Looks like I have a pretty clear understanding what a Monoid is in Haskell, but last time I heard about something called a free monoid.
What is a free monoid and how does it relate to a monoid?
Can you provide an example in Haskell?
As you already know, a monoid is a set with an element e and an operation <> satisfying
e <> x = x <> e = x (identity)
(x<>y)<>z = x<>(y<>z) (associativity)
Now, a free monoid, intuitively, is a monoid which satisfies only those equations above, and, obviously, all their consequences.
For instance, the Haskell list monoid ([a], [], (++)) is free.
By contrast, the Haskell sum monoid (Sum Int, Sum 0, \(Sum x) (Sum y) -> Sum (x+y)) is not free, since it also satisfies additional equations. For instance, it's commutative
x<>y = y<>x
and this does not follow from the first two equations.
Note that it can be proved, in maths, that all the free monoids are isomorphic to the list monoid [a]. So, "free monoid" in programming is only a fancy term for any data structure which 1) can be converted to a list, and back, with no loss of information, and 2) vice versa, a list can be converted to it, and back, with no loss of information.
In Haskell, you can mentally substitute "free monoid" with "list-like type".
In a programming context, I usually translate free monoid to [a]. In his excellent series of articles about category theory for programmers, Bartosz Milewski describes free monoids in Haskell as the list monoid (assuming one ignores some problems with infinite lists).
The identity element is the empty list, and the binary operation is list concatenation:
Prelude Data.Monoid> mempty :: [Int]
[]
Prelude Data.Monoid> [1..3] <> [7..10]
[1,2,3,7,8,9,10]
Intuitively, I think of this monoid to be 'free' because it a monoid that you can always apply, regardless of the type of value you want to work with (just like the free monad is a monad you can always create from any functor).
Additionally, when more than one monoid exists for a type, the free monoid defers the decision on which specific monoid to use. For example, for integers, infinitely many monoids exist, but the most common are addition and multiplication.
If you have two (or more integers), and you know that you may want to aggregate them, but you haven't yet decided which type of aggregation you want to apply, you can instead 'aggregate' them using the free monoid - practically, this means putting them in a list:
Prelude Data.Monoid> [3,7]
[3,7]
If you later decide that you want to add them together, then that's possible:
Prelude Data.Monoid> getSum $ mconcat $ Sum <$> [3,7]
10
If, instead, you wish to multiply them, you can do that as well:
Prelude Data.Monoid> getProduct $ mconcat $ Product <$> [3,7]
21
In these two examples, I've deliberately chosen to elevate each number to a type (Sum, Product) that embodies a more specific monoid, and then use mconcat to perform the aggregation.
For addition and multiplication, there are more succinct ways to do this, but I did it that way to illustrate how you can use a more specific monoid to interpret the free monoid.
A free monoid is a specific type of monoid. Specifically, it’s the monoid you get by taking some fixed set of elements as characters and then forming all possible strings from those elements. Those strings, with the underlying operation being string concatenation, form a monoid, and that monoid is called the free monoid.
A monoid (M,•,1) is a mathematical structure such that:
M is a set
1 is a member of M
• : M * M -> M
a•1 = a = 1•a
Given elements a, b and c in M, we have a•(b•c) = (a•b)•c.
A free monoid on a set M is a monoid (M',•,0) and function e : M -> M' such that, for any monoid (N,*,1), given a (set) map f : M -> N we can extend this to a monoid morphism f' : (M',•,0) -> (N,*,1), i.e
f a = f' (e a)
f' 0 = 1
f' (a•b) = (f' a) • (f' b)
In other words, it is a monoid that does nothing special.
An example monoid is the integers with the operation being addition and the identity being 0. Another monoid is sequences of integers with the operation being concatenation and the identity being the empty sequence. Now the integers under addition is not a free monoid on the integers. Consider the map into sequences of integers taking n to (n). Then for this to be free we would need to extend this to a map taking n + m to (n,m), i.e. it must take 0 to (0) and to (0,0) and to (0,0,0) and so on.
On the other hand if we try to look at sequences of integers as a free monoid on the integers, we see that it seems to work in this case. The extension of the map into the integers with addition is one that takes the sum of a sequence (with the sum of () being 0).
So what is the free monoid on a set S? Well one thing we could try is just arbitrary binary trees of S. In a Haskell type this would look like:
data T a = Unit | Single a | Conc (T a) (T a)
And it would have an identity of Unit, e = Single and (•) = Conc.
And we can write a function to show how it is free:
-- here the second argument represents a monoid structure on b
free :: (a -> b) -> (b -> b -> b, b) -> T a -> b
free f ((*),zero) = f' where
f' (Single a) = f a
f' Unit = zero
f' (Conc a b) = f' a * f' b
It should be quite obvious that this satisfies the required laws for a free monoid on a. Except for one: T a is not a monoid because it does not quite satisfy laws 4 or 5.
So now we should ask if we can make this into a simpler free monoid, ie one that is an actual monoid. The answer is yes. One way is to observe that Conc Unit a and Conc a Unit and Single a should be the same. So let’s make the first two types unrepresentable:
data TInner a = Single a | Conc (TInner a) (TInner a)
data T a = Unit | Inner (TInner a)
A second observation we can make is that there should be no difference between Conc (Conc a b) c and Conc a (Conc b c). This is due to law 5 above. We can then flatten our tree:
data TInner a = Single a | Conc (a,TInner a)
data T a = Unit | Inner (TInner a)
The strange construction with Conc forces us to only have a single way to represent Single a and Unit. But we see we can merge these all together: change the definition of Conc to Conc [a] and then we can change Single x to Conc [x], and Unit to Conc [] so we have:
data T a = Conc [a]
Or we can just write:
type T a = [a]
And the operations are:
unit = []
e a = [a]
(•) = append
free f ((*),zero) = f' where
f' [] = zero
f' (x:xs) = f x * f' xs
So in Haskell, the list type is called the free monoid.
