Lambda Calculus! Gunnar Gotshalks! LC-1

Transcription

1 Lambda Calculus! LC-1

2 λ Calculus History! Developed by Alonzo Church during mid 1930 s! One fundamental goal was to describe what can be computed.! Full definition of λ-calculus is equivalent in power to a Turing machine!» Turing machines and λ-calculus are alternate descriptions of our understanding of what is computable! LC-2

3 λ Calculus History 2! In the mid to late 1950 s, John McCarthy developed Lisp!» A programming language based on λ-calculus!» Implementation includes syntactic sugar! > functions and forms that do not add to the power of what we can compute but make programs simpler and easier to understand! LC-3

4 λ Calculus Basis! Mathematical theory for anonymous functions!» functions that have not been bound to names! Present a subset of full definition to present the flavour! Notation and interpretation scheme identifies!» functions and their application to operands! > argument-parameter binding!» Clearly indicates which variables are free and which are bound! LC-4

5 Bound and Free Variables! Bound variables are similar to local variables in Java function (or any procedural language)!» Changing the name of a bound variable (consistently) does not change the semantics (meaning) of a function! Free variables are similar to global variables in Java function (or any procedural language)!» Changing the name of a free variable normally changes the semantics (meaning) of a function.! LC-5

6 Consider following expression!» ( u + 1 ) ( u 1 )!» is u bound or free?! Disambiguate the expression with the following λ-function!» ( λ u. ( u + 1 ) ( u 1 ) )! defining form!» Clearly indicates that u is a bound variable! bound variables! λ functions 1! Note the parallel with programming language functions!» functionname ( arguments ) { function definition }! It seems obvious now but that is because programming languages developed out of these mathematical notions LC-6

7 λ functions 2! Consider the following expression!» ( u + a ) ( u + b )! Can have any of the following functions, depending on what you mean!» ( λ u. ( u + a ) ( u + b ) )! > u is bound, a and b are free (defined in the enclosing context)!» ( λ u b. ( u + a ) ( u + b ) )! > u and b are bound, a is free!» ( λ u a b. ( u + a ) ( u + b ) )! > u, a and b are all bound, no free variables in the expression! LC-7

8 Function application! Functions are applied to arguments in a list immediately following the l-function!» { λ u. ( u + 1 ) ( u + 2 ) } [ 3 ]! > 3 ==> u then ==> (3 + 1) (3 + 2) ==> 20!» { λ u. ( u + a ) ( u + b ) } [ 7 1 ]! > 7 1 ==> u then ==> ( 6 + a ) ( 6 + b ) and no further in this context!» {λ u v. ( u v ) ( u + v ) } [ 2p + q, 2p - q ]! > ==> ( (2p+q) (2p - q) ) ( (2p + q) + (2p q) )! > Can pass expressions to a variable! Can use different bracketing symbols for visual clarity; they all mean the same thing.! LC-8

9 Using auxiliary definitions! Build up longer definitions with auxiliary definitions!» Define u / ( u + 5 )!where u = a ( a + 1 )! where a = 7 3! { λ u. u / ( u + 5 ) } [ { λ a. a ( a + 1 ) } [ 7 3 ] ]! > Note the nested function definition and argument application! ==> { λ u. u / ( u + 5 ) } [ 4 ( ) ]! ==> { 20 / ( ) }! ==> 0.8! LC-9

10 Functions are Variables! Define f ( 3 ) + f ( 5 ) where f ( x ) = a x ( a + x ) where a = 4! { λ f. f (3) + f (5) } [ { λ a. { λ x. a x ( a + x ) } } [ 4 ] ]! Arguments must be evaluated first! ==> { λ f. f (3) + f (5) } [ { λ x. 4 x ( 4 + x ) } ]! ==> { λ x. 4 x (4 + x ) } (3) + { λ x. 4 x (4 + x ) } (5)! ==> 4 * 3 ( ) + 4 * 5 ( ) ==> 264! LC-10

11 Lamba notation in Lisp! Lambda expressions are a direct analogue of λ-calculus expressions!» They are the basis of Lisp functions a modified syntax to simplify the interpreter! For example! ( defun double ( x ) ( + x x ) )! > is the named version of the following unnamed lambda expression! ( lambda ( x ) ( + x x ) ) { λ x. ( x + x ) }! > Note the similar syntax with λ-calculus and the change to prefix, from infix, to permit a uniform syntax for functions of any number of arguments! LC-11

12 Anonymous functions! Recall in the abstraction for sumint we defined support functions to handle each case! (defun double (int) (+ int int))! (defun square (int) (* int int))! (defun identity (int) int)! This adds additional symbols we may not want, especially if the function is to be used only once.! Using lambda we get the same effect without adding symbols! (sumint # (lambda (int) (+ int int)) 10)! (sumint # (lambda (int) (* int int)) 10)! (sumint # (lambda (int) int) 10)! LC-12

13 The function function! What is the meaning of # in the following! (sumint # (lambda (int) (+ int int)) 10)! It is a short hand!» # (...) ==> (function (...))! One of its attributes is it works like quote, in that its argument is not evaluated, thus, in this simple context the following will also work! (sumint (lambda (int) (+ int int)) 10)! Later we will see another attribute of function that makes it different from quote.! Whenever a function is to be quoted use # in place of! LC-13

