Data Structures. Chapter 04. 3/10/2016 Md. Golam Moazzam, Dept. of CSE, JU

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1 Data Structures Chapter 04 1

2 Linear Array A linear array is a list of finite number N of homogeneous data elements such that: The elements of the array are referenced respectively by an index set consisting of N consecutive numbers. The elements of the array are stored respectively in successive memory locations. The number N of elements is called the length or size of the array. 2

3 Linear Array Three numbers define an array: Lower Bound, Upper Bound, and Size. The lower bound is the smallest subscript you can use in the array (usually 0). The upper bound is the largest subscript you can use in the array. The size / length of the array refers to the number of elements in the array. It can be computed as Length= Upper bound - Lower bound + 1 Let the array name be A. Then the elements of A are: a 1, a 2,.., a n. Or by the bracket notation: A[1], A[2], A[3],., A[n] The number k in A[k] is called a subscript and A[k] is called a subscripted variable. 3

4 Linear Array Example: A linear array DATA consisting of six elements: DATA DATA [1] = 247 DATA [2] = 56 DATA [3] = 429 DATA [4] = 135 DATA [5] = 87 DATA [6] = 156 4

5 Linear Array Example: An automobile company uses an array AUTO to record the number of auto mobile sold each year from 1932 through LB = 1932, UB = 1984 AUTO[k] = Number of auto mobiles sold in the year k. Length = UB LB+1 = =55 5

6 Operations on Linear Array Traversing: means to visit all the elements of the array in an operation is called traversing. Insertion: means to put values into an array. Deletion / Remove: to delete a value from an array. Sorting: Re-arrangement of values in an array in a specific order (Ascending / Descending) is called sorting. Searching: The process of finding the location of a particular element in an array is called searching. There are two popular searching techniques/mechanisms: Linear search and binary search. 6

7 Representation of Linear Array in Memory Let LA be a linear array in the memory of the computer. The memory of the computer is a sequence of addressed locations. The computer does not need to keep track of the address of every element of LA, but needs to keep track only of the first element of LA, denoted by Base (LA) called the base address of LA. Using this address Base (LA), the computer calculates the address of any element of LA by the following formula: LOC(LA[k]) = Base(LA) + w(k lower bound) Where w is the number of words per memory cell for the array LA 7

8 Representation of Linear Array in Memory Example: An automobile company uses an array AUTO to record the number of auto mobiles sold each year from 1932 through Suppose AUTO appears in memory as pictured in Fig. That is, Base(AUTO) = 200, and w = 4 words per memory cell, LB=1932, UB=1984. Then, LOC(AUTO[1932]) = 200, LOC(AUTO[1933]) =204 LOC(AUTO[1934]) = 208 The address of the array element for the year K = 1965 can be obtained by using: LOC(AUTO[1965]) = Base(AUTO) + w(1965 LB) =200+4( ) = AUTO 8

9 Solved Problems 4.1: Consider the linear arrays AAA(5:50), BBB(-5:10) and CCC(18). (a) Find the number of elements in each array. (b) Suppose Base(AAA)=300 and w=4 words per memory cell for AAA. Find the address of AAA[15], AAA[35] and AAA[55]. Solution: (a) The number of elements is equal to the length; hence use the formula Length = UB LB+1 Accordingly, Length(AAA) = =46 Length(BBB) =10-(-5)+1=16 Length(CCC) = =18 9

10 Solved Problems 4.1: Consider the linear arrays AAA(5:50), BBB(-5:10) and CCC(18). (a) Find the number of elements in each array. (b) Suppose Base(AAA)=300 and w=4 words per memory cell for AAA. Find the address of AAA[15], AAA[35] and AAA[55]. Solution: (b) Use the formula, LOC(AAA[K]) = Base(AAA) + w(k LB) Hence, LOC (AAA[15])=300+4(15-5)=340 LOC(AAA[35])=300+4(35-5)=420 AAA[55] is not an element of AAA, since 55 exceeds UB=50 10