I've been going through the Typeclassopedia to learn the type classes. I'm stuck understanding Alternative (and MonadPlus, for that matter).
The problems I'm having:
the 'pedia says that "the Alternative type class is for Applicative functors which also have a monoid structure." I don't get this -- doesn't Alternative mean something totally different from Monoid? i.e. I understood the point of the Alternative type class as picking between two things, whereas I understood Monoids as being about combining things.
why does Alternative need an empty method/member? I may be wrong, but it seems to not be used at all ... at least in the code I could find. And it seems not to fit with the theme of the class -- if I have two things, and need to pick one, what do I need an 'empty' for?
why does the Alternative type class need an Applicative constraint, and why does it need a kind of * -> *? Why not just have <|> :: a -> a -> a? All of the instances could still be implemented in the same way ... I think (not sure). What value does it provide that Monoid doesn't?
what's the point of the MonadPlus type class? Can't I unlock all of its goodness by just using something as both a Monad and Alternative? Why not just ditch it? (I'm sure I'm wrong, but I don't have any counterexamples)
Hopefully all those questions are coherent ... !
Bounty update: #Antal's answer is a great start, but Q3 is still open: what does Alternative provide that Monoid doesn't? I find this answer unsatisfactory since it lacks concrete examples, and a specific discussion of how the higher-kindedness of Alternative distinguishes it from Monoid.
If it's to combine applicative's effects with Monoid's behavior, why not just:
liftA2 mappend
This is even more confusing for me because many Monoid instances are exactly the same as the Alternative instances.
That's why I'm looking for specific examples that show why Alternative is necessary, and how it's different -- or means something different -- from Monoid.
To begin with, let me offer short answers to each of these questions. I will then expand each into a longer detailed answer, but these short ones will hopefully help in navigating those.
No, Alternative and Monoid don’t mean different things; Alternative is for types which have the structure both of Applicative and of Monoid. “Picking” and “combining” are two different intuitions for the same broader concept.
Alternative contains empty as well as <|> because the designers thought this would be useful, and because this gives rise to a monoid. In terms of picking, empty corresponds to making an impossible choice.
We need both Alternative and Monoid because the former obeys (or should) more laws than the latter; these laws relate the monoidal and applicative structure of the type constructor. Additionally, Alternative can’t depend on the inner type, while Monoid can.
MonadPlus is slightly stronger than Alternative, as it must obey more laws; these laws relate the monoidal structure to the monadic structure in addition to the applicative structure. If you have instances of both, they should coincide.
Doesn’t Alternative mean something totally different from Monoid?
Not really! Part of the reason for your confusion is that the Haskell Monoid class uses some pretty bad (well, insufficiently general) names. This is how a mathematician would define a monoid (being very explicit about it):
Definition. A monoid is a set M equipped with a distinguished element ε ∈ M and a binary operator · : M × M → M, denoted by juxtaposition, such that the following two conditions hold:
ε is the identity: for all m ∈ M, mε = εm = m.
· is associative: for all m₁,m₂,m₃ ∈ M, (m₁m₂)m₃ = m₁(m₂m₃).
That’s it. In Haskell, ε is spelled mempty and · is spelled mappend (or, these days, <>), and the set M is the type M in instance Monoid M where ....
Looking at this definition, we see that it says nothing about “combining” (or about “picking,” for that matter). It says things about · and about ε, but that’s it. Now, it’s certainly true that combining things works well with this structure: ε corresponds to having no things, and m₁m₂ says that if I glom m₁ and m₂’s stuff together, I can get a new thing containing all their stuff. But here’s an alternative intuition: ε corresponds to no choices at all, and m₁m₂ corresponds to a choice between m₁ and m₂. This is the “picking” intuition. Note that both obey the monoid laws:
Having nothing at all and having no choice are both the identity.
If I have no stuff and glom it together with some stuff, I end up with that same stuff again.
If I have a choice between no choice at all (something impossible) and some other choice, I have to pick the other (possible) choice.
Glomming collections together and making a choice are both associative.
If I have three collections of things, it doesn’t matter if I glom the first two together and then the third, or the last two together and then the first; either way, I end up with the same total glommed collection.
If I have a choice between three things, it doesn’t matter if I (a) first choose between first-or-second and third and then, if I need to, between first and second, or (b) first choose between first and second-or-third and then, if I need to, between second and third. Either way, I can pick what I want.
(Note: I’m playing fast and loose here; that’s why it’s intuition. For instance, it’s important to remember that · need not be commutative, which the above glosses over: it’s perfectly possible that m₁m₂ ≠ m₂m₁.)
Behold: both these sorts of things (and many others—is multiplying numbers really either “combining” or “picking”?) obey the same rules. Having an intuition is important to develop understanding, but it’s the rules and definitions that determine what’s actually going on.
And the best part is that these both of these intuitions can be interpreted by the same carrier! Let M be some set of sets (not a set of all sets!) containing the empty set, let ε be the empty set ∅, and let · be set union ∪. It is easy to see that ∅ is an identity for ∪, and that ∪ is associative, so we can conclude that (M,∅,∪) is a monoid. Now:
If we think about sets as being collections of things, then ∪ corresponds to glomming them together to get more things—the “combining” intuition.
