With respect to Listing 1, how would I go about proving the type level axiom
(t a) = (t (getUI (t a)))
holds?
Listing 1
data Continuant a = Continuant a deriving (Show,Eq)
class UI a where -- ...
instance UI Int where -- ...
class Category t where
getUI :: (UI a) => (t a) -> a
instance Category Continuant where
getUI (Continuant a) = a
-- Does axiom (t a) = (t (getUI(t a))) holds for given types?
test :: Int -> Bool
test x = (Continuant x) == (Continuant (getUI (Continuant x)))
The code is based on a paper where it is stated:
For all implementations of getUI one may require that the axiom (t a)
= (t (getUI (t a))) holds. This must be proven to hold for every specific type class instance declaration. For finite types this can be
done by a program that enumerates all possibilities. For infinite
types this must be done manually via proofs by induction.
My current intuition is that the test function in some way satisfies the axiom, but I do not think that it amounts to a proof.
This question follows on from a previous question.
To prove this, just start with one side of the equation and rewrite until you get to the other side. I like to start with the more complicated side.
when x :: Int,
Continuant (getUI (Continuant x))
-- ^^^^^^^^^^^^^^^^^^^^
-- by definition of getUI in Category Continuant Int
= Continuant x
That was easy! This does count as a proof (mind, not a formally verified one -- Haskell is not powerful enough to express term-level proofs. But it's so trivial it wouldn't be worth the boilerplate in agda.).
I was a bit bewildered by the phrasing of this axiom, since it seems to be mixing up types and terms quite a lot. Skimming the paper, it seems like this is only intended to work for simple single-constructor newtypes, thus this mixing is justified (still odd). Anyway, it seems like the paper doesn't have the Category class parameterized on a: i.e. instead of
class Category t a where ...
it would be
class Category t where ...
which makes more sense to me, that the class describes polymorphic wrappers, rather than a possibly different description of how it wraps each individual type (especially since it appears that the axiom requires the implementation to be the same no matter what a you pick!).
Related
With respect Listing 1, it is required that the type level axiom
(t a) = (t (getUI(t a)))
should derivable as a theorem for every specific type class instance.
Does the compilation of the test function prove the type level axiom holds for the particular types in the program? Does the compilation provide an example of Theorems for free!?
Listing 1
{-# LANGUAGE MultiParamTypeClasses #-}
data Continuant a = Continuant a deriving (Show,Eq)
class UI a where
instance UI Int where
class Category t a where
getUI :: (UI a) => (t a) -> a
instance Category Continuant Int where
getUI (Continuant a) = a
-- axiom (t a) = (t (getUI(t a))) holds for given types?
test :: Int -> Bool
test x = (Continuant x) == (Continuant (getUI (Continuant x)))
Additional Context
The code is based on a paper where it is stated:
For all implementations of getUI one may require that the axiom (t a)
= (t (getUI (t a))) holds. This must be proven to hold for every specific type class instance declaration. For finite types this can be
done by a program that enumerates all possibilities. For infinite
types this must be done manually via proofs by induction.
Listing 1 was my attempt at a proof. In the light of Solution 1, perhaps I mistakenly thought this required Theorems for free!
No, the fact that it's a class gives you far too much leeway for the type alone to guarantee that axiom. For example, the following alternate instance typechecks but violates your axiom:
instance Category Continuant Int where
getUI _ = 42
(Being completely explicit, our adversary could choose for example t=Continuant and a=6*9 to violate the axiom.) This abuses the fact that the class instantiator gets to choose the contained type. We can remove that ability by removing that argument to the class:
class Category t where
getUI :: UI a => t a -> a
Alas, even this is not enough. We can write
data Pair a = Pair a a
and then the free theorem tells us that for instance Category Pair, we must write one of the following two definitions:
getUI (Pair x y) = x
-- OR
getUI (Pair x y) = y
Whichever we choose, our adversary can choose a t which shows us that our axiom is wrong.
