Here's what we did in seminar on Monday 9/13,

Sometimes these notes will expand on things mentioned only briefly in class, or discuss useful tangents that didn't even make it into class. This present page expands on a lot, and some of this material will be reviewed next week.

# Linguistic and Philosophical Applications of the Tools We'll be Studying

Explanation of the "Damn" example shown in class

# Basics of Lambda Calculus

The lambda calculus we'll be focusing on for the first part of the course has no types. (Some prefer to say it instead has a single type---but if you say that, you have to say that functions from this type to this type also belong to this type. Which is weird... In fact, though, such types are studied, under the name "recursive type." More about these later in the seminar.)

Here is its syntax:

Variables: x, y, z...

Each variable is an expression. For any expressions M and N and variable a, the following are also expressions:

Abstract: (λa M)

We'll tend to write (λa M) as just (\a M), so we don't have to write out the markup code for the λ. You can yourself write (λa M) or (\a M) or (lambda a M).

Application: (M N)

Examples of expressions:

x
(y x)
(x x)
(\x y)
(\x x)
(\x (\y x))
(x (\x x))
((\x (x x)) (\x (x x)))


The lambda calculus has an associated proof theory. For now, we can regard the proof theory as having just one rule, called the rule of beta-reduction or "beta-contraction". Suppose you have some expression of the form:

((\a M) N)


that is, an application of an abstract to some other expression. This compound form is called a redex, meaning it's a "beta-reducible expression." (\a M) is called the head of the redex; N is called the argument, and M is called the body.

The rule of beta-reduction permits a transition from that expression to the following:

M [a:=N]


What this means is just M, with any free occurrences inside M of the variable a replaced with the term N.

What is a free occurrence?

An occurrence of a variable a is bound in T if T has the form (\a N).

If T has the form (M N), any occurrences of a that are bound in M are also bound in T, and so too any occurrences of a that are bound in N.

An occurrence of a variable is free if it's not bound.

For instance:

T is defined to be (x (\x (\y (x (y z)))))

The first occurrence of x in T is free. The \x we won't regard as containing an occurrence of x. The next occurrence of x occurs within a form that begins with \x, so it is bound as well. The occurrence of y is bound; and the occurrence of z is free.

Here's an example of beta-reduction:

((\x (y x)) z)


beta-reduces to:

(y z)


We'll write that like this:

((\x (y x)) z) ~~> (y z)


Different authors use different notations. Some authors use the term "contraction" for a single reduction step, and reserve the term "reduction" for the reflexive transitive closure of that, that is, for zero or more reduction steps. Informally, it seems easiest to us to say "reduction" for one or more reduction steps. So when we write:

M ~~> N


We'll mean that you can get from M to N by one or more reduction steps. Hankin uses the symbol → for one-step contraction, and the symbol ↠ for zero-or-more step reduction. Hindley and Seldin use ⊳1 and ⊳.

When M and N are such that there's some P that M reduces to by zero or more steps, and that N also reduces to by zero or more steps, then we say that M and N are beta-convertible. We'll write that like this:

M <~~> N


This is what plays the role of equality in the lambda calculus. Hankin uses the symbol = for this. So too do Hindley and Seldin. Personally, I keep confusing that with the relation to be described next, so let's use this notation instead. Note that M <~~> N doesn't mean that each of M and N are reducible to each other; that only holds when M and N are the same expression. (Or, with our convention of only saying "reducible" for one or more reduction steps, it never holds.)

In the metatheory, it's also sometimes useful to talk about formulas that are syntactically equivalent before any reductions take place. Hankin uses the symbol ≡ for this. So too do Hindley and Seldin. We'll use that too, and will avoid using = when discussing the metatheory. Instead we'll use <~~> as we said above. When we want to introduce a stipulative definition, we'll write it out longhand, as in:

T is defined to be (M N).

We'll regard the following two expressions:

(\x (x y))

(\z (z y))


as syntactically equivalent, since they only involve a typographic change of a bound variable. Read Hankin section 2.3 for discussion of different attitudes one can take about this.

Note that neither of those expressions are identical to:

(\x (x w))


because here it's a free variable that's been changed. Nor are they identical to:

(\y (y y))


because here the second occurrence of y is no longer free.

