Data Structures. Chapter 06. 3/10/2016 Md. Golam Moazzam, Dept. of CSE, JU

Transcription

1 Data Structures Chapter 06 1

2 Stacks A stack is a list of elements in which an element may be inserted or deleted only at one end, called the top of the stack. This means that elements are removed from a stack in the reverse order of that in which they were inserted into the stack. Two basic operations associated with stacks: Push is the term used to insert an element into a stack. Pop is the term used to delete an element from a stack. 2

3 Diagrams of Stacks 3

4 Procedure 6.1: Write a procedure that pushes an ITEM onto a stack. PUSH (STACK, TOP, MAXSTK, ITEM) This procedure pushes an ITEM onto a stack. 1. If TOP = MAXSTK, then: Print: OVERFLOW and Return. 2. Set TOP = TOP Set STACK[TOP] = ITEM. 4. Return. 4

5 Procedure 6.2: Write a procedure that deletes the top element of a stack. POP (STACK, TOP, ITEM) This procedure deletes the top element of STACK and assigns it to the variable ITEM. 1. If TOP = 0, then: Print: UNDERFLOW, and Return. 2. Set ITEM = STACK[TOP]. 3. Set TOP = TOP Return. 5

6 Solved Problem 6.1: Consider the following stack of characters, where STACK is allocated N=8 memory cells: STACK: A, C, D, F, K, -, -, - Describe the stack as the following operations take place: i) POP (STACK, ITEM) v) POP (STACK, ITEM) ii) POP (STACK, ITEM) vi) PUSH(STACK, R) iii) PUSH(STACK, L) vii) PUSH(STACK, S) iv) PUSH(STACK, P) Solution: viii) POP (STACK, ITEM) i) STACK: A, C, D, F, -, -, -, - v) STACK: A, C, D, L, -, -, -, - ii) STACK: A, C, D, -, -, -, -, - vi) STACK: A, C, D, L, R, -, -, - iii) STACK: A, C, D, L, -, -, -, - vii) STACK: A, C, D, L, R, S, -, - iv) STACK: A, C, D, L, P, -, -, - viii) STACK: A, C, D, L, R, -, -, - 6

7 ARITHMETIC EXPRESSIONS; POLISH NOTATION Infix notation: The operator symbol is placed between its two operands. Example: A+B, A-B, A*B, A/B Prefix notation: The operator symbol is placed before its two operands. Example: +AB, -AB, *AB, /AB This notation is also known as Polish notation, named after the Polish mathematician Jan Lukasiewicz. The fundamental property of Polish notation is that the order in which the operations are to be performed is completely determined by the positions of the operators and operands in the expression. Accordingly, one never needs parentheses when writing expressions in Polish notation. 7

8 ARITHMETIC EXPRESSIONS; POLISH NOTATION Postfix notation: The operator symbol is placed after its two operands. Example: AB+, AB-, AB*, AB/ This notation is also known as Reverse Polish notation. One never needs parentheses to determine the order of the operations in any arithmetic expression written in reverse Polish notation. The computer usually evaluates an arithmetic expression written in infix notation in two steps. First, it converts the expression to postfix notation, and then it evaluates the postfix expression. In each step, the stack is the main tool that is used to accomplish the given task. 8

9 Example 6.7: Transform the following arithmetic infix expression Q into its equivalent postfix expression P: Q: A + (B*C - (D /E F)*G)*H Solution: Push ( onto stack and then add ) to the end of Q. Thus, Q becomes Q: A + ( B * C - ( D / E F ) * G ) * H ) Start scanning. The following table shows the status of STACK and of the string P as each element of Q is scanned. 9

10 Example 6.7: Transform the following arithmetic infix expression Q into its equivalent postfix expression P: Q: A + (B*C - (D /E F)*G)*H) Solution: Symbol Scanned STACK Expression, P A ( A + ( + A ( ( + ( A B ( + ( A B * ( + ( * A B C ( + ( * A B C - ( + ( - A B C * ( ( + ( - ( A B C * 10

