Objectives. Chapter 23 Sorting. Why study sorting? What data to sort? Insertion Sort. CS1: Java Programming Colorado State University

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Chapter 3 Sorting Objectives To study and analyze time complexity of various sorting algorithms ( 3. 3.7). To design, implement, and analyze insertion sort ( 3.). To design, implement, and analyze bubble sort ( 3.3). To design, implement, and analyze sort ( 3.4). CS1: Java Programming Colorado State University Original slides by Daniel Liang Modified slides by Chris Wilcox 1 Why study sorting? Sorting is a classic subject in computer science. There are three reasons for studying sorting algorithms. First, sorting algorithms illustrate many creative approaches to problem solving and these approaches can be applied to solve other problems. Second, sorting algorithms are good for practicing fundamental programming techniques using selection statements, loops, methods, and arrays. Third, sorting algorithms are excellent examples to demonstrate algorithm performance. What data to sort? The data to be sorted might be integers, doubles, characters, or objects. 7.8, Sorting Arrays, presented selection sort and insertion sort for numeric values. The selection sort algorithm was extended to sort an array of objects in 11.5.7, Example: Sorting an Array of Objects. The Java API contains several overloaded sort methods for sorting primitive type values and objects in the java.util.arrays and java.util.collections class. For simplicity, this section assumes: data to be sorted are integers, data are sorted in ascending order, and data are stored in an array. The programs can be easily modified to sort other types of data, to sort in descending order, or to sort data in an ArrayList or a LinkedList. 3 4 The insertion sort algorithm sorts a list of values by repeatedly inserting an unsorted element into a sorted sublist until the whole list is sorted. Insertion Sort int[] mylist = {, 9, 5, 4, 8, 1, 6; // Unsorted Step 1: Initially, the sorted sublist contains the 9 5 4 8 1 6 first element in the list. Insert 9 into the sublist. Step: The sorted sublist is {, 9. Insert 5 into 9 5 4 8 1 6 the sublist. Step 3: The sorted sublist is {, 5, 9. Insert 4 5 9 4 8 1 6 into the sublist. Step 4: The sorted sublist is {, 4, 5, 9. Insert 8 4 5 9 8 1 6 into the sublist. animation Insertion Sort Animation http://www.cs.armstrong.edu/liang/animation/web/insertio nsort.html Step 5: The sorted sublist is {, 4, 5, 8, 9. Insert 1 into the sublist. 4 5 8 9 1 6 Step 6: The sorted sublist is {1,, 4, 5, 8, 9. Insert 6 into the sublist. 1 4 5 8 9 6 Step 7: The entire list is now sorted. 1 4 5 6 8 9 5 6

animation Insertion Sort How to Insert? int[] mylist = {, 9, 5, 4, 8, 1, 6; // Unsorted 9 5 4 8 1 6 9 5 4 8 1 6 5 9 4 8 1 6 4 5 9 8 1 6 4 5 8 9 1 6 1 4 5 8 9 6 1 4 5 6 8 9 The insertion sort algorithm sorts a list of values by repeatedly inserting an unsorted element into a sorted sublist until the whole list is sorted. [0] [1] [] [3] [4] [5] [6] list 5 9 4 [0] [1] [] [3] [4] [5] [6] list 5 9 [0] [1] [] [3] [4] [5] [6] list 5 9 [0] [1] [] [3] [4] [5] [6] list 4 5 9 Step 1: Save 4 to a temporary variable currentelement Step : Move list[] to list[3] Step 3: Move list[1] to list[] Step 4: Assign currentelement to list[1] 7 8 From Idea to Solution From Idea to Solution for (int i = 1; i < list.length; i++) { insert list[i] into a sorted sublist list[0..i] so that list[0..i] is sorted for (int i = 1; i < list.length; i++) { insert list[i] into a sorted sublist list[0..i] so that list[0..i] is sorted list[0] Expand list[0] list[1] list[0] list[1] list[] list[0] list[1] list[] list[3] list[0] list[1] list[] list[3]... double currentelement = list[i]; int k; for (k = i - 1; k >= 0 && list[k] > currentelement; k--) { list[k + 1] = list[k]; // Insert the current element into list[k + 1] list[k + 1] = currentelement; InsertSort 9 10 9 5 4 8 1 5 9 4 8 1 5 4 9 8 1 5 4 8 9 1 5 4 8 1 9 4 5 8 1 9 4 5 8 1 9 4 5 1 8 9 Bubble Sort 5 4 8 1 9 4 5 1 8 9 4 1 5 8 9 1 4 5 8 9 4 5 1 8 9 1 4 5 8 9 4 1 5 8 9 Bubble Sort Animation http://www.cs.armstrong.edu/liang/animation/web/bubblesort.html (a) 1st pass (b) nd pass (c) 3rd pass (d) 4th pass (e) 5th pass Bubble sort time: O(n ) ( n ) + ( n - ) +... + + 1 = - BubbleSort 11 1

