Computer Engineering Made Easy: 2013

Wednesday, 18 December 2013

Semester 8 Syllabus for Computer Engineering Mumbai University (MU)

Download Sem 8 Syllabus for Computer Engineering

Friday, 8 November 2013

Water Jug Problem Java Code

Use the DOWNLOAD BOX to download the file 'wj2.java'.

View the Java Code for Water Jug Problem.

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Wumpus World Problem Java Code

Use the DOWNLOAD BOX provided in the blog to download the file 'wumpus.java'.

View the Java Code for Wumpus World Problem.

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Friday, 18 October 2013

Semester 7 : Digital Signal and Image Processing (DSIP) MATLAB codes

The MATLAB codes are available in the 'dsip.zip' file.
Download this file from the download box given in the blog.

It contains MATLAB codes for :-
1.IGS
2.Ideal High pass filter
3.Gaussian High pass filter
4.Butterworth High pass filter
5.Ideal Low pass filter
6.Gaussian Low pass filter
7.Butterworth Low pass filter

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Friday, 20 September 2013

Mobile Computing Project : WML (Wireless Markup Language) mini project on 'E-Commerce website' free download

This is a mini project in wml on an online shop. You can Donload the project from the DOWNLOAD BOX.Download the file 'wmlproject.zip' from the download box.

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Monday, 16 September 2013

Robotics and AI : Wumpus world problem Java Code/Program.

Given below is the java program for "Wumpus world problem".
The program gives a perfect implementation of this problem.It accepts the environment from the user dynamically, and then the agent tries to explore the best possible path to reach the gold, preventing the pits and the wumpus.
This is a perfect code with sample outputs for two scenarios/environments.

VIEW OUR SPECIAL PAGE ON, " ROBOTICS AND ARTIFICIAL INTELLIGENCE " FOR SEM 7 STUDENTS - Robotics and Artificial Intelligence

// Aim:To implement Wumpus world problem in Java
//Program:

import java.util.*;
class Environment
{
Scanner scr=new Scanner(System.in);
//char w[][]=new char[5][5];
int np;     //number of pits
int wp,gp; // wumpus position   gold position
int pos[]; // position of pits
int b_pos[]=new int[20];
int s_pos[]=new int[20];
void accept(String w[][])
{
for(int i=0;i<20;++i)
{
b_pos[i]=-1;
s_pos[i]=-1;
}

for(int i=0;i<5;++i)
for(int j=0;j<5;++j)
w[i][j]="";

int count=1;
System.out.println("\n\n********* Wumpus World Problem *********\n-by Aditya Mandhare.\n");

System.out.println("The positions are as follows.");
for(int i=1;i<=4;++i)
{
System.out.println("\n-----------------------------------------------------------------");
System.out.print("|\t");
for(int j=1;j<=4;++j)
System.out.print((count++)+"\t|\t");
}
System.out.println("\n-----------------------------------------------------------------");
System.out.println("\nAgent start position: 13");
w[4][1]="A";
System.out.println("\nEnter the number of pits.");
np=scr.nextInt();
pos=new int[np];
System.out.println("Positions of pit, gold and wumpus should not overlap.");
System.out.println("Enter the position of pits.");
for(int i=0;i<np;++i)
{
pos[i]=scr.nextInt();
show_sense(pos[i],1,w);
}
System.out.println("Enter the position of wumpus.");
wp=scr.nextInt();
show_sense(wp,2,w);

System.out.println("Enter the position of gold.");
gp=scr.nextInt();

insert(w);
}

void insert(String w[][])
{
int temp=0;
int count=0;
int flag1=0,flag2=0;
for(int i=0;i<np;++i)
{
temp=pos[i];
count=0;
for(int j=1;j<=4;++j)
{
for(int k=1;k<=4;++k)
{
++count;
if(count==temp)
w[j][k]+="P";
else
if(count==gp && flag1==0)
{
w[j][k]+="G";
flag1=1;
}
else
if(count==wp && flag2==0)
{
w[j][k]+="W";
flag2=1;
}
}
}
}

display(w);
}

void show_sense(int a,int b,String w[][])
{
int t1,t2,t3,t4;
t1=a-1;
t2=a+1;
t3=a+4;
t4=a-4;

if(a==5 || a==9)
t1=0;
if(a==8 || a==12)
t2=0;
if(a==4)
t2=0;
if(a==13)
t1=0;

if(t3>16)
t3=0;
if(t4<0)
t4=0;

//int temp[]=new int[4];

if(b==1)
{b_pos[0]=t1;b_pos[1]=t2;b_pos[2]=t3;b_pos[3]=t4;}
else
if(b==2)
{s_pos[0]=t1;s_pos[1]=t2;s_pos[2]=t3;s_pos[3]=t4;}

int temp1,count;

for(int i=0;i<4;++i)
{
if(b==1)
temp1=b_pos[i];
else
temp1=s_pos[i];
count=0;
for(int j=1;j<=4;++j)
{
for(int k=1;k<=4;++k)
{
++count;
if(count==temp1 && b==1 && !w[j][k].contains("B"))
{
w[j][k]+="B";
}
else
if(count==temp1 && b==2 && !w[j][k].contains("S"))
w[j][k]+="S";
}
}
}
 
//display(w);
}
void display(String w[][])
{
System.out.println("\nThe environment for problem is as follows.\n");
for(int i=1;i<=4;++i)
{
System.out.println("\n-----------------------------------------------------------------");
System.out.print("|\t");
for(int j=1;j<=4;++j)
System.out.print(w[i][j]+"\t|\t");
}
System.out.println("\n-----------------------------------------------------------------");
}


}

class tiles
{
int safe=0;
int unsafe=0;
int wump=0;
int pit=0;
int gold=0;
int doubt_pit=0;
int doubt_wump=0;
String env;
int num=0;
int br=0;
int bl=0;
int bu=0;
int bd=0;
int visited=0;
int l,r,u,d;
String back="";
tiles(String s,int n)
{
env=s;
num=n;
l=r=u=d=0;
if(n==9 || n==5)
bl=1;

if(n==8 || n==12)
br=1;

if(n==1)
{
bu=1;bl=1;
}
if(n==13)
{
bd=1;bl=1;
}
if(n==4)
{
bu=1;br=1;
}
if(n==16)
{
bd=1;br=1;
}

}

int sense()
{
if(env.contains("B"))
return 1;
else
if(env.contains("S"))
return 2;
else
if(env.contains("G"))
return 3;
if(env.contains("W"))
return 4;
else
return 0;
}

}


class wumpus 
{
static int scream=0;
static int score=0;
static int complete=0;

static boolean check(tiles t)
{
int temp=t.sense();
if(temp==1 || temp==2)
return false;

return true;
}
public static void main(String args[])
{
Scanner scr=new Scanner(System.in);
Environment e=new Environment();
String w[][]=new String[5][5];
e.accept(w);
System.out.println("\n\nFinding the solution...");

tiles t[]=new tiles[17];
int c=1;
out:for(int i=1;i<5;++i)
{
for(int j=1;j<5;++j)
{
if(c>16)
break out;
t[c]=new tiles(w[i][j],c);
++c;
}
}

t[13].safe=1;
t[13].visited=1;

int pos=13;
int condition;
int limit=0;
String temp1,temp2;
do
{
++limit;
condition=-1;

if(t[pos].env.contains("G"))
{
complete=1;
System.out.println("Gold Found!!");
break;
}

if(t[pos].br!=1 && t[pos].r!=1 && t[pos+1].doubt_pit<1 && t[pos+1].doubt_wump<1 && t[pos+1].pit!=1 && t[pos+1].wump!=1 && !(t[pos].back.contains("r") && (t[pos].l!=1 || t[pos].u!=1 || t[pos].d!=1) && check(t[pos]) ))
{
////////////

temp1="l";
///////////
t[pos].r=1;
++pos;
System.out.println("\nfront pos="+pos);
++score;
//t[pos].visited=1;
////////////////
t[pos].back+=temp1;
////////////////
condition=t[pos].sense();
if(condition==3)
{complete=1;break;}
else
if(condition==1 && t[pos].visited==0)
{
if(t[pos].br!=1 && t[pos+1].safe!=1)
t[pos+1].doubt_pit+=1;
if(t[pos].bu!=1 && (pos-4)>=1 && t[pos-4].safe!=1)
t[pos-4].doubt_pit+=1;
if(t[pos].bl!=1 && t[pos-1].safe!=1 )
t[pos-1].doubt_pit+=1;
if(t[pos].bd!=1 && (pos+4)<=16 && t[pos+4].safe!=1)
t[pos+4].doubt_pit+=1;

t[pos].safe=1;
}
else
if(condition==2 && t[pos].visited==0)
{
if(t[pos].br!=1 && t[pos+1].safe!=1)
t[pos+1].doubt_wump+=1;
if(t[pos].bu!=1 && (pos-4)>=1 &&  t[pos-4].safe!=1)
t[pos-4].doubt_wump+=1;
if(t[pos].bl!=1 && t[pos-1].safe!=1)
t[pos-1].doubt_wump+=1;
if(t[pos].bd!=1 && (pos+4)<=16 &&  t[pos+4].safe!=1)
t[pos+4].doubt_wump+=1;

t[pos].safe=1;
}
/*else
if(condition==4)
{
score=score+100;
t[pos].safe=1;
}*/
else
if(condition==0)
t[pos].safe=1;


t[pos].visited=1;
}
else
if(t[pos].bl!=1 && t[pos].l!=1 && t[pos-1].doubt_pit<1 && t[pos-1].doubt_wump<1 && t[pos-1].pit!=1 && t[pos-1].wump!=1 && !(t[pos].back.contains("l") && (t[pos].r!=1 || t[pos].u!=1 || t[pos].d!=1) && check(t[pos]) ))
{
////////////////////
temp1="r";
///////////////////
t[pos].l=1;
pos=pos-1;
System.out.println("\nback pos= "+pos);
++score;
//t[pos].visited=1;

//////////////////////

t[pos].back+=temp1;
/////////////////////


condition=t[pos].sense();
if(condition==3)
{complete=1;break;}
else
if(condition==1 && t[pos].visited==0)
{
if(t[pos].br!=1 && t[pos+1].safe!=1)
t[pos+1].doubt_pit+=1;
if(t[pos].bu!=1 && (pos-4)>=1 &&  t[pos-4].safe!=1)
t[pos-4].doubt_pit+=1;
if(t[pos].bl!=1 && t[pos-1].safe!=1)
t[pos-1].doubt_pit+=1;
if(t[pos].bd!=1 && (pos+4)<=16 &&  t[pos+4].safe!=1)
t[pos+4].doubt_pit+=1;


t[pos].safe=1;
}
else
if(condition==2 && t[pos].visited==0)
{
if(t[pos].br!=1 && t[pos+1].safe!=1)
t[pos+1].doubt_wump+=1;
if(t[pos].bu!=1 && (pos-4)>=1 &&  t[pos-4].safe!=1)
t[pos-4].doubt_wump+=1;
if(t[pos].bl!=1 && t[pos-1].safe!=1)
t[pos-1].doubt_wump+=1;
if(t[pos].bd!=1 && (pos+4)<=16 &&  t[pos+4].safe!=1)
t[pos+4].doubt_wump+=1;

t[pos].safe=1;
}
else
if(condition==0)
t[pos].safe=1;

t[pos].visited=1;


}
else
if(t[pos].bu!=1 && t[pos].u!=1 && (pos-4)>=1 &&  t[pos-4].doubt_pit<1 && t[pos-4].doubt_wump<1 && t[pos-4].pit!=1 && t[pos-1].wump!=1 && !(t[pos].back.contains("u") && (t[pos].l!=1 || t[pos].r!=1 || t[pos].d!=1) && check(t[pos])  ))
{
/////////////////////

temp1="d";
/////////////////////
t[pos].u=1;
pos=pos-4;
System.out.println("\nUp pos= "+pos);
++score;
//t[pos].visited=1;

///////////////////////
t[pos].back+=temp1;
/////////////////////
condition=t[pos].sense();
if(condition==3)
{complete=1;break;}
else
if(condition==1 && t[pos].visited==0)
{
if(t[pos].br!=1 && t[pos+1].safe!=1)
t[pos+1].doubt_pit+=1;
if(t[pos].bu!=1 && (pos-4)>=1 &&  t[pos-4].safe!=1)
t[pos-4].doubt_pit+=1;
if(t[pos].bl!=1 && t[pos-1].safe!=1)
t[pos-1].doubt_pit+=1;
if(t[pos].bd!=1 && (pos+4)<=16 &&  t[pos+4].safe!=1)
t[pos+4].doubt_pit+=1;


t[pos].safe=1;
}
else
if(condition==2 && t[pos].visited==0)
{
if(t[pos].br!=1 && t[pos+1].safe!=1)
t[pos+1].doubt_wump+=1;
if(t[pos].bu!=1 && (pos-4)>=1 &&  t[pos-4].safe!=1)
t[pos-4].doubt_wump+=1;
if(t[pos].bl!=1 && t[pos-1].safe!=1)
t[pos-1].doubt_wump+=1;
if(t[pos].bd!=1 && (pos+4)<=16 &&  t[pos+4].safe!=1)
t[pos+4].doubt_wump+=1;

t[pos].safe=1;
}
else
if(condition==0)
t[pos].safe=1;

t[pos].visited=1;
}
else
if(t[pos].bd!=1 && t[pos].d!=1 && (pos+4)<=16 &&  t[pos+4].doubt_pit<1 && t[pos+4].doubt_wump<1 && t[pos+4].pit!=1 && t[pos+4].wump!=1)
{
/////////////////
temp1="u";
////////////////
t[pos].d=1;
pos=pos+4;
System.out.println("\ndown pos= "+pos);
++score;
//t[pos].visited=1;

