Lecture E Introduction to Algorithms Unit E1 – Basic Algorithms Lesson E – Introduction to Algorithms Slide 1 of 53. Lesson or Unit Topic or Objective Demonstrate the notion of an algorithm using two classic ones Lesson E – Introduction to Algorithms Slide 2 of 53. Computational problems • A computational problem specifies an input-output relationship What does the input look like? What should the output be for each input? • Example: Input: an integer number N Output: Is the number prime? • Example: Input: A list of names of people Output: The same list sorted alphabetically • Example: Input: A picture in digital format Output: An English description of what the picture shows Lesson E – Introduction to Algorithms Slide 3 of 53. Algorithms • An algorithm is an exact specification of how to solve a computational problem • An algorithm must specify every step completely, so a computer can implement it without any further “understanding” • An algorithm must work for all possible inputs of the problem. • Algorithms must be: Correct: For each input produce an appropriate output Efficient: run as quickly as possible, and use as little memory as possible – more about this later • There can be many different algorithms for each computational problem. Lesson E – Introduction to Algorithms Slide 4 of 53. Describing Algorithms • Algorithms can be implemented in any programming language • Usually we use “pseudo-code” to describe algorithms Testing whether input N is prime: For j = 2 .. N-1 If j|N Output “N is composite” and halt Output “N is prime” • In this course we will just describe algorithms in Java Lesson E – Introduction to Algorithms Slide 5 of 53. Greatest Common Divisor • The first algorithm “invented” in history was Euclid’s algorithm for finding the greatest common divisor (GCD) of two natural numbers • Definition: The GCD of two natural numbers x, y is the largest integer j that divides both (without remainder). I.e. j|x, j|y and j is the largest integer with this property. • The GCD Problem: Input: natural numbers x, y Output: GCD(x,y) – their GCD Lesson E – Introduction to Algorithms Slide 6 of 53. Euclid’s GCD Algorithm public static int gcd(int x, int y) { while (y!=0) { int temp = x%y; x = y; y = temp; } return x; } Lesson E – Introduction to Algorithms Slide 7 of 53. Euclid’s GCD Algorithm – sample run while int x = y = } (y!=0) { temp = x%y; y; temp; Example: Computing GCD(48,120) After After After After After 0 1 2 3 4 rounds round rounds rounds rounds temp -72 48 24 0 x 72 120 72 48 24 y 120 72 48 24 0 Output: 24 Lesson E – Introduction to Algorithms Slide 8 of 53. Correctness of Euclid’s Algorithm • Theorem: When Euclid’s GCD algorithm terminates, it • • • • • returns the mathematical GCD of x and y. Notation: Let g be the GCD of the original values of x and y. Loop Invariant Lemma: For all k 0, The values of x, y after k rounds of the loop satisfy GCD(x,y)=g. Proof of lemma: next slide. Proof of Theorem: The method returns when y=0. By the loop invariant lemma, at this point GCD(x,y)=g. But GCD(x,0)=x for every integer x (since x|0 and x|x). Thus g=x, which is the value returned by the code. Still Missing: The algorithm always terminates. Lesson E – Introduction to Algorithms Slide 9 of 53. Proof of Lemma • Loop Invariant Lemma: For all k 0, The values of x, y after k rounds of the loop satisfy GCD(x,y)=g. • Proof: By induction on k. For k=0, x and y are the original values so clearly GCD(x,y)=g. Induction step: Let x, y denote that values after k rounds and x’, y’ denote the values after k+1 rounds. We need to show that GCD(x,y)=GCD(x’,y’). According to the code: x’=y and y’=x%y, so the lemma follows from the following mathematical lemma. • Lemma: For all integers x, y: GCD(x, y) = GCD(x%y, y) • Proof: Let x=ay+b, where y>b 0. I.e. x%y=b. (1) Since g|y, and g|x, we also have g|(x-ay), I.e. g|b. Thus GCD(b,y) g = GCD(x,y). (2) Let g’=GCD(b,y), then g’|(x-ay) and g’|y, so we also have g’|x. Thus GCD(x,y) g’=GCD(b,y). Lesson E – Introduction to Algorithms Slide 10 of 53. Termination of Euclid’s Algorithm • Why does this algorithm terminate? After any iteration we have that x > y since the new value of y is the remainder of division by the new value of x. In further iterations, we replace (x, y) with (y, x%y), and x%y < x, thus the numbers decrease in each iteration. Formally, the value of xy decreases each iteration (except, maybe, the first one). When it reaches 0, the algorithm must terminate. public static int gcd(int x, int y) { while (y!