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SIMPLE Sorting

Learn about different sorting algorithms and how they can be used to order elements in an array. This includes internal sorting methods like selection sort, bubble sort, and insertion sort, as well as external sorting methods for large data sets. See examples and code implementations in C++.

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SIMPLE Sorting

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  1. SIMPLE Sorting • Sorting is a typical operation to put the elements in an array in order. • Internal Sorts [for small data sets] selection bubble (exchange) • External Sorts [for large data sets]

  2. Find smallest element, and put at the head of the list,repeat with remainder of list 21 13 9 15 17 9 13 21 15 17 9 13 21 15 17 9 13 15 21 17 9 13 15 17 21 Selection Sort index (k) sm_index 0 2 swap 21, 9 1 1 swap 13, 13 2 3 swap 21, 15 3 4 swap 21, 17

  3. Selection Sort void sort(double [5]); void swap(double [5], int, int); // prototypes void main(void) { int index; double my_list[ ] = {21, 13, 9, 15, 17}; sort(my_list); // function call cout<<"\nThe sorted array is: \n"; for(index=0; index<5; index++) cout<<'\t'<<my_list[index]<<endl; }

  4. Selection Sort void sort(double testArray[5]) { int n, k, sm_index, moves=0; double smallest; for(k=0; k<4; k++) // size-1 = number of passes { smallest=testArray[k]; sm_index=k; for(n=k+1; n<5; n++) // size = # elem. to look at if(testArray[n]<smallest) { smallest=testArray[n]; sm_index=n; } swap(testArray, sm_index, k); // call to swap() } }

  5. 21 13 9 25 17 13 21 9 25 17 13 9 21 25 17 13 9 21 25 17 Bubble Sort • Put smaller first • Put smaller first • No change • Put smaller first

  6. 13 9 21 17 25 9 13 21 17 25 9 13 21 17 25 9 13 17 21 25 Bubble Sort • Begin again and put smaller first • No change • Put smaller first

  7. A Bubble Sort Function void bubble_sort(int array[ ], int length) { int j, k, flag=1, temp; for(j=1; j<=length && flag; j++) { flag=0; // false for(k=0; k < (length-j); k++) { if (array[k+1] > array[k]) // > low to high { temp=array[k+1]; // swap array[k+1]= array[k]; array[k]=temp; flag=1; // indicates a swap } } } }// has occurred

  8. Insertion sort: Pick up the cards one at a time. When a card is picked up put it in the “already picked up list” in its correct position. *

  9. G D Z F B E 0 1 2 3 4 5 Index number G D Z F B E 0 1 2 3 4 5 D G Z F B E 0 1 2 3 4 5 D GZ F B E 0 1 2 3 4 5 D F G Z B E 0 1 2 3 4 5 B D F G Z E 0 1 2 3 4 5 B D E F G Z 0 1 2 3 4 5

  10. InsertionSort( int a[], int n) { int I; loc; temp; for (I = 1; I < n; I++) { temp = a[I]; loc = I; while (loc && (a[loc-1] > temp)) { a[loc] = a[loc-1]; -- loc; } a[loc] = temp; } GET NEXT ELEMENT TO INSERT Find its place in list -- keep moving items down until the correct locations is found

  11. O(n) Good for mostly sorted lists Average case = O(n2) Insertion Sort Analysis: Worst case: O(n2) 1+2+3+4+5….comparisons and moves for (I = 1; I < n; I++) { temp = a[I]; loc = I; while (loc && (a[loc-1] > temp)) { a[loc] = a[loc-1]; -- loc; } a[loc] = temp; BEST CASE? *

  12. Using Pointers for Sorting In the algorithms looked at the data is physically re-arranged

  13. Quicksort algorithm • Split list into two “halves” one greater than the other • now split each of these two lists

  14. Elements………….. j Less than j Greater than j e s <s >s < e >e m c g w <c >c <g >g <m >m <w >w a b c d D e f F g h i j k l m o q s u v w y z

  15. Partition • Divide the list into two halves: left and right where the left half is smaller than the right: • ----> use a “pivot” elements less than the pivot go left, elements greater than the pivot go right

  16. left right 98 32 45 99 101 73 67 Pivot = 98 Invariant: elements to the left of “left” are smaller, and to the right of “right” are bigger.

  17. left right 98 32 45 99 101 73 67 Pivot = 98 left left left right 67 32 45 99 101 73 67 Invariant: elements to the left of “left” are smaller, and to the right of “right” are bigger.

  18. left right 67 32 45 99 101 73 99 left right 98 32 45 99 101 73 67 Pivot = 98 left right 67 32 45 99 101 73 67 Invariant: elements to the left of “left” are smaller, and to the right of “right” are bigger.

  19. left right 67 32 45 99 101 73 99 left right 67 32 45 73 101 73 99 right left 67 32 45 73 98 101 99 left right 98 32 45 99 101 73 67 Pivot = 98 left right 67 32 45 99 101 73 67 Invariant: elements to the left of “left” are smaller, and to the right of “right” are bigger.

