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Data Structures in Java Lecture 15: Sorting II 11/11/2015 Daniel - PowerPoint PPT Presentation

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Data Structures in Java Lecture 15: Sorting II 11/11/2015 Daniel Bauer 1 Quick Sort Another divide-and-conquer algorithm. Pick any pivot element v. Partition the array into elements x v and x v. Recursively sort the

  1. Partitioning the Array ¡ ¡ ¡ ¡public ¡static ¡void ¡quicksort(Integer[] ¡a, ¡int ¡left, ¡int ¡right) ¡{ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡if ¡(right ¡> ¡left) ¡{ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡int ¡v ¡= ¡find_pivot_index(a, ¡left, ¡right); ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡int ¡i ¡= ¡0; ¡ ¡ ¡int ¡j ¡= ¡right-­‑1; ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ // ¡move ¡pivot ¡to ¡the ¡end ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡Integer ¡tmp ¡= ¡a[v]; ¡a[v] ¡= ¡a[right]; ¡a[right] ¡= ¡tmp; ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡while ¡(true) ¡{ ¡ // ¡partition ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡while ¡(a[++i] ¡< ¡v) ¡{}; ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡while ¡(a[++j] ¡> ¡v) ¡{}; ¡ O(N) ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡if ¡(i ¡>= ¡j) ¡break; ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡tmp ¡= ¡a[i]; ¡a[i] ¡= ¡a[j]; ¡a[j] ¡= ¡tmp; ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡} ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ // ¡move ¡pivot ¡back ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡tmp ¡= ¡a[i]; ¡a[i] ¡= ¡a[right]; ¡a[right] ¡= ¡tmp; ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ //recursively ¡sort ¡both ¡partitions ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡quicksort(a,left, ¡i-­‑1); ¡ ¡quicksort(a,i+1, ¡right); ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡} ¡ ¡ ¡ ¡ ¡} 12

  2. Quick Sort: Worst Case • Running time depends on the how the pivot partitions the array. • Worst case: Pivot is always the smallest or largest element. One of the partitions is empty! 34 8 64 2 51 32 21 1 . 
 . 
 13 .

  3. Quick Sort: Worst Case • Running time depends on the how the pivot partitions the array. • Worst case: Pivot is always the smallest or largest element. One of the partitions is empty! 34 8 64 2 51 32 21 1 1 34 8 64 2 51 21 32 . 
 . 
 13 .

  4. Quick Sort: Worst Case • Running time depends on the how the pivot partitions the array. • Worst case: Pivot is always the smallest or largest element. One of the partitions is empty! 34 8 64 2 51 32 21 1 1 34 8 64 2 51 21 32 2 34 8 64 51 32 21 . 
 . 
 13 .

  5. Quick Sort: Worst Case 34 8 64 2 51 32 21 1 1 34 8 64 2 51 32 21 8 2 34 64 51 32 21 ⁞ 51 64 51 64 T(1) = 1 14

  6. Quick Sort: Worst Case 34 8 64 2 51 32 21 1 1 34 8 64 2 51 32 21 8 2 34 64 51 32 21 ⁞ 51 64 T(2) = T(1) + 2 51 64 T(1) = 1 Time for partitioning 14

  7. Quick Sort: Worst Case 34 8 64 2 51 32 21 1 1 34 8 64 2 51 32 21 8 2 34 64 51 32 21 T(N-2) = T(N-3) + (N-2) ⁞ 51 64 T(2) = T(1) + 2 51 64 T(1) = 1 Time for partitioning 14

  8. Quick Sort: Worst Case 34 8 64 2 51 32 21 1 T(N-1) = T(N-2) + (N-1) 1 34 8 64 2 51 32 21 8 2 34 64 51 32 21 T(N-2) = T(N-3) + (N-2) ⁞ 51 64 T(2) = T(1) + 2 51 64 T(1) = 1 Time for partitioning 14

  9. Quick Sort: Worst Case 34 8 64 2 51 32 21 1 T(N) = T(N-1) + N T(N-1) = T(N-2) + (N-1) 1 34 8 64 2 51 32 21 8 2 34 64 51 32 21 T(N-2) = T(N-3) + (N-2) ⁞ 51 64 T(2) = T(1) + 2 51 64 T(1) = 1 Time for partitioning 14

  10. Quick Sort: Worst Case 15

  11. Quick Sort: Worst Case 15

  12. Quick Sort: Worst Case 15

  13. Quick Sort: Worst Case 15

  14. Quick Sort: Worst Case 15

  15. Quick Sort: Best Case • Best case: Pivot is always the median element. 
 Both partitions have about the same size. 34 8 64 2 51 32 21 1 16

