Lecture 8 Cons car cdr sdr wdr CS

Lecture 8: Cons car cdr sdr wdr CS 200: Computer Science David Evans University of Virginia 3 February 2003 CS 200 Spring 2003 http: //www. cs. virginia. edu/~evans Computer Science

Menu • History of Scheme –LISP • Lists • List Recursion 3 February 2003 CS 200 Spring 2003 2

Confusion Is Good! It means you are learning new ways of thinking. 3 February 2003 CS 200 Spring 2003 3

History of Scheme • Scheme [1975] – Guy Steele and Gerry Sussman – Originally “Schemer” – “Conniver” [1973] and “Planner” [1967] • Based on LISP – John Mc. Carthy (late 1950 s) • Based on Lambda Calculus – Alonzo Church (1930 s) – Last few lectures in course 3 February 2003 CS 200 Spring 2003 4

LISP “Lots of Insipid Silly Parentheses” “LISt Processing language” Lists are pretty important – hard to write a useful Scheme program without them. 3 February 2003 CS 200 Spring 2003 5

Making Lists 3 February 2003 CS 200 Spring 2003 6

Making a Pair > (cons 1 2) (1. 2) 1 2 constructs a pair 3 February 2003 CS 200 Spring 2003 7

Splitting a Pair > (car (cons 1 2)) 1 > (cdr (cons 1 2)) 2 cons 1 2 car cdr car extracts first part of a pair cdr extracts second part of a pair 3 February 2003 CS 200 Spring 2003 8

Why “car” and “cdr”? • Original (1950 s) LISP on IBM 704 – Stored cons pairs in memory registers – car = “Contents of the Address part of the Register” – cdr = “Contents of the Decrement part of the Register” (“could-er”) • Doesn’t matters unless you have an IBM 704 • Think of them as first and rest (define first car) (define rest cdr) 3 February 2003 CS 200 Spring 2003 9

Implementing cons, car and cdr • Using PS 2: (define cons make-point) (define car x-of-point) (define cdr y-of-point) • As we implemented make-point, etc. : (define (cons a b) (lambda (w) (if (w) a b))) (define (car pair) (pair #t) (define (cdr pair) (pair #f) 3 February 2003 CS 200 Spring 2003 10

Pairs are fine, but how do we make threesomes? 3 February 2003 CS 200 Spring 2003 11

Threesome? (define (threesome a b c) (lambda (w) (if (= w 0) a (if (= w 1) b c)))) (define (first t) (t 0)) (define (second t) (t 1)) (define (third t) (t 2)) Is there a better way of thinking about our triple? 3 February 2003 CS 200 Spring 2003 12

Triple • A triple is just a pair where one of the parts is a pair! (define (triple a b c) (cons a (cons b c))) (define (t-first t) (car t)) (define (t-second t) (car (cdr t))) (define (t-third t) (cdr t))) 3 February 2003 CS 200 Spring 2003 13

Quadruple • A quadruple is a pair where the second part is a triple (define (quadruple a b c d) (cons a (triple b c d))) (define (q-first q) (car q)) (define (q-second q) (t-first (cdr t))) (define (q-third t) (t-second (cdr t))) (define (q-fourth t) (t-third (cdr t))) 3 February 2003 CS 200 Spring 2003 14

Multuples • A quintuple is a pair where the second part is a quadruple • A sextuple is a pair where the second part is a quintuple • A septuple is a pair where the second part is a sextuple • An octuple is group of octupi • A list (any length tuple) is a pair where the second part is a …? 3 February 2003 CS 200 Spring 2003 15

Lists List : : = (cons Element List) A list is a pair where the second part is a list. One little problem: how do we stop? This only allows infinitely long lists! 3 February 2003 CS 200 Spring 2003 16

From Lecture 6 Recursive Transition Networks ORNATE NOUN begin ARTICLE ADJECTIVE NOUN end ORNATE NOUN : : = ARTICLE ADJECTIVE NOUN ORNATE NOUN : : = ARTICLE ADJECTIVE ADJECTIVE ADJECTIVE NOUN 3 February 2003 CS 200 Spring 2003 17

Recursive Transition Networks ORNATE NOUN begin ARTICLE ADJECTIVE NOUN end ORNATE NOUN : : = ARTICLE ADJECTIVES NOUN ADJECTIVES : : = ADJECTIVES : : = 3 February 2003 CS 200 Spring 2003 18

Lists List : : = (cons Element List) List : : = It’s hard to write this! A list is either: a pair where the second part is a list or, empty 3 February 2003 CS 200 Spring 2003 19

Null List : : = (cons Element List) List : : = null A list is either: a pair where the second part is a list or, empty (null) 3 February 2003 CS 200 Spring 2003 20

List Examples > null () > (cons 1 null) (1) > (list? null) #t > (list? (cons 1 2)) #f > (list? (cons 1 null)) #t 3 February 2003 CS 200 Spring 2003 21

More List Examples > (list? (cons 1 (cons 2 null))) #t > (car (cons 1 (cons 2 null))) 1 > (cdr (cons 1 (cons 2 null))) (2) 3 February 2003 CS 200 Spring 2003 22

List Recursion 3 February 2003 CS 200 Spring 2003 23

Defining Recursive Procedures 1. Be optimistic. – Assume you can solve it. – If you could, how would you solve a bigger problem. 2. Think of the simplest version of the problem, something you can already solve. (This is the base case. ) 3. Combine them to solve the problem. 3 February 2003 CS 200 Spring 2003 24

Defining Recursive Procedures on Lists 1. Be optimistic. Be very optimistic – Assume you can solve it. – If you could, how would you solve a bigger For lists, assume we can solve problem. it for the cdr 2. Think of the simplest version of the problem, something you can already For lists, the simplest version is solve. usually null (the zero-length list) 3. Combine them to solve the problem. 3 February 2003 Combine something on the car of the list with the recursive. CSevaluation 200 Spring 2003 on the cdr. Remember 25 to test null? before using car or cdr.

Defining Sumlist (define sumlist (lambda (lst) (if (null? lst) > (sumlist (list 1 2 3 4)) 10 > (sumlist null) 0 0 ( + (car lst) (sumlist (cdr lst)) 3 February 2003 CS 200 Spring 2003 26

Defining Productlist (define productlist (lambda (lst) (if (null? lst) > (productlist (list 1 2 3 4)) 24 > (productlist null) 1 1 (* 3 February 2003 (car lst) (sumlist (cdr lst)) CS 200 Spring 2003 27

Defining Length (define length (lambda (lst) (if (null? lst) > (length (list 1 2 3 4)) 4 > (length null) 0 0 ( + (car 1 lst) (length (cdr lst)) 3 February 2003 CS 200 Spring 2003 28

Defining insertl (define insertl (lambda (lst f stopval) (if (null? lst) stopval (f (car lst) (insertl (cdr lst) f stopval))))) 3 February 2003 CS 200 Spring 2003 29

Definitions (define (sumlist lst) (insertl lst + 0)) (define insertl (lambda (lst f stopval) (if (null? lst) stopval (f (car lst) (insertl (cdr lst) f stopval))))) (define (productlist lst) (insertl lst * 1)) (define (length lst) (insertl lst (lambda (head rest) (+ 1 rest)) 0)) 3 February 2003 CS 200 Spring 2003 30

Charge • Next Time: lots more things you can do with lists (including the peg board puzzle!) • PS 3 Out Today – Use lists to make fractals – You have seen everything you need for it after today – Due next week Wednesday 3 February 2003 CS 200 Spring 2003 31