April 2011 ~ Ajinkya Mandhare

Friday, 29 April 2011

Tagged under: lexical analyzer, Programs, Source codes, TE VI SEM

Lexical Analyzer

LEXICAL ANALYZER
Download the program for Lexical Analyzer from the download box alongside

Anonymous 12:57 pm No comment

Tagged under: TE VI SEM

OOSE Practicals

Question Bank for OOSE Practical Exam

1. Airline Reservation System

2. Railway Reservation System

3. Nokia Gallery

4. Banking System

5. Big Bazaar

6. Bus Reservation System

7. Movie Ticket Reservation

8. Hospital Management System

9. Star Plus Serial Schedule Management System

10. SUC Event Management

11. Hotel Management

12. IPL Matches

Following diagrams will be asked based on above systems…

1. Class Diagram

2. Sequence Diagram

3. Collaboration Diagram

4. State Transition Diagram

5. Activity Diagram

6. Use Case Diagram

Only System name would be given…

Example: Class diagram for Hospital Management System.

Note: Students first have to write the problem statement.. Then draw d diagram on answer sheet nd then perform pracs..

Anonymous 12:38 pm 2 comments

Thursday, 28 April 2011

Tagged under: Android operating system, featured, PPT, Seminar, TE VI SEM

ANDROID SEMINAR

Android Seminar

Anonymous 5:54 pm No comment

Tagged under: Assemblers, system programming and compiler construction, TE VI SEM

Assemblers (Pass1)

Programs for Pass1 assembler:
1. Pass 1(a):

import java.io.*;
import java.util.*;
class Pass1a
{public static void main(String args[])throws IOException
{
int i=0;
FileWriter fw=new FileWriter("Input.txt");
Scanner src=new Scanner(System.in);
System.out.println("Enter code");
String code=src.next();
char code1[]=new char[500];
code1=code.toCharArray();
int n=code.length();

for( i=0;i<n;++i)
{
while((code1[i]!='|')&&i<n-1)
{
if(code1[i]==';')
{fw.write("\n");
++i;
continue;

}
fw.write(code1[i]);
++i;
}
fw.write(" ");
}

fw.close();
FileReader fr=new FileReader("Input.txt");
BufferedReader br=new BufferedReader(fr);
String input;
while((input=br.readLine())!=null)
System.out.println(input);
String token="";
i=0;
String prg[][] = new String[30][4];
int lines=0;
input="";
FileReader fr1=new FileReader("Input.txt");
BufferedReader br1=new BufferedReader(fr1);

while((input=br1.readLine())!=null)
{
int l=input.length();
i=0;
token = "";

while(i<l)
{
		if(input.charAt(i)!=' ')
			token+=input.charAt(i);
		else break;
			i++;
}

if(token=="") token="   ";
	prg[lines][0]=token;

token = "";

while(i<l)
		if(input.charAt(i)==' ')
		i++;
		else break;

while(i<l)
	{
		if(input.charAt(i)!=' ')
			token+=input.charAt(i);
		else break;
		i++;
	}

if(token=="") token=" ";
	prg[lines][1]=token;

while(i<l)
		if(input.charAt(i)==' ') 
		i++;
		else break;

token = "";

while(i<l)
	{
		if(input.charAt(i)!=' ' && input.charAt(i)!=',')
			token+=input.charAt(i);
		else break;
		i++;
	}

if(token=="") token=" ";
	prg[lines][2]=token;

while(i<l)
		if(input.charAt(i)==' ' || input.charAt(i)==',') i++;
		else break;

token = "";

while(i<l)
	{
		if(input.charAt(i)!=' ')
			token+=input.charAt(i);
		else break;
		i++;
	}

if(token=="") token="";
	prg[lines][3]=token;

lines++;
	i=l;

}
int j;

System.out.println("\n");
for(i=0;i<lines;i++)
{
	for(j=0;j<4;j++)
		System.out.print(prg[i][j]+"\t");
	System.out.println();
}