Our choice Adversary's choice
getUI (Pair x y) = x t y = Pair 42 y; a = 6*9
getUI (Pair x y) = y t x = Pair x 42; a = 6*9
Okay, this abuses the fact that the class instantiator gets to choose t. What if we removed that ability...?
class Category where
getUI :: UI a => t a -> a
This restricts the instantiator of Category quite a lot. Too much, in fact: getUI can't be implemented except as an infinite loop or the like.
I suspect it will be very difficult to encode the axiom you wish for as a type which can only be inhabited by things that satisfy it.
Take for example the class Functor:
class Functor a
instance Functor Maybe
Here Maybe is a type constructor.
But we can do this in two other ways:
Firstly, using multi-parameter type classes:
class MultiFunctor a e
instance MultiFunctor (Maybe a) a
Secondly using type families:
class MonoFunctor a
instance MonoFunctor (Maybe a)
type family Element
type instance Element (Maybe a) a
Now there's one obvious advantage of the two latter methods, namely that it allows us to do things like this:
instance Text Char
Or:
instance Text
type instance Element Text Char
So we can work with monomorphic containers.
The second advantage is that we can make instances of types that don't have the type parameter as the final parameter. Lets say we make an Either style type but put the types the wrong way around:
data Silly t errorT = Silly t errorT
instance Functor Silly -- oh no we can't do this without a newtype wrapper
Whereas
instance MultiFunctor (Silly t errorT) t
works fine and
instance MonoFunctor (Silly t errorT)
type instance Element (Silly t errorT) t
is also good.
Given these flexibility advantages of only using complete types (not type signatures) in type class definitions, is there any reason to use the original style definition, assuming you're using GHC and don't mind using the extensions? That is, is there anything special you can do putting a type constructor, not just a full type in a type class that you can't do with multi-parameter type classes or type families?
Your proposals ignore some rather important details about the existing Functor definition because you didn't work through the details of writing out what would happen with the class's member function.
class MultiFunctor a e where
mfmap :: (e -> ??) -> a -> ????
instance MultiFunctor (Maybe a) a where
mfmap = ???????
An important property of fmap at the moment is that its first argument can change types. fmap show :: (Functor f, Show a) => f a -> f String. You can't just throw that away, or you lose most of the value of fmap. So really, MultiFunctor would need to look more like...
class MultiFunctor s t a b | s -> a, t -> b, s b -> t, t a -> s where
mfmap :: (a -> b) -> s -> t
instance (a ~ c, b ~ d) => MultiFunctor (Maybe a) (Maybe b) c d where
mfmap _ Nothing = Nothing
mfmap f (Just a) = Just (f a)
Note just how incredibly complicated this has become to try to make inference at least close to possible. All the functional dependencies are in place to allow instance selection without annotating types all over the place. (I may have missed a couple possible functional dependencies in there!) The instance itself grew some crazy type equality constraints to allow instance selection to be more reliable. And the worst part is - this still has worse properties for reasoning than fmap does.
Supposing my previous instance didn't exist, I could write an instance like this:
instance MultiFunctor (Maybe Int) (Maybe Int) Int Int where
mfmap _ Nothing = Nothing
mfmap f (Just a) = Just (if f a == a then a else f a * 2)
This is broken, of course - but it's broken in a new way that wasn't even possible before. A really important part of the definition of Functor is that the types a and b in fmap don't appear anywhere in the instance definition. Just looking at the class is enough to tell the programmer that the behavior of fmap cannot depend on the types a and b. You get that guarantee for free. You don't need to trust that instances were written correctly.
Because fmap gives you that guarantee for free, you don't even need to check both Functor laws when defining an instance. It's sufficient to check the law fmap id x == x. The second law comes along for free when the first law is proven. But with that broken mfmap I just provided, mfmap id x == x is true, even though the second law is not.
As the implementer of mfmap, you have more work to do to prove your implementation is correct. As a user of it, you have to put more trust in the implementation's correctness, since the type system can't guarantee as much.
If you work out more complete examples for the other systems, you find that they have just as many issues if you want to support the full functionality of fmap. And this is why they aren't really used. They add a lot of complexity for only a small gain in utility.
