1 מבוא מורחב למדעי המחשב בשפת scheme תרגול 9. 2 outline mutable list...

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מבוא מורחב למדעי המחשב Scheme בשפת

9תרגול

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Outline

• Mutable list structure

• RPN calculator

• Vectors and sorting

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(define (set-to-wow! x)  (set-car! (car x) 'wow)  x)

(define x (list 'a 'b))

(define z1 (cons x x))

(set-to-wow! z1)

((wow b) wow b)

eq? point to the same object, equal? same content

(eq? (car z1) (cdr z1)) is true

(eq? (car z2) (cdr z2)) is false

(define z2 

(cons (list 'a 'b) 

(list 'a 'b)))

(set-to-wow! z2)

((wow b) a b)

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Map without list copying(define (map! f s) (if (null? s) 'done (begin (set-car! s (f (car s))) (map! f (cdr s)))))

(define s '(1 2 3 4))(map! square s) => dones => (1 4 9 16)

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append!

(define (append! x y)(set-cdr! (last-pair x) y)x)

where:(define (last-pair x)(if (null? (cdr x))

x(last-pair (cdr x))))

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append vs. append!

(define x ‘(a b))

(define y ‘(c d))

(define z (append x y))

z ==> (a b c d)

(cdr x) ==> ?

(define w (append! x y))

w ==> (a b c d)

(cdr x) ==> ?

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(define (last-pair x)

(if (null? (cdr x))

x

(last-pair (cdr x))))

(define (make-cycle x)

(set-cdr! (last-pair x) x)

x)

(define z (make-cycle (list 'a 'b 'c)))

Cycle

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What happens?

(last-pair z) =>

a b c

z

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Examine the list and determine whether it contains a cycle, whether a program that tried to find the end of the list by taking successive cdrs would go into an infinite loop

(define (is-cycle? c) (define (loop fast slow) (cond ((null? fast) #f) ((null? (cdr fast)) #f) ((eq? (cdr fast) slow) #t) (else (loop (cddr fast)

(cdr slow))))) (loop c c))

is-cycle?

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Bounded CounterA bounded counter can be incremented or decremented in

steps of 1, until upper and lower bounds are reached.

syntax: (make-bounded-counter init bottom top)

example:(define c (make-bounded-counter 3 1 5))(counter-inc! c) 4(counter-inc! c) 5(counter-inc! c) 5(counter-dec! c) 4

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List Implementation

Constructor:

(define (make-bounded-counter init bottom top)

(list init bottom top))

Selectors:

(define (counter-value c) (car c))

(define (counter-bottom c) (cadr c))

(define (counter-top c) (caddr c))

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List Implementation – cont.Mutators:

(define (counter-inc! c) (if (< (counter-value c) (counter-top c)) (set-car! c (+ 1 (counter-value c)))) (counter-value c))

(define (counter-dec! c) (if (> (counter-value c) (counter-bottom c)) (set-car! c (- (counter-value c) 1))) (counter-value c))

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Polish Notation• PostFix Notation: operands before operators• Expressions with binary operators can be written without

Parentheses• Apply operators, from left to right, on the last two numbers

– 5 7 4 - 2 * +– 5 3 2 * +– 5 6 +– 11

• Stack implementation:– Init: Empty stack– Number: insert! Into stack– Operator: apply on 2 top stack elements and insert!

result into stack

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Simulation(calc)574-TOP= 32*TOP= 6+TOP= 11exitok

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Read & Eval

• (read) - returns an expression from the user• (eval exp) - evaluates an expression

• We will use these functions in:– (perform m) - receives a symbol of an operation,

applies it on the two top numbers in the stack, and returns the result to the stack

– (iter) - reads an input from the user. If it is a number it is pushed to the stack, if it is an operator we call perform

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perform

(define (perform m)

(let ((arg2 ((stk 'top))))

((stk 'delete!))

(let ((arg1 ((stk 'top))))

((stk 'delete!))

((stk 'insert!) ((eval m) arg1 arg2)))))

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iter(define (iter) (let ((in (read))) (if (eq? in 'exit) 'ok (begin (cond ((number? in) ((stk 'insert!) in))

(else (perform in) (display "TOP= ") (display ((stk 'top))) (newline))) (iter)))))

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Calc

(define (calc)

(let ((stk (make-stack)))

(define (perform m) ...)

