Lecture 16: Big-Oh Notation CSC 212 – Data Structures Lecture 16: Big-Oh Notation

Question of the Day What is smallest number of matchsticks moved to make this equation valid (using roman numerals) + =

Algorithm Analysis Approximate time to run a program with n inputs on 1GHz machine: n = 10 n = 50 n = 100 n = 1000 n = 106 O(n log n) 35 ns 200 ns 700 ns 10000 ns 20 ms O(n2) 100 ns 2500 ns 1 ms 17 min O(n5) 0.1 ms 0.3 s 10.8 s 11.6 days 3x1013 years O(2n) 1000 ns 13 days 4 x 1014 years Too long! O(n!) 4 ms

Big-Oh Notation Only consider very, very large data sets Nobody cares when entire program takes 2 minutes Leading multipliers are ignored O(5n) = O(2n) = O(n) Often an implementation issue, anyway Only keep the largest terms O(n5 + n2) = O(n5) What is extra 17 min. after 3x1013 years?

Big-Oh Notation Number of primitive operations executed Assignments, method calls, arithmetic operations, comparisons, indexing into arrays, following a reference, … Good, rough approximation of time Useful to compare algorithms Really good at eliminating algorithms

Big-Oh Rules of Thumb Primitive statements take O(1) time Loop from 1 - n Runs O(n) times, when adding constant amount Runs O(log n) times when multiplying by constant Multiply complexities of nested loops Add complexities of consecutive loops

Big-Oh Measures Worst-Case int power(int a, int b > 0) if a == 0 && b == 0 return -1; exp = 1; loop from 1 to b exp *= a; return exp; power takes O(n) in most cases Is O(1) when a & b are both 0, however Would still call this algorithm O(n)

Not All Nesting Is The Same public static int sum(int[][] a) { int total = 0; for (int i = 0; i < a.length; i++) { for (int j = 0; j < a[i].length; j++) { total += a[i][j]; } } return total; } Despite nested loops, this runs in O(n) time n here would be entries in entire array

Handling Method Calls Method call is O(1) operation, … … but method may take more time Must consider the complexity of the entire process Including complexity of all the methods called

Methods Calling Methods public static int sumOdds(int m) { int total = 0; for (int i = 1; i <= m; i+=2) { total += i; } return total; } public static void oddSeries(int n) { for (int i = 1; i < n; i++) { System.out.println(i + “ ” + sumOdds(n)); } oddSeries calls sumOdds n times Each call does O(n) work, takes O(n2) total time!

Justifying an Answer Important to explain your answer Saying method is O(n) does not make it O(n) Especially true for tricky & recursive methods Process can derive difficult answer May simplify the big-Oh computation Discover that big-Oh is less than thought Does not need to be formal proof But must explain your answer

Big-Oh Notation Ignoring recursive call, runs in O(1) time public static int factorial(int n) { if (n <= 2) { return 1; } else { return n * factorial(n - 1); } } Ignoring recursive call, runs in O(1) time There can at most n - 1 calls to factorial Therefore total time is n - 1 * O(1) = O(n)

Big-Oh Notation This function takes O(2n) time int fibonacci(int k) if k < 1 return k else return fibonacci(k-1) + fibonacci(k - 2) This function takes O(2n) time k is 2, makes 2 calls k is 3, makes 4 calls - 3: fibonacci(2) & 1: fibonacci(1) k is 4, makes 8 calls - 5: fibonacci(3) & 3: fibonacci(2) k is n, makes 2n-1 = 2-1 * 2n = O(2n) calls

Your Turn Get back into groups and do activity

Midterm Overview Midterm results were very mixed Max score = 99 Mean score = 79 Standard deviation = 14.03 May apply curve at end of term Will not curve any single item Some of you may need to work harder, come and ask me for help as we go on Everyone still has passing grade, however

Before Next Lecture… Finish week #6 assignment Finish lab #5 Programming assignment #2 next week