Well, for one thing the traditional functor class is just much simpler. That alone is a valid reason to prefer it, even though this is Haskell and not Python. And it also represents the mathematical idea better of what a functor is supposed to be: a mapping from objects to objects (f :: *->*), with extra property (->Constraint) that each (forall (a::*) (b::*)) morphism (a->b) is lifted to a morphism on the corresponding object mapped to (-> f a->f b). None of that can be seen very clearly in the * -> * -> Constraint version of the class, or its TypeFamilies equivalent.
On a more practical account, yes, there are also things you can only do with the (*->*)->Constraint version.
In particular, what this constraint guarantees you right away is that all Haskell types are valid objects you can put into the functor, whereas for MultiFunctor you need to check every possible contained type, one by one. Sometimes that's just not possible (or is it?), like when you're mapping over infinitely many types:
data Tough f a = Doable (f a)
| Tough (f (Tough f (a, a)))
instance (Applicative f) = Semigroup (Tough f a) where
Doable x <> Doable y = Tough . Doable $ (,)<$>x<*>y
Tough xs <> Tough ys = Tough $ xs <> ys
-- The following actually violates the semigroup associativity law. Hardly matters here I suppose...
xs <> Doable y = xs <> Tough (Doable $ fmap twice y)
Doable x <> ys = Tough (Doable $ fmap twice x) <> ys
twice x = (x,x)
Note that this uses the Applicative instance of f not just on the a type, but also on arbitrary tuples thereof. I can't see how you could express that with a MultiParamTypeClasses- or TypeFamilies-based applicative class. (It might be possible if you make Tough a suitable GADT, but without that... probably not.)
BTW, this example is perhaps not as useless as it may look – it basically expresses read-only vectors of length 2n in a monadic state.
The expanded variant is indeed more flexible. It was used e.g. by Oleg Kiselyov to define restricted monads. Roughly, you can have
class MN2 m a where
ret2 :: a -> m a
class (MN2 m a, MN2 m b) => MN3 m a b where
bind2 :: m a -> (a -> m b) -> m b
allowing monad instances to be parametrized over a and b. This is useful because you can restrict those types to members of some other class:
import Data.Set as Set
instance MN2 Set.Set a where
-- does not require Ord
return = Set.singleton
instance Prelude.Ord b => MN3 SMPlus a b where
-- Set.union requires Ord
m >>= f = Set.fold (Set.union . f) Set.empty m
Note than because of that Ord constraint, we are unable to define Monad Set.Set using unrestricted monads. Indeed, the monad class requires the monad to be usable at all types.
Also see: parameterized (indexed) monad.
The reason why Set is not a functor is given here. It seems to boil down to the fact that a == b && f a /= f b is possible. So, why doesn't Haskell have as standard an alternative to Eq, something like
class Eq a => StrongEq a where
(===) :: a -> a -> Bool
(/==) :: a -> a -> Bool
x /== y = not (x === y)
x === y = not (x /== y)
for which instances are supposed to obey the laws
∀a,b,f. not (a === b) || (f a === f b)
∀a. a === a
∀a,b. (a === b) == (b === a)
and maybe some others? Then we could have:
instance StrongEq a => Functor (Set a) where
-- ...
Or am I missing something?
Edit: my problem is not “Why are there types without an Eq instance?”, like some of you seem to have answered. It's the opposite: “Why are there instances of Eq that aren't extensionally equal? Why are there too many Eq instances?”, combined with “If a == b does imply extensional equality, why is Set not an instance of Functor?”.
Also, my instance declaration is rubbish (thanks #n.m.). I should have said:
newtype StrongSet a = StrongSet (Set a)
instance Functor StrongSet where
fmap :: (StrongEq a, StrongEq b) => (a -> b) -> StrongSet a -> StrongSet b
fmap (StrongSet s) = StrongSet (map s)
instance StrongEq a => Functor (Set a) where
This makes sense neither in Haskell nor in the grand mathematical/categorical scheme of things, regardless of what StrongEq means.
