### Why isn't ListT [] a monad?

Well I'm back from my vacation. But this isn't my personal blog so I think it's time to dive right in. Anything to take my mind off these mosquito bites...

So consider the free semiring R generated by some set S. In other words, S consists of finite sums and products of 0, 1 and elements of S and where the only simplification rules allowed are

For example, consider a term like ab+c(ef+gh). We can use distributivity to multiply this out to ab+cef+cgh. There's not much else we can do to simplify this. Now notice that if you multiply out all of the brackets wherever you can, ie. use distributivity until you can no longer, you end up with an expression that is a sum of terms and each term is a product of generators from S. (The sum of terms may be empty, in which case it's zero, and each of the terms may be a product of zero generators, in which case it is 1). This allows us to write elements of R in a canonical form: as a set of terms where each term is an ordered list of generators. For example, we could write ab+cef+cgh as {[a,b],[c,e,f],[c,g,h]}. Notice how the commutativity of addition is represented by the fact that we're using a

In Haskell it's more convenient to work with lists. Se we'll represent our running example as [['a','b'],['c','e','f'],['c','g','h']]. So if S is the Haskell type corresponding to our set of generators, then [[S]] can be thought of as the free semiring generated by elements of S, with the proviso that we consider two elements equal if one can be reordered to the other. Note that []=0 and [[]] = 1.

Now suppose that S is itself a free semiring generated by T. Then R = [[ [[T]] ]], modulo ordering. If you think about it, there's a nice way to use the algebraic structure to 'flatten' an element of [[ [[T]] ]] down to an element of [[T]]. R is freely generated by elements of S, in other words it consists of sums of products of elements of S. But the elements of S are themselves sums and products. So we have sums of

Look to you like a monad? Of course it does! :-) But what monad is it?

First, let's implement the map

We can test it out

It multiplies out exactly the way we want.

Now compare with running this:

In other words, apart from the

And now I can answer my original question. In a semiring, addition is commutative, so the order of terms doesn't matter. But in

Here's why: consider the expression ((a+b)+(a+b))*((a+b)+(a+b)). Multiply out the inner parentheses first and we get

Now multiply out the outer parentheses first and we get

The terms are coming out in a different order. Essentially distributivity doesn't work in a semiring with non-commutative addition.

This translates directly into failure of one of the monad laws. First write our expression as an object of type

u = a+b

v = (a+b)+(a+b)

w = ((a+b)+(a+b))*((a+b)*(a+b))

Working outwards from the inner parentheses we get:

(Note how I use

Evaluate

I think this is now a fairly complete analysis of what

Note that the reason

I actually figured out all of this stuff before this. I realised that the trees I was scribbling on the backs of my envelopes to represent elements of semirings could actually be generalised to just about any kind of tree, so I wrote about the general case first.

I prefer my example of

So consider the free semiring R generated by some set S. In other words, S consists of finite sums and products of 0, 1 and elements of S and where the only simplification rules allowed are

a+0=a, a+b=b+a, a+(b+c)=(a+b)+c

a1=1, a(bc) = (ab)c

a(b+c) = ab+ac and (a+b)c = ac+bc.

For example, consider a term like ab+c(ef+gh). We can use distributivity to multiply this out to ab+cef+cgh. There's not much else we can do to simplify this. Now notice that if you multiply out all of the brackets wherever you can, ie. use distributivity until you can no longer, you end up with an expression that is a sum of terms and each term is a product of generators from S. (The sum of terms may be empty, in which case it's zero, and each of the terms may be a product of zero generators, in which case it is 1). This allows us to write elements of R in a canonical form: as a set of terms where each term is an ordered list of generators. For example, we could write ab+cef+cgh as {[a,b],[c,e,f],[c,g,h]}. Notice how the commutativity of addition is represented by the fact that we're using a

*set*of terms, but each term is a*list*of generators because we're making no assumption about the commutativity of multiplication.In Haskell it's more convenient to work with lists. Se we'll represent our running example as [['a','b'],['c','e','f'],['c','g','h']]. So if S is the Haskell type corresponding to our set of generators, then [[S]] can be thought of as the free semiring generated by elements of S, with the proviso that we consider two elements equal if one can be reordered to the other. Note that []=0 and [[]] = 1.

