and unroll as necessary:

    predChurch = \n -> Church $ 
        \f -> \x -> unChurch n (\g -> \h -> h (g f)) (\u -> x) (\u -> u)
    

    Same with subChurch:

    subChurch = \m -> \n -> unChurch n predChurch m
    

    But we don't need type signatures anymore -- there is enough information in the roll/unroll to infer types again.

    I always recommend newtypes when creating a new abstraction. Regular type synonyms are pretty rare in my code.

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