11 Algorithm 4.1: (Traversing a Linear Array) Here LA is a linear array with lower boundary LB and upper boundary UB. This algorithm traverses LA applying an operation Process to each element of LA. 1. [Initialize counter.] Set K=LB. 2. Repeat Steps 3 and 4 while K UB. 3. [Visit element.] Apply PROCESS to LA[K]. 4. [Increase counter.] Set k=k+1. [End of Step 2 loop.] 5. Exit. 11

12 Algorithm 4.1: (Traversing a Linear Array) Here LA is a linear array with lower boundary LB and upper boundary UB. This algorithm traverses LA applying an operation Process to each element of LA. 1. Repeat for K=LB to UB Apply PROCESS to LA[K]. [End of loop]. 2. Exit. 12

13 Insertion into Arrays Inserting an element somewhere in the middle of the array require that each subsequent element be moved downward to new locations to accommodate the new element and keep the order of the other elements. Algorithm 4.2: (Inserting into a linear array) Here LA is linear array with N elements and K is a positive integer such that K<=N. This algorithm inserts an element ITEM into the K-th position in LA. Step 1. Step 2. Step 3. Step 4. Step 5. Step 6. Step 7. Set J:=N Repeat Steps 3 and 4 while J>=K Set LA [J+1]: =LA [J] Set J:=J-1 [End of step 2 loop] Set LA [K]: =ITEM Set N:=N+1 Exit 13

14 Deletion of elements from linear arrays Deleting an element somewhere in the middle of the array requires that each subsequent element be moved one location upward in order to fill up the array. Algorithm 4.3: (Deleting from a linear array) Here LA is a linear array with N elements and K is a positive integer such that K N. This algorithm deletes the K-th element from LA. Step 1. Set ITEM: = LA [K] Step 2. Repeat for J=K to N-1 Step 3. Step 4. Set LA [J]: =LA [J+1] [End of loop] Set N:=N-1 Exit 14

15 Bubble Sort Suppose the following numbers are stored in an array A: A: 32, 51, 27, 85, 66, 23, 13, 57 Apply the bubble sort to the array A. Show each pass separately. Pass 1: (a) Compare A 1 and A 2. Since 32< 51, the list is not altered. A: 32, 51, 27, 85, 66, 23, 13, 57 (b) Compare A 2 and A 3. Since 51>27, interchange 51 and 27. A: 32, 27, 51, 85, 66, 23, 13, 57 (c) Compare A 3 and A 4. Since 51<85, the list is not altered. A: 32, 27, 51, 85, 66, 23, 13, 57 (d) Compare A 4 and A 5. Since 85>66, interchange 85 and 66. A: 32, 27, 51, 66, 85, 23, 13, 57 15

16 Bubble Sort Suppose the following numbers are stored in an array A: A: 32, 51, 27, 85, 66, 23, 13, 57 Apply the bubble sort to the array A. Show each pass separately. Pass 1: (e) Compare A 5 and A 6. Since 85>23, interchange 85 and 23. A: 32, 27, 51, 66, 23, 85, 13, 57 (f) Compare A 6 and A 7. Since 85>13, interchange 85and 13. A: 32, 27, 51, 66, 23, 13, 85, 57 (g) Compare A 7 and A 8. Since 85>57, interchange 85 and 57. A: 32, 27, 51, 66, 23, 13, 57, 85 16

17 Bubble Sort Suppose the following numbers are stored in an array A: A: 32, 51, 27, 85, 66, 23, 13, 57 Apply the bubble sort to the array A. Show each pass separately. After Pass 1: A: 32, 27, 51, 66, 23, 13, 57, 85 Pass 2: 27, 32, 51, 66, 23, 13, 57, 85 27, 32, 51, 66, 23, 13, 57, 85 27, 32, 51, 66, 23, 13, 57, 85 27, 32, 51, 23, 66, 13, 57, 85 27, 32, 51, 23, 13, 66, 57, 85 27, 32, 51, 23, 13, 57, 66, 85 17