There is plenty of discussion of this, and the fine points of how substitution works, in Hankin and in various of the tutorials we've linked to about the lambda calculus. We expect you have a good intuitive understanding of what to do already, though, even if you're not able to articulate it rigorously.

## Shorthand

The grammar we gave for the lambda calculus leads to some verbosity. There are several informal conventions in widespread use, which enable the language to be written more compactly. (If you like, you could instead articulate a formal grammar which incorporates these additional conventions. Instead of showing it to you, we'll leave it as an exercise for those so inclined.)

Parentheses Outermost parentheses around applications can be dropped. Moreover, applications will associate to the left, so M N P will be understood as ((M N) P). Finally, you can drop parentheses around abstracts, but not when they're part of an application. So you can abbreviate:

(\x (x y))


as:

\x (x y)


but you should include the parentheses in:

(\x (x y)) z


and:

z (\x (x y))


Dot notation Dot means "put a left paren here, and put the right paren as far the right as possible without creating unbalanced parentheses". So:

\x (\y (x y))


can be abbreviated as:

\x (\y. x y)


and that as:

\x. \y. x y


This:

\x. \y. (x y) x


abbreviates:

\x (\y ((x y) x))


This on the other hand:

(\x. \y. (x y)) x


abbreviates:

((\x (\y (x y))) x)


Merging lambdas An expression of the form (\x (\y M)), or equivalently, (\x. \y. M), can be abbreviated as:

(\x y. M)


Similarly, (\x (\y (\z M))) can be abbreviated as:

(\x y z. M)


## Lambda terms represent functions

The untyped lambda calculus is Turing complete: all (recursively computable) functions can be represented by lambda terms. For some lambda terms, it is easy to see what function they represent:

(\x x) represents the identity function: given any argument M, this function simply returns M: ((\x x) M) ~~> M.

(\x (x x)) duplicates its argument: ((\x (x x)) M) ~~> (M M)

(\x (\y x)) throws away its second argument: (((\x (\y x)) M) N) ~~> M

and so on.

It is easy to see that distinct lambda expressions can represent the same function, considered as a mapping from input to outputs. Obviously:

(\x x)


and:

(\z z)


both represent the same function, the identity function. However, we said above that we would be regarding these expressions as synactically equivalent, so they aren't yet really examples of distinct lambda expressions representing a single function. However, all three of these are distinct lambda expressions:

(\y x. y x) (\z z)

(\x. (\z z) x)

(\z z)


yet when applied to any argument M, all of these will always return M. So they have the same extension. It's also true, though you may not yet be in a position to see, that no other function can differentiate between them when they're supplied as an argument to it. However, these expressions are all syntactically distinct.

The first two expressions are convertible: in particular the first reduces to the second. So they can be regarded as proof-theoretically equivalent even though they're not syntactically identical. However, the proof theory we've given so far doesn't permit you to reduce the second expression to the third. So these lambda expressions are non-equivalent.

There's an extension of the proof-theory we've presented so far which does permit this further move. And in that extended proof theory, all computable functions with the same extension do turn out to be equivalent (convertible). However, at that point, we still won't be working with the traditional mathematical notion of a function as a set of ordered pairs. One reason is that the latter but not the former permits many uncomputable functions. A second reason is that the latter but not the former prohibits functions from applying to themselves. We discussed this some at the end of Monday's meeting (and further discussion is best pursued in person).

# Booleans and pairs

Our definition of these is reviewed in Assignment1.

It's possible to do the assignment without using a Scheme interpreter, however you should take this opportunity to get Scheme installed on your computer, and get started learning Scheme. It will help you test out proposed answers to the assignment.

There's also a (slow, bare-bones, but perfectly adequate) version of Scheme available for online use at http://tryscheme.sourceforge.net/.

# Declarative/functional vs Imperatival/dynamic models of computation

Many of you, like us, will have grown up thinking the paradigm of computation is a sequence of changes. Let go of that. It will take some care to separate the operative notion of "sequencing" here from other notions close to it, but once that's done, you'll see that languages that have no significant notions of sequencing or changes are Turing complete: they can perform any computation we know how to describe. In itself, that only puts them on equal footing with more mainstream, imperatival programming languages like C and Java and Python, which are also Turing complete. But further, the languages we want you to become familiar with can reasonably be understood to be more fundamental. They embody the elemental building blocks that computer scientists use when reasoning about and designing other languages.