14 Recursion! Recursion with lambda functions uses labels to temporarily name a function! The following is a general λ-calculus template.! > The name is in scope within the entire body but is out of scope outside of the lambda expression.! { label name ( lambda arguments. body_references_name ) }! In Lisp can use labels to define a mutually recursive set of functions! ( labels (list of named lambda expressions) sequence of forms using the temporarily named functions )! LC-14

15 Example 1 of recursion! A recursive multiply that uses only addition.! > The temporary function is called mult! > Use quote not function using eval! (eval '(labels! ((mult (k n)! (cond ((zerop n) 0)! (t (+ k (mult k (1- n))))! )))! (mult 2 3)! )! )! LC-15

16 Example 2 of recursion! rectimes computes k * n by supplying the paramters to a unary function that is a variation of example 1.! (defun rectimes (k n)! (labels (( temp (n)! (cond ((zerop n) 0)! ( t (+ k (temp (1- n))))! )))! (temp n)! ))! LC-16

17 Churchʼs Lambda Calculus! The preceding description is not exactly what Church developed! He worked with one parameter! Result of a function is a function!» In lambda calculus numbers are functions! LC-17

18 Function notation! Do not need parenthesis for one argument functions.!» f ( x ) = f x!» g ( f ( x ) ) = g (f x)! > Cannot have g f x as that could mean (g f) x! > Could have g f x! LC-18

19 Function notation 2! For multiple argument functions convert to single argument functions!» f ( x, y ) = ( f x ) y! > (f x) returns a one parameter function that is applied to y!» f ( x, y, z ) = ( ( f x ) y ) z!»! LC-19

20 Function definition examples! Use λ to denote the parameters of a function!» { λ u. u * v + 2*a }!» { λ u v. u * v + 2*a } = { λ u. [ λ v. u * v + 2*a ] }! > Can see how multiple parameters are equivalent to nested one parameter functions! > At this level of definition, we simplify!» { λ v a. u * v + 2*a }!» { λ u v a. u * v + 2*a }!» { λ x. f x }!!! LC-20

21 Function application! Consider the following!» g = { λ x. (1 / 6) x^3 }!» g 3 = (1 / 6) 3^3 = 9 / 6!» g (a + 1) = (1 / 6) (a + 1)^3 = (1 / 6) a^3 + (1 / 2) a ^ 2 + (1 / 2) a + (1 / 6)! LC-21

22 Function application 2! Functions can be parameters!» { λ f. f x } g f g = g x! Can abstract out the x!» { λ f. [ λ x. f x ] } g = { λ f x. f x } g!simplify = { λ x. g x }! Apply the function!» { λ x. g x } a = g a! As a consequence g = { λ x. g x }!! LC-22

23 Function application 3! A more complex example!» { λ f x. f (f x) } g = { λ x. g (g x) } = g (g x)! Church associated this function with the number 2! > Numbers are functions!» 2 = { λ f x. f (f x) }!» 2 g = g (g x)!g applied twice LC-23

24 Definition of Natural Numbers! The following shows how the natural numbers can be defined!» 0 = { λ f x. x }!» 1 = { λ f x. f x }!» 2 = { λ f x. f (f x) }!» 3 = { λ f x. f (f (f x)) }!» 4 = { λ f x. f (f (f (f x))) }!»! LC-24

25 Add 1 to a number! Define add_1 as follows!» add_1 = { λ a b c. b ( (a b) c ) }! Apply add_1 to the number (i.e. function)!!3 = { λ f x. f (f (f x) ) }!» add_1 3 = { λ b c. b ( (3 b) c ) } = { λ b c. b ( { λ x. b (b (b x) ) } c ) } = { λ b c. b ( b (b (b c) ) ) } = 4! LC-25

26 Multiply a number by two! Define mult_2 as follows!» mult_2 = { λ a b c. (a b) ( (a b) c ) }! Apply multi_2 to the number (i.e. function)!!3 = { λ f x. f (f (f x) ) }!» mult_2 3 = { λ b c. (3 b) ( (3 b) c ) } = { λ b c. (3 b) ({ λ x. b (b (b x) ) } c ) } = { λ b c. (3 b) ( b (b (b c) ) ) } = { λ b c. { λ x. b (b (b x) ) } ( b (b (b c) ) ) } = { λ b c. b (b (b (b (b (b c) ) ) ) ) } = 6! LC-26

27 Basic arithmetic operator definitions! Addition operator p + q!» add = { λ p q x y. ( (p x) (q x) ) y ) }! > Exercise show that (add 2) 3 = = 5 Multiplication operator p q!» mult = { λ p q x. p (q x) }! > Exercise show that (mult 2) 3 = 2 3 = 6! Power operator p q!» pow = { λ p q. p q }! LC-27

28 (pow 2) 3 = 9!» (pow 2) 3 = ({ λ p q. p q } 2) 3 = { λ q. 2 q } 3 = { λ q. { λ f x. f (f x) } g } 3 = { λ q x. q (q x) } 3 = { λ x. 3 (3 x) } = { λ x. { λ f y. f (f (f y) ) } (3 x) } = { λ x. { λ y. (3 x) ((3 x) ((3 x) y) ) } } = { λ x y. (3 x) ((3 x) (x (x (x y) ) ) ) } } = { λ x y. (3 x) (x ( x (x (x (x (x y) ) ) ) ) ) } = { λ x y. x (x (x (x ( x (x (x (x (x y) ) ) ) ) ) ) ) } = 9 = 3 2 LC-28