On 9/8/10 4:57 AM, Patrick LeBoutillier wrote:
Hi all,
I've been reading the book "To Mock A Mockingbird" by Robert Smullyan recently and I don't get the part about the birds at all. I think I understand that a bird is a function, but what would it's type be in Haskell?
It says that birds hear a call and answer by another call, so I thought String -> String or something like that. But if you have a mockingbird that can answer what a bird would answer to itself, that means that it's input must contain the other bird... Do all the birds have the same type?
Thanks,
Patrick
Patrick, It's a cool book, well worth reading. A bird is a function of a single argument, which must be another bird, and it returns yet another bird. So the only kind of data are functions of a single argument which return the same kinds of functions. This is called "combinatory logic" (CL), and is equivalent to what's called untyped lambda calculus (LC). CL terms can be converted to LC terms and vice-versa. Haskell itself is based on a system similar to LC (but much richer), and not all of the terms in CL can be written in Haskell. You can play with them in ghci: ghci> let i x = x ghci> :t i i :: forall t: t -> t ghci> let k x y = x ghci> :t k k :: forall t t1. t -> t1 -> t ghci> let s x y z = (x z) (y z) ghci> :t s s :: forall t t1 t2. (t -> t1 -> t2) -> (t -> t1) -> t -> t2 s, k, and i are three basic combinators. In fact, you can get i from s and k: ghci> let i2 = s k k ghci> :t i2 i2 :: forall t. t -> t i and i2 are the identity function (called "id" in Haskell). k is called "const" in Haskell. s is a generalized function application function. However, some simple combinations of s, k, and i can't be typed in Haskell: ghci> :t s i i <interactive>:1:4: Occurs check: cannot construct the infinite type: t1 = t1 -> t2 Probable cause: `i' is applied to too few arguments In the second argument of `s', namely `i' In the expression: s i i That's because (s i i) is equivalent to this function: ghci> let m z = z z This is Smullyan's Mockingbird function. Haskell doesn't allow functions with so-called "infinite types" like this, even though they can be useful (Haskell provides other ways to get what infinite types would give you). There is a long, fascinating road ahead of you if you decide to continue studying this material :-) Mike