In many languages you have type constructors; given a type A and a type constructor Lift , you get a new type Lift<A> . A functor is a type constructor together with a function

lift: (A -> B) -> (Lift<A> -> Lift<B>)

that preserves composition and identities. If h is the composition of two other functions g and f

h (a) = g (f (a)),

then lift (h) is the composition of lift (g) and lift (f)

lift (h) (la) = lift (g) (lift (f) (la)),

where the variable la has the type Lift<A> . Similarly, if h is the identity function on variables of type A

h (a: A) = a,

then lift (h) will be the identity on variables of type Lift<A>

lift (h) (la) = la.

Examples:

Multiplication

Lift<> adjoins an extra integer to any type: Lift<A> = Pair<A, int> The function lift() pairs up f with the identity function on integers: lift (f) = (f, id)

adjoins an extra integer to any type: Concatenation

Lift<> adjoins an extra string to any type: Lift<A> = Pair<A, string> The function lift() pairs up f with the identity function on strings: lift (f) = (f, id)

adjoins an extra string to any type: Composition

Let Env be a type representing the possible states of the environment and Effect = Env -> Env Also, we’ll be explicit in the type of the identity function id<A>: A -> A id<A> (a) = a, so one possible Effect is id<Env> , the “trivial side-effect”. Then Lift<> adjoins an extra side-effect to any type: Lift<A> = Pair<A, Effect> The function lift() pairs up f with the identity on side-effects: lift (f) = (f, id<Effect>)

Let be a type representing the possible states of the environment and Lists

The previous three examples used the Pair type constructor to adjoin an extra value. This functor is slightly different. Here, Lift<> takes any type A to a list of A ‘s: Lift<A> = List<A> The function lift() is the function map(): lift = map

The previous three examples used the Pair type constructor to adjoin an extra value. This functor is slightly different. Here, takes any type to a list of ‘s: Double negation, or the continuation passing transform

In a nice type system, there’s the Unit type, with a single value, and there’s also the Empty type, with no values (it’s “uninhabited”). The only function of type X -> Empty is the identity function id<Empty> . This means that we can think of types as propositions, where a proposition is true if it’s possible to construct a value of that type. We interpret the arrow as implication, and negation can be defined as “arrowing into Empty “: let F = Empty and T = F -> F . Then T -> F = F (since T -> F is uninhabited) and T is inhabited since we can construct the identity function of type F -> F. Functions correspond to constructive proofs. “Negation” of a proof is changing it into its contrapositive form: If A then B => If NOT B then NOT A. Double negation is doing the contrapositive twice: IF A then B => If NOT NOT A then NOT NOT B. Here, Lift<> is double negation: Lift<A> = (A -> F) -> F. The function lift takes a proof to its double contrapositive: lift: (A -> B) -> ((A -> F) -> F) -> ((B -> F) -> F) lift (f) (k1) (k2) = k1 (lambda (a) { k2 (f (a)) })

Monads

A monad is a functor together with two functions

m: Lift<Lift<A>> -> Lift<A> e: A -> Lift<A>

satisfying some conditions I’ll get to in a minute.

Examples:

Multiplication

If you adjoin two integers, m() multiplies them to get a single integer: m: Pair<Pair<A, int>, int> -> Pair<A, int> m (a, i, j) = (a, i * j). The function e() adjoins the multiplicative identity, or “unit”: e: A -> Pair<A, int> e (a) = (a, 1)

If you adjoin two integers, multiplies them to get a single integer: Concatenation

If you adjoin two strings, m() concatenates them to get a single string: m: Pair<Pair<A, string>, string> -> Pair<A, string> m (a, s, t) = (a, s + t). The function e() adjoins the identity for concatenation, the empty string: e: A -> Pair<A, string> e (a) = (a, "")

If you adjoin two strings, concatenates them to get a single string: Composition

If you adjoin two side-effects, m() composes them to get a single effect: m: Pair<Pair<A, Effect>, Effect> -> Pair<A, Effect> m (a, s, t) = (a, t o s), where (t o s) (x) = t (s (x)). The function e() adjoins the identity for composition, the identity function on Env : e: A -> Pair<A, Effect> e (a) = (a, id<Env>)

If you adjoin two side-effects, composes them to get a single effect: Lists

If you have two layers of lists, m() flattens them to get a single layer: m: List<List<A>> -> List<A> m = flatten The function e() makes any element of A into a singleton list: e: A -> List<A> e (a) = [a]

