-<!-- λ Λ ∀ ≡ α β γ ρ ω Ω -->
+<!-- λ Λ ∀ ≡ α β γ ρ ω Ω ○ μ η δ ζ ξ ⋆ ★ • ∙ ● 𝟎 𝟏 𝟐 𝟘 𝟙 𝟚 𝟬 𝟭 𝟮 -->
<!-- Loved this one: http://www.stephendiehl.com/posts/monads.html -->
Introducing Monads
of monads, followed by instances of monads from the philosophical and
linguistics literature.
-> <small>After you've read this once and are coming back to re-read it to try to digest the details further, the "endofunctors" that slogan is talking about are a combination of our boxes and their associated maps. Their "monoidal" character is captured in the Monad Laws, where a "monoid"---don't confuse with a mon*ad*---is a simpler algebraic notion, meaning a universe with some associative operation that has an identity. For advanced study, here are some further links on the relation between monads as we're working with them and monads as they appear in category theory:
+> <small>After you've read this once and are coming back to re-read it to try to digest the details further, the "endofunctors" that slogan is talking about are a combination of our boxes and their associated maps. Their "monoidal" character is captured in the Monad Laws, where a "monoid"---don't confuse with a mon*ad*---is a simpler algebraic notion, meaning a universe with some associative operation that has an identity. For advanced study, here are some further links on the relation between monads as we're working with them and monads as they appear in Category Theory:
[1](http://en.wikipedia.org/wiki/Outline_of_category_theory)
[2](http://lambda1.jimpryor.net/advanced_topics/monads_in_category_theory/)
[3](http://en.wikibooks.org/wiki/Haskell/Category_theory)
P_3 ≡ ∀α. α -> α
P_4 ≡ ∀α. α -> β
-etc.
+and so on.
A *box type* will be a type expression that contains exactly one free
type variable. (You could extend this to expressions with more free variables; then you'd have
The idea is that whatever type the free type variable `α` might be instantiated to,
we will have a "type box" of a certain sort that "contains" values of type `α`. For instance,
-if `α list` is our box type, and `α` is the type `int`, then in this context, `int list`
+if `α list` is our box type, and `α` instantiates to the type `int`, then in this context, `int list`
is the type of a boxed integer.
-Warning: although our initial motivating examples are readily thought of as "containers" (lists, trees, and so on, with `α`s as their "elements"), with later examples we discuss it will be less natural to describe the boxed types that way. For example, where `R` is some fixed type, `R -> α` is a box type.
+Warning: although our initial motivating examples are readily thought of as "containers" (lists, trees, and so on, with `α`s as their "elements"), with later examples we discuss it will be less natural to describe the boxed types that way. For example, where `R` is some fixed type, `R -> α` will be one box type we work extensively with.
Also, for clarity: the *box type* is the type `α list` (or as we might just say, the `list` type operator); the *boxed type* is some specific instantiation of the free type variable `α`. We'll often write boxed types as a box containing what the free
-type variable instantiates to. So if our box type is `α list`, and `α` instantiates to the specific type `int`, we would write:
+type variable instantiates to. So if our box type is `α list`, and `α` instantiates to the specific type `int`, we write:
<code><u>int</u></code>
<code>P -> <u>Q</u></code>
That is, they are functions from values of one type `P` to a boxed type `Q`, for some choice of type expressions `P` and `Q`.
-For instance, the following are Kleisli arrows:
+For instance, the following are Kleisli arrow types:
<code>int -> <u>bool</u></code>
In the first, `P` has become `int` and `Q` has become `bool`. (The boxed type <code><u>Q</u></code> is <code><u>bool</u></code>).
-Note that the left-hand schema `P` is permitted to itself be a boxed
-type. That is, if `α list` is our box type, and `P` is to boxed type
-`int list`, we can write the boxed type that has `P` as its left-hand
-side as
+Note that either of the schemas `P` or `Q` are permitted to themselves be boxed
+types. That is, if `α list` is our box type, we can write the second type as:
<code><u>int</u> -> <u>int list</u></code>
-If it's clear that we're uniformly talking about the same box type (in
-this example, `α list`), we can equivalently write
+And also what the rhs there is a boxing of is itself a boxed type (with the same kind of box):, so we can write it as:
-<code><u>int</u> -> <u><u>int</u></u></code>
+<code><u>int</u> -> <span class="box2">int</span></code>
+
+We have to be careful though not to to unthinkingly equivocate between different kinds of boxes.