If we think about sets as representing possible actions, then ∪ corresponds to increasing your pool of possible actions to pick from—the “picking” intuition.
And this is exactly what’s going on with [] in Haskell: [a] is a Monoid for all a, and [] as an applicative functor (and monad) is used to represent nondeterminism. Both the combining and the picking intuitions coincide at the same type: mempty = empty = [] and mappend = (<|>) = (++).
So the Alternative class is just there to represent objects which (a) are applicative functors, and (b) when instantiated at a type, have a value and a binary function on them which follow some rules. Which rules? The monoid rules. Why? Because it turns out to be useful :-)
Why does Alternative need an empty method/member?
Well, the snarky answer is “because Alternative represents a monoid structure.” But the real question is: why a monoid structure? Why not just a semigroup, a monoid without ε? One answer is to claim that monoids are just more useful. I think many people (but perhaps not Edward Kmett) would agree with this; almost all of the time, if you have a sensible (<|>)/mappend/·, you’ll be able to define a sensible empty/mempty/ε. On the other hand, having the extra generality is nice, since it lets you place more things under the umbrella.
You also want to know how this meshes with the “picking” intuition. Keeping in mind that, in some sense, the right answer is “know when to abandon the ‘picking’ intuition,” I think you can unify the two. Consider [], the applicative functor for nondeterminism. If I combine two values of type [a] with (<|>), that corresponds to nondeterministically picking either an action from the left or an action from the right. But sometimes, you’re going to have no possible actions on one side—and that’s fine. Similarly, if we consider parsers, (<|>) represents a parser which parses either what’s on the left or what’s on the right (it “picks”). And if you have a parser which always fails, that ends up being an identity: if you pick it, you immediately reject that pick and try the other one.
All this said, remember that it would be entirely possible to have a class almost like Alternative, but lacking empty. That would be perfectly valid—it could even be a superclass of Alternative—but happens not to be what Haskell did. Presumably this is out of a guess as to what’s useful.
Why does the Alternative type class need an Applicative constraint, and why does it need a kind of * -> *? … Why not just [use] liftA2 mappend?
Well, let’s consider each of these three proposed changes: getting rid of the Applicative constraint for Alternative; changing the kind of Alternative’s argument; and using liftA2 mappend instead of <|> and pure mempty instead of empty. We’ll look at this third change first, since it’s the most different. Suppose we got rid of Alternative entirely, and replaced the class with two plain functions:
fempty :: (Applicative f, Monoid a) => f a
fempty = pure mempty
(>|<) :: (Applicative f, Monoid a) => f a -> f a -> f a
(>|<) = liftA2 mappend
We could even keep the definitions of some and many. And this does give us a monoid structure, it’s true. But it seems like it gives us the wrong one . Should Just fst >|< Just snd fail, since (a,a) -> a isn’t an instance of Monoid? No, but that’s what the above code would result in. The monoid instance we want is one that’s inner-type agnostic (to borrow terminology from Matthew Farkas-Dyck in a very related haskell-cafe discussion which asks some very similar questions); the Alternative structure is about a monoid determined by f’s structure, not the structure of f’s argument.
Now that we think we want to leave Alternative as some sort of type class, let’s look at the two proposed ways to change it. If we change the kind, we have to get rid of the Applicative constraint; Applicative only talks about things of kind * -> *, and so there’s no way to refer to it. That leaves two possible changes; the first, more minor, change is to get rid of the Applicative constraint but leave the kind alone:
class Alternative' f where
empty' :: f a
(<||>) :: f a -> f a -> f a
The other, larger, change is to get rid of the Applicative constraint and change the kind:
class Alternative'' a where
empty'' :: a
(<|||>) :: a -> a -> a
In both cases, we have to get rid of some/many, but that’s OK; we can define them as standalone functions with the type (Applicative f, Alternative' f) => f a -> f [a] or (Applicative f, Alternative'' (f [a])) => f a -> f [a].
Now, in the second case, where we change the kind of the type variable, we see that our class is exactly the same as Monoid (or, if you still want to remove empty'', Semigroup), so there’s no advantage to having a separate class. And in fact, even if we leave the kind variable alone but remove the Applicative constraint, Alternative just becomes forall a. Monoid (f a), although we can’t write these quantified constraints in Haskell, not even with all the fancy GHC extensions. (Note that this expresses the inner-type–agnosticism mentioned above.) Thus, if we can make either of these changes, then we have no reason to keep Alternative (except for being able to express that quantified constraint, but that hardly seems compelling).
So the question boils down to “is there a relationship between the Alternative parts and the Applicative parts of an f which is an instance of both?” And while there’s nothing in the documentation, I’m going to take a stand and say yes—or at the very least, there ought to be. I think that Alternative is supposed to obey some laws relating to Applicative (in addition to the monoid laws); in particular, I think those laws are something like
Right distributivity (of <*>): (f <|> g) <*> a = (f <*> a) <|> (g <*> a)
Right absorption (for <*>): empty <*> a = empty
Left distributivity (of fmap): f <$> (a <|> b) = (f <$> a) <|> (f <$> b)
Left absorption (for fmap): f <$> empty = empty
These laws appear to be true for [] and Maybe, and (pretending its MonadPlus instance is an Alternative instance) IO, but I haven’t done any proofs or exhaustive testing. (For instance, I originally thought that left distributivity held for <*>, but this “performs the effects” in the wrong order for [].) By way of analogy, though, it is true that MonadPlus is expected to obey similar laws (although there is apparently some ambiguity about which). I had originally wanted to claim a third law, which seems natural:
Left absorption (for <*>): a <*> empty = empty
However, although I believe [] and Maybe obey this law, IO doesn’t, and I think (for reasons that will become apparent in the next couple of paragraphs) it’s best not to require it.