If you have two layers of lists, flattens them to get a single layer: Double negation, or the continuation passing transform

If you have a quadruple negation, m() reduces it to a double negation: m: ((((A -> F) -> F) -> F) -> F) -> ((A -> F) -> F) m (k1) (k2) = k1 (lambda (k3) { k3 (k2) }) The function e() is just reverse application: e: A -> (A -> F) -> F e (a) (k) = k (a)

The conditions that e and m have to satisfy are that m is associative and e is a left and right unit for m . In other words, assume we have

llla: Lift<Lift<Lift<A>>> la: Lift<A>

Then

m (lift (m) (llla)) = m (m (llla))

and

m (e (la)) = m (lift (e) (la)) = la

Examples:

Multiplication:

There are two different ways we can use lifting with these two extra functions e() and m() . The first is applying lift() to them. When we apply lift to m() , it acts on three integers instead of two; but because lift (m) = (m, id), it ignores the third integer: lift (m) (a, i, j, k) = (a, i * j, k). Similarly, lifting e() will adjoin the multiplicative unit, but will leave the last integer alone: lift (e) = (e, id) lift (e) (a, i) = (a, 1, i) The other way to use lifting with m() and e() is to apply Lift<> to their input types. This specifies A as Pair<A', int> , so the first integer gets ignored: m (a, i, j, k) = (a, i, j * k) e (a, i) = (a, i, 1) Now when we apply m() to all of these, we get the associativity and unit laws. For associativity we get m (lift (m) (a, i, j, k)) = m(a, i * j, k) = (a, i * j * k) m (m (a, i, j, k)) = m(a, i, j * k) = (a, i * j * k) and for unit, we get m (lift (e) (a, i)) = m (a, 1, i) = (a, 1 * i) = (a, i) m (e (a, i)) = m (a, i, 1) = (a, i * 1) = (a, i)

There are two different ways we can use lifting with these two extra functions and . The first is applying to them. When we apply to , it acts on three integers instead of two; but because Concatenation

There are two different ways we can use lifting with these two extra functions e() and m() . The first is applying lift() to them. When we apply lift to m() , it acts on three strings instead of two; but because lift (m) = (m, id), it ignores the third string: lift (m) (a, s, t, u) = (a, s + t, u). Similarly, lifting e() will adjoin the empty string, but will leave the last string alone: lift (e) = (e, id) lift (e) (a, s) = (a, "", s) The other way to use lifting with m() and e() is to apply Lift<> to their input types. This specifies A as Pair<A', string> , so the first string gets ignored: m (a, s, t, u) = (a, s, t + u) e (a, s) = (a, s, 1) Now when we apply m() to all of these, we get the associativity and unit laws. For associativity we get m (lift (m) (a, s, t, u)) = m(a, s + t, u) = (a, s + t + u) m (m (a, s, t, u)) = m(a, s, t + u) = (a, s + t + u) and for unit, you get m (lift (e) (a, s)) = m (a, "", s) = (a, "" + s) = (a, s) m (e (a, s)) = m (a, s, "") = (a, s + "") = (a, s)

There are two different ways we can use lifting with these two extra functions and . The first is applying to them. When we apply lift to , it acts on three strings instead of two; but because Composition

There are two different ways we can use lifting with these two extra functions e() and m() . The first is applying lift() to them. When we apply lift to m() , it acts on three effects instead of two; but because lift (m) = (m, id<Effect>), it ignores the third effect: lift (m) (a, s, t, u) = (a, t o s, u). Similarly, lifting e() will adjoin the identity function, but will leave the last string alone: lift (e) = (e, id<Effect>) lift (e) (a, s) = (a, id<Env>, s) The other way to use lifting with m() and e() is to apply Lift<> to their input types. This specifies A as Pair<A', Effect> , so the first effect gets ignored: m (a, s, t, u) = (a, s, u o t) e (a, s) = (a, s, id<Env>) Now when we apply m() to all of these, we get the associativity and unit laws. For associativity we get m (lift (m) (a, s, t, u)) = m(a, t o s, u) = (a, u o t o s) m (m (a, s, t, u)) = m(a, s, u o t) = (a, u o t o s) and for unit, you get m (lift (e) (a, s)) = m (a, id<Env>, s) = (a, s o id<Env>) = (a, s) m (e (a, s)) = m (a, s, id<Env>) = (a, id<Env> o s) = (a, s)

