1. 1 Designing algorithms There are many ways to design an algorithm. Insertion sort uses an incremental approach: having sorted the sub-array A[1…j - 1], we insert the single element A[ j] into its proper place, yielding the sorted sub-array A[1…j]. Another approach to design is the divide-and-conquer approach which has a recursive structure to solve a given problem;  they break the problem into several sub-problems that are similar to the original problem but smaller in size,  solve the sub-problems recursively,  and then combine these solutions to create a solution to the original problem.

2. 2 The divide-and-conquer approach Recursive in structure  Divide the problem into several smaller sub- problems that are similar to the original but smaller in size  Conquer the sub-problems by solving them recursively. If they are small enough, just solve them in a straightforward manner.  Combine the solutions to create a solution to the original problem

3. 3 An Example: Merge Sort Divide: Divide the n-element sequence to be sorted into two subsequences of n/2 elements each Conquer: Sort the two subsequences recursively using merge sort. Combine: Merge the two sorted subsequences to produce the sorted answer.

4. 4 Merge Sort To sort n numbers  if n = 1 done!  recursively sort 2 lists of numbers  n/2  and  n/2  elements  merge 2 sorted lists in O(n) time Strategy  break problem into similar (smaller) subproblems  recursively solve subproblems  combine solutions to answer

5. 5 Merge Sort cont. [8, 3, 13, 6, 2, 14, 5, 9, 10, 1, 7, 12, 4] [8, 3, 13, 6, 2, 14, 5][9, 10, 1, 7, 12, 4] [8, 3, 13, 6][2, 14, 5] [8, 3][13, 6] [8][3][13][6] [2, 14][5] [2][14] [9, 10, 1][7, 12, 4] [9, 10][1] [9][10] [7, 12][4] [7][12]

6. 6 Merge Sort cont. [3, 8][6, 13] [3, 6, 8, 13] [8][3][13][6] [2, 14] [2, 5, 14] [2, 3, 5, 6, 8, 13, 14] [5] [2][14] [9, 10] [1, 9, 10] [1] [9][10] [7, 12] [4, 7, 12] [1, 4, 7, 9, 10,12] [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13,14] [4] [7][12]

7. 7 Merge Sort Procedure To sort the entire sequence A ={A[1], A[2],..., A[ n]}, we make the initial call MERGE-SORT( A, 1, length[ A]), where length[ A] = n. The procedure MERGE-SORT(A, p, r) sorts the elements in the sub-array A[ p…r]. The divide step simply computes an index q that partitions A[ p…r] into two sub-arrays: A[ p…q], containing n/2 elements, and A[ q + 1…r], containing n/2 elements.

8. 8

9. 9 Merge algorithm

10. 10 Merge Sort The key operation of the merge sort algorithm is the merging of two sorted sequences in the "combine" step. To perform the merging, we use an auxiliary procedure MERGE(A, p, q, r), where A is an array and p, q, and r are indices numbering elements of the array such that p ≤ q < r. The procedure assumes that the sub-arrays A[ p…q] and A[ q + 1…r] are in sorted order. It merges them to form a single sorted sub-array that replaces the current sub-array A[ p…r].

11. 11 Merge algorithm cont.  The operation of lines 10-17 in the call MERGE(A, 9, 12, 16).

12. 12 Merge algorithm cont. The operation of lines 10-17 in the call MERGE(A, 9, 12, 16)

13. 13 Analysis of Merge Sort StatementEffort So T(n) =  (1) when n = 1, and 2T(n/2) +  (n) when n > 1 MergeSort(A, left, right) {T(n) if (left < right) {  (1) mid = floor((left + right) / 2);  (1) MergeSort(A, left, mid);T(n/2) MergeSort(A, mid+1, right);T(n/2) Merge(A, left, mid, right);  (n) }

14. 14 Analysis of Merge Sort Divide: computing the middle takes O(1) Conquer: solving 2 sub-problem takes 2T(n/2) Combine: merging n-element takes O(n) Total: T(n) = O(1) if n = 1 T(n) = 2T(n/2) + O(n) + O(1) if n > 1  T(n) = O(n lg n) Solving this recurrence (how?) gives T(n) = O(n lg n) This expression is a recurrence To simplify the analysis we assume that the original problem size is a power of 2. Each divide step then yields two subsequences of size exactly n/2.

15. 15 Analysis of Merge Sort cont. Assume n=2 k for k>=1 T(n) = 2 T(n/2) + bn + c T(n/2) = 2T((n/2) /2) + b(n/2) + c = 2[2T(n/4) + b(n/2) + c] + bn +c = 4 T(n/4)+ bn +2c +bn +c =4 T(n/4) + 2bn+ (1 + 2) c = 2 2 T(n/2 2 )+2bn+(2 0 +2 1 ) = 4 [2T((n/4)/2) + b(n/4) + c] +2bn + (1+2)c =8 T(n/8) + 3bn+ (1+2+4)c =2 3 T(n/2 3 ) + 3bn+ (2 0 +2 1 +2 2 )c =2 k T(n/2 k ) +kbn+(2 0 +2 1 +…+2 k-1 )c T(1) = a, since n=2 k  log n = log2 k = k T(n) = 2 k. a + k bn + (2 0 +2 1 +…+2 k-1 ) c, but = b. n log n + (a + c) n – c = O (n log n)Worst case