  20. int partition(int num[], int left, int right) {int pivot_val, temp; Pivot_val = num[left]; // "capture" the pivot value, which frees up one slot while (left < right){ // scan from right to left while(num[right] >= pivot_val && left < right) // skip over larger or equal val right--; if (right != left) { num[left] = num[right]; // move the higher value left++; } // scan from left to right while (num[left] <= pivot_val && left < right) // skip over smaller or equal val left++; if (right != left) { num[right] = num[left]; // move lower value into the available slot right--; } } num[left] = pivot_val; // move pivot into correct position return left; } // return the pivot index

  21. #include <iostream> using namespace std; int partition(int [], int, int); // function prototype int main() { const int NUMEL = 7; int nums[NUMEL] = {98,32,45,99,101,73,67}; int i, pivot; pivot = partition(nums, 0, NUMEL-1); cout << "\nThe returned pivot index is " << pivot; cout << "\nThe list is now in the order:\n"; for (i = 0; i < NUMEL; i++) cout << " " <<nums[i]; cout << endl; return 0; } The pivot index is 4 the list is now in the order: 67 32 45 73 98 101 99 *

  22. Quicksort algorithm must call the partition algorithm until the “list is sorted”, I.e. on each half recursively until the “halves” are too small Quicksort algorithm: pick pivot & partition list call quicksort(left half) call quicksort (right half)

  23. Elements………….. Quicksort(left) j Quicksort(right) Less than j Greater than j Q(left/left) e Q(left/rgt) Q(rgt/left) s Q(rgt/rgt) <s >s < e >e m c g w <c >c <g >g <m >m <w >w a b c d D e f F g h i j k l m o q s u v w y z 14 calls to Quicksort

  24. left right 67 32 45 99 101 73 99 left right 98 32 45 99 101 73 67 Pivot = 98 left left left right 67 32 45 99 101 73 67 left right 67 32 45 73 101 73 99 After the partition with the pivot=98 98 is in its final position right left 67 32 45 73 98 101 99

  25. 67 32 45 73 98 101 99 Result of 1st partition 2nd and 3rd partition 4th partition 45 32 67 73 98 99 101 32 45

  26. void quicksort(int num[], int lower, int upper) { int i, j, pivot; pivot = partition(num,lower, upper); if (lower < pivot) quicksort(num, lower, pivot - 1); if (upper > pivot) quicksort(num, pivot + 1, upper); return; }

  27. #include <iostream> using namespace std; void quicksort(int [], int, int); // function prototypes int partition(int [], int, int); int main() { const int NUMEL = 7; int nums[NUMEL] = {67,32,45,73,98,101,99}; int i; quicksort(nums, 0, NUMEL-1); cout << "\nThe sorted list, in ascending order, is:\n"; for (i = 0; i < NUMEL; i++) cout << " " <<nums[i]; cout << endl; return 0; }

  28. Quicksort(nums, 0, 6) pivot (after partition = 2) 67 32 45 73 98 101 99 45 32 67 73 98 101 99 Quicksort(nums, 0, 1) pivot (after partition = 1) 45 32 32 45 Quicksort(nums, 0, 0) pivot (after partition = 0) 73 98 101 99 32 Quicksort(nums, 3,6) pivot (after partition = 3) Quicksort(nums, 4,6) pivot (after partition = 4) 98 101 99 101 99 99 101 Quicksort(nums, 5,6) pivot (after partition = 6) Quicksort(nums, 5,5) pivot (after partition = 5) 99

  29. IN CLASS Assignment: Given the list of numbers below, write out the steps for the quicksort algorithm: 12 14 3 6 56 2 10 25 89 8

  30. 12 14 3 6 56 2 10 25 89 8 Piv=12 8 10 3 6 2 12 56 25 89 14 Piv=8 Piv=56 2 6 3 8 10 12 14 25 56 89 Piv=2 Piv=14 2 6 3 81012 14 25 5689 Piv=6 2 3 6 81012 14255689

  31. Analysis of Quicksort: void quicksort(int num[], int lower, int upper) { int i, j, pivot; pivot = partition(num,lower, upper); if (lower < pivot) quicksort(num, lower, pivot - 1); if (upper > pivot) quicksort(num, pivot + 1, upper); Takes O(upper-lower) since every element must be compared to the pivot

  32. Best case time: Time(Quicksort(n)) = Partition(n) + Quicksort(n/2)+Quicksort(n/2) O(n) Partition(n/2) + Quicksort(n/4) + Quicksort(n/4) + Partition(n/2) + Quicksort(n/4) + Quicksort(n/4) = O(n) + 2*O(n/2) + 4*(Quicksort(n/4) … = O(n) + O(n) + O(n) Log(n) times….. = O(nlogn)

  33. Worst case time: 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 N times = O(n2)

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