Jim offered the metaphor: think of imperatival languages, which include "mutation" and "side-effects" (we'll flesh out these keywords as we proceeed), as the pâté of computation. We want to teach you about the meat and potatoes, where as it turns out there is no sequencing and no changes. There's just the evaluation or simplification of complex expressions.

Now, when you ask the Scheme interpreter to simplify an expression for you, that's a kind of dynamic interaction between you and the interpreter. You may wonder then why these languages should not also be understood imperatively. The difference is that in a purely declarative or functional language, there are no dynamic effects in the language itself. It's just a static semantic fact about the language that one expression reduces to another. You may have verified that fact through your dynamic interactions with the Scheme interpreter, but that's different from saying that there are dynamic effects in the language itself.

What the latter would amount to will become clearer as we build our way up to languages which are genuinely imperatival or dynamic.

Many of the slogans and keywords we'll encounter in discussions of these issues call for careful interpretation. They mean various different things.

For example, you'll encounter the claim that declarative languages are distinguished by their referential transparency. What's meant by this is not always exactly the same, and as a cluster, it's related to but not the same as this means for philosophers and linguists.

The notion of function that we'll be working with will be one that, by default, sometimes counts as non-identical functions that map all their inputs to the very same outputs. For example, two functions from jumbled decks of cards to sorted decks of cards may use different algorithms and hence be different functions.

It's possible to enhance the lambda calculus so that functions do get identified when they map all the same inputs to the same outputs. This is called making the calculus extensional. Church called languages which didn't do this intensional. If you try to understand that kind of "intensionality" in terms of functions from worlds to extensions (an idea also associated with Church), you may hurt yourself. So too if you try to understand it in terms of mental stereotypes, another notion sometimes designated by "intension."

It's often said that dynamic systems are distinguished because they are the ones in which order matters. However, there are many ways in which order can matter. If we have a trivalent boolean system, for example---easily had in a purely functional calculus---we might choose to give a truth-table like this for "and":

true and true   = true
true and *      = *
true and false  = false
* and true      = *
* and *         = *
* and false     = *
false and true  = false
false and *     = false
false and false = false


And then we'd notice that * and false has a different intepretation than false and *. (The same phenomenon is already present with the material conditional in bivalent logics; but seeing that a non-symmetric semantics for and is available even for functional languages is instructive.)

Another way in which order can matter that's present even in functional languages is that the interpretation of some complex expressions can depend on the order in which sub-expressions are evaluated. Evaluated in one order, the computations might never terminate (and so semantically we interpret them as having "the bottom value"---we'll discuss this). Evaluated in another order, they might have a perfectly mundane value. Here's an example, though we'll reserve discussion of it until later:

(\x. y) ((\x. x x) (\x. x x))


Again, these facts are all part of the metatheory of purely functional languages. But there is a different sense of "order matters" such that it's only in imperatival languages that order so matters.

x := 2
x := x + 1
x == 3


Here the comparison in the last line will evaluate to true.

x := x + 1
x := 2
x == 3


Here the comparison in the last line will evaluate to false.

One of our goals for this course is to get you to understand what is that new sense such that only so matters in imperatival languages.

Finally, you'll see the term dynamic used in a variety of ways in the literature for this course:

• dynamic versus static typing

• dynamic versus lexical scoping

• dynamic versus static control operators

• finally, we're used ourselves to talking about dynamic versus static semantics

For the most part, these uses are only loosely connected to each other. We'll tend to use "imperatival" to describe the kinds of semantic properties made available in dynamic semantics, languages which have robust notions of sequencing changes, and so on.

To read further about the relation between declarative or functional programming, on the one hand, and imperatival programming on the other, you can begin here:

# Map

 Scheme (functional part) OCaml (functional part) C, Java, Python Scheme (imperative part) OCaml (imperative part) untyped lambda calculus combinatorial logic --------------------------------------------------- Turing complete --------------------------------------------------- more advanced type systems, such as polymorphic types simply-typed lambda calculus (what linguists mostly use)

# Rosetta Stone

Here's how it looks to say the same thing in various of these languages.