//////////////////

t[pos].back+=temp1;
//////////////////
condition=t[pos].sense();
if(condition==3)
{complete=1;break;}
else
if(condition==1 && t[pos].visited==0)
{
if(t[pos].br!=1 && t[pos+1].safe!=1)
t[pos+1].doubt_pit+=1;
if(t[pos].bu!=1 && (pos-4)>=1 &&  t[pos-4].safe!=1)
t[pos-4].doubt_pit+=1;
if(t[pos].bl!=1 && t[pos-1].safe!=1)
t[pos-1].doubt_pit+=1;
if(t[pos].bd!=1 && (pos+4)<=16 &&  t[pos+4].safe!=1)
t[pos+4].doubt_pit+=1;

t[pos].safe=1;
}
else
if(condition==2 && t[pos].visited==0)
{
if(t[pos].br!=1 && t[pos+1].safe!=1)
t[pos+1].doubt_wump+=1;
if(t[pos].bu!=1 && (pos-4)>=1 &&  t[pos-4].safe!=1)
t[pos-4].doubt_wump+=1;
if(t[pos].bl!=1 && t[pos-1].safe!=1)
t[pos-1].doubt_wump+=1;
if(t[pos].bd!=1 && (pos+4)<=16 &&  t[pos+4].safe!=1)
t[pos+4].doubt_wump+=1;

t[pos].safe=1;
}
else
if(condition==0)
t[pos].safe=1;

t[pos].visited=1;
}
else
if(limit>50)
{
int temp3=pos;
int flag_1=0,flag2=0,flag3=0,flag4=0;

System.out.println("\nCurrently at position "+temp3+".\nThinking....");

//if(!(t[pos].back.contains("r") && (t[pos].l!=1 || t[pos].u!=1 || t[pos].d!=1) && check(t[pos]) ))
while(t[pos].visited==1 && t[pos].br!=1)
{
++pos;
++score;
}


if(t[pos].pit==1 || t[pos].wump==1 || (t[pos].br==1 && t[pos].visited==1 && t[pos].safe!=1))
{
//System.out.println("\nUnsuccessful at pos "+pos);
pos=temp3;
//System.out.println("\nBack at pos "+pos);
flag_1=1;
} 

if(flag_1==0)
t[pos].back+="l";

//if(!(t[pos].back.contains("u") && (t[pos].l!=1 || t[pos].r!=1 || t[pos].d!=1) && check(t[pos])  ))
while(pos+4>=1 && t[pos].bu!=1 && t[pos].visited==1)
{
pos-=4;
++score;
}

if(t[pos].pit==1 || t[pos].wump==1 || (t[pos].bu==1 && t[pos].visited==1  && t[pos].safe!=1))
{
//System.out.println("\nUnsuccessful at pos "+pos);
pos=temp3;
//System.out.println("\nBack at pos "+pos);
flag3=1;
} 

if(flag3==0)
t[pos].back+="d";

//if(!(t[pos].back.contains("l") && (t[pos].r!=1 || t[pos].u!=1 || t[pos].d!=1) && check(t[pos]) ))
while(t[pos].visited==1 && t[pos].bl!=1)
{
--pos;
++score;
}

if(t[pos].pit==1 || t[pos].wump==1 || (t[pos].bl==1 && t[pos].visited==1 && t[pos].safe!=1))
{
//System.out.println("\nUnsuccessful at pos "+pos);
pos=temp3;
//System.out.println("\nBack at pos "+pos);
flag2=1;
} 

if(flag2==0)
t[pos].back+="r";



//if(!(t[pos].back.contains("d") && (t[pos].l!=1 || t[pos].r!=1 || t[pos].u!=1) && check(t[pos])  ))
while(pos+4<=16 && t[pos].bd!=1 && t[pos].visited==1)
{
pos+=4;
++score;
}

if(t[pos].pit==1 || t[pos].wump==1 || (t[pos].bd==1 && t[pos].visited==1 && t[pos].safe!=1))
{
//System.out.println("\nUnsuccessful at pos "+pos);
pos=temp3;
//System.out.println("\nBack at pos "+pos);
flag4=1;
} 
//t[pos-1].r=0;
//++pos;
//if(!t[pos].env.contains("P") && !t[pos].env.contains("W"))

if(flag4==0)
t[pos].back+="u";

t[pos].safe=1;
t[pos].visited=1;
System.out.println("reached at position "+pos);
limit=0;
}
if(t[pos].env.contains("W") && scream!=1)
{
score+=100;
scream=1;
t[pos].safe=1;
System.out.println("\n\nWumpus killed >--0-->");
t[pos].env.replace("W"," ");
for(int l=1;l<=16;++l)
{
t[l].doubt_wump=0;
t[l].env.replace("S"," ");
}
}

if(t[pos].env.contains("P"))
{
score+=50;
t[pos].pit=1;
System.out.println("\n\nFallen in pit of position "+pos+".");
}

for(int k=1;k<=16;++k)
{
if(t[k].doubt_pit==1 && t[k].doubt_wump==1)
{
t[k].doubt_pit=0;
t[k].doubt_wump=0;
t[k].safe=1;
}
}

for(int y=1;y<=16;++y)
{
if(t[y].doubt_wump>1)
{
t[y].wump=1;
for(int h=1;h<=16;++h)
{
if(h!=y)
{
t[h].doubt_wump=0;
t[h].env.replace("S"," ");
}
}

}

}

///////////////////////////
for(int y=1;y<=16;++y)
{
if(t[y].doubt_pit>1)
{t[y].pit=1;
//System.out.println("\nPit confirmed at position "+y);
}
}
///////////////////////////


try{Thread.sleep(200);}catch(Exception p){}

}
while(complete==0);

if(complete==1)
{
//score=score*2;
//if(scream==1)
//score-=100;

score*=-1;
score+=1000;
}
System.out.println("The score of the agent till he reaches gold is "+score+".\nNow he will return back following the best explored path.");

}
}

/* OUTPUT 1

C:\Users\Aditya\Desktop\aditya>javac wumpus.java

C:\Users\Aditya\Desktop\aditya>java wumpus


********* Wumpus World Problem *********


The positions are as follows.

-----------------------------------------------------------------
|       1       |       2       |       3       |       4       |
-----------------------------------------------------------------
|       5       |       6       |       7       |       8       |
-----------------------------------------------------------------
|       9       |       10      |       11      |       12      |
-----------------------------------------------------------------
|       13      |       14      |       15      |       16      |
-----------------------------------------------------------------

Agent start position: 13

Enter the number of pits.
3
Positions of pit, gold and wumpus should not overlap.
Enter the position of pits.
4 7 15
Enter the position of wumpus.
5
Enter the position of gold.
6

The environment for problem is as follows.


-----------------------------------------------------------------
|       S       |               |       B       |       P       |
-----------------------------------------------------------------
|       W       |       BSG     |       P       |       B       |
-----------------------------------------------------------------
|       S       |               |       B       |               |
-----------------------------------------------------------------
|       A       |       B       |       P       |       B       |
-----------------------------------------------------------------


Finding the solution...

front pos=14

back pos= 13

Up pos= 9

front pos=10

front pos=11

back pos= 10

Up pos= 6
Gold Found!!
The score of the agent till he reaches gold is 993.
Now he will return back following the best explored path.



OUTPUT 2

C:\Users\Aditya\Desktop\aditya>javac wumpus.java

C:\Users\Aditya\Desktop\aditya>java wumpus


********* Wumpus World Problem *********


The positions are as follows.

-----------------------------------------------------------------
|       1       |       2       |       3       |       4       |
-----------------------------------------------------------------
|       5       |       6       |       7       |       8       |
-----------------------------------------------------------------
|       9       |       10      |       11      |       12      |
-----------------------------------------------------------------
|       13      |       14      |       15      |       16      |
-----------------------------------------------------------------

Agent start position: 13

Enter the number of pits.
2
Positions of pit, gold and wumpus should not overlap.
Enter the position of pits.
5 11
Enter the position of wumpus.
6
Enter the position of gold.
4

The environment for problem is as follows.


-----------------------------------------------------------------
|       B       |       S       |               |       G       |
-----------------------------------------------------------------
|       SP      |       BW      |       BS      |               |
-----------------------------------------------------------------
|       B       |       BS      |       P       |       B       |
-----------------------------------------------------------------
|       A       |               |       B       |               |
-----------------------------------------------------------------


Finding the solution...

front pos=14

front pos=15

back pos= 14

Up pos= 10

down pos= 14

Currently at position 14.
Thinking....
reached at position 16

Currently at position 16.
Thinking....
reached at position 12

Up pos= 8

back pos= 7

front pos=8

Up pos= 4
The score of the agent till he reaches gold is 988.
Now he will return back following the best explored path.

*/

VIEW OUR SPECIAL PAGE FOR SEM 7 STUDENTS - sem 7 java codes

Thursday, 5 September 2013

Sem 4 & Sem 7 (AOAD & RAI) : N queen Algorithm Java Code

import java.util.*;
class nqueen
{
static int n=10;
static int board[]=new int[n];
static boolean place(int row,int column)
{
for(int i=1;i<=row-1;++i)
{
if(column==board[i])
return false;
else 
if(abs(column-board[i])==abs(row-i))
return false;
}
return true;
}
static int abs(int arg)
{
if(arg<0)
return arg*-1;
else
if(arg==0)
return 0;
else
return arg;
}
static void queen(int row,int n)
{

for(int column=1;column<=n;++column)
{
if(place(row,column))
{
board[row]=column;
if(row==n)
display();
else
queen(row+1,n);
}
}
}

static void display()
{
System.out.println();
System.out.print("Pattern ");
for(int i=1;i<=n;++i)
System.out.print(" "+board[i]);
System.out.println();
int pos;
System.out.print(" ");
for(int i=1;i<=n;++i)
System.out.print(" "+i);
System.out.println();
for(int i=1;i<=n;++i)
{
pos=board[i];
System.out.print(" "+i);
for(int j=1;j<=n;++j)
{
if(j==pos)
System.out.print(" Q");
else
System.out.print(" -");
}
System.out.println();
}
}

public static void main(String args[])
{
Scanner scr=new Scanner(System.in);
System.out.println("**********N Queen problem*************");
System.out.println("Enter the number of queens");
n=scr.nextInt();
/* IF there are 'n' queens there are n rows and n columns */
queen(1,n);
}
}

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Tuesday, 27 August 2013

Water Jug Problem Java program

The java code for Water Jug Problem is given below.There are many solutions for it run the code to find all.