18 Bubble Sort Suppose the following numbers are stored in an array A: A: 32, 51, 27, 85, 66, 23, 13, 57 Apply the bubble sort to the array A. Show each pass separately. After Pass 1: A: 32, 27, 51, 66, 23, 13, 57, 85 After Pass 2: A: 27, 32, 51, 23, 13, 57, 66, 85 Pass 3: 27, 32, 51, 23, 13, 57, 66, 85 27, 32, 51, 23, 13, 57, 66, 85 27, 32, 23, 51, 13, 57, 66, 85 27, 32, 23, 13, 51, 57, 66, 85 27, 32, 23, 13, 51, 57, 66, 85 18

19 Bubble Sort Suppose the following numbers are stored in an array A: A: 32, 51, 27, 85, 66, 23, 13, 57 Apply the bubble sort to the array A. Show each pass separately. After Pass 1: A: 32, 27, 51, 66, 23, 13, 57, 85 After Pass 2: A: 27, 32, 51, 23, 13, 57, 66, 85 After Pass 3: A: 27, 32, 23, 13, 51, 57, 66, 85 Pass 4: 27, 32, 23, 13, 51, 57, 66, 85 27, 23, 32, 13, 51, 57, 66, 85 27, 23, 13, 32, 51, 57, 66, 85 27, 23, 13, 32, 51, 57, 66, 85 19

20 Bubble Sort Suppose the following numbers are stored in an array A: A: 32, 51, 27, 85, 66, 23, 13, 57 Apply the bubble sort to the array A. Show each pass separately. After Pass 1: A: 32, 27, 51, 66, 23, 13, 57, 85 After Pass 2: A: 27, 32, 51, 23, 13, 57, 66, 85 After Pass 3: A: 27, 32, 23, 13, 51, 57, 66, 85 After Pass 4: A: 27, 23, 13, 32, 51, 57, 66, 85 Pass 5: 23, 27, 13, 32, 51, 57, 66, 85 23, 13, 27, 32, 51, 57, 66, 85 23, 13, 27, 32, 51, 57, 66, 85 20

21 Bubble Sort Suppose the following numbers are stored in an array A: A: 32, 51, 27, 85, 66, 23, 13, 57 Apply the bubble sort to the array A. Show each pass separately. After Pass 1: A: 32, 27, 51, 66, 23, 13, 57, 85 After Pass 2: A: 27, 32, 51, 23, 13, 57, 66, 85 After Pass 3: A: 27, 32, 23, 13, 51, 57, 66, 85 After Pass 4: A: 27, 23, 13, 32, 51, 57, 66, 85 After Pass 5: A: 23, 13, 27, 32, 51, 57, 66, 85 Pass 6: 13, 23, 27, 32, 51, 57, 66, 85 13, 23, 27, 32, 51, 57, 66, 85 21

22 Bubble Sort Suppose the following numbers are stored in an array A: A: 32, 51, 27, 85, 66, 23, 13, 57 Apply the bubble sort to the array A. Show each pass separately. After Pass 1: A: 32, 27, 51, 66, 23, 13, 57, 85 After Pass 2: A: 27, 32, 51, 23, 13, 57, 66, 85 After Pass 3: A: 27, 32, 23, 13, 51, 57, 66, 85 After Pass 4: A: 27, 23, 13, 32, 51, 57, 66, 85 After Pass 5: A: 23, 13, 27, 32, 51, 57, 66, 85 After Pass 6: A: 13, 23, 27, 32, 51, 57, 66, 85 Pass 7: 13, 23, 27, 32, 51, 57, 66, 85 22

23 Bubble Sort Suppose the following numbers are stored in an array A: A: 32, 51, 27, 85, 66, 23, 13, 57 Apply the bubble sort to the array A. Show each pass separately. After Pass 1: A: 32, 27, 51, 66, 23, 13, 57, 85 After Pass 2: A: 27, 32, 51, 23, 13, 57, 66, 85 After Pass 3: A: 27, 32, 23, 13, 51, 57, 66, 85 After Pass 4: A: 27, 23, 13, 32, 51, 57, 66, 85 After Pass 5: A: 23, 13, 27, 32, 51, 57, 66, 85 After Pass 6: A: 13, 23, 27, 32, 51, 57, 66, 85 After Pass 7: A: 13, 23, 27, 32, 51, 57, 66, 85 Sorted List: A: 13, 23, 27, 32, 51, 57, 66, 85 23

