Solved PAPER Computer Science-XII. Part -I. Answer all questions

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1 ISC Solved PAPER 2017 Computer Science-XII Fully Solved (Question-Answer) Time : 3 hrs Max. Marks : 100 General Instructions 1. Answer all questions in Part I (compulsory) and seven questions from Part II. 2. Attempt any seven questions from Part II, choosing three questions from Section A, two from Section B and two from Section C. 3. The intended marks for questions or parts of question are given in brackets [ ]. Part -I (30 Marks) Answer all questions While answering questions in this Part, indicate briefly your working and reasoning, wherever required. 1. (a) State the law represented by the following proposition and prove it with the help of a truth table P V P = P (b) State the Principle of Duality. (c) Find the complement of the following Boolean expression using De Morgan s law. F( a, b, c) = ( b + c) + a (d) Draw the logic diagram and truth table for a 2 input XNOR gate. (e) If (~ P => Q) then write its (i) Inverse (ii) Converse

2 Ans. (a) P V P = P proposition represents the idempotent law. Truth table for P VP = P P P P = P V P Hence, P = P V P (b) Principle of Duality It is states that a boolean relation can be derived by (i) changing every OR (+) by AND (.) (ii) changing every AND (.) by OR (+) (iii) changing every 0 by 1 and 1 by 0 (c) Given expression is, F (a, b, c) =(b + c) + a The complement of F i.e. F F = [(b + c) + a] = (b + c). a = (b ). c. a = b.c. a = a b c (d) Logic diagram for a 2 input XNOR gate : Proved A B Y Y = A B = A B + AB Truth table for a 2 input XNOR gate A B Y (e) (i) To form the inverse of the statement, take the negation of both the antecedent and the consequent. So, ~ P Q Here, P is a antecedent and Q is a consequent Hence, inverse would be ~ (~ P) ~ Q P ~ Q (ii) To form the converse of the statement, interchange the antecedent to consequent. Hence, converse would be Q ~ P 2. (a) What is an interface? How is it different from a class? (b) Convert the following infix expression to postfix form P * Q/R + (S + T) (c) A matrix P[15] [10] is stored with each element required 8 bytes of storage. If the base address at P[0][0] is 1400, determine the address at P[10][7] when the matrix is stored in Row Major Wise.

3 Ans. (d) (i) What is the worst case complexity of the following code segment for (int x = 1; x <=a; x++) statements; for (int y = 1; y <=b; y++) for (int z = 1; z <= c; z++ ) statements; (ii) How would the complexity change if all the three loops went to N instead of a, b and c? (e) Differentiate between a constructor and a method of a class. (a) Interfaces are basically a collection of method which are public and abstract by default. These interfaces do not have any method body. Syntax interface interface_name returntype method_name(arg_list); Differences between interface and class are as follows Interface The members of an interface are always declared as constant i.e. their values are final. It cannot be used to declare objects. It can only be inherited by a class (b) Given, infix expression : P * Q/R + (S + T) Class The members of a class can be declared as constant or variables. It can be instantiated by declaring objects. First, we eliminate bracket content and then convert it into postfix expression = P*Q/R + (ST+) = PQ*/R + (ST+) = PQ*R/+(ST+) = PQ*R/(ST+)+ The postfix expression is, = PQ*R/ST++ (c) Given, B =1400, W = 8, L r = 0, L c = 0, U r = 15, U c = 10, I = 10, J = 7 For row major wise, A[I][J] = B + W [ U (I L ) + (J L )] r r c A[ 10][7] = [ 15 ( 10 0) + ( 7 0)] = [ ] = [ ] = = = 2656 (d) (i) In the given code, first loop will execute c times and its sequence will take constant time t 1. So, execution time will be c* t 1. Second loop is nested loop. The outer loop of second loop will execute a times and inner loop will execute b times and sequence of statements will take constance time i.e t 2. So, execution time will be a* b + t 2. Hence, total time = c* t + a * b + t 2 2 Worst case complexity = o( c + ab)

4 (ii) If all the loop execute N times, the body of the loop get executed N 1times, the condition will evaluate to false and loop terminated without executing the body. So, the time complexity for such code would be : Total time = c * N = cn i.e. O(N) (e) Differences between constructor and method of a class are as follows : It has no return type Constructor Constructor name should be same as the class name. Constructor is automatically called upon object creation. It has a return type. Method Method name can be any valid identifier. Method are invoked explicitly. 3. The following function magicfun() is a part of some class. What will the function magicfun() return, when the value of n=7 and n=10, respectively? Show the dry run/working: int magicfun(int n) if (n==0) return 0; else return magicfun(n/2) * 10 + (n % 2); Ans. When n=7 then, magicfun (3) 10 + (1) Return 111 Call magicfun (1) * 10+(1) Return 11 Call magicfun (0) * 10 + (1) Return 1 Hence, function magicfun( ) return 111, when the value of n=7 when n = 10 then, magicfun (5)*10+(0) Return 1010 Call magicfun (2)*10+(1) Return 101 Call magicfun (1)*10+(0) Return 10 Return 1 Call magicfun (0)*10+(1) Hence, function magicfun( ) return 1010, when the value of n=10.

