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Chapter 2: Lambda Calculus

Programming Distributed Computing Systems: A Foundational Approach

Carlos VarelaRensselaer Polytechnic Institute

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Mathematical FunctionsTake the mathematical function:

f(x) = x2

f is a function that maps integers to integers:

f: Z Z

We apply the function f to numbers in its domain to obtain a number in its range, e.g.:

f(-2) = 4

Function

Domain Range

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Function CompositionGiven the mathematical functions:

f(x) = x2 , g(x) = x+1

f g is the composition of f and g:

f g (x) = f(g(x))

f g (x) = f(g(x)) = f(x+1) = (x+1)2 = x2 + 2x + 1 g f (x) = g(f(x)) = g(x2) = x2 + 1

Function composition is therefore not commutative. Function composition can be regarded as a (higher-order) function with the following type:

: (Z Z) x (Z Z) (Z Z)

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Lambda Calculus (Church and Kleene 1930’s)

A unified language to manipulate and reason about functions.

Givenf(x) = x2

x. x2

represents the same f function, except it is anonymous.

To represent the function evaluation f(2) = 4, we use the following -calculus syntax:

(x. x2 2) 22 4

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Lambda Calculus Syntax and Semantics

The syntax of a -calculus expression is as follows:

e ::= v variable| v.e functional abstraction| (e e) function application

The semantics of a -calculus expression is as follows:

(x.E M) E{M/x}

where we alpha-rename the lambda abstraction E if necessary to avoid capturing free variables in M.

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Currying

The lambda calculus can only represent functions of one variable.It turns out that one-variable functions are sufficient to represent multiple-variable functions, using a strategy called currying.

E.g., given the mathematical function: h(x,y) = x+y of type h: Z x Z Z

We can represent h as h’ of type: h’: Z Z ZSuch that

h(x,y) = h’(x)(y) = x+y For example,

h’(2) = g, where g(y) = 2+y

We say that h’ is the curried version of h.

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Function Composition in Lambda Calculus

S: x.x2 (Square)I: x.x+1 (Increment)

C: f.g.x.(f (g x)) (Function Composition)

((C S) I)

((f.g.x.(f (g x)) x.x2) x.x+1) (g.x.(x.x2 (g x)) x.x+1)

x.(x.x2 (x.x+1 x)) x.(x.x2 x+1)

x.x+12

Recall semantics rule:

(x.E M) E{M/x}

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Free and Bound Variables

The lambda functional abstraction is the only syntactic construct that binds variables. That is, in an expression of the form:

v.e

we say that free occurrences of variable v in expression e are bound. All other variable occurrences are said to be free.

E.g.,

(x.y.(x y) (y w))

Free Variables

Bound Variables

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-renaming

Alpha renaming is used to prevent capturing free occurrences of variables when reducing a lambda calculus expression, e.g.,

(x.y.(x y) (y w))y.((y w) y)

This reduction erroneously captures the free occurrence of y.

A correct reduction first renames y to z, (or any other fresh variable) e.g.,

(x.y.(x y) (y w)) (x.z.(x z) (y w))

z.((y w) z)

where y remains free.

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Order of Evaluation in the Lambda Calculus

Does the order of evaluation change the final result?Consider:

x.(x.x2 (x.x+1 x))

There are two possible evaluation orders:

x.(x.x2 (x.x+1 x)) x.(x.x2 x+1)

x.x+12

and:x.(x.x2 (x.x+1 x))

x.(x.x+1 x)2

x.x+12

Is the final result always the same?

Recall semantics rule:

(x.E M) E{M/x}

Applicative Order

Normal Order

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Church-Rosser TheoremIf a lambda calculus expression can be evaluated in two different ways and both ways terminate, both ways will yield the same result.

e

e1 e2

e’

Also called the diamond or confluence property.

Furthermore, if there is a way for an expression evaluation to terminate, using normal order will cause termination.

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Order of Evaluation and Termination

Consider:(x.y (x.(x x) x.(x x)))

There are two possible evaluation orders:

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

and:(x.y (x.(x x) x.(x x)))

y

In this example, normal order terminates whereas applicative order does not.

Recall semantics rule:

(x.E M) E{M/x}

Applicative Order

Normal Order

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Combinators

A lambda calculus expression with no free variables is called a combinator. For example:

I: x.x (Identity)App: f.x.(f x) (Application)C: f.g.x.(f (g x)) (Composition)L: (x.(x x) x.(x x)) (Loop)Cur: f.x.y.((f x) y) (Currying)Seq: x.y.(z.y x) (Sequencing--normal order)ASeq: x.y.(y x) (Sequencing--applicative order)

where y denotes a thunk, i.e., a lambda abstraction wrapping the second expression to evaluate.

