why study sorting? Sorting is a classic subject in computer science. There are three reasons for studying sorting algorithms.
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1 Chapter 5 Sortig IST311 - CIS65/506 Clevelad State Uiversity Prof. Victor Matos Adapted from: Itroductio to Java Programmig: Comprehesive Versio, Eighth Editio by Y. Daiel Liag
2 why study sortig? Sortig is a classic subject i computer sciece. There are three reasos for studyig sortig algorithms. First, sortig algorithms illustrate may creative approaches to problem solvig ad these approaches ca be applied to solve other problems. Secod, sortig algorithms are good for practicig fudametal programmig techiques usig selectio statemets, loops, methods, ad arrays. Third, sortig algorithms are excellet examples to demostrate algorithm performace.
3 what data to sort? For simplicity, this sectio assumes: 1. data to be sorted are itegers,. data are sorted i ascedig order, ad 3. data are stored i a array. The programs ca be easily modified to sort other types of data, to sort i descedig order, or to sort data i a ArrayList or a LikedList. 3
4 Bubble Sort 1. The bubble sort algorithm makes several passes through the array.. O each pass, successive eighborig pairs are compared. list[i] list[i+1] 3. If a pair is i decreasig order, its values are swapped; otherwise, the values remai uchaged. 4. The techique is called a bubble sort or sikig sort, because the smaller values gradually bubble their way to the top ad the larger values sik to the bottom. 4
5 Bubble Sort public class BubbleSort { /** Bubble sort method */ public static void bubblesort(it[] list) { boolea eednextpass = true; for (it k = 1; k < list.legth && eednextpass; k++) { // Array may be sorted ad ext pass ot eeded eednextpass = false; for (it i = 0; i < list.legth - k; i++) { if (list[i] > list[i + 1]) { // Swap list[i] with list[i + 1] it temp = list[i]; list[i] = list[i + 1]; list[i + 1] = temp; } } } } } eednextpass = true; // Next pass still eeded 5
6 Bubble Sort K elemets are i their correct positio. temp k list[0] list[1] list[i] list[i+1] list[-] list[-1] if ( list[i] > list[i+1] ) swap(list, i, i+1) 6
7 Bubble Sort (a) 1st pass (b) d pass (c) 3rd pass (d) 4th pass (e) 5th pass Bubble sort time: O( ) T ( ) ( 1) ( )
8 Merge Sort Algorithm The merge sort algorithm ca be described recursively as follows: 1. The algorithm divides the array ito two halves ad applies merge sort o each half recursively.. After the two halves are sorted, merge them. 8 8
9 Merge Sort Algorithm (1 of ) public class MergeSort { /** Divide & Coquer method for sortig the umbers i give array */ public static void mergesort(it[] list) { if (list.legth > 1) { // Merge sort the first half it[] firsthalf = ew it[list.legth / ]; System.arraycopy(list, 0, firsthalf, 0, list.legth / ); mergesort(firsthalf); // Merge sort the secod half it secodhalflegth = list.legth - list.legth / ; it[] secodhalf = ew it[secodhalflegth]; System.arraycopy(list, list.legth /, secodhalf, 0, secodhalflegth); mergesort(secodhalf); } } // Merge firsthalf with secodhalf it[] temp = merge(firsthalf, secodhalf); System.arraycopy(temp, 0, list, 0, temp.legth); 9 9
10 Merge Sort Algorithm ( of ) 10 } /** Merge two sorted lists */ private static it[] merge(it[] list1, it[] list) { it[] temp = ew it[list1.legth + list.legth]; it curret1 = 0; // Curret idex i list1 it curret = 0; // Curret idex i list it curret3 = 0; // Curret idex i temp while (curret1 < list1.legth && curret < list.legth) { if (list1[curret1] < list[curret]) temp[curret3++] = list1[curret1++]; else temp[curret3++] = list[curret++]; } while (curret1 < list1.legth) temp[curret3++] = list1[curret1++]; while (curret < list.legth) temp[curret3++] = list[curret++]; retur temp; }//merge 10
11 Merge Sort split split split merge merge divide coquer merge
12 Merge Two Sorted Lists curret1 curret curret1 curret curret1 curret curret3 (a) After movig 1 to temp curret3 (b) After movig all the elemets i list to temp curret3 (c) After movig 9 to temp 1