And indeed, it appears that Edward Kmett has some slides where he espouses a similar view; to get into that, we’ll need to take brief digression involving some more mathematical jargon. The final slide, “I Want More Structure,” says that “A Monoid is to an Applicative as a Right Seminearring is to an Alternative,” and “If you throw away the argument of an Applicative, you get a Monoid, if you throw away the argument of an Alternative you get a RightSemiNearRing.”
Right seminearrings? “How did right seminearrings get into it?” I hear you cry. Well,
Definition. A right near-semiring (also right seminearring, but the former seems to be used more on Google) is a quadruple (R,+,·,0) where (R,+,0) is a monoid, (R,·) is a semigroup, and the following two conditions hold:
· is right-distributive over +: for all r,s,t ∈ R, (s + t)r = sr + tr.
0 is right-absorbing for ·: for all r ∈ R, 0r = 0.
A left near-semiring is defined analogously.
Now, this doesn’t quite work, because <*> is not truly associative or a binary operator—the types don’t match. I think this is what Edward Kmett is getting at when he talks about “throw[ing] away the argument.” Another option might be to say (I’m unsure if this is right) that we actually want (f a, <|>, <*>, empty) to form a right near-semiringoid, where the “-oid” suffix indicates that the binary operators can only be applied to specific pairs of elements (à la groupoids). And we’d also want to say that (f a, <|>, <$>, empty) was a left near-semiringoid, although this could conceivably follow from the combination of the Applicative laws and the right near-semiringoid structure. But now I’m getting in over my head, and this isn’t deeply relevant anyway.
At any rate, these laws, being stronger than the monoid laws, mean that perfectly valid Monoid instances would become invalid Alternative instances. There are (at least) two examples of this in the standard library: Monoid a => (a,) and Maybe. Let’s look at each of them quickly.
Given any two monoids, their product is a monoid; consequently, tuples can be made an instance of Monoid in the obvious way (reformatting the base package’s source):
instance (Monoid a, Monoid b) => Monoid (a,b) where
mempty = (mempty, mempty)
(a1,b1) `mappend` (a2,b2) = (a1 `mappend` a2, b1 `mappend` b2)
Similarly, we can make tuples whose first component is an element of a monoid into an instance of Applicative by accumulating the monoid elements (reformatting the base package’s source):
instance Monoid a => Applicative ((,) a) where
pure x = (mempty, x)
(u, f) <*> (v, x) = (u `mappend` v, f x)
However, tuples aren’t an instance of Alternative, because they can’t be—the monoidal structure over Monoid a => (a,b) isn’t present for all types b, and Alternative’s monoidal structure must be inner-type agnostic. Not only must b be a monad, to be able to express (f <> g) <*> a, we need to use the Monoid instance for functions, which is for functions of the form Monoid b => a -> b. And even in the case where we have all the necessary monoidal structure, it violates all four of the Alternative laws. To see this, let ssf n = (Sum n, (<> Sum n)) and let ssn = (Sum n, Sum n). Then, writing (<>) for mappend, we get the following results (which can be checked in GHCi, with the occasional type annotation):
Right distributivity:
(ssf 1 <> ssf 1) <*> ssn 1 = (Sum 3, Sum 4)
(ssf 1 <*> ssn 1) <> (ssf 1 <*> ssn 1) = (Sum 4, Sum 4)
Right absorption:
mempty <*> ssn 1 = (Sum 1, Sum 0)
mempty = (Sum 0, Sum 0)
Left distributivity:
(<> Sum 1) <$> (ssn 1 <> ssn 1) = (Sum 2, Sum 3)
((<> Sum 1) <$> ssn 1) <> ((<> Sum 1) <$> ssn 1) = (Sum 2, Sum 4)
Left absorption:
(<> Sum 1) <$> mempty = (Sum 0, Sum 1)
mempty = (Sum 1, Sum 1)
Next, consider Maybe. As it stands, Maybe’s Monoid and Alternative instances disagree. (Although the haskell-cafe discussion I mention at the beginning of this section proposes changing this, there’s an Option newtype from the semigroups package which would produce the same effect.) As a Monoid, Maybe lifts semigroups into monoids by using Nothing as the identity; since the base package doesn’t have a semigroup class, it just lifts monoids, and so we get (reformatting the base package’s source):
instance Monoid a => Monoid (Maybe a) where
mempty = Nothing
Nothing `mappend` m = m
m `mappend` Nothing = m
Just m1 `mappend` Just m2 = Just (m1 `mappend` m2)
On the other hand, as an Alternative, Maybe represents prioritized choice with failure, and so we get (again reformatting the base package’s source):
instance Alternative Maybe where
empty = Nothing
Nothing <|> r = r
l <|> _ = l
And it turns out that only the latter satisfies the Alternative laws. The Monoid instance fails less badly than (,)’s; it does obey the laws with respect to <*>, although almost by accident—it comes form the behavior of the only instance of Monoid for functions, which (as mentioned above), lifts functions that return monoids into the reader applicative functor. If you work it out (it’s all very mechanical), you’ll find that right distributivity and right absorption for <*> all hold for both the Alternative and Monoid instances, as does left absorption for fmap. And left distributivity for fmap does hold for the Alternative instance, as follows:
f <$> (Nothing <|> b)
= f <$> b by the definition of (<|>)
= Nothing <|> (f <$> b) by the definition of (<|>)
= (f <$> Nothing) <|> (f <$> b) by the definition of (<$>)
f <$> (Just a <|> b)
= f <$> Just a by the definition of (<|>)
= Just (f a) by the definition of (<$>)
= Just (f a) <|> (f <$> b) by the definition of (<|>)
= (f <$> Just a) <|> (f <$> b) by the definition of (<$>)
However, it fails for the Monoid instance; writing (<>) for mappend, we have:
(<> Sum 1) <$> (Just (Sum 0) <> Just (Sum 0)) = Just (Sum 1)
((<> Sum 1) <$> Just (Sum 0)) <> ((<> Sum 1) <$> Just (Sum 0)) = Just (Sum 2)
Now, there is one caveat to this example. If you only require that Alternatives be compatibility with <*>, and not with <$>, then Maybe is fine. Edward Kmett’s slides, mentioned above, don’t make reference to <$>, but I think it seems reasonable to require laws with respect to it as well; nevertheless, I can’t find anything to back me up on this.
