Suppose I have
getXY :: X a -> Y a
setXY :: C a => X a -> Y a -> X a
Is it possible to write something lens-like that imposes C a only when setting? I'm interested in both Van Laarhoven and profunctor-style optics.
I can't say for certain, but with exactly the types you've given, I don't see how.
That said, are you being overly restrictive in your types? Could it be possible that setXY might be allowed to have the type:
setXY :: C b => X a -> Y b -> X b
In other words, the constraint C matters for the second argument and the output but not for the input? If so, then you can make the following lens:
lensXY :: C b => Lens (X a) (X b) (Y a) (Y b)
lensXT = lens getXY setXY
This is still a somewhat strange lens. For instance, say you have a value of type x :: X Int, but Int is not an instance of C. Then, you'll get an error if you try x ^. lensXY. Funny enough, this isn't a flaw with the lens so much as with ^., which has an overly restrictive type.
Now consider if we redefine ^. as follows:
(^.) :: s -> Lens s t a b -> a
s ^. l = getConst (l Const s)
Note that the function definition is exactly the same as the previous definition, but we've changed the type signature (it's more general for lenses but more constrained for everything else). Now, we can almost execute x ^. lensXY, but the problem now is that GHC doesn't know what types to use for t and b. Specifically, it needs a type that satisfies the constraint C. So, let's make one final assumption: that there is an instance C Bool. Then, we can write x ^. lensXY #Bool, and it type-checks and extracts the Y a value even though a is not an instance of C.
On the other hand, using %~ or similar still demands the class constraint.
So, I'm trying to write my own replacement for Prelude, and I have (^) implemented as such:
{-# LANGUAGE RebindableSyntax #-}
class Semigroup s where
infixl 7 *
(*) :: s -> s -> s
class (Semigroup m) => Monoid m where
one :: m
class (Ring a) => Numeric a where
fromIntegral :: (Integral i) => i -> a
fromFloating :: (Floating f) => f -> a
class (EuclideanDomain i, Numeric i, Enum i, Ord i) => Integral i where
toInteger :: i -> Integer
quot :: i -> i -> i
quot a b = let (q,r) = (quotRem a b) in q
rem :: i -> i -> i
rem a b = let (q,r) = (quotRem a b) in r
quotRem :: i -> i -> (i, i)
quotRem a b = let q = quot a b; r = rem a b in (q, r)
-- . . .
infixr 8 ^
(^) :: (Monoid m, Integral i) => m -> i -> m
(^) x i
| i == 0 = one
| True = let (d, m) = (divMod i 2)
rec = (x*x) ^ d in
if m == one then x*rec else rec
(Note that the Integral used here is one I defined, not the one in Prelude, although it is similar. Also, one is a polymorphic constant that's the identity under the monoidal operation.)
Numeric types are monoids, so I can try to do, say 2^3, but then the typechecker gives me:
*AlgebraicPrelude> 2^3
<interactive>:16:1: error:
* Could not deduce (Integral i0) arising from a use of `^'
from the context: Numeric m
bound by the inferred type of it :: Numeric m => m
at <interactive>:16:1-3
The type variable `i0' is ambiguous
These potential instances exist:
instance Integral Integer -- Defined at Numbers.hs:190:10
instance Integral Int -- Defined at Numbers.hs:207:10
* In the expression: 2 ^ 3
In an equation for `it': it = 2 ^ 3
<interactive>:16:3: error:
* Could not deduce (Numeric i0) arising from the literal `3'
from the context: Numeric m
bound by the inferred type of it :: Numeric m => m
at <interactive>:16:1-3
The type variable `i0' is ambiguous
These potential instances exist:
instance Numeric Integer -- Defined at Numbers.hs:294:10
instance Numeric Complex -- Defined at Numbers.hs:110:10
instance Numeric Rational -- Defined at Numbers.hs:306:10
...plus four others
(use -fprint-potential-instances to see them all)
* In the second argument of `(^)', namely `3'
In the expression: 2 ^ 3
In an equation for `it': it = 2 ^ 3
I get that this arises because Int and Integer are both Integral types, but then why is it that in normal Prelude I can do this just fine? :
Prelude> :t (2^)
(2^) :: (Num a, Integral b) => b -> a
Prelude> :t 3
3 :: Num p => p
Prelude> 2^3
8
Even though the signatures for partial application in mine look identical?
