Revised 12600 Sorting Selection Sort Cmput 115 Lecture

Revised 1/26/00 Sorting - Selection Sort Cmput 115 - Lecture 10 Department of Computing Science University of Alberta ©Duane Szafron 2000 Some code in this lecture is based ©Duane Szafron 1999 on code from the book: Java Structures by Duane A. Bailey or the companion structure package

2 About This Lecture l In this lecture we will learn about a sorting algorithm called the Selection Sort. l We will study its implementation and its time and space complexity. ©Duane Szafron 1999

3 Outline l Selection Sort Algorithm l Selection Sort - Arrays l Time and Space Complexity of Selection Sort l Selection Sort - Vectors ©Duane Szafron 1999

4 The Sort Problem l Given a collection, with elements that can be compared, put the elements in increasing or decreasing order. 0 1 2 3 4 5 6 7 8 60 30 10 20 40 90 70 80 50 0 1 2 3 4 5 6 7 8 10 20 30 40 50 60 70 80 90 ©Duane Szafron 1999

5 Selection Sort Algorithm 1 l Find the index of the largest element and exchange references with the element at the last index. largest last 0 1 2 3 4 5 6 7 8 60 30 10 20 40 90 70 80 50 0 1 2 3 4 5 6 7 8 60 30 10 20 40 50 70 80 90 l Decrement the last index. ©Duane Szafron 1999

6 Selection Sort Algorithm 2 l Find the index of the largest element and exchange references with the element at the last index. 0 1 2 3 4 5 6 7 8 60 30 10 20 40 50 70 80 90 0 1 2 3 4 5 6 7 8 No exchanges 60 30 10 20 40 50 70 80 90 necessary l Decrement the last index. ©Duane Szafron 1999

7 Selection Sort Algorithm 3 l Find the index of the largest element and exchange references with the element at the last index. 0 1 2 3 4 5 6 7 8 60 30 10 20 40 50 70 80 90 0 1 2 3 4 5 6 7 8 No exchanges 60 30 10 20 40 50 70 80 90 necessary l Decrement the last index. ©Duane Szafron 1999

8 Selection Sort Algorithm 4 l Find the index of the largest element and exchange references with the element at the last index. 0 1 2 3 4 5 6 7 8 60 30 10 20 40 50 70 80 90 0 1 2 3 4 5 6 7 8 50 30 10 20 40 60 70 80 90 l Decrement the last index. ©Duane Szafron 1999

9 Selection Sort Algorithm 5 l Find the index of the largest element and exchange references with the element at the last index. 0 1 2 3 4 5 6 7 8 50 30 10 20 40 60 70 80 90 0 1 2 3 4 5 6 7 8 40 30 10 20 50 60 70 80 90 l Decrement the last index. ©Duane Szafron 1999

10 Selection Sort Algorithm 6 l Find the index of the largest element and exchange references with the element at the last index. 0 1 2 3 4 5 6 7 8 40 30 10 20 50 60 70 80 90 0 1 2 3 4 5 6 7 8 20 30 10 40 50 60 70 80 90 l Decrement the last index. ©Duane Szafron 1999

11 Selection Sort Algorithm 7 l Find the index of the largest element and exchange references with the element at the last index. 0 1 2 3 4 5 6 7 8 20 30 10 40 50 60 70 80 90 0 1 2 3 4 5 6 7 8 20 10 30 40 50 60 70 80 90 l Decrement the last index. ©Duane Szafron 1999

12 Selection Sort Algorithm 8 l Find the index of the largest element and exchange references with the element at the last index. 0 1 2 3 4 5 6 7 8 20 10 30 40 50 60 70 80 90 0 1 2 3 4 5 6 7 8 10 20 30 40 50 60 70 80 90 l Decrement the last index. ©Duane Szafron 1999