Now suppose that S is itself a free semiring generated by T. Then R = [[ [[T]] ]], modulo ordering. If you think about it, there's a nice way to use the algebraic structure to 'flatten' an element of [[ [[T]] ]] down to an element of [[T]]. R is freely generated by elements of S, in other words it consists of sums of products of elements of S. But the elements of S are themselves sums and products. So we have sums of

*products*of*sums*of products. I emphasised two words in that sentence. This is because R contains subparts that are products of sums. If we multiply these out in the usual way, we get sums of sums of products of products. Group the sums and products together, and we get back to sums of products. Here's an example: any element of T trivially gives an element of S. So if a,b,c,e,f,g and h are all in T, then a,b,c and ef+gh are all in S and hence are generators of R, so ab+c(ef+gh) is in R. Multiply out and we obviously have the element ab+cef+cgh of S. It's not hard to see how this generalises to map any element of S back down to T. So we have mapsT -> [[T]](by trivial inclusion of generators)[[ [[T]] ]] -> [[T]](by above argument)

Look to you like a monad? Of course it does! :-) But what monad is it?

First, let's implement the map

`[[ [[T]] ]] -> [[T]]`. After tinkering I came up with

> import Control.Monad.List

> flatten x = concatMap (map concat . sequence) x

We can test it out

> x = [[ [['a']],[['b']] ], [ [['c']],[['e','f'],['g','h']] ]]

> test1 = flatten x

It multiplies out exactly the way we want.

Now compare with running this:

> x' = ListT [[ ListT [['a']],ListT [['b']] ],

> [ ListT [['c']],ListT [['e','f'],['g','h']] ]]

> test2 = runListT $ join x'

In other words, apart from the

`ListT`fluff,`join`for`ListT []`is`flatten`. So if we consider lists as a mechanism to represent sets,`ListT []`is revealed as the monad of semirings. I'm not sure, but I think that historically this is where monads originally came from. Certainly there are many papers on the relationship between monads and algebraic structures like semirings.And now I can answer my original question. In a semiring, addition is commutative, so the order of terms doesn't matter. But in

`ListT []`, we're using lists, and order does matter in a list. So if we*do*take order into account, then really`ListT []`is the monad of semirings where both addition and multiplication is non-commutative. And here's the problem: in general, there is no such thing as a freely generated semiring with non-commutative addition.Here's why: consider the expression ((a+b)+(a+b))*((a+b)+(a+b)). Multiply out the inner parentheses first and we get

(a+b+a+b)*(a+b+a+b)

= a*a+a*b+a*a+a*b+…

Now multiply out the outer parentheses first and we get

(a+b)*(a+b)+(a+b)*(a+b)+(a+b)*(a+b)+(a+b)*(a+b)

= a*a+a*b+b*a+b*b+…

The terms are coming out in a different order. Essentially distributivity doesn't work in a semiring with non-commutative addition.

This translates directly into failure of one of the monad laws. First write our expression as an object of type

`ListT []`:u = a+b

> u = ListT [["a"],["b"]]

v = (a+b)+(a+b)

> v = ListT [[u],[u]]

w = ((a+b)+(a+b))*((a+b)*(a+b))

> w = ListT [[v,v]]

`join`multiplies out parentheses. So working from the outer parentheses inwards we can use:

> expanded1 = join $ join w

> go1 = runListT expanded1

Working outwards from the inner parentheses we get:

> expanded2 = join $ fmap join w

> go2 = runListT expanded2

(Note how I use

`fmap`to 'duck down' through the outer layer of parentheses so as to multiply out each of the subexpressions first.)Evaluate

`go1`and`go2`and you'll see that they corresponds to the two ways of multiplying out that I gave above. And more importantly, the values of`expanded1`and`expanded2`aren't equal, meaning that`join . join = join . fmap join`isn't satisfied. You may recognise this: it's one of the monad laws. (At least it's one of the monad laws written the way category theorists write them.) So we ain't got no monad.I think this is now a fairly complete analysis of what

`ListT []`is all about. So one obvious remaining question is: where do games come into all this? The answer is that games form a semiring in a way that I haven't seen documented anywhere (though is surely common knowledge). I'd explain but I've run out of time...Note that the reason

`ListT []`isn't a monad is that`[]`isn't commutative, in some sense. This has been observed many times in the past. Two papers mentioning this are this and this.I actually figured out all of this stuff before this. I realised that the trees I was scribbling on the backs of my envelopes to represent elements of semirings could actually be generalised to just about any kind of tree, so I wrote about the general case first.