VIEW OUR SPECIAL PAGE ON, " ROBOTICS AND ARTIFICIAL INTELLIGENCE " FOR SEM 7 STUDENTS - Robotics and Artificial Intelligence

//Aim:To implement Water jug problem in Java.
//Program:
import java.util.*;
class wj2
{
static int gallon1,gallon2;
static int wastage;
static int steps;
static int rules[]=new int[15];
static int reset=0;
static int ct=0;
static int incr;
static int tgallon1,tgallon2;
static int check()
{
if(steps>10 || reset==1)
return 1;
else 
if(gallon1==2)
return 2;
else
return 0;
}

static void empty1()
{
wastage=wastage+gallon1;
gallon1=0;
} 
static void empty2()
{
wastage=wastage+gallon2;
gallon2=0;
}
static void fill1()
{
gallon1=gallon1+4;
}
static void fill2()
{
gallon2+=3;
}
static void gallon1_to_gallon2()
{
int temp=0;
temp=3-gallon2;
gallon2=3;
gallon1=gallon1-temp;
}
static void gallon2_to_gallon1()
{
int temp=0;
temp=4-gallon1;
gallon1=4;
gallon2=gallon2-temp;
}
static void empty1_fill2()
{
gallon2=gallon2+gallon1;
gallon1=0;
}
static void empty2_fill1()
{
gallon1=gallon1+gallon2;
gallon2=0;
}


static void call(int n)
{

tgallon1=gallon1;
tgallon2=gallon2;

try
{
//Thread.sleep(1000);
}
catch(Exception e)
{
System.out.println("Exception in thread");
}
if(steps>10)
{
System.out.println("ERROR num of steps greater than 10");
}

++steps;
if(n==1)
fill1();
//empty1();
else
if(n==2)
fill2();
//empty2();
else
if(n==3)
empty1();
//fill1();
else
if(n==4)
empty2();
//fill2();
else
if(n==5)
gallon1_to_gallon2();
else
if(n==6)
gallon2_to_gallon1();
else
if(n==7)
empty1_fill2();
else
if(n==8)
empty2_fill1();

if(gallon1>4 || gallon1<0 || gallon2>3 || gallon2<0)
reset=1;

}

static void repeat(int rules[])
{
gallon1=0;
gallon2=0;
wastage=0;
steps=0;

System.out.println("\nGallon1\tGallon2\tRule");
System.out.println(gallon1+"\t"+gallon2+"\t");
call(rules[0]);
if(rules[0]!=0)
System.out.println(gallon1+"\t"+gallon2+"\t"+rules[0]);
call(rules[1]);
if(rules[1]!=0)
System.out.println(gallon1+"\t"+gallon2+"\t"+rules[1]);
call(rules[2]);
if(rules[2]!=0)
System.out.println(gallon1+"\t"+gallon2+"\t"+rules[2]);
call(rules[3]);
if(rules[3]!=0)
System.out.println(gallon1+"\t"+gallon2+"\t"+rules[3]);
call(rules[4]);
if(rules[4]!=0)
System.out.println(gallon1+"\t"+gallon2+"\t"+rules[4]);
call(rules[5]);
if(rules[5]!=0)
System.out.println(gallon1+"\t"+gallon2+"\t"+rules[5]);
call(rules[6]);
if(rules[6]!=0)
System.out.println(gallon1+"\t"+gallon2+"\t"+rules[6]);
call(rules[7]);
if(rules[7]!=0)
System.out.println(gallon1+"\t"+gallon2+"\t"+rules[7]);
call(rules[8]);
if(rules[8]!=0)
System.out.println(gallon1+"\t"+gallon2+"\t"+rules[8]);
call(rules[9]);
if(rules[9]!=0)
System.out.println(gallon1+"\t"+gallon2+"\t"+rules[9]);
}
public static void main(String args[])
{
Scanner scr=new Scanner(System.in);
System.out.println("**********This is waterjug *************");
System.out.println("\nRules:\n1.Fill Gallon 1.\n2.Fill Gallon 2.\n3.Empty gallon 1.\n4.Empty Gallon 2.\n5.Gallon 1 to Gallon 2 till full.\n6.Gallon 2 to Gallon 1 till full.\n7.Empty Gallon 1 and add to Gallon 2.\n8.Empty Gallon 2 and add to Gallon 1.");
System.out.println("\nThe Optimal solution is the one with minimum steps and minimum wastage.\n");
int flag=0;
int count1=0;

flag=0;
count1=0;
wastage=0;
steps=0;
for(int i=0;i<10;++i)
rules[i]=0;
reset=0;
int temp1=0;
incr=0;
/************Data for testing the functionality****************/
/*
rules[0]=1;rules[1]=5;rules[2]=3;rules[3]=8;rules[4]=2;rules[5]=6;rules[6]=3;rules[7]=8;rules[8]=0;rules[9]=0;
repeat(rules);
System.out.println("\nWastage= "+wastage);
rules[0]=2;rules[1]=8;rules[2]=2;rules[3]=6;rules[4]=3;rules[5]=8;rules[6]=0;rules[7]=0;rules[8]=0;rules[9]=0;
repeat(rules);
System.out.println("\nWastage= "+wastage);
rules[0]=1;rules[1]=5;rules[2]=4;rules[3]=7;rules[4]=1;rules[5]=5;rules[6]=0;rules[7]=0;rules[8]=0;rules[9]=0;
repeat(rules);
System.out.println("\nWastage= "+wastage);
*/

l1:for(int i=0;i<8;++i)
{
l2:for(int j=0;j<8;++j)
{
l3:for(int k=0;k<8;++k)
{
l4:for(int l=0;l<8;++l)
{
l5:for(int m=0;m<8;++m)
{
l6:for(int n=0;n<8;++n)
{
l7:for(int o=0;o<8;++o)
{
l8:for(int p=0;p<8;++p)
{
l9:for(int q=0;q<8;++q)
{
l10:for(int r=0;r<8;++r)
{
reset=0;
call(i);
call(j);
call(k);
call(l);
call(m);
call(n);
call(o);
call(p);
call(q);
call(r);
if(gallon1==2 && reset==0)
{
try
{
Thread.sleep(10000);
rules[0]=i;rules[1]=j;rules[2]=k;rules[3]=l;rules[4]=m;rules[5]=n;rules[6]=o;rules[7]=p;rules[8]=q;rules[9]=r;
System.out.println("\nFound a pattern !!!!\n");
repeat(rules);
int count2=0;

for(int z=0;z<10;++z)
{
if(rules[z]!=0)
{
//System.out.print(" "+rules[z]);
++count2;
}
}

System.out.println("\nNumber of Steps= "+(count2+1));
System.out.println("\nWastage= "+wastage);
System.out.println("\n\n\nPlease wait for another solution.....");

Thread.sleep(10000);
}
catch(Exception e)
{}
}

gallon1=0;
gallon2=0;
steps=0;
reset=0;
wastage=0;
}
}
}
}
}
}
}
}
}
}

}
}
/*OUTPUT
C:\Users\Aditya\Desktop\aditya>javac wj2.java

C:\Users\Aditya\Desktop\aditya>java wj2
**********This is waterjug *************

Rules:
1.Fill Gallon 1.
2.Fill Gallon 2.
3.Empty gallon 1.
4.Empty Gallon 2.
5.Gallon 1 to Gallon 2 till full.
6.Gallon 2 to Gallon 1 till full.
7.Empty Gallon 1 and add to Gallon 2.
8.Empty Gallon 2 and add to Gallon 1.

The Optimal solution is the one with minimum steps and minimum wastage.


Found a pattern !!!!


Gallon1 Gallon2 Rule
0       0
4       0       1
1       3       5
0       3       3
3       0       8
3       3       2
4       2       6
0       2       3
2       0       8

Number of Steps= 9

Wastage= 5


Please wait for another solution.....

Found a pattern !!!!


Gallon1 Gallon2 Rule
0       0
0       3       2
3       0       8
3       3       2
4       2       6
0       2       3
2       0       8

Number of Steps= 7

Wastage= 4



Please wait for another solution.....

Found a pattern !!!!



Gallon1 Gallon2 Rule
0       0
4       0       1
1       3       5
1       0       4
0       1       7
4       1       1
2       3       5

Number of Steps= 7

Wastage= 3


Please wait for another solution.....

*/

Thursday, 1 August 2013

System Security : C code for Playfair Cipher (Should be executed in Linux with gcc.)

The Playfair cipher or Playfair square is a manual symmetric encryption technique and was the first literal digraph substitution cipher. The scheme was invented in 1854 by Charles Wheatstone, but bears the name of Lord Playfair who promoted the use of the cipher.

The technique encrypts pairs of letters (digraphs), instead of single letters as in the simple substitution cipher and rather more complex Vigenère ciphersystems then in use. The Playfair is thus significantly harder to break since the frequency analysis used for simple substitution ciphers does not work with it. Frequency analysis can still be undertaken, but on the 600 possible digraphs rather than the 26 possible monographs. The frequency analysis of digraphs is possible, but considerably more difficult – and it generally requires a much larger ciphertext in order to be useful.

//Aim:To implement Playfair cipher in C.

//Program:

#include<stdio.h>

int check(char table[5][5],char k)
{
int i,j;
for(i=0;i<5;++i)
for(j=0;j<5;++j)
{
if(table[i][j]==k)
return 0;
}
return 1;
}

void main()
{
int i,j,key_len;
char table[5][5];
for(i=0;i<5;++i)
for(j=0;j<5;++j)
table[i][j]='0';

printf("**********Playfair Cipher************\n\n");

printf("Enter the length of the Key. ");
scanf("%d",&key_len);

char key[key_len];

printf("Enter the Key. ");
for(i=-1;i<key_len;++i)
{
scanf("%c",&key[i]);
if(key[i]=='j')
key[i]='i';
}

int flag;
int count=0;

// inserting the key into the table
for(i=0;i<5;++i)
{
for(j=0;j<5;++j)
{
flag=0;
while(flag!=1)
{
if(count>key_len)
goto l1;

flag=check(table,key[count]);
++count;
}// end of while
table[i][j]=key[(count-1)];
}// end of inner for
}// end of outer for


l1:printf("\n");

int val=97;
//inserting other alphabets
for(i=0;i<5;++i)
{
for(j=0;j<5;++j)
{
if(table[i][j]>=97 && table[i][j]<=123)
{}
else
{
flag=0;
while(flag!=1)
{
if('j'==(char)val)
++val;
flag=check(table,(char)val);
++val;
}// end of while
table[i][j]=(char)(val-1);
}//end of else
}// end of inner for
}// end of outer for

printf("The table is as follows:\n");
for(i=0;i<5;++i)
{
for(j=0;j<5;++j)
{
printf("%c ",table[i][j]);
}
printf("\n");
}

int l=0;
printf("\nEnter the length length of plain text.(without spaces) ");
scanf("%d",&l);

printf("\nEnter the Plain text. ");
char p[l];
for(i=-1;i<l;++i)
{
scanf("%c",&p[i]);
}

for(i=-1;i<l;++i)
{
if(p[i]=='j')
p[i]='i';
}

printf("\nThe replaced text(j with i)");
for(i=-1;i<l;++i)
printf("%c ",p[i]);

count=0;
for(i=-1;i<l;++i)
{
if(p[i]==p[i+1])
count=count+1;
}

printf("\nThe cipher has to enter %d bogus char.It is either 'x' or 'z'\n",count);

int length=0;
if((l+count)%2!=0)
length=(l+count+1);
else
length=(l+count);


printf("\nValue of length is %d.\n",length);
char p1[length];

//inserting bogus characters.
char temp1;
int count1=0;
for(i=-1;i<l;++i)
{
p1[count1]=p[i];
if(p[i]==p[i+1])
{
count1=count1+1;
if(p[i]=='x')
p1[count1]='z';
else
p1[count1]='x';
}
count1=count1+1;
}

//checking for length

char bogus;
if((l+count)%2!=0)
{
if(p1[length-1]=='x')
p1[length]='z';
else
p1[length]='x';
}

printf("The final text is:");
for(i=0;i<=length;++i)
printf("%c ",p1[i]);

char cipher_text[length];
int r1,r2,c1,c2;
int k1;

for(k1=1;k1<=length;++k1)
{
for(i=0;i<5;++i)
{
for(j=0;j<5;++j)
{
if(table[i][j]==p1[k1])
{
r1=i;
c1=j;
}
else
if(table[i][j]==p1[k1+1])
{
r2=i;
c2=j;
}
}//end of for with j
}//end of for with i

if(r1==r2)
{
cipher_text[k1]=table[r1][(c1+1)%5];
cipher_text[k1+1]=table[r1][(c2+1)%5];
}

else
if(c1==c2)
{
cipher_text[k1]=table[(r1+1)%5][c1];
cipher_text[k1+1]=table[(r2+1)%5][c1];
}
else
{
cipher_text[k1]=table[r1][c2];
cipher_text[k1+1]=table[r2][c1];
}

k1=k1+1;
}//end of for with k1

printf("\n\nThe Cipher text is:\n ");
for(i=1;i<=length;++i)
printf("%c ",cipher_text[i]);

}

/*OUTPUT
[aditya@localhost Desktop]$ gcc playfair.c
[aditya@localhost Desktop]$ ./a.out

**********Playfair Cipher************

Enter the length of the Key. 15
Enter the Key. playfairexample

The table is as follows:
p l a y f 
i r e x m 
b c d g h 
k n o q s 
t u v w z 

Enter the length length of plain text.(without spaces) 25

Enter the Plain text. hidethegoldinthetreestump

The replaced text(j with i)
 h i d e t h e g o l d i n t h e t r e e s t u m p 

The cipher has to enter 1 bogus char.It is either 'x' or 'z'

Value of length is 26.
The final text is:
 h i d e t h e g o l d i n t h e t r e x e s t u m p 

The Cipher text is:
 b m o d z b x d n a b e k u d m u i x m m o u v i f 

*/

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System Security : C code for Transposition Cipher (should be executed in linux with gcc)

In cryptography, a transposition cipher is a method of encryption by which the positions held by units of plaintext (which are commonly characters or groups of characters) are shifted according to a regular system, so that the ciphertext constitutes a permutation of the plaintext. That is, the order of the units is changed. Mathematically a bijective function is used on the characters' positions to encrypt and an inverse function to decrypt.

