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))

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 11

(= 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 12

(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 13

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 14

(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) ) 15

Normal Distribution The formula of the normal probability density function is based on two parameters: the mean (μ) and the standard deviation (σ). Its formula is: 16

Normal Distribution 17

Normal Distribution The cumulative density function: 18

Standard Normal Distribution If μ=0 and σ=1 then we call it Standard Normal Distribution. The formula of the standard normal probability density function is: The cumulative density function is defined: Note: F(x)=Φ((x-μ)/σ) 19

Standard Normal Distribution Let’s first recall two general functions: ; General sum procedure (define (sum term a next b) (if (> a b) 0 (+ (term a) (sum term (next a) next b)))) ; General Integral function (define (integral f a b) (define dx 1.0e-4) (* (sum f a (lambda (x) (+ x dx)) b) dx)) 20

Write a procedure std-normal that computes the standard normal cumulative function. Assume that Φ(x)=0 for x 6. (define (std-normal x) (let ((pi ) (e ) (sigma 6)) (define (phi x) ) (cond ((< x -sigma) ) ((> x sigma) ) (else ))) Standard Normal Distribution (/ (expt e (* (- 0.5) (* x x))) (sqrt (* 2 pi))) 0 1 (integral phi (- sigma) x) 21

Write a procedure normalize which, given a function foo and two parameters a, b returns the function G such that G(x) = foo((x – a)/b). That is, you should have: ((normalize square 5 2) 1) ==> 4 (define (normalize foo a b) (lambda (x) (if (= b 0) (display "Error in normalize: Divide by zero") ))) Standard Normal Distribution (foo (/ (- x a) b)) 22

Write a procedure normal that computes the normal cumulative function for any μ and σ. Remember: F(x)=Φ((x-μ)/σ) (define (normal x miu sigma) ) Normal Distribution ((normalize std-normal miu sigma) x) 23

Accelerating Computations 24

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? 25

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? 26

Double Steps Fibonacci Transformation: b a a+b 2a+b 3a+2b … Double Transformation: 27

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 28

Squaring Transformations General Linear Transformation: Squared: 29

Iterative Algorithm Initialize: Stop condition: If count=0 return b Step count is oddcount is even 30

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!): 31

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)) 32