  16. Quick Sort: Best Case • Best case: Pivot is always the median element. 
 Both partitions have about the same size. 34 8 64 2 51 32 21 1 8 2 1 21 34 64 51 32 16

  17. Quick Sort: Best Case • Best case: Pivot is always the median element. 
 Both partitions have about the same size. 34 8 64 2 51 32 21 1 8 2 1 21 34 64 51 32 1 2 8 16

  18. Quick Sort: Best Case • Best case: Pivot is always the median element. 
 Both partitions have about the same size. 34 8 64 2 51 32 21 1 8 2 1 21 34 64 51 32 1 2 8 34 32 51 64 16

  19. Quick Sort: Best Case • Best case: Pivot is always the median element. 
 Both partitions have about the same size. 34 8 64 2 51 32 21 1 T(N) = 2 T(N/2) +N 8 2 1 21 34 64 51 32 1 2 8 34 32 51 64 (we ignore the pivot element, so this overestimates the running time slightly) 17

  20. Quick Sort: Best Case • Best case: Pivot is always the median element. 
 Both partitions have about the same size. 34 8 64 2 51 32 21 1 T(N) = 2 T(N/2) +N 8 2 1 21 34 64 51 32 T(N/2) = 2 T(N/4) +N/2 1 2 8 34 32 51 64 (we ignore the pivot element, so this overestimates the running time slightly) 17

  21. Quick Sort: Best Case • Best case: Pivot is always the median element. 
 Both partitions have about the same size. 34 8 64 2 51 32 21 1 T(N) = 2 T(N/2) +N 8 2 1 21 34 64 51 32 T(N/2) = 2 T(N/4) +N/2 ⁞ 1 2 8 34 32 51 64 T(1) = 1 (we ignore the pivot element, so this overestimates the running time slightly) 17

  22. Quick Sort: Best Case (note that this is the same analysis as for Merge Sort) 18

  23. Quick Sort: Best Case (note that this is the same analysis as for Merge Sort) 18

  24. Quick Sort: Best Case (note that this is the same analysis as for Merge Sort) 18

  25. Quick Sort: Best Case assume (note that this is the same analysis as for Merge Sort) 18

  26. Quick Sort: Best Case assume (note that this is the same analysis as for Merge Sort) 18

  27. Quick Sort: Best Case assume (note that this is the same analysis as for Merge Sort) 18

  28. Choosing the Pivot 19

  29. Choosing the Pivot • Ideally we want to choose the median in each partition, but we don’t know where it is! 19

  30. Choosing the Pivot • Ideally we want to choose the median in each partition, but we don’t know where it is! • Computing the pivot should be a constant time operation. 19

  31. Choosing the Pivot • Ideally we want to choose the median in each partition, but we don’t know where it is! • Computing the pivot should be a constant time operation. • Choosing the element at the beginning/end/middle is a terrible idea! 
 Better: Choose a random element. 19

  32. Choosing the Pivot • Ideally we want to choose the median in each partition, but we don’t know where it is! • Computing the pivot should be a constant time operation. • Choosing the element at the beginning/end/middle is a terrible idea! Better: Choose a random element. • Good approximation for median: “Median-of-three” 19

  33. Choosing a Pivot: 
 Median of Three Choose the median of array[0], array[n]m and array[n/2]. 34 8 64 2 51 32 21 1 20

  34. Choosing a Pivot: 
 Median of Three Choose the median of array[0], array[n]m and array[n/2]. 34 8 64 2 51 32 21 1 1 2 34 8 64 51 32 21 20

  35. Choosing a Pivot: 
 Median of Three Choose the median of array[0], array[n]m and array[n/2]. 34 8 64 2 51 32 21 1 1 2 34 8 64 51 32 21 8 32 21 34 64 51 20

  36. Choosing a Pivot: 
 Median of Three Choose the median of array[0], array[n]m and array[n/2]. 34 8 64 2 51 32 21 1 1 2 34 8 64 51 32 21 8 32 21 34 64 51 8 21 32 20

  37. Median of Three ¡ ¡ ¡ ¡public ¡static ¡int ¡find_pivot_index(Integer[] ¡a, ¡int ¡left, ¡int ¡right) ¡{ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡int ¡center ¡= ¡( ¡left ¡+ ¡right ¡) ¡/ ¡2; ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡Integer ¡tmp; ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡if ¡(a[center] ¡< ¡a[left]) ¡{ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡tmp ¡= ¡a[center]; ¡a[center] ¡= ¡a[left]; ¡a[left] ¡= ¡tmp;} ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡if ¡(a[right] ¡< ¡a[left]) ¡{ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡tmp ¡= ¡a[right]; ¡a[right] ¡= ¡a[left]; ¡a[left] ¡= ¡tmp;} ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡if ¡(a[right] ¡< ¡a[center]) ¡{ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡tmp ¡= ¡a[right]; ¡a[right] ¡= ¡a[center]; ¡a[center] ¡= ¡tmp;} ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡return ¡center; ¡ ¡ ¡ ¡ ¡} 21

  38. 
 