24 Algorithm 4.4: (Bubble Sort) BUBBLE (DATA, N) Here DATA is an Array with N elements. This algorithm sorts the elements in DATA. 1. Repeat Steps 2 and 3 for K = 1 to N-1 2. Set PTR: =1 3. Repeat while PTR<=N-K 4. Exit (a) If DATA[PTR]>DATA[PTR+1], then: Interchange DATA[PTR] and DATA[PTR+1] [End of if structure] (b) Set PTR: =PTR+1 [End of inner loop] [End of Step 1 Outer loop] 24

25 Complexity of the bubble sort algorithm The time for a sorting algorithm is measured in terms of the number of comparisons. The number f(n) of comparisons in the bubble sort is easily computed. Specifically, there are n-1 comparisons during first pass, which places the largest element in the last position, there are n-2 comparisons in the second pass, which places the second largest element in the next to last position, and so on. Thus, f(n) = (n-1) + (n-2) = n(n-1)/2=n 2 /2+O(n) = O(n 2 ) In other words, the time required to execute bubble sort algorithm is proportional to n 2, where n is the number of input items. 25

26 Solved Problem 4.9: Using the bubble sort algorithm, find the number C of comparisons and the number D of interchanges which alphabetize the n=6 letters in PEOPLE. Pass 1: P E O P L E, E P O P L E, E O P P L E E O P P L E, E O P L P E, E O P L E P Pass 2: E O P L E P, E O P L E P, E O P L E P E O L P E P, E O L E P P Pass 3: E O L E P P, E O L E P P, E L O E P P E L E O P P Pass 4: E L E O P P, E L E O P P, E E L O P P Pass 5: E E L O P P, E E L O P P 26

27 Solved Problem 4.9: Using the bubble sort algorithm, find the number C of comparisons and the number D of interchanges which alphabetize the n=6 letters in PEOPLE. Since n=6, the number of comparisons, C = =15 and the number of interchanges, D = 9. 27

28 Binary Search Example 4.9: Suppose the following sorted 13 elements are stored in an array A: 11, 22, 30, 33, 40, 44, 55, 60, 66, 77, 80, 88, 99 Now apply the binary search to array A for i) ITEM = 40 and ii) ITEM= 85. Solution: i) For ITEM = BEG = 1 and END =13 MID = INT [(1+13)/2] = 7. So, A [MID] = 55 ITEM 2. Since 40<55, BEG=1 and END = MID-1 = 7-1 = 6 MID = INT [(1+6)/2] = 3. So, A[MID] = 30 ITEM 3. Since 40>30, BEG = MID+1 = 3+1= 4 and END = 6 MID= INT [(4+6)/2] = 5. So, A[MID] = 40 = ITEM We have found ITEM in location LOC = MID = 5 28

29 Binary Search Example 4.9: Suppose the following sorted 13 elements are stored in an array A: 11, 22, 30, 33, 40, 44, 55, 60, 66, 77, 80, 88, 99 Now apply the binary search to array A for i) ITEM = 40 and ii) ITEM= 85. Solution: ii) For ITEM = BEG = 1 and END =13 MID = INT [(1+13)/2] = 7. So, A [MID] = 55 ITEM 2. Since 85>55, BEG=MID+1=7+1=8 and END = 13 MID = INT [(8+13)/2] = 10. So, A[MID] = 77 ITEM 3. Since 85>77, BEG = MID+1 = 10+1= 11 and END = 13 MID= INT [(11+13)/2] = 12. So, A[MID] = 88 ITEM 29