Thus, we can conclude that being an Alternative is a stronger requirement than being a Monoid, and so it requires a different class. The purest example of this would be a type with an inner-type agnostic Monoid instance and an Applicative instance which were incompatible with each other; however, there aren’t any such types in the base package, and I can’t think of any. (It’s possible none exist, although I’d be surprised.) Nevertheless, these inner-type gnostic examples demonstrate why the two type classes must be different.
What’s the point of the MonadPlus type class?
MonadPlus, like Alternative, is a strengthening of Monoid, but with respect to Monad instead of Applicative. According to Edward Kmett in his answer to the question “Distinction between typeclasses MonadPlus, Alternative, and Monoid?”, MonadPlus is also stronger than Alternative: the law empty <*> a, for instance, doesn’t imply that empty >>= f. AndrewC provides two examples of this: Maybe and its dual. The issue is complicated by the fact that there are two potential sets of laws for MonadPlus. It is universally agreed that MonadPlus is supposed to form a monoid with mplus and mempty, and it’s supposed to satisfy the left zero law, mempty >>= f = mempty. Hhowever, some MonadPlusses satisfy left distribution, mplus a b >>= f = mplus (a >>= f) (b >>= f); and others satisfy left catch, mplus (return a) b = return a. (Note that left zero/distribution for MonadPlus are analogous to right distributivity/absorption for Alternative; (<*>) is more analogous to (=<<) than (>>=).) Left distribution is probably “better,” so any MonadPlus instance which satisfies left catch, such as Maybe, is an Alternative but not the first kind of MonadPlus. And since left catch relies on ordering, you can imagine a newtype wrapper for Maybe whose Alternative instance is right-biased instead of left-biased: a <|> Just b = Just b. This will satisfy neither left distribution nor left catch, but will be a perfectly valid Alternative.
However, since any type which is a MonadPlus ought to have its instance coincide with its Alternative instance (I believe this is required in the same way that it is required that ap and (<*>) are equal for Monads that are Applicatives), you could imagine defining the MonadPlus class instead as
class (Monad m, Alternative m) => MonadPlus' m
The class doesn’t need to declare new functions; it’s just a promise about the laws obeyed by empty and (<|>) for the given type. This design technique isn’t used in the Haskell standard libraries, but is used in some more mathematically-minded packages for similar purposes; for instance, the lattices package uses it to express the idea that a lattice is just a join semilattice and a meet semilattice over the same type which are linked by absorption laws.
The reason you can’t do the same for Alternative, even if you wanted to guarantee that Alternative and Monoid always coincided, is because of the kind mismatch. The desired class declaration would have the form
class (Applicative f, forall a. Monoid (f a)) => Alternative''' f
but (as mentioned far above) not even GHC Haskell supports quantified constraints.
Also, note that having Alternative as be a superclass of MonadPlus would require Applicative being a superclass of Monad, so good luck getting that to happen. If you run into that problem, there’s always the WrappedMonad newtype, which turns any Monad into an Applicative in the obvious way; there’s an instance MonadPlus m => Alternative (WrappedMonad m) where ... which does exactly what you’d expect.
import Data.Monoid
import Control.Applicative
Let's trace through an example of how Monoid and Alternative interact with the Maybe functor and the ZipList functor, but let's start from scratch, partly to get all the definitions fresh in our minds, partly to stop from switching tabs to bits of hackage all the time, but mainly so I can run this past ghci to correct my typos!
(<>) :: Monoid a => a -> a -> a
(<>) = mappend -- I'll be using <> freely instead of `mappend`.
Here's the Maybe clone:
data Perhaps a = Yes a | No deriving (Eq, Show)
instance Functor Perhaps where
fmap f (Yes a) = Yes (f a)
fmap f No = No
instance Applicative Perhaps where
pure a = Yes a
No <*> _ = No
_ <*> No = No
Yes f <*> Yes x = Yes (f x)
and now ZipList:
data Zip a = Zip [a] deriving (Eq,Show)
instance Functor Zip where
fmap f (Zip xs) = Zip (map f xs)
instance Applicative Zip where
Zip fs <*> Zip xs = Zip (zipWith id fs xs) -- zip them up, applying the fs to the xs
pure a = Zip (repeat a) -- infinite so that when you zip with something, lengths don't change
Structure 1: combining elements: Monoid
Maybe clone
First let's look at Perhaps String. There are two ways of combining them. Firstly concatenation
(<++>) :: Perhaps String -> Perhaps String -> Perhaps String
Yes xs <++> Yes ys = Yes (xs ++ ys)
Yes xs <++> No = Yes xs
No <++> Yes ys = Yes ys
No <++> No = No
Concatenation works inherently at the String level, not really the Perhaps level, by treating No as if it were Yes []. It's equal to liftA2 (++). It's sensible and useful, but maybe we could generalise from just using ++ to using any way of combining - any Monoid then!