 Analyzing Quick Sort • Worst case running time: 
 • Best and average case (random pivot): 
 • Is QuickSort stable? 
 • Space requirement? 
 22

  39. 
 
 Analyzing Quick Sort • Worst case running time: 
 • Best and average case (random pivot): 
 • Is QuickSort stable? 
 No. Partitioning can change order of elements. 
 (but can make QuickSort stable). • Space requirement? 
 22

  40. 
 
 Analyzing Quick Sort • Worst case running time: 
 • Best and average case (random pivot): 
 • Is QuickSort stable? 
 No. Partitioning can change order of elements. 
 (but can make QuickSort stable). • Space requirement? 
 In-place O(1), but the method activation stack grows with the running time. O(N) 22

  41. Comparison-Based Sorting Algorithms T Worst T Best T Avg Space Stable? ✓ Insertion Sort ✗ Shell Sort ✗ Heap Sort ✓ Merge Sort ✗ Quick Sort gray entries: not shown in class *depends on increment sequence 23

  42. Comparison-Based Sorting Algorithms T Worst T Best T Avg worst case lower Space Stable? bound on comparison based general sorting! ✓ Insertion Sort Can we do better if we make some assumptions? ✗ Shell Sort ✗ Heap Sort ✓ Merge Sort ✗ Quick Sort gray entries: not shown in class *depends on increment sequence 23

  43. Bucket Sort • Assume we know there are M possible values. • Keep an array count of length M . • Scan through the input array A and for each i increment count [ A i ]. 
 A 1 8 2 3 2 4 6 1 count 0 0 0 0 0 0 0 0 0 0 0 1 2 3 4 5 6 7 8 9 24

  44. Bucket Sort • Assume we know there are M possible values. • Keep an array count of length M . • Scan through the input array A and for each i increment count [ A i ]. 
 A 1 8 2 3 2 4 6 1 count 0 1 0 0 0 0 0 0 0 0 0 1 2 3 4 5 6 7 8 9 25

  45. Bucket Sort • Assume we know there are M possible values. • Keep an array count of length M . • Scan through the input array A and for each i increment count [ A i ]. 
 A 1 8 2 3 2 4 6 1 count 0 1 0 0 0 0 0 0 1 0 0 1 2 3 4 5 6 7 8 9 26

  46. Bucket Sort • Assume we know there are M possible values. • Keep an array count of length M . • Scan through the input array A and for each i increment count [ A i ]. 
 A 1 8 2 3 2 4 6 1 count 0 1 1 0 0 0 0 0 1 0 0 1 2 3 4 5 6 7 8 9 27

  47. Bucket Sort • Assume we know there are M possible values. • Keep an array count of length M . • Scan through the input array A and for each i increment count [ A i ]. 
 A 1 8 2 3 2 4 6 1 count 0 1 1 1 0 0 0 0 1 0 0 1 2 3 4 5 6 7 8 9 28

  48. Bucket Sort • Assume we know there are M possible values. • Keep an array count of length M . • Scan through the input array A and for each i increment count [ A i ]. 
 A 1 8 2 3 2 4 6 1 count 0 1 2 1 0 0 0 0 1 0 0 1 2 3 4 5 6 7 8 9 29

  49. Bucket Sort • Assume we know there are M possible values. • Keep an array count of length M . • Scan through the input array A and for each i increment count [ A i ]. 
 A 1 8 2 3 2 4 6 1 count 0 1 2 1 1 0 0 0 1 0 0 1 2 3 4 5 6 7 8 9 30

  50. Bucket Sort • Assume we know there are M possible values. • Keep an array count of length M . • Scan through the input array A and for each i increment count [ A i ]. 
 A 1 8 2 3 2 4 6 1 count 0 1 2 1 1 0 1 0 1 0 0 1 2 3 4 5 6 7 8 9 31

  51. Bucket Sort • Assume we know there are M possible values. • Keep an array count of length M . • Scan through the input array A and for each i increment count [ A i ]. 
 A 1 8 2 3 2 4 6 1 count 0 2 2 1 1 0 1 0 1 0 0 1 2 3 4 5 6 7 8 9 32

  52. Bucket Sort • Assume we know there are M possible values. • Keep an array count of length M . • Scan through the input array A and for each i O(N) increment count [ A i ]. 
 A 1 8 2 3 2 4 6 1 count 0 2 2 1 1 0 1 0 1 0 0 1 2 3 4 5 6 7 8 9 32