In Haskell, Functor requires a type constructor of kind * -> *. The arrow reflects the fact that in category theory, a functor is a kind of mapping. [] and (the hypothetical) Set are such type constructors. [a] and Set a have kind * and cannot be functors.
In Haskell, it is hard to define Set such that it can be made into a Functor because equality cannot be sensibly defined for some types no matter what. You cannot compare two things of type Integer->Integer, for example.
Let's suppose there is a function
goedel :: Integer -> Integer -> Integer
goedel x y = -- compute the result of a function with
-- Goedel number x, applied to y
Suppose you have a value s :: Set Integer. What fmap goedel s should look like? How do you eliminate duplicates?
In your typical set theory equality is magically defined for everything, including functions, so Set (or Powerset to be precise) is a functor, no problem with that.
Since I'm not a category theorist, I'll try to write a more concrete/practical explanation (i.e., one I can understand):
The key point is the one that #leftaroundabout made in a comment:
== is supposed to
witness "equivalent by all observable means" (that doesn't necessarily
require a == b must hold only for identical implementations; but
anything you can "officially" do with a and b should again yield
equivalent results. So unAlwaysEq should never be exposed in the first
place). If you can't ensure this for some type, you shouldn't give it
an Eq instance.
That is, there should be no need for your StrongEq because that's what Eq is supposed to be already.
Haskell values are often intended to represent some sort of mathematical or "real-life" value. Many times, this representation is one-to-one. For example, consider the type
data PlatonicSolid = Tetrahedron | Cube |
Octahedron | Dodecahedron | Icosahedron
This type contains exactly one representation of each Platonic solid. We can take advantage of this by adding deriving Eq to the declaration, and it will produce the correct instance.
In many cases, however, the same abstract value may be represented by more than one Haskell value. For example, the red-black trees Node B (Node R Leaf 1 Leaf) 2 Leaf and Node B Leaf 1 (Node R Leaf 2 Leaf) can both represent the set {1,2}. If we added deriving Eq to our declaration, we would get an instance of Eq that distinguishes things we want to be considered the same (outside of the implementation of the set operations).
It's important to make sure that types are only made instances of Eq (and Ord) when appropriate! It's very tempting to make something an instance of Ord just so you can stick it in a data structure that requires ordering, but if the ordering is not truly a total ordering of the abstract values, all manner of breakage may ensue. Unless the documentation absolutely guarantees it, for example, a function called sort :: Ord a => [a] -> [a] may not only be an unstable sort, but may not even produce a list containing all the Haskell values that go into it. sort [Bad 1 "Bob", Bad 1 "James"] can reasonably produce [Bad 1 "Bob", Bad 1 "James"], [Bad 1 "James", Bad 1 "Bob"], [Bad 1 "James", Bad 1 "James"], or [Bad 1 "Bob", Bad 1 "Bob"]. All of these are perfectly legitimate. A function that uses unsafePerformIO in the back room to implement a Las Vegas-style randomized algorithm or to race threads against each other to get an answer from the fastest may even give different results different times, as long as they're == to each other.
tl;dr: Making something an instance of Eq is a way of making a very strong statement to the world; don't make that statement if you don't mean it.
Your second Functor instance also doesn't make any sense. The biggest reason why Set can't be a Functor in Haskell is fmap can't have constraints. Inventing different notions of equality as StrongEq doesn't change the fact that you can't write those constraints on fmap in your Set instance.
fmap in general shouldn't have the constraints you need. It makes perfect sense to have functors of functions, for example (without it the whole notion of using Applicative to apply functions inside a functor breaks down), and functions can't be members of Eq or your StrongEq in general.
fmap can't have extra constraints on only some instances, because of code like this:
fmapBoth :: (Functor f, Functor g) => (a -> b, c -> d) -> (f a, g c) -> (f b, g d)
fmapBoth (h, j) (x, y) = (fmap h x, fmap j y)
This code claims to work regardless of the functors f and g, and regardless of the functions h and j. It has no way of checking whether one of the functors is a special one that has extra constraints on fmap, nor any way of checking whether one of the functions it's applying would violate those constraints.
