A couple weeks ago I was working on a project using Conal Elliott's uniform-pair library and
noticed it had a curious `Monad`

instance, which I've reproduced below.

```
data Pair a = Pair a a
instance Monad Pair where
return a = Pair a a
m >>= f = joinP (f <$> m)
joinP :: Pair (Pair a) -> Pair a
joinP (Pair (Pair a _) (Pair _ d)) = Pair a d
```

I was especially curious about why `joinP`

chose the first element of the first pair and the second element of
the second pair. My initial guess was that it was determined by the `Functor`

instance which would've looked something
like..

```
fmapP :: (a -> b) -> Pair a -> Pair b
fmapP f (Pair x y) = Pair (f x) (f y)
```

For `Monad`

to be consistent with `Functor`

the follow equation should hold..

```
fmapP f p = p >>= (return . f)
```

..but this didn't really help.

```
(Pair x y) >>= (return . f)
= joinP ((return . f) <$> Pair x y)
= joinP (Pair (Pair (f x) (f y)) (Pair (f x) (f y)))
```

Taking either element of the outer pair would've been consistent with the `Functor`

instance, as would taking the first
element of the first pair and the second element of the second pair.

A couple days later I was talking with Conal about it and he hinted at using the fact that uniform pairs are
representable functors. For a functor to be representable in Haskell^{1} means it is isomorphic to
the set of functions from `X`

, for some fixed `X`

(this "set of functions from X" is also known as the reader monad).
For uniform pairs, `X = Bool`

. Indeed, the following functions are mutual inverses.

```
to :: Pair a -> Bool -> a
to (Pair x _) False = x
to (Pair _ y) True = y
from :: (Bool -> a) -> Pair a
from f = Pair (f False) (f True)
```

To prove that a functor `f`

is representable in Haskell is to implement the `Representable`

type class. The
following is reproduced from the representable-functors package.

```
class Representable f where
index :: f a -> Key f -> a
tabulate :: (Key f -> a) -> f a
```

The `Key f`

refers to the fixed `X`

mentioned above, so `Key Pair = Bool`

. Substituing `Bool`

for `Key f`

reveals
signatures matching the `to`

(`index`

) and `from`

(`tabulate`

) functions^{2}.

As it turns out every `Representable`

has a canonical monadic return and bind, defined as:

```
returnRep :: Representable f => a -> f a
returnRep = tabulate . const
bindRep :: Representable f => f a -> (a -> f b) -> f b
bindRep m f = tabulate (\a -> index (f (index m a)) a)
```

Let's see what this looks like for `Pair`

. First let's do some substitution on `returnRep`

:

```
returnRep :: a -> Pair a
returnRep a
= (tabulate . const) a
= tabulate (const a)
= Pair (const a False) (const a True) -- Pair's tabulate = from
= Pair a a
```

That matches our `return`

definition above. Now let's do the same for `bindRep`

:

```
bindRep :: Pair a -> (a -> Pair b) -> Pair b
bindRep (Pair x y) f
= tabulate (\a -> index (f (index (Pair x y) a)) a)
= tabulate g -- call the lambda 'g'
= Pair (g False) (g True) -- Pair's tabulate = from
```

Now substituting `False`

and `True`

into the lambda:

```
-- g False
= index (f (index (Pair x y) False)) False
= index (f x) False -- Pair's index = to
= first element of f x -- Pair's index = to
-- g True
= index (f (index (Pair x y) True)) True
= index (f y) True -- Pair's index = to
= second element of f y -- Pair's index = to
```

Thus:

```
bindRep (Pair x y) f
= Pair a d -- where Pair (Pair a _) (Pair _ d)
-- ^ f x ^ f y
```

The same as `joinP`

above.

This is awesome. By starting with the *meaning* of his data type, Conal discovered the only
natural type class instance consistent with the meaning. While in this case I started with the instance and worked
my way back, I believe the more useful and consistent approach is to think hard about your data type's
denotation and work your way forward.

^{1}

Specifically I mean the $Hask$ category with types as objects and functions as arrows.

^{2}

In general the type class law for `Representable`

requires `index`

and `tabulate`

to be mutual inverses.