There are two different ways we can use lifting with these two extra functions and . The first is applying to them. When we apply lift to , it acts on three effects instead of two; but because Lists

There are two different ways we can use lifting with these two extra functions e() and m() . The first is applying lift() to them. When we apply lift to m() , it acts on three layers instead of two; but because lift (m) = map (m), it ignores the third (outermost) layer: lift (m) ([[[a, b, c], [], [d, e]], [[]], [[x], [y, z]]]) = [[a, b, c, d, e], [], [x, y, z]] Similarly, lifting e() will make singletons, but will leave the outermost layer alone: lift (e) ([a, b, c]) = [[a], [b], [c]] The other way to use lifting with m() and e() is to apply Lift<> to their input types. This specifies A as List<A'> , so the *innermost* layer gets ignored: m ([[[a, b, c], [], [d, e]], [[]], [[x], [y, z]]]) = [[a, b, c], [], [d, e], [], [x], [y, z]] e ([a, b, c]) = [[a, b, c]] Now when we apply m() to all of these, we get the associativity and unit laws. For associativity we get m (lift (m) ([[[a, b, c], [], [d, e]], [[]], [[x], [y, z]]])) = m([[a, b, c, d, e], [], [x, y, z]]) = [a, b, c, d, e, x, y, z] m (m ([[[a, b, c], [], [d, e]], [[]], [[x], [y, z]]])) = m([[a, b, c], [], [d, e], [], [x], [y, z]]) = [a, b, c, d, e, x, y, z] and for unit, we get m (lift (e) ([a, b, c])) = m ([[a], [b], [c]]) = [a, b, c] m (e ([a, b, c])) = m ([[a], [b], [c]]) = [a, b, c]

Monads in Haskell style, or “Kleisli arrows”

Given a monad (Lift, lift, m, e) , a Kleisli arrow is a function

f: A -> Lift<B>,

so the e() function in a monad is already a Kleisli arrow. Given

g: B -> Lift<C>

we can form a new Kleisli arrow

(g >>= f): A -> Lift<C> (g >>= f) (a) = m (lift (g) (f (a))).

The operation >>= is called “bind” by the Haskell crowd. You can think of it as composition for Kleisli arrows; it’s associative, and e() is the identity for bind. e() is called “return” in that context. Sometimes code is less complicated with bind and return instead of m and e .

If we have a function f: A -> B , we can turn it into a Kleisli arrow by precomposing with e() :

(e o f): A -> Lift<B> (e o f) (a) = e (f (a)) = return (f (a)).

Example:

Double negation, or the continuation passing style transform

We’re going to (1) show that the CPS transform of a function takes a continuation and applies that to the result of the function. We’ll also (2) show that for two functions r, s , CPS (s o r) = CPS (s) >>= CPS (r), (1) To change a function f: A -> B into a Kleisli arrow (i.e. continuized function) CPS (f): A -> (B -> X) -> X , we just compose with e —or in the language of Haskell, we return the result: CPS (f) (a) (k) = return (f (a)) (k) = (e o f) (a) (k) = e (f (a)) (k) = k (f (a)) (2) Given two Kleisli arrows f: A -> (B -> F) -> F and g: B -> (C -> F) -> F, we can bind them: (g >>= f) (a) (k) = m (lift (g) (f (a))) (k) // defn of bind = lift (g) (f (a)) (lambda (k3) { k3 (k) }) // defn of m = f (a) (lambda (b) { (lambda (k3) { k3 (k) }) (g (b)) }) // defn of lift = f (a) (lambda (b) { g (b) (k) }), // application which is just what we wanted. In particular, if f and g are really just continuized functions f = (e o r) g = (e o s) then (g >>= f) (a) (k) = f (a) (lambda (b) { g (b) (k) }) // by above = (e o r) (a) (lambda (b) { (e o s) (b) (k) }) // defn of f and g = (e o r) (a) (lambda (b) { k (s (b)) }) // defn of e = (e o r) (a) (k o s) // defn of composition = (k o s) (r (a)) // defn of e = k (s (r (a))) // defn of composition = (e o (s o r)) (a) (k) // defn of e = CPS (s o r) (a) (k) // defn of CPS so CPS (s) >>= CPS (r) = CPS (s o r).