Re: [Haskell-cafe] Infinite types should be optionally allowed

Sorry for bringing back an ancient thread but I'd still like to understand this better. It is still not obvious to me why typechecking infinite types is so hard. Is determining type 'equivalence' the hard part? or is that a separate issue? I wrote a simple infinite type inferer and made an attempt at an algorithm that can answer your question. The algorithm works like this: Given two types *a* and *b*: unify *a* with * b*, if the resulting type is 'equivalent' to the original *a,* then *b* must be (I think) at least as general as *a*. To determine equivalence, I start with the head of both types (which are represented as graphs) and check to see if the constructors are the same. If they are then I set those two nodes 'equal' and recurse with the children. It's a 'destructive' algorithm that effectively 'zips' the two graphs together. It returns false if it encounters two constructors that are different. My current algorithm says that neither of the types you gave is strictly more general than the other, which I'm guessing is probably not true. I'm curious what the correct answer is and would appreciate someone pointing out the flaw in my reasoning/code :) My test code is on github here: https://github.com/jvranish/InfiniteTypes Also, is there a book you'd recommend that would explain this in further detail? Thanks, - Job On Mon, Feb 16, 2009 at 5:16 PM, Luke Palmer wrote:

On Sat, Feb 14, 2009 at 2:06 PM, Job Vranish wrote:

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I'm pretty sure that the problem is decidable, at least with haskell 98 types (other type extensions may complicate things a bit). It ends up being a graph unification algorithm. I've tried some simple algorithms and they seem to work.

What do you mean by "the inference engine is only half of the story"? From what I understand, the inference engine infers types via unification, if the types unify, then the unified types are the inferred types, if the types don't unify, then type check fails. Am I missing/misunderstanding something?

Sorry it took me so long to respond. It took a while to formulate this example.

Here are two (convoluted) functions, passed to the fixtypes inference engine:

Expr> y (b (c i) (c (b b (b c (c i))))) (fix b . (a -> b -> (a -> c -> d) -> d) -> c) -> c Expr> y (b (c i) (b (c (b b (b c (c i)))) (b (c i) k))) (fix c . ((a -> ((b -> c) -> d) -> (a -> d -> e) -> e) -> f) -> f)

These are somewhat complex types; sorry about that. But here's a challenge: is one of these types more general than the other? For example, if you wrote the first term and gave the second signature, should it typecheck? If you figure it out, can you give an algorithm for doing so?

I'm not going to say how I came up with these functions, because that would give away the answer :-)

Luke

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I almost think that the problem might be solvable by just generating the appropriate newtype whenever an infinite type shows up, and doing the wrapping/unwrapping behind the scenes. This would be a hacked up way to do it, but I think it would work.

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On Fri, Feb 13, 2009 at 4:04 PM, Luke Palmer wrote:

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On Fri, Feb 13, 2009 at 3:13 PM, Job Vranish

wrote:

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There are good reasons against allowing infinite types by default (mostly, that a lot of things type check that are normally not what we want). An old haskell cafe conversation on the topic is here:

http://www.nabble.com/There%27s-nothing-wrong-with-infinite-types!-td7713737...http://www.nabble.com/There%27s-nothing-wrong-with-infinite-types%21-td77137...

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However, I think infinite types should be allowed, but only with an explicit type signature. In other words, don't allow infinite types to be inferred, but if they are specified, let them pass. I think it would be very hard to shoot yourself in the foot this way.

Oops! I'm sorry, I completely misread the proposal. Or read it correctly, saw an undecidability hiding in there, and got carried away.

What you are proposing is called equi-recursive types, in contrast to

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more popular iso-recursive types (which Haskell uses). There are

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undecidable problems with equi-recursive types, but there are ways to

On Fri, Feb 13, 2009 at 6:09 PM, Luke Palmer wrote: the plentiful pull

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it off. The question is whether these ways play nicely with Haskell's type system.

But because of the fundamental computational problems associated, there needs to be a great deal of certainty that this is even possible before considering its language design implications.

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That inference engine seems to be a pretty little proof-of-concept, doesn't it? But it is sweeping some very important stuff under the

carpet.

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The proposal is to infer the type of a term, then check it against an annotation. Thus every program is well-typed, but it's the compiler's

job

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to check that it has the type the user intended. I like the idea.

But the inference engine is only half of the story. It does no type checking. Although checking is often viewed as the easier of the two problems, in this case it is not. A term has no normal form if and only if its type is equal to (forall a. a). You can see the problem here.

Luke

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Newtype is the standard solution to situations where you really need an infinite type, but in some cases this can be a big annoyance. Using newtype sacrifices data type abstraction and very useful type classes like Functor. You can use multiparameter type classes and functional dependencies to recover some of the lost abstraction, but then type checking becomes harder to reason about and the code gets way more ugly (If you doubt, let me know, I have some examples). Allowing infinite types would fix this.

I'm imagining a syntax something like this: someFunctionThatCreatesInfiniteType :: a -> b | b = [(a, b)]

Thoughts? Opinions? Am I missing anything obvious?

- Job _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe