Monday, March 12, 2012

ISC Computer Science Theory Paper - Structure

ISC Computer Science Theory Paper

PART A
(Attempt all questions)
Ques 1. Questions based on Boolean Algebra, Prepositional Logic
(5 sub-parts * 2 = 10 marks)
Ques 2. Questions based on Array address calculation, Implementation of Stacks in Algebraic
expression, interface, dynamic binding, exception, computation complexity
Ques 3. Output question have 2 parts of 5 marks each (Atleast one based on recursion)

PART 2
Section A (Attempt any three questions)

Ques 4. Question based on K-Map along with logic gate circuit (1 SOP and 1 POS)
Ques 5. Question based situation - Draw truth table, Reduce using K-Map and draw logic circuit
Ques 6. Question based on Hardware - Half and Full Adder, Decoder, Encoder and Multiplexer
Ques 7. Question on Truth table, Prepositional logic and Laws of Boolean Algebra

Section B (Attempt any two questions)
Ques 8. Object passing / general programs
Ques 9. Strings / Array
Ques 10. Recursion

Section C (Attempt any two questions)
Ques 11. Inheritance
Ques 12. Stacks / Queue
Ques 13. Binary Tree (Tree traversal) and Algorithm on Linked List

Thursday, February 23, 2012

Wednesday, February 22, 2012

MOBIUS function

The MOBIUS function M( N) for a natural number N is defined as follows:

M( N) = 1 if N = 1
= 0 if any prime factor of N is contained in N more than once
= ( -1)^ P if N is a product of p distinct prime factors
Example :

M( 78) = -1 ( for 78 = 2 * 3 * 13 M( 78) = ( -1)^ 3 = -1 )
M( 34) = 1 ( for 34 = 2 * 17 M( 34) = ( -1) ^2 = 1 )
M( 12) = 0 ( for 12 = 2 * 2 * 3 M( 12) = 0 for 2 appears two times)
M( 17) = -1 ( for 17 = 17 M( 17) = ( -1)^ 1 = -1 )
Write a program to input natural number N and output M( N).

import java.io.*;
class MobiusFn
{
BufferedReader ob = new BufferedReader(new InputStreamReader(System.in));
int n,a,p,c,q;
public MobiusFn()
{
a=2;
p=0;
c=0;
}
public void input() throws IOException
{
System.out.println("Enter a number ");
n=Integer.parseInt(ob.readLine());
}
public int primefac(int x)
{
if(x==1) return 1;
while(x>1)
{
c=0;
while(x%a==0)
{
c++;
p++;
System.out.println("Prime Fator is "+a);
x=x/a;
}
if(c>1) return 0;
a++;
}
q=(int)Math.pow(-1,p);
return q;
}
public void display()
{
int r=primefac(n);
if(r==0) System.out.println("Prime factor of N is contained in N more than once. Mobius function returns " + r );
else System.out.println("Mobius function returns value " + r);
}
public static void main(String arg[]) throws IOException
{
MobiusFn obj = new MobiusFn();
obj.input();
obj.display();
}
}

Number combination - Expected question 2012

for Input N=5
Output : Given number 5
Combinations :
1 1 1 1 1
1 1 1 2
1 2 2
1 1 3
1 4
2 3

import java.io.*;

class numcombi

{

String str="";

int a=0;

int b=0;

public void combi() throws IOException

{

BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

System.out.print("Enter a number : ");

int num = Integer.parseInt(br.readLine());

System.out.println("Combinations :");

for(int i=1;i<=num/2;i++)

{

str="";

extract(num,i);

}

}

public void extract(int n, int m)

{

a=m;

str+=" "+a;

b=n-a;

System.out.println(str+" "+b);

if(b<=m)

{

return;

}

extract(b,m);

}

public static void main(String arg[]) throws IOException

{

numcombi obj= new numcombi();

obj.combi();

}

}

Sunday, February 19, 2012

Sum of two Binary numbers

import java.util.*;
class BinarySum
{
public int n,m,p,q,r;
public void input()
{
// two binary numbers added by converting into decimal
// and sum then converted into binary
Scanner br = new Scanner(System.in);
System.out.println(" Input a binary number n");
n=br.nextInt();
System.out.println(" Input a binary number m");
m=br.nextInt();
}
public void calculate()
{
p=bintodec(n);
q=bintodec(m);
if(p== -1 || q== -1)
System.out.println("INVALID BINARY VALUE ");
else
{
System.out.println("Binary number " + n + " in decimal is " + p);
System.out.println("Binary number " + m + " in decimal is " + q);
r=sumbin(p+q);
System.out.println("Sum of two binary number " + n + " and " + m + " is " + r);
}
}
public int bintodec(int x)
{
int a,s=0, i=0;
while(x>0)
{
a=x%10;
if(a>1) return -1;
s = s + (int)Math.pow(2,i)*a;
x=x/10;
i++;
}
return s;
}
public int sumbin(int x)
{
int a,s=0, i=0;
while(x>0)
{
a=x%2;
s = s + (int)Math.pow(10,i)*a;
x=x/2;
i++;
}
return s;
}
public static void main(String arg[])
{
BinarySum ob= new BinarySum();
ob.input();
ob.calculate();
}
}

