Section Analysis of Algorithm with Big Ohpreview of Section

Section Analysis of Algorithm with Big Oh Preview of section I. Code demo Binary Search and run CompareSearches.java with Sequential Search II. As a section, count the instructions together III. In teams of two, answer questions 1..4 IV. Go over answers

I. Write method public int binarySearch(Integer[], int n, Integer search) to return the integer index of search or -1 of search is not in the array. Program to run with the method written as a method stub is located here. Run with n = 10,000 50,000 150,000. http://www.cs.arizona.edu/~mercer/Sections/CompareSearches.java

II. Count the instructions. The answer should be a polynomial (several answers are possible depending on how things are counted).

double[] data = { 76.0, 91.0, 100.0, 62.0, 89.0 };

int n = data.length;

int indexOfSmallest = 0;

for (int left = 0; left < n - 1; left++) {

indexOfSmallest = left;

for (int index = left + 1; index < n; index++) { if (data[index] < data[indexOfSmallest]) indexOfSmallest = index; }

double temp = data[left]; data[left] = data[indexOfSmallest]; data[indexOfSmallest] = temp; }

III. In teams of two, answers questions 1..4

1. Sort these functions by order of growth from highest to lowest. Write 1 above the fastest growing function, 2 above the second fastest growing function, and 8 above the slowest growing function. (Note: log24 = 2, log28 = 3 and log216 = 4)

2 n n 3 100*n n 1000 2 10 n 2*n log2n

2. Which term dominates this function when n gets really big, n2, 10n, or 100? ______

n2 + 10n + 100

3. When n=500 in the function above, what percentage of the function is the term 100? _____ Before answering question 4, here is a review of order of magnitudes in increasing growth rates: O(1) O(log n) O(n) O(n log n) O(n2) O(n3) O(2n)

… and some example big-O run times:

O(1) O(log n) O(n) int sum = 0; for(int j = n; j > 0; j = j / 2) for(int j = 1; j <= n; j++) sum += j; sum += j;

O(n2) product O(n) sum for(int r = 0; r < n; r++) for(int j = 1; j <= n; j++) for(int c = 0; c < n; c++) sum += j; sum += a[r][c]; for(int j = 1; j <= 2*n; j++) sum += j;

And some big-O style guidelines

 use highest order term (drop lower order terms)  Drop coefficients and drop constants  Use the tightest upper bound. It could be said that most algorithms run O(2n), a function that grow very fast. But use the tightest upper bound instead.

4. Determine the runtimes of the following loops expressed in big-O notation a. ______d. ______int sum = 0; for(int j = 0; j < n; j++) int n = 1000000; sum++; for(int k = 0; k < n; k++) sum--; b. ______int sum = 0; e. ______for (int j = 0; j < n; j++) for(int j = 0; j < n; j++) for (int k = 0; k < n; k++) System.out.println(x [ j ]); sum += j * k; c. ______for (int j = 0; j < n; j++) f. ______for (int k = 0; k < n; k++) for(int j = 1; j < n; j = 2 * j) for (int l = 0; l < n; l++) sum += j; sum += j * k * l;

IV Go over answers