*AlgebraicPrelude> :t (2^)
(2^) :: (Numeric m, Integral i) => i -> m
*AlgebraicPrelude> :t 3
3 :: Numeric a => a
How would I make it so that 2^3 would in fact work, and thus give 8?
A Hindley-Milner type system doesn't really like having to default anything. In such a system, you want types to be either properly fixed (rigid, skolem) or properly polymorphic, but the concept of “this is, like, an integer... but if you prefer, I can also cast it to something else” as many other languages have doesn't really work out.
Consequently, Haskell sucks at defaulting. It doesn't have first-class support for that, only a pretty hacky ad-hoc, hard-coded mechanism which mainly deals with built-in number types, but fails at anything more involved.
You therefore should try to not rely on defaulting. My opinion is that the standard signature for ^ is unreasonable; a better signature would be
(^) :: Num a => a -> Int -> a
The Int is probably controversial – of course Integer would be safer in a sense; however, an exponent too big to fit in Int generally means the results will be totally off the scale anyway and couldn't feasibly be calculated by iterated multiplication; so this kind of expresses the intend pretty well. And it gives best performance for the extremely common situation where you just write x^2 or similar, which is something where you very definitely don't want to have to put an extra signature in the exponent.
In the rather fewer cases where you have a concrete e.g. Integer number and want to use it in the exponent, you can always shove in an explicit fromIntegral. That's not nice, but rather less of an inconvenience.
As a general rule, I try to avoid† any function-arguments that are more polymorphic than the results. Haskell's polymorphism works best “backwards”, i.e. the opposite way as in dynamic language: the caller requests what type the result should be, and the compiler figures out from this what the arguments should be. This works pretty much always, because as soon as the result is somehow used in the main program, the types in the whole computation have to be linked to a tree structure.
OTOH, inferring the type of the result is often problematic: arguments may be optional, may themselves be linked only to the result, or given as polymorphic constants like Haskell number literals. So, if i doesn't turn up in the result of ^, avoid letting in occur in the arguments either.
†“Avoid” doesn't mean I don't ever write them, I just don't do so unless there's a good reason.
I am trying to decipher the record syntax in haskell for newtype and my understanding breaks when there is a function inside newtype. Consider this simple example
newtype C a b = C { getC :: (a -> b) -> a }
As per my reasoning C is a type which accepts a function and a parameter in it's constructor.
so,
let d1 = C $ (2 *) 3
:t d1 also gives
d1 :: Num ((a -> b) -> a) => C a b
Again to check this I do :t getC d1, which shows this
getC d1 :: Num ((a -> b) -> a) => (a -> b) -> a
Why the error if I try getC d1? getC should return the function and it's parameter or at least apply the parameter.
I can't have newtype C a b = C { getC :: (a->b)->b } deriving (Show), because this won't make sense!
It's always good to emphasise that Haskell has two completely separate namespaces, the type language and the value language. In your case, there's
A type constructor C :: Type -> Type -> Type, which lives in the type language. It takes two types a, b (of kind Type) and maps them to a type C a b (also of kind Type)†.
A value constructor C :: ((a->b) -> a) -> C a b, which lives in the value language. It takes a function f (of type (a->b) -> a) and maps it to a value C f (of type C a b).
Perhaps it would be less confusing if you had
newtype CT a b = CV ((a->b) -> a)
but because for a newtype there is always exactly one value constructor (and exactly one type constructor) it makes sense to name them the same.
CV is a value constructor that accepts one function, full stop. That function will have signature (a->b) -> a, i.e. its argument is also a function, but as far as CT is concerned this doesn't really matter.