//Aim:To implement Transposition Cipher in C.
//Program:

#include<stdio.h>
int check(int x,int y)
{
int a,b,c;

if(x%y==0)
return 0;

a=x/y;
b=y*(a+1);
c=b-x;
return c;
}
void main()
{
int l1,i,d,j;

printf("\nEnter the length of the key. ");
scanf("%d",&l1);
int sequence[l1];
printf("\nEnter the sequence key. ");
for(i=0;i<l1;++i)
{
scanf("%d",&sequence[i]);
}

int order[l1];
for(i=1;i<=l1;++i)
{
for(j=0;j<l1;++j)
{
if(sequence[j]==i)
order[i-1]=j;
}
}

printf("\nEnter the depth. ");
scanf("%d",&d);

int l2;
printf("\nEnter the length of String without spaces . ");
scanf("%d",&l2);
int temp1=check(l2,l1);

int r=(l2+temp1)/l1;

char p[l2+temp1];
char p1[r][l1];
//char p2[r][l1];
if(temp1>0)
printf("\nYou need to enter %d bogus characters.So enter total %d characters. ",temp1,(l2+temp1));
else
printf("\nEnter the string. ");

for(i=-1;i<(l2+temp1);++i)
{
scanf("%c",&p[i]);
}
int count=0;
while(d>0)
{
count=0;

for(i=0;i<r;++i)
{
for(j=0;j<l1;++j)
{
p1[i][j]=p[count];
count=count+1;
}
}

printf("\n\n\n");
for(i=0;i<r;++i)
{
for(j=0;j<l1;++j)
{
printf("%c ",p1[i][j]);
}
printf("\n");
}

count=0;
for(i=0;i<l1;++i)
{
for(j=0;j<r;++j)
{
p[count]=p1[j][order[i]];
count=count+1;
}
}

for(i=0;i<(l2+temp1);++i)
printf("%c ",p[i]);

d=d-1;
}
}

/* OUTPUT
[aditya@localhost Desktop]$ gcc transposition.c
[aditya@localhost Desktop]$ ./a.out

Enter the length of the key. 7

Enter the sequence key. 4 3 1 2 5 6 7

Enter the depth. 2

Enter the length of String without spaces . 23

You need to enter 5 bogus characters.So enter total 28 characters. attackpostponeduntiltwoamxyz



a t t a c k p 
o s t p o n e 
d u n t i l t 
w o a m x y z 

t t n a a p t m t s u o a o d w c o i x k n l y p e t z 


t t n a a p t 
m t s u o a o 
d w c o i x k 
n l y p e t z 

n s c y a u o p t t w l t m d n a o i e p a x t t o k z 

*/

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Thursday, 25 July 2013

System Security : C code for Vigenere Cipher (should be executed in linux with gcc).

//Aim:To implement vigenere cipher in C.
//Program:
#include<stdio.h>

void main()
{
int i;
int kl;
int pl;
printf("Enter the length of the key stream. ");
scanf("%d",&kl);
printf("Enter the length of the plain text stream.(Without spaces) ");
scanf("%d",&pl);

char p[pl];
char k[kl];

printf("\nEnter the Key. ");
for(i=-1;i<kl;++i)
scanf("%c",&k[i]);

printf("\nEnter the Plain text. ");
for(i=-1;i<pl;++i)
scanf("%c",&p[i]);

int s[3][pl];

for(i=0;i<pl;++i)
{
if(65<=p[i] && p[i]<=91)
s[0][i]=p[i]%65;
else
s[0][i]=p[i]%97;
}

for(i=0;i<pl;++i)
printf("%d ",s[0][i]);
int count=0;
while(count<pl)
{
for(i=0;i<kl;++i)
{
if(65<=k[i] && k[i]<=91)
s[1][count+i]=k[i]%65;
else
s[1][count+i]=k[i]%97;
}
count=count+kl;
}
printf("\n");
for(i=0;i<pl;++i)
printf("%d ",s[1][i]);

printf("\n");
for(i=0;i<pl;++i)
printf("%d ",s[2][i]);

printf("\n\nThe cipher text is: ");
char cipher[kl];
for(i=0;i<pl;++i)
{
s[2][i]=(s[0][i]+s[1][i])%26;
cipher[i]=(char)(s[2][i]+65);
printf("%c ",cipher[i]);
}

}

/*Output

lab2@lab2-OptiPlex-330:~/Desktop$ gcc vigenere.c
lab2@lab2-OptiPlex-330:~/Desktop$ ./a.out
Enter the length of the key stream. 6
Enter the length of the plain text stream.(Without spaces) 14

Enter the Key. pascal

Enter the Plain text. sheislistening

18 7 4 8 18 11 8 18 19 4 13 8 13 6 
15 0 18 2 0 11 15 0 18 2 0 11 15 0 

The cipher text is: H H W K S W X S L G N T C G 
*/

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Wednesday, 24 July 2013

System Security : C code for Hill Cipher (should be executed in linux with gcc).

//Aim:To implement Hill Cipher in C
//Program:

#include<stdio.h>
int check(int x)
{
if(x%3==0)
return 0;

int a=x/3;
int b=3*(a+1);
int c=b-x;

return c;
}

void main()
{
int l,i,j;
int temp1;
int k[3][3];
int p[3][1];
int c[3][1];
char ch;
printf("\nThe cipher has a key of length 9. ie. a 3*3 matrix.\nEnter the 9 character key. ");

for(i=0;i<3;++i)
{
for(j=0;j<3;++j)
{
scanf("%c",&ch);
if(65<=ch && ch<=91)
k[i][j]=(int)ch%65;
else
k[i][j]=(int)ch%97;
}
}

for(i=0;i<3;++i)
{
for(j=0;j<3;++j)
{
printf("%d ",k[i][j]);
}
printf("\n");
}


printf("\nEnter the length of string to be encoded(without spaces). ");
scanf("%d",&l);

temp1=check(l);

if(temp1>0)
printf("You have to enter %d bogus characters.",temp1);

char pi[l+temp1];
printf("\nEnter the string. ");
for(i=-1;i<l+temp1;++i)
{
scanf("%c",&pi[i]);
}


int temp2=l;
int n=(l+temp1)/3;
int temp3;
int flag=0;
int count;
printf("\n\nThe encoded cipher is : ");

while(n>0)
{
count=0;
for(i=flag;i<flag+3;++i)
{
if(65<=pi[i] && pi[i]<=91)
temp3=(int)pi[i]%65;
else
temp3=(int)pi[i]%97;

p[count][0]=temp3;
count=count+1;
}

int k1;
for(i=0;i<3;++i)
c[i][0]=0;

for(i=0;i<3;++i)
{
for(j=0;j<1;++j)
{
for(k1=0;k1<3;++k1)
c[i][j]+=k[i][k1]*p[k1][j];
}
}

for(i=0;i<3;++i)
{
c[i][0]=c[i][0]%26;  
printf("%c ",(char)(c[i][0]+65));
}

n=n-1;
flag=flag+3;
}
}

/*OUTPUT
[adi@localhost Desktop]$ gcc hill.c
[adi@localhost Desktop]$ ./a.out

The cipher has a key of length 9. ie. a 3*3 matrix.
Enter the 9 character key. GYBNQKURP
6 24 1 
13 16 10 
20 17 15 

Enter the length of string to be encoded(without spaces). 6

Enter the string. actCAt


The encoded cipher is : P O H F I N 


[adi@localhost Desktop]$ gcc hill.c
[adi@localhost Desktop]$ ./a.out

The cipher has a key of length 9. ie. a 3*3 matrix.
Enter the 9 character key. GYBNQKURP
6 24 1 
13 16 10 
20 17 15 

Enter the length of string to be encoded(without spaces). 7
You have to enter 2 bogus characters.
Enter the string. actcatact


The encoded cipher is : P O H F I N P O H  

*/

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Tuesday, 23 July 2013

Robotics & AI: Java code for Homogeneous Composite Transformation Matrix (HCTM) finding fixed , mobile co-ordinates.

//Aim: To implement fixed and mobile co-ordinates in JAVA 
//Program
import java.util.*;
class htmc
{
static int m1[][]={ {3,0,0},{0,2,1},{0,-1,2}};
static int m2[][]={ {2,0,-1},{0,3,0},{1,0,2}};
static int m3[][]={ {2,1,0},{-1,2,0},{0,0,3}};
static double pm[][]=new double[3][1];
static double pf[][]=new double[3][1];
static double p[][]=new double[3][3];

static int a,t;
static void mat_mul()
{
for(int i=0;i<3;++i)
pf[i][0]=0;


for(int i=0;i<3;++i)
{
for(int j=0;j<1;++j)
{
for(int k=0;k<3;++k)
pf[i][j]+=p[i][k]*pm[k][j];
}
}
}

static void get_m1()
{
double temp;
for(int i=0;i<3;++i)
{
for(int j=0;j<3;++j)
{
if(m1[i][j]==2)
p[i][j]=Math.cos(Math.toRadians(t));
else
if(m1[i][j]==3)
p[i][j]=1;
else
{
temp=90+m1[i][j]*t;
p[i][j]=Math.cos(Math.toRadians(temp));
}
}
}
}
static void get_m2()
{
double temp;
for(int i=0;i<3;++i)
{
for(int j=0;j<3;++j)
{
if(m2[i][j]==2)
p[i][j]=Math.cos(Math.toRadians(t));
else
if(m2[i][j]==3)
p[i][j]=1;
else
{
temp=90+m2[i][j]*t;
p[i][j]=Math.cos(Math.toRadians(temp));
}
}
}
}
static void get_m3()
{
double temp;
for(int i=0;i<3;++i)
{
for(int j=0;j<3;++j)
{
if(m3[i][j]==2)
p[i][j]=Math.cos(Math.toRadians(t));
else
if(m3[i][j]==3)
p[i][j]=1;
else
{
temp=90+m3[i][j]*t;
p[i][j]=Math.cos(Math.toRadians(temp));
}
}
}

}

public static void main(String args[])
{
Scanner scr=new Scanner(System.in);
System.out.println("Enter [P]m values. ");
for(int i=0;i<3;++i)
{
pm[i][0]=scr.nextDouble();
}
System.out.println("Rotation about which axis? 1,2,3 ");
a=scr.nextInt();

System.out.println("Enter the angle ");
t=scr.nextInt();

if(a==1)
get_m1();
else
if(a==2)
get_m2();
else
get_m3();

mat_mul();

System.out.println("The [P]f matrix is as follows. ");
for(int i=0;i<3;++i)
System.out.println(pf[i][0]);

int temp1;
for(int i=0;i<3;++i)
{
for(int j=0;j<3;++j)
{
p[i][j]=(p[i][j]*100);
temp1=(int)p[i][j];
p[i][j]=temp1/100.0;
}
}
System.out.println("\nThe final HTMC matrix is as follows.");
for(int i=0;i<3;++i)
{
for(int j=0;j<3;++j)
{
System.out.print(p[i][j]+"\t");
}
System.out.print(pm[i][0]+"\n");
}
System.out.print("\n0\t0\t0\t1");

}
}

/*OUTPUT
C:\Users\Aditya\Desktop\aditya>javac htmc.java
C:\Users\Aditya\Desktop\aditya>java htmc

Enter [P]m values.
2 0 3
Rotation about which axis? 1,2,3
1
Enter the angle
60
The [P]f matrix is as follows.
2.0
-2.598076211353316
1.5000000000000007

The final HTMC matrix is as follows.
1.0     0.0     0.0     2.0
0.0     0.5     -0.86   0.0
0.0     0.86    0.5     3.0

0       0       0       1
*/

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Thursday, 18 July 2013

System Security : C code for Additive cipher / Shift cipher / Caeser cipher ( can be executed in Linux )

In cryptography, a Caesar cipher, also known as Caesar's cipher, the shift cipher, Caesar's code or Caesar shift, is one of the simplest and most widely known encryption techniques. It is a type of substitution cipher in which each letter in the plain text is replaced by a letter some fixed number of positions down the alphabet. For example, with a left shift of 3, D would be replaced by A, E would become B, and so on. The method is named after Julius Caeser, who used it in his private correspondence.
VIEW OUR SPECIAL PAGE FOR SEM 7 STUDENTS - sem 7 java codes

Given below is the C code for the same.

