What are the laws for Eq1, Eq2, Ord1, and Ord2?

[Gershom B]

So looking at the “bad” examples they both seem to only be bad for Eq2, and be bad because they “abuse” a design choice in the class.

In particular, we have

class Eq2 f where
    liftEq2 :: (a -> b -> Bool) -> (c -> d -> Bool) -> f a c -> f b d -> Bool Source#

  — Lift equality tests through the type constructor.

  — The function will usually be applied to equality functions, but the more general type ensures that the implementation uses them to compare elements of the first container with elements of the second.

This is to say that the two types, `a` and `b`, are intended to be equal. However, to prevent implementations from making the error of comparing two values from the first container, they are made distinct. So in the intended usage, this is not called directly but rather via eq2.

eq2 :: (Eq2 f, Eq a, Eq b) => f a b -> f a b -> Bool

What David has pointed out, I think, is that this clever trick, which prevents some bad implementations, is open to abuse in the case of _good_ implementations, since the assumption of the two types being equal is necessary for some structural invariants to hold. (In particular, if the `a` and `b` differ, they will have different ordering properties).

So I see two options here, neither of which involves removing any instances.

A) change the types in the Eq2 and Ord2 classes to not use this clever trick. I think this should not break anything, since according to the documentation, nothing should be making use of the possibility of these types varying.

B) change the documentation of the classes to make explicit that the laws are _only_ expected to apply when the types are equal, and when they are not, the behavior is undefined.

Cheers,
Gershom


On March 15, 2018 at 7:57:34 PM, Oleg Grenrus (oleg.grenrus at iki.fi) wrote:

My hand-wavy intuition:  

Eq witnesses an equality. a -> a -> Bool type isn't enough to guarantee  
that,  
so we have laws.  

The Eq1 class has member  

    liftEq :: (a -> b -> Bool) -> f a -> f b -> Bool  

the type `a -> b -> Bool` is not an equality decision procedure,  
but rather an isomorphism one.  

It looks like for me, that liftEq gives a natural transformation from  
`Iso a b` to `Iso (f a) (f b)`. Does this make sense?  

The liftEq2 is the same. We have to think Map as not of  

  Hask * Hask -> Hask (wrong)  

Functor but  

  OrdHask * Hask -> Hask (OrdHask: Ord-types and monotone functions)  

Then, comparing Map k () with Map (Down k) () with \k (Down k') = k == k'  
is not well defined, as that "isomorphism candidate" doesn't respect  
ordering.  
(as an example, for k = Integer, \k (Down k') = k == negate k' would  
work, or  
even \n m -> n == succ m, for n m :: Integer)  

Similar argument can be used on (Eq k, Hashable k) requirement of  
HashMap k v,  
with constraints giving an additional Hashable structure. The motivational  
example is being able to compare  

    HashMap Text Foo   and   HashMap BusinessId Foo'  

directly, where  
   
    newtype BusinessId = BusinessId Text deriving newtype (Eq, Hashable)  


As Hashable is somewhat arbitrary, I cannot think of other use-cases  
(preserving Hashable is next to impossible otherwise then via newtype  
deriving).  However, for Map there might be A, B such that `Map A v` and  
`Map  
B v` are comparable, i.e. there is monotone f :: A -> B, and  

    compareAB a b = f a `compare` b  

which we can use as a first argument for liftCompare2 (Ord2). (There is  
mapKeysMonotonic which can be used to achieve the same effect!  
unordered-containers could have similar unsafe function too).  That said, I  
didn't have such use case myself (yet!?).  

When writing the instances, I was thinking what's (a -> b -> Ordering),  
and at this point I have to wave hands even more. I don't know how to  
complete  
the following sentence  

    The ??? to total order is as isomorphism to equality.  

As a side-note: we have  
- `~` witnessing nominal equality  
- `Coercible` witnessing structural equality  
- It would be very cool to being to say `CoercibleInstance Hashable`,  
which is  
  satisfied when `Hashable` is `newtype` derived. Is it roles, refining  
kinds,  
  or something else, I don't know. But I feel it would be useful.  It can be  
  used to give `mapKeysMonotonic` a safe type (at least in some cases!)  
  Maybe if `Constraint ~ Type`, we could simply require `Coercible (Hashable  
  k) (Hashable k')`, but how it would fit the roles?  