}}

2. Pass 1(b):

import java.io.*;
import java.util.*;
class Pass1b
{public static void main(String args[])throws IOException
{
int i=0;
FileWriter fw=new FileWriter("Input.txt");
Scanner src=new Scanner(System.in);
System.out.println("Enter code");
//int n=src.nextInt();
//St s[]=new St[n];
String code=src.next();
char code1[]=new char[500];
code1=code.toCharArray();
//for(int i=0;i<10;++i)
//System.out.println(code1[i]);
int n=code.length();
System.out.println(n);

for( i=0;i<n;++i)
{
while((code1[i]!='|')&&i<n-1)
{
if(code1[i]==';')
{fw.write("\n");
++i;
continue;

}
//System.out.println(i);
fw.write(code1[i]);
++i;
}
fw.write(" ");
}

fw.close();
FileReader fr=new FileReader("Input.txt");
BufferedReader br=new BufferedReader(fr);
String input;
while((input=br.readLine())!=null)
System.out.println(input);
//fr.close();
String token="";
i=0;
String prg[][] = new String[30][4];
int lines=0;
input="";
FileReader fr1=new FileReader("Input.txt");
BufferedReader br1=new BufferedReader(fr1);

while((input=br1.readLine())!=null)
{
int l=input.length();
i=0;
token = "";

while(i<l)
{
		if(input.charAt(i)!=' ')
			token+=input.charAt(i);
		else break;
			i++;
}

if(token=="") token=" ";
	prg[lines][0]=token;

token = "";

while(i<l)
		if(input.charAt(i)==' ') i++;
		else break;

while(i<l)
	{
		if(input.charAt(i)!=' ')
			token+=input.charAt(i);
		else break;
		i++;
	}

if(token=="") token=" ";
	prg[lines][1]=token;

while(i<l)
		if(input.charAt(i)==' ') i++;
		else break;

token = "";

while(i<l)
	{
		if(input.charAt(i)!=' ' && input.charAt(i)!=',')
			token+=input.charAt(i);
		else break;
		i++;
	}

if(token=="") token=" ";
	prg[lines][2]=token;

while(i<l)
		if(input.charAt(i)==' ' || input.charAt(i)==',') i++;
		else break;

token = "";

while(i<l)
	{
		if(input.charAt(i)!=' ')
			token+=input.charAt(i);
		else break;
		i++;
	}

if(token=="") token="";
	prg[lines][3]=token;

lines++;
	i=l;

}
int j;
System.out.println("\n");
for(i=0;i<lines;i++)
{
	for(j=0;j<4;j++)
		System.out.print(prg[i][j]+"\t");
	System.out.println();
}

String mnemonics[] = new String[10];
int mlen[] = new int[10];
int mno = 0;

mnemonics[mno]="L";
mlen[mno++]=4;
mnemonics[mno]="A";
mlen[mno++]=4;
mnemonics[mno]="ST";
mlen[mno++]=4;

String pseudo[] = new String[10];
int pno=0;

pseudo[pno++]="START";
pseudo[pno++]="BALR";
pseudo[pno++]="USING";
pseudo[pno++]="DC";
pseudo[pno++]="DS";
pseudo[pno++]="END";

System.out.println();
System.out.println("-------------------------------------------");
System.out.println("Mnemonics opcode Table is : ");
System.out.println();

System.out.println("Mnemonics\tLength");
System.out.println();

for(i=0;i<mno;i++)
{
	System.out.print(mnemonics[i]+"\t\t  "+mlen[i]);
	System.out.println();
}

System.out.println();
System.out.println("-------------------------------------------");
System.out.println("Pseudo opcode Table is : ");
System.out.println();

System.out.println("Psuedo\tRoutine Adderess");
System.out.println("Opcode");
System.out.println();

for(i=0;i<pno;i++)
{
	System.out.print(pseudo[i]+"\t\t"+i);
	System.out.println();
}

String symbol[] = new String[20];
int sym_value[] = new int[20];
int symno = 0;

int lc=0;
for(i=0;i<lines;i++)
{
	token = prg[i][0];
	if(token!=" ")
	{
		symbol[symno] = token;
		sym_value[symno++] = lc;
	}

token = prg[i][1];

if(token.equalsIgnoreCase("DC")) lc=lc+4;
	else if(token.equalsIgnoreCase("DS")) lc=lc+4;
	else
	{
		for(j=0;j<mno;j++)
		{
			if(token.equals(mnemonics[j])) 
			{
				lc=lc+mlen[j];
				break;
			}
		}
	}

}

System.out.println();
System.out.println("-------------------------------------------");
System.out.println("Symbol Table is : ");
System.out.println();

System.out.println("Symbol\tValue");
System.out.println();

for(i=0;i<symno;i++)
{
	System.out.print(symbol[i]+"\t"+sym_value[i]);
	System.out.println();
}
}}

3. Pass 1(c):

import java.io.*;
import java.util.*;
class Pass1c
{public static void main(String args[])throws IOException
{
int i=0;
FileWriter fw=new FileWriter("Input.txt");
Scanner src=new Scanner(System.in);
System.out.println("Enter code with | as space & ; for next line");
String code=src.next();
char code1[]=new char[500];
code1=code.toCharArray();
int n=code.length();

for( i=0;i<n;++i)
{
while((code1[i]!='|')&&i<n-1)
{
if(code1[i]==';')
{fw.write("\n");
++i;
continue;

}
fw.write(code1[i]);
++i;
}
fw.write(" ");
}

while((input=br1.readLine())!=null)
{
int l=input.length();
i=0;
token = "";

while(i<l)
{
		if(input.charAt(i)!=' ')
			token+=input.charAt(i);
		else break;
			i++;
}

if(token=="") token=" ";
	prg[lines][0]=token;

token = "";

while(i<l)
		if(input.charAt(i)==' ') i++;
		else break;

while(i<l)
	{
		if(input.charAt(i)!=' ')
			token+=input.charAt(i);
		else break;
		i++;
	}

if(token=="") token=" ";
	prg[lines][1]=token;

while(i<l)
		if(input.charAt(i)==' ') i++;
		else break;

token = "";

while(i<l)
	{
		if(input.charAt(i)!=' ' && input.charAt(i)!=',')
			token+=input.charAt(i);
		else break;
		i++;
	}

if(token=="") token=" ";
	prg[lines][2]=token;

while(i<l)
		if(input.charAt(i)==' ' || input.charAt(i)==',') i++;
		else break;

token = "";

while(i<l)
	{
		if(input.charAt(i)!=' ')
			token+=input.charAt(i);
		else break;
		i++;
	}

if(token=="") token="";
	prg[lines][3]=token;

lines++;
	i=l;

}
int j;
System.out.println("\n");
for(i=0;i<lines;i++)
{
	for(j=0;j<4;j++)
		System.out.print(prg[i][j]+"\t");
	System.out.println();
}

String mnemonics[] = new String[10];
int mlen[] = new int[10];
int mno = 0;

mnemonics[mno]="L";
mlen[mno++]=4;
mnemonics[mno]="A";
mlen[mno++]=4;
mnemonics[mno]="ST";
mlen[mno++]=4;

String pseudo[] = new String[10];
int pno=0;

pseudo[pno++]="START";
pseudo[pno++]="BALR";
pseudo[pno++]="USING";
pseudo[pno++]="DC";
pseudo[pno++]="DS";
pseudo[pno++]="END";

System.out.println();
System.out.println("-------------------------------------------");
System.out.println("Mnemonics opcode Table is : ");
System.out.println();

System.out.println("Mnemonics\tLength");
System.out.println();

for(i=0;i<mno;i++)
{
	System.out.print(mnemonics[i]+"\t\t  "+mlen[i]);
	System.out.println();
}

System.out.println();
System.out.println("-------------------------------------------");
System.out.println("Pseudo opcode Table is : ");
System.out.println();

System.out.println("Psuedo\tRoutine Adderess");
System.out.println("Opcode");
System.out.println();

for(i=0;i<pno;i++)
{
	System.out.print(pseudo[i]+"\t\t"+i);
	System.out.println();
}

String symbol[] = new String[20];
int sym_value[] = new int[20];
int symno = 0;
String literal[] = new String[20];
int lit_value[] = new int[20];
int litno = 0;
String lit="";
int lc=0;
for(i=0;i<lines;i++)
{
	token = prg[i][0];
	lit=prg[i][2];
	if(lit!="")
	{
	if(lit.charAt(0)=='=')
	{
   		literal[litno++] = lit;
 	}
	}

token = prg[i][1];