I prefer my example of

`ListT []`failing to be a monad to the examples given here. The latter make use of the IO monad so they aren't quite as 'pure'.Labels: haskell, mathematics

## 11 Comments:

So basically, if I read the argument correctly, you would get a monad if you instead of using ListT [] constructed a SetT-monad, using some sort of set implementation (sorted lists, balanced tree, whatever), and then used SetT [] for the free semiring?

Either it's SetT [] or ListT Set, I can never figure out which way round monad transformers go in my head.

BUT, you can't write a Set monad in Haskell. Here's a discussion about it. In fact, many of the standard mathematical examples of monads can't be implemented nicely in Haskell.

There's no problem making Set a monad if you make some other sacrifices:

module Set(Set, union, intersection, member) where

data Set a = Set { unSet :: [a] }

instance Monad Set where

return x = Set [x]

Set xs >>= f = Set $ xs >>= unSet . f

union :: Set a -> Set a -> Set a

union (Set xs) (Set ys) = Set (xs ++ ys)

intersection :: (Eq a) => Set a -> Set a -> Set a

intersection (Set xs) (Set ys) = Set $ filter (`elem` ys) xs

member :: (Eq a) => a -> Set a -> Bool

member x (Set xs) = x `elem` xs

You can do this many ways. It's only the member function that needs to have a context. (Which is just an indication that it's only member that needs to do any real work.)

But that's not a very satisfying Set monad for all sorts of reasons, not not just the fact that you can't use it to build free semirings out of sets and lists like I feel you ought to be able to.

So from an implementation perspective my Set type really sucks.

But from a mathematical point of view I don't know what you think is wrong with them. They obey all the set axioms you expect, and I don't see why you can't build free semirings with them.

This is for the abstract type Set, of course. The concrete type Set has no nice set properties.

The obvious problem is that you can't compare these sets using ==. The second problem is that Set ought to be a commutative monad, which it isn't. If it were commutative, we'd have a pretty construction for free semirings from ListT Set, which we don't. Maybe you have a suggestion for another nice way to make this construction?

The suckiness of the implementation is also pretty bad for real applications - and that is actually the real reason I care about this. A while back I tried writing some quantum computation code (which is just a fancy way of saying I wrote some linear algebra code) where I tried to implement vector spaces as a monad. I used something similar to your Set construction where I delayed doing actual work on my vector representation until the last moment (ie. at 'wavefunction collapse'). But for long computations you end up building large intermediate objects, exactly analogous to the way you can find yourself building large 'redundant' lists to represent small sets. Admittedly, none of the examples I tested produced intermediates that caused a performance problem, even for a modest PC, but it's still ugly carrying around all that extra baggage.

Ignore my complaint about the lack of ==.

Hey, I think you're right and that Set is probably is a perfectly decent commutative monad with a suitable definition of Eq. I think ListT Set probably does do what I want, although I haven't tested it yet. I remembered having trouble writing a Set monad ages ago but I realise now that my problems were only about efficiency, not about whether or not it was fundamentally possible.

Yes, Eq would have to defined to preserve the abstraction. E.g., by set differences both ways being empty. And set difference is easy.

I'm a fair bit late here, but here's my $0.02 (or less):

It seems to me a valid perspective on the "Set monad" and related issues would be to accept that the given definitions are monads, within the context of the Eq instance. The Monad instance may not be able to constrain the application of the Monad operations, but if (==) isn't defined for "Set a", it's essentially outside the scope of the monad, despite the fact that the typechecker won't slap your hand for it. So, the Set type itself wouldn't qualify as a monad, but its quotient by (==) would.

In any case, I think it's important to distinguish between a failure to construct a type-safe implementation of a monad and a failure to construct a monad. Just because it doesn't fit into the Monad type class (or obey the laws) doesn't mean it can't be one, given the right chance.

A typo:

>a1=1

I'm guessing you meant to write a0=0, a1=a.

Post a Comment

## Links to this post:

Create a Link

<< Home