//Aim: To implement additive cipher in C
//Program:
#include<stdio.h>

void main()
{

char c[20];
int a1[20];
int a2[20];
int n=0;
int k=0;

printf("\nPlease Enter the key value: ");
scanf("%d",&k);

printf("\nHow many letters in the text? : ");
scanf("%d",&n);

int i;
printf("\nPlease enter a word to be encoded:");

for(i=-1;i<n;++i)
{
scanf("%c",&c[i]);
}

int temp=0;
int j=0;
int count=0;
for(i=0;i<=n;++i)
{
l1:temp=(int)c[i];

if(c[i]==' ')
{
i=i+1;
count=count+1;
goto l1;
}

if(65<=temp && temp<=91)
a1[j]=temp%65;
else
if(97<=temp<=123)
a1[j]=temp%97;

a2[j]=(a1[j]+k)%26;
++j;
}

n=n-count;
printf("\nThe encoded cipher is: ");
char c2[20];
for(i=0;i<n;++i)
{
c2[i]=(char)(a2[i]+97);
printf("%c",c2[i]);
}

printf("\n\nDecoding it receiver side ");

int a3[20];
for(i=0;i<n;++i)
a3[i]=(int)(c2[i]-97);

for(i=0;i<n;++i)
{
if(a3[i]>=k)
a3[i]=a3[i]-k;
else
a3[i]=a3[i]-15+26;
}

printf("\nThe text is: ");
for(i=0;i<n;++i)
{
printf("%c",(char)(a3[i]+97));
}

}

/*OUTPUT

Please Enter the key value: 15

How many letters in the text? : 5

Please enter a word to be encoded:hello

The encoded cipher is: wtaad

Decoding it receiver side 
The text is: hello

Please Enter the key value: 15

How many letters in the text? : 12

Please enter a word to be encoded:good morning

The encoded cipher is: vddsbdgcxcv

Decoding it receiver side 
The text is: goodmorning 
*/

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Tuesday, 4 June 2013

SEM 7 [Revised] Syllabus for Computer engineering (MU) Free Download

Use the download box to download the Syllabus copy for SEM 7.

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Everything about SEM 6 (Computer Engineering)

This is an extremely important and useful post for sem 6 computer enginering students.
Download the "Everything for SEM 6.zip" file from the download box in the blog.

It contains all the assignments, experiments, important notes, for the following,
1.OOSE
2.AMP
3.SPCC
4.ACN
5.DWM
6.SEMINAR( topic:web personalization)

Download this file and you can complete all your term work with great ease and feel free to comment if you find it useful.So just download the file "Everything for SEM 6.zip" file from the download box.
Best of luck for your SEM 6 !!
The blog also provides Java codes for all experiments of sem 6.

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Tuesday, 14 May 2013

Brute Force Algorithm for mining Association Rules

923NNTB75NZ5

//AIM:To implement Brute Force Algorithm for mining association rule.
//PROGRAM
import java.util.*;
class transaction
{
int tid;
int item[];
transaction(int a,int b)
{
tid=a;
item=new int[b];
}
}
class brute
{
static int n=5;
static int p=1;
static int flag=0;
static int found=0;
static int s=0;
static void check(int a[],int l,transaction t[])
{
int c1=0,c2=0;
loop1:for(int i=0;i<4;++i)
{
c1=0;
loop2:for(int j=0;j<l;++j)
{
loop3:for(int k=0;k<t[i].item.length;++k)
if(t[i].item[k]==a[j])
{
++c1;
}
}
if(c1==l)
++c2;
}

if(c2>=s){for(int i=0;i<l;++i)System.out.print(" "+a[i]);System.out.print("\t"+c2);System.out.println();++found;}
}

public static void main(String args[])
{
Scanner scr=new Scanner(System.in);
System.out.println("\nBrute force Algorithm\n");
System.out.println("You have 4 transactions:- T0,T1,T2,T3.\n\n");
System.out.println("You have 5 items with id:- 0,1,2,3,4.\n");
System.out.println("Enter items in each transaction");
transaction t[]=new transaction[4];
int count;
for(int i=0;i<4;++i)
{
count=0;
System.out.print("Enter no. of items in transaction T"+i+"  ");
count=scr.nextInt();
t[i]=new transaction(i,count);
System.out.print("Enter items in transaction T"+i+"  ");
for(int k=0;k<count;++k)
t[i].item[k]=scr.nextInt();
}
System.out.println("Enter support");
s=scr.nextInt();
System.out.println("\nTid\tItems\n");
for(int i=0;i<4;++i)
{
System.out.print("T"+i+"\t");
for(int k=0;k<t[i].item.length;++k)
System.out.print(" "+t[i].item[k]);

System.out.println();
}

int temp[]={-1,-1,-1,-1,-1};
do
{
++flag;
found=0;
int f=0;
for(int i=0;i<=n-p;++i)
{
if(flag==1)
{
if(f==0)
{
System.out.println("\n1-frequent set");
System.out.println("itemset\tsup");
}
f=1;
temp[0]=i;
check(temp,1,t);
}
else
for(int j=i+1;j<=n-(p-1);++j)
{
if(flag==2)
{
if(f==0)
{
System.out.println("\n2-frequent set");
System.out.println("itemset\tsup");
}
f=1;
temp[0]=i;
temp[1]=j;
check(temp,2,t);
}
else
for(int k=j+1;k<=n-(p-2);++k)
{
if(flag==3)
{
if(f==0)
{
System.out.println("\n3-frequent set");
System.out.println("itemset\tsup");
}
f=1;
temp[0]=i;
temp[1]=j;
temp[2]=k;
check(temp,3,t);
}
else
for(int l=k+1;l<=n-(p-3);++l)
{
if(flag==4)
{
if(f==0)
{
System.out.println("\n4-frequent set");
System.out.println("itemset\tsup");
}
f=1;
temp[0]=i;
temp[1]=j;
temp[2]=k;
temp[3]=l;
check(temp,4,t);
}
else
for(int m=l+1;m<=n-(p-4);++m)
{
if(f==0)
{
System.out.println("\n5-frequent set");
System.out.println("itemset\tsup");
}
f=1;
temp[0]=i;
temp[1]=j;
temp[2]=k;
temp[3]=l;
temp[4]=m;
check(temp,5,t);
}
}
}
}
}
++p;
if(found==0)
System.out.println("not found.");
}//end of do
while(found>0);
System.out.println("Association rules will be calculated from "+(flag-1)+"-frequent set.");
}
}

/*OUTPUT

C:\Users\Aditya\Desktop\aditya>javac brute.java

C:\Users\Aditya\Desktop\aditya>java brute

Brute force Algorithm

You have 4 transactions:- T0,T1,T2,T3.


You have 5 items with id:- 0,1,2,3,4.

Enter items in each transaction
Enter no. of items in transaction T0  3
Enter items in transaction T0  0 2 3
Enter no. of items in transaction T1  3
Enter items in transaction T1  1 2 4
Enter no. of items in transaction T2  4
Enter items in transaction T2  0 1 2 4
Enter no. of items in transaction T3  2
Enter items in transaction T3  1 4
Enter support
2

Tid     Items

T0       0 2 3
T1       1 2 4
T2       0 1 2 4
T3       1 4

1-frequent set
itemset sup
 0      2
 1      3
 2      3
 4      3

2-frequent set
itemset sup
 0 2    2
 1 2    2
 1 4    3
 2 4    2

3-frequent set
itemset sup
 1 2 4  2

4-frequent set
itemset sup
not found.
Association rules will be calculated from 3-frequent set.

*/

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Saturday, 27 April 2013

Web Personalization Report and Power Point Presentation

View Paper on Web Personalization

Use Download Box in my blog to download the "Power Point Presentation.

Abstract:
Web personalization is the process of customizing the content and the structure of a Web site to the specific and individual needs of each user, without requiring from them to ask for it explicitly.This can be achieved by taking advantage of the user’s navigational behaviour, as it can be revealed through the processing of the Web usage logs, as well as the user’s characteristics and interests. Suchinformation can be further analysed in association with the content of a Web site, resulting to improvement of the system performance, users’ retention and/or site modification.The overall process of Web personalization consists of five modules, namely: user profiling, log analysis and Web usage mining, information acquisition, content management and Web sitepublishing.

Web personalization is a domain that has been recently gaining great momentum not only in the research area, where many research teams have addressed this problem from different perspectives, but also in the industrial area, where there exists a variety of tools and applications addressing one or more modules of the personalization process. Enterprises expect that by exploiting the information hidden in their Web server logs they could discover the interactions between their Web site visitors and the products offered through their Web site. Using such information, they can optimise their site in order to increase sales and ensure customer retention. Apart from Web usage mining, user profiling techniques are also employed in order to form a complete customer profile. Lately, there is an effort to incorporate web content in the recommendation process, in order to enhance the effectiveness of personalization. Thus web personalization has made web activities user-centric and has made the users an integrated part of the web environment.

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Branch prediction logic (JAVA CODE)

//Aim:To implement branch prediction logic in java
import java.util.*;
class branch
{
public static void main(String args[])
{
Scanner scr=new Scanner(System.in);
int stage,inst,br;
int cycle1,cycle2;
System.out.println("Enter num of stages ");
stage=scr.nextInt();
System.out.println("Enter num of instructions ");
inst=scr.nextInt();
System.out.println("Enter num of branch instructions ");
br=scr.nextInt();

cycle1=stage+(inst-1);
cycle2=cycle1+(br*(stage-1));

System.out.println("Num of cycles without branching= "+cycle1);
System.out.println("Num of cycles with branching= "+cycle2);
}
}
/*OUTPUT
C:\Users\Aditya\Desktop\aditya>javac branch.java

C:\Users\Aditya\Desktop\aditya>java branch
Enter num of stages
5
Enter num of instructions
5
Enter num of branch instructions
1
Num of cycles without branching= 9
Num of cycles with branching= 13
*/

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RAID 0 and RAID 1 JAVA Implementation

RAID has seven levels.It stands for Redundand Array Of Independent Disk.The code in JAVA gives the implementation of RAID 0 followed by a seperate code for RAID 1

// Aim:To implement RAID 0 in java
import java.util.*;
import java.io.*;
class raid0
{
public static void main(String args[]) throws Exception
{
Scanner scr=new Scanner(System.in);
InputStreamReader isr=new InputStreamReader(System.in);
BufferedReader br=new BufferedReader(isr);
char r[][]=new char[4][4];
System.out.println("RAID 0 Non-redundant");
System.out.println("Enter 16 characters or numbers");
String s=scr.nextLine();
int c=0;
for(int i=0;i<4;++i)
{
for(int j=0;j<4;++j)
{
r[i][j]=s.charAt(c);
c=c+2;
}
}
int choice=0;
do
{
System.out.println("Enter your choice: 1.display 2.update 3.delete 4.exit");
choice=scr.nextInt();
switch(choice)
{
case 1:
System.out.println("Disk0\tDisk1\tDisk2\tDisk3");
for(int i=0;i<4;++i)
{
for(int j=0;j<4;++j)
{
System.out.print(r[i][j]+"\t");
}
System.out.println();
}
break;
case 2:
int pos=0;
String s1;
char val;
System.out.println("Enter position to be updated(start pos=0) ");
pos=scr.nextInt();
System.out.println("Enter value to be inserted ");
s1=br.readLine();
val=s1.charAt(0);
int count=0;
for(int i=0;i<4;++i)
{
for(int j=0;j<4;++j)
{
if(count==pos)
{
r[i][j]=val;
}
++count;
}
}
System.out.println("Value at position "+pos+" has been updated");
break;
case 3:
System.out.println("Enter position to be deleted(start pos=0) ");
int pos1=scr.nextInt();
int count1=0;
for(int i=0;i<4;++i)
{
for(int j=0;j<4;++j)
{
if(count1==pos1)
{
r[i][j]=' ';
}
++count1;
}
}
System.out.println("Value at position "+pos1+" has been deleted.");
break;
}
}
while(choice!=4);

}
}

/* OUTPUT
C:\Users\Aditya\Desktop\aditya>java raid0
RAID 0 Non-redundant
Enter 16 characters or numbers
1 2 3 a b c 4 5 6 d e f 7 8 9 g
Enter your choice: 1.display 2.update 3.delete 4.exit
1
Disk0   Disk1   Disk2   Disk3
1       2       3       a
b       c       4       5
6       d       e       f
7       8       9       g
Enter your choice: 1.display 2.update 3.delete 4.exit
2
Enter position to be updated(start pos=0)
4
Enter value to be inserted
T
Value at position 4 has been updated
Enter your choice: 1.display 2.update 3.delete 4.exit
1
Disk0   Disk1   Disk2   Disk3
1       2       3       a
T       c       4       5
6       d       e       f
7       8       9       g
Enter your choice: 1.display 2.update 3.delete 4.exit
3
Enter position to be deleted(start pos=0)
12
Value at position 12 has been deleted.
Enter your choice: 1.display 2.update 3.delete 4.exit
1
Disk0   Disk1   Disk2   Disk3
1       2       3       a
T       c       4       5
6       d       e       f
        8       9       g
Enter your choice: 1.display 2.update 3.delete 4.exit
4
*/



// Aim:To implement RAID 1 in java
import java.util.*;
import java.io.*;
class raid1
{
public static void main(String args[]) throws Exception
{
Scanner scr=new Scanner(System.in);
InputStreamReader isr=new InputStreamReader(System.in);
BufferedReader br=new BufferedReader(isr);
char r1[][]=new char[4][4];
char r2[][]=new char[4][4];

