# A Non-Unique Monad Instance

by Justin Le ♦

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Just stopping in for a short post before continuing with a long-overdue series or two :) This post is a bit of a short fun one that describes a quest I had, and hopefully some useful extra ideas I found along the way.

Soon after I discovered Haskell, one question has plagued my mind. Day and night, I wondered…

Are there any Haskell types with more than one unique

`Monad`

instance?

This was a question that was pretty simple…so simple that I was sure many people had already asked and answered this. But I couldn’t really find any answers and nobody I asked at the time could really give me one either, so this soon embedded itself as a pretty deep mystery to my psyche.

The background?

## Functor and Applicative

All `Functor`

instances, if they exist, are *unique* for the type. The type uniquely determines the instance. There is only one possible `Functor`

instance for `[]`

, one possible `Functor`

instance for `Maybe`

, `Either`

, etc.

This fact is taken advantage of by GHC to allow you to derive, for some types, a `Functor`

instance automatically.

```
ghci> :set -XDeriveFunctor
ghci> data Foo a = Bar [a] (Maybe (Foo a)) | Baz (Either String a) (Foo a)
ghci> let x = Bar [1, 3] (Just (Baz (Right 4) (Bar [10] Nothing)))
ghci> fmap (*2) x
Bar [2, 6] (Just (Baz (Right 8) (Bar [20] Nothing)))
```

There is no other possible `Functor`

instance for that data type. Go ahead, try :D

```
data Foo a = Bar [a] (Maybe (Foo a)) | Baz (Either String a) (Foo a)
instance Functor Foo where
fmap f (Bar xs y) = Bar (fmap f xs) (fmap f y)
fmap f (Baz x fy) = Baz (fmap f x) (fmap f fy)
```

However, this is not the case for `Applicative`

. Everyone knows of course about the normal (cartesian product) Applicative instance and the zippy Applicative instance for list:

```
instance Applicative [] where
pure x = [x]
fs <*> xs = [ f x | f <- fs, x <- xs ]
instance Applicative [] where
pure = repeat
fs <*> xs = zipWith ($) fs xs
```

What is also fairly established is that every *noncommutative* `Applicative`

instance also has a “flipped” version:

```
-- a flipped IO Applicative
data FlipIO a = FlipIO { runFlipIO :: IO a }
instance Applicative FlipIO where
pure x = FlipIO (pure x)
fi <*> xi = FlipIO $ do
x <- runFlipIO xi
f <- runFlipIO fi -- note the backwards effects
return (f x)
```

```
data State s a = State { runState :: s -> (a, s) }
-- the normal instance
instance Applicative (State s) where
pure x = State $ \s0 -> (x, s0)
fs <*> xs = State $ \s0 -> let (f, s1) = runState fs s0
(x, s2) = runState xs s1
in (f x, s2)
-- the flipped instance
instance Applicative (State s) where
pure x = State $ \s0 -> (x, s0)
fs <*> xs = State $ \s0 -> let (x, s1) = runState xs s0
(f, s2) = runState fs s1
in (f x, s2)
```

```
ghci> liftA2 (,) getLine getLine
> hello -- asking for the first field
> world -- asking for the second field
("hello", "world")
ghci> runFlipIO $ liftA2 (,) (FlipIO getLine) (FlipIO getLine)
> hello -- asking for the second field
> world -- asking for the first field
("world", "hello")
```

Every non-commutative `Applicative`

admits an alternative instance where “flipping” the order of the “effects” is also a valid `Applicative`

instance. So, not `Maybe`

or `Either`

, but `State`

, `[]`

, and `IO`

.

```
-- free "flipped" Applicative instance
data Flipped f a = Flipped { runFlipped :: f a }
-- instance where (<*>) is the same, but the order of effects is switched
instance Applicative f => Applicative (Flipped f) where
pure = Flipped . pure
Flipped f <*> Flipped x = Flipped $ liftA2 (flip ($)) x f
```

Cool. Types that have `Functor`

instances only have one. Types that have `Applicative`

instances very often have more than one.

So, the obvious next question is…what about `Monad`

s? Is a `Monad`

instance uniquely determined by its type?

## Monad

The answer wasn’t that simple, for me. Yes, *most* `Applicative`

s in the wild are non-unique, and there was a generating rule. But not so for `Monad`

s. You can’t have a `Monad`

where the effects are switched, because for `(>>=)`

, you need the effects of the first action in order to even decide what the effects of the next action are.

I vaguely remember from my past two data types that are very similar yet have very different `Monad`

and `Applicative`

instances: (finite) lists, (infinite) streams. From the outset, the two have almost identical structure. A `Stream`

is just a list with no `[]`

/nil:

The `Functor`

instance is identical:

And the (only??) `Applicative`

instance is the `ZipList`

instance for lists:

The `Monad`

instance is however very different from that of lists:

```
instance Monad Stream where
return x = x :~ return x
xs >>= f = join' (fmap f xs)
where
join' :: Stream (Stream a) -> Stream a
join' ((x :~ _) :~ yss) = x :~ join' (fmap tail' yss)
tail' (_ :~ xs) = xs
```

The `Monad`

instance itself is actually interesting enough to write about. It all revolves around `join`

, where `join`

takes a stream of streams and creates a stream *of the diagonals*. So it takes the first element of the first stream, the second element of the second stream, the third element of the third stream, etc.