11 Example 6.7: Transf orm..... postfix expression P: Q: A + (B*C - (D /E F)*G)*H) Symbol Scanned STACK Expression, P D ( + ( - ( A B C * D / ( + ( - ( / A B C * D E ( + ( - ( / A B C * D E ( + ( - ( / A B C * D E F ( + ( - ( / A B C * D E F ) ( + ( - A B C * D E F / * ( + ( - * A B C * D E F / G ( + ( - * A B C * D E F / G ) ( + A B C * D E F / G * - * ( + * A B C * D E F / G * - H ( + * A B C * D E F / G * - H ) A B C * D E F / G * - H * + 11

12 Solved Problem 6.10: Transform the following arithmetic infix expression Q into its equivalent postfix expression P: Q: ((A + B)*D) (E-F) Solution: Push ( onto stack and then add ) to the end of Q. Thus, Q becomes Q: ((A + B) * D) (E-F) ) Start scanning. The following table shows the status of STACK and of the string P as each element of Q is scanned. 12

13 Solved Problem 6.10: Transform... postfix expression P: Q: ((A + B)*D) (E-F)) Symbol Scanned STACK Expression, P ( ( ( ( ( ( ( A ( ( ( A + ( ( ( + A B ( ( ( + A B ) ( ( A B + * ( ( * A B + D ( ( * A B + D ) ( A B + D * ( A B + D * ( ( ( A B + D * E ( ( A B + D * E - ( ( - A B + D * E F ( ( - A B + D * E F ) ( A B + D * E F - ) A B + D * E F - 13

14 Algorithm 6.6: Write an algorithm that transforms the infix expression into its equivalent postfix expression. POLISH (Q, P) Suppose Q is an arithmetic expression written in infix notation. This algorithm finds the equivalent postfix expression P. 1. Push ( onto STACK and add ) to the end of Q. 2. Scan Q from left to right and repeat steps 3 to 6 for each element of Q until the STACK is empty. 3. If an operand is encountered, add it to P. 4. If a left parenthesis is encountered, push it onto STACK. 14

15 Algorithm 6.6: Write an algorithm that transforms the infix expression into its equivalent postfix expression. 5. If an operator is encountered, then (a) Repeatedly pop from STACK and add to P each operator (on the top of STACK) which has the same precedence as or higher precedence than. (b) Add to STACK. [End of if structure] 6. If a right parenthesis is encountered, then (a) Repeatedly pop from STACK and add to P each operator (on the top of STACK) until a left parenthesis is encountered. (b) Remove the left parenthesis. [End of if structure] [End of Step 2 loop] 3/10/ Exit. Md. Golam Moazzam, Dept. of CSE, JU 15

16 Example 6.6: Find the value of the following arithmetic expression P written in postfix notation: P: 5, 6, 2, +, *, 12, 4, /, - Solution: - First we add a sentinel right parenthesis at the end of P: P: 5, 6, 2, +, *, 12, 4, /, -, ) (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) - Start scanning from left to right. The following table shows the contents of STACK as each element of P is scanned. The final number in STACK, 37, which is assigned to VALUE when the sentinel ")" is scanned, is the value of P. 16

17 Example 6.6: Find the value of the following arithmetic expression P written in postfix notation: P: 5, 6, 2, +, *, 12, 4, /, -, ) (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) Solution: Symbol Scanned (1) 5 (2) 6 (3) 2 (4) + (5) * (6) 12 (7) 4 (8) / (9) - (10) ) 5 5, 6 5, 6, 2 5, , 12 40, 12, 4 40, 3 37 STACK 17