System.out.println("RAID 1 Mirrored");
System.out.println("Enter 16 characters or numbers");
String s=scr.nextLine();
int c=0;
for(int i=0;i<4;++i)
{
for(int j=0;j<4;++j)
{
r1[i][j]=s.charAt(c);
r2[i][j]=r1[i][j];
c=c+2;
}
}

int choice=0;
do
{
System.out.println("Enter your choice: 1.display 2.update 3.delete 4.exit");
choice=scr.nextInt();
switch(choice)
{
case 1:
System.out.println("Disk0\tDisk1\tDisk2\tDisk3\t\tMD0\tMD1\tMD2\tMD3");
for(int i=0;i<4;++i)
{
for(int j=0;j<4;++j)
{
System.out.print(r1[i][j]+"\t");
}
System.out.print("|\t");
for(int j=0;j<4;++j)
{
System.out.print(r2[i][j]+"\t");
}
System.out.println();
}
break;
case 2:
int pos=0;
String s1;
char val;
System.out.println("Enter position to be updated(start pos=0) ");
pos=scr.nextInt();
System.out.println("Enter value to be inserted ");
s1=br.readLine();
val=s1.charAt(0);
int count=0;
for(int i=0;i<4;++i)
{
for(int j=0;j<4;++j)
{
if(count==pos)
{
r1[i][j]=val;
r2[i][j]=val;
}
++count;
}
}
System.out.println("Value at position "+pos+" has been updated");
break;
case 3:
System.out.println("Enter position to be deleted(start pos=0) ");
int pos1=scr.nextInt();
int count1=0;
for(int i=0;i<4;++i)
{
for(int j=0;j<4;++j)
{
if(count1==pos1)
{
r1[i][j]=' ';
r2[i][j]=' ';
}
++count1;
}
}
System.out.println("Value at position "+pos1+" has been deleted.");
break;
}
}
while(choice!=4);

}
}
/*OUTPUT
C:\Users\Aditya\Desktop\aditya>java raid1
RAID 1 Mirrored
Enter 16 characters or numbers
1 2 3 4 5 6 7 8 9 a b c d e f g
Enter your choice: 1.display 2.update 3.delete 4.exit
1
Disk0   Disk1   Disk2   Disk3           MD0     MD1     MD2     MD3
1       2       3       4       |       1       2       3       4
5       6       7       8       |       5       6       7       8
9       a       b       c       |       9       a       b       c
d       e       f       g       |       d       e       f       g
Enter your choice: 1.display 2.update 3.delete 4.exit
2
Enter position to be updated(start pos=0)
4
Enter value to be inserted
R
Value at position 4 has been updated
Enter your choice: 1.display 2.update 3.delete 4.exit
1
Disk0   Disk1   Disk2   Disk3           MD0     MD1     MD2     MD3
1       2       3       4       |       1       2       3       4
R       6       7       8       |       R       6       7       8
9       a       b       c       |       9       a       b       c
d       e       f       g       |       d       e       f       g
Enter your choice: 1.display 2.update 3.delete 4.exit
3
Enter position to be deleted(start pos=0)
12
Value at position 12 has been deleted.
Enter your choice: 1.display 2.update 3.delete 4.exit
1
Disk0   Disk1   Disk2   Disk3           MD0     MD1     MD2     MD3
1       2       3       4       |       1       2       3       4
R       6       7       8       |       R       6       7       8
9       a       b       c       |       9       a       b       c
        e       f       g       |               e       f       g
Enter your choice: 1.display 2.update 3.delete 4.exit
4
*/

VIEW OUR SPECIAL PAGE FOR SEM 7 STUDENTS - sem 6 java codes

4 stage floating point addition pipeline (JAVA CODE)

The 4 stages in floating point addition pipeline are:
1.Exponent comparer
2.Mantissa shifter
3.Mantissa adder
4.Normalize result

// Aim:To Implement 4 stage floating point addition pipeline
import java.util.*;
class fp // class of floating pt. nos.
{
double man;
int exp;
}
class fp_add
{
static int compare_exp(fp f1,fp f2) // compares the exponent
{
if(f1.exp<f2.exp)
{
diff=f2.exp-f1.exp; // claculates difference in exponent
return 1; // represents first number
}
else
if(f2.exp<f1.exp)
{
diff=f1.exp-f2.exp;
return 2; // represents second number
}
else
return 0; // if both exponents are same
}

static fp shift_man(fp f) // shifts the mantissa by using "diff" value
{
for(int i=0;i<diff;++i)
{
f.man/=10;
f.exp+=1;
}
return f;
}

static fp normalize(fp f) // normalize final result
{
while(f.man>=10)
{
f.man/=10;
f.exp+=1; // exponent is -ve and hence adding 1 will increase its val ex. -5+1=-4
}
return f;
}

static int diff; // difference in exponent values
public static void main(String args[])
{
Scanner scr=new Scanner(System.in);

/* creating objects for two floating pt. nos. and result */
fp f1=new fp();
fp f2=new fp();
fp r=new fp();

System.out.println("Enter mantissa and exponent of x ");
f1.man=scr.nextDouble();
f1.exp=scr.nextInt();
System.out.println("Enter mantissa and exponent of y ");
f2.man=scr.nextDouble();
f2.exp=scr.nextInt();

int choice=compare_exp(f1,f2);
System.out.print("STAGE 1(exponent comparer): The num with smaller exponent is ");
if(choice==1)System.out.println("x  \n");else if(choice==2)System.out.println("y  \n");else System.out.println("Both are same.\n");

System.out.print("STAGE 2(mantissa shifter): ");

if(choice==1) // choice is 1 hence first mantissa has to be shifted
{
f1=shift_man(f1);
System.out.println("shifted mantissa "+f1.man+"\n");
}
else
if(choice==2) // choice is 2 hence second mantissa has to be shifted
{
f2=shift_man(f2);
System.out.println("shifted mantissa "+f2.man+"\n");
}
else // equal exponents
System.out.println("no shifting required\n");

System.out.print("STAGE 3(mantissa adder): ");
r.man=f1.man+f2.man;
System.out.println("added mantissa "+r.man+"\n");
r.exp=f1.exp; // both f1 and f2 NOW have same exponents.Hence assign any one's exponent.  

r=normalize(r); // call the method to normalize final result.
System.out.println("STAGE 4(normalize result):");
System.out.println("The normalized result is "+r.man+" * 10^"+r.exp);

int c=-1*r.exp;

/* Additional code to represent answer in different form ex 3.52*10^-3 will be 0.00352
(int i=0;i<c;++i)
r.man=r.man/10;
System.out.println("ie. The result is "+r.man);*/
}
}
/*OUTPUT
C:\Users\Aditya\Desktop\aditya>java fp_add
Enter mantissa and exponent of x
8.5 -3
Enter mantissa and exponent of y
2.345 -5
STAGE 1(exponent comparer): The num with smaller exponent is y

STAGE 2(mantissa shifter): shifted mantissa 0.023450000000000002

STAGE 3(mantissa adder) added mantissa 8.52345

STAGE 4(normalize result):
The normalized result is 8.52345 * 10^-3
*/

VIEW OUR SPECIAL PAGE FOR SEM 7 STUDENTS - sem 6 java codes

Data Dependency Hazard (RAW, WAR, WAW) JAVA CODE

The Data dependency hazards are of 4 types.

1.RAR :Read After Read (This is not harmful as only read operations are performed.)

2.RAW :Read After Write(This may create problems and needs to be handled.)

3.WAR :Write After Read(A naming conflict. Not hazardous if sequential execution takes place)

4.WAW : Write After Write(A naming conflict. Not hazardous if sequential execution takes place)

//Aim:To check Data Dependency Hazards in Instructions using JAVA.
import java.util.*;
class hazard
{
static void accept()
{
Scanner scr=new Scanner(System.in);
System.out.print("\nFirst Instruction:\nOpcode ");
s[0][0]=scr.nextLine();
System.out.print("\nDomain: ");
s[0][2]=scr.nextLine();
System.out.print("\nRange: ");
s[0][1]=scr.nextLine();
System.out.print("\nSecond Instruction:\nOpcode ");
s[1][0]=scr.nextLine();
System.out.print("\nDomain: ");
s[1][2]=scr.nextLine();
System.out.print("\nRange: ");
s[1][1]=scr.nextLine();
}
static int check()
{
if(s[0][1].equals(s[1][2]))
return 1;
else
if(s[0][2].equals(s[1][1]))
return 2;
else
if(s[0][1].equals(s[1][1]))
return 3;
else
return 0;
}
static String s[][]=new String[2][3];
public static void main(String args[])
{
Scanner scr=new Scanner(System.in);
int ch=0;
do
{
int c=0;
accept();
/*Display Instructions*/
System.out.println(s[0][0]+" "+s[0][1]+" "+s[0][2]);
System.out.println(s[1][0]+" "+s[1][1]+" "+s[1][2]);

c=check();

if(c==1)
System.out.println("RAW hazard exists");
else
if(c==2)
System.out.println("WAR hazard exists");
else
if(c==3)
System.out.println("WAW hazard exists");
else
System.out.println("There are No Data Dependency Hazards");

System.out.println("Do you wish to continue?(1/0)");
ch=scr.nextInt();
}
while(ch==1);
}
}
/*OUTPUT
C:\Users\Aditya\Desktop\aditya>java hazard

First Instruction:
Opcode LOAD

Domain: a

Range: r1

Second Instruction:
Opcode ADD

Domain: r1

Range: r2
LOAD r1 a
ADD r2 r1
RAW hazard exists
Do you wish to continue?(1/0)
1

First Instruction:
Opcode MUL

Domain: r2

Range: r1

Second Instruction:
Opcode ADD

Domain: r3

Range: r2
MUL r1 r2
ADD r2 r3
WAR hazard exists
Do you wish to continue?(1/0)
1

First Instruction:
Opcode MUL

Domain: r2

Range: r1

Second Instruction:
Opcode ADD

Domain: r4

Range: r1
MUL r1 r2
ADD r1 r4
WAW hazard exists
Do you wish to continue?(1/0)
1

First Instruction:
Opcode ADD

Domain: r1

Range: r2

Second Instruction:
Opcode MUL

Domain: r3

Range: r4
ADD r2 r1
MUL r4 r3
There are No Data Dependency Hazards
Do you wish to continue?(1/0)
0
*/

VIEW OUR SPECIAL PAGE FOR SEM 7 STUDENTS - sem 6 java codes

Tuesday, 23 April 2013

Easy Solution for Sem 6 DWDM pracs(Java Code)

Use Download box to download programs for decision tree and bayesian algo

VIEW OUR SPECIAL PAGE FOR SEM 7 STUDENTS - sem 6 java codes

Wednesday, 20 March 2013

SEM 6 COMP(MU):UML diagrams(online job portal) for OOSE practicals

USE CASE

COLLABORATION

SEQUENCE

STATE

ACTIVITY

CLASS

COMPONENT

DEPLOYMENT

Thursday, 28 February 2013

K-means Clustering Algorithm: Java Code

The Java Code for K-means clustering is given below:

//Aim:To implement Kmeans clustering algorithm.
//Program
import java.util.*;
class k_means
{
static int count1,count2,count3;
static int d[];
static int k[][];
static int tempk[][];
static double m[];
static double diff[];
static int n,p;

static int cal_diff(int a) // This method will determine the cluster in which an element go at a particular step.
{
int temp1=0;
for(int i=0;i<p;++i)
{
if(a>m[i])
diff[i]=a-m[i];
else
diff[i]=m[i]-a;
}
int val=0;
double temp=diff[0];
for(int i=0;i<p;++i)
{
if(diff[i]<temp)
{
temp=diff[i];
val=i;
}
}//end of for loop
return val;
}

static void cal_mean() // This method will determine intermediate mean values
{
for(int i=0;i<p;++i)
m[i]=0; // initializing means to 0
int cnt=0;
for(int i=0;i<p;++i)
{
cnt=0;
for(int j=0;j<n-1;++j)
{
if(k[i][j]!=-1)
{
m[i]+=k[i][j];
++cnt;
}}
m[i]=m[i]/cnt;
}
}

static int check1() // This checks if previous k ie. tempk and current k are same.Used as terminating case.
{
for(int i=0;i<p;++i)
for(int j=0;j<n;++j)
if(tempk[i][j]!=k[i][j])
{
return 0;
}
return 1;
}

public static void main(String args[])
{
Scanner scr=new Scanner(System.in);
/* Accepting number of elements */
System.out.println("Enter the number of elements ");
n=scr.nextInt();
d=new int[n];
/* Accepting elements */
System.out.println("Enter "+n+" elements: ");
for(int i=0;i<n;++i)
d[i]=scr.nextInt();
/* Accepting num of clusters */
System.out.println("Enter the number of clusters: ");
p=scr.nextInt();
/* Initialising arrays */
k=new int[p][n];
tempk=new int[p][n];
m=new double[p];
diff=new double[p];
/* Initializing m */
for(int i=0;i<p;++i)
m[i]=d[i];

int temp=0;
int flag=0;
do
{
for(int i=0;i<p;++i)
for(int j=0;j<n;++j)
{
k[i][j]=-1;
}
for(int i=0;i<n;++i) // for loop will cal cal_diff(int) for every element.
{
temp=cal_diff(d[i]);
if(temp==0)
k[temp][count1++]=d[i];
else
if(temp==1)
k[temp][count2++]=d[i];
else
if(temp==2)
k[temp][count3++]=d[i]; 
}
cal_mean(); // call to method which will calculate mean at this step.
flag=check1(); // check if terminating condition is satisfied.
if(flag!=1)
/*Take backup of k in tempk so that you can check for equivalence in next step*/
for(int i=0;i<p;++i)
for(int j=0;j<n;++j)
tempk[i][j]=k[i][j];