}

for(i=0;i<litno;i++)
{
lit=literal[i];

if(lit.charAt(1)=='F'||lit.charAt(1)=='f')
	{
   		lit_value[i] = lc;
		lc=lc+4;
	}

if(lit.charAt(1)=='H'||lit.charAt(1)=='h')
	{
   		lit_value[i] = lc;
		lc=lc+2;
	}

if(lit.charAt(1)=='D'||lit.charAt(1)=='d')
	{
   		lit_value[i] = lc;
		lc=lc+8;
	}
	
}
System.out.println();
System.out.println("-------------------------------------------");
System.out.println("Literal Table is : ");
System.out.println();

System.out.println("Literal\tValue");
System.out.println();

for(i=0;i<litno;i++)
{
	System.out.print(literal[i]+"\t"+lit_value[i]);
	System.out.println();
}

}}

NOTE:While entering the code use '|' as space & ';' for next line.

Anonymous 1:07 pm No comment

Wednesday, 20 April 2011

Tagged under: Android operating system, featured, Report, Seminar, TE VI SEM

Seminar Report

Android Operating System

Final Android

Anonymous 9:46 pm No comment

Thursday, 14 April 2011

Tagged under: TE VI SEM

ACN Practical exam list

ACN Practical Exam 2011

1. Install Web Server

Site Name:
Contents of the site:

2. Implement Virtual Hosting (name based)

IP Address:

Site Name:
Contents of the Site:

3. Implement Virtual Hosting (IP based)

IP Address:
Site Name:
Contents of the Site:

4. Analysis of packet captured (UDP)

Mention the port numbers of the captured packets

5. Analysis of packet captured (TCP)

Mention the flag status of the captured packets

6. Analysis of packet captured (ARP)

Mention the IP address of the source and destination hosts

7. Analysis of packet captured (IP)

Mention the IP address of the source and destination hosts

8. Implement Sliding Window Protocol (Go-Back-N)

9. Implement Sliding Window Protocol (Selective Repeat)

10. Implement Group Chat Server (2 Clients)

11. Implement Group Chat Server (3 Clients)

12. Implement multicast routing protocol DVMRP with following topology and groups

Node ____ is the source.

CBR traffic starts at ______ time.

Node ____ joins the group at _______ time.

Node ____ leaves the group at _______ time.

Anonymous 9:43 am No comment

Friday, 8 April 2011

Tagged under: Advanced Microprocessors, Superscaler Architecture, TE VI SEM

SUPERSCALAR ARCHITECTURE in terms of Pipelining

Superscaler Program

/***********To Study The SUPERSCALAR ARCHITECTURE in terms of Pipelining****************/

import java.util.*;
class  superscalar
{
	public static void main(String[] args) 
	{
		int n,i,j,k;
		Scanner sc=new Scanner(System.in);
		System.out.print("Enter number of instructions: ");
		n=sc.nextInt();
		if(n<=0)
		{
		System.out.println("No instructions available for execution!!!");
		System.exit(0);
		}
		System.out.println("Fetch\tDecode\tExecute\tWrite");
		for(i=1;i<=n;i++)
		{
			if(i%2==0)
			      even(i);
			else
			       odd(i);   
		}
		
		for(i=n-1;i>=(n-5)&&i>0;i--)
		{
			System.out.print("   \t"+i);
			i--;
		}
		System.out.println();
		for(i=n;i>=(n-4)&&i>0;i--)
		{
			System.out.print("   \t"+i);
			i--;
		}
		System.out.println();

System.out.print("   \t");
		for(i=n-1;i>=(n-3)&&i>0;i--)
		{
			System.out.print("   \t"+i);
			i--;
		}
		System.out.println();
		
