This document provides instructions for formatting a term paper. It includes sections for the title page, table of contents, introduction, proposed system description, system requirements, requirement analysis, system design, source code, testing, and future scope. It then provides a sample term paper formatted according to these instructions on the topic of a tic-tac-toe game. The paper includes an introduction describing the existing system and proposed improvements. It details the system requirements, design including flowcharts, and provides the source code for the tic-tac-toe game.

1. FORMAT OF TERM PAPER
1. Title page of project report
2. Table of contents
3. Introduction
4. Proposed system
i. Description
ii. System requirements
5. Requirement Analysis
6. System Design
7. Source code
8. Testing
9. Future scope of project

2. OF
(TIC TAC TOE)
Submitted in the partial fulfillment of the Degree of Bachelor of Technology
(Integrated)
In
Computer Science and Engineering
SUBMITTED BY:- GUIDED BY:
Name SANDEEP YADAV Miss Sukhdilpreet Kaur
Regd. No.10805155
RollnoR246B54
SUBMITTED TO
Department of Computer Science and Engineering Lovely Professional University
Phagwara

3. ACKNOWLEDGEMENT
I take this opportunity to present my votes of thanks to all those guidepost who really
acted as lightening pillars to enlighten our way throughout this project that has led to
successful and satisfactory completion of this study.
We are really grateful to our HOD Mr. Rohit Dhand for providing us with an
opportunity to undertake this project in this university and providing us with all the
facilities. We are highly thankful to Miss Sukhdilpreet Kaur for her active support,
valuable time and advice, whole-hearted guidance, sincere cooperation and pains-taking
involvement during the study and in completing the assignment of preparing
the said project within the time stipulated.
Lastly, We are thankful to all those, particularly the various friends , who have been
instrumental in creating proper, healthy and conductive environment and including
new and fresh innovative ideas for us during the project, their help, it would have
been extremely difficult for us to prepare the project in a time bound framework.
Name ANIL KUMAR
Regd. No.10802020
Roll No.R246B55

4. TABLE OF CONTENTS
1. Introduction
2. Proposed system
i. Description
ii. System requirements
3. Requirement Analysis
4. System Design
5. Source code
6. Testing
7. Future scope of project

5. INTRODUCTION
In the existing system, most of the records are maintained on paper. It becomes very
inconvenient to modify the data. In the existing system, here is a possibility that the same
data in different registers may have different values which means the entries of the same
data do not match. This inconsistent state does not supply the concrete information which
poses a problem in the case information related to particular search record.
Our project is very useful. User is no longer required to check his register in search of
records, as now it can be searched over the software by choosing some options. The user
need not to type in most of the information. He/she is just required to enter the desired
options. On the whole it liberates the user from keeping lengthy manual records. In a
nutshell, it abates the work load of an organization.
In today’s world, no one likes to perform calculations on calculator or manually when
computer is there. Every one wants his/her work to be done by computer automatically
and displaying the result for further manipulations.

6. PROPOSED SYSTEM
The following documentation is a project the “Name of the term paper allotted”. It is a
detailed summary of all the drawbacks of the old system and how the new proposed
system overcomes these shortcomings. The new system takes into account the various
factors while designing a new system. It keeps into the account the Economical
bandwidth available for the new system. The foremost thing that is taken care of is the
Need and Requirements of the User.
DESCRIPTION
Before developing software we keep following things in mind that we can develop
powerful and quality software
PROBLEM STATEMENT
o Problem statement was to design a module:
o Which is user friendly
o Which will restrict the user from accessing other user’s data.
o Which will help user in viewing his data and privileges.
o Which will help the administrator to handle all the changes.
FUNCTIONS TO BE PROVIDED:
The system will be user friendly and completely menu driven so that the users shall have
no problem in using all options.
o The system will be efficient and fast in response.
o The system will be customized according to needs.
SYSTEM REQUIRMENTS
Operating system: MS Windows XP or Windows Vista
Language: C Language
Processor: Pentium IV Processor
RAM: 512 MB
Hard disk: 5 GB

