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Divide and Conquer: Basic Algorithm Design Techniques

Learn about the divide and conquer algorithm design technique, including dynamic programming and greedy approaches. Explore how to break down complex problems into smaller sub-problems and combine their solutions to solve the original problem.

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Divide and Conquer: Basic Algorithm Design Techniques

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  1. Lecture 2: Divide and Conquer

  2. Basic Algorithm Design Techniques • Divide and conquer • Dynamic Programming • Greedy • Common Theme: To solve a large, complicated problem, break it into many smaller sub-problems.

  3. Divide and Conquer • Idea: Break the problem into several unrelated sub-problems, then combine the solutions to solve the original problem. • Recipe: • 1. (divide) Divide the problem into sub-problems • 2. (conquer) Solve the sub-problems recursively. • 3. (merge) Combine the solutions.

  4. Example 1 Merge Sort • Goal: Sort an array of numbers in ascending order. • Input: Array a[] = {6, 2, 4, 1, 5, 3, 7, 8} • Algorithm: MergeSort(a[]) IF Length(a) < 2 THENRETURN a. (Base Case) Partition a[] evenly into two arrays b[], c[]. (Divide) b[] = MergeSort(b[]) c[] = MergeSort(c[]) (Conquer) RETURN Merge(b[], c[]). (Merge)

  5. Merge Sort: Example: {6, 2, 4, 1, 5, 3, 7, 8} {1, 2, 3, 4, 5, 6, 7, 8} {5, 3, 7, 8} {3, 5, 7, 8} {6, 2, 4, 1} {1, 2, 4, 6} {7, 8} {7, 8} {5, 3} {3, 5} {4, 1} {1, 4} {6, 2}{2, 6}

  6. Key Design Step: How to Merge • {1, 2, 4, 6} {3, 5, 7, 8} • Idea: Smallest element must be between the first elements of the two arrays. • Determine the smallest, remove it and continue. Merge(b[], c[]) a[] = empty i = 1, j = 1 WHILEi <= Length(b[]) OR j <= Length(c[]) IF b[i] < c[j] THEN a.append(b[i]); i = i+1 ELSE a.append(c[j]); j = j+1 RETURN a[] Careful when you actually write a program (i,j can be out of bound)

  7. Example 2 Counting Inversions • Goal: Given array a[], count the number of pairs (i,j) such that i < j but a[i] > a[j]. • Input: Array a[] = {6, 2, 4, 1, 5, 3, 7, 8} • Answer = 9

  8. Designing the algorithm • Recall Recipe: • Recipe: • 1. (divide) Divide the problem into sub-problems • 2. (conquer) Solve the sub-problems recursively. • 3. (merge) Combine the solutions.

  9. Failed Attempt: • Partition a[] evenly to two arrays b[], c[]. • Count Inversions recursively. • Combine results • a[] = {6, 2, 4, 1, 5, 3, 7, 8}, 9 inversionsb[] = {6, 2, 4, 1}, 5 inversionsc[] = {5, 3, 7, 8}, 1 inversion • #inversion = #inversion in b + #inversion in c + #inversion between b[], c[] Can take O(n2) time!

  10. How to count inversions more efficiently? • b[] = {6, 2, 4, 1}, 5 inversionsc[] = {5, 3, 7, 8}, 1 inversion • For each number in b, want to know how many numbers in c are smaller.6: 2, 2: 0, 4: 1, 1: 0, result = 2+0+1+0 = 3. • Much easier if c[] is sorted! • Idea: Can sort the arrays when we are counting inversions.

  11. Sort&Count Sort&Count(a[]) IF Length(a) < 2 THENRETURN (a[], 0). (Base Case) Partition a[] evenly into two arrays b[], c[]. (Divide) (b[], count_b) = Sort&Count(b[]) (c[], count_c) = Sort&Count(c[]) (Conquer) (a[], count_bc) = Merge&Count(b[], c[]). (Merge) RETURN a[], count_b+count_c+count_bc.

  12. Merge&Count • b[] = {1, 2, 4, 6} c[] = {3, 5, 7, 8} • {1, 2, 3, 4, 5, 6, 7, 8} • When 4 goes into the array, pointer in c[] is pointing at 5 (2nd element), so 4 is larger than 1 element in c[]. • When 6 goes into the array, pointer in c[] is pointing at 7 (3rd element), so 6 is larger than 2 elements in c[].

  13. Merge&Count Merge&Count(b[], c[]) a[] = empty i = 1, j = 1 count = 0 WHILEi <= Length(b[]) OR j <= Length(c[]) IF b[i] < c[j] THEN a.append(b[i]); i = i+1 count = count + (j - 1) ELSE a.append(c[j]); j = j+1 RETURN a[], count The i-th element of a[] is smaller than c[j], but larger than c[1,2,…, j-1]

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