		System.out.print("   \t");
		for(i=n;i>=(n-2)&&i>0;i--)
		{
			System.out.print("   \t"+i);
			i--;
		}
		System.out.println();

if((n-1)>0)
			System.out.println("   \t\t\t"+(n-1));
		System.out.println("   \t\t\t"+n);
	}

static void even(int i)
	{
		int count=0,l;

for(l=i;l>=2;l--)
		{
			if(count==4)
			{
//                                System.out.println();
				break;
			}
			if(l%2==0)
			{
				System.out.print(l+"   \t");
				count++;
			}
		}
		System.out.println();
	}

static void odd(int i)
	{
		int count=0,l;
		for(l=i;l>=1;l--)
		{
			if(count==4)
			{
//                                System.out.println();
				break;
			}
			if(l%2==1)
			{
				System.out.print(l+"   \t");
				count++;
			}
		}
		System.out.println();
	}
}
/******************OUTPUT*****************
Enter number of instructions: 5
Fetch   Decode  Execute Write
1
2
3       1
4       2
5       3         1
        4         2
        5         3       1
                  4       2
                  5       3
                          4
                          5
*/

Anonymous 8:04 pm No comment

Tagged under: Data and Web Mining, TE VI SEM

Data & Web Mining

DWM Programs in java
Download the Data & Web Mining programs from the download box alongside.

Anonymous 8:02 pm No comment

Wednesday, 6 April 2011

Tagged under: Advanced Microprocessors, Arithmetic Pipelining, TE VI SEM

Arithmetic Pipelining for multiplicaion

ARITHMETIC PIPELINING
This the final prog for arithmetic pipelining. It works upto 8bit X 8bit multiplication. There were some minor changes in the datatypes of some variables as compared to the previous prog.

//Arithmetic Pipelining Program by Ajinkya
#include<iostream.h>
#include<math.h>
#include<conio.h>
#include<stdio.h>

int a1[20],a2[20],x[20],p[8][20],q[8][20],q2[8][20],i,j,k;
long q3[8][20];
int a,b;
void dectobin(int a, int a1[]);
void csa(int p[8][20]);
void cpa(int p[8][20]);

void main()
{
for(i=0;i<20;++i)
a1[i]=a2[i]=x[i]=0;
for(i=0;i<8;++i)
for(j=0;j<20;++j)
{p[i][j]=0;
q2[i][j]=0;
q3[i][j]=0;
}
clrscr();
cout<<"Enter multiplicand";
cin>>a;
cout<<"Enter multiplier";
cin>>b;
cout<<"A= ";
dectobin(a,a1);
for(i=0;i<8;++i)
cout<<a1[i];
cout<<endl<<"B= ";
dectobin(b,a2);
for(i=0;i<8;++i)
cout<<a2[i];
cout<<endl;
for(i=0;i<8;++i)
{
for(j=15;j>16-i-1;--j)
p[i][j]=0;
if(a2[i]==1)
{
k=0;
for(j=i;j<16;++j)
p[i][16-j-1]=a1[k++];
}
else
{for(j=i;j<16;++j)
p[i][16-j-1]=0;
}
}
for(i=0;i<8;++i)
{
cout<<"P"<<i<<" = ";
for(j=0;j<16;++j)
cout<<p[i][j];
cout<<endl;
}
i=0;
j=0;

csa(p);
i=i+3;
j++;
csa(p);
int z;
for(j=0;j<3;++j)
for(z=0;z<16;++z)
{if(j==0||j==1)
 {p[2*j][z]=q[j][z];
  p[2*j+1][z]=q2[j][z];
 }
 else
 {p[2*j][z]=p[3*j][z];
  p[2*j+1][z]=p[3*j+1][z];
 }
}
i=0;
j=0;
csa(p);
i=i+3;
++j;
csa(p);
for(j=0;j<3;++j)
for(z=0;z<16;++z)
{if(j==0||j==1)
 {p[2*j][z]=q[j][z];
  p[2*j+1][z]=q2[j][z];
 }
}
i=0;
j=0;
csa(p);
for(j=0;j<2;++j)
for(z=0;z<16;++z)
{if(j==0)
 {p[2*j][z]=q[j][z];
  p[2*j+1][z]=q2[j][z];
 }
 else
 p[2*j][z]=p[3*j][z];
}
i=0;j=0;
csa(p);
for(j=0;j<1;++j)
for(z=0;z<16;++z)
{p[2*j][z]=q[j][z];
 p[2*j+1][z]=q2[j][z];
}
i=0;j=0;
cpa(p);

getch();
}
void dectobin(int a, int a1[])
{i=0;
while(a>0)
{
x[i]=a%2;
a=a/2;
++i;
}
--i;
j=0;
while(i>=0)
{
a1[i]=x[i];