13 Merge Sort Time Let T() deote the time required for sortig a array of elemets usig merge sort. Without loss of geerality, assume is a power of. The merge sort algorithm splits the array ito two subarrays, sorts the subarrays usig the same algorithm recursively, ad the merges the subarrays. So, T( ) T( ) T( ) mergetime 13
14 Merge Sort Time T( ) T( ) T( ) mergetime The first T(/) is the time for sortig the first half of the array ad the secod T(/) is the time for sortig the secod half. To merge two subarrays, it takes (a) at most -1 comparisos to compare the elemets from the two subarrays, ad (b) (b) moves to trasfer elemets to the temporary array. So, the total time is -1. Therefore, 14
15 Merge Sort Time 15 ) log ( 1 log 1 )log ( 1) ( )log ( ) ( log, 1) ( ) ( ) ( 1) ( ) 3( ) ( ) ( 3 4 1) ) ( ( 3 ) ( ) ( 1 ) ( 1 1) ) 4 ( ( 1 ) ( ) ( log log log log O T k assume k T T T T T T T T T k k k k k k
16 Quick Sort QuickSort works as follows: 1. The algorithm selects a elemet, called the pivot, i the array.. Divide the array ito two parts such that all the elemets i the first part are less tha or equal to the pivot ad all the elemets i the secod part are greater tha the pivot. 3. Recursively apply the quick sort algorithm to the first part ad the the secod part. pivot < Pivot > pivot 16 First part Secod part
17 Quick Sort public class QuickSort { // helper method public static void quicksort(it[] list) { quicksort(list, 0, list.legth - 1); } 1 of private static void quicksort(it[] list, it first, it last) { if (last > first) { it pivotidex = partitio(list, first, last); quicksort(list, first, pivotidex - 1); } quicksort(list, pivotidex + 1, last); } 17
18 Quick Sort /** Partitio the array list[first..last] */ private static it partitio(it[] list, it first, it last) { it pivot = list[first]; // Choose the first elemet as the pivot it low = first + 1; // Idex for forward search it high = last; // Idex for backward search while (high > low) { // Search forward from left while (low <= high && list[low] <= pivot) low++; // Search backward from right while (low <= high && list[high] > pivot) high--; } // Swap two elemets i the list if (high > low) { it temp = list[high]; list[high] = list[low]; list[low] = temp; } while (high > first && list[high] >= pivot) high--; 18 } } // Swap pivot with list[high] if (pivot > list[high]) { list[first] = list[high]; list[high] = pivot; retur high; } else { retur first; }
19 Quick Sort pivot (a) The origial array pivot pivot (b)the origial array is partitioed pivot pivot (c) The partial array ( ) is partitioed (d) The partial array (0 1 3) is partitioed 1 3 (e) The partial array ( 1 3) is partitioed 19
20 Quick Sort First/Pvt: 0 Low: High: 9 left First/Pvt: 0 Low: High: 7 right First/Pvt: 0 Low: High: 7 swap First/Pvt: 0 Low: 4 High: 7 left First/Pvt: 0 Low: 4 High: 6 right First/Pvt: 0 Low: 4 High: 6 swap First/Pvt: 0 Low: 6 High: 6 left First/Pvt: 0 Low: 6 High: 5 adjust First/Pvt: 0 Low: 6 High: 5 swap pivot PIVOT= 5 First/Pvt: 0 Low: 5 High: 4 left First/Pvt: 0 Low: 5 High: 4 right First/Pvt: 0 Low: 5 High: 4 swap pivot PIVOT= 4 First/Pvt: 0 Low: 1 High: 3 left First/Pvt: 0 Low: 1 High: 0 right PIVOT= 0 First/Pvt: 1 Low: 3 High: 3 left First/Pvt: 1 Low: 3 High: right First/Pvt: 1 Low: 3 High: swap pivot PIVOT= First/Pvt: 6 Low: 7 High: 9 left First/Pvt: 6 Low: 7 High: 9 right First/Pvt: 6 Low: 7 High: 9 swap First/Pvt: 6 Low: 9 High: 9 left First/Pvt: 6 Low: 9 High: 8 right First/Pvt: 6 Low: 9 High: 8 swap pivot PIVOT= 8 First/Pvt: 6 Low: 7 High: 6 adjust PIVOT= 6 First/Pvt: 6 Low: 7 High: 6 adjust
21 Partitio pivot low high pivot low high pivot low high pivot low high pivot low high pivot low high pivot (a) Iitialize pivot, low, ad high (b) Search forward ad backward (c) 9 is swapped with 1 (d) Cotiue search (e) 8 is swapped with 0 (f) whe high < low, search is over (g) pivot is i the right place 1 The idex of the pivot is retured
22 Quick Sort Time To partitio a array of elemets, it takes -1 comparisos (pivot vs. all other) ad moves i the worst case. So, the time required for partitio is O().