Really, it's kind of wrong that data and newtype declarations use a = symbol, because it doesn't mean the things on the left and right are “the same” – can't, because they don't even belong to the same language. There's an alternative syntax which expresses the relation better:
{-# LANGUAGE GADTs #-}
import Data.Kind
data CT :: Type -> Type -> Type where
CV :: ((a->b) -> a) -> CT a b
As for that value you tried to construct
let d1 = CV $ (\x->(2*x)) 3
here you did not pass “a function and a parameter” to CV. What you actually did‡ was, you applied the function \x->2*x to the value 3 (might as well have written 6) and passed that number to CV. But as I said, CV expects a function. What then happens is, GHC tries to interpret 6 as a function, which gives the bogus constraint Num ((a->b) -> a). What that means is: “if (a->b)->a is a number type, then...”. Of course it isn't a number type, so the rest doesn't make sense either.
†It may seem redundant to talk of “types of kind Type”. Actually, when talking about “types” we often mean “entities in the type-level language”. These have kinds (“type-level types”) of which Type (the kind of (lifted) value-level values) is the most prominent, but not the only one – you can also have type-level numbers and type-level functions – C is indeed one.Note that Type was historically written *, but this notation is deprecated because it's inconsistent (confusion with multiplication operator).
‡This is because $ has the lowest precedence, i.e. the expression CV $ (\x->(2*x)) 3 is actually parsed as CV ((\x->(2*x)) 3), or equivalently let y = 2*3 in CV y.
As per my reasoning C is a type which accepts a function and a parameter
How so? The constructor has only one argument.
Newtypes always have a single constructor with exactly one argument.
The type C, otoh, has two type parameters. But that has nothing to do with the number of arguments you can apply to the constructor.
So I understand the basic algebraic interpretation of types:
Either a b ~ a + b
(a, b) ~ a * b
a -> b ~ b^a
() ~ 1
Void ~ 0 -- from Data.Void
... and that these relations are true for concrete types, like Bool, as opposed to polymorphic types like a. I also know how to translate type signatures with polymorphic types into their concrete type representations by just translating the Church encoding according to the following isomorphism:
(forall r . (a -> r) -> r) ~ a
So if I have:
id :: forall a . a -> a
I know that it does not mean id ~ a^a, but it actually means:
id :: forall a . (() -> a) -> a
id ~ ()
~ 1
Similarly:
pair :: forall r . (a -> b -> r) -> r
pair ~ ((a, b) -> r) - > r
~ (a, b)
~ a * b
Which brings me to my question. What is the "algebraic" interpretation of this rule:
(forall r . (a -> r) -> r) ~ a
For every concrete type isomorphism I can point to an equivalent algebraic rule, such as:
(a, (b, c)) ~ ((a, b), c)
a * (b * c) = (a * b) * c
a -> (b -> c) ~ (a, b) -> c
(c^b)^a = c^(b * a)
But I don't understand the algebraic equality that is analogous to:
(forall r . (a -> r) -> r) ~ a
This is the famous Yoneda lemma for the identity functor.
Check this post for a readable introduction, and any category theory textbook for more.
Briefly, given f :: forall r. (a -> r) -> r you can apply f id to get an a, and conversely, given x :: a you can take ($x) to get forall r. (a -> r) -> r.
These operations are mutually inverse. Proof:
Obviously ($x) id == x. I will show that
($(f id)) == f,
since functions are equal when they are equal on all arguments, let's take x :: a -> r and show that
($(f id)) x == f x i.e.
x (f id) == f x.
Since f is polymorphic, it works as a natural transformation; this is the naturality diagram for f:
f_A
Hom(A, A) → A
(x.) ↓ ↓ x
Hom(A, R) → R
f_R
So x . f == f . (x.).
Plugging identity, (x . f) id == f x. QED
(Rewritten for clarity)
There seem to be two parts to your question. One is implied and is asking what the algebraic interpretation of forall is, and the other is asking about the cont/Yoneda transformation, which sdcvvc's answer already covered pretty well.
I'll try to address the algebraic interpretation of forall for you. You mention that A -> B is B^A but I'd like to take that a step further and expand it out to B * B * B * ... * B (|A| times). Although we do have exponentiation as a notation for repeated multiplication like that, there's a more flexible notation, ∏ (uppercase Pi) representing arbitrary indexed products. There are two components to a Pi: the range of values we want to multiply over, and the expression that we're multiplying out. For example, at the value level, you might express the factorial function as fact i = ∏ [1..i] (λx -> x).