(define (iter) ... )

(iter)))

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VectorsConstructors:(vector v1 v2 v3 . . .)(make-vector size init)

Selector:(vector-ref vec place)

Mutator:(vector-set! vec place value)

Other functions:(vector-length vec)

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Example - accumulating

(define (accumulate-vec op base vec)

(define (helper from to)

(if (> from to) base

(op (vector-ref vec from)

(helper (+ from 1) to))))

(helper 0 (- (vector-length vec) 1)))

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Bucket Sort• Problem: Sorting numbers that distribute “uniformly”

across a given interval (for example, [0,1) ).• Observation: The number of elements that fall within

a sub-interval is proportional to the sub-interval’s size• Idea:

– Divide into sub-intervals (buckets)– Throw each number into appropriate bucket– Sort within each bucket (any sorting method)– Adjoin all sorted buckets

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Example

If we want ~3 numbers in each bucket, we need 3-4 buckets. Using 3 buckets we have:

Bucket 0.000 - 0.333: 0.31 0.10 0.05 0.23

Bucket 0.333 – 0.667: 0.44 0.56

Bucket 0.667 – 1 : 0.72 0.89 0.97 0.68

Sorting:0.31 0.44 0.72 0.89 0.10 0.05 0.97 0.23 0.56 0.68

Now sort!Bucket 0.000 - 0.333: 0.05 0.10 0.23 0.31 Bucket 0.333 – 0.667: 0.44 0.56Bucket 0.667 – 1 : 0.68 0.72 0.89 0.97

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Analysis

• Observation: If number of buckets is proportional to the number of elements, then the number of elements in each bucket is more or less the same, and bounded by a constant.

• Dividing into buckets: O(n)• Sorting one bucket: O(1)• Sorting all buckets: O(n)• Adjoining sorted sequences: O(n)

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Implementation

• A bucket is a list• The set of buckets is a vector• The elements are sorted while inserted into the buckets

(very similar so insertion sort):

(define (insert x s) (cond ((null? s) (list x)) ((< x (car s)) (cons x s)) (else (cons (car s) (insert x (cdr s))))))

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Implementation – for-each

• Scheme primitive• Similar to map, without returning a value• Useful for side-effects

(define (for-each proc lst)

(if (null? lst) ‘done

(begin (proc (car lst))

(for-each proc

(cdr lst)))))

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Implementation – cont.

(define (bucket-sort s) (let* ((size 5) (n (ceiling (/ (length s) size))) (buckets (make-vector n null)))

(define (insert! x) (let ((bucket (inexact->exact (floor (* n x))))) (vector-set! buckets bucket (insert x (vector-ref buckets bucket)))))

(for-each insert! s)

(accumulate-vec append null buckets)))

Streams

3.5, pages 316-352

definitions file on web

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cons, car, cdr

(define s (cons 9 (begin (display 7) 5)))-> prints 7

The display command is evaluated whileevaluating the cons.

(car s) -> 9(cdr s) -> 5

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cons-stream, stream-car, stream-cdr

(define s (cons-stream 9 (begin (display 7) 5)))

Due to the delay of the second argument, cons-stream does not activate the display command

(stream-car s) -> 9(stream-cdr s) -> prints 7 and returns 5 stream-cdr activates the display whichprints 7, and then returns 5.

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List enumerate(define (enumerate-interval low high)

(if (> low high) nil

(cons low

(enumerate-interval

(+ low 1) high))))

(enumerate-interval 2 8)

-> (2 3 4 5 6 7 8)

(car (enumerate-interval 2 8))

-> 2

(cdr (enumerate-interval 2 8))

-> (3 4 5 6 7 8)30

Stream enumerate(define (stream-enumerate-interval low high)

(if (> low high) the-empty-stream

(cons-stream low

(stream-enumerate-interval

(+ low 1) high))))

(stream-enumerate-interval 2 8)

-> (2 . #<promise>)

(stream-car (stream-enumerate-interval 2 8))

-> 2

(stream-cdr (stream-enumerate-interval 2 8))

-> (3 . #<promise>)31

List map

(map <proc> <list>)

(define (map proc s) (if (null? s) nil (cons (proc (car s)) (map proc (cdr s)))))

(map square (enumerate-interval 2 8))