(<++>) :: Monoid a => Perhaps a -> Perhaps a -> Perhaps a
Yes xs <++> Yes ys = Yes (xs `mappend` ys)
Yes xs <++> No = Yes xs
No <++> Yes ys = Yes ys
No <++> No = No
This monoid structure for Perhaps tries to work as much as possible at the a level. Notice the Monoid a constraint, telling us we're using structure from the a level. This isn't an Alternative structure, it's a derived (lifted) Monoid structure.
instance Monoid a => Monoid (Perhaps a) where
mappend = (<++>)
mempty = No
Here I used the structure of the data a to add structure to the whole thing. If I were combining Sets, I'd be able to add an Ord a context instead.
ZipList clone
So how should we combine elements with a zipList? What should these zip to if we're combining them?
Zip ["HELLO","MUM","HOW","ARE","YOU?"]
<> Zip ["this", "is", "fun"]
= Zip ["HELLO" ? "this", "MUM" ? "is", "HOW" ? "fun"]
mempty = ["","","","",..] -- sensible zero element for zipping with ?
But what should we use for ?. I say the only sensible choice here is ++. Actually, for lists, (<>) = (++)
Zip [Just 1, Nothing, Just 3, Just 4]
<> Zip [Just 40, Just 70, Nothing]
= Zip [Just 1 ? Just 40, Nothing ? Just 70, Just 3 ? Nothing]
mempty = [Nothing, Nothing, Nothing, .....] -- sensible zero element
But what can we use for ? I say that we're meant to be combining elements, so we should use the element-combining operator from Monoid again: <>.
instance Monoid a => Monoid (Zip a) where
Zip as `mappend` Zip bs = Zip (zipWith (<>) as bs) -- zipWith the internal mappend
mempty = Zip (repeat mempty) -- repeat the internal mempty
This is the only sensible way of combining the elements using a zip - so it's the only sensible monoid instance.
Interestingly, that doesn't work for the Maybe example above, because Haskell doesn't know how to combine Ints - should it use + or *? To get a Monoid instance on numerical data, you wrap them in Sum or Product to tell it which monoid to use.
Zip [Just (Sum 1), Nothing, Just (Sum 3), Just (Sum 4)] <>
Zip [Just (Sum 40), Just (Sum 70), Nothing]
= Zip [Just (Sum 41),Just (Sum 70), Just (Sum 3)]
Zip [Product 5,Product 10,Product 15]
<> Zip [Product 3, Product 4]
= Zip [Product 15,Product 40]
Key point
Notice the fact that the type in a Monoid has kind * is exactly what allows us to put the Monoid a context here - we could also add Eq a or Ord a. In a Monoid, the raw elements matter. A Monoid instance is designed to let you manipulate and combine the data inside the structure.
Structure 2: higher-level choice: Alternative
A choice operator is similar, but also different.
Maybe clone
(<||>) :: Perhaps String -> Perhaps String -> Perhaps String
Yes xs <||> Yes ys = Yes xs -- if we can have both, choose the left one
Yes xs <||> No = Yes xs
No <||> Yes ys = Yes ys
No <||> No = No
Here there's no concatenation - we didn't use ++ at all - this combination works purely at the Perhaps level, so let's change the type signature to
(<||>) :: Perhaps a -> Perhaps a -> Perhaps a
Yes xs <||> Yes ys = Yes xs -- if we can have both, choose the left one
Yes xs <||> No = Yes xs
No <||> Yes ys = Yes ys
No <||> No = No
Notice there's no constraint - we're not using the structure from the a level, just structure at the Perhaps level. This is an Alternative structure.
instance Alternative Perhaps where
(<|>) = (<||>)
empty = No
ZipList clone
How should we choose between two ziplists?
Zip [1,3,4] <|> Zip [10,20,30,40] = ????
It would be very tempting to use <|> on the elements, but we can't because the type of the elements isn't available to us. Let's start with the empty. It can't use an element because we don't know the type of the elements when defining an Alternative, so it has to be Zip []. We need it to be a left (and preferably right) identity for <|>, so
Zip [] <|> Zip ys = Zip ys
Zip xs <|> Zip [] = Zip xs
There are two sensible choices for Zip [1,3,4] <|> Zip [10,20,30,40]:
Zip [1,3,4] because it's first - consistent with Maybe
Zip [10,20,30,40] because it's longest - consistent with Zip [] being discarded
Well that's easy to decide: since pure x = Zip (repeat x), both lists might be infinite, so comparing them for length might never terminate, so it has to be pick the first one. Thus the only sensible Alternative instance is:
instance Alternative Zip where
empty = Zip []
Zip [] <|> x = x
Zip xs <|> _ = Zip xs
This is the only sensible Alternative we could have defined. Notice how different it is from the Monoid instance, because we couldn't mess with the elements, we couldn't even look at them.
Key Point
Notice that because Alternative takes a constructor of kind * -> * there is no possible way to add an Ord a or Eq a or Monoid a context. An Alternative is not allowed to use any information about the data inside the structure. You cannot, no matter how much you would like to, do anything to the data, except possibly throw it away.
Key point: What's the difference between Alternative and Monoid?
