מבוא מורחב למדעי המחשב בשפת Scheme תרגול 4

Outline High order procedures –Finding Roots –Compose Functions Accelerating Computations –Fibonacci 2

Finding roots of equations 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 3

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 4

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 5

Example b a 6

Example (continued) b a And again and again… 7

(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 )))))) x (search f a x) (search f x b) x Scheme implementation 8 Complexity?

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) We need to define (define (close-enough? x y) (< (abs (- x y)) 0.001)) 9

sin(x)=0, x  (2,4) (half-interval-method ) x 3 -2x-3=0, x  (1,2) (half-interval-method ) Examples: sin … (lambda (x) (- (* x x x) (* 2 x) 3))

11 (define (double f) (lambda (x) (f (f x)))) (define (inc x) (+ x 1)) ((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 Double

12 (((double double) inc) 5) => 9 (double double) = double twice ((double double) inc) = (double (double inc)) inc = x + 1 ((double double) inc) = x + 4 (((double double) square) 5) => square = x^2 (double square) = x^4 ((double double) square) = x^16 Double

13 (((double (double double)) inc) 5) => 21 (double double) = double twice (double (double double) = twice, then twice again (4 times) inc = x + 1 ((double (double double)) inc) = x + 16 Double Double Double

14 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 (= 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

16 (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

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 (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) )

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 19

Accelerating Computations 20

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? 21

Slow vs Fast Expt Slow (linear) –b 0 =1 –b n =b  b n-1 Fast (logarithmic) –b n =(b 2 ) n/2 if n is even –b n =b  b n-1 if n is odd Can we do the same with Fibonacci? 22

Double Steps Fibonacci Transformation: b a a+b 2a+b 3a+2b … Double Transformation: 23

A Squaring Algorithm If we can square (or multiply) linear transformations, we have an algorithm: –Apply T n on (a,b), where: –T n =(T 2 ) n/2 If n is even –T n =T  T n-1 If n is odd 24

Squaring Transformations General Linear Transformation: Squared: 25

Iterative Algorithm Initialize: Stop condition: If count=0 return b Step count is oddcount is even 26

Representing Transformations We need to remember x, y, z, w Fibonacci Transformations belong to a simpler family: T 01 is the basic Fibonacci transformation Squaring (verify on your own!): 27

Implementation (finally) (define fib n) (fib-iter 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)) 28