13 Selection Sort Algorithm 9 l Stop. 0 1 2 3 4 5 6 7 8 10 20 30 40 50 60 70 80 90 ©Duane Szafron 1999

14 Sorting Using Arrays l Often when we are sorting elements in an Array, the physical size of the array is larger than its logical size (the number of “real” elements it contains). l To accommodate this, we add the logical size of the array as an extra parameter to our sorting methods. l If we know the logical size and physical size are the same, we can omit this extra parameter. Why wouldn’t we expect them to always be the same? ©Duane Szafron 1999

15 Selection Sort Code - Arrays public static void selection. Sort(Comparable an. Array[ ], int size) { // pre: 0 <= size <= an. Array. length // post: values in an. Array are in ascending order int max. Index; //index of largest object index; //index of last unsorted element } for (index = size - 1; index > 0; index--) { max. Index = this. get. Max. Index(an. Array, index); this. swap(an. Array, index, max. Index); } ©Duane Szafron 1999 code based on Bailey pg. 82

16 Method - get. Max. Index() - Arrays public static int get. Max. Index(Comparable an. Array[], int last) { // pre: 0 <= last < an. Array. length // post: return the index of the max value in the // given Array up to the last index int max. Index; //index of largest object index; //index of inspected element max. Index = last; for (index = last - 1; index >= 0; index--) { if (an. Array[index]. compare. To(an. Array[max. Index]) > 0) max. Index = index; } } ©Duane Szafron 1999 code based on Bailey pg. 82

17 Counting Comparisons How many comparison operations are required for a selection sort of an n-element collection? l The sort method executes get. Largest in a loop for the indexes: k = n-1, n-2, … 1. l Each time get. Largest is executed for an index, k, it does a comparison in a loop for the indexes: k-1, k-2, … 0. This is k comparisons l The total number of comparisons is: (n-1) + (n-2) + … + 1= (1 + 2 + … n) - 1 = l n(n + 1) - 1 = O(n 2). 2 ©Duane Szafron 1999

18 Counting Assignments l How many assignment operations are required for a selection sort of an n-element collection? l The only time we do an assignment is in a reference exchange which takes three assignments. l The sort method executes swap() in a loop for the indexes: k = n-1, n-2, … 1. l The total number of assignments is: 3*(n-1) = O(n). ©Duane Szafron 1999

19 Time Complexity of Selection Sort l The number of comparisons and assignments is independent of the data (the initial order of elements doesn’t matter). l Therefore, the best, best average and worst case time complexities of the Selection Sort are all O(n 2). ©Duane Szafron 1999

20 Space Complexity of Selection Sort l Besides the collection itself, the only extra storage for this sort is the single temp reference used in the swap method. l Therefore, the space complexity of Selection Sort is O(n). ©Duane Szafron 1999

21 Sorting Using Vectors l When we are sorting elements in a Vector, the Vector knows both its logical and physical size so we don’t need any extra parameters in the sort method. l However, we must access the elements in a Vector using the message element. At() and we must cast the elements as we access them. l We must modify elements in a Vector using the message set. Element. At(). l Using access methods adds execution time. ©Duane Szafron 1999

22 Selection Sort Code - Vectors public static void selection. Sort(Vector a. Vector) { // post: values in a. Vector are in ascending order int max. Index; //index of largest object index; //index of last unsorted element } for (index = a. Vector. size()-1; index > 0; index--) { max. Index = this. get. Max. Index(a. Vector, index); this. swap(a. Vector, index, max. Index); } ©Duane Szafron 1999

23 Method - get. Max. Index() - Vectors public static int get. Max. Index(Vector a. Vector, int last) { // pre: 0 <= last < a. Vector. size // post: return the index of the max value in the // given Vector up to the last index int max. Index; //index of largest object index; //index of inspected element max. Index = last; for (index = last - 1; index >= 0; index--) { if (((Comparable) a. Vector. element. At(index)). compare. To( (Comparable) a. Vector. element. At(max. Index)) > 0) max. Index = index; } } ©Duane Szafron 1999