Not a lot - they're both monoids, but to summarise the last two sections:
Monoid * instances make it possible to combine internal data. Alternative (* -> *) instances make it impossible. Monoid provides flexibility, Alternative provides guarantees. The kinds * and (* -> *) are the main drivers of this difference. Having them both allows you to use both sorts of operations.
This is the right thing, and our two flavours are both appropriate. The Monoid instance for Perhaps String represents putting together all characters, the Alternative instance represents a choice between Strings.
There is nothing wrong with the Monoid instance for Maybe - it's doing its job, combining data.
There's nothing wrong with the Alternative instance for Maybe - it's doing its job, choosing between things.
The Monoid instance for Zip combines its elements. The Alternative instance for Zip is forced to choose one of the lists - the first non-empty one.
It's good to be able to do both.
What's the Applicative context any use for?
There's some interaction between choosing and applying. See Antal S-Z's laws stated in his question or in the middle of his answer here.
From a practical point of view, it's useful because Alternative is something that is used for some Applicative Functors to choose. The functionality was being used for Applicatives, and so a general interface class was invented. Applicative Functors are good for representing computations that produce values (IO, Parser, Input UI element,...) and some of them have to handle failure - Alternative is needed.
Why does Alternative have empty?
why does Alternative need an empty method/member? I may be wrong, but it seems to not be used at all ... at least in the code I could find. And it seems not to fit with the theme of the class -- if I have two things, and need to pick one, what do I need an 'empty' for?
That's like asking why addition needs a 0 - if you want to add stuff, what's the point in having something that doesn't add anything? The answer is that 0 is the crucual pivotal number around which everything revolves in addition, just like 1 is crucial for multiplication, [] is crucial for lists (and y=e^x is crucial for calculus). In practical terms, you use these do-nothing elements to start your building:
sum = foldr (+) 0
concat = foldr (++) []
msum = foldr (`mappend`) mempty -- any Monoid
whichEverWorksFirst = foldr (<|>) empty -- any Alternative
Can't we replace MonadPlus with Monad+Alternative?
what's the point of the MonadPlus type class? Can't I unlock all of its goodness by just using something as both a Monad and Alternative? Why not just ditch it? (I'm sure I'm wrong, but I don't have any counterexamples)
You're not wrong, there aren't any counterexamples!
Your interesting question has got Antal S-Z, Petr Pudlák and I delved into what the relationship between MonadPlus and Applicative really is. The answer,
here
and here
is that anything that's a MonadPlus (in the left distribution sense - follow links for details) is also an Alternative, but not the other way around.
This means that if you make an instance of Monad and MonadPlus, it satisfies the conditions for Applicative and Alternative anyway. This means if you follow the rules for MonadPlus (with left dist), you may as well have made your Monad an Applicative and used Alternative.
If we remove the MonadPlus class, though, we remove a sensible place for the rules to be documented, and you lose the ability to specify that something's Alternative without being MonadPlus (which technically we ought to have done for Maybe). These are theoretical reasons. The practical reason is that it would break existing code. (Which is also why neither Applicative nor Functor are superclasses of Monad.)
Aren't Alternative and Monoid the same? Aren't Alternative and Monoid completely different?
the 'pedia says that "the Alternative type class is for Applicative functors which also have a monoid structure." I don't get this -- doesn't Alternative mean something totally different from Monoid? i.e. I understood the point of the Alternative type class as picking between two things, whereas I understood Monoids as being about combining things.
Monoid and Alternative are two ways of getting one object from two in a sensible way. Maths doesn't care whether you're choosing, combining, mixing or blowing up your data, which is why Alternative was referred to as a Monoid for Applicative. You seem to be at home with that concept now, but you now say
for types that have both an Alternative and a Monoid instance, the instances are intended to be the same
I disagree with this, and I think my Maybe and ZipList examples are carefully explained as to why they're different. If anything, I think it should be rare that they're the same. I can only think of one example, plain lists, where this is appropriate. That's because lists are a fundamental example of a monoid with ++, but also lists are used in some contexts as an indeterminate choice of elements, so <|> should also be ++.
Summary
We need to define (instances that provide the same operations as) Monoid instances for some applicative functors, that genuinely combine at the applicative functor level, and not just lifting lower level monoids. The example error below from litvar = liftA2 mappend literal variable shows that <|> cannot in general be defined as liftA2 mappend; <|> works in this case by combining parsers, not their data.
If we used Monoid directly, we'd need language extensions to define the instances. Alternative is higher kinded so you can make these instances without requiring language extensions.
Example: Parsers
Let's imagine we're parsing some declarations, so we import everything we're going to need
import Text.Parsec
import Text.Parsec.String
import Control.Applicative ((<$>),(<*>),liftA2,empty)
import Data.Monoid
import Data.Char
and think about how we'll parse a type. We choose simplistic:
data Type = Literal String | Variable String deriving Show
examples = [Literal "Int",Variable "a"]
Now let's write a parser for literal types:
literal :: Parser Type
literal = fmap Literal $ (:) <$> upper <*> many alphaNum
Meaning: parse an uppercase character, then many alphaNumeric characters, combine the results into a single String with the pure function (:). Afterwards, apply the pure function Literal to turn those Strings into Types. We'll parse variable types exactly the same way, except for starting with a lowercase letter:
variable :: Parser Type
variable = fmap Variable $ (:) <$> lower <*> many alphaNum
That's great, and parseTest literal "Bool" == Literal "Bool" exactly as we'd hoped.
Question 3a: If it's to combine applicative's effects with Monoid's behavior, why not just liftA2 mappend
Edit:Oops - forgot to actually use <|>!