18 Algorithm 6.5: Write an algorithm that finds the value of an arithmetic expression written in postfix notation. This algorithm finds the VALUE of an arithmetic expression P written in postfix notation. 1. Add a right parenthesis ) at the end of P. 2. Scan P from left to right and repeat Steps 3 and 4 for each element of P until the sentinel ")" is encountered. 3. If an operand is encountered, put it on STACK. 4. If an operator is encountered, then: a) Remove the two top elements of STACK, where A is the top element and B is the next-to-top element. b) Evaluate B A. c) Place the result of (b) back on STACK. [End of If structure] [End of Step 2 loop.] 5. Set VALUE equal to the top element on STACK. 6. Exit. 18

19 Quicksort, An Application of Stacks The following numbers are stored in an array A: A: 44, 33, 11, 55, 77, 90, 40, 60, 99, 22, 88, 66 Now apply Quick sort algorithm to find the final position of the first number. Solution: A: Two Stacks for keeping array boundary values LOWER UPPER 19

20 Quicksort, An Application of Stacks Consider the list: Lower boundary=1 and Upper boundary= 12 Start scanning from right to left. Stop at the first number less than 44 and interchange LOWER UPPER LOWER UPPER 20

21 Quicksort, An Application of Stacks Start scanning from left to right. Stop at the first number greater than 44 and interchange LOWER UPPER LOWER UPPER 21

22 Quicksort, An Application of Stacks Sublist 1 44 is in its final Sublist 2 position. LOWER UPPER Push the boundary values of each sub-list onto the stacks. 6 1 LOWER 12 4 UPPER 22

23 Quicksort, An Application of Stacks Consider the sub-list: LOWER=6 and UPPER= 12 Start scanning from right to left. Stop at the first number less than 90 and interchange LOWER UPPER Scan from left to right. Scan from right to left. Scan from left to right is in its Sublist final position. 23

24 Quicksort, An Application of Stacks Sublist 90 is in its final position. 1 4 LOWER UPPER Push the boundary values of the sub-list onto the stacks. 6 1 LOWER 10 4 UPPER 24

25 Quicksort, An Application of Stacks Sublist Consider the sub-list: LOWER=6 and UPPER= 10 Repeat the above steps is in its final position LOWER 10 4 UPPER LOWER UPPER

26 Quicksort, An Application of Stacks Sublist1 66 is in its final position. Sublist2 1 4 LOWER UPPER Push the boundary values of each sub-list onto the stacks LOWER UPPER 26

27 Quicksort, An Application of Stacks Sublist1 66 is in its final position. Sublist2 6 1 LOWER UPPER Consider the sub-list: LOWER=9 and UPPER= 10 Repeat the above steps LOWER 7 4 UPPER 88 is in its final position. 27

28 Quicksort, An Application of Stacks Consider the sub-list: LOWER=6 and UPPER=7 Repeat the above steps LOWER 7 4 UPPER LOWER UPPER 55 is in its final position. 28

29 Quicksort, An Application of Stacks LOWER UPPER Consider the sub-list: LOWER=1 and UPPER=4 Repeat the above steps LOWER UPPER 29

30 Quicksort, An Application of Stacks is in its final position. Sublist LOWER UPPER Push the boundary values of the sub-list onto the stacks. 3 4 LOWER UPPER 30

31 Quicksort, An Application of Stacks LOWER UPPER Consider the sub-list: LOWER=3 and UPPER=4 Repeat the above steps is in its final position. LOWER UPPER 31

32 Quicksort, An Application of Stacks Now the stacks are empty. That is, no sub-lists exist. LOWER UPPER Finally, the list becomes: A:

33 Complexity of the Quicksort Algorithm i) Average case: O(n log n) ii) Worst case: O(n log n) f(n) = n + (n-1) = n (n+1) / 2 = O(n 2 ) 33

34 Solved Problem 6.12: Suppose S is the following list of 14 alphabetic characters: S: D A T A S T R U C T U R E S Use the Quick sort algorithm to find the final position of the first character D D A T A S T R U C T U R E S C A T A S T R U D T U R E S C A D A S T R U T T U R E S C A A D S T R U T T U R E S D is in its final position. 34