5 Part - II (70 Marks) Answer six questions in this part, choosing two questions from Section A, two from Section B and two Section C. Section - A Answer any three questions 4. (a) Given the Boolean function F(A, B, C, D) = Σ (2, 3, 4, 5, 6, 7, 8, 10, 11) (i) Reduce the above expression by using 4-variable Karnaugh map, showing the various groups (i.e. octal, quads and pairs). (ii) Draw the logic gate diagram for the reduced expression. Assume that the variable and their complements are available as inputs. (b) Given the Boolean function F(P, Q, R, S) = π (0, 1, 2, 4, 5, 6, 8, 10) (i) Reduce the above expression by using 4-variable Karnaugh map, showing the various groups (i.e. octal, quads and pairs). (ii) Draw the logic gate diagram for the reduced expression. Assume that the variable and their complements are available as inputs. Ans. Given Boolean function F(A, B, C, D) = Σ (2, 3, 4, 5, 6, 7, 8, 9, 10, 11) (i) K-map CD AB AB CD CD CD CD AB AB AB There are 2 quads and 1 pair Quad 1( m4 + m5 + m6 + m7 ) = AB Quad 2 ( m + m + m + m ) = BC Pair ( m + m ) = A B D 8 10 Hence, the final reduced expression F ( A, B, C, D) = AB + BC + A B D

6 (ii) Logic gate diagram for F ( A, B, C, D) = AB + BC + A B D A A B B C C D D AB BC F=AB + BC + ABD ABD (b) Given Boolean function F( P, Q, R, S )= π (0, 1, 2, 4, 5, 6, 8, 10) (i) K-map RS PQ R+S R+S R+S R+S P+Q P+Q P+Q P+Q There are 3 quads Quad 1 ( m + m + m + m ) = P + R Quad 2 ( m + m + m + m ) = P + S Quad 3 ( m + m + m + m ) = Q + S Hence, the final reduced expression F( P, Q, R, S ) = ( P + R). ( P + S ). ( Q + S ) (ii) Logic gate diagram for F( P, Q, R, S ) = ( P + R).( P + S ).( Q + S ) P Q R S P+ R P+ S F= ( P+ R). ( P+ S). ( Q+ S) Q+ S

7 5. (a) A school intends to select candidates for an Inter-School Essay Competition as per the criteria given below The student has participated in an earlier competition and is very creative. Or The student is very creative and has excellent general awareness, but has not participated in any competition earlier. Or The student has excellent general awareness and has won prize in an inter-house competition. The inputs are INPUTS A B C D participated in competition earlier is very creative won prize in an inter-house competition has excellent general awareness Ans. (In all the above cases 1 indicates yes and 0 indicates no). Output : X [1 indicates yes, 0 indicates no for all cases] Draw the truth table for the inputs and outputs given above and write the POS expression for X (A, B, C, D). (b) State the application of a Half Adder. Draw the truth table and circuit diagram for a Half Adder. (c) Convert the following Boolean expression into its canonical POS form F (A, B, C) = (B + C ). (A + B) (a) Truth table for given inputs in below A B C D X

8 The POS expression for above truth table is (b) Application of a Half Adder X( A, B, C, D) = ( A + B + C + D). ( A + B+ C + D).( A + B + C + D) ( A + B + C + D). ( A + B + C + D). ( A + B + C + D) ( A + B + C + D). ( A + B + C + D) The ALU (Arithmetic Logic Circuit) of a computer uses half adder to compute the binary addition operation on two bits. Truth table for a half adder Logic/Circuit diagram X Y S C Sum ( S ) = XY + XY = X Y Carry ( C) = XY X Y S = X + Y C = X Y Ans. (c) Given Boolean expression is F( A, B, C) = ( B + C ). ( A + B) = ( A. A + B + C ). ( A + B + CC ) [by complement law] = ( A + B + C )( A + B + C ) ( A + B + C )( A + B + C ) = ( A + B + C ) ( A + B + C )( A + B + C) [by distributive law] 6. (a) What is a Multiplexer? How is it different from a decoder? Draw the circuit diagram for a 8 : 1 Multiplexer. (b) Prove the Boolean expression using Boolean laws. Also, mention the law used at each step. F = (x + z) + [(y + z). (x + y)] = 1 (c) Define maxterms and minterms. Find the maxterm and minterm when P = 0, Q = 1, R = 1 and S = 0 (a) Multiplexer It is a combinational circuit that selects binary information from one of many input lines and directs it to a single output line. A multiplexer is also called a data selector. Circuit diagram for a 8 : 1multiplexer