--i;
}
}
void csa(int p[8][20])
{
cout<<"CSA";
int z;
cout<<"Exor"<<endl;
 for(z=0;z<16;++z)
 {if((p[i][z]==p[i+1][z]==p[i+2][z]==0)||(p[i][z]==p[i+1][z]==1&&p[i+2][z]==0)||(p[i][z]==0&&p[i+1][z]==p[i+2][z]==1)||(p[i+1][z]==0&&p[i][z]==p[i+2][z]==1))
 q[j][z]=0;
 else
 q[j][z]=1;
 cout<<q[j][z];
 }

q2[j][15]=0;
 int r=0;
cout<<"Carry"<<endl;
 for(z=15;z>0;--z)
 {
 if((p[i][z]==0&&p[i+1][z]==0&&p[i+2][z]==0)||(p[i][z]==0&&p[i+1][z]==0)||(p[i+1][z]==0&&p[i+2][z]==0)||(p[i][z]==0&&p[i+2][z]==0))
{ //cout<<endl<<p[i][z]<<p[i+1][z]<<p[i+2][z];

q2[j][z-1]=0;
}  else
  q2[j][z-1]=1;

}

for(z=0;z<16;++z)
   cout<<q2[j][z];
   cout<<endl;

}

void cpa(int p[8][20])
{
cout<<"CPA";
cout<<endl<<"P= ";
   int z;
   int c=0;
   for(z=15;z>=0;--z)
   {if(q[j][z]==0&&q2[j][z]==0&&c==0)
    {q3[j][z]=0;
    }
    else
    {if((q[j][z]==1&&q2[j][z]==0&&c==0)||(q2[j][z]==1&&q[j][z]==0&&c==0)||(q[j][z]==0&&q2[j][z]==0&&c==1))
    { q3[j][z]=1;
       c=0;
     }
     else
     {if((q[j][z]==1&&q2[j][z]==1&&c==0)||(q[j][z]==1&&q2[j][z]==0&&c==1)||(q[j][z]==0&&q2[j][z]==1&&c==1))
      {q3[j][z]=0;
	c=1;
       }
      else
      {q3[j][z]=1;
	c=1;
       }
     }
    }
   }
   if(c==1)
   q3[j][16]=1;
cout<<"Sum: ";
   for(z=0;z<16;++z)
   cout<<q3[j][z];
   cout<<endl;
cout<<"Carry: "<<c;
for(z=0;z<16;++z)
{q3[j][z]=abs(q3[j][z])*pow(2,15-z);
}
float ans=0;
for(z=0;z<16;++z)
ans=ans+q3[j][z];
cout<<endl<<"Ans "<<ans;

}
/*******OUTPUT**************
Enter Multiplicand 181
Enter Multiplier 147

A=10110101
B=10010011

P0=000000010110101
P1=000000111101010
P2=000000000000000
P3=000000000000000
P4=000101101010000
P5=000000000000000
P6=000000000000000
P7=101101010000000