23 Quick Sort: Worst-Case Time I the worst case, each time the pivot divides the array ito oe big subarray with the other empty. The size of the big subarray is oe less tha the oe before it is divided. The algorithm requires O( ) time: ( 1) ( )... 1 O( ) 3
24 Quick Sort: Best-Case Time 4 I the best case, each time the pivot divides the array ito two parts of about the same size. Let T() deote the time required for sortig a array of elemets usig quicksort. Therefore, ) log ( ) ( ) ( ) ( ) ( O T T T T
25 Quick Sort: Average-Case Time O the average, each time the pivot will ot divide the array ito two parts of the same size or oe empty part. Statistically, the sizes of the two parts are very close. So the average time is O(log). The exact average-case aalysis is beyod the scope of this book. 5
26 Heaps Heap is a useful data structure for desigig efficiet sortig algorithms ad priority queues. A heap is a biary tree with the followig properties: 1. It is a complete biary tree.. Each ode is greater tha or equal to ay of its childre. a > b a > c either (b c) or (c b) 6
27 Complete Biary Tree A biary tree is complete if every level of the tree is full except that the last level may ot be full ad all the leaves o the last level are placed i left-most mode. For example, i the figure below, the biary trees i (a) ad (b) are complete, but the biary trees i (c) ad (d) are ot complete. Further, the biary tree i (a) is a heap, but the biary tree i (b) is ot a heap, because the root (39) is less tha its right child (4) (a) (b) (c) (d) 7
28 Represetig a Heap Usig a Array Node positio Paret Left Child Right Child i ( i 1) / i + 1 i + Example. The ode for elemet 39 is at positio 4, so its left child (elemet 14) is at.. 9 (*4+1), its right child (elemet 33) is at 10 (*4+), ad its paret (elemet 4) is at 1 ((4-1)/). 6 [0] [1] [] [3] [4] [5] [6] [7] [8] [9] [10][11][1][13] paret left right
29 Addig Elemets to a Heap Addig 3, 5, 1, 19, 11, ad to a heap, iitially empty (a) After addig 3 (b) After addig 5 (c) After addig (d) After addig (e) After addig (f) After addig 9
30 Addig Elemets to a Heap 1. Add ew ode to the ed of the heap. Rebuild heap as follows Let last ode be the curret ode; While (the curret ode is greater tha its paret) { Swap the curret ode with its paret; } 30
31 Rebuild the heap after addig a ew ode Addig 88 to the heap (a) Add 88 to a heap (b) After swappig 88 with (b) After swappig 88 with 31
32 Removig the Root ad Rebuild the Tree Remove the root (maximum value i the heap) Rebuild heap as follows: Move the last ode to replace the root; Let the ew root be the curret ode; While(the curret ode has childre ad the curret ode is smaller tha oe of its childre) { Swap the curret ode with the larger childre; } 3
33 Removig the Root ad Rebuild the Tree 6 Removig root 6 from the heap
34 Removig the Root ad Rebuild the Tree Move 9 to root
35 Removig the Root ad Rebuild the Tree Swap 9 with
36 Removig the Root ad Rebuild the Tree Swap 9 with
37 Removig the Root ad Rebuild the Tree Swap 9 with
38 The Heap Class Heap<E> -list: java.util.arraylist<e> +Heap() +Heap(objects: E[]) +add(ewobject: E): void +remove(): E +getsize(): it Creates a default empty heap. Creates a heap with the specified objects. Adds a ew object to the heap. Removes the root from the heap ad returs it. Returs the size of the heap. 38
39 Heap Sort public class HeapSort { /** Heap sort method */ public static <E exteds Comparable> void heapsort(e[] list) { // Create a Heap of itegers Heap<E> heap = ew Heap<E>(); // Add elemets to the heap for (it i = 0; i < list.legth; i++) heap.add(list[i]); } } // Remove root elemet from the heap for (it i = list.legth - 1; i >= 0; i--) list[i] = heap.remove(); 39 Note: Heap is ot a primitive type, you must provide your ow versio.