So I do agree, that Eq1 f has functorial feel and it's "natural",  
this is what its type tells. But if you imply that we should add  
`Functor` or  
`Bifunctor` super-classes, then I disagree, based on above arguments.  
Yes, the functions are unsafe to use (to get meaningful results),  
but I see a value for them.  

An example of legitimate (in above sence) instance of Eq1 which isn't a  
functor is Eq1 Set. As long as `eq :: a -> b -> Bool` respects the  
ordering of  
`a` and `b`, we can compare `Set a` with `Set b`.  
<Insert a riddle about comparing sets of apples and oranges,  
if one can compare apples with oranges, so fruit comparison diagram  
commutes>  

Cheers, Oleg  


> I was looking at the Eq2 instances for Data.Map and Data.HashMap, and  
> the Eq1 instances for Set and HashSet, and I realized that they're a  
> bit ... weird. My instinct is to remove these instances immediately,  
> but I figure I should first check with the community (and Oleg, who  
> added them to begin with) to make sure they don't make some sort of  
> sense I don't understand. In particular, these instances compare keys  
> in the order in which they appear in the container. That order may be  
> completely unrelated to the given key comparison function.  
>  
> Data.Map example: suppose I have a Map k () and a Map (Down k) (),  
> where Down is from Data.Ord. If I call liftEq2 (\x (Down y) -> x == y)  
> (==), I will get what looks to me like a totally meaningless result.  
>  
> Data.HashMap example: suppose I have a HashMap Int () and a Hashmap  
> String (). If I call liftEq2 (\x y -> show x == y) (==), that won't  
> return True even if the strings in the second map are actually the  
> results of applying show to the numbers in the first map.  
>  
> Intuitively, I think Eq1 f only makes sense if the shape of f x does  
> not depend on the values of type x, and Eq2 p only makes sense if the  
> shape of p x y does not depend on the values of types x and y. Is  
> there a way to formalize this intuition with class laws? I believe  
> that in the case of a Functor, parametricity will guarantee that  
>  
>   liftEq eq (f <$> xs) (g <$> ys) == liftEq (\x y -> eq (f x) (g y)) xs ys  
>  
> Are there *any* legitimate instances of Eq1 or Ord1 that are not  
> Functors? Are there *any* legitimate instances of Eq2 or Ord2 that are  
> not Bifunctors? My intuition says no. We might wish we could write  
> instances for unboxed arrays or vectors, but I believe that is totally  
> impossible anyway.  
>  
> David  


On 15.03.2018 23:48, David Feuer wrote:  
> I was looking at the Eq2 instances for Data.Map and Data.HashMap, and  
> the Eq1 instances for Set and HashSet, and I realized that they're a  
> bit ... weird. My instinct is to remove these instances immediately,  
> but I figure I should first check with the community (and Oleg, who  
> added them to begin with) to make sure they don't make some sort of  
> sense I don't understand. In particular, these instances compare keys  
> in the order in which they appear in the container. That order may be  
> completely unrelated to the given key comparison function.  
>  
> Data.Map example: suppose I have a Map k () and a Map (Down k) (),  
> where Down is from Data.Ord. If I call liftEq2 (\x (Down y) -> x == y)  
> (==), I will get what looks to me like a totally meaningless result.  
>  
> Data.HashMap example: suppose I have a HashMap Int () and a Hashmap  
> String (). If I call liftEq2 (\x y -> show x == y) (==), that won't  
> return True even if the strings in the second map are actually the  
> results of applying show to the numbers in the first map.  
>  
> Intuitively, I think Eq1 f only makes sense if the shape of f x does  
> not depend on the values of type x, and Eq2 p only makes sense if the  
> shape of p x y does not depend on the values of types x and y. Is  
> there a way to formalize this intuition with class laws? I believe  
> that in the case of a Functor, parametricity will guarantee that  
>  
> liftEq eq (f <$> xs) (g <$> ys) == liftEq (\x y -> eq (f x) (g y)) xs ys  
>  
> Are there *any* legitimate instances of Eq1 or Ord1 that are not  
> Functors? Are there *any* legitimate instances of Eq2 or Ord2 that are  
> not Bifunctors? My intuition says no. We might wish we could write  
> instances for unboxed arrays or vectors, but I believe that is totally  
> impossible anyway.  
>  
> David  


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