Saying that Set is a Functor in Haskell, is saying that there is a (lawful) operation fmap :: (a -> b) -> Set a -> Set b, with that exact type. That is precisely what Functor means. fmap :: (Eq a -> Eq b) => (a -> b) -> Set a -> Set b is not an example of such an operation.
It is possible, I understand, to use the ConstraintKinds GHC extendsion to write a different Functor class that permits constraints on the values which vary by Functor (and what you actually need is an Ord constraint, not just Eq). This blog post talks about doing so to make a new Monad class which can have an instance for Set. I've never played around with code like this, so I don't know much more than that the technique exists. It wouldn't help you hand off Sets to existing code that needs Functors, but you should be able to use it instead of Functor in your own code if you wish.
This notion of StrongEq is tough. In general, equality is a place where computer science becomes significantly more rigorous than typical mathematics in the kind of way which makes things challenging.
In particular, typical mathematics likes to talk about objects as though they exist in a set and can be uniquely identified. Computer programs usually deal with types which are not always computable (as a simple counterexample, tell me what the set corresponding to the type data U = U (U -> U) is). This means that it may be undecidable as to whether two values are identifiable.
This becomes an enormous topic in dependently typed languages since typechecking requires identifying like types and dependently typed languages may have arbitrary values in their types and thus need a way to project equality.
So, StrongEq could be defined over a restricted part of Haskell containing only the types which can be decidably compared for equality. We can consider this a category with the arrows as computable functions and then see Set as an endofunctor from types to the type of sets of values of that type. Unfortunately, these restrictions have taken us far from standard Haskell and make defining StrongEq or Functor (Set a) a little less than practical.
In Scrap your boilerplate reloaded, the authors describe a new presentation of Scrap Your Boilerplate, which is supposed to be equivalent to the original.
However, one difference is that they assume a finite, closed set of "base" types, encoded with a GADT
data Type :: * -> * where
Int :: Type Int
List :: Type a -> Type [a]
...
In the original SYB, type-safe cast is used, implemented using the Typeable class.
My questions are:
What is the relationship between these two approaches?
Why was the GADT representation chosen for the "SYB Reloaded" presentation?
[I am one of the authors of the "SYB Reloaded" paper.]
TL;DR We really just used it because it seemed more beautiful to us. The class-based Typeable approach is more practical. The Spine view can be combined with the Typeable class and does not depend on the Type GADT.
The paper states this in its conclusions:
Our implementation handles the two central ingredients of generic programming differently from the original SYB paper: we use overloaded functions with
explicit type arguments instead of overloaded functions based on a type-safe
cast 1 or a class-based extensible scheme [20]; and we use the explicit spine
view rather than a combinator-based approach. Both changes are independent
of each other, and have been made with clarity in mind: we think that the structure of the SYB approach is more visible in our setting, and that the relations
to PolyP and Generic Haskell become clearer. We have revealed that while the
spine view is limited in the class of generic functions that can be written, it is
applicable to a very large class of data types, including GADTs.
Our approach cannot be used easily as a library, because the encoding of
overloaded functions using explicit type arguments requires the extensibility of
the Type data type and of functions such as toSpine. One can, however, incorporate Spine into the SYB library while still using the techniques of the SYB
papers to encode overloaded functions.
So, the choice of using a GADT for type representation is one we made mainly for clarity. As Don states in his answer, there are some obvious advantages in this representation, namely that it maintains static information about what type a type representation is for, and that it allows us to implement cast without any further magic, and in particular without the use of unsafeCoerce. Type-indexed functions can also be implemented directly by using pattern matching on the type, and without falling back to various combinators such as mkQ or extQ.
Fact is that I (and I think the co-authors) simply were not very fond of the Typeable class. (In fact, I'm still not, although it is finally becoming a bit more disciplined now in that GHC adds auto-deriving for Typeable, makes it kind-polymorphic, and will ultimately remove the possibility to define your own instances.) In addition, Typeable wasn't quite as established and widely known as it is perhaps now, so it seemed appealing to "explain" it by using the GADT encoding. And furthermore, this was the time when we were also thinking about adding open datatypes to Haskell, thereby alleviating the restriction that the GADT is closed.