=0) { int temp = x%y; x = y; y = temp; } return x; } Lesson E – Introduction to Algorithms Slide 11 of 53. Square Roots • The problem we want to address is to compute the square root of a real number. • When working with real numbers, we can not have complete precision. The inputs will be given in finite precision The outputs should only be computed approximately • The square root problem: Input: a positive real number x, and a precision requirement Output: a real number r such that |r-x| Lesson E – Introduction to Algorithms Slide 12 of 53. Square Root Algorithm public static double sqrt(double x, double epsilon){ double low = 0; double high = x>1 ? x : 1; while (high-low > epsilon) { double mid = (high+low)/2; if (mid*mid > x) high = mid; else low = mid; } return low; } Lesson E – Introduction to Algorithms Slide 13 of 53. Binary Search Algorithm – sample run while (high-low > epsilon) { double mid = (high+low)/2; if (mid*mid > x) high = mid; else low = mid; } Example: Computing sqrt(2) with precision 0.05: mid mid*mid After 0 rounds --After 1 round 1 1 After 2 rounds 1.5 2.25 After 3 rounds 1.25 1.56.. After 4 rounds 1.37.. 1.89.. After 5 rounds 1.43.. 2.06.. After 6 rounds 1.40.. 1.97.. low 0 1 1 1.25 1.37.. 1.37.. 1.40.. high 2 2 1.5 1.5 1.5 1.43.. 1.43.. Output: 1.40… Lesson E – Introduction to Algorithms Slide 14 of 53. Correctness of Binary Search Algorithm • Theorem: When the algorithm terminates it returns a value r that satisfies |r-x|. • Loop invariant lemma: For all k 0, The values of low, high after k rounds of the loop satisfy: low x high. • Proof of Lemma: For k=0, clearly low=0 x high=max(x,1). Induction step: The code only sets low=mid if mid x, and only sets high=mid if mid>x. • Proof of Theorem: The algorithm terminates when high-low, and returns low. At this point, by the lemma: low x high low+. Thus |low-x|. • Missing Part: Does the algorithm always terminate? How Fast? We will deal with this later. Lesson E – Introduction to Algorithms Slide 15 of 53. In General… • This type of binary search can be used to find the roots of any continuous function f. • Mean Value Theorem: if f(low)<0 and f(high)>0 then for some low<x<high, f(x)=0. • In our case, to find 2, we solved f ( x ) x 2 2 0 Lesson E – Introduction to Algorithms Slide 16 of 53. Lecture E Introduction to Algorithms Unit E1 – Basic Algorithms Lesson E – Introduction to Algorithms Slide 17 of 53. Lecture E Introduction to Algorithms Unit E2 – Running Time Analysis Lesson E – Introduction to Algorithms Slide 18 of 53. Lesson or Unit Topic or Objective Analysis of Running Times of Algorithms Lesson E – Introduction to Algorithms Slide 19 of 53. How fast will your program run? • The running time of your program will depend upon: The algorithm The input Your implementation of the algorithm in a programming language The compiler you use The OS on your computer Your computer hardware Maybe other things: temperature outside; other programs on your computer; … • Our Motivation: analyze the running time of an algorithm as a function of only simple parameters of the input. Lesson E – Introduction to Algorithms Slide 20 of 53. Basic idea: counting operations • Each algorithm performs a sequence of basic operations: Arithmetic: Comparison: Assignment: Branching: … (low + high)/2 if ( x > 0 ) … temp = x while ( true ) { … } • Idea: count the number of basic operations performed on the input. • Difficulties: Which operations are basic? Not all operations take the same amount of time. Operations take different times with different hardware or compilers Lesson E – Introduction to Algorithms Slide 21 of 53. Testing operation times on your system import java.util.