-> (4 9 16 25 36 49 64)

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Stream map

(map <proc> <stream>)

(define (stream-map proc s) (if (stream-null? s) the-empty-stream (cons-stream (proc (stream-car s)) (stream-map proc (stream-cdr s)) )))

(stream-map square (stream-enumerate-interval 2 8))-> (4 . #<promise>)

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List of squares

(define squares

(map square

(enumerate-interval 2 8)))

squares

-> (4 9 16 25 36 49 64)

(car squares)

-> 4

(cdr squares)

-> (9 16 25 36 49 64)34

Stream of squares

(define stream-squares

(stream-map square

(stream-enumerate-interval 2 8)))

stream-squares

-> (4 . #<promise>)

(stream-car stream-squares)

-> 4

(stream-cdr stream-squares)

-> (9 . #<promise>)35

List reference

(define (list-ref s n)

(if (= n 0) (car s)

(list-ref (cdr s) (- n 1))))

(define squares

(map square

(enumerate-interval 2 8)))

(list-ref squares 3)

-> 25

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Stream reference

(define (stream-ref s n)

(if (= n 0) (stream-car s)

(stream-ref (stream-cdr s) (- n 1))))

(define stream-squares

(stream-map square

(stream-enumerate-interval 2 8)))

(stream-ref stream-squares 3)

-> 25

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List filter

(filter <predicate> <list>)

(define (filter pred s) (cond ((null? s) nil) ((pred (car s)) (cons (car s) (filter pred (cdr s)))) (else (filter pred (cdr s)))))

(filter even? (enumerate-interval 1 20))-> (2 4 6 8 10 12 14 16 18 20)

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Stream filter(stream-filter <predicate> <stream>)

(define (stream-filter pred s) (cond ((stream-null? s) the-empty-stream) ((pred (stream-car s)) (cons-stream (stream-car s) (stream-filter pred (stream-cdr s)))) (else (stream-filter pred (stream-cdr s))))))(stream-filter even?

(stream-enumerate-interval 1 20)) -> (2 . #<promise>)

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Generalized list map

(generalized-map <proc> <list1> … <listn>)

(define (generalized-map proc . arglists) (if (null? (car arglists)) nil (cons (apply proc (map car arglists)) (apply generalized-map (cons proc (map cdr arglists))))))

(generalized-map + squares squares squares)

-> (12 27 48 75 108 147 192)

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Generalized stream map

(generalized-stream-map <proc> <stream1> … <streamn>)

(define (generalized-stream-map proc . argstreams) (if (stream-null? (car argstreams)) the-empty-stream (cons-stream (apply proc (map stream-car argstreams)) (apply generalized-stream-map (cons proc (map stream-cdr argstreams))))))

(generalized-stream-map + stream-squares stream-squares stream-squares)-> (12 . #<promise>)

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List for each

(define (for-each proc s)

(if (null? s) 'done

(begin

(proc (car s))

(for-each proc (cdr s)))))

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Stream for each

(define (stream-for-each proc s) (if (stream-null? s) 'done (begin (proc (stream-car s)) (stream-for-each proc (stream-cdr s)))))

useful for viewing (finite!) streams(define (display-stream s) (stream-for-each display s))

(display-stream (stream-enumerate-interval 1 20)) -> prints 1 … 20 done

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Lists(define sum 0)(define (acc x) (set! sum (+ x sum)) sum)

(define s (map acc (enumerate-interval 1 20)))s -> (1 3 6 10 15 21 28 36 45 55 66 78 91 105 120 136 153 171 190 210) sum -> 210(define y (filter even? s))y -> (6 10 28 36 66 78 120 136 190 210)

sum -> 210

(define z (filter (lambda (x) (= (remainder x 5) 0)) s))z -> (10 15 45 55 105 120 190 210) sum -> 210

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(list-ref y 7)

-> 136 sum -> 210

(display z)

-> prints (10 15 45 55 105 120 190 210)

sum -> 210

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Streams

(define sum 0)(define (acc x) (set! sum (+ x sum)) sum)

(define s (stream-map acc (stream-enumerate-interval 1 20)))s -> (1 . #<promise>) sum -> 1

(define y (stream-filter even? s))y -> (6 . #<promise>) sum -> 6

(define z (stream-filter (lambda (x) (= (remainder x 5) 0)) s))z -> (10 . #<promise>) sum -> 10

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(stream-ref y 7)

-> 136 sum -> 136

(display-stream z)

-> prints 10 15 45 55 105 120 190 210 done

sum -> 210

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Defining streams implicitlyby delayed evaluation

Suppose we needed an infinite list of Dollars.We can(define bill-gates (cons-stream ‘dollar bill-gates))

If we need a Dollar we can take the car

(stream-car bill-gates) -> dollar

The cdr would still be an infinite list of

Dollars.