7. REQUIREMENT ANALYSIS
This process is adopted when management of the system development, Personnel decide
that the particular system needs improvement. The system development life cycle is the
set of activities, carried out by the analyst, designers and users to develop and implement
a system. The systems that are present in the nature follow common life cycle pattern.
For example consider the raining system. Initially the rain falls into the river, river flows
into sea, the sea water evaporates to form vapors, the vapors form clouds which again
bring rain. Similarly consider a man made system initially a system is analyzed, designed
and made operational by the efforts of system analysis. After successful operation or a
number of users, the system becomes less and less effective by change in the
environment. So these changes have to be incorporated in to the system by minor
modifications. So the general activities from the life cycle of the system are given below:
· Select ion and identification of the system to be studied
· Preliminary study
· Defining the system
· Design and development of the system
· Implementation of the system

8. SYSTEM DESIGN
Then we began with the design phase of the system. System design is a solution, a “HOW
TO” approach to the creation of a new system. It translates system requirements into
ways by which they can be made operational. It is a translational from a user oriented
document to a document oriented programmers. For that, it provides the understanding
and procedural details necessary for the implementation. Here we use Flowchart to
supplement the working of the new system. The system thus made should be reliable,
durable and above all should have least possible maintenance costs. It should overcome
all the drawbacks of the Old existing system and most important of all meet the user
requirements.
START
WELCOME TO GAME TIC TAC TOE
Enter
your
choice
?
START CHOICE QUIT
Do you want to
continue?
STOP

9. SOURCE CODE
#include<iostream.h>
#include<stdio.h>
#include<conio.h>
#include<graphics.h>
#include<stdlib.h>
#include<string.h>
#include<dos.h>
void*message;
int select(int mult)
{
union REGS inregs, outregs ;
int bli=1,use=1,key=34,i;
settextstyle(2,0,5);
while(key!=28)
{
if(bli>0)
{
use=bli;
setfillstyle(1,0);
bli=0-bli;
}
else if(bli<0)
{
use=0-bli;
setfillstyle(1,8);
bli=0-bli;
}
floodfill(221,111+use*40,15);
delay(100);
if(bli<0)
{
key=kbhit();
if(kbhit())
{
inregs.h.ah = 0 ;
int86(22, &inregs, &outregs) ;
key=outregs.h.ah;
}
}

10. if((key==72)&&(use>1))
{
bli=use-1;
}
if((key==80)&&(use<mult))
{
bli=use+1;
}
}
if(bli<0)
bli=0-bli;
return(bli);
}
void box(char mes[50])
{ putimage(5,5,message,0);
settextstyle(0,0,1);
outtextxy(20,30,mes);
}
void draw(char mn[3][3])
{
char as[3][3][3];
char num[9][3];
for(int i=0;i<10;i++)
{ strcpy(num[i]," ");
num[i][0]=char(49+i);
}
for(i=0;i<3;i++)
for(int j=0;j<3;j++)
strcpy(as[i][j]," ");
for(i=0;i<3;i++)
{ for(j=0;j<3;j++)
{as[i][j][0]=mn[i][j];}
}
clearviewport();
setcolor(15);
rectangle(0,0,639,479);
setfillstyle(1,8);
settextstyle(0,0,1);
for(i=0;i<3;i++)
{
rectangle(192,117+i*85,267,192+i*85);
outtextxy(260,185+i*85,num[0+i*3]);
rectangle(277,117+i*85,352,192+i*85);
outtextxy(345,185+i*85,num[1+i*3]);
rectangle(362,117+i*85,437,192+i*85);
outtextxy(430,185+i*85,num[2+i*3]);

11. }
floodfill(500,430,15);
setcolor(15);
settextstyle(1,0,4);
for(i=0;i<3;i++)
{
outtextxy(221,135+i*85,as[i][0]);
outtextxy(306,135+i*85,as[i][1]);
outtextxy(391,135+i*85,as[i][2]);
}
}
void main()
{
clrscr();
int gd=DETECT,gm;
initgraph(&gd,&gm,"");
message=malloc(imagesize(5,5,634,55));
setcolor(15);
rectangle(5,5,634,55);
setfillstyle(1,RED);
floodfill(30,30,15);
outtextxy(10,10,"Message:-");
getimage(5,5,634,55,message);
char col[3][3],input,madu,comps,hums,mess[70]={"computer has selected the
symbol . Press any key to continue.."};
int
exii,dang[8],my[8],hard,many,result,guess=7,bre,mad=2,count=0,dont=0,play[8],p,q
,end=0,note,inpu,first,use;
do
{guess=7;mad=2;count=0;dont=0;end=0;result=0;
for(int i=0;i<8;i++)
play[i]=0;
many=0;exii=1;
clearviewport();
setcolor(15);
rectangle(0,0,639,479);
rectangle(20,320,620,460);
rectangle(220,150,390,180);
rectangle(240,155,370,175);
setfillstyle(1,8);
floodfill(100,100,15);
setcolor(15);
settextstyle(4,0,4);
outtextxy(200,50,"TIC TAC TOE");
settextstyle(3,0,1);
outtextxy(40,290,"How to play :-");