Going back to the world of types, we can view the exponentiation operator in the A -> B ~ B^A correspondence as a Pi: B^A ~ ∏ A (λ_ -> B). This says that we're defining an A-ary product of Bs, such that the Bs cannot depend on the particular A we've chosen. Sure, it's equivalent to plain exponentiation, but it lets us move up to cases in which there is a dependence.
In the most general case, we get dependent types, like what you see in Agda or Coq: in Agda syntax, replicate : Bool -> ((n : Nat) -> Vec Bool n) is one possible application of a Pi type, which could be expressed more explicitly as replicate : Bool -> ∏ Nat (Vec Bool), or further as replicate : ∏ Bool (λ_ -> ∏ Nat (Vec Bool)).
Note that as you might expect from the underlying algebra, you can fuse both of the ∏s in the definition of replicate above into a single ∏ ranging over the cartesian product of the domains: ∏ Bool (\_ -> ∏ Nat (Vec Bool)) is equivalent to ∏ (Bool, Nat) (λ(_, n) -> Vec Bool n) just like it would be at the "value level". This is simply uncurrying from the perspective of type theory.
I do realize your question was about polymorphism, so I'll stop going on about dependent types, but they are relevant: forall in Haskell is roughly equivalent to a ∏ with a domain over the type (kind) of types, *. Indeed, the function-like behavior of polymorphism can be observed directly in GHC core, which types them as capital lambdas (Λ). As such, a polymorphic type like forall a. a -> a is actually just ∏ * (Λ a -> (a -> a)) (using the Λ notation now that we distinguish between types and values), which can be expanded out to the infinite product (Bool -> Bool, Int -> Int, () -> (), (Int -> Bool) -> (Int -> Bool), ...) for every possible type. Instantiation of the type variable is simply projecting out the suitable element from the *-ary product (or applying the type function).
Now, for the big piece I missed in my original version of this answer: parametricity. Parametricity can be described in several different ways, but none of the ones I know of (viewing types as relations, or (di)naturality in category theory) really has a very algebraic interpretation. For our purposes, though, it boils down to something fairly simple: you can't pattern-match on *. I know that GHC lets you do that at the type level with type families, but you can only cover a finite chunk of * when doing that, so there are necessarily always points at which your type family is undefined.
What this means, from the point of view of polymorphism, is that any type function F we write in ∏ * F must either be constant (i.e., completely ignore the type it was polymorphic over) or pass the type through unchanged. Thus, ∏ * (Λ _ -> B) is valid because it ignores its argument, and corresponds to forall a. B. The other case is something like ∏ * (Λ x -> Maybe x), which corresponds to forall a. Maybe a, which doesn't ignore the type argument, but only "passes it through". As such, a ∏ A that has an irrelevant domain A (such as when A = *) can be seen as more of an A-ary indexed intersection (picking the common elements across all instantiations of the index), rather than a product.
Crucially, at the value level, the rules of parametricity prevent any funny behavior that might suggest the types are larger than they really are. Because we don't have typecase, we can't construct a value of type forall a. B that does something different based on what a was instantiated to. Thus, although the type is technically a function * -> B, it is always a constant function, and is thus equivalent to a single value of B. Using the ∏ interpretation, it is indeed equivalent to an infinite *-ary product of Bs, but those B values must always be identical, so the infinite product is effectively as big as a single B.
Similarly, although ∏ * (Λ x -> (x -> x)) (a.k.a., forall a. a -> a) is technically equivalent to an infinite product of functions, none of those functions can inspect the type, so all are constrained to only return their input value and not do any funny business like (+1) : Int -> Int when instantiated to Int. Because there is only one (assuming a total language) function that can't inspect the type of its argument but must return a value of that same type, the infinite product is thus just as large as a single value.