35 Recursion A procedure is called a recursive procedure if it contains a Call statement to itself. A recursive procedure must have the following two properties: There must be certain criteria, called base criteria, for which the procedure does not call itself. Each time the procedure does call itself, it must be closer to the base criteria. 35

36 Recursion Example 6.9: Calculate 4! using the recursive definition. Solution: This calculation requires the following nine steps: 1) 4! = 4. 3! 2) 3! = 3. 2! 3) 2! = 2. 1! 4) 1! = 1. 0! 5) 0! = 1 6) 1! = 1. 1 = 1 7) 2! = 2. 1 =2 8) 3! = 3. 2 = 6 9) 4!=4. 6 = 24 36

37 Procedure 6.9A: Write a procedure that calculates N! FACTORIAL(FACT, N) This procedure calculates N! and returns the value in the variable FACT. 1. If N = 0, then: Set FACT := 1, and Return. 2. Set FACT := Repeat for K = 1 to N. Set FACT := K*FACT. [End of loop.] 4. Return. 37

38 Procedure 6.9B: Write a recursive procedure that calculates N! FACTORIAL(FACT, N) This procedure calculates N! and returns the value in the variable FACT. 1. If N = 0, then: Set FACT := 1, and Return. 2. Call FACTORIAL(FACT, N 1). 3. Set FACT := N*FACT. 4. Return. 38

39 Fibonacci Sequence The Fibonacci sequence (usually denoted by F 0, F 1, F 2,..) is as follows: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55,... That is, F 0 = 0 and F 1 = 1 and each succeeding term is the sum of the two preceding terms. Definition: (Fibonacci Sequence) a) If n = 0 or n = 1, then F n = n. b) If n > 1, then F n = F n-2 + F n-1 39

40 Fibonacci Sequence Procedure 6.10: Write a recursive procedure that finds the nth Fibonacci number. FIBONACCI(F1B, N) This procedure calculates F N and returns the value in the first parameter FIB. 1. If N = 0 or N = 1, then: Set FIB := N, and Return. 2. Call FIBONACCI(F1BA, N - 2). 3. Call FIBONACCI(FIBB, N - 1). 4. Set FIB := F1BA + FIBB. 5. Return. 40

41 Ackermann Function The Ackermann function is a function with two arguments each of which can be assigned any nonnegative integer: 0, 1, 2,... This function is defined as follows: Definition 6.3: (Ackermann Function) a) If m = 0, then A(m, n) = n + 1. b) If m 0 but n = 0, then A(m, n) = A(m - 1, 1). c) If m 0 and n 0, then A(m, n) = A(m - 1, A(m, n - 1)) 41

42 Solved Problem 6.18: Define Ackermann function. Find the value of A(1, 3). 1) A(1, 3) = A(0, A(1, 2)) 2) A(1, 2) = A(0, A(1, 1)) 3) A(1, 1) = A(0, A(1, 0)) 4) A(1, 0) = A(0, 1) 5) A(0, 1) = 1+1 = 2 6) A(1, 0) = 2 7) A(1, 1) = A(0, 2) 8) A(0, 2) = =3 9) A(1, 1) = 3 10) A(1, 2) = A(0, 3) 11) A(0, 3) = = 4 12) A(1, 2) = 4 13) A(1, 3) = A(0, 4) 14) A(0, 4) = = 5 15) A(1, 3) = 5 42

43 Solved Problem 6.15: Let a and b denote positive integers. Suppose a function Q is defined recursively as follows: Solution: 0 Q( a, b) Q( a b, b) 1 i) Find the value of Q(2, 3) and Q(14, 3). ii) What does this function do? Find Q(5861, 7). i) Q(2, 3) = 0, Since 2 < 3. Q(14, 3) = Q(11, 3) + 1 = [Q(8, 3) + 1] + 1 = Q(8, 3) + 2 = [Q(5, 3) + 1] + 2 = Q(5, 3) + 3 = [Q(2, 3) + 1] + 3 = Q(2, 3) + 4 = = 4 if a b if b a 43

