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The Lambda Calculus The Lambda Calculus Stephen A. Edwards Columbia University Fall 2012 Lambda Expressions Function application written in prefix form. “Add four and five” is ( 4 5) Å Evaluation: select a redex and evaluate it: ( ( 5 6) ( 8 3)) ( 30 ( 8 3)) Å ¤ ¤ ! Å ¤ ( 30 24) ! Å 54 ! Often more than one way to proceed: ( ( 5 6) ( 8 3)) ( ( 5 6) 24) Å ¤ ¤ ! Å ¤ ( 30 24) ! Å 54 ! Simon Peyton Jones, The Implementation of Functional Programming Languages, Prentice-Hall, 1987. Function Application and Currying Function application is written as juxtaposition: f x Every function has exactly one argument. Multiple-argument functions, e.g., , are represented by currying, named after Haskell Å Brooks Curry (1900–1982). So, ( x) Å is the function that adds x to its argument. Function application associates left-to-right: ( 3 4) (( 3) 4) Å Æ Å 7 ! Lambda Abstraction The only other thing in the lambda calculus is lambda abstraction: a notation for defining unnamed functions. (¸x . x 1) Å ( ¸ x . x 1 ) Å """""" That function of x that adds x to 1 The Syntax of the Lambda Calculus expr ::= expr expr ¸ variable . expr j constant j variable j (expr) j Constants are numbers and built-in functions; variables are identifiers. Function application binds more tightly than ¸: ¸x.f g x ¡¸x.(f g)x¢ Æ Beta-Reduction Evaluation of a lambda abstraction—beta-reduction—is just substitution: (¸x . x 1) 4 ( 4 1) Å ! Å 5 ! The argument may appear more than once (¸x . x x) 4 ( 4 4) Å ! Å 8 ! or not at all (¸x . 3) 5 3 ! Beta-Reduction Fussy, but mechanical. Extra parentheses may help. µ³ ³ ´´ ¶ (¸x . ¸y . x y) 3 4 ¸x . ¸y . ¡( x) y¢ 3 4 Å Æ Å µ ¶ ¸y . ¡( 3) y¢ 4 ! Å ¡( 3) 4¢ ! Å 7 ! Functions may be arguments (¸f . f 3) (¸x . x 1) (¸x . x 1) 3 Å ! Å ( 3 1) ! Å 4 ! Free and Bound Variables (¸x . x y) 4 Å Here, x is like a function argument but y is like a global variable. Technically, x occurs bound and y occurs free in (¸x . x y) Å However, both x and y occur free in ( x y) Å Beta-Reduction More Formally (¸x . E) F E 0 !¯ where E 0 is obtained from E by replacing every instance of x that appears free in E with F . The definition of free and bound mean variables have scopes. Only the rightmost x appears free in (¸x . ( x 1)) x 3 Å ¡ so (¸x .(¸x . ( x 1)) x 3) 9 (¸ x . ( x 1)) 9 3 Å ¡ ! Å ¡ ( 9 1) 3 !Å ¡ 8 3 !Å 11 ! Another Example ³ ´ ³ ´ ¸x . ¸y . x ¡(¸x . x 3) y¢ 5 6 ¸y . 5 ¡(¸x . x 3) y¢ 6 Å ¡ ! Å ¡ 5 ¡(¸x . x 3) 6¢ !Å ¡ 5 ( 6 3) !Å ¡ 5 3 !Å 8 ! Alpha-Conversion One way to confuse yourself less is to do ®-conversion: renaming a ¸ argument and its bound variables. Formal parameters are only names: they are correct if they are consistent. (¸x .(¸x . ( x 1)) x 3) 9 (¸x .(¸y . ( y 1)) x 3) 9 Å ¡ $ Å ¡ ((¸y . ( y 1)) 9 3) ! Å ¡ ( ( 9 1) 3) ! Å ¡ ( 8 3) ! Å 11 ! You’ve probably done this before in C or Java: int add(int x, int y) int add(int a, int b) { { return x + y; return a + b; } $ } Beta-Abstraction and Eta-Conversion Running ¯-reduction in reverse, leaving the “meaning” of a lambda expression unchanged, is called beta abstraction: 4 1 (¸x . x 1) 4 Å Ã Å Eta-conversion is another type of conversion that leaves “meaning” unchanged: (¸x . 1 x) ( 1) Å $´ Å Formally, if F is a function in which x does not occur free, (¸x . F x) F $´ int f(int y) { ... } int g(int x){ return f(x); } g(w); can be replaced with f(w) à Reduction Order The order in which you reduce things can matter. (¸x . ¸y . y) ¡(¸z . z z)(¸z . z z)¢ Two things can be reduced: (¸z . z z)(¸z . z z) (¸x . ¸y . y)( ) ¢¢¢ However, (¸z . z z)(¸z . z z) (¸z . z z)(¸z . z z) ! (¸x . ¸y . y)( ) (¸y . y) ¢¢¢ ! Normal Form A lambda expression that cannot be ¯-reduced is in normal form. Thus, ¸y . y is the normal form of (¸x . ¸y . y) ¡(¸z . z z)(¸z . z z)¢ Not everything has a normal form. E.g., (¸z . z z)(¸z . z z) can only be reduced to itself, so it never produces an non-reducible expression. Normal Form Can a lambda expression have more than one normal form? Church-Rosser Theorem I: If E E , then there ex- 1 $ 2 ists an expression E such that E E and E E. 1 ! 