Now let's combine these two parsers using Alternative:
types :: Parser Type
types = literal <|> variable
This can parse any Type: parseTest types "Int" == Literal "Bool" and parseTest types "a" == Variable "a".
This combines the two parsers, not the two values. That's the sense in which it works at the Applicative Functor level rather than the data level.
However, if we try:
litvar = liftA2 mappend literal variable
that would be asking the compiler to combine the two values that they generate, at the data level.
We get
No instance for (Monoid Type)
arising from a use of `mappend'
Possible fix: add an instance declaration for (Monoid Type)
In the first argument of `liftA2', namely `mappend'
In the expression: liftA2 mappend literal variable
In an equation for `litvar':
litvar = liftA2 mappend literal variable
So we found out the first thing; the Alternative class does something genuinely different to liftA2 mappend, becuase it combines objects at a different level - it combines the parsers, not the parsed data. If you like to think of it this way, it's combination at the genuinely higher-kind level, not merely a lift. I don't like saying it that way, because Parser Type has kind *, but it is true to say we're combining the Parsers, not the Types.
(Even for types with a Monoid instance, liftA2 mappend won't give you the same parser as <|>. If you try it on Parser String you'll get liftA2 mappend which parses one after the other then concatenates, versus <|> which will try the first parser and default to the second if it failed.)
Question 3b: In what way does Alternative's <|> :: f a -> f a -> f a differ from Monoid's mappend :: b -> b -> b?
Firstly, you're right to note that it doesn't provide new functionality over a Monoid instance.
Secondly, however, there's an issue with using Monoid directly:
Let's try to use mappend on parsers, at the same time as showing it's the same structure as Alternative:
instance Monoid (Parser a) where
mempty = empty
mappend = (<|>)
Oops! We get
Illegal instance declaration for `Monoid (Parser a)'
(All instance types must be of the form (T t1 ... tn)
where T is not a synonym.
Use -XTypeSynonymInstances if you want to disable this.)
In the instance declaration for `Monoid (Parser a)'
So if you have an applicative functor f, the Alternative instance shows that f a is a monoid, but you could only declare that as a Monoid with a language extension.
Once we add {-# LANGUAGE TypeSynonymInstances #-} at the top of the file, we're fine and can define
typeParser = literal `mappend` variable
and to our delight, it works: parseTest typeParser "Yes" == Literal "Yes" and parseTest typeParser "a" == Literal "a".
Even if you don't have any synonyms (Parser and String are synonyms, so they're out), you'll still need {-# LANGUAGE FlexibleInstances #-} to define an instance like this one:
data MyMaybe a = MyJust a | MyNothing deriving Show
instance Monoid (MyMaybe Int) where
mempty = MyNothing
mappend MyNothing x = x
mappend x MyNothing = x
mappend (MyJust a) (MyJust b) = MyJust (a + b)
(The monoid instance for Maybe gets around this by lifting the underlying monoid.)
Making a standard library unnecessarily dependent on language extensions is clearly undesirable.
So there you have it. Alternative is just Monoid for Applicative Functors (and isn't just a lift of a Monoid). It needs the higher-kinded type f a -> f a -> f a so you can define one without language extensions.
Your other Questions, for completeness:
Why does Alternative need an empty method/member?
Because having an identity for an operation is sometimes useful.
For example, you can define anyA = foldr (<|>) empty without using tedious edge cases.
what's the point of the MonadPlus type class? Can't I unlock all of its goodness by just using something as both a Monad and Alternative?
No. I refer you back to the question you linked to:
Moreover, even if Applicative was a superclass of Monad, you'd wind up needing the MonadPlus class anyways, because obeying empty <*> m = empty isn't strictly enough to prove that empty >>= f = empty.
....and I've come up with an example: Maybe. I explain in detail, with proof in this answer to Antal's question. For the purposes of this answer, it's worth noting that I was able to use >>= to make the MonadPlus instance that broke the Alternative laws.
Monoid structure is useful. Alternative is the best way of providing it for Applicative Functors.
I won't cover MonadPlus because there is disagreement about its laws.
After trying and failing to find any meaningful examples in which the structure of an Applicative leads naturally to an Alternative instance that disagrees with its Monoid instance*, I finally came up with this:
Alternative's laws are more strict than Monoid's, because the result cannot depend on the inner type. This excludes a large number of Monoid instances from being Alternatives.
These datatypes allow partial (meaning that they only work for some inner types) Monoid instances which are forbidden by the extra 'structure' of the * -> * kind. Examples:
the standard Maybe instance for Monoid assumes that the inner type is Monoid => not an Alternative
ZipLists, tuples, and functions can all be made Monoids, if their inner types are Monoids => not Alternatives
sequences that have at least one element -- cannot be Alternatives because there's no empty:
data Seq a
= End a
| Cons a (Seq a)
deriving (Show, Eq, Ord)
On the other hand, some data types cannot be made Alternatives because they're *-kinded:
unit -- ()
Ordering
numbers, booleans
My inferred conclusion: for types that have both an Alternative and a Monoid instance, the instances are intended to be the same. See also this answer.
excluding Maybe, which I argue doesn't count because its standard instance should not require Monoid for the inner type, in which case it would be identical to Alternative
I understood the point of the Alternative type class as picking between two things, whereas I understood Monoids as being about combining things.
If you think about this for a moment, they are the same.
The + combines things (usually numbers), and it's type signature is Int -> Int -> Int (or whatever).
The <|> operator selects between alternatives, and it's type signature is also the same: take two matching things and return a combined thing.