This is actually a special case of the `Monad`

instance for all fixed-sized ordered containers. A length 5 vector, for example, will have the same `Applicative`

and `Monad`

instance as described here: `(<*>)`

with “zipping”, and `join`

with grabbing the diagonal of the 5-vector of 5-vectors.

This was a promising lead, but, it doesn’t take *too* much thought to see that neither lists nor `Stream`

are appropriate for *both* instances.

**Aside**

In case you were wondering, here is an elaboration :D

Fixed length vectors can’t have the normal list Applicative instance at all, unless they are of size 0 or 1. That’s because the result after

`(<*>)`

, the resulting list’s length is the product of the original lists. So you can forget the`Monad`

instance, too.Streams give you no luck, either. The easiest way to see is by considering the analogous

`Monad`

instance, where`join`

is the straight-up concatenation.`m >>= return == m`

is clearly violated. If`m`

is an infinite list,`fmap return`

gives you an infinite list of infinite lists, “joining”/concatenating them back will just give you an infinite list of the first item in`m`

.To put succincently, for

`Stream`

,`concat == head`

.Lists can have the

`Applicative`

instance fine, but not the`Monad`

instance. Here we assume that zipping and “getting the diagonal” go only as “far as possible”, and stop when one of the lists is too short.This one is a little trickier, but the weakness is when you have lists of lists of lists of different lengths.

`ghci> let counterexample = [[[1]], [[], [2,3]]] ghci> join counterexample [[1], [2,3]] ghci> join. join $ counterexample [1,3] ghci> fmap join counterexample [[1], []] ghci> join . fmap join $ counterexample [1]`

For a monad, joining the inner layer and then joining it all should be the same as joining it all and joining it all. The order of the joining shouldn’t count. We can see this in the more haskelly monad laws by noting:

So, dead end here.

So I didn’t really have any leads at that point; I tried a couple of other paths but nothing really panned out. So I shelved it for a while.

## Revelation

Several centuries later^{1}, the final revelation came as many revelations do in Haskell — from a hint by Edward Kmett. He pointed out something interesting regarding a `Monad`

instance that I had yet to notice:

This is the classic “Writer” monad instance, which is literally about as old as monads in functional programming is.

The key is that the `Monad`

instance of `(w,)`

depends on the `Monoid`

instance of `w`

. This is the “log”, so to speak. You need a `Monoid`

instance in order to make the `Monad`

instance…and the behavior of the `Monad`

instance is directly determined by the behavior of the `Monoid`

instance of `w`

.

And…`Monoid`

instances in Haskell are rarely ever unique! A different `Monoid`

instance would create a very different `Monad`

instance for the same type!

So, by factoring out the dependency on an external `Monoid`

instance, you get…

```
data Two a = One a | Two a
instance Functor Two where
fmap f (One a) = One (f a)
fmap f (Two a) = Two (f a)
```

and…voila! There it is!

This type is basically equivalent to `(Bool, a)`

. And `Bool`

has multiple `Monoid`

s on it. Instead of requiring an outside `Monoid`

instance, we can encode the instance directly into the behavior of `(>>=)`

. And here we go!

Our instances are basically the `Writer`

instance for `(Bool, a)`

, with different `Monoid`

instances for `Bool`

.

The first instance:

```
instance Applicative Two where
pure = One
One f <*> One x = One (f x)
One f <*> Two x = Two (f x)
Two f <*> One x = Two (f x)
Two f <*> Two x = Two (f x)
instance Monad Two where
return = One
One x >>= f = f x
Two x >>= f = case f x of
One y -> Two y
Two y -> Two y
```

Which represents the monoids formed by `(&&)`

with `True`

or by `(||)`

with `False`

(depending on which one you pick as `True`

and which one you pick as `False`

; the two instances are isomorphic)

The second:

```
instance Applicative Two where
pure = One
One f <*> One x = One (f x)
One f <*> Two x = Two (f x)
Two f <*> One x = Two (f x)
Two f <*> Two x = One (f x)
instance Monad Two where
return = One
One x >>= f = f x
Two x >>= f = case f x of
One y -> Two y
Two y -> One y
```

Which represents the monoid formed by `(/=)`

(or “XOR”) with `False`

.

And there you go. One type, two possible unique, non-isomorphic `Monad`

instances.

**Aside**

One interesting thing to note is that the Monad instance for `(->) a`

requires no monoid constraint, and the Monad instance for `(,) a`

*does*.

Interestingly enough, if we look at *comonads*, the Comonad instance for `(->) a`

*does* require a monoid constraint on `a`

(so for example there are many unique Comonad instances for things isomorphic to `(->) a`

where `a`

has more than one Monoid instance) and and the Comonad instance for `(,) a`

*does not* require a monoid constraint on `a`

.

Is there some duality at play here?

The answer is, apparently, yes! But according to Edward Kmett, it is one that is pretty hard to arrive at and a big headache and overall not worth the time to dig into. So you’re going to have to take my second-hand word for it.

More accurately, “about a year”↩