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Chapter 2: Lambda Calculus

Chapter 2: Lambda Calculus. Programming Distributed Computing Systems: A Foundational Approach Carlos Varela Rensselaer Polytechnic Institute. Mathematical Functions. Take the mathematical function: f(x) = x 2 f is a function that maps integers to integers: f: Z  Z

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Chapter 2: Lambda Calculus

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  1. Chapter 2: Lambda Calculus Programming Distributed Computing Systems: A Foundational Approach Carlos Varela Rensselaer Polytechnic Institute C. Varela

  2. Mathematical Functions Take 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 Range Domain C. Varela

  3. 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

  4. 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

  5. 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. C. Varela

  6. 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 typeh: 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

  7. 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+1x))  x.(x.x2 x+1)  x.x+12 Recall semantics rule: (x.E M)  E{M/x} C. Varela

  8. 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 Free Variables C. Varela

  9. -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

  10. 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+1x))  x.(x.x2 x+1)  x.x+12 and: x.(x.x2 (x.x+1 x))  x.(x.x+1x)2  x.x+12 Is the final result always the same? Recall semantics rule: (x.E M)  E{M/x} Applicative Order Normal Order C. Varela

  11. 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

  12. 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 C. Varela

  13. Combinators A lambda calculus expression with no freevariables 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. C. Varela

  14. 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

  15. 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

  16. 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 C. Varela

  17. -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. α→ C. Varela

  18. b-reduction b→ (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 freey remains free. α→ b→ C. Varela

  19. h-conversion h→ 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) α→ b→ h→ C. Varela

  20. Recursion Combinator (Y or rec) Suppose we want to express a factorial function in the l 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. C. Varela

  21. 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

  22. 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 C. Varela

  23. Recursion Combinator (Y or rec) (f X) = X The X that solves this equation is the following: X: (lx.(g.n.(if (= n 0) 1 (* n (g (- n 1)))) ly.((x x) y)) lx.(g.n.(if (= n 0) 1 (* n (g (- n 1)))) ly.((x x) y))) C. Varela

  24. Recursion Combinator (Y or rec) X can be defined as (Y f), where Y is the recursion combinator. Y: lf.(lx.(f ly.((x x) y)) lx.(f ly.((x x) y))) Y: lf.(lx.(f (x x)) lx.(f (x x))) You get from the normal order to the applicative order recursion combinator by h-expansion (h-conversion from right to left). Applicative Order Normal Order C. Varela

  25. Natural Numbers in Lambda Calculus |0|: x.x(Zero) |1|: x.x.x (One) … |n+1|: x.|n| (N+1) s: ln.lx.n (Successor) (s 0) (n.x.nx.x)  x.x.x Recall semantics rule: (x.E M)  E{M/x} C. Varela

  26. Booleans and Branching (if) in l Calculus |true|: x.y.x(True) |false|: x.y.y (False) |if|: b.lt.le.((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.ly.x a) e) b)  ((x.ly.x a) b) • (y.a b) • a Recall semantics rule: (x.E M)  E{M/x} C. Varela

  27. Exercises • PDCS Exercise 2.11.1 (page 31). • PDCS Exercise 2.11.2 (page 31). • PDCS Exercise 2.11.5 (page 31). • PDCS Exercise 2.11.6 (page 31). C. Varela

  28. Exercises • PDCS Exercise 2.11.7 (page 31). • PDCS Exercise 2.11.9 (page 31). • PDCS Exercise 2.11.10 (page 31). • Prove that your addition operation is correct using induction. • PDCS Exercise 2.11.11 (page 31). • PDCS Exercise 2.11.12 (page 31). C. Varela

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