CSA
Exor: 000000111011111
Carry: 0000000001000000

CSA
Exor: 0000101101010000
Carry: 0000000000000000

CSA
Exor: 0000101011001111
Carry: 0000001010100000

CSA
Exor: 0101101010000000
Carry: 0000000000000000

CSA
Exor: 0101001011101111
Carry: 0001010100000000

CSA
Exor: 0100011111101111
Carry: 0010000000000000

CPA
Sum= 0110011111101111
Carry=0
 
P=0110011111101111
Ans=26607
***************************/

Anonymous 8:42 pm No comment

Saturday, 2 April 2011

Tagged under: K-mean, Nearest Neighbour, TE VI SEM

K-Mean & Nearest Neighbour Programs

Kmean & Nearest Neighbour Prog in java

1. Kmean:

#include<iostream.h>
#include<conio.h>
#include<math.h>
void main()
{
 clrscr();
 int m1,m2,i,j,n,x,y,sum1,sum2,k;
 int d[100],k1[50],k2[50];
 cout<<"Enter the no. of items in cluster:";
 cin>>n;
 cout<<"Enter the items of the cluster:";
 for(i=0;i<n;i++)
 cin>>d[i];
 for(i=0;i<n;i++)
 {
  k1[i]=k2[i]=0;
 }
 m1=d[0];
 m2=d[1];
 k=1;
 while(1)
 {
 x=0;
 y=0;
 for(i=0;i<n;i++)
 {
  if(abs(m1-d[i]) <= abs(m2-d[i]))
  {
   k1[x++]=d[i];
  }
  else
  k2[y++]=d[i];
 }
 cout<<"\n\n\n** Iteration:"<<k<<" **\n"<<endl;
 cout<<"cluster1={";
 for(i=0;i<x;i++)
 {
 cout<<k1[i]<<",";
 }
 cout<<"}"<<endl;
 cout<<"cluster2={";
 for(i=0;i<y;i++)
 {
 cout<<k2[i]<<",";
 }
 cout<<"}"<<endl;
 sum1=sum2=0;
 for(i=0;i<x;i++)
 sum1=sum1+k1[i];
 sum1=sum1/x;
 for(i=0;i<y;i++)
 sum2=sum2+k2[i];
 sum2=sum2/y;
 if(sum1==m1 && sum2==m2)
 break;
 else
 {
  m1=sum1;
  m2=sum2;
 }
 k++;
 }
 getch();
}

2. Nearest Neighbour:

#include<iostream.h>
#include<conio.h>
void main()
{
 clrscr();
 int m,i,j,n,t,p,l,iflag,jflag;
 int a[10][10];
 char k[10],k1[10],k2[10];
 cout<<"Enter the no. of tuples:";
 cin>>n;
 cout<<"Enter the name of the nodes:";
 for(i=0;i<n;i++)
 {
  cin>>k[i];
 }
 cout<<"Enter the adjacent matrix:\n\n";
 cout<<" ";
 for(i=0;i<n;i++)
 cout<<" "<<k[i];
 cout<<endl;
 for(i=0;i<n;i++)
 {
  cout<<endl<<k[i]<<" ";
  for(j=0;j<n;j++)
  cin>>a[i][j];
 }
 cout<<endl;
 t=1;
 while(1)
 {
 cout<<"** Threshold="<<t<<" **\n\n";
 m=0;
 iflag=1;
 jflag=1;
 for(i=0;i<n;i++)
 for(j=0;j<n;j++)
 {
  if(a[i][j]<=t && a[i][j]>0 && i<j)
  {
   cout<<k[i]<<" is connected to "<<k[j]<<" having distance of "<<a[i][j]<<endl;
   for(l=0;l<m;l++)
   {
    if(k[i]==k1[l])
    {
    iflag=0;
    break;
    }
    else
    {
    iflag=1;
    }
   }
   for(l=0;l<m;l++)
   {
    if(k[j]==k1[l])
    {
    jflag=0;
    break;
    }
    else
    {
    jflag=1;
    }
   }
   if(iflag==1)
   k1[m++]=k[i];
   if(jflag==1)
   k1[m++]=k[j];
  }
 }
 p=0;
 for(i=0;i<n;i++)
 {
  if(k[i]!=k1[i])
  {
   k2[p++]=k[i];
  }
 }
 cout<<"k1={";
 for(i=0;i<m;i++)
 cout<<k1[i]<<",";
 cout<<"}\n";
 cout<<"k2={";
 for(i=0;i<p;i++)
 cout<<k2[i]<<",";
 cout<<"}\n";
 if(m==n)
 break;
 t++;
 cout<<"\n\n";
}
 getch();
}

Anonymous 11:10 am No comment

Tagged under: Android operating system, TE VI SEM

Seminar

ANDROID report
Download the report from the download box alongside

Points for seminar:
1) intro
2) Innovative features
3) open source

4) arch