Merge Sort Merge Sort 9 5 4 8 1 6 7 9 5 4 8 1 6 7 9 5 4 8 1 6 7 9 5 4 8 1 6 7 divide Sort(list): firsthalf = Sort(firstHalf); secondhalf = Sort(secondHalf); list = (firsthalf, secondhalf); 9 4 5 1 8 6 7 4 5 9 conquer 1 4 5 6 7 8 9 MergeSort 13 14 Merge Two Sorted Lists Merge Sort Time 4 5 9 1 current (a) After moving 1 to temp 4 5 9 current 1 4 5 6 7 8 (b) After moving all the elements in list to temp to temp 4 5 9 current 1 4 5 6 7 8 9 (c) After moving 9 to temp Let T(n) denote the time required for sorting an array of n elements using sort. Without loss of generality, assume n is a power of. The sort algorithm s the array into two subarrays, sorts the subarrays using the same algorithm recursively, and then s the subarrays. So, = T( ) + T( ) + time Animation for Merging Two Sorted Lists = T( ) + T( ) + O( n) 15 16 Merge Sort Time The first T(n/) is the time for sorting the first half of the array and the second T(n/) is the time for sorting the second half. To two subarrays, it takes at most n comparisons to compare the elements from the two subarrays and n moves to move elements to the temporary array. So, the total time is n. Therefore, n n = T ( ) + n = (T ( ) + ) + n = T ( ) + n - + n 4 k n k = T ( ) + n - +... + n - + n k n = T ( ) + n - +... + n - + n = n + n - + 1 = n + 1 = O( n ) Quick Sort Quick sort, developed by C. A. R. Hoare (196), works as follows: The algorithm selects an element, called the, in the array. Divide the array into two parts such that all the elements in the first part are less than or equal to the and all the elements in the second part are greater than the. Recursively apply the quick sort algorithm to the first part and then the second part. 17 18

5 9 3 8 4 0 1 6 7 4 1 3 0 5 8 9 6 7 0 1 3 4 Quick Sort (a) The original array (b)the original array is (c) The partial array (4 1 3 0) is 0 1 3 (d) The partial array (0 1 3) is Partition 5 9 3 8 4 0 1 6 7 Animation for partition 5 9 3 8 4 0 1 6 7 5 1 3 8 4 0 9 6 7 5 1 3 8 4 0 9 6 7 5 1 3 0 4 8 9 6 7 5 1 3 0 4 8 9 6 7 (a) Initialize, low, and high (b) Search forward and backward (c) 9 is swapped with 1 (d) Continue search (e) 8 is swapped with 0 (f) when high < low, search is over 1 3 (e) The partial array ( 1 3) is QuickSort 4 1 3 0 5 8 9 6 7 The index of the is returned (g) is in the right place 19 0 Quick Sort Time To partition an array of n elements, it takes n comparisons and n moves in the worst case. So, the time required for partition is O(n). Worst-Case Time In the worst case, each time the divides the array into one big subarray with the other empty. The size of the big subarray is one less than the one before divided. The algorithm requires time: O( n ) ( n ) + ( n - ) +... + + 1= O( n ) 1 Best-Case Time Average-Case Time In the best case, each time the divides the array into two parts of about the same size. Let T(n) denote the time required for sorting an array of elements using quick sort. So, = T ( ) + T ( ) + n = O( n) On the average, each time the will not divide the array into two parts of the same size nor one empty part. Statistically, the sizes of the two parts are very close. So the average time is O(n). The exact average-case analysis is beyond the scope of this book. 3 4

Computational Complexity (Big O) T(n)=O(1) T(n)=O(log n) T(n)=O(n) T(n)=O(nlog n) T(n)=O(n ) T(n)=O(n 3 ) // constant time // logarithmic // linear // linearithmic // quadratic // cubic Complexity Examples http://bigocheatsheet.com/ 5 6 Complexity Examples Why does it matter? http://bigocheatsheet.com/ 7 Algorithm 10 0 50 100 1,000 10,000 100,000 O(1) <1 s <1 s <1 s <1 s <1 s <1 s <1 s O(log(n)) <1 s <1 s <1 s <1 s <1 s <1 s <1 s O(n) <1 s <1 s <1 s <1 s <1 s <1 s <1 s O(n*log(n)) <1 s <1 s <1 s <1 s <1 s <1 s <1 s O(n ) <1 s <1 s <1 s <1 s <1 s s 3 m O(n 3 ) <1 s <1 s <1 s <1 s 0 s 6 h 3 d O( n ) <1 s <1 s 60 d O(n!) <1 s O( ) 3 m 8