System.out.println("\n\nAt this step");
System.out.println("\nValue of clusters");
for(int i=0;i<p;++i)
{
System.out.print("K"+(i+1)+"{ ");
for(int j=0;k[i][j]!=-1 && j<n-1;++j)
System.out.print(k[i][j]+" ");
System.out.println("}");
}//end of for loop
System.out.println("\nValue of m ");
for(int i=0;i<p;++i)
System.out.print("m"+(i+1)+"="+m[i]+"  ");

count1=0;count2=0;count3=0;
}
while(flag==0);

System.out.println("\n\n\nThe Final Clusters By Kmeans are as follows: ");
for(int i=0;i<p;++i)
{
System.out.print("K"+(i+1)+"{ ");
for(int j=0;k[i][j]!=-1 && j<n-1;++j)
System.out.print(k[i][j]+" ");
System.out.println("}");
}
}
}
/*
OUTPUT
C:\Users\Aditya\Desktop\aditya>javac k_means.java
C:\Users\Aditya\Desktop\aditya>java k_means
Enter the number of elements
8
Enter 8 elements:
2 3 6 8 12 15 18 22
Enter the number of clusters:
3

At this step
Value of clusters
K1{ 2 }
K2{ 3 }
K3{ 6 8 12 15 18 22 }
Value of m
m1=2.0  m2=3.0  m3=13.5

At this step
Value of clusters
K1{ 2 }
K2{ 3 6 8 }
K3{ 12 15 18 22 }
Value of m
m1=2.0  m2=5.666666666666667  m3=16.75

At this step
Value of clusters
K1{ 2 3 }
K2{ 6 8 }
K3{ 12 15 18 22 }
Value of m
m1=2.5  m2=7.0  m3=16.75

At this step
Value of clusters
K1{ 2 3 }
K2{ 6 8 }
K3{ 12 15 18 22 }
Value of m
m1=2.5  m2=7.0  m3=16.75

The Final Clusters By Kmeans are as follows:
K1{ 2 3 }
K2{ 6 8 }
K3{ 12 15 18 22 } */

VIEW OUR SPECIAL PAGE FOR SEM 7 STUDENTS - sem 6 java codes

Friday, 22 February 2013

Internal Forwarding Techniques(Java Code):Store Fetch,Fetch Fetch,Store Store Forwarding

To download the program use the download box

AIM:To study Internal Forwarding Techniques.
//Program:
import java.util.*;
class internal_forward
{
static void menu()
{
System.out.println("Menu:\nRegisters: R1 R2 R3\nMemory locations: M1 M2 M3\nSample instruction:MOV M1,R1\n");
}
static void accept(String s[][])
{
Scanner scr=new Scanner(System.in);
System.out.print("\nEnter the two intructions:");
System.out.print("\nFirst Instruction:\nSource: ");
s[0][1]=scr.nextLine();
System.out.print("\nDestination: ");
s[0][0]=scr.nextLine();
System.out.print("\nSecond instruction:\nSource: ");
s[1][1]=scr.nextLine();
System.out.print("\nDestination: ");
s[1][0]=scr.nextLine();
}
static int check1(String s[][])
{
if(s[0][0].equals(s[1][1]))
{
System.out.println("\nStore Fetch Forwarding can be used");
System.out.println("MOV "+s[0][0]+" "+s[0][1]+"\nMOV "+s[1][0]+" "+s[0][1]);
return 1;
}
return 0;
}
static int check2(String s[][])
{
if(s[0][1].equals(s[1][1]))
{
System.out.println("\nFetch Fetch Forwarding can be used");
System.out.println("MOV "+s[0][0]+" "+s[0][1]+"\nMOV "+s[1][0]+" "+s[0][0]);
return 2;
}
return 0;
}
static int check3(String s[][])
{
if(s[0][0].equals(s[1][0]))
{
System.out.println("\nStore Store Forwarding can be used");
System.out.println("MOV "+s[0][0]+" "+s[1][1]);
return 3;
}
return 0;
}
public static void main(String args[])
{
Scanner scr=new Scanner(System.in);
System.out.println("Three Internal forwarding techniques:\n1.Store Fetch Forwarding\n2.Fetch Fetch Forwarding\n3.Store Store Forwarding\n\n");
String s[][]=new String[2][2];
int choice=1;
int flag=0;
do
{
flag=0;
menu();
accept(s);
System.out.println("Your intructions are as follows:\nMOV "+s[0][0]+" "+s[0][1]+"\nMOV "+s[1][0]+" "+s[1][1]);
flag=check1(s);
if(flag==0)
flag=check2(s);
if(flag==0)
flag=check3(s);
if(flag==0)
System.out.println("Your instructions are OK.No need to apply any of the 3 internal forwarding techniques.");
System.out.println("Do you wish to continue?(1/0)");
choice=scr.nextInt();
}
while(choice==1);
}
}

/*OUTPUT
C:\Users\Aditya\Desktop\aditya>javac internal_forward.java
C:\Users\Aditya\Desktop\aditya>java internal_forward
Three Internal forwarding techniques:
1.Store Fetch Forwarding
2.Fetch Fetch Forwarding
3.Store Store Forwarding

Menu:
Registers: R1 R2 R3
Memory locations: M1 M2 M3
Sample instruction:MOV M1,R1

Enter the two intructions:
First Instruction:
Source: R1
Destination: M1

Second instruction:
Source: M1
Destination: R2

Your intructions are as follows:
MOV M1 R1
MOV R2 M1

Store Fetch Forwarding can be used
MOV M1 R1
MOV R2 R1
Do you wish to continue?(1/0)
1

Menu:
Registers: R1 R2 R3
Memory locations: M1 M2 M3
Sample instruction:MOV M1,R1

Enter the two intructions:
First Instruction:
Source: M2
Destination: R1

Second instruction:
Source: M2
Destination: R2

Your intructions are as follows:
MOV R1 M2
MOV R2 M2

Fetch Fetch Forwarding can be used
MOV R1 M2
MOV R2 R1
Do you wish to continue?(1/0)
1

Menu:
Registers: R1 R2 R3
Memory locations: M1 M2 M3
Sample instruction:MOV M1,R1

Enter the two intructions:
First Instruction:
Source: R1
Destination: M3

Second instruction:
Source: R3
Destination: M3

Your intructions are as follows:
MOV M3 R1
MOV M3 R3

Store Store Forwarding can be used
MOV M3 R3
Do you wish to continue?(1/0)
1

Menu:
Registers: R1 R2 R3
Memory locations: M1 M2 M3
Sample instruction:MOV M1,R1

Enter the two intructions:
First Instruction:
Source: R1
Destination: M2

Second instruction:
Source: M1
Destination: R3

Your intructions are as follows:
MOV M2 R1
MOV R3 M1
Your instructions are OK.No need to apply any of the 3 internal forwarding techniques.
Do you wish to continue?(1/0)
0
*/

VIEW OUR SPECIAL PAGE FOR SEM 7 STUDENTS - sem 6 java codes

Wednesday, 13 February 2013

Decision Tree Algorithm:(Classification Problem Java Code)

Java Program for Decision Tree Algorithm.

import java.util.*;
//Aim: To implement Decision Tree algorithm in java
class tree
{
public double hd;
public double p;
char ch;
String parent;
tree(char c,int x)
{
ch=c;
if(x==1)
parent="outlook";
else if(x==2)
parent="temperature";
else if(x==3)
parent="humidity";
else if(x==4)
parent="windy";
}
} 
class d_tree
{
static char outlook[]={'S','S','O','R','R','R','O','S','S','R','S','O','O','R'};
static char temperature[]={'B','B','B','M','C','C','C','M','C','M','M','M','B','M'};
static char humidity[]={'H','H','H','H','N','N','N','H','N','N','N','H','N','H'};
static char windy[]={'F','T','F','F','F','T','T','F','F','F','T','T','F','T'};
static char class1[]={'N','N','P','P','P','N','P','N','P','P','P','P','P','N'};
static double G1,G2,G3,G4;
static double HD;
static double play=9.0/14.0;
static double nplay=5.0/14.0;
static double temp1,temp2;
static int row=0,column=0;
static char classify[][]=new char[10][10];

static tree cal_hd(tree t,int choice)
{
double count1=0,count2=0;
if(choice==1)
{for(int i=0;i<14;++i)
{if(t.ch==outlook[i] && class1[i]=='P')
++count1;
else
if(t.ch==outlook[i] && class1[i]=='N')
++count2;
}}

if(choice==2)
{for(int i=0;i<14;++i)
{if(t.ch==temperature[i] && class1[i]=='P')
++count1;
else
if(t.ch==temperature[i] && class1[i]=='N')
++count2;
}}

if(choice==3)
{for(int i=0;i<14;++i)
{if(t.ch==humidity[i] && class1[i]=='P')
++count1;
else
if(t.ch==humidity[i] && class1[i]=='N')
++count2;
}}

if(choice==4)
{for(int i=0;i<14;++i)
{if(t.ch==windy[i] && class1[i]=='P')
++count1;
else
if(t.ch==windy[i] && class1[i]=='N')
++count2;
}}

temp1=count1/(count1+count2);
temp2=count2/(count1+count2);
t.p=(count1+count2)/14;

if(temp1==0 || temp2==0)
t.hd=0;
else
t.hd=temp1*(Math.log(1/temp1)/Math.log(10))+temp2*(Math.log(1/temp2)/Math.log(10));

return t;
}

static int check_if_equal(char ca[][],int cnt)
{
int count1=0,count2=0,count3=0;
char c1=ca[0][0],c2=ca[0][1],c3=ca[0][2];

for(int i=0;i<cnt;++i)
{
if(ca[i][0]==c1)
++count1;

if(ca[i][1]==c2)
++count2;

if(ca[i][2]==c3)
++count3;
}

if(count1==cnt)
return 0;
else
if(count2==cnt)
return 1;
else
if(count3==cnt)
return 2;
else
return -1;
}

static void compute(char ch)
{
char ca1[][]=new char[10][10];
char ca2[][]=new char[10][10];
char ca3[][]=new char[10][10];
int cnt1=0;
int cnt2=0;
for(int i=0;i<14;++i)
{
if(outlook[i]==ch)
{
ca1[cnt1][0]=temperature[i];
ca1[cnt1][1]=humidity[i];
ca1[cnt1][2]=windy[i];
ca1[cnt1][3]=class1[i];
++cnt1;
}
}

int check1=0,check2=0;

for(int i=0;i<cnt1;++i)
{
if(ca1[i][3]=='P')
++check1;
else
if(ca1[i][3]=='N')
++check2;
if(check1==cnt1)
{
classify[row][column++]=ch;
classify[row][column++]='P';
System.out.print("--------->PLAY");
return;
}
else
if(check2==cnt1)
{
classify[row][column++]=ch;
classify[row][column++]='N';
System.out.print("--------->NO PLAY");
return;
}
}

//end of for loop

/*System.out.println("The array containing outlook value ch");
for(int i=0;i<9;++i)
{
for(int j=0;j<9;++j)
{
System.out.print(" "+ca1[i][j]);
}
System.out.println();
}
*/



//The array containing outlook value ch and class1=P
cnt1=0;
for(int i=0;i<10;++i)
{
if(ca1[i][3]=='P')
{
ca2[cnt1][0]=ca1[i][0];
ca2[cnt1][1]=ca1[i][1];
ca2[cnt1][2]=ca1[i][2];
ca2[cnt1][3]=ca1[i][3];
++cnt1;
}
}

classify[row][column++]=ch;

int z=check_if_equal(ca2,cnt1);

//System.out.println("The value of Z is "+z);
if(z==0)
{
System.out.print("\n--------->TEMPERATURE("+ca2[0][0]+")--------->PLAY\n");
classify[row][column++]=ca2[0][0];
classify[row][column++]='P';
}
else
if(z==1)
{
System.out.print("\n--------->HUMIDITY("+ca2[0][1]+")--------->PLAY\n");
classify[row][column++]=ca2[0][1];
classify[row][column++]='P';
}
else
if(z==2)
{
System.out.print("\n--------->WINDY("+ca2[0][2]+")--------->PLAY\n");
classify[row][column++]=ca2[0][2];
classify[row][column++]='P';
}

/*
System.out.println("The array containing ch and play");
for(int i=0;i<9;++i)
{
for(int j=0;j<9;++j)
{
System.out.print(" "+ca2[i][j]);
}
System.out.println();
}
*/