(stream-cdr bill-gates)->(dollar . #<promise>)48

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Infinite Streams

Formulate rules defining

infinite series

wishful thinking is key

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1,ones

(define ones

(cons-stream 1 ones))

1,1,1,… = ones =

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2,twos(define twos (cons-stream 2 twos))

ones + onesadding two infinite series of ones(define twos (stream-map + ones ones))

2 * oneselement-wise operations on an infinite series of ones(define twos (stream-map (lambda (x) (* 2 x)) ones)) or (+ x x)

2,2,2,… = twos =

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1,2,3,… = integers =

1,ones + integers

1,1,1…

1,2,3,…

2,3,4,…

(define integers

(cons-stream 1

(stream-map + ones integers)))

+

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0,1,1,2,3,… = fibs =

0,1,fibs + (fibs from 2nd position) 0,1,1,2,… 1,1,2,3,… 1,2,3,5,…

(define fibs (cons-stream 0 (cons-stream 1 (stream-map + fibs (stream-cdr fibs)))))

+

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1,doubles + doubles

1,2,4,8,…

1,2,4,8,…

2,4,8,16,…

(define doubles (cons-stream 1

(stream-map + doubles doubles)))

1,2,4,8,… = doubles =

+

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1,2 * doubles

(define doubles (cons-stream 1

(stream-map (lambda (x) (* 2 x)) doubles)))

or (+ x x)

1,2,4,8,… = doubles =

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1,factorials * integers from 2nd

position 1, 1*2, 1*2*3,… 2, 3, 4,… 1*2,1*2*3,1*2*3*4,…

(define factorials (cons-stream 1 (stream-map * factorials (stream-cdr integers))))

1,1x2,1x2x3,... = factorials =

x

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(1),(1 2),(1 2 3),… = runs =

(1), append runs with a list of integers from 2nd position

(1), (1 2), (1 2 3),… (2), (3), (4),… (1 2),(1 2 3),(1 2 3 4),…

(define runs (cons-stream (list 1) (stream-map append runs (stream-map list (stream-cdr integers)))))

append

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a0,a0+a1,a0+a1+a2,… = partial sums =

a0,partial sums + (stream from 2nd pos) a0, a0+a1, a0+a1+a2,… a1, a2, a3,… a0+a1,a0+a1+a2,a0+a1+a2+a3,…

(define (partial-sums a) (cons-stream (stream-car a) (stream-map + (partial-sums a) (stream-cdr a))))

+

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Partial Sums (cont.)(define (partial-sums a)(define sums

(cons-stream (stream-car a) (stream-map + sums (stream-cdr a)))) sums)

This implementation is more efficient since it uses the stream itself rather than recreating it recursively

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RepeatInput: procedure f of one argument, number of repetitionsOutput: f*…*f, n times

(define (repeated f n) (if (= n 1) f (compose f (repeated f (- n 1)))))

(define (compose f g) (lambda (x) (f (g x))))

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Repeat streamf,f*f,f*f*f,… = repeat =f,compose f,f,f,… with repeat

(define f-series (cons-stream f f-series))

(define stream-repeat (cons-stream f (stream-map compose f-series stream-repeat)))

We would like f to be a parameter61

Repeat streamf,f*f,f*f*f,… = repeat =f,compose f,f,f,… with repeat

(define (repeated f) (define f-series (cons-stream f f-series)) (define stream-repeat (cons-stream f (stream-map compose f-series stream-repeat))) stream-repeat)

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Interleave

1,1,1,2,1,3,1,4,1,5,1,6,…(interleave ones integers)

s0,t0,s1,t1,s2,t2,… interleave =s0,interleave (t, s from 2nd position)(define (interleave s t) (if (stream-null? s) t (cons-stream (stream-car s) (interleave t (stream-cdr s)))))

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