9 Boolean function for a 8 : 1 multiplexer F = S S S I + S S S I + S S1S I S 2S1S 0I4 S 2S1S 0I5 S 2S1S 0I + S S S I 6 From the expression the logic diagram is formed as follows. S 2 S 1 S I 0 I 1 I 2 I 3 I 4 F I 5 I 6 I 7 Difference between Multiplexer and Decoder Differences between multiplexer and decoder are as follows Multiplexer A digital multiplexer is a combinational circuit that selects binary information from one of many input lines and directs it to a single output line. It is called as data selector, since it selects one of many inputs. A multiplexer is used to share several signals to one device like A/D convertor. e.g.4-to-1 line multiplexer. (b) Given Boolean expression is Hence, LHS = RHS F = ( x + z) + [( y + z) ( x + y)] = 1 LHS ( x + z) + [( y + z). ( x + y)] Decoders A decoder is a combinational circuit that coverts binary information from n input line to a maximum of 2 n unique output lines. This decoder is called n-to-m line decoder where m 2 n. Decoders are used to keep abstraction and modularity in hardware design. e.g. 3-to-8 decoder. = ( x + z) + ( y + z) + ( x + y ) [Using Demorgan s law] = ( x + z) + ( y ). z + ( x ). y [Using Demorgan s law] = ( x + z) + y. z + x. y [Using Involution law] = x + z+ yz + xy = x ( y + y ) + z( y + y ) + yz + xy [Using Complementary law] = x y + x y + yz + y z + yz + xy = x y + y ( x + x ) + y( z + z ) + y z [Using Complementary law] = x y + y + y + y z = y( x + 1) + y ( 1 + z) [Using Postulates] = y + y = 1 = RHS [Using Postulates]

10 (c) Minterms The variables are combined to form a standard product when we take AND operation between the variables. These AND terms are known as minterms. Maxterms The variables are combined to form a standard sum when we take OR operation between the variables. These OR terms are known as maxterms. When, P = 0, Q = 1, R = 1and S = 0 Maxterm P + Q + R + S Minterm PQRS Section - B Answer any two questions Each program should be written in such a way that it clearly depicts the logic of the problem. This can be achieved by using mnemonic names and comments in the program. (Flowcharts and Algorithm are not required) The programs must be written in Java 7. A class Palin has been defined to check whether a positive number is a Palindrome number or not. The number N is palindrome if the original number and its reverse are same. Some of the members of the class are give below; Class Name Data members/instance variables num revnum Methods/Member functions Palin() void accept() int reverse(int y) void check() Plain integer to store the number integer to store the reverse of the number constructor to initialize data members with legal initial values to accept the number reverses the parameterized argument y and stores it in revnum using recursive technique checks whether the number is a Palindrome by invoking the function reverse() and display the result with an appropriate message Ans. Specify the class Palin giving the details of the constructor(), void accept(), int reverse(int) and void check(). Define the main() function to create an object and call the functions accordingly to enable the task. import java.io.*; class Palin int num; int revnum; int reverse = 0; Palin()

11 num=0; revnum=0; void accept() throws IOException InputStreamReader IR=new InputStreamReader(System.in); BufferedReader br=new BufferedReader(IR); System.out.println( Enter the number ); num=integer.parseint(br.readline()); int sum=0,rem; int reverse(int y) if(y>0) rem=y%10; sum=sum*10+rem; reverse(y/10); else return sum; return sum; void check() revnum=reverse(num); if(num==revnum) System.out.println(num+" is a palindrome number"); else System.out.println(num+" is not a palindrome number"); public static void main(string args[]) throws IOException Palin P = new Palin(); P.accept(); P.check(); 8. A class Adder has been defined to add any two accepted time Example : Time A - 6 hours 35 minutes Time B - 7 hours 45 minutes Their sum is - 14 hours 20 minutes (where 60 minutes = 1 hour)