1. Function application and parentheses

In Scheme and the lambda calculus, the functions you're applying always go to the left. So you write (foo 2) and also (+ 2 3).

Mostly that's how OCaml is written too:

foo 2


But a few familiar binary operators can be written infix, so:

2 + 3


You can also write them operator-leftmost, if you put them inside parentheses to help the parser understand you:

( + ) 2 3


I'll mostly do this, for uniformity with Scheme and the lambda calculus.

In OCaml and the lambda calculus, this:

foo 2 3


means the same as:

((foo 2) 3)


These functions are "curried". foo 2 returns a 2-fooer, which waits for an argument like 3 and then foos 2 to it. ( + ) 2 returns a 2-adder, which waits for an argument like 3 and then adds 2 to it. For further reading:

2. Currying

In Scheme, on the other hand, there's a difference between ((foo 2) 3) and (foo 2 3). Scheme distinguishes between unary functions that return unary functions and binary functions. For our seminar purposes, it will be easiest if you confine yourself to unary functions in Scheme as much as possible.

Scheme is very sensitive to parentheses and whenever you want a function applied to any number of arguments, you need to wrap the function and its arguments in a parentheses. So you have to write (foo 2); if you only say foo 2, Scheme won't understand you.

Scheme uses a lot of parentheses, and they are always significant, never optional. Often the parentheses mean "apply this function to these arguments," as just described. But in a moment we'll see other constructions in Scheme where the parentheses have different roles. They do lots of different work in Scheme.

3. Binding suitable values to the variables three and two, and adding them.

In Scheme:

(let* ((three 3))
(let* ((two 2))
(+ three two)))


Most of the parentheses in this construction aren't playing the role of applying a function to some arguments---only the ones in (+ three two) are doing that.

In OCaml:

let three = 3 in
let two = 2 in
( + ) three two


In the lambda calculus:

Here we're on our own, we don't have predefined constants like + and 3 and 2 to work with. We've got to build up everything from scratch. We'll be seeing how to do that over the next weeks.

But supposing you had constructed appropriate values for + and 3 and 2, you'd place them in the ellided positions in:

(((\three (\two ((... three) two))) ...) ...)


In an ordinary imperatival language like C:

int three = 3;
int two = 2;
three + two;

4. Mutation

In C this looks almost the same as what we had before:

int x = 3;
x = 2;


Here we first initialize x to hold the value 3; then we mutate x to hold a new value.

In (the imperatival part of) Scheme, this could be done as:

(let ((x (box 3)))
(set-box! x 2))


In general, mutating operations in Scheme are named with a trailing !. There are other imperatival constructions, though, like (print ...), that don't follow that convention.

In (the imperatival part of) OCaml, this could be done as:

let x = ref 3 in
x := 2


Of course you don't need to remember any of this syntax. We're just illustrating it so that you see that in Scheme and OCaml it looks somewhat different than we had above. The difference is much more obvious than it is in C.

In the lambda calculus:

Sorry, you can't do mutation. At least, not natively. Later in the term we'll be learning how in fact, really, you can embed mutation inside the lambda calculus even though the lambda calculus has no primitive facilities for mutation.

5. Anonymous functions

Functions are "first-class values" in the lambda calculus, in Scheme, and in OCaml. What that means is that they can be arguments to, and results of, other functions. They can be stored in data structures. And so on. To read further:

We'll begin by looking at what "anonymous" functions look like. These are functions that have not been bound as values to any variables. That is, there are no variables whose value they are.

In the lambda calculus:

(\x M)


---where M is any simple or complex expression---is anonymous. It's only when you do:

((\y N) (\x M))


that (\x M) has a "name" (it's named y during the evaluation of N).

