Lambda Calculus. Lecture 4 CS /26/10

Transcription

1 Lambda Calculus Lecture 4 CS /26/10

2 Pure (Untyped) Lambda Calculus The only value is a function Variables denote functions Functions always take functions as arguments Functions always return functions as results Minimalist Can express essentially all modern programming constructs Can apply syntactic reasoning techniques (e.g. operational semantics) to understand behavior.

3 Scope The λ abstraction λ x. t binds variable x. The scope of the binding is t. Occurrences of x that are not within the scope of an abstraction binding x are said to be free: λ x. λ y. (x y z) λ x. ((λ y. z y) y) Occurrences of x that are within the scope of an abstraction binding x are said to be bound by the abstraction.

4 Free Variables Intuitively, the free variables of an expression are non-local variables. Define FV(M) formally thus: FV(x) = {x} FV(M 1 M 2 ) = FV(M 1 ) U FV(M 2 ) FV(λ x. M) = FV(M) {x} Free variables become bound after substitution. But, if proper care is not taken, this may lead to unexpected results: (λx.λy. y x) y = λy. y y) We say that term M is α-congruent to N if N results from M by a series of changes to bound variables: λx. (x y) α-congruent to λz. (z y) not α-congruent to λy. (y y) λx.x (λx.x) α-congruent to λx.x (λx.x) and α-congruent to λx.x (λx.x )

5 Substitution λx.m α-congruent to λy.m[y/x] if y is not free or bound in M. Define this more generally: Let x be a variable, and M and N expressions. Then [M/x]N is the expression N : N is a variable: N = x then N = M N x then N = N N is an application (Y Z): N = ([M/x]Y) ([M/x]Z)

6 Substitution (cont) N is an abstraction λy.y: y = x then N = N y x then: x does not occur free in Y or if y does not occur free in M: N = λy.[m/x]y x does occur free in Y and y does occur free in M: N = λz.[m/x]([z/y]y) for fresh z

7 Example (λp.(λq.(λp.p( p q))(λr. (+ p r)))(+ p 4)) 2 [(+ p 4]/q]((λp.p(p q))(λr. (+ p r))) ([(+ p 4)/q](λp.p(p q)))([(+ p 4)/q](λr. (+ p r)) (by case 2) ([(+ p 4)/q](λp.p(p q)))(λr.(+ p r)) (by case since q does not occur free in (+ p r) (λa.[(+ p 4)/q]([a/p](p(p q))))(λr. (+ p r)) (by case 3.3.2) (λa.a (a (+ p 4)))(λr. (+ p r)) (p. (a.a (a (+ p 4)))(r. (+ p r))) 2

8 Operational Semantics Values: λ x. t Computation rule: ((λ x. t) v) t[v/x] Congruence rules t 1 t 1 (t 1 t 2 ) (t 1 t 2 ) t 2 t 2 (v t 2 ) (v t 2 ) The first computation rule is referred to as the β-substitution or β-conversion rule. ((λ x. t 1 ) t 2 ) is called a β-redex. The last congruence rule is referred as the η-conversion rule. (λ x. (t x)) where x not in FV(t) is an η-redex x not free in t λ x. (t x) t η-conversion related to notion of function extensionality. Why?

9 Multiple arguments The λ calculus has no built-in support to handle multiple arguments. However, we can interpret λ terms that when applied yield another λ term as effectively providing the same effect: Example: double λ f. λ x. (f (f x)) We can think of double as a two-argument function. Representing a multi-argument function in terms of single-argument higher-order functions is known as currying.

10 Programming Examples: Booleans true λ t. λ f. t false λ t. λ f. f (true v w) ((λ t.λ f. t) v) w) ((λ f. v) w) v (false v w) ((λ t.λ f. f) v) w) ((λ f. f) w) w

11 Booleans (cont) not λ b. b false true The function that returns true if b is false, and false if b is true. and λ b. λ c. b c false The function that given two Boolean values (v and w) returns w if v is true and false if v is false. Thus, (and v w) yields true only if both v and w are true.

12 Pairs!!!!!!!!!!!!!

13 Numbers (Church Numerals)!!!!!!

14 Numbers!!!!!!!

15 Example!!!!!!!!!!!!!!!!

16 Recursion and Divergence Consider the application: Ω ((λ x. (x x)) (λ x. (x x))) Ω evaluates to itself in one step. It has no normal form. Y A lambda term is in normal form if it does not contain any redex (i.e., a term that is subject to β-reduction) Now, consider: Y ((λ x. (f (x x))) (λ x. (f (x x)))) (f ((λ x. (f (x x))) (λ x. (f (x x))))) (f (f (λ x. (f (x x))) (λ x. (f (x x))))))... (f (f (... (f (λ x. (f (x x))) (λ x. (f (x x)))...)))