The meaning of a combinator is always the same independently of its context.

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Combinators in Functional Programming Languages

Most functional programming languages have a syntactic form for lambda abstractions. For example the identity combinator:

x.x

can be written in Oz as follows:

fun {$ X} X end

and it can be written in Scheme as follows:

(lambda(x) x)

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Currying Combinator in Oz

The currying combinator can be written in Oz as follows:

fun {$ F}fun {$ X}

fun {$ Y} {F X Y}

endend

end

It takes a function of two arguments, F, and returns its curried version, e.g.,

{{{Curry Plus} 2} 3} 5

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Normal vs Applicative Order of Evaluation and Termination

Consider:(x.y (x.(x x) x.(x x)))

There are two possible evaluation orders:

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

and:(x.y (x.(x x) x.(x x)))

y

In this example, normal order terminates whereas applicative order does not.

Recall semantics rule:

(x.E M) E{M/x}

Applicative Order

Normal Order

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-renaming

Alpha renaming is used to prevent capturing free occurrences of variables when beta-reducing a lambda calculus expression.

In the following, we rename x to z, (or any other fresh variable):

(x.(y x) x)

(z.(y z) x)

Only bound variables can be renamed. No free variables can be captured (become bound) in the process. For example, we cannot alpha-rename x to y.

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-reduction

(x.E M) E{M/x}

Beta-reduction may require alpha renaming to prevent capturing free variable occurrences. For example:

(x.y.(x y) (y w))

(x.z.(x z) (y w))

z.((y w) z)

Where the free y remains free.

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-conversion

x.(E x) E

if x is not free in E.

For example:(x.y.(x y) (y w))

(x.z.(x z) (y w))

z.((y w) z)

(y w)

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Recursion Combinator (Y or rec)

Suppose we want to express a factorial function in the calculus.

1 n=0 f(n) = n! =

n*(n-1)! n>0

We may try to write it as:

f: n.(if (= n 0)1(* n (f (- n 1))))

But f is a free variable that should represent our factorial function.

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Recursion Combinator (Y or rec)

We may try to pass f as an argument (g) as follows:

f: g.n.(if (= n 0)1(* n (g (- n 1))))

The type of f is:

f: (Z Z) (Z Z)

So, what argument g can we pass to f to get the factorial function?

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Recursion Combinator (Y or rec)

f: (Z Z) (Z Z)

(f f) is not well-typed.

(f I) corresponds to:

1 n=0 f(n) =

n*(n-1) n>0

We need to solve the fixpoint equation:

(f X) = X

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Recursion Combinator (Y or rec)

(f X) = X

The X that solves this equation is the following:

X: (x.(g.n.(if (= n 0) 1 (* n (g (- n 1))))

y.((x x) y))x.(g.n.(if (= n 0)

1 (* n (g (- n 1))))

y.((x x) y)))

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Recursion Combinator (Y or rec)

X can be defined as (Y f), where Y is the recursion combinator.

Y: f.(x.(f y.((x x) y))x.(f y.((x x) y)))

Y: f.(x.(f (x x))x.(f (x x)))

You get from the normal order to the applicative order recursion combinator by -expansion (-conversion from right to left).

Applicative Order

Normal Order

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Natural Numbers in Lambda Calculus

|0|: x.x (Zero)|1|: x.x.x (One)…|n+1|: x.|n| (N+1)

s: n.x.n (Successor)

(s 0)

(n.x.n x.x)

x.x.x

Recall semantics rule:

(x.E M) E{M/x}

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Booleans and Branching (if) in Calculus

|true|: x.y.x (True)|false|: x.y.y (False)

|if|: b.t.e.((b t) e) (If)

(((if true) a) b)

(((b.t.e.((b t) e) x.y.x) a) b) ((t.e.((x.y.x t) e) a) b)

(e.((x.y.x a) e) b) ((x.y.x a) b)

(y.a b)a

Recall semantics rule:

(x.E M) E{M/x}

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Exercises

20. PDCS Exercise 2.11.1 (page 31).21. PDCS Exercise 2.11.2 (page 31).22. PDCS Exercise 2.11.5 (page 31).23. PDCS Exercise 2.11.6 (page 31).

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Exercises

24.PDCS Exercise 2.11.7 (page 31).25.PDCS Exercise 2.11.9 (page 31).26.PDCS Exercise 2.11.10 (page 31).27.Prove that your addition operation is correct using induction.28.PDCS Exercise 2.11.11 (page 31).29.PDCS Exercise 2.11.12 (page 31).