Download - Lecture #7
מבוא מורחב1
Lecture #7
מבוא מורחב2
The rational number abstraction
Wishful thinking:
(make-rat <n> <d>) Creates a rational number <r>
(numer <r>) Returns the numerator of <r>
(denom <r>) Returns the denominator of <r>
Constructor:
Selectors:
מבוא מורחב3
A contract
(numer (make-rat <n> <d>))
(denom (make-rat <n> <d>))= <n>
<d>
How do we do it ?
מבוא מורחב5
Using pairs
(define (make-rat n d) (cons n d))
(define (numer x) (car x))
(define (denom x) (cdr x))
מבוא מורחב6
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))))
AbstractionViolation
מבוא מורחב7
Reducing to lowest terms
(define (make-rat n d) (let ((g (gcd n d))) (cons (/ n g) (/ d g))))
מבוא מורחב8
Abstraction barriers
Programs that use rational numbers
add-rat sub-rat …….
make-rat numer denom
car cdr cons
מבוא מורחב9
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))
מבוא מורחב10
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))))
מבוא מורחב11
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))
מבוא מורחב12
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”
מבוא מורחב13
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)
3
2
1
מבוא מורחב14
Box and pointer diagram
4
1
3
2
(cons (cons 1 (cons 2 3)) 4)
מבוא מורחב15
Lists
1 3 2
(cons 1 (cons 3 (cons 2 ‘())))
מבוא מורחב16
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.
מבוא מורחב17
Lists
•(list <el1> <el2> ... <eln>)
• Same as
(cons <el1>
(cons <el2> (cons … (cons <eln> ‘()))))
<el1> <el2> <eln>
…
מבוא מורחב18
Lists
• Predicates
; null? anytype -> boolean(null? <z>) ==> #t if <z> evaluates to empty list
; pair? anytype -> boolean(pair? <z>) ==> #t if <z> evaluates to a pair
; atom? anytype -> boolean(define (atom? z)
(and (not (pair? z)) (not (null? z))))
מבוא מורחב19
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
מבוא מורחב20
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
2 3 4
מבוא מורחב21
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
מבוא מורחב22
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
מבוא מורחב23
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
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)
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 (n2)
מבוא מורחב27
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)
מבוא מורחב28
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
מבוא מורחב29
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 ()))))
מבוא מורחב30
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 '()))