Here are some examples of values of these Kleisli arrow types, where the box type is `α list`, and the Kleisli arrow types are <code>int -> <u>int</u></code> (that is, `int -> int list`) or <code>int -> <u>bool</u></code>:
<code>m$ or mapply (/εm@plai/): <u>P -> Q</u> -> <u>P</u> -> <u>Q</u></code>
-> We'll use `m$` as an infix operator, reminiscent of `$` which is just ordinary function application (also expressed by mere juxtaposition). In the class presentation Jim called `m$` `●`. In Haskell, it's called `Control.Monad.ap` or `Control.Applicative.<*>`.
+> We'll use `m$` as a left-associative infix operator, reminiscent of (the right-associative) `$` which is just ordinary function application (also expressed by mere left-associative juxtaposition). In the class presentation Jim called `m$` `●`. In Haskell, it's called `Control.Monad.ap` or `Control.Applicative.<*>`.
<code><=< or mcomp : (Q -> <u>R</u>) -> (P -> <u>Q</u>) -> (P -> <u>R</u>)</code>
> In Haskell, this is `Control.Monad.join`. In other theoretical contexts it is sometimes called `μ`.
-Haskell uses the symbol `>>=` but calls it "bind". This is not well chosen from the perspective of formal semantics, but it's too deeply entrenched to change. We've at least preprended an `m` to the front of "bind".
+Haskell uses the symbol `>>=` but calls it "bind". This is not well chosen from the perspective of formal semantics, but it's too deeply entrenched to change. We've at least preprended an "m" to the front of "bind".
Haskell's names "return" and "pure" for `mid` are even less well chosen, and we think it will be clearer in our discussion to use a different name. (Also, in other theoretical contexts this notion goes by other names, anyway, like `unit` or `η` --- having nothing to do with `η`-reduction in the Lambda Calculus.)
1. <code>mid (id : P->P) : <u>P</u> -> <u>P</u></code> is a left identity for `m$`, that is: `(mid id) m$ xs = xs`
2. `mid (f a) = (mid f) m$ (mid a)`
- 3. The `map2`ing of composition onto boxes `fs` and `gs` of functions, when `m$`'d to a box `xs` of arguments == the `m$`ing of `fs` to the `m$`ing of `gs` to xs: `((mid ○) m$ fs m$ gs) m$ xs = fs m$ (gs m$ xs)`.
- 4. When the arguments are `mid`'d, the order of `m$`ing doesn't matter: `fs m$ (mid x) = (mid ($ x)) m$ fs`. In examples we'll be working with at first, order _never_ matters; but down the road, sometimes it will. This Law states a class of cases where it's guaranteed not to.
- 5. A consequence of the laws already stated is that when the functions are `mid`'d, the order of `m$`ing doesn't matter either: TODO
-
+ 3. The `map2`ing of composition onto boxes `fs` and `gs` of functions, when `m$`'d to a box `xs` of arguments == the `m$`ing of `fs` to the `m$`ing of `gs` to xs: `(mid (○) m$ fs m$ gs) m$ xs = fs m$ (gs m$ xs)`.
+ 4. When the arguments (the right-hand operand of `m$`) are an `mid`'d value, the order of `m$`ing doesn't matter: `fs m$ (mid x) = mid ($x) m$ fs`. (Though note that it's `mid ($x)`, or `mid (\f. f x)` that gets `m$`d onto `fs`, not the original `mid x`.) Here's an example where the order *does* matter: `[succ,pred] m$ [1,2] == [2,3,0,1]`, but `[($1),($2)] m$ [succ,pred] == [2,0,3,1]`. This Law states a class of cases where the order is guaranteed not to matter.
+ 5. A consequence of the laws already stated is that when the _left_-hand operand of `m$` is a `mid`'d value, the order of `m$`ing doesn't matter either: `mid f m$ xs == map (flip ($)) xs m$ mid f`.
+
+<!-- Probably there's a shorter proof, but:
+ mid T m$ xs m$ mid f
+== mid T m$ ((mid id) m$ xs) m$ mid f, by 1
+== mid (○) m$ mid T m$ mid id m$ xs m$ mid f, by 3
+== mid ($id) m$ (mid (○) m$ mid T) m$ xs m$ mid f, by 4
+== mid (○) m$ mid ($id) m$ mid (○) m$ mid T m$ xs m$ mid f, by 3
+== mid ((○) ($id)) m$ mid (○) m$ mid T m$ xs m$ mid f, by 2
+== mid ((○) ($id) (○)) m$ mid T m$ xs m$ mid f, by 2
+== mid id m$ mid T m$ xs m$ mid f, by definitions of ○ and $
+== mid T m$ xs m$ mid f, by 1
+== mid ($f) m$ (mid T m$ xs), by 4
+== mid (○) m$ mid ($f) m$ mid T m$ xs, by 3
+== mid ((○) ($f)) m$ mid T m$ xs, by 2
+== mid ((○) ($f) T) m$ xs, by 2
+== mid f m$ xs, by definitions of ○ and $ and T == flip ($)
+-->
* ***Monad*** (or "Composables") A MapNable box type is a *Monad* if there
is in addition an associative `mcomp` having `mid` as its left and
As hinted in last week's homework and explained in class, the operations available in a Mappable system exactly preserve the "structure" of the boxed type they're operating on, and moreover are only sensitive to what content is in the corresponding original position. If you say `map f [1,2,3]`, then what ends up in the first position of the result depends only on how `f` and `1` combine.