44 Solved Problem 6.15: Let a and b denote positive integers. Suppose a function Q is defined recursively as follows: 0 Q( a, b) Q( a b, b) 1 if a b if b a i) Find the value of Q(2, 3) and Q(14, 3). ii) What does this function do? Find Q(5861, 7). Solution: ii) Each time b is subtracted from a and the values of Q is increased by 1. Hence Q(a, b) finds the quotient when a is divided by b. Thus, Q(5861, 7) =

45 Solved Problem 6.16: Let n denote a positive integer. Suppose a function L is defined recursively as follows: 0 L( n) L( i) Find L(25). n / 2 ii) What does this function do? 1) if if n 1 n 1 Solution: i) L(25) = L(12) + 1 = [L(6) + 1] + 1 = L(6) + 2 = [L(3) + 1] + 2 = L(3) + 3 = [L(1) + 1] + 3 = L(1) + 4 = = 4 45

46 Solved Problem 6.16: Let n denote a positive integer. Suppose a function L is defined recursively as follows: 0 L( n) L( i) Find L(25). n / 2 ii) What does this function do? 1) if if n 1 n 1 Solution: ii) Each time n is divided by 2 and the value of L is increased by 1. Hence L is the greatest integer such that 2 L n Accordingly, the function finds L log 2 n 46

47 Towers of Hanoi Consider three pegs labeled A, B, and C. Suppose on peg A, there are placed a finite number n of disks with decreasing size. The object of the game is to move the disks from peg A to peg C using peg B as an auxiliary. The rules of the game are as follows: Only one disk may be moved at a time. Specially, only the top disk on any peg may be moved to any other peg. At no time can a larger disk be placed on a smaller disk. 47

48 Draw a schematic diagram of the recursive solution to Towers of Hanoi problem for 1 disk. OR, How many moves will it take to transfer 1 disk from the left post to the right post? Solution: 1 disk: 1 move Move 1: A C 48

49 Draw a schematic diagram of the recursive solution to Towers of Hanoi problem for 2 disks. OR, How many moves will it take to transfer 2 disks from the left post to the right post? Solution: No. of moves= 2 n -1 = = 3 Moves: A B A C and B C 49

50 Draw a schematic diagram of the recursive solution to Towers of Hanoi problem for 3 disks. OR, How many moves will it take to transfer 3 disks from the left post to the right post? Solution: No. of moves= 2 n -1 = = 7 Moves: A C, A B, C B, A C, B A, B C, and A C 50

51 Towers of Hanoi problem: Recursive Solution Procedure 6.11: Write a recursive procedure that gives a solution to the Tower of Hanoi problem for N disks. TOWER(N, BEG, AUX, END) This procedure gives a recursive solution to the Towers of Hanoi problem for N disks. 1. If N = 1, then: a) Write: BEG END. b) Return. [End of If structure.] 2. Call TOWER(N - 1, BEG, END, AUX). 3. Write: BEG END. 4. Call TOWER(N - 1, AUX, BEG, END). 5. Return. 51

52 Draw a schematic diagram of the recursive solution to Towers of Hanoi problem for n = 4 disks. 52

53 Draw a schematic diagram of the recursive solution to Towers of Hanoi problem for n = 4 disks. Solution: No. of moves= 2 n -1 = = 15 Moves: A B A C B C A B C A C B A B A C B C B A C A B C A B A C B C 53

54 Towers of Hanoi: Applications Frequently used in psychological research on problem solving. There also exists a variant of this task called Tower of London for neuropsychological diagnosis and treatment of executive functions. Also used as Backup rotation scheme when performing computer data Backups where multiple tapes/media are involved. Tower of Hanoi is popular for teaching recursive algorithms to beginning programming students. Tower of Hanoi is also used as a test by neuropsychologists trying to evaluate frontal lobe (The frontal lobes are considered our emotional control center and home to our personality. )deficits. 54

