1. Sorting divide and conquer

2. Divide-and-conquer  a recursive design technique  solve small problem directly  divide large problem into two subproblems, each approximately half the size of the original problem  solve each subproblem with a recursive call  combine the solutions of the two subproblems to obtain a solution of the larger problem

3. Divide-and-conquer sorting General algorithm:  divide the elements to be sorted into two lists of (approximately) equal size  sort each smaller list (use divide-and-conquer unless the size of the list is 1)  combine the two sorted lists into one larger sorted list

4. Divide-and-conquer sorting  Design decisions: How is list partitioned into two lists? How are two sorted lists combined? Common techniques:  Mergesort trivial partition, merge to combine  Quicksort sophisticated partition, no work to combine

5. Mergesort  divide array into first half and last half  sort each subarray with recursive call  merge together two sorted subarrays use a temporary array to do the merge

6. mergesort public static void mergesort( int[] data, int first, int n) { int n1, n2; // sizes of subarrays if (n>1) { n1 = n / 2; n2 = n – n1; mergesort(data, first, n1); mergesort(data, first+n1, n2); merge(data, first, n1, n2); }

7. merge - 1 public static void merge(int [] data, int first, int n1, int n2) { int[] temp = new int[n1+n2]; int i; int c, c1, c2 = 0; while ((c1<n1)&&(c2<n2)) { if (data[first+c1] < data[first+n1+c2]) temp[c++]=data[first+(c1++)]; else temp[c++]=data[first+n1+(c2++)]; }  }

8. merge -2 public static void merge(int [] data, int first, int n1, int n2) {  while (c1 < n1) temp[c++]=data[first+(c1++); while (c2 < n2 ) temp[c++]=data[first+n1+(c2++)]; for (i=0; i<n1+n2; i++) data[first+i] = temp[i]; }

9. Analysis of Mergesort  depth of recursion: Θ (log n)  no. of operations needed to merge at each level of recursion: Θ (n)  overall time: Θ (n log n) best, average, and worst cases

10. Advantages of Mergesort conceptually simple suited to sorting linked lists of elements because merge traverses each linked list suited to sorting external files; divides data into smaller files until can be stored in array in memory stable performance

11. sorting huge files  sort smaller subfiles

12. sorting huge files  sort smaller subfiles

13. sorting huge files  merge subfiles

14. sorting huge files  merge subfiles

15. sorting huge files  merge subfiles

16. sorting huge files  and merge subfiles into one file

17. sorting huge files: k-way merge  merge k subfiles

18. Quicksort  choose a key to be the “pivot” key design decision: how to choose pivot  divide the list into two sublists: 1) keys less than or equal to pivot 2) keys greater than pivot  sort each sublist with a recursive call  first sorted sublist, pivot, second sorted sublist already make one large sorted array

19. quicksort public static void quicksort( int[] data, int first, int n) { int n1, n2; int pivotIndex; if (n>1) { pivotIndex=partition(data, first, n); n1 = pivotIndex – first; n2 = n – n1 – 1; quicksort(data, first, n1); quicksort(data, pivotIndex+1, n2); }

20. partitioning the data set public static int partition (int[] data, int first, int n) { int pivot = data[first]; //first array element as pivot int tooBigIndex = first+1; int tooSmallIndex = first+n-1; while (tooBigIndex <= tooSmallIndex) {while ((tooBigIndex < first+n-1) && (data[tooBigIndex]< pivot)) tooBigIndex++; while (data[tooSmallIndex]> pivot) tooSmallIndex--; if (tooBigIndex < tooSmallIndex) swap(data, tooBigIndex++, tooSmallIndex--); } data[first]=data[tooSmallIndex]; data[tooSmallIndex]=pivot; return tooSmallIndex; } implies swapping elements equal to the pivot! why?

21. Choosing pivot  How does method choose pivot? - first element in subarray is pivot  When will this lead to very poor partitioning? - if data is sorted or nearly sorted.  Alternative strategy for choosing pivot? 1.middle element of subarray 2.look at three elements of the subarray, and choose the middle of the three values.

22. Pivot as median of three elements  advantages: reduces probability of worst case performance removes need for sentinel to stop looping

23. Additional improvements  instead of stopping recursion when subarray is 1 element, use 10-15 elements as stopping case, and sort small subarray without recursion (eg. insertionsort )  as above, but don’t sort each small subarray. At end, use insertionsort, which is efficient for nearly sorted arrays

24. Additional improvements  remove recursion – manage the stack directly (in an int array of indices) and always stack the larger subarray

25. non-recursive quicksort (pseudocode not optimized) public static void quicksort(int[] data, int first, int n) { int n1, n2; int pivotIndex; Stack stack; stack.push(first); stack.push(n); while (!stack.empty()) { n = stack.pop(); first = stack.pop(); pivotIndex=partition(first, n); n1 = pivotIndex – first; n2 = n – n1 – 1; if(n2 > 1) { stack.push(pivot+1); stack.push(n2);} if(n1 > 1) { stack.push(first); stack.push(n1);} }

26. non-recursive quicksort (some optimization) public static void quicksort(int[] data, int first, int n) { int n1, n2; int pivotIndex; Stack stack; stack.push(first); stack.push(n); while (!stack.empty()) { n = stack.pop(); first = stack.pop(); pivotIndex=partition(first, n); n1 = pivotIndex – first; n2 = n – n1 – 1; if(n2 > 1) { stack.push(pivot+1); stack.push(n2);} if(n1 > 1) { stack.push(first); stack.push(n1);} } inline method code replace stack object with array avoid consecutive pushing and popping sort short subarrays by insertionsort

27. Analysis of Quicksort  depth of recursion: Θ (log n) on average Θ (n) worst case  no. of operations needed to partition at each level of recursion: Θ (n)  overall time: Θ (n log n) on average Θ (n 2 ) worst case

28. Advantages of Quicksort in-place (doesn’t require temporary array) worst case time extremely unlikely usually best algorithm for large arrays

29. Quicksort tuned for robustness and speed  no internal method calls – local stack  insertion sort for small sets  swap of elements equal to pivot  pivot as median of three

30. Radix Sort sorting by digits or bits

31. Straight radix sort – sort key parts: right to left 35 29 56 26 38 32 41 21 39 46 22 21 41 22 32 35 46 26 56 38 39 29 41 21 32 22 35 56 26 46 38 29 39 29 26 22 21 39 38 35 32 46 41 56 21 22 26 29 32 35 38 39 41 46 56 tail queue head 0

32. Straight radix sort pseudocode a[n] array of integers, length n, D decimal digits q[10] queues for (p=0; p<D; p++) for (i=0; i<n; i++) q[(a[i]/10^p)%10].insert(a[i]) i=0; for (j=0; j<10; j++) while (q[j].notEmpty()) a[i] = q[j].remove() i++

33. Straight radix sort performance  O(nD) D is number of digits  n ≈ 10 D  D ≈ log 10 n  O( n log n )

34. Radix sort  needs significant extra space – queues  only known O(n log n) sorting method for humans  famous TV sorting method…

35. IBM 082 card sorter 1949

36. IBM (Hollerith) punch card

37. IBM 80 card sorter 1925

38. Herman Hollerith’s tabulator 1890