Two decimal numbers converted into binary and sum in binary form

import java.util.*;
class BinaryAdd
{
public int n,m,p,q,r;
public void input()
{
// two decimal numbers converted into binary
// and sum is also converted into binary
Scanner br = new Scanner(System.in);
System.out.println(" Input a number n");
n=br.nextInt();
System.out.println(" Input a number m");
m=br.nextInt();
}
public void calculate()
{
p=decbin(n);
q=decbin(m);
System.out.println("Binary value for " + n + " is " + p);
System.out.println("Binary value for " + m + " is " + q);
r=sumbin(m+n);
System.out.println("Sum of two binary number " + p + " and " + q + " is " + r);
}
public int decbin(int x)
{
int a,s=0, i=0;
while(x>0)
{
a=x%2;
s = s + (int)Math.pow(10,i)*a;
x=x/2;
i++;
}
return s;
}
public int sumbin(int x)
{
int a,s=0, i=0;
while(x>0)
{
a=x%2;
s = s + (int)Math.pow(10,i)*a;
x=x/2;
i++;
}
return s;
}
public static void main(String arg[])
{
BinaryAdd ob= new BinaryAdd();
ob.input();
ob.calculate();
}
}

Decimal number to binary conversion

import java.util.*;
class dectobin
{
public int n,m,s,i,a;
public void input()
{
Scanner br = new Scanner(System.in);
System.out.println(" Input anumber n");
n=br.nextInt();
}
public void calculate()
{
s=0;
i=0;
m=n;
while(n>0)
{
a=n%2;
s = s + (int)Math.pow(10,i)*a;
n=n/2;
i++;
}
System.out.println("Binary value for " + m + " is " + s);
}
public static void main(String arg[])
{
dectobin ob= new dectobin();
ob.input();
ob.calculate();
}
}

ISC Practical ( Expected )

ISC Practical ( Expected )

1. Binary Conversions
2. Hamming Distance
3. Mobius function
4. Sorting Rows / Columns of 2D Array
5. Finding maximum and minimum value of a 2D Array
6. Reversing words of a sentence
7. Sorting of words of a sentence

Sort words of a sentence

Read a single sentence which terminates with a full stop(.). The words are to be separated with a single blank space and are in lower case. Arrange the words contained in the sentence according to the length of the words in ascending order. If two words are of the same length then the word occurring first in the input sentence should come first. For both, input and output the sentence must begin in upper case.

Test your program for the following data and some random data.

INPUT : The lines are printed in reverse order.

OUTPUT : In the are lines order printed reverse.

INPUT : Print the sentence in ascending order.

OUTPUT : In the print order sentence ascending.


import java.io.*;

import java.util.*;

class sortlength

{

public static void main(String args[])throws IOException

{

String t,a;

int i, k,j=0, x=0, l, l1, l2, l3;

BufferedReader ob=new BufferedReader(new InputStreamReader(System.in));

System.out.println("Enter the sentence");

a=ob.readLine();

a=a.toLowerCase();

a=a.substring(0,a.length()-1);

StringTokenizer str=new StringTokenizer(a);

l=str.countTokens();

String z[ ]=new String[l];

for(i=0;i

z[j++]=str.nextToken( );

for(i=0;i

{

for(k=0;k<(l-1);k++)

{ l2=z[k].length();

l3=z[k+1].length();

if(l2>l3)

{ t=z[k];

z[k]=z[k+1];

z[k+1]=t;

}

}

}

System.out.print((char)(z[0].charAt(0)-32)+z[0].substring(1)+" ");

for(i=1;i

System.out.print(z[i]+" ");

System.out.print(z[j-1]+".");

}

}

Friday, February 17, 2012

Program to convert a number in to words

Input a number - 568
Output - Five Hundred Sixty Eight


import java.util.*;
class numtoword
{
public static void main(String arg[])
{
String a[]={"", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Ninteen"};
String b[]={"", "", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninty"};
Scanner ob=new Scanner(System.in);
int n,m,p,q,r;
System.out.println("Enter a number less than 1000");
n=ob.nextInt();
if(n<1000)
{
m=n/100;
p=n%100;
if(m>0) System.out.print(a[m] + " Hundred ");
else System.out.print("");
if(p<20) System.out.println(a[p]);
else
{
q=p%10;
r=p/10;
System.out.println(b[r] + " " + a[q]);
}
}
}
}