In Scheme, the same thing is written:

(lambda (x) M)


Not very different, right? For example, if M stands for (+ 3 x), then here is an anonymous function that adds 3 to whatever argument it's given:

(lambda (x) (+ 3 x))


In OCaml, we write our anonymous function like this:

fun x -> ( + ) 3 x

6. Supplying an argument to an anonymous function

Just because the functions we built aren't named doesn't mean we can't do anything with them. We can give them arguments. For example, in Scheme we can say:

((lambda (x) (+ 3 x)) 2)


The outermost parentheses here mean "apply the function (lambda (x) (+ 3 x)) to the argument 2, or equivalently, "give the value 2 as an argument to the function (lambda (x) (+ 3 x)).

In OCaml:

(fun x -> ( + ) 3 x) 2

7. Binding variables to values with "let"

Let's go back and re-consider this Scheme expression:

(let* ((three 3))
(let* ((two 2))
(+ three two)))


Scheme also has a simple let (without the *), and it permits you to group several variable bindings together in a single let- or let*-statement, like this:

(let* ((three 3) (two 2))
(+ three two))


Often you'll get the same results whether you use let* or let. However, there are cases where it makes a difference, and in those cases, let* behaves more like you'd expect. So you should just get into the habit of consistently using that. It's also good discipline for this seminar, especially while you're learning, to write things out the longer way, like this:

(let* ((three 3))
(let* ((two 2))
(+ three two)))


However, here you've got the double parentheses in (let* ((three 3)) ...). They're doubled because the syntax permits more assignments than just the assignment of the value 3 to the variable three. Myself I tend to use [ and ] for the outer of these parentheses: (let* [(three 3)] ...). Scheme can be configured to parse [...] as if they're just more (...).

It was asked in seminar if the 3 could be replaced by a more complex expression. The answer is "yes". You could also write:

(let* [(three (+ 1 2))]
(let* [(two 2)]
(+ three two)))


It was also asked whether the (+ 1 2) computation would be performed before or after it was bound to the variable three. That's a terrific question. Let's say this: both strategies could be reasonable designs for a language. We are going to discuss this carefully in coming weeks. In fact Scheme and OCaml make the same design choice. But you should think of the underlying form of the let-statement as not settling this by itself.

Repeating our starting point for reference:

(let* [(three 3)]
(let* [(two 2)]
(+ three two)))


Recall in OCaml this same computation was written:

let three = 3 in
let two = 2 in
( + ) three two

8. Binding with "let" is the same as supplying an argument to a lambda

The preceding expression in Scheme is exactly equivalent to:

(((lambda (three) (lambda (two) (+ three two))) 3) 2)


The preceding expression in OCaml is exactly equivalent to:

(fun three -> (fun two -> ( + ) three two)) 3 2


Read this several times until you understand it.

9. Functions can also be bound to variables (and hence, cease being "anonymous").

In Scheme:

(let* [(bar (lambda (x) B))] M)


then wherever bar occurs in M (and isn't rebound by a more local let or lambda), it will be interpreted as the function (lambda (x) B).

Similarly, in OCaml:

let bar = fun x -> B in
M


This in Scheme:

(let* [(bar (lambda (x) B))] (bar A))


as we've said, means the same as:

((lambda (bar) (bar A)) (lambda (x) B))


which beta-reduces to:

((lambda (x) B) A)


and that means the same as:

(let* [(x A)] B)


in other words: evaluate B with x assigned to the value A.

Similarly, this in OCaml:

let bar = fun x -> B in
bar A


is equivalent to:

(fun x -> B) A


and that means the same as:

let x = A in
B

10. Pushing a "let"-binding from now until the end

What if you want to do something like this, in Scheme?

(let* [(x A)] ... for the rest of the file or interactive session ...)


or this, in OCaml:

let x = A in
... for the rest of the file or interactive session ...


Scheme and OCaml have syntactic shorthands for doing this. In Scheme it's written like this:

(define x A)
... rest of the file or interactive session ...


In OCaml it's written like this:

let x = A;;
... rest of the file or interactive session ...


It's easy to be lulled into thinking this is a kind of imperative construction. But it's not! It's really just a shorthand for the compound let-expressions we've already been looking at, taking the maximum syntactically permissible scope. (Compare the "dot" convention in the lambda calculus, discussed above. I'm fudging a bit here, since in Scheme (define ...) is really shorthand for a letrec epression, which we'll come to in later classes.)