*; public class PerformanceEvaluation { public static void main(String[] args) { int i=0; double d = 1.618; SimpleObject o = new SimpleObject(); final int numLoops = 1000000; long startTime = System.currentTimeMillis();; for (i=0 ; i<numLoops ; i++){ // put here a command to be timed } long endTime = System.currentTimeMillis(); long duration = endTime - startTime; double iterationTime = (double)duration / numLoops; System.out.println("duration: "+duration); System.out.println("sec/iter: "+iterationTime); }} class SimpleObject { private int x=0; public void m() { x++; } } Lesson E – Introduction to Algorithms Slide 22 of 53. Sample running times of basic Java operations Operation Loop Body nSec/iteration Sys1 Loop Overhead ; Sys2 196 10 Double division d = 1.0 / d; 400 77 Method call o.m(); 372 93 o=new SimpleObject(); 1080 110 Object Construction Sys1: PII, 333MHz, jdk1.1.8, -nojit Sys2: PIII, 500MHz, jdk1.3.1 Lesson E – Introduction to Algorithms Slide 23 of 53. Asymptotic running times • Operation counts are only problematic in terms of constant factors. • The general form of the function describing the running time is invariant over hardware, languages or compilers! public static int myMethod(int N){ int sq = 0; for(int j=0; j<N ; j++) for(int k=0; k<N ; k++) sq++; return sq; } • Running time is “about” 2 . • We use “Big-O” notation, and say that the running time is O(N 2 ) N Lesson E – Introduction to Algorithms Slide 24 of 53. Asymptotic behavior of functions Lesson E – Introduction to Algorithms Slide 25 of 53. Mathematical Formalization • Definition: Let f and g be functions from the natural numbers to the natural numbers. We write f=O(g) if there exists a constant c such that for all n: f(n) cg(n). f=O(g) c n: f(n) cg(n) • This is a mathematically formal way of ignoring constant factors, and looking only at the “shape” of the function. • f=O(g) should be considered as saying that “f is at most g, up to constant factors”. • We usually will have f be the running time of an algorithm and g a nicely written function. E.g. The running time of the previous algorithm was O(N^2). Lesson E – Introduction to Algorithms Slide 26 of 53. Asymptotic analysis of algorithms • We usually embark on an asymptotic worst case analysis of the running time of the algorithm. • Asymptotic: Formal, exact, depends only on the algorithm Ignores constants Applicable mostly for large input sizes • Worst Case: Bounds on running time must hold for all inputs. Thus the analysis considers the worst-case input. Sometimes the “average” performance can be much better Real-life inputs are rarely “average” in any formal sense Lesson E – Introduction to Algorithms Slide 27 of 53. The running time of Euclid’s GCD Algorithm • How fast does Euclid’s algorithm terminate? After the first iteration we have that x > y. In each iteration, we replace (x, y) with (y, x%y). In an iteration where x>1.5y then x%y < y < 2x/3. In an iteration where x 1.5y then x%y y/2 < 2x/3. Thus, the value of xy decreases by a factor of at least 2/3 each iteration (except, maybe, the first one). public static int gcd(int x, int y) { while (y!=0) { int temp = x%y; x = y; y = temp; } return x; } Lesson E – Introduction to Algorithms Slide 28 of 53. The running time of Euclid’s Algorithm • Theorem: Euclid’s GCD algorithm runs it time O(N), where N is the input length (N=log2x + log2y). • Proof: Every iteration of the loop (except maybe the first) the value of xy decreases by a factor of at least 2/3. Thus after k+1 iterations the k value of xy is at most ( 2 / 3 ) the original value. Thus the algorithm must terminate when k satisfies: xy ( 2 / 3 ) k 1 (for the original values of x, y). Thus the algorithm runs for at most 1 log 3 / 2 xy iterations. Each iteration has only a constant L number of operations, thus the total number of operations is at most (1 log 3 / 2 xy ) L Formally, (1 log 3 / 2 xy ) L L (1 2 log 2 x 2 log 2 y ) 3 LN Thus the running time is O(N). Lesson