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מבוא מורחב 1 Lecture #7. מבוא מורחב 2 The rational number abstraction Wishful thinking: (make-rat ) Creates a rational number (numer ) Returns the numerator.

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Presentation on theme: "מבוא מורחב 1 Lecture #7. מבוא מורחב 2 The rational number abstraction Wishful thinking: (make-rat ) Creates a rational number (numer ) Returns the numerator."— Presentation transcript:

1 מבוא מורחב 1 Lecture #7

2 מבוא מורחב 2 The rational number abstraction Wishful thinking: (make-rat ) Creates a rational number (numer ) Returns the numerator of (denom ) Returns the denominator of Constructor: Selectors:

3 מבוא מורחב 3 A contract (numer (make-rat )) (denom (make-rat )) = How do we do it ?

4 מבוא מורחב 4 Pairs (cons cells) (cons ) ==> ;type: x, x  Pair Where evaluates to a value, and evaluates to a value Returns a pair whose car-part is and whose cdr-part is (car ) ==> ;type: Pair  type of car part Returns the car-part of the pair (cdr ) ==> ;type: Pair  type of cdr part Returns the cdr-part of the pair

5 מבוא מורחב 5 Using pairs (define (make-rat n d) (cons n d)) (define (numer x) (car x)) (define (denom x) (cdr x))

6 מבוא מורחב 6 Alternative implementation (define (add-rat x y) (make-rat (+ (* (numer x) (denom y)) (* (numer y) (denom x))) (* (denom x) (denom y)))) (define (add-rat x y) (cons (+ (* (car x) (cdr y)) (* (car y) (cdr x))) (* (cdr x) (cdr y)))) Abstraction Violation

7 מבוא מורחב 7 Reducing to lowest terms (define (make-rat n d) (let ((g (gcd n d))) (cons (/ n g) (/ d g))))

8 מבוא מורחב 8 Abstraction barriers Programs that use rational numbers add-rat sub-rat ……. make-rat numer denom car cdr cons

9 מבוא מורחב 9 How can we implement pairs ? (define (cons x y) (lambda (m) (cond ((= m 0) x) ((= m 1) y) (else (error "Argument not 0 or 1 -- CONS" m)))))) (define (car z) (z 0)) (define (cdr z) (z 1))

10 מבוא מורחב 10 The pair constructed by (cons 6 9) is the procedure (lambda (m) (cond ((= m 0) 6) ((= m 1) 9) (else (error “Argument not 0 or 1 -- CONS” m))))

11 מבוא מורחב 11 In the book: (define (cons x y) (define (dispatch m) (cond ((= m 0) x) ((= m 1) y) (else (error "Argument not 0 or 1 -- CONS" m)))) dispatch) (define (car z) (z 0)) (define (cdr z) (z 1))

12 מבוא מורחב 12 Compound data Ideally want the result of this “gluing” to have the property of closure: “the result obtained by creating a compound data structure can itself be treated as a primitive object and thus be input to the creation of another compound object”

13 מבוא מורחב 13 Box and pointer diagram Note how pairs have the property of closure – we can use the result of a pair as an element of a new pair: (cons (cons 1 2) 3) 32 1

14 מבוא מורחב 14 Box and pointer diagram 4 1 3 2 (cons (cons 1 (cons 2 3)) 4)

15 מבוא מורחב 15 Lists 1 3 2 (cons 1 (cons 3 (cons 2 ‘())))

16 מבוא מורחב 16 lists A list is either ‘() (The empty list) A pair whose cdr is a list. Note that lists are closed under operations of cons and cdr.

17 מבוא מורחב 17 Lists (list... ) Same as (cons (cons (cons … (cons ‘())))) …

18 מבוא מורחב 18 Lists Predicates ; null? anytype -> boolean (null? ) ==> #t if evaluates to empty list ; pair? anytype -> boolean (pair? ) ==> #t if evaluates to a pair ; atom? anytype -> boolean (define (atom? z) (and (not (pair? z)) (not (null? z))))

19 מבוא מורחב 19 Lists -- examples (define one-to-four (list 1 2 3 4)) one-to-four ==> (1 2 3 4) (1 2 3 4) ==> error (car one-to-four) ==> (car (cdr one-to-four)) ==> 1 2 ( cadr one-to-four) ==> 2 ( caddr one-to-four) ==> 3