55 Queues A queue is a linear list of elements in which deletions can take place only at one end, called the front, and insertions can take place only at the other end, called the rear. Queues are also called first-in first-out (FIFO) lists, since the first element in a queue will be the first element out of the queue. In other words, the order in which elements enter a queue is the order in which they leave. Example: A line of people waiting at a bus stop form a queue, where the first person in line is the first person to board the bus. A line of people waiting to be served by an ATM machine. Collection of documents sent to a shared printer. 55

56 Representation of Queues Queues may be represented in the computer by a linear array QUEUE and two pointer variables: FRONT, containing the location of the front element of the queue; and REAR, containing.the location of the rear element of the queue. FRONT = NULL indicates that the queue is empty. The following Figure indicates the way elements will be deleted from the queue and the way new elements will be added to the queue. Whenever an element is deleted from the queue, the value of FRONT is increased by 1. FRONT := FRONT + 1 Similarly, whenever an element is added to the queue, the value of REAR is increased by 1. REAR := REAR

57 Representation of Queues 57

58 Procedure 6.13: Write a procedure that inserts an ITEM onto a queue. QINSERT (QUEUE, N, FRONT, REAR, ITEM) This procedure inserts an element ITEM into a queue. 1. If FRONT = 1 and REAR = N, or if FRONT = REAR + 1, then Write: OVERFLOW, and Return. 2. If FRONT = NULL, then Set FRONT = 1 and REAR = 1 Else if REAR = N, then Set REAR = 1 Else Set REAR = REAR + 1 [End of If structure.] 3. Set QUEUE [REAR] = ITEM 4. Return. 58

59 Procedure 6.14: Write a procedure that deletes an element from a queue. QDELETE(QUEUE, N, FRONT, REAR, ITEM) This procedure deletes an element from a queue and assigns it to the variable ITEM. 1. If FRONT:= NULL, then: Write: UNDERFLOW, and Return. 2. Set ITEM:= QUEUE[FRONT]. 3. If FRONT= REAR, then: Set FRONT := NULL and REAR := NULL. Else if FRONT = N, then: Else: Set FRONT := 1. Set FRONT := FRONT + 1 [End of If structure.] 4. Return. 59

60 Solved Problem 6.22: Consider the following queue of characters, where QUEUE is a circular array which is allocated six memory cells: FRONT=2, REAR=4 QUEUE: -, A, C, D, -, - Describe the queue as the following operations take place: i) F is added to the queue. vi) Two letters are deleted. ii) Two letters are deleted. iii) K, L, and M are added to the queue. iv) Two letters are deleted. vii) S is added to the queue. viii) Two letters are deleted. ix) One letter is deleted. v) R is added to the queue. x) One letter is deleted. Solution: i) FRONT=2, REAR=5 QUEUE: -, A, C, D, F, - ii) FRONT=4, REAR=5 QUEUE: -, -, -, D, F, - iii) FRONT=4, REAR=2 QUEUE: L, M, -, D, F, K 60

61 Solved Problem 6.22: Consider the following queue of characters, where QUEUE is a circular array which is allocated six memory cells: FRONT=2, REAR=4 QUEUE: -, A, C, D, -, - Describe the queue as the following operations take place: i) F is added to the queue. vi) Two letters are deleted. ii) Two letters are deleted. iii) K, L, and M are added to the queue. iv) Two letters are deleted. vii) S is added to the queue. viii) Two letters are deleted. ix) One letter is deleted. v) R is added to the queue. x) One letter is deleted. Solution: iv) FRONT=6, REAR=2 QUEUE: L, M, -, -, -, K v) FRONT=6, REAR=3 QUEUE: L, M, R, -, -, K vi) FRONT=2, REAR=3 QUEUE: -, M, R, -, -, - 61