30 Binary Search Example 4.9: Suppose the following sorted 13 elements are stored in an array A: 11, 22, 30, 33, 40, 44, 55, 60, 66, 77, 80, 88, 99 Now apply the binary search to array A for i) ITEM = 40 and ii) ITEM= 85. Solution: ii) For ITEM = Since 85<88, BEG = 11 and END = MID-1 =12-1=11 MID= INT [(11+11)/2] = 11. So, A[MID] = 80 ITEM 5. Since 85>80, BEG = MID + 1 = =12 and END = 11 Since BEG > END, ITEM does not belong to A. 30

31 Binary Search Data Structures- Chapter 4 Algorithm 4.6: (Binary Search) BINARY (DATA, LB, UB, ITEM, LOC) Here DATA is a sorted array with lower bound LB and upper bound UB, and ITEM is a given item of information. The variables BEG, END and MID denote respectively the beginning, end and middle locations of a segment of elements of DATA. This algorithm finds the location LOC of ITEM in DATA or sets LOC=NULL. 31

32 Binary Search 1. Set BEG=LB, END=UB and MID=INT ((BEG+END)/2). 2. Repeat Steps 3 and 4 while BEG END and DATA[MID] ITEM 3. If ITEM < DATA[MID], then Else Set END= MID - 1 Set BEG= MID + 1 [End of If structure] 4. Set MID= INT((BEG+END)/2) [End of Step 2 loop] 5. If ITEM = DATA[MID], then Else Set LOC= MID Set LOC= NULL [End of If structure] 6. Exit. 32

33 Complexity of the Binary Search Algorithm The Complexity of the Binary Search algorithm is measured by the maximum (worst case) number of Comparisons it performs for searching operations. The searched array is divided by 2 for each comparison/iteration. Therefore, we require at most f(n) comparisons to locate ITEM: 2 f(n) > n or equivalently, f(n) = log 2 n + 1 Approximately, equal to log 2 n. Example: If a given sorted array contains 1024 elements, then the maximum number of comparisons required is: log 2 (1024) = 10 (only 10 comparisons are enough) 33

34 Limitations of the Binary Search Algorithm The algorithm requires two conditions: The list must be sorted. One must have direct access to the middle element in any sub list. 34

35 Solved Problem 4.11: Suppose multidimensional arrays A and B are declared using A(-2:2, 2:22) and B(1:8, -5:5, -10;5) (a) Find the length of each dimension and the number of elements in A and B. (b) Consider the element B[3, 3, 3] in B. Find the effective indices E 1, E 2, E 3 and the address of the element, assuming Base(B)=400 and there are w=4 words per memory location. Solution: (a) Length = UB LB +1 Hence, the lengths of the dimensions of A are: L 1 = 2-(-2)+1=5 and L 2 = =21 Accordingly, A has 5.21=105 elements. The lengths of the dimensions of B are: L 1 = 8-1+1=8, L 2 = 5-(-5)+1=11, and L 3 =5-(-10)+1=16 Therefore, B has =1408 elements. 35

36 Solved Problem 4.11: Suppose multidimensional arrays A and B are declared using A(-2:2, 2:22) and B(1:8, -5:5, -10;5) (b) Consider the element B[3, 3, 3] in B. Find the effective indices E 1, E 2, E 3 and the address of the element, assuming Base(B)=400 and there are w=4 words per memory location. Solution: (b) The effective index E i is obtained from E i =k i LB, where k i is the given index and LB is the lower bound. Hence, E 1 =3-1=2 E 2 =3-(-5)=8 E 3 =3-(-10)=13 The address depends on whether the programming language stores B in rowmajor order or column-major order. Assuming B is the stored in column major order, we get: E 3 L 2 =13.11=143 E 3 L 2 + E 2 =143+8=151 (E 3 L 2 + E 2 )L 1 =151.8=1208 (E 3 L 2 + E 2 )L 1 +E 1 =1208+2=1210 Therefore, LOC(B[3,3,3])=400+4(1210)= =

DATA STRUCTURES/UNIT 3

DATA STRUCTURES/UNIT 3 UNIT III SORTING AND SEARCHING 9 General Background Exchange sorts Selection and Tree Sorting Insertion Sorts Merge and Radix Sorts Basic Search Techniques Tree Searching General Search Trees- Hashing.