Now, about your direct question on (forall r . (a -> r) -> r) ~ a. First, let's express your ~ operator more formally. It's really isomorphism, so we need two functions going back and forth, and an argument that they're inverses.
data Iso a b = Iso
{ to :: a -> b
, from :: b -> a
-- proof1 :: forall x. to (from x) == x
-- proof2 :: forall x. from (to x) == x
}
and now we express your original question in more formal terms. Your question amounts to constructing a term of the following (impredicative, so GHC has trouble with it, but we'll survive) type:
forall a. Iso (forall r. (a -> r) -> r) a
Which, using my earlier terminology, amounts to ∏ * (Λ a -> Iso (∏ * (Λ r -> ((a -> r) -> r))) a). Once again we have an infinite product that can't inspect its type argument. By handwaving, we can argue that the only possible values considering the parametricity rules (the other two proofs are respected automatically) for to and from are ($ id) and flip id.
If this feels unsatisfying, it's probably because the algebraic interpretation of forall didn't really add anything to the proof. It's really just plain old type theory, but I hope I was able to provide something that feels a little less categorical than the Yoneda form of it. It's worth noting that we don't actually need to use parametricity to write proof1 and proof2 above, though. Parametricity only enters the picture when we want to state that ($ id) and flip id are our only options for to and from (which we can't prove in Agda or Coq, for that reason).
To (attempt to) answer the actual question (which is less interesting than the answers to the broader issues raised), the question is ill formed because of a "type error"
Either ~ (+)
(,) ~ (*)
(->) b ~ flip (^)
() ~ 1
Void ~ 0
These all map types to integers, and type constructors to functions on naturals. In a sense, you have a functor from the category of types to the category of naturals. In the other direction, you "forget" stuff, since the types preserve algebraic structure while the naturals throw it away. I.e. given Either () () you can get a unique natural, but given that natural, you can get many types.
But this is different:
(forall r . (a -> r) -> r) ~ a
It maps a type to another type! It is not part of the above functor. It's just an isomorphism within the category of types. So let's give that a different symbol, <=>
Now we have
(forall r . (a -> r) -> r) <=> a
Now you note that we can not only send types to nats and arrows to arrows, but also some isomorphisms to other isomorphisms:
(a, (b, c)) <=> ((a, b), c) ~ a * (b * c) = (a * b) * c
But something subtle is going on here. In a sense, the latter isomorphism on pairs is true because the algebraic identity is true. This is to say that the "isomorphism" in the latter simply means that the two types are equivalent under the image of our functor to the nats.
The former isomorphism we need to prove directly, which is where we start to get to the underlying question -- is given our functor to the nats, what does forall r. map to? But the answer is that forall r. is neither a type, nor a meaningful arrow between types.
By introducing forall, we have moved away from first order types. There's no reason to expect that forall should fit in our above Functor, and indeed, it doesn't.
So we can explore, as others have above, why the isomorphism holds (which is itself very interesting) -- but in doing so we've abandoned the algebraic core of the question. A question which can be answered, I think, is, given the category of higher-order types and constructors as arrows between them, what is there meaningful Functor to?
Edit:
So now I have another approach which shows why adding polymorphism makes things go nuts. We start by asking a simpler question -- does a given polymorphic type have zero or more than zero inhabitants? This is the type inhabitation problem, and winds up being, via Curry-Howard, a problem in modified realizability, since it's the same thing as asking if a formula in some logic is realizable in an appropriate computational model. Now as that page explains, this is decidable in the simply typed lambda calculus but is PSPACE-complete. But once we move to anything more complicated, by adding polymorphism for example and going to System F, then it goes to undecidable!
So, if we can't decide if an arbitrary type is inhabited at all, then we clearly can't decide how many inhabitants it has!
It's an interesting question. I don't have a full answer, but this was too long for a comment.
The type signature (forall r. (a -> r) -> r) can be expressed as me saying
For any type r that you care to name, if you give me a function that takes a and produces an r, then I will give you back an r.
Now, this has to work for any type r, but it can be a specific type a. So the way for me to pull of this neat trick is to have an a sitting around somewhere, that I feed to the function (which produces an r for me) and then I hand that r back to you.