  53. Bucket Sort • Then iterate through count.F or each i write count [ i ] copies of i to A . 
 A count 0 2 2 1 1 0 1 0 1 0 0 1 2 3 4 5 6 7 8 9 33

  54. Bucket Sort • Then iterate through count.F or each i write count [ i ] copies of i to A . 
 A 1 1 count 0 2 2 1 1 0 1 0 1 0 0 1 2 3 4 5 6 7 8 9 34

  55. Bucket Sort • Then iterate through count.F or each i write count [ i ] copies of i to A . 
 A 1 1 2 2 count 0 2 2 1 1 0 1 0 1 0 0 1 2 3 4 5 6 7 8 9 35

  56. Bucket Sort • Then iterate through count.F or each i write count [ i ] copies of i to A . 
 A 1 1 2 2 3 count 0 2 2 1 1 0 1 0 1 0 0 1 2 3 4 5 6 7 8 9 36

  57. Bucket Sort • Then iterate through count.F or each i write count [ i ] copies of i to A . 
 A 1 1 2 2 3 4 count 0 2 2 1 1 0 1 0 1 0 0 1 2 3 4 5 6 7 8 9 37

  58. Bucket Sort • Then iterate through count.F or each i write count [ i ] copies of i to A . 
 A 1 1 2 2 3 4 count 0 2 2 1 1 0 1 0 1 0 0 1 2 3 4 5 6 7 8 9 38

  59. Bucket Sort • Then iterate through count.F or each i write count [ i ] copies of i to A . 
 A 1 1 2 2 3 4 6 count 0 2 2 1 1 0 1 0 1 0 0 1 2 3 4 5 6 7 8 9 39

  60. Bucket Sort • Then iterate through count.F or each i write count [ i ] copies of i to A . 
 A 1 1 2 2 3 4 6 count 0 2 2 1 1 0 1 0 1 0 0 1 2 3 4 5 6 7 8 9 40

  61. Bucket Sort • Then iterate through count.F or each i write count [ i ] copies of i to A . 
 A 1 1 2 2 3 4 6 8 count 0 2 2 1 1 0 1 0 1 0 0 1 2 3 4 5 6 7 8 9 41

  62. Bucket Sort • Then iterate through count.F or each i write count [ i ] copies of i to A . 
 A 1 1 2 2 3 4 6 8 count 0 2 2 1 1 0 1 0 1 0 0 1 2 3 4 5 6 7 8 9 42

  63. Bucket Sort • Then iterate through count.F or each i write O(M) count [ i ] copies of i to A . 
 A 1 1 2 2 3 4 6 8 count 0 2 2 1 1 0 1 0 1 0 0 1 2 3 4 5 6 7 8 9 42

  64. Bucket Sort • Then iterate through count.F or each i write O(M) count [ i ] copies of i to A . 
 Total time for Bucket Sort:O(N +M) A 1 1 2 2 3 4 6 8 count 0 2 2 1 1 0 1 0 1 0 0 1 2 3 4 5 6 7 8 9 42

  65. Radix Sort • Generalization of Bucket sort for Large M. • Assume M contains all base b numbers up to b p -1 
 (e.g. all base-10 integers up to 10 3 ) • Do p passes over the data, using Bucket Sort for each digit. • Bucket sort is stable! 06 4 00 8 21 6 51 2 02 7 72 9 00 0 00 1 34 3 12 5 43

  66. Radix Sort 06 4 00 8 21 6 51 2 02 7 72 9 00 0 00 1 34 3 12 5 0 1 2 • Bucket sort according 3 to least significant digit. 4 064 5 6 7 8 9 44

  67. Radix Sort 06 4 00 8 21 6 51 2 02 7 72 9 00 0 00 1 34 3 12 5 0 1 2 • Bucket sort according 3 to least significant digit. 4 064 5 6 7 008 8 9 45

  68. Radix Sort 06 4 00 8 21 6 51 2 02 7 72 9 00 0 00 1 34 3 12 5 0 1 2 • Bucket sort according 3 to least significant digit. 4 064 5 216 6 7 008 8 9 46

  69. Radix Sort 06 4 00 8 21 6 51 2 02 7 72 9 00 0 00 1 34 3 12 5 0 1 512 2 • Bucket sort according 3 to least significant digit. 4 064 5 216 6 7 008 8 9 47

  70. Radix Sort 06 4 00 8 21 6 51 2 02 7 72 9 00 0 00 1 34 3 12 5 0 1 512 2 • Bucket sort according 3 to least significant digit. 4 064 5 216 6 027 7 008 8 9 48

  71. Radix Sort 06 4 00 8 21 6 51 2 02 7 72 9 00 0 00 1 34 3 12 5 0 1 512 2 • Bucket sort according 3 to least significant digit. 4 064 5 216 6 027 7 008 8 9 729 49

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