12. outtextxy(35,330,"In this Game, you may select your symbol. You musttryattain");
outtextxy(26,350,"three of your symbols in a line. if you suceed you are the
winner.");
outtextxy(35,370," But at the same time you should prevent the computer from");
outtextxy(35,390,"getting three of its symbols in a line. To play enterthe number");
outtextxy(32,410,"associated with the place where you want to play.Press any key");
outtextxy(35,430,"to start");
settextstyle(2,0,6);
outtextxy(258,155,"Start Game");
select(1);
hard=2;
for(int j=0;j<8;j++)
{dang[j]=0;my[j]=0;play[j]=0;}
for(j=0;j<3;j++)
{
for(int k=0;k<3;k++)
col[j][k]=' ';
}
draw(col);
box("Please type in your symbol");
hums=getche();
if((hums!='X')&&(hums!='x')
)
comps='X';
else
comps='0';
mess[33]=comps;
box(mess);
getch();
randomize();
first=(int(rand()%100));
if(hard==2)guess=(int(rand()%100));
else guess=5;
if((first%4)>=2)
{use=guess%3;box("Computer has the first chance to play!");}
else
{use=3;
mad=0;box("You have the first chance to play!");}
delay(2000);
do
{f
or(int j=0;j<8;j++)
{dang[j]=0;my[j]=0;}
count++;
mad++;bre=0;
if((end!=1)&&(mad!=1))

13. { switch(use)
{
case 0:{ switch(count)
{
case 1: col[2][2]=comps;break;
case 2: {if(col[1][1]==hums)
{col[0][0]=comps;play[0]=1;}
else if((col[2][0]==hums)||(col[2][1]==hums))
{col[0][2]=comps;play[1]=1;}
else if((col[0][1]==hums))
{col[0][2]=comps;play[3]=1;}
else if((col[1][0]==hums))
{col[2][0]=comps;play[4]=1;}
else if((col[0][2]==hums)||(col[1][2]==hums))
{col[2][0]=comps;play[2]=1;}
else if (col[0][0]==hums)
{col[0][2]=comps;play[3]=1;}
else dont=1;
}break;
case 3:{if(play[0]==1)
dont=1;
else if((play[1]==1)&&(col[1][2]==hums))
{col[0][0]=comps;}
else if((play[2]==1)&&(col[2][1]==hums))
{col[0][0]=comps;}
else if((play[3]==1)&&((col[2][1]==hums)||(col[1][2]==hums)))
{col[2][0]=comps;}
else if((play[4]==1)&&(col[2][1]==hums))
{col[0][2]=comps;}
else
dont=1;
}break;
case 4:dont=1;break;
}
}break;
case 1:{switch(count)
{ case 1:col[0][1]=comps;break;
case 2:{if(col[2][0]==hums)
col[0][0]=comps;
else if(col[1][0]==hums)
col[0][0]=comps;
else if(col[0][2]==hums)
col[1][0]=comps;
else if(col[1][2]==hums)
col[0][2]=comps;

14. else if(col[0][0]==hums)
col[1][2]=comps;
else if(col[2][2]==hums)
col[0][2]=comps;
else if(q<=1)
col[2][2]=comps;
else
col[2][0]=comps;
} break;
case 3:dont=1;
}
}break;
case 2:{switch(count)
{ case 1:col[1][1]=comps;break;
case 2:dont=1;
}
}break;
case 3:{dont=1;
}break;
} if(dont==1)
{f
or(int i=0,l=2;i<3;i++,l--)
{ if(col[i][i]==hums)
dang[0]++;
else if(col[i][i]==comps)
my[0]++;
if(col[i][l]==hums)
dang[1]++;
else if(col[i][l]==comps)
my[1]++;
}f
or(j=0;j<3;j++)
{
for(int k=0;k<3;k++)
{
if(col[j][k]==hums)
dang[j+2]++;
else if(col[j][k]==comps)
my[j+2]++;
if(col[k][j]==hums)
dang[j+5]++;
else if(col[k][j]==comps)
my[j+5]++;