But if I have an a sitting around, I could give it to you:
If you give me a 1, I'll give you an a.
which corresponds to the type signature 1 -> a or simply a. By this informal argument we have
(forall r. (a -> r) -> r) ~ a
The next step would be to generate the corresponding algebraic expression, but I'm not clear on how the algebraic quantities interact with the universal quantification. We may need to wait for an expert!
A few links to the nLab:
Universal quantifier, corresponds to dependent product.
Existential quantifier, corresponds to dependent sum (dependent coproduct).
Thus, in settings of category theory:
Type | Modeled¹ as | In category
-------------------+---------------------------+-------------
Unit | Terminal object | CCC
Bottom | Initial object |
Record | Product |
Union | Sum (coproduct) |
Function | Exponential |
-------------------+---------------------------+-------------
Dependent product² | Right adjoint to pullback | LCCC
Dependent sum | Left adjoint to pullback |
¹) in appropriate category ─ CCC for total and non-polymorphic subset of Haskell (link), CPO for non-total traits of Haskell (link), LCCC for dependently typed languages.
²) forall quantification is a special case of dependent product:
∀(x :: *). y[x] ~ ∏(x : Set)y[x]
where Set is the universe of all small types.
In Haskell we are given the ability to combine constraints on types with a logical and.
Consider the following
type And (a :: Constraint) b = (a, b)
or more complicatedly
class (a, b) => And a b
instance (a, b) => And a b
I want to know how to logically or two constraints together in Haskell.
My closest attempt is this, but it doesn't quite work. In this attempt I reify type constraints with tags and than dereify them with implicit parameters.
data ROr a b where
L :: a => ROr a b
R :: b => ROr a b
type Or a b = (?choose :: ROr a b)
y :: Or (a ~ Integer) (Bool ~ Integer) => a
y = case ?choose of
L -> 4
x :: Integer
x = let ?choose = L in y
It almost works, but the user has to apply the final part, and the compiler should do that for me. As well, this case does not let one choose a third choice when both constraints are satisfied.
How can I logically or two constraints together?
I believe that there is no way to automatically pick an ROr a b; it would violate the open world assumption if, e.g. b was satisfied, but later a was satisfied as well; any conflict resolution rule would necessarily cause the addition of an instance to change the behaviour of existing code.
That is, picking R when b is satisfied but a is not breaks the open world assumption, because it involves deciding that an instance is not satisfied;1 even if you added a "both satisfied" constructor, you would be able to use it to decide whether an instance is not present (by seeing if you get an L or an R).
Therefore, I do not believe that such an or constraint is possible; if you can observe which instance you get, then you can create a program whose behaviour changes by adding an instance, and if you can't observe which instance you get, then it's pretty useless.
1 The difference between this and normal instance resolution, which can also fail, is that normally, the compiler cannot decide that a constraint is satisfied; here, you're asking the compiler to decide that the constraint cannot be satisfied. A subtle but important difference.
I came here to answer your question on the cafe. Not sure the q here is the same, but anyway ...
a type class with three parameters.
class Foo a b c | a b -> c where
foo :: a -> b -> c
instance Foo A R A where ...
instance Foo R A A where ...
In addition to the functional dependency I'd like to express that at least one of the parameters a and b is c,
import Data.Type.Equality
import Data.Type.Bool
class ( ((a == c) || (b == c)) ~ True)
=> Foo a b c | a b -> c where ...
You'll need a bunch of extensions switched on. In particular UndecidableSuperClasses, because the type family calls in the class constraint are opaque as far as GHC can see.
Your q here
How can I logically or two constraints together?
Is far more tricky. For the type equality approach, == uses a Closed Type Family. So you could write a Closed Type Family returning kind Constraint, but I doubt there's a general solution. For your Foo class:
type family AorBeqC a b c :: Constraint where
AorBeqC a b a = ()
AorBeqC a b c = (b ~ c)
class AorBeqC a b c => Foo a b c | a b -> c where ...
It's likely to have poor and non-symmetrical type improvement behaviour: if GHC can see that a, c are apart, it'll go to the second equation and use (b ~ c) to improve either; if it can't see they're apart nor that they're unifiable, it'll get stuck.
In general, as #ehird points out, you can't test whether a constraint is not satisfiable. Type equality is special.