מבוא מורחב למדעי המחשב בשפת Scheme

מבוא מורחב למדעי המחשב Schemeבשפת

4תרגול

Outline

• Repeated f• Accelerating computations

– Fibonacci

• Let and let*• Recursion examples

– Palindrome?– Log

• Order of growth

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3

(= n 1) f

(repeated f (- n 1))

f(x), f(f(x)), f(f(f(x))), …apply f, n times

(define (repeated f n) (if

(compose f )))

((repeated inc 5) 100) => 105((repeated square 2) 5) => 625

Repeated f

(define (compose f g)

(lambda (x) (f (g x)))) Compose now

Execute later

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(define (repeated f n)

(lambda (x) (repeated-iter f n x)))

Repeated f - iterative

(define (repeated-iter f n x) (if (= n 1)

(f x) (repeated-iter f (- n 1) (f x))))

Do nothing until called later

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Repeated f – Iterative II

(define (repeated f n)

(define (repeated-iter count accum)

(if (= count n)

accum

(repeated-iter (+ count 1)

(compose f accum))))

(repeated-iter 1 f))

Compose now

Execute later

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(define (smooth f) (let ((dx 0.1))

))

(define (average x y z) (/ (+ x y z) 3))

(lambda (x) (average (f (- x dx)) (f x)

(f (+ x dx))))

Smooth a function f:g(x) = (f(x – dx) + f(x) + f(x + dx)) / 3

((repeated smooth n) f)

Repeatedly smooth a function

(define (repeated-smooth f n) )

AcceleratingComputations

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Iterative Fibonacci(define (fib n)

(define (fib-iter a b count)

(if (= count 0)

b

(fib-iter (+ a b) a (- count 1)))

(fib-iter 1 0 n))

• Computation time: (n)• Much better than Recursive implementation, but…• Can we do better?

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Slow vs Fast Expt

• Slow (linear)– b0=1– bn=bbn-1

• Fast (logarithmic)– bn=(b2)n/2 if n is even– bn=bbn-1 if n is odd

• Can we do the same with Fibonacci?

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Double Steps

• Fibonacci Transformation:

ab

baaT

0 1 1 2 3 5 8 13 21 b a a+b 2a+b 3a+2b …

• Double Transformation:

bab

baaT

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A Squaring Algorithm

• If we can square (or multiply) linear transformations, we have an algorithm:– Apply Tn on (a,b), where:– Tn=(T2)n/2 If n is even– Tn=TTn-1 If n is odd

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Squaring Transformations• General Linear Transformation:

wbzab

ybxaaT wzyx ,,,

• Squared:

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2

2

2,,,

wyzzwxzywxyyzx

wzyx

T

bwyz

azwxz

wbzaw

ybxazb

bywxy

ayzx

wbzay

ybxaxa

T

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Iterative Algorithm• Initialize:

ab

baaTncountba 01

• Stop condition: If count=0 return b

• Step

2/1

,, 2

countcountcountcount

TTbaTba

count is odd count is even

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Representing Transformations

• We need to remember x, y, z, w• Fibonacci Transformations belong to a simpler

family:

aqbpb

apaqbqaTpq

• T01 is the basic Fibonacci transformation• Squaring (verify on your own!):

222 2

2

qpqqppq TT

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Implementation (finally)(define fib n)

(fib-iter 1 0 0 1 n))

(define (fib-iter a b p q count)

(cond ((= count 0) b)

((even? count)

(fib-iter a

b

(/ count 2)

(else (fib-iter

p

q

(- count 1))))

(+ (square p) (square q))

(+ (* 2 p q) (square q))

(+ (* b q) (* a q) (* a p))

(+ (* b p) (* a q))

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The syntactic sugar “Let”(define (f x y)

( define (f-helper a b) *( +( x (square a))

*( y b) *( a b)))

( f-helper (+ 1 (* x y)) -( 1 y)))

(define (f x y)(( lambda (a b) *( +( x (square a))

*( y b) *( a b)))

+( 1 *( x y)) -( 1 y)))

(define (f x y)( let ((a (+ 1 (* x y)))

( b (- 1 y))) *( +( x (square a))

*( y b) *( a b))))

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The syntactic sugar “Let”(Let ((<var1> <exp1>)

<( var2> <exp2)>

..

<( varn> <expn))>

< body)>

((lambda (<var1> ….. <varn>)

< body)>

< exp1>

< exp2>

…

< expn)>

Is defined to be equivalent to:

bindings

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let*

(let* ((<var1> <exp1>)…(<varn> <expn>))

<body>)

Is equivalent to

(let ((<var1> <exp1>)) (let* ((<var2> <exp2>)… (<varn> <expn>)) <body>))

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let vs. let*

(let ((x 2) (y 3)) (let ((x 7)

(z (+ x y))) (* z x))) ==> 35

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let vs. let*

(let ((x 2) (y 3)) (let* ((x 7)

(z (+ x y))) (* z x))) ==> 70

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palindrome?

• Palindromes are (positive) numbers that read the same in both directions (e.g. left to right and right to left).

• Write a procedure (palindrome? x) that gets an integer as parameter and returns true (#t) if the number is a palindrome and false (#f) otherwise.

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Examples

> (palindrome? 1234567)#f> (palindrome? 1234321)#t> (palindrome? 56865)#t

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Useful functions

(define (least-significant x) (remainder x 10))

(define (remove-least-significant x) (quotient x 10))

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Implementation

1. Construct a new number by reversing the digits of the input (x)

2. Test whether these two numbers are equal.

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reverse-num

(define (reverse-num x) (define (helper x factor)(if (< x 10) x (+ (* ( ) factor)

(helper ( ) (/ factor 10)))))

(define (factor x) (if (< x 10)

1 (* 10 (factor ( )))))

(helper x (factor x)))

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reverse-num

(define (reverse-num x) (define (helper x factor)(if (< x 10) x (+ (* (least-significant x) factor)

(helper ( ) (/ factor 10)))))

(define (factor x) (if (< x 10)

1 (* 10 (factor ( )))))


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reverse-num


(helper (remove-least-significant x) (/ factor 10))))) (define (factor x)

(if (< x 10) 1

(* 10 (factor ( )))))


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reverse-num


(helper (remove-least-significant x) (/ factor 10))))) (define (factor x)

(if (< x 10) 1 (* 10 (factor (remove-least-significant x)))))


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palindrome?

(define (palindrome? x) (if (< x 0)

#f (= x (reverse-num x))))

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log

Consider the function defined as follows: lg(1) = 0lg(n) = lg( n/2 ) + 1, n > 1⌊ ⌋ *Do not use mathematical functions not shown here, such as expt or sqrt. *You may use (floor x) to compute x .⌊ ⌋

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Recursive process

(define (lg n) (if (= n 1)

0(+ (lg (floor (/ n 2))) 1)))

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Iterative process

(define (lg n) (define (helper n result)

(if (= n 1)result(helper ( ) ( ))))

(helper n 0))

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Iterative process


(if (= n 1)result(helper (floor (/ n 2))

( )))) (helper n 0))

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Iterative process


(if (= n 1)result(helper (floor (/ n 2)) (+ result 1))))

(helper n 0))

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Order of Growth

• For each of the following statements, determine whether it is true or false. If it is true, provide a proof, if false, provide a counterexample.

• – Θ(2^n) = Θ(3^n)– 2^(3*lg n+2) = O(n^3)

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Order of Growth

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מבוא מורחב למדעי המחשב בשפת Scheme