2 ! Corollary. No expression may have two distinct normal forms. Proof. Assume E and E are distinct normal forms for E: E E 1 2 $ 1 and E E . So E E and by the Church-Rosser Theorem I, there $ 2 1 $ 2 must exist an F such that E F and E F . However, since E 1 ! 2 ! 1 and E are in normal form, E F E , a contradiction. 2 1 Æ Æ 2 Normal-Order Reduction Not all expressions have normal forms, but is there a reliable way to find the normal form if it exists? Church-Rosser Theorem II: If E E and E is in normal form, 1 ! 2 2 then there exists a normal order reduction sequence from E1 to E2. Normal order reduction: reduce the leftmost outermost redex. Normal-Order Reduction µ³ ´ ¶ ¸x . ¡(¸w . ¸z . w z) 1¢ ¡(¸x . x x)(¸x . x x)¢ ¡(¸y . y 1) ( 2 3)¢ Å Å Å leftmost outermost ¸x ¸y leftmost innermost ¸x ¸x 3 ¸w 1 1 + 2 x x x x + y ¸z z + w Boolean Logic in the Lambda Calculus “Church Booleans” true ¸x . ¸y . x Æ false ¸x . ¸y . y Æ Each is a function of two arguments: true is “select first;” false is “select second.” If-then-else uses its predicate to select then or else: ifelse ¸p . ¸a . ¸b . p a b Æ E.g., ifelse true 42 58 true 42 58 Æ (¸x . ¸y . x) 42 58 ! (¸y . 42) 58 ! 42 ! Boolean Logic in the Lambda Calculus Logic operators can be expressed with if-then-else: and ¸p . ¸q . p q p Æ or ¸p . ¸q . p p q Æ not ¸p . ¸a . ¸b . p b a Æ and true false (¸p . ¸q . p q p) true false Æ true false true ! (¸x . ¸y . x) false true ! false ! not true (¸p . ¸a . ¸b . p b a) true Æ ¸a . ¸b . true b a !¯ ¸a . ¸b . b !¯ ¸x . ¸y . y !® false Æ Arithmetic: The Church Numerals 0 ¸f . ¸x . x Æ 1 ¸f . ¸x . f x Æ 2 ¸f . ¸x . f (f x) Æ 3 ¸f . ¸x . f ¡f (f x)¢ Æ I.e., for n 0,1,2,..., n f x f (n)(x). The successor function: Æ Æ succ ¸n . ¸f . ¸x . f (n f x) Æ succ 2 ¡¸n . ¸f . ¸x . f (n f x)¢ 2 Æ ¸f . ¸x . f (2 f x) ! ³ ´ ¸f . ¸x . f ¡¸f . ¸x . f (f x)¢ f x Æ ¸f . ¸x . f ¡ f (f x)¢ ! 3 Æ Adding Church Numerals Finally, we can add: plus ¸m.¸n.¸f .¸x. m f ( n f x) Æ plus 3 2 ¡¸m.¸n.¸f .¸x. m f ( n f x)¢ 3 2 Æ ¸f .¸x. 3 f ( 2 f x) ! ¸f .¸x. f (f (f (2 f x))) ! ¸f .¸x. f (f (f (f ( f x)))) ! 5 Æ (m) (n) (m n) Not surprising since f f f Å ± Æ Multiplying Church Numerals mult ¸m.¸n.¸f .m (n f ) Æ mult 2 3 ¡¸m.¸n.¸f .m (n f )¢ 2 3 Æ ¸f .2 (3 f ) ! ¸f . 2 (¸x. f (f (f x))) Æ ¸f . 2 (¸y. f (f (f y))) $® ¸f . ¸x.(¸y. f (f (f y))) ((¸y. f (f (f y))) x) ! ¸f . ¸x.(¸y. f (f (f y))) (f (f (f x))) ! ¸f . ¸x. f (f (f (f (f (f x))) )) ! 6 Æ The predecessor function is trickier since there aren’t negative numbers. Recursion Where is recursion in the lambda calculus? µ ³ ´¶ fac ¸n . if ( n 0) 1 n ¡fac ( n 1)¢ Æ Æ ¤ ¡ This does not work: functions are unnamed in the lambda calculus. But it is possible to express recursion as a function. fac (¸n . ... fac ...) Æ (¸f .(¸n .... f ...)) fac ï H fac Æ That is, the factorial function, fac, is a fixed point of the (non-recursive) function H: H ¸f . ¸n . if ( n 0) 1 ( n (f ( n 1))) Æ Æ ¤ ¡ Recursion Let’s invent a function Y that computes fac from H, i.e., fac YH: Æ fac H fac Æ YH H (YH) Æ fac 1 YH 1 Æ H (YH) 1 Æ (¸f . ¸n . if ( n 0) 1 ( n (f ( n 1)))) (YH) 1 Æ Æ ¤ ¡ (¸n . if ( n 0) 1 ( n ((YH)( n 1)))) 1 ! Æ ¤ ¡ if ( 1 0) 1 ( 1 ((YH)( 1 1))) ! Æ ¤ ¡ 1 (YH 0) ! ¤ 1 (H (YH) 0) Æ ¤ 1 ((¸f . ¸n . if ( n 0) 1 ( n (f ( n 1)))) (YH) 0) Æ ¤ Æ ¤ ¡ 1 ((¸n . if ( n 0) 1 ( n (YH ( n 1)))) 0) ! ¤ Æ ¤ ¡ 1 (if ( 0 0) 1 ( 0 (YH ( 0 1)))) ! ¤ Æ ¤ ¡ 1 1 ! ¤ 1 ! The Y Combinator Here’s the eye-popping part: Y can be a simple lambda expression. Y Æ ¸f . ¡¸x . f (x x)¢ ¡¸x . f (x x)¢ Æ ³ ´ YH ¸f .
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