12 The details of the members of the class are given below Class Name Data members/instance variable Adder a[ ] integer array to hold two elements (hours and minutes) Member functions/methods Adder() void readtime() void addtime(adder X, Adder Y) void disptime() constructor to assign 0 to the array elements to enter the elements of the array adds the time of the two parameterized objects X and Y and stores the sum in the current calling object displays the array elements with an appropriate message (i.e. hours = and minutes = ) Ans. Specify the class Adder giving details of the constructor( ), void readtime( ), void addtime(adder, Adder) and void disptime( ). Define the main( ) function to create objects and call the functions accordingly to enable the task. import java.io.*; class Adder int a[] = new int[2]; Adder() a[0]=0; a[1]=0; void readtime() throws IOException InputStreamReader IR=new InputStreamReader(System.in); BufferedReader br=new BufferedReader(IR); System.out.println("Enter the Time in Hours and Minutes"); a[0]=integer.parseint(br.readline()); a[1]=integer.parseint(br.readline()); void addtime(adder X, Adder Y) a[0]=x.a[0]+y.a[0]; a[1]=x.a[1]+y.a[1]; if(a[1]>=60) a[0]=a[0]+a[1]/60; a[1]=a[1]%60;

13 void disptime() System.out.println(a[0]+" hours "+a[1]+" minutes"); public static void main(string args[]) throws IOException Adder A = new Adder(); Adder B = new Adder(); Adder C = new Adder(); A.readtime(); B.readtime(); C.addtime(A, B); System.out.print("Time A - "); A.disptime(); System.out.print("Time B - "); B.disptime(); System.out.print("Their sum is - "); C.disptime(); 9. A class SwapSort has been defined to perform string related operations on a word input. Some of the members of the class are as follows Class Name Data members/instance variables wrd len swapwrd sortwrd Member functions\methods SwapSort( ) void readword( ) void swapchar( ) void sortword( ) void display( ) to store a word SwapSort integer to store length of the word to store the swapped word to store the sorted word default constructor to initialize data members with legal initial values to accepts a word in UPPER CASE to interchange/swap the first and last characters of the word in wrd and stores the new word in swapwrd sorts the character of the original word in alphabetical order and stores it in sortwrd displays the original word, swapped word and the sorted word. Specify the class SwapSort, giving the details of the constructor( ), void readword( ), void swapchar( ), void sortword( ) and void display( ). Define the main( ) function to create an object and call the functions accordingly to enable the task.

14 Ans. import java.io.*; public class SwapSort char [] wrd; int len; char [] swapwrd; char [] sortwrd; SwapSort() len=0; void readword() throws IOException InputStreamReader IR=new InputStreamReader(System.in); BufferedReader br=new BufferedReader(IR); System.out.println( Enter the word in UPPER CASE ); String t=br.readline(); len=t.length(); wrd = new char[len]; for(int i=0; i<len;i++) wrd[i]=t.charat(i); swapwrd = new char[len]; sortwrd = new char[len]; void swapchar() for(int i=0;i<len;i++) if(i==0) swapwrd[i]=wrd[len-1]; else if(i==len-1) swapwrd[len-1]=wrd[0]; else swapwrd[i]=wrd[i]; void sortword() for(int i=0;i<len;i++) sortwrd[i]=wrd[i]; for(int i=0;i<len;i++) for(int j=0;j<len-i-1;j++) if(sortwrd[j]>sortwrd[j+1])

15 char t = sortwrd[j]; sortwrd[j]=sortwrd[j+1]; sortwrd[j+1]=t; void display() System.out.print("Original Word : "); for(int i=0;i<len;i++) System.out.print(wrd[i]); System.out.println(); System.out.print("Swapped Word : "); for(int i=0;i<len;i++) System.out.print(swapwrd[i]); System.out.println(); System.out.print("Sorted Word : "); for(int i=0;i<len;i++) System.out.print(sortwrd[i]); System.out.println(); public static void main(string args[]) throws IOException SwapSort S = new SwapSort(); S.readword(); S.swapchar(); S.sortword(); S.display(); Section - C Answer any two questions Each program should be written in such a way that it clearly depicts the logic of the problem stepwise. This can be achieved by using comments in the program and mnemonic names or pseudo codes for algorithms. The programs must be written in Java and the algorithms must be written in general/standard form, wherever required/specified. (Flowcharts are not required) 10. A super class Product has been defined to store the details of a product sold by a wholesaler to a retailer. Define a sub class Sales to compute the total amount paid by the retailer with or without fine along with service tax. Some of the members of both the classes are given below