17 Normal forms and order of evaluation No expression can be converted to two distinct normal forms (Church-Rosser Theorem 1) Is there an order of evaluation guaranteed to terminate whenever a particular expression is reducible to normal form? Normal-order: leftmost, outermost reduction: no expression in the argument position of a redex is reduced until the redex is reduced If there is a reduction from A to B and B is in normal form, then there exists a normal order reduction from A to B (Church- Rosser Theorem 2)

18 Recursion The previous definition applies f an infinite number of times Basis for iterated application But, how can we slow its rate of unfolding? Consider: Ω v (λ y. ((λ x. (λ y. (x x y))) (λ x. (λ y. (x x y))) y)) Ω v is in normal form. However, if it is applied to an argument it diverges.

19 Recursion (cont) (Ω v v) ((λ y. ((λ x. (λ y. (x x y))) (λ x. (λ y. (x x y))) y) v) ((λ x. (λ y. (x x y))) (λ x. (λ y. (x x y))) y) v)...

20 Recursion (cont) Now, consider Z f (λ y. ((λ x. (f (λ y. (x x y)))) (λ x. (f (λ y. (x x y)))) y)) If we apply Z f to an argument: ((λ y. ((λ x. (f (λ y. (x x y)))) (λ x. (f (λ y. (x x y)))) y) v) (f (λ y. ((λ x. (f (λ y. (x x y)))) (λ x. (f (λ y. (x x y)))) y)) v) Since the arguments to f are all values, this expression is equivalent to: f Z f v

21 Recursion (cont) How do we apply these insights? f λ fact. λ n. if n = 0 then 1 else n * (fact (n 1)) We can use Z f to turn f into a real factorial function:

22 Fixpoints Z f 3 f Z f 3 (λ fact. λ n....) Z f 3 if 3 = 0 then 1 else 3 * (Z f 2) 3 * (f Z f 2)... We ll stop when n = 0

23 Fixpoints Define Z = λ f. Z f Now, Z defines a fixpoint for any f: Z λ f. (λ y. ((λ x. (f (λ y. (x x y)))) (λ x. (f (λ y. (x x y)))) y)) Z computes the least fixpoint of a function.

24 Fixpoints and order of evaluation Consider an alternative definition: Y h.(x.h( x x))(x.h(x x)) What happens if we apply Y to f (the factorial functional) with argument 3? Under normal-order evaluation: Y f (x. f(x x))(x. f(x x)) 3 f ((λx. f(x x))(λx. f(x x))) 3 What happens under applicative-order?

25 Naming and substitution Although we claimed that lambda calculus essentially manipulates functions (it does), we ve spent a lot of time thinking about variables substitutions free variables equivalence upto renaming Implementations must consider these issues seriously Rename bound variables when performing substitutions with fresh names. Impose a condition that all bound variables be distinct from each other, and other free variables. Derive a canonical representation that does not require renaming at all.

26 Terms and Contexts De Brujin indices: Have variable occurrences point directly to their binders rather than referring to them by name. Do so by replacing variable occurrences with numbers: number k stands for the variable bound by the k th enclosing λ-term Example: λ x. λ y. x (y x) λ. λ. 1 (0 1) Similar to static offsets in an activation record or display.

27 Examples identity λ x. x λ.0 true λ x. λ y. x λ.λ. 1 false λ x. λ y. y λ.λ. 0 two λ s. λ z. s (s z) λ. λ. (1 (1 0))

28 Contexts How do we replace free variables with their binders? Assume an ordered context listing all free variables that can occur, and map free variables to their index in this context (counting right to left) Context: a, b a 1, b 0 λ x. a λ. 2 λ x. b λ. 1 λ x.b (λ y. a) λ.1(λ.3)

29 Shifting and substitution When substituting into a λ term, indices must be adjusted: λ y. x [ z/x] in context x,y,z [2 0] λ. 2 λ. [3 1] 3 λ.1 Key point: context becomes longer when substituting inside an abstraction. Need to be careful to adjust free variables, not bound ones. shift(d,c)(k) = k if k < c k + d if k c shift(d,c)(λ.t) = (λ.shift(d,c+1)(t)) shift(d,c)(t 1 t 2 ) = (shift(d,c)(t 1 ))(shift(d,c)(t 2 ))

30 Example shift(2,0)(λ.λ. 1 (0 2)) λ.λ. 1 (0 4) shift(2,0)(λ. 0 1 (λ )) λ. 0 3 (λ )

31 Substitution [ j s] k = s if k = j k otherwise [j s](λ.t ) = λ. [j+1 shift(1,0)s] t [j s](t 1 t 2 ) = ([j s) t 1 ) ([j s] t 2 ) Beta-reduction: (λ t) v shift(-1,0)([0 shift(1,0)(v)] t)

32 Examples Assume context <a,b> Then, a 1, b 0 [a / b] b x. y. b [0 1] a x. y. a [(a ( z. a)) / b] (b ( x.b)) [0 (1 (. 2))] (0. 1) 1 (. 2) (. (2 (. 3))) (a ( z. a)) ( x. (a ( z. a)))

33 Examples [a/b] (λ b. (b a)) [0 1] (λ. (0 2)) (λ. (0 2)) (λ b. (b a)) [a/b] (λ a. (b a)) [0 1] λ. (1 0) λ. (2 0) (λ a. a a )