//The array containing outlook value ch and class1=N
cnt1=0;
for(int i=0;i<10;++i)
{
if(ca1[i][3]=='N')
{
ca3[cnt1][0]=ca1[i][0];
ca3[cnt1][1]=ca1[i][1];
ca3[cnt1][2]=ca1[i][2];
ca3[cnt1][3]=ca1[i][3];
++cnt1;
}}

++row;
column=0;
classify[row][column++]=ch;
z=check_if_equal(ca3,cnt1);
if(z==0)
{
System.out.print("--------->TEMPERATURE("+ca3[0][0]+")--------->NO PLAY");
classify[row][column++]=ca3[0][0];
classify[row][column++]='N';
}
else
if(z==1)
{
System.out.print("--------->HUMIDITY("+ca3[0][1]+")--------->NO PLAY");
classify[row][column++]=ca3[0][1];
classify[row][column++]='N';
}
else
if(z==2)
{
System.out.print("--------->WINDY("+ca3[0][2]+")--------->NO PLAY");
classify[row][column++]=ca3[0][2];
classify[row][column++]='N';
}

/*System.out.println("The array containing ch and not play");
for(int i=0;i<9;++i)
{
for(int j=0;j<9;++j)
{
System.out.print(" "+ca3[i][j]);
}
System.out.println();
}
*/

}
public static void main(String args[])
{
Scanner scr=new Scanner(System.in);
HD=play*(Math.log(1/play)/Math.log(10))+nplay*(Math.log(1/nplay)/Math.log(10));
System.out.println("\nThe value of H(D) is "+HD);

tree sunny=new tree('S',1);
tree overcast=new tree('O',1);
tree rain=new tree('R',1);
tree hot=new tree('B',2); 
tree mild=new tree('M',2);
tree cool=new tree('C',2);
tree high=new tree('H',3); 
tree normal=new tree('N',3);
tree tru=new tree('T',4);
tree fal=new tree('F',4);

sunny=cal_hd(sunny,1);
overcast=cal_hd(overcast,1);
rain=cal_hd(rain,1);
hot=cal_hd(hot,2);
mild=cal_hd(mild,2);
cool=cal_hd(cool,2);
high=cal_hd(high,3);
normal=cal_hd(normal,3);
tru=cal_hd(tru,4);
fal=cal_hd(fal,4);

G1=HD-(sunny.p*sunny.hd+overcast.p*overcast.hd+rain.p*rain.hd);
G2=HD-(hot.p*hot.hd+mild.p*mild.hd+cool.p*cool.hd);
G3=HD-(high.p*high.hd+normal.p*normal.hd);
G4=HD-(tru.p*tru.hd+fal.p*fal.hd);

int gflag=0;

System.out.println("\nGain(D,Outlook)= "+G1 +"\nGain(D,Temperature)= "+G2+"\nGain(D,Humidity)= "+G3+"\nGain(D,Windy)= "+G4);
System.out.print("\nThe Splitting factor is ");
if(G1>G2 && G1>G3 && G1>G4)
System.out.println("Outlook");
else
if(G2>G1 && G2>G3 && G2>G4)
System.out.println("Temperature");
else
if(G3>G2 && G3>G1 && G3>G4)
System.out.println("Humidity");
else
if(G4>G2 && G4>G3 && G4>G1)
System.out.println("Windy");

System.out.println("\nThe Tree is as follows:-");
System.out.print("\nOUTLOOK(S)");
compute('S');
++row;
column=0;
System.out.print("\n\n\nOUTLOOK(O)");
compute('O');
++row;
column=0;
System.out.print("\n\n\nOUTLOOK(R)");
compute('R');

char input[]=new char[4];
String s; 

System.out.println("\nMenu:\n\nOutlook: Sunny=S Overcast=O Rainy=R\n\nTemperature: Hot=B Medium=M Cool=C\n\nHumidity: High=H Normal=N\n\nWindy: True=T False=F");

System.out.println("\n\nEnter your new tuple to be classified ");
System.out.print("\nOutlook(S/O/R)= ");
s=scr.nextLine();
input[0]=s.charAt(0);
System.out.print("\nTemperature(B/M/C)= ");
s=scr.nextLine();
input[1]=s.charAt(0);
System.out.print("\nHumidity(H/N)= ");
s=scr.nextLine();
input[2]=s.charAt(0);
System.out.print("\nWindy(T/F)= ");
s=scr.nextLine();
input[3]=s.charAt(0);

/* Display classify
for(int i=0;i<9;++i)
{
for(int j=0;j<9;++j)
{
System.out.print(" "+classify[i][j]);
}
System.out.println();
}
*/
System.out.print("Your input is ");
for(int i=0;i<4;++i)
System.out.print(" "+input[i]);
int inc1=0;
do
{
if(input[0]==classify[inc1][0])
{
if(classify[inc1][1]=='P')
   {
    System.out.println("\n\nTuple classified as PLAY");
    break;
   }
else
if(input[1]==classify[inc1][1] || input[2]==classify[inc1][1] || input[3]==classify[inc1][1])
{
   if(classify[inc1][2]=='P')
   {
   System.out.println("\n\nTuple classified as PLAY");
   break;
   }
   else
   if(classify[inc1][2]=='N')
   {
   System.out.println("\n\nTuple classified as NO PLAY");
   break;
   }
}
}
++inc1;
}
while(true);
}
}

/*OUTPUT1

C:\Users\Aditya\Desktop\aditya>java d_tree

The value of H(D) is 0.2830542780615224

Gain(D,Outlook)= 0.07427909717678749
Gain(D,Temperature)= 0.00879686881360553
Gain(D,Humidity)= 0.04570704031674405
Gain(D,Windy)= 0.014487679755121663

The Splitting factor is Outlook

The Tree is as follows:-

OUTLOOK(S)
--------->HUMIDITY(N)--------->PLAY
--------->HUMIDITY(H)--------->NO PLAY


OUTLOOK(O)--------->PLAY


OUTLOOK(R)
--------->WINDY(F)--------->PLAY
--------->WINDY(T)--------->NO PLAY
Menu:

Outlook: Sunny=S Overcast=O Rainy=R

Temperature: Hot=B Medium=M Cool=C

Humidity: High=H Normal=N

Windy: True=T False=F


Enter your new tuple to be classified

Outlook(S/O/R)= S

Temperature(B/M/C)= M

Humidity(H/N)= N

Windy(T/F)= T
Your input is  S M N T

Tuple classified as PLAY

------------------------------------------------------------------------

OUTPUT2

C:\Users\Aditya\Desktop\aditya>java d_tree

The value of H(D) is 0.2830542780615224

Gain(D,Outlook)= 0.07427909717678749
Gain(D,Temperature)= 0.00879686881360553
Gain(D,Humidity)= 0.04570704031674405
Gain(D,Windy)= 0.014487679755121663

The Splitting factor is Outlook

The Tree is as follows:-

OUTLOOK(S)
--------->HUMIDITY(N)--------->PLAY
--------->HUMIDITY(H)--------->NO PLAY


OUTLOOK(O)--------->PLAY


OUTLOOK(R)
--------->WINDY(F)--------->PLAY
--------->WINDY(T)--------->NO PLAY
Menu:

Outlook: Sunny=S Overcast=O Rainy=R

Temperature: Hot=B Medium=M Cool=C

Humidity: High=H Normal=N

Windy: True=T False=F


Enter your new tuple to be classified

Outlook(S/O/R)= S

Temperature(B/M/C)= M

Humidity(H/N)= H

Windy(T/F)= T
Your input is  S M H T

Tuple classified as NO PLAY

*/

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Monday, 4 February 2013

Classification problem:Bayesian Classification(Java Code)

Java code for Bayesian Classification is as follows

import java.util.*;
class btheo
{
static char outlook[]={'S','S','O','R','R','R','O','S','S','R','S','O','O','R'};
static char temperature[]={'H','H','H','M','C','C','C','M','C','M','M','M','H','M'};
static char humidity[]={'P','P','P','P','N','N','N','P','N','N','N','P','N','P'};
static char windy[]={'F','T','F','F','F','T','T','F','F','F','T','T','F','T'};
static char class1[]={'N','N','P','P','P','N','P','N','P','P','P','P','P','N'};

static double prob[][]=new double[4][2];

static double pp=9.0/14.0;
static double npp=5.0/14.0;

static int flag=0;
static int flag1=0;

static double play_N=1;
static double notplay_N=1;

static void cal_N(int a)
{
  if(a==1)
  {
    for(int i=0;i<4;++i)
    play_N*=prob[i][0];

    play_N*=pp;
  //System.out.println("\nValue of N of play \n"+play_N);
  }
  else
  {
    for(int i=0;i<4;++i)
    notplay_N*=prob[i][1];

    notplay_N*=npp;
  //System.out.println("\nValue of N of No play \n"+notplay_N);
  }
}

static double cal_play_prob(char ch)
{
double prob=0;
double count=0;

if(flag==0)
{
for(int i=0;i<14;++i)
if(outlook[i]==ch && class1[i]=='P')
++count;

prob=count/9.0;

flag=1;
}
else if(flag==1)
{
for(int i=0;i<14;++i)
if(temperature[i]==ch && class1[i]=='P')
++count;

prob=count/9.0;

flag=2;
}
else if(flag==2)
{
for(int i=0;i<14;++i)
if(humidity[i]==ch && class1[i]=='P')
++count;

prob=count/9.0;

flag=3;
}
else
if(flag==3)
{
for(int i=0;i<14;++i)
if(windy[i]==ch && class1[i]=='P')
++count;

prob=count/9.0;
}
return prob;
}






static double cal_noplay_prob(char ch)
{
double prob=0;
double count=0;

if(flag1==0)
{
for(int i=0;i<14;++i)
if(outlook[i]==ch && class1[i]=='N')
++count;

prob=count/5.0;

flag1=1;
}
else if(flag1==1)
{
for(int i=0;i<14;++i)
if(temperature[i]==ch && class1[i]=='N')
++count;

prob=count/5.0;

flag1=2;
}
else if(flag1==2)
{
for(int i=0;i<14;++i)
if(humidity[i]==ch && class1[i]=='N')
++count;

prob=count/5.0;

flag1=3;
}
else
if(flag1==3)
{
for(int i=0;i<14;++i)
if(windy[i]==ch && class1[i]=='N')
++count;

prob=count/5.0;
}
return prob;
}
public static void main(String args[])
{
Scanner scr=new Scanner(System.in);
System.out.println("Table\n");
System.out.println("Outlook\t   Temperature\t    Humidity\t      Windy     \tClass");
for(int i=0;i<14;++i)
{
System.out.print(outlook[i]+"\t\t"+temperature[i]+"\t\t"+humidity[i]+"\t\t"+windy[i]+"\t\t"+class1[i]);
System.out.println();
}
System.out.println("Menu:\nOutlook: Sunny=S Overcast=O Rain=R\tTemperature: Hot=H Mild=M Cool=C\n");
System.out.println("Humidity: Peak=P Normal=N\t\tWindy: True=T False=F\n\nYour input should belong to one of these classes.\n");
System.out.println("class1: Play=P   class2:Not Play=NP");
System.out.println("\nEnter your input: example. t={rain,hot,peak,false} input will be R,H,P,F");
String s=scr.nextLine();
char ch;
int count=0;
for(int i=0;i<8;i+=2)
{
ch=s.charAt(i);
prob[count][0]=cal_play_prob(ch);
prob[count][1]=cal_noplay_prob(ch);
++count;
}

cal_N(1);
cal_N(2);

double pt=play_N+notplay_N;

double prob_of_play=0;
double prob_of_noplay=0;

prob_of_play=play_N/pt;
prob_of_noplay=notplay_N/pt;

System.out.println("\nProbability of play "+prob_of_play);
System.out.println("\nProbability of NO play "+prob_of_noplay );

if(prob_of_play>prob_of_noplay)
System.out.println("\nThe new tuple classified under \"PLAY\" category.Hence there will be play!!!");
else
System.out.println("\nThe new tuple classified under \"NO PLAY\" category.Hence there will be NO play.");

}
}

/* output

C:\Users\Aditya\Desktop\aditya>javac btheo.java

C:\Users\Aditya\Desktop\aditya>java btheo
Table

Outlook    Temperature      Humidity          Windy             Class
S               H               P               F               N
S               H               P               T               N
O               H               P               F               P
R               M               P               F               P
R               C               N               F               P
R               C               N               T               N
O               C               N               T               P
S               M               P               F               N
S               C               N               F               P
R               M               N               F               P
S               M               N               T               P
O               M               P               T               P
O               H               N               F               P
R               M               P               T               N
Menu:
Outlook: Sunny=S Overcast=O Rain=R      Temperature: Hot=H Mild=M Cool=C

Humidity: Peak=P Normal=N               Windy: True=T False=F

Your input should belong to one of these classes.

class1: Play=P   class2:Not Play=NP

Enter your input: example. t={rain,hot,peak,false} input will be R,H,P,F
R,H,P,F

Probability of play 0.3665689149560117

Probability of NO play 0.6334310850439884

The new tuple classified under "NO PLAY" category.Hence there will be NO play.

*/

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