20 מבוא מורחב 20 Common Pattern #1: cons’ing up a list (define (enumerate-interval from to) (if (> from to) nil (cons from (enumerate-interval (+ 1 from) to)))) (e-i 2 4) (if (> 2 4) nil (cons 2 (e-i (+ 1 2) 4))) (if #f nil (cons 2 (e-i 3 4))) (cons 2 (e-i 3 4)) (cons 2 (cons 3 (e-i 4 4))) (cons 2 (cons 3 (cons 4 (e-i 5 4)))) (cons 2 (cons 3 (cons 4 nil))) (cons 2 (cons 3 )) (cons 2 ) ==> (2 3 4) 4 3 4 23 4

21 מבוא מורחב 21 Common Pattern #2: cdr’ing down a list (define (list-ref lst n) (if (= n 0) (car lst) (list-ref (cdr lst) (- n 1)))) (define squares (list 1 4 9 16 25)) (list-ref squares 3) ==> 16

22 מבוא מורחב 22 Common Pattern #2: cdr’ing down a list (define (length lst) (if (null? lst) 0 (+ 1 (length (cdr lst))))) (define odds (list 1 3 5 7)) (length odds) ==> 4

23 מבוא מורחב 23 Iterative length (define (length items) (define (length-iter a count) (if (null? a) count (length-iter (cdr a) (+ 1 count)))) (length-iter items 0))

24 24 Cdr’ing and Cons’ing Examples (define (append list1 list2) (cond ((null? list1) list2) ; base (else (cons (car list1) ; recursion (append (cdr list1) list2))))) (append (list 1 2) (list 3 4)) (cons 1 (append (2) (3 4))) (cons 1 (cons 2 (append () (3 4)))) (cons 1 (cons 2 (3 4))) ==> (1 2 3 4) T(n) =  (n)

25 25 Cdr’ing and Cons’ing Examples (define (copy lst) (if (null? lst) nil ; base case (cons (car lst) ; recursion (copy (cdr lst))))) (append (list 1 2) (list 3 4)) ==> (1 2 3 4) Strategy: “copy” list1 onto front of list2. (define (append list1 list2) (cond ((null? list1) list2) ; base (else (cons (car list1) ; recursion (append (cdr list1) list2)))))

26 26 Reverse (reverse (list 1 2 3 4)) ==> (4 3 2 1) (define (reverse lst) (cond ((null? lst) lst) (else (append (reverse (cdr lst)) (list (car lst))))))) (reverse (list 1 2 3)) (append (reverse (2 3)) (1)) (append (append (reverse (3)) (2)) (1)) (append (append (append (reverse ()) (3)) (2)) (1)) Append: T(n) = c*n   (n) Reverse: T(n) = c*(n-1) + c*(n-2) … c*1   (n 2 )

27 מבוא מורחב 27 Shall we have some real fun.. Lets write a procedure scramble.. (scramble (list 0 1 0 2)) ==> (0 0 0 1) (scramble (list 0 0 0 2 3 1 0 0 8 1)) ==> (0 0 0 0 0 3 0 0 0 8) (scramble (list 0 1 2 3 4 5 6 7)) ==> (0 0 0 0 0 0 0 0) (scramble (list 0 1 2 1 2 1 2 1 2)) ==> ( 0 0 0 2 2 2 2 2 2)

28 מבוא מורחב 28 Ok ok Each number in the argument is treated as backword index from its own position to a point earlier in the tup. The result at each position is found by counting backward from the current position according to this index ==> No number can be greater than its index

29 מבוא מורחב 29 tup = (0 1 2 1) rev-pre = () (cons 0 …. tup = (1 2 1) rev-pre = (0) (cons 0 (cons 0 tup = (2 1) rev-pre = (1 0) (cons 0 (cons 0 (cons 0 tup = (1) rev-pre = (2 1 0) (cons 0 (cons 0 (cons 0 (cons 2 tup = ( ) rev-pre = (1 2 1 0) (cons 0 (cons 0 (cons 0 (cons 2 ()))))

30 מבוא מורחב 30 list-ref scramble-i (define (scramble-i tup rev-pre) (if (null? tup) '() (cons ( (cons (car tup) rev-pre) (car tup)) ( (cdr tup) (cons (car tup) rev-pre))))) (define (scramble tup) (scramble-b tup '()))


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