16 Class Name Data members/instance variables name code amount Member functions/methods Product(String n, int c, double p) void show() Class Name Data members/instance variables day tax totamt Member functions/methods Sales ( ) void compute( ) void show( ) Product stores the name of the product integer to store the product code stores the total sale amount of the product (in decimals) parameterized constructor to assign data members name=n, code=c and amount=p displays the details of the data members Sales stores number of days taken to pay the sale amount to store the services tax (in decimals) to store the total amount (in decimals) paramaterized constructor to assign values to data members to both the classes calculates the services 12.4% of the actual sale amount calculates the 2.5% of the actual sale amount only if the amount paid by the retailer to the wholesaler exceeds 30 days calculates the total amount paid by the retailer as (actual sale amount + service tax + fine) displays the data members of super class and the total amount Ans. Assume that the super class Product has been defined. Using the concept of inheritance, specify the class Sales giving the details of the constructor( ), void compute( ) and void show( ). The super class, main function and algorithm need NOT be written. class Sales extends Product int day; double tax; double totamt; Sales (String n, int c; double p, int d) Super (n, c, p); day = d; tax = 0.0; totamt = 0.0;

17 void compute( ) tax = (amount*12.4)/100; int fine = 0; if (day > 30) fine = (amount * 2.5)/100; totamt = amount + tax + fine; void show( ) System.out.println("Name of the product:" + name); System.out.println("Product code :" + code); System.out.println("Sale amount of the product :" +amount); System.out.println("Total amount :" + totamt); 11. Queue is an entity which can hold a maximum of 100 integers. The queue enables the user to add integers from the rear and remove integers from the front. Define a class Queue with the following details Class Name Product Ans. Data members/instance variables Que[ ] size front rear Member functions Queue (int mm) void addele(int v) int delele( ) void display( ) array to hold the integer elements stores the size of the array to point the index of the front to point the index of the rear constructor to initialize the data size = mm, front = 0, rear = 0 to add integer from the rear if possible else display the message Overflow returns elements from front if present, otherwise displays the message Underflow and return-9999 displays the array elements Specify the class Queue giving details of ONLY the functions void addele(int) and int delele ( ). Assume that the other functions have been defined. The main function and algorithm need NOT be written. class Queue void addele (int v) if (rear = = 1) front = 0; rear = 0; Que [rear] = v;

18 else if (rear + 1 < Que.length) Que [++rear] = v; else System.out.println("Overflow"); int delele ( ) int element; if (front = = 0) System.out.println("Underflow"); return 9999; else if (front = = rear) element = Que [front]; front = rear = 0; return element; else return (Que [front ++]); ; 12. (a) A linked list is formed from the objects of the class Node. The class structure of the Node is given below : class Node int num; Node next; Write an Algorithm OR a Method to count the nodes that contain only odd integers from an existing linked list and returns the count. The method declaration is as follows : int CountOdd(Node startptr) (b) Answer the following questions from the diagram of a Binary Tree given below M N G W Y Z D F R

19 Ans. (i) Write the postorder traversal of the above tree structure. (ii) State the level numbers of the nodes N and R if the root is at 0 (zero) level. (iii) List the internal nodes of the right sub-tree. (a) Algorithm (b) CountOdd (Node startptr) Step 1 Set count = 0 (Initialise counter) Step 2 Set num = startptr (Initialise pointer) Step 3 Repeat steps 4 and 5 while num! = NULL Step 4 If num % 2! = 0 then count = count + 1 Step 5 num = next [num] /* updates pointer to point to next node */ Step 6 Return count Or Method int CountOdd(Node startptr) int count = 0 ; num = startptr; while (num! = NULL) if (num%2!= 0_ count = count + 1; num = next [num]; return count; (i) Post order traversal of tree is (ii) = NGM = W Y N Z D G M = W F Y N R Z D G M M Level 0 N G Level 1 W Y Z D Level 2 F R Level 3 Level number of the node N =1 Level number of the node R = 3 (iii) Internal nodes of the right sub-tree = G, Z

ISC 2017 COMPUTER SCIENCE (THEORY) Md. Zeeshan Akhtar

ISC 2017 COMPUTER SCIENCE (THEORY) Md. Zeeshan Akhtar ISC 27 COMPUTER SCIENCE (THEORY) Md. Zeeshan Akhtar Part I (2 marks) Answer all questions. While answering questions in this Part, indicate briefly your working and reasoning, wherever required. Question