1. CSE 2010:
Algorithms and Data Structures
Running time analysis of mergesort

2. current next current next current next p = 1 r = 8 q = 4 p = 1 r = 4
MERGE-SORT( A, 1, 8 )
current
next
MERGE-SORT( A, 1, 4 )
p = 1
r = 4
q = 2
current
next
MERGE-SORT( A, 1, 2 )
p = 1
r = 2
q = 1
current
next
MERGE-SORT( A, 1, 1)
p = 2
r = 2

3. The divide-and-conquer approach
Constant time if problem size is small enough
What are the values for a and b in merge sort?
a: number of subproblems
T(n/b): time to solve subproblem of size n/b.
Time to divide the problem into subproblems
Time to merge the partial solutions into the solution of the original problems.

4. The divide-and-conquer: worst-case analysis
For merge sort, c is equal to 1.
The sum D(n) + C(n) is Theta(n) + Theta(1), which in turn is Theta(n).

5. The divide-and-conquer: worst-case analysis
For merge sort, c is equal to 1.
The sum D(n) + C(n) is Theta(n) + Theta(1), which in turn is Theta(n).

6. Time required to solve the problems of size 1, also the time per array element of the divide and combine steps.
logn + 1 levels cn
Let’s re-rewrite the recurrence equation.
Height is logn. So, total time is of order n log n.
Each level contributes with a cost of cn. So, total cost is cn long + cn
Log n + 1, Base case is log 1 = 0. So, the correct number of levels is log n + 1.
cn/2
cn/4
2 x (cn/2) = cn
4 x (cn/4) = cn cn c

7. a >= 1 (must be contant): Number of smaller subproblems
f(n) is the combination time must be positive
a >= 1 (must be contant): Number of smaller subproblems
Each subproblem is 1/b the size of the original problem,
with b > 1.

8. We need to memorize 3 cases

9. Example 1
1st Step:
f(n) = (1/2)n^2+n which is polynomial. So, we are okay, and d = 2.
a = 1
b = 2
2nd Step:
Compare the values of a and b^d.

10. Example 2

11. Example 3

12. Example 4

13. Example 5

14. Example 6
Binary search

15. Example 7

16. Example 8

17. Example 9
The combination time f(n) must be positive.