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Chapter 2: Lambda Calculus Programming Distributed Computing Systems: A Foundational Approach Carlos Varela Rensselaer Polytechnic Institute C. Varela 1 Mathematical Functions Take the mathematical function: f(x) = x2 f is a function that maps integers to integers: Function f: Z Z Domain Range We apply the function f to numbers in its domain to obtain a number in its range, e.g.: f(-2) = 4 C. Varela 2 Function Composition Given 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) C. Varela 3 Lambda Calculus (Church and Kleene 1930’s) A unified language to manipulate and reason about functions. Given f(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 C. Varela 4 Lambda Calculus Syntax and Semantics The syntax of a -calculus expression is as follows: e ::= | | v v.e (e e) variable functional abstraction 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. C. Varela 5 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: of type h(x,y) = x+y h: Z x Z Z We can represent h as h’ of type: h’: Z Z Z Such 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. C. Varela 6 Function Composition in Lambda Calculus S: I: x.x2 x.x+1 (Square) (Increment) C: f.g.x.(f (g x)) (Function Composition) Recall semantics rule: ((C S) I) (x.E M) E{M/x} ((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 C. Varela 7 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)) Bound Variables C. Varela Free Variables 8 -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. C. Varela 9 Order of Evaluation in the Lambda Calculus Does the order of evaluation change the final result? Consider: Recall semantics rule: 2 x.(x.x (x.x+1 x)) (x.E M) E{M/x} There are two possible evaluation orders: and: x.(x.x2 (x.x+1 x)) x.(x.x2 x+1) x.x+12 Applicative Order x.(x.x2 (x.x+1 x)) x.(x.x+1 x)2 x.x+12 Normal Order Is the final result always the same? C. Varela 10 Church-Rosser Theorem If 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. C. Varela 11 Order of Evaluation and Termination Consider: (x.y (x.(x x) x.(x x))) Recall semantics rule: There are two possible evaluation orders: and: (x.E M) E{M/x} (x.y (x.(x x) x.(x x))) (x.y (x.(x x) x.(x x))) Applicative Order (x.y (x.(x x) x.(x x))) y Normal Order In this example, normal order terminates whereas applicative order does not. C. Varela 12 Combinators A lambda calculus expression with no free variables is called a combinator. For example: I: App: C: L: Cur: Seq: ASeq: x.x f.x.(f x) f.g.x.(f (g x)) (x.(x x) x.(x x)) f.x.y.((f x) y) x.y.(z.y x) x.y.(y x) (Identity) (Application) (Composition) (Loop) (Currying) (Sequencing--normal order) (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. C. Varela 13 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) C. Varela 14 Currying Combinator in Oz The currying combinator can be written in Oz as follows: fun {$ F} fun {$ X} fun {$ Y} {F X Y} end end end It takes a function of two arguments, F, and returns its curried version, e.g., {{{Curry Plus} 2} 3} 5 C. Varela 15 Normal vs Applicative Order of Evaluation and Termination Consider: (x.y (x.(x x) x.(x x))) Recall semantics rule: There are two possible evaluation orders: and: (x.E M) E{M/x} (x.y (x.(x x) x.(x x))) (x.y (x.(x x) x.(x x))) Applicative Order (x.y (x.(x x) x.(x x))) y Normal Order In this example, normal order terminates whereas applicative order does not. C. Varela 16 -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) α → (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. C. Varela 17 b-reduction → (x.E M) b E{M/x} Beta-reduction may require alpha renaming to prevent capturing free variable occurrences. For example: (x.y.(x y) (y w)) α → → b (x.z.(x z) (y w)) z.((y w) z) Where the free y remains free. C. Varela 18 h-conversion → x.(E x) h E if x is not free in E. For example: (x.y.(x y) (y w)) α → → b → h (x.z.(x z) (y w)) z.((y w) z) (y w) C. Varela 19 Recursion Combinator (Y or rec) Suppose we want to express a factorial function in the calculus. 1 n=0 n*(n-1)! n>0 f(n) = n! = 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. C. Varela 20 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? C. Varela 21 Recursion Combinator (Y or rec) f: (Z Z) (Z Z) (f f) is not well-typed. (f I) corresponds to: 1 n=0 n*(n-1) n>0 f(n) = We need to solve the fixpoint equation: (f X) = X C. Varela 22 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))) C. Varela 23 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))) Applicative Order Normal Order You get from the normal order to the applicative order recursion combinator by h-expansion (h-conversion from right to left). C. Varela 24 Natural Numbers in Lambda Calculus |0|: |1|: … |n+1|: x.x x.x.x (One) x.|n| (N+1) s: n.x.n (Successor) (Zero) (s 0) (n.x.n x.x) Recall semantics rule: (x.E M) E{M/x} x.x.x C. Varela 25 Booleans and Branching (if) in Calculus |true|: |false|: x.y.x x.y.y (False) |if|: b.t.e.((b t) e) (If) (True) Recall semantics rule: (((if true) a) b) (x.E M) E{M/x} (((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 C. Varela 26 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). C. Varela 27 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). C. Varela 28