E – Introduction to Algorithms Slide 29 of 53. Running time of Square root algorithm • The value of (high-low) decreases by a factor of exactly 2 each iteration. It starts at max(x,1), and the algorithm terminates when it public static double sqrt(double x, double epsilon){ goes below . double low = 0; • Thus the number of double high = x>1 ? x : 1; while (high-low > epsilon) { iterations is at most double mid = (high+low)/2; if (mid*mid > x) high = mid; else low = mid; log 2 (max( x ,1) / ) • The running time is O (log x log 1 ) } return low; } Lesson E – Introduction to Algorithms Slide 30 of 53. Newton-Raphson Algorithm public static double sqrt(double x, double epsilon){ double r = 1; while ( Math.abs(r - x/r) > epsilon) r = (r + x/r)/2; return r; } Lesson E – Introduction to Algorithms Slide 31 of 53. Newton-Raphson – sample run while ( Math.abs(r - x/r) > epsilon) r = (r + x/r)/2; Example: Computing sqrt(2) with precision 0.01: After 0 rounds After 1 round After 2 rounds r 1 1.5 1.41.. x/r 2 1.33.. 1.41.. Output: 1.41… Lesson E – Introduction to Algorithms Slide 32 of 53. Analysis of Running Time • Correctness is clear since for every r the square root of x is between and r and x/r. • Here we will analyze the running time only for 1<x<2 • Denote: r ' ( r x r ) / 2 r 2r x x 4r x 4 r ' x (r x / r ) / 4 x 2 2 2 2 4r • • • 2 (r x) 2 4r 2 2 , where n r x after n loops At the beginning 0 1 , and 1 1 / 4 n 2 In general we have that n 2 2 At the end it suffices that n , since | r x | | r x | Thus the algorithm terminates when n log log 1 • Thus n • 2 2 n 1 2 Lesson E – Introduction to Algorithms Slide 33 of 53. In General… • The Newton-Raphson method can be used to find the roots of any differentiable function f. 2 • In our case, to find 2, we solved f ( r ) r 2 0 2 f (r ) r 2 r2 r • So, r' r r f ' (r ) 2r 2 Lesson E – Introduction to Algorithms Slide 34 of 53. Lecture E Introduction to Algorithms Unit E2 – Running Time Analysis Lesson E – Introduction to Algorithms Slide 35 of 53. Lecture E Introduction to Algorithms Unit E3 - Recursion Lesson E – Introduction to Algorithms Slide 36 of 53. Lesson or Unit Topic or Objective Using and understanding Recursion Lesson E – Introduction to Algorithms Slide 37 of 53. Designing Algorithms • There is no single recipe for inventing algorithms • There are basic rules: Understand your problem well – may require much mathematical analysis! Use existing algorithms (reduction) or algorithmic ideas • There is a single basic algorithmic technique: Divide and Conquer • In its simplest (and most useful) form it is simple induction In order to solve a problem, solve a similar problem of smaller size • The key conceptual idea: Think only about how to use the smaller solution to get the larger one Do not worry about how to solve to smaller problem (it will be solved using an even smaller one) Lesson E – Introduction to Algorithms Slide 38 of 53. Recursion • A recursive method is a method that contains a call to itself • Technically: All modern computing languages allow writing methods that call themselves We will discuss how this is implemented later • Conceptually: This allows programming in a style that reflects divide-n-conquer algorithmic thinking At the beginning recursive programs are confusing – after a while they become clearer than non-recursive variants Lesson E – Introduction to Algorithms Slide 39 of 53. Factorial public static void Factorial { public static void main() { System.out.println(“5!=“ + factorial(5)); } public static long factorial(int n){ if (n == 0) return 1; else return n * factorial(n-1); } } Lesson E – Introduction to Algorithms Slide 40 of 53. Elements of a recursive program • Basis: a case that can be answered without using further recursive calls In our case: if (n==0) return 1; • Creating the smaller problem, and invoking a recursive call on it In our case: factorial(n-1) • Finishing to solve the original problem In our case: return n * /*solution of recursive call*/ Lesson E – Introduction to Algorithms Slide 41 of 53. Tracing the factorial method System.out.println(“5!