11. Some shorthand

OCaml permits you to abbreviate:

let bar = fun x -> B in
M


as:

let bar x = B in
M


It also permits you to abbreviate:

let bar = fun x -> B;;


as:

let bar x = B;;


Similarly, Scheme permits you to abbreviate:

(define bar (lambda (x) B))


as:

(define (bar x) B)


and this is the form you'll most often see Scheme definitions written in.

However, conceptually you should think backwards through the abbreviations and equivalences we've just presented.

(define (bar x) B)


just means:

(define bar (lambda (x) B))


which just means:

(let* [(bar (lambda (x) B))] ... rest of the file or interactive session ...)


which just means:

(lambda (bar) ... rest of the file or interactive session ...) (lambda (x) B)


or in other words, interpret the rest of the file or interactive session with bar assigned the function (lambda (x) B).

You can override a binding with a more inner binding to the same variable. For instance the following expression in OCaml:

let x = 3 in
let x = 2 in
x


will evaluate to 2, not to 3. It's easy to be lulled into thinking this is the same as what happens when we say in C:

int x = 3;
x = 2;


but it's not the same! In the latter case we have mutation, in the former case we don't. You will learn to recognize the difference as we proceed.

The OCaml expression just means:

(fun x -> ((fun x -> x) 2) 3)


and there's no more mutation going on there than there is in:

∀x. (F x or ∀x (not (F x)))


When a previously-bound variable is rebound in the way we see here, that's called shadowing: the outer binding is shadowed during the scope of the inner binding.

## Some more comparisons between Scheme and OCaml

• Simple predefined values

Numbers in Scheme: 2, 3
In OCaml: 2, 3

Booleans in Scheme: #t, #f
In OCaml: true, false

The eighth letter in the Latin alphabet, in Scheme: #\h
In OCaml: 'h'

• Compound values

These are values which are built up out of (zero or more) simple values.

Ordered pairs in Scheme: '(2 . 3) or (cons 2 3)
In OCaml: (2, 3)

Lists in Scheme: '(2 3) or (list 2 3)
In OCaml: [2; 3]
We'll be explaining the difference between pairs and lists next week.

The empty list, in Scheme: '() or (list)
In OCaml: []

The string consisting just of the eighth letter of the Latin alphabet, in Scheme: "h"
In OCaml: "h"

A longer string, in Scheme: "horse"
In OCaml: "horse"

A shorter string, in Scheme: ""
In OCaml: ""

## What "sequencing" is and isn't

We mentioned before the idea that computation is a sequencing of some changes. I said we'd be discussing (fragments of, and in some cases, entire) languages that have no native notion of change.

Neither do they have any useful notion of sequencing. But what this would be takes some care to identify.

First off, the mere concatenation of expressions isn't what we mean by sequencing. Concatenation of expressions is how you build syntactically complex expressions out of simpler ones. The complex expressions often express a computation where a function is applied to one (or more) arguments,

Second, the kind of rebinding we called "shadowing" doesn't involve any changes or sequencing. All the precedence facts about that kind of rebinding are just consequences of the compound syntactic structures in which it occurs.

Third, the kinds of bindings we see in:

(define foo A)
(foo 2)


Or even:

(define foo A)
(define foo B)
(foo 2)


don't involve any changes or sequencing in the sense we're trying to identify. As we said, these programs are just syntactic variants of (single) compound syntactic structures involving lets and lambdas.

Since Scheme and OCaml also do permit imperatival constructions, they do have syntax for genuine sequencing. In Scheme it looks like this:

(begin A B C)


In OCaml it looks like this:

begin A; B; C end


Or this:

(A; B; C)


In the presence of imperatival elements, sequencing order is very relevant. For example, these will behave differently:

(begin (print "under") (print "water"))

(begin (print "water") (print "under"))


And so too these:

begin x := 3; x := 2; x end

begin x := 2; x := 3; x end


However, if A and B are purely functional, non-imperatival expressions, then:

begin A; B; C end


just evaluates to C (so long as A and B evaluate to something at all). So:

begin A; B; C end


contributes no more to a larger context in which it's embedded than C does. This is the sense in which functional languages have no serious notion of sequencing.

We'll discuss this more as the seminar proceeds.