Complexity Analysis (Part II )

1. Complexity Analysis (Part II) • Asymptotic Analysis. • Big-O Notation: More Details. • Order Arithmetic Rules. • Big-O Computation: Improved Guide lines.

2. Require Time n0 Input size Asymptotic Analysis. • The Big-O notation is typically written as: f(n) = O(g(n)). • It reads: “f(n) is of the order of g(n)”or “f(n) is proportional to g(n)”. • If f(n) = O(g(n)), then there exists c and no such that: f(n) < cg(n),n > n0 • The above equation states that as n increases, the algorithm complexity grows NO faster than a constant multiple ofg(n).

3. Big-O Notation: Some Details • Let's consider: f(n) = 5n + 3. • A possible bounding function is: g(n) = n4. 5n+3 = O(n4) • However, it is expected that the function g(n) should be of order (highest power) as small as possible. 5n+3 = O(n3) 5n+3 = O(n2)

4. Big-O Notation: Some Examples • Consider the following complexity function f(n) = 100 n. • Then, choosing: c = 100, no = 0 and g(n) = n will satisfy: f(n) < 100n • Then, if we have the complexity f(n) = 5n + 3.In this case, the following choice: c = 5, no = 0 and g(n) = n will satisfy: f(n) < 5n • Let assume now that the complexity is: f(n) = 3n2 - 100. • The choice of: c = 3, no = 0 and g(n) = n2 is possible to have: f(n) < 3n2

5. Big-O Notation: Simple Rule • Simple Rule: Drop out lower terms, constant terms and constant factors. SOME EXAMPLE • If f(n) = 100 n, then:g(n) = n or 100 n is O(n). • If f(n) = 5 n + 3, then:g(n) = n. Or 5 n + 3 is O(n). • If f(n) = 8 n2 log n + 5 n2 + n, then: g(n) = n2 log n. Or 8 n2 log n + 5 n2 + n is O( n2 log n).

6. Big-O Notation: Limitation • The big-O notation has some problems. • Let’s assume that we have the following complexity function: • We can find so many g(n) limits that can be considered as a valid bound of the complexity f(n). • So one cannot decide which g(n) will be the best for the algorithm at hand. Different choices of c and no are possible.

7. Big-O Notation: Arithmetic Rule • Multiplicative constants: O(k*f(n)) = O(f(n)) • Addition rule: O(f(n)+g(n)) = O(max[f(n),g(n)]) • Multiplication rule: O(f(n)*g(n)) = O(f(n)) * O(g(n)) SOME EXAMPLE • O(1000n) = O(n). Use multiplicative constants rule. • O(n2+3n+2) = O(n2). Use addition rule. • O((n3-1) (n log n + n + 2)) = O(n4log n). How??..

8. Computing Big-O Notation: Guideline • Loops such as for, while, and do-while: • The number of operations is equal to the number of iterations (e.g., n) times all the statements inside the for loop. • Nested loops: • The number of statements in all the loops times the product of the sizes of all the loops. • Consecutive statements: • Use the addition rule of order arithmetic: O(f(n)+g(n))=(max[f(n), g(n)]). • if/else and if/else if statement: • The number of operations is equal to running time of the condition evaluation and the maximum of running time of the if and else clauses. So, the complexity is: O(Cond) + O(max[if, else])

9. Computing Big-O Notation: Guideline (Contd.) • switch statements: Take the complexity of the most expensive case (with the highest number of operations). • Methods call: First, evaluate the complexity of the method being called. • Recursive methods: If it is a simple recursion, convert it to a for loop.

10. Computing Big-O Notation: Guideline (Contd.) • Let’s look at the for loop case: 1 for(int I = 0; I < n; i++) 2 sum += A[i]; 1 + (n + 1) + 2n + n = 4 n + 2 = O(n)

11. Computing Big-O Notation: Guideline (Contd.) • Now Let’s consider an example of nested loops: 1 for(int I = 0; I < n; i++) • for(int j = 0; j < n; j++) • sum+= B[i][j]; 4n2+ 4n +2 = O(n2)

12. Computing Big-O Notation: Guideline (Contd.) • An example of consecutive statements case is shown below: 1 for(int i = 0; i < n; i++) 2 a[i] = 0; 3 for(int i = 0; i < n; i++) 4 for(int j = 0; j < n; j++) { 5 sum = i + j; 6 size += 1; • } // End of inner loop. O(n +n2 ) = O(n2)

13. Computing Big-O Notation: Guideline (Contd.) • An example of switch statements: 1 char key; 2 int[] X = new int[5]; 3 int[][] Y = new int[10][10]; 4 ........ 5 switch(key) { 6 case 'a': 7 for(int i = 0; i < X.length; i++) 8 sum += X[i]; 9 break; 10 case 'b': 11 for(int i = 0; i < Y.length; j++) 12 for(int j = 0; j < Y[0].length; j++) 13 sum += Y[i][j]; 14 break; 15 } // End of switch block o(n) o(n2) o(n2)

14. Computing Big-O Notation: Guideline (Contd.) • Look at the case of selection block (if-elseif) shown below: 1 char key; 2 int[][] A = new int[5][5]; 3 int[][] B = new int[5][5]; 4 int[][] C = new int[5][5]; 5 ........ 6 if(key == '+') { 7 for(int i = 0; i < n; i++) 8 for(int j = 0; j < n; j++) 9 C[i][j] = A[i][j] + B[i][j]; 10 } // End of if block 11 else if(key == 'x') 12 C = matrixMult(A, B); 13 else 14 System.out.println("Error! Enter '+' or 'x'!"); O(n2) O(n3) O(n3) O(n)

15. O(1) • Sometimes ifelse statements must carefully checked: O(ifelse) = O(Condition)+ Max[O(if), O(else)] 1 int[] integers = new int[10]; 2 ........ 3 if(hasPrimes(integers) == true) 4 integers[0] = 20; 5 else • integers[0] = -20; 1 public boolean hasPrimes(int[] arr) { 2 for(int i = 0; i < arr.length; i++) 3 .......... 4 .......... 5 .......... 6 } // End of hasPrimes() O(1) O(ifelse) = O(Condition) = O(n)

16. Drill Questions • Consider the function f(n) = 3 n2 - n + 4. Show that: f(n) = O(n2). • Consider the functions f(n) = 3 n2 - n + 4 and g(n) = n log n + 5. Show that: f(n) + g(n) = O(n2). • Consider the functions f(n) =√ n and g(n) = log n. Show that: f(n) + g(n) =O(√n). • Indicate whether f(n) = O(g(n)) when we have f(n) = 10n and g(n) = n2 - 10n. • Indicate whether f(n) = O(g(n)) when we have f(n) = n3 and g(n) = n2 log n.