15. }
}f
or(int j=0;j<8;j++)
{ if((my[j]==3)||(dang[j]==3)||(count==5))
end=1;
if((dang[j]==2)&&(my[j]!=0))
dang[j]=0;
if((my[j]==2)&&(dang[j]==0))
{my[j]=3;bre=1;}
} if(bre==1)
{for(j=0;j<8;j++)
dang[j]=0;
} if((dang[0]==2)||(my[0]==3))
{
for(int i=0;i<3;i++)
{ if(col[i][i]==' ')
col[i][i]=comps;
}}
else if((dang[1]==2)||(my[1]==3))
{
for(int i=0,l=2;i<3;i++,l--)
{ if(col[i][l]==' ')
col[i][l]=comps;
}}
else
if((dang[2]==2)||(my[2]==3)||(dang[3]==2)||(my[3]==3)||(dang[4]==2)||(my[4
]==3))
{f
or(j=0;j<3;j++)
{if((dang[j+2]==2)||(my[j+2]==3))
for(int k=0;k<3;k++)
{if(col[j][k]==' ')
{col[j][k]=comps;bre=1;}}
}
}
else
if((dang[5]==2)||(my[5]==3)||(dang[6]==2)||(my[6]==3)||(dang[7]==2)||(my[7
]==3))
{f
or(int j=0;j<3;j++)
{if((dang[j+5]==2)||(my[j+5]==3))

16. for(int k=0;k<3;k++)
{if(col[k][j]==' ')
{col[k][j]=comps;bre=1;}}
}
}
else if(col[1][1]==' ')
col[1][1]=comps;
else if((use==2)&&(col[2][2]==' '))
col[2][2]=comps;
else if((use==2)&&(col[0][2]==' '))
col[0][2]=comps;
else
if((((col[0][0]==hums)&&(col[2][2]==hums))||((col[0][2]==hums)&&(col[2]
[0]==hums)))&&(col[1][2]==' '))col[1][2]=comps;
else
if((col[1][1]!=hums)&&((col[0][0]==hums)||(col[2][2]==hums))&&((col[0]
[1]==hums)||(col[1][2]==hums))&&(col[0][2]==' '))col[0][2]=comps;
else
if((col[1][1]!=hums)&&((col[0][0]==hums)||(col[2][2]==hums))&&((col[1]
[0]==hums)||(col[2][1]==hums))&&(col[2][0]==' '))col[2][0]=comps;
else
if((col[1][1]!=hums)&&((col[0][2]==hums)||(col[2][0]==hums))&&((col[2]
[1]==hums)||(col[1][2]==hums))&&(col[2][2]==' '))col[2][2]=comps;
else
if((col[1][1]!=hums)&&((col[0][2]==hums)||(col[2][0]==hums))&&((col[0]
[1]==hums)||(col[1][0]==hums))&&(col[0][0]==' '))col[0][0]=comps;
else if((col[1][1]!=comps)&&(col[2][2]==' '))
col[2][2]=comps;
else if((col[1][1]!=comps)&&(col[0][2]==' '))
col[0][2]=comps;
else if(col[0][0]==' ')
col[0][0]=comps;
else if(col[2][2]==' ')
col[2][2]=comps;
else if(col[0][1]==' ')
col[0][1]=comps;
else if(col[1][2]==' ')
col[1][2]=comps;
else if(col[0][2]==' ')
col[0][2]=comps;
else if(col[2][0]==' ')
col[2][0]=comps;
else if(col[1][0]==' ')
col[1][0]=comps;
else if(col[2][1]==' ')

17. col[2][1]=comps;
}f
or(int i=0;i<8;i++)
{if(my[i]==3)
end=1;
}}s
tar:
draw(col);
box(" ");
if(end!=1)
{
box("play");
madu=getche();
if((int(madu)<49)||(int(madu)>57))
{box("INVALID ENTRY!");for(long double jk=0;jk<99999999;jk++);
goto star;}
inpu=int(madu)-48;
p=(inpu-1)/3;
switch(inpu%3)
{case 0:q=2;break;
case 1:q=0;break;
case 2:q=1;break;
}
if(col[p][q]!=' ')
{box("Space is already occupied!");for(long double
jk=0;jk<99999999;jk++);goto star;}
col[p][q]=hums;
}f
or(j=0;j<8;j++)
{dang[j]=0;my[j]=0;}
for(int i=0,l=2;i<3;i++,l--)
{ if(col[i][i]==hums)
dang[0]++;
else if(col[i][i]==comps)
my[0]++;
if(col[i][l]==hums)
dang[1]++;
else if(col[i][l]==comps)
my[1]++;
}f
or(j=0;j<3;j++)
{
for(int k=0;k<3;k++)
{

18. if(col[j][k]==hums)
dang[j+2]++;
else if(col[j][k]==comps)
my[j+2]++;
if(col[k][j]==hums)
dang[j+5]++;
else if(col[k][j]==comps)
my[j+5]++;
}
}f
or(j=0;j<8;j++)
{if((my[j]==3)||(dang[j]==3))
end=1;
} }while((end!=1));
draw(col);
for(int asd=0;asd<6;asd++)
{many=many+1;
if((my[0]==3)||(dang[0]==3))
{exii=0;
if(many%2==1)
for(int m=0,n=0;m<3;m++,n++)
{ setfillstyle(1,BLUE);floodfill(193+m*85,118+n*85,15); }
else
for(int m=0,n=0;m<3;m++,n++)
{ setfillstyle(1,BLACK);floodfill(193+m*85,118+n*85,15); }}
else if((my[1]==3)||(dang[1]==3))
{exii=0;
if(many%2==1)
for(int m=0,n=2;m<3;m++,n--)
{ setfillstyle(1,BLUE);floodfill(193+m*85,118+n*85,15); }
else
for(int m=0,n=2;m<3;m++,n--)
{ setfillstyle(1,BLACK);floodfill(193+m*85,118+n*85,15); }}
else for(j=2;j<8;j++)
{if(((my[j]==3)||(dang[j]==3))&&(j<5))
{exii=0;
if(many%2==1)
for(int m=0,n=j-2;m<3;m++)
{ setfillstyle(1,BLUE);floodfill(193+m*85,118+n*85,15); }
else
for(int m=0,n=j-2;m<3;m++)
{ setfillstyle(1,BLACK);floodfill(193+m*85,118+n*85,15); }}
else if((my[j]==3)||(dang[j]==3))
{exii=0;
if(many%2==1)

19. for(int m=0,n=j-5;m<3;m++)
{ setfillstyle(1,BLUE);floodfill(193+n*85,118+m*85,15); }
else
for(int m=0,n=j-5;m<3;m++)
{ setfillstyle(1,BLACK);floodfill(193+n*85,118+m*85,15); }}
}
for(long double jk=0;jk<9999999;jk++);
if(exii==1)break;
}f
or(int m=0;m<8;m++)
{ if(my[m]==3)
result=1;
}
for(m=0;m<8;m++)
{ if(dang[m]==3)
result=2; }
switch(result)
{case 1:box("You loose! Want to try again(y/n)");break;
case 2:box("You win! Want to try again(y/n)");break;
default:box("The game is draw! Want to try again(y/n)");break;
} input=getche();
}while(input=='Y'||input=='y');
clearviewport();
setlinestyle(3,4,6);
outtextxy(80,150,"This game is developed by SUNDEEP YADAV");
outtextxy(170,175,"from the PROGRAMME B.TECH-M.TECH-C.S.E(246)");
delay(4000);
for(long double mas=0;mas<=99999999;mas++);
exit(0);
}

20. TESTING
Testing is the major control measure used during software development. Its basic
function is to detect errors in the software. During requirement analysis and design, the
output is a document that is usually textual and no executable. After the coding phase,
computer programs are available that can be executed for testing purpose. This implies
that testing not only, has to uncover errors introduced during coding, but also errors
introduced during previous phase. Thus the goal of testing is to uncover the requirements,
design and coding errors in the programs. So after testing the outputs of my project are as
follows:
** Paste the outputs of your project

21. FUTURE SCOPE OF THE PROJECT
Our project will be able to implement in future after making some changes and
modifications as we make our project at a very low level. So the modifications that can
be done in our project are:
In future one change can be done by adding the fingerprints of the persons of which the
address is entered.And one more major change which can be done in this project is that to
add the snaps of the person of which the address is entered.We can also add or subtract
details of the individual.