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מבוא מורחב למדעי המחשב בשפת Scheme

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מבוא מורחב למדעי המחשב בשפת Scheme. תרגול 3. Outline. High order procedures Finding Roots Compose Functions Accelerating Computations Fibonacci. Input : Continuous function f(x) a , b such that f(a)<0<f(b) Goal : Find a root of the equation f(x)=0

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### מבוא מורחב למדעי המחשב בשפת Scheme

תרגול 3

Outline
• High order procedures
• Finding Roots
• Compose Functions
• Accelerating Computations
• Fibonacci

Input:

• Continuous function f(x)
• a, b such that f(a)<0<f(b)
• Goal: Find a root of the equation f(x)=0
• Relaxation: Settle with a “close-enough” solution

Finding roots of equations

General Binary Search
• Search space: set of all potential solutions
• e.g. every real number in the interval [a b] can be a root
• Divide search space into halves
• e.g. [a (a+b)/2) and [(a+b)/2 b]
• Identify which half has a solution
• e.g. r is in [a (a+b)/2) or r is in [(a+b)/2 b]
• Find the solution recursively in reduced search space
• [a (a+b)/2)
• Find solution for small search spaces
• e.g. if abs(a-b)<e, r=(a+b)/2
Back to finding roots
• Theorem: if f(a)<0<f(b) and f(x) is continuous, then there is a point c(a,b) such that f(c)=0
• Note: if a>b then the interval is (b,a)
• Half interval method
• Divide (a,b) into 2 halves
• If interval is small enough – return middle point
• Otherwise, use theorem to select correct half interval
• Repeat iteratively
Example (continued)

a

b

And again and again…

Scheme implementation

x

(search f a x)

(search f x b)

x

(define (search f a b)

(let ((x (average a b)))

(if (close-enough? a b)

(let ((y (f x)))

(cond ((positive? y) )

((negative? y) )

(else ))))))

Complexity?

We need to define

(define (close-enough? x y)

(< (abs (- x y)) 0.001))

Determine positive and negative ends

(define (half-interval-method f a b)

(let ((fa (f a))

(fb (f b)))

(cond

((and )

(search f a b))

((and )

(search f b a))

(else

(display “values are not of opposite signs”)))

))

(negative? fa) (positive? fb)

(negative? fb) (positive? fa)

Examples:

sin(x)=0, x(2,4)

(half-interval-method 2.0 4.0)

x3-2x-3=0, x(1,2)

(half-interval-method

1.0 2.0)

sin

3.14111328125…

(lambda (x) (- (* x x x) (* 2 x) 3))

1.89306640625

(define (double f)

(lambda (x) (f (f x))))

(define (inc x) (+ x 1))

Double

((double inc) 5)

=> 7

inc(x) = x + 1

f(f(x)) = x + 2

((double square) 5)

=> 625

square(x) = x^2

f(f(x)) = x^4

11

(((double double) inc) 5) => x +4

9

(((double double) square) 5) => x16

=> 152587890625= 516

Double Double

12

inc = x + 1

(double (double(double inc)))

= x + 8

((double (double double)) inc)

= x + 16

(((double (double double)) inc) 5)

21

Double Double Double

13

Double Double Double
• Let D = double
• ((double (double double)) = D(D(D))= D(D◦D)=(D◦D)◦(D◦D)= D◦D◦D◦D=D4
• D4 (f)=D3 (f2)= D2 (f4)=D(f8)=f8◦f8=f16
Compose

Compose f(x), g(x) to f(g(x))

(define (compose f g)

(lambda (x) (f (g x))))

(define (inc x) (+ x 1))

((compose inc square) 3)

10

((compose square inc) 3)

16

15

f(x), f(f(x)), f(f(f(x))), …

apply f, n times

Repeated f

Compose now

Execute later

(define (compose f g)

(lambda (x) (f (g x))))

(= n 1)

f

(repeated f (- n 1))

(define (repeated f n)

(if

(compose f )))

((repeated inc 5) 100)

=> 105

((repeated square 2) 5)

=> 625

16

Repeated f - iterative

Do nothing until called later

(define (repeated-iter f n x)

(if (= n 1)

(f x)

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

(define (repeated f n)

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

17

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

18

Smooth a function f:

g(x) = (f(x – dx) + f(x) + f(x + dx)) / 3

Repeatedly smooth a function

(define (repeated-smooth f n)

)

(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))))

((repeated smooth n) f)

19

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?
Slow vs Fast Expt
• Slow (linear)
• b0=1
• bn=bbn-1
• Fast (logarithmic)
• bn=(b2)n/2if n is even
• bn=bbn-1if n is odd
• Can we do the same with Fibonacci?

b a a+b 2a+b 3a+2b …

Double Steps
• Fibonacci Transformation:
• 0 1 1 2 3 5 8 13 21
• Double Transformation:
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=TTn-1 If n is odd
Squaring Transformations
• General Linear Transformation:
• Squared:
Iterative Algorithm
• Initialize:
• Stop condition: If count=0 return b
• Step

count is odd

count is even

Representing Transformations
• We need to remember x, y, z, w
• Fibonacci Transformations belong to a simpler family:
• T01 is the basic Fibonacci transformation
• Squaring (verify on your own!):
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))