So, to summarize: If you actually need dynamic type information only for a closed universe, I'd always go for the GADT, because you can use pattern matching to define type-indexed functions, and you do not have to rely on unsafeCoerce nor advanced compiler magic. If the universe is open, however, which is quite common, certainly for the generic programming setting, then the GADT approach might be instructive, but isn't practical, and using Typeable is the way to go.
However, as we also state in the conclusions of the paper, the choice of Type over Typeable isn't a prerequisite for the other choice we're making, namely to use the Spine view, which I think is more important and really the core of the paper.
The paper itself shows (in Section 8) a variation inspired by the "Scrap your Boilerplate with Class" paper, which uses a Spine view with a class constraint instead. But we can also do a more direct development, which I show in the following. For this, we'll use Typeable from Data.Typeable, but define our own Data class which, for simplicity, just contains the toSpine method:
class Typeable a => Data a where
toSpine :: a -> Spine a
The Spine datatype now uses the Data constraint:
data Spine :: * -> * where
Constr :: a -> Spine a
(:<>:) :: (Data a) => Spine (a -> b) -> a -> Spine b
The function fromSpine is as trivial as with the other representation:
fromSpine :: Spine a -> a
fromSpine (Constr x) = x
fromSpine (c :<>: x) = fromSpine c x
Instances for Data are trivial for flat types such as Int:
instance Data Int where
toSpine = Constr
And they're still entirely straightforward for structured types such as binary trees:
data Tree a = Empty | Node (Tree a) a (Tree a)
instance Data a => Data (Tree a) where
toSpine Empty = Constr Empty
toSpine (Node l x r) = Constr Node :<>: l :<>: x :<>: r
The paper then goes on and defines various generic functions, such as mapQ. These definitions hardly change. We only get class constraints for Data a => where the paper has function arguments of Type a ->:
mapQ :: Query r -> Query [r]
mapQ q = mapQ' q . toSpine
mapQ' :: Query r -> (forall a. Spine a -> [r])
mapQ' q (Constr c) = []
mapQ' q (f :<>: x) = mapQ' q f ++ [q x]
Higher-level functions such as everything also just lose their explicit type arguments (and then actually look exactly the same as in original SYB):
everything :: (r -> r -> r) -> Query r -> Query r
everything op q x = foldl op (q x) (mapQ (everything op q) x)
As I said above, if we now want to define a generic sum function summing up all Int occurrences, we cannot pattern match anymore, but have to fall back to mkQ, but mkQ is defined purely in terms of Typeable and completely independent of Spine:
mkQ :: (Typeable a, Typeable b) => r -> (b -> r) -> a -> r
(r `mkQ` br) a = maybe r br (cast a)
And then (again exactly as in original SYB):
sum :: Query Int
sum = everything (+) sumQ
sumQ :: Query Int
sumQ = mkQ 0 id
For some of the stuff later in the paper (e.g., adding constructor information), a bit more work is needed, but it can all be done. So using Spine really does not depend on using Type at all.
Well, obviously the Typeable use is open -- new variants can be added after the fact, and without modifying the original definitions.
The important change though is that in that TypeRep is untyped. That is, there is no connection between the runtime type , TypeRep, and the static type it encodes. With the GADT approach we can encode the mapping between a type a and its Type, given by the GADT Type a.
We thus bake in evidence for the type rep being statically linked to its origin type, and can write statically typed dynamic application (for example) using Type a as evidence that we have a runtime a.
In the older TypeRep case, we have no such evidence and it comes down to runtime string equality, and a coerce and hope for the best through fromDynamic.
Compare the signatures:
toDyn :: Typeable a => a -> TypeRep -> Dynamic
versus GADT style:
toDyn :: Type a => a -> Type a -> Dynamic
I can't fake my type evidence, and I can use that later when reconstructing things, to e.g. lookup the type class instances for a when all I have is a Type a.