62 Solved Problem 6.22: Consider the following queue of characters, where QUEUE is a circular array which is allocated six memory cells: FRONT=2, REAR=4 QUEUE: -, A, C, D, -, - Describe the queue as the following operations take place: ii) Two letters are deleted. iii) K, L, and M are added to the queue. iv) Two letters are deleted. vii) S is added to the queue. viii) Two letters are deleted. ix) One letter is deleted. v) R is added to the queue. x) One letter is deleted. Solution: vii) FRONT=2, REAR=4 QUEUE: -, M, R, S, -, - viii) FRONT=4, REAR=4 QUEUE: -, -, -, S, -, - ix) FRONT=0, REAR=0 QUEUE: -, -, -, -, -, - x) Queue is empty. Underflow has occurred. 62

63 DEQUES A deque is a linear list in which elements can be added or removed at either end but not in the middle. The term deque is a contraction of the name double-ended queue. Deque is maintained in memory by a circular array DEQUE with pointers LEFT and RIGHT, which point to the two ends of the deque. Elements extend from the left end to the right end in the array. Figure pictures two deques, each with 4 elements maintained in an array with N = 8 memory locations. LEFT = NULL indicates that a deque is empty. 63

64 DEQUES 64

65 DEQUES There are two variations of a deque: An input-restricted deque An input-restricted deque is a deque which allows insertions at only one end of the list but allows deletions at both ends of the list. An output-restricted deque An output-restricted deque is a deque which allows deletions at only one end of the list but allows insertions at both ends of the list. 65

66 Solved Problem 6.24: Consider the following deque of characters, where DEQUE is a circular array which is allocated six memory cells: LEFT=2, RIGHT=4 DEQUE: -, A, C, D, -, - Describe the deque as the following operations take place: i) F is added to the right of the deque. ii) Two letters on the right are deleted. iii) K, L, and M are added to the left of the deque. iv) One letter on the left is deleted. v) R is added to the left of the deque. vi) S is added to the right of the deque. vii) T is added to the right of the deque. 66

67 Solved Problem 6.24:... LEFT=2, RIGHT=4 DEQUE: -, A, C, D, -, - i) F is added to the right of the deque. ii) Two letters on the right are deleted. iii) K, L, and M are added to the left of the deque. iv) One letter on the left is deleted. v) R is added to the left of the deque. Solution: i) LEFT=2, RIGHT=5 DEQUE: -, A, C, D, F, - ii) LEFT=2, RIGHT=3 DEQUE: -, A, C, -, -, - iii) LEFT=5, RIGHT=3 DEQUE: K, A, C, -, M, L iv) LEFT=6, RIGHT=3 DEQUE: K, A, C, -, -, L v) LEFT=5, RIGHT=3 DEQUE: K, A, C, -, R, L 67

68 Solved Problem 6.24:... LEFT=5, RIGHT=3 DEQUE: K, A, C, -, R, L vi) S is added to the right of the deque. vii) T is added to the right of the deque. Solution: vi) LEFT=5, RIGHT=4 DEQUE: K, A, C, S, R, L vii) Since LEFT = RIGHT + 1, the deque is full. That is, OVERFLOW has occurred. 68

69 Priority Queues A priority queue is a collection of elements such that each element has been assigned a priority and such that the order in which the elements are deleted and processed comes from the following rules: An element of higher priority is processed before any element of lower priority. Two elements with the same priority are processed according to the order in which they were added to the queue. 69

70 One-Way List Representation of a Priority Queues One way to maintain a priority queue in memory is by means of a oneway list as follows: a) Each node contains three items of information: An information field INFO, a priority number PRN and a link number LINK. b) A node X proceeds a node Y in the list: When X has higher priority than Y OR When both have the same priority but X was added to the list before Y. 70

71 One-Way List Representation of a Priority Queues Start PRN AAA 1 BBB 2 CCC 2 DDD 4 EEE 5 FFF 6 71