40 Heap Sort Time Let h deote the height for a heap of elemets. Sice a heap is a complete biary tree, the first level has 1 ode, the secod level has odes, the k th level has (k-1) odes, the (h-1)th level has (h-) odes, ad the h th level has at least oe ode ad at most (h-1) odes. Therefore, the height of a Heap is O(log ). Sice the add method traces a path from a leaf to a root, it takes at most h steps to add a ew elemet to the heap. The it takes O(log) to create a heap holdig elemets. Similar reasoig for the remove method, yieldig O(log). Therefore Heap-Sort requires O(log) time. 40
41 Bucket Sort ad Radix Sort The lower boud for geeral sortig algorithms is O(log). So, o sortig algorithms based o comparisos ca perform better tha O(log). However, if the keys are small itegers, you ca use bucket sort without havig to compare the keys. I geeral Bucket Sort complexity is the order of O(K) where K depeds o the umber of buckets ad the size of the key 41
42 Radix Sort Uses 10 buckets ad a list of iput keys. 1. The i th digit of a key idicates the host bucket.. Begi with least sigificat key digit ad place list items i correspodig buckets. 3. Remove data from buckets to recostruct list. 4. Repeat with ext least sigificat digit util all key positios have bee used. 4 The cost of sortig is T() = kc where k is the umber of digits i the largest of the key values, ad c is the card s processig time (read, compare, move to pocket). Therefore the complexity of Radix Sort is O() 4
43 Radix Sort IBM 83 Card Sortig Machie
44 Radix Sort Example: List = { 331, 454, 30, 034, 343, 045, 345, 059, 453, 345, 31, 009 } Bucket Last radix Remove 30, 331, 31, 343, 453, 454, 034, 045, 345, 059, 009 Secod radix Remove 009, 30, 331, 31, 034, 343, 045, 345, 453, 454, 059 Third radix Remove 009, 034, 045, 059, 30, 31, 331, 343, 345, 453,
45 Exteral Sort To sort data stored i a exteral file, you may first brig data to the memory, the sort it iterally. However, if the file is too large, all data i the file caot be brought to memory at oe time. 45
46 Exteral Sort - Phase I Repeatedly brig data from the file to a array, sort the array usig a iteral sortig algorithm, ad output the data from the array to a temporary file. Program Origial file Array Temporary file S 1 S S k 46
47 Phase II Merge a pair of sorted segmets (e.g., S1 with S, S3 with S4,..., ad so o) ito a larger sorted segmet ad save the ew segmet ito a ew temporary file. Cotiue the same process util oe sorted segmet results. S 1 S S 3 S 4 S 5 S 6 S 7 S 8 S 1, S merged S 3, S 4 merged S 5, S 6 merged S 7, S 8 merged S 1, S, S 3, S 4 merged S 5, S 6, S 7, S 8 merged Merge Merge Merge 47 S 1, S, S 3, S 4, S 5, S 6, S 7, S 8 merged
48 Implemetig Phase II Each merge step merges two sorted segmets to form a ew larger segmet. The ew segmet doubles the umber elemets. The umber of segmets is reduced by half after each merge step. A segmet is too large to be brought to a array i memory. To implemet a merge step, 1. Copy half umber of segmets from file f1.dat to a temporary file f.dat.. The merge the first remaiig segmet i f1.dat with the first segmet i f.dat ito a temporary file amed f3.dat. 48
49 Implemetig Phase II S 1 S S 3 S 4 S 5 S 6 S 7 S 8 f1.dat Copy to f.dat S 1 S S 3 S 4 f.dat S 1, S 5 merged S, S 6 merged S 3, S 7 merged S 4, S 8 merged f3.dat 49
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