=“ + factorial(5)) 5 * factorial(4) 4 * factorial(3) 3 * factorial(2) 2 * factorial(1) 1 * factorial(0) return 1 return 1 return 2 return 6 return 24 return 120 Lesson E – Introduction to Algorithms Slide 42 of 53. Correctness of factorial method • Theorem: For every positive integer n, factorial(n) returns the value n!. • Proof: By induction on n: • Basis: for n=0, factorial(0) returns 1=0!. • Induction step: When called on n>1, factorial calls factorial(n-1), which by the induction hypothesis returns (n-1)!. The returned value is thus n*(n-1)!=n!. Lesson E – Introduction to Algorithms Slide 43 of 53. Raising to power – take 1 public static double power(double x, long n) { if (n == 0) return 1.0; return x * power(x, n-1); } Lesson E – Introduction to Algorithms Slide 44 of 53. Running time analysis • Simplest way to calculate the running time of a recursive program is to add up the running times of the separate levels of recursion. • In the case of the power method: There are n+1 levels of recursion • power(x,n), power(x,n-1), power(x, n-2), … power(x,0) Each level takes O(1) steps Total time = O(n) Lesson E – Introduction to Algorithms Slide 45 of 53. Raising to power – take 2 public static double power(double x, long n) { if (n == 0) return 1.0; if (n%2 == 0) { double t = power(x, n/2); return t*t; } return x * power(x, n-1); } Lesson E – Introduction to Algorithms Slide 46 of 53. Analysis • Theorem: For any x and positive integer n, the power method returns x n . • Proof: by complete induction on n. Basis: For n=0, we return 1. If n is even, we return power(x,n/2)*power(x,n/2). By the induction n/2 n/2 2 n hypothesis power(x,n/2) returns x , so we return ( x ) x . If n is odd, we return x*power(x,n-1). By the induction hypothesis power(x,n-1) returns x n 1 , so we return x x n 1 x n . • The running time is now O(log n): After 2 levels of recursion n has decreased by a factor of at least two (since either n or n-1 is even, in which case the recursive call is with n/2) Thus we reach n==0 after at most 2log2n levels of recursion Each level still takes O(1) time. Lesson E – Introduction to Algorithms Slide 47 of 53. Reverse public class Reverse { static InputRequestor in = new InputRequestor(): public static void main(String[] args) { printReverse(); } public static void printReverse() { int j = in.requestInt(“Enter another Number”+ ” (0 for end of list):”); if (j!=0){ printReverse(); System.out.println(j); } } } Lesson E – Introduction to Algorithms Slide 48 of 53. Recursive Definitions • Many things are defined recursively. • Fibonaci Numbers: 1, 1, 2, 3, 5, 8, 13, 21, … fn = fn-1 + fn-2 • Arithmetic Expressions. E.g. 2+3*(5+(3-4)) A number is an expression For any expression E: (E) is an expression For any two expressions E1, E2: E1+E2, E1-E2, E1*E2, E1/E2 are expressions • Fractals • In such cases recursive algorithms are very natural Lesson E – Introduction to Algorithms Slide 49 of 53. Fibonaci Numbers public class Fibonaci { public static void main(String[] args) for(int j = 1 ; j<20 ; j++) System.out.println(fib(j)); } { public static int fib (int n) { if (n <= 1) return 1; return fib(n-1) + fib(n-2); } } Lesson E – Introduction to Algorithms Slide 50 of 53. TurtleFractal public class TurtleFractal { static Turtle turtle = new Turtle(); public static void main(String[] args) turtle.tailDown(); drawFractal(500,4); } public static void drawFractal(int if (level==0) turtle.moveForward(length); else { drawFractal(length/3, level-1) turtle.turnLeft(60); drawFractal(length/3, level-1) turtle.turnRight(120); drawFractal(length/3, level-1) turtle.turnLeft(60); drawFractal(length/3, level-1) } }} { length, int level){ ; ; ; ; Lesson E – Introduction to Algorithms Slide 51 of 53. FractalTurtle output Lesson E – Introduction to Algorithms Slide 52 of 53. Lecture E Introduction to Algorithms Unit E3 - Recursion Lesson E – Introduction to Algorithms Slide 53 of 53.

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# Lecture E Introduction to Algorithms