-For MapNable operations, on the other hand, the structure of the result may instead by a complex function of the structure of the original arguments. But only of their structure, not of their contents. And if you say `map2 f [10,20] [1,2,3]`, what ends up in the first position of the result depends only on how `f` and `10` and `1` combine.
+For MapNable operations, on the other hand, the structure of the result may instead be a complex function of the structure of the original arguments. But only of their structure, not of their contents. And if you say `map2 f [10,20] [1,2,3]`, what ends up in the first position of the result depends only on how `f` and `10` and `1` combine.
With `map`, you can supply an `f` such that `map f [3,2,0,1] == [[3,3,3],[2,2],[],[1]]`. But you can't transform `[3,2,0,1]` to `[3,3,3,2,2,1]`, and you can't do that with MapNable operations, either. That would involve the structure of the result (here, the length of the list) being sensitive to the content, and not merely the structure, of the original.
-For Monads (Composables), you can perform more radical transformations of that sort. For example, `join (map (\x. dup x x) [3,2,0,1])` would give us `[3,3,3,2,2,1]` (for a suitable definition of `dup`).
+For Monads (Composables), on the other hand, you can perform more radical transformations of that sort. For example, `join (map (\x. dup x x) [3,2,0,1])` would give us `[3,3,3,2,2,1]` (for a suitable definition of `dup`).
<!--
Some global transformations that we work with in semantics, like Veltman's test functions, can't directly be expressed in terms of the primitive Monad operations? For example, there's no `j` such that `xs >>= j == mzero` if `xs` anywhere contains the value `1`.
* <code>mzero</code> is a value of type <code><u>α</u></code> that is exemplified by `Nothing` for the box type `Maybe α` and by `[]` for the box type `List α`. It has the behavior that `anything m$ mzero == mzero == mzero m$ anything == mzero >>= anything`. In Haskell, this notion is called `Control.Applicative.empty` or `Control.Monad.mzero`.
-* Haskell has a notion `>>` definable as `\u v. map (const id) u m$ v`, or as `u >> v == u >>= const v`. This is often useful, and `u >> v` won't in general be identical to just `v`. For example, using the box type `List α`, `[1,2,3] >> [4,5] == [4,5,4,5,4,5]`. But in the special case of `mzero`, it is a consequence of what we said above that `anything >> mzero == mzero`. Haskell also calls `>>` `Control.Applicative.*>`.
+* Haskell has a notion `>>` definable as `\u v. map (const id) u m$ v`, or as `\u v. u >>= const v`. This is often useful, and `u >> v` won't in general be identical to just `v`. For example, using the box type `List α`, `[1,2,3] >> [4,5] == [4,5,4,5,4,5]`. But in the special case of `mzero`, it is a consequence of what we said above that `anything >> mzero == mzero`. Haskell also calls `>>` `Control.Applicative.*>`.
* Haskell has a correlative notion `Control.Applicative.<*`, definable as `\u v. map const u m$ v`. For example, `[1,2,3] <* [4,5] == [1,1,2,2,3,3]`. You might expect Haskell to call `<*` `<<`, but they don't. They used to use `<<` for `flip (>>)` instead, but now they seem not to use `<<` anymore.
In terms of the box analogy, the Identity box type is a completely invisible box. With the following
definitions:
- mid ≡ \p. p
- mcomp ≡ \f g x.f (g x)
+ mid ≡ \p. p, that is, our familiar combinator I
+ mcomp ≡ \f g x. f (g x), that is, ordinary function composition (○) (aka the B combinator)
Identity is a monad. Here is a demonstration that the laws hold:
`j 7` produced `[49, 14]`, which after being fed through `k` gave us `[49, 50, 14, 15]`.