I'm currently in Chapter 8 of Learn you a Haskell, and I've reached the section on the Functor typeclass. In said section the author gives examples of how different types could be made instances of the class (e.g Maybe, a custom Tree type, etc.) Seeing this, I decided to (for fun and practice) try implementing an instance for the Data.Set type; in all of this ignoring Data.Set.map, of course.
The actual instance itself is pretty straight-forward, and I wrote it as:
instance Functor Set.Set where
fmap f empty = Set.empty
fmap f s = Set.fromList $ map f (Set.elems s)
But, since I happen to use the function fromList this brings in a class constraint calling for the types used in the Set to be Ord, as is explained by a compiler error:
Error occurred
ERROR line 4 - Cannot justify constraints in instance member binding
*** Expression : fmap
*** Type : Functor Set => (a -> b) -> Set a -> Set b
*** Given context : Functor Set
*** Constraints : Ord b
See: Live Example
I tried putting a constraint on the instance, or adding a type signature to fmap, but neither succeeded (both were compiler errors as well.)
Given a situation like this, how can a constraint be fulfilled and satisfied? Is there any possible way?
Thanks in advance! :)
Unfortunately, there is no easy way to do this with the standard Functor class. This is why Set does not come with a Functor instance by default: you cannot write one.
This is something of a problem, and there have been some suggested solutions (e.g. defining the Functor class in a different way), but I do not know if there is a consensus on how to best handle this.
I believe one approach is to rewrite the Functor class using constraint kinds to reify the additional constraints instances of the new Functor class may have. This would let you specify that Set has to contain types from the Ord class.
Another approach uses only multi-parameter classes. I could only find the article about doing this for the Monad class, but making Set part of Monad faces the same problems as making it part of Functor. It's called Restricted Monads.
The basic gist of using multi-parameter classes here seems to be something like this:
class Functor' f a b where
fmap' :: (a -> b) -> f a -> f b
instance (Ord a, Ord b) => Functor' Data.Set.Set a b where
fmap' = Data.Set.map
Essentially, all you're doing here is making the types in the Set also part of the class. This then lets you constrain what these types can be when you write an instance of that class.
This version of Functor needs two extensions: MultiParamTypeClasses and FlexibleInstances. (You need the first extension to be able to define the class and the second extension to be able to define an instance for Set.)
Haskell : An example of a Foldable which is not a Functor (or not Traversable)? has a good discussion about this.
This is impossible. The purpose of the Functor class is that if you have Functor f => f a, you can replace the a with whatever you like. The class is not allowed to constrain you to only return this or that. Since Set requires that its elements satisfy certain constraints (and indeed this isn't an implementation detail but really an essential property of sets), it doesn't satisfy the requirements of Functor.
There are, as mentioned in another answer, ways of developing a class like Functor that does constrain you in that way, but it's really a different class, because it gives the user of the class fewer guarantees (you don't get to use this with whatever type parameter you want), in exchange for becoming applicable to a wider range of types. That is, after all, the classic tradeoff of defining a property of types: the more types you want to satisfy it, the less they must be forced to satisfy.
(Another interesting example of where this shows up is the MonadPlus class. In particular, for every instance MonadPlus TC you can make an instance Monoid (TC a), but you can't always go the other way around. Hence the Monoid (Maybe a) instance is different from the MonadPlus Maybe instance, because the former can restrict the a but the latter can't.)
You can do this using a CoYoneda Functor.
{-# LANGUAGE GADTs #-}
data CYSet a where
CYSet :: (Ord a) => Set.Set a -> (a -> b) -> CYSet b
liftCYSet :: (Ord a) => Set.Set a -> CYSet a
liftCYSet s = CYSet s id
lowerCYSet :: (Ord a) => CYSet a -> Set.Set a
lowerCYSet (CYSet s f) = Set.fromList $ map f $ Set.elems s
instance Functor CYSet where
fmap f (CYSet s g) = CYSet s (f . g)
main = putStrLn . show
$ lowerCYSet
$ fmap (\x -> x `mod` 3)
$ fmap abs
$ fmap (\x -> x - 5)
$ liftCYSet $ Set.fromList [1..10]
-- prints "fromList [0,1,2]"