מבוא מורחב למדעי המחשב בשפת Scheme

1. מבוא מורחב למדעי המחשב בשפת Scheme תרגול 3

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

3. 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

4. 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

5. 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

6. Example a b

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

8. 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?

9. 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)

10. 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

11. (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

12. (((double double) inc) 5) => x +4 9 (((double double) square) 5) => x16 => 152587890625= 516 Double Double 12

13. 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

14. 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

15. 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

16. 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

17. 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

18. 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

19. 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

20. AcceleratingComputations

21. 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?

22. 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?

23. b a a+b 2a+b 3a+2b … Double Steps • Fibonacci Transformation: • 0 1 1 2 3 5 8 13 21 • Double Transformation:

24. 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

25. Squaring Transformations • General Linear Transformation: • Squared:

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

27. 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!):

28. 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))