-Contrast that to `m$` (`mapply`, which operates not on two *box-producing functions*, but instead on two *values of a boxed type*, one containing functions to be applied to the values in the other box, via some predefined scheme. Thus:
+Contrast that to `m$` (`mapply`), which operates not on two *box-producing functions*, but instead on two *values of a boxed type*, one containing functions to be applied to the values in the other box, via some predefined scheme. Thus:
let js = [(\a->a*a),(\a->a+a)] in
let xs = [7, 5] in
mapply js xs ==> [49, 25, 14, 10]
-As we illustrated in class, there are clear patterns shared between lists and option types and trees, so perhaps you can see why people want to figure out the general structures. But it probably isn't obvious yet why it would be useful to do so. To a large extent, this will only emerge over the next few classes. But we'll begin to demonstrate the usefulness of these patterns by talking through a simple example, that uses the monadic functions of the Option/Maybe box type.
+The question came up in class of when box types might fail to be Mappable, or Mappables might fail to be MapNables, or MapNables might fail to be Monads.
+
+For the first failure, we noted that it's easy to define a `map` operation for the box type `R -> α`, for a fixed type `R`. You `map` a function of type `P -> Q` over a value of the boxed type <code><u>P</u></code>, that is `R -> P`, by just returning a function that takes some `R` as input, first supplies it to your `R -> P` value, and then supplies the result to your `map`ped function of type `P -> Q`. (We will be working with this Mappable extensively; in fact it's not just a Mappable but more specifically a Monad.)
+
+But if on the other hand, your box type is `α -> R`, you'll find that there is no way to define a `map` operation that takes arbitrary functions of type `P -> Q` and values of the boxed type <code><u>P</u></code>, that is `P -> R`, and returns values of the boxed type <code><u>Q</u></code>.
+
+For the second failure, that is cases of Mappables that are not MapNables, we cited box types like `(R, α)`, for arbitrary fixed types `R`. The `map` operation for these is defined by `map f (r,a) = (r, f a)`. For certain choices of `R` these can be MapNables too. The easiest case is when `R` is the type of `()`. But when we look at the MapNable Laws, we'll see that they impose constraints we cannot satisfy for *every* choice of the fixed type `R`. Here's why. We'll need to define `mid a = (r0, a)` for some specific `r0` of type `R`. Then the MapNable Laws will entail:
+
+ 1. (r0,id) m$ (r,x) == (r,x)
+ 2. (r0,f x) == (r0,f) m$ (r0,x)
+ 3. (r0,(○)) m$ (r'',f) m$ (r',g) m$ (r,x) == (r'',f) m$ ((r',g) m$ (r,x))
+ 4. (r'',f) m$ (r0,x) == (r0,($x)) m$ (r'',f)
+ 5. (r0,f) m$ (r,x) == (r,($x)) m$ (r0,f)
+
+Now we are not going to be able to write a `m$` function that inspects the second element of its left-hand operand to check if it's the `id` function; the identity of functions is not decidable. So the only way to satisfy Law 1 will be to use the first element of the right-hand operand (`r`) at least in those cases when the first element of the left-hand operand is `r0`. But then that means that the result of the lhs of Law 5 will also have a first element of `r`; so, turning now to the rhs of Law 5, we see that `m$` must use the first element of its _left_-hand operand (here again `r`) at least in those cases when the first element of its right-hand operand is `r0`. If our `R` type has a natural *monoid* structure, we could just let `r0` be the monoid's identity, and have `m$` combine other `R`s using the monoid's operation. Alternatively, if the `R` type is one that we can safely apply the predicate `(r0==)` to, then we could define `m$` something like this:
+
+ let (m$) (r1,f) (r2,x) = ((if r0==r1 then r2 else if r0==r2 then r1 else ...), ...)
+
+But for some types neither of these will be the case. For function types, as we already mentioned, `==` is not decidable. If the functions have suitable types, they do form a monoid with `○` as the operation and `id` as the identity; but many function types won't be such that arbitrary functions of that type are composable. So when `R` is the type of functions from `int`s to `bool`s, for example, we won't have any way to write a `m$` that satisfies the constraints stated above.
+
+For the third failure, that is examples of MapNables that aren't Monads, we'll just state that lists where the `map2` operation is taken to be zipping rather than taking the Cartesian product (what in Haskell are called `ZipList`s), these are claimed to exemplify that failure. But we aren't now in a position to demonstrate that to you.
## Safe division ##
+As we discussed in class, there are clear patterns shared between lists and option types and trees, so perhaps you can see why people want to figure out the general structures. But it probably isn't obvious yet why it would be useful to do so. To a large extent, this will only emerge over the next few classes. But we'll begin to demonstrate the usefulness of these patterns by talking through a simple example, that uses the monadic functions of the Option/Maybe box type.
+
Integer division presupposes that its second argument
(the divisor) is not zero, upon pain of presupposition failure.
Here's what my OCaml interpreter says: