cn lab manual

AIM:- Implement the data link layer framing methods such as Character Stuffing.

PROGRAM:

#include<stdio.h>

#include<conio.h>

#include<string.h>

void main()

{

int i=0;j=0;n,pos;

char a[20],a[50],ch;

clrscr();

printf(“enter string :”);

scanf(“%s”,&a);

n=strlen[a];

printf(“enter position:”);

scanf(“%d”,&pos);

if(pos>n)

{

printf(“invalid position,enter again”);

scanf(“%d”,&pos);

}

printf(“enter the character”);

ch=getche()

b[0]=’d’;

b[1]=’l’;

KITS ENGG COLLEGE Page No:

b[2]=’e’;

b[3]=’s’;

b[4]=’t’;

b[5]=’x’;

j=6;

while(i<n)

{

if I = = position-1

{

b[j]=’d’;

b[j+1]=’l’;

b[j=2]=’e’;

b[j+3]=’ch’;

b[j+4]=’d’;

b[j+5]=’l’;

b[j+6]=’e’;

then

j=j+7;

}

If a[j]=d;

b[j+1]=l;

b[j+2]=e;

j=j+3;

}


b[j]=a[i];

I++;

J++;

}

b[j]=’d’;

b[j+1]=’l’;

b[j+2]=’e’;

b[j+3]=’e’;

b[j+4]=’t’;

b[j+5]=’x’;

b[j+6]= \0;

printf(“frame after stuffing is:”);

printf(“%s”,b);

getch();

}


Input:

Enter string : a s d l e f g h

Enter position: 8

invalid position

Enter again 3

Enter the character k

Output:

Frame after stuffing is:

d l e s t x a s d l e k d l e d le e f g h d l e e t x


AIM:- Implement the data link layer framing methods such as Bit Stuffing.

PROGRAM:

#include<stdio.h>

#include<conio.h>

#include<string.h>

void main()

{

int n,i,j,c,count,a[20],b[30];

clrscr();

printf(“enter the frame length”);

scanf(“%d”,&n);

printf(“enter input frame is 1s,0s only”);

for(i=0;i<n;i++)

scanf(“%d”,&a[i]);

i=0;

count=1;

j=0;

while(i<n)

{

If(a[i]==1)

b[j]=a[i];

for(k=i+1;a[k]==1&&k<n&&count<5;k++)


{

j++;

b[j]=a[k];

count++;

if(count ==5)

{

J++;

b[j]=0;

}

I=k;

}

}

else

{

b[j]=a[i];

}

I++;

J++;

}

printf(“after stuffing the frame is:”);

for(i=0;i<j;i++)

printf(“%d”,b[j]);

getch();

}


Output:

Enter frame length:10

Enter input frame (0’s,1’s)

1010111111

After stuffing the frame:1010111111


AIM:-Implement On A Data Set Of Characters the Cyclic Redundancy Check

PROGRAM:

#include<stdio.h>

#include<conio.h>

int gen[4],gen1,frl,rem[4];

void main( )

{

int I,j,fr[8],dupfr[11],recfr[11],then,flag;

clrscr( );

flr=8;

gen1=4;

printf(“enter frame:”);

for(i=0;i<fr1;i++)

{

scanf(“%d”,&fr[i]);

dupfr[i]=fr[i];

}

printf(“enter generator;”);

for(i=0;i<gen1;i++)

scanf(“%d”,&gen[i]);

tlen=fr1+genl-1;

for(i=fr1;i<tlen;i++)


{

dupfr[i]=0;

}

remainder(dupfr);

for(i=0;i<fr1;i++)

{

recfr[i]=fr[i];

}

for(i=fr1;j=1;j<gen1;i++,j++)

{

recfr[i]=rem[j];

}

remainder(recfr);

flag=0;

for(i=0;i<4;i++)

{

if(rem[i]!=0)

flag++;

}

if(flag==0)

{

printf(“frame received correctly”);

}

else


{

printf(“the received frame is wrong”);

}

getch( );

}

remainder (int fr[])

{

int k,kl,I,j;

for(k=0;k<fr1;k++)

{

if(fr[k]==1)

{

kl=k;

for(i=0,j=k,i<gen1;i++,j++)

{

rem[i]=fr[j]^gen[i];

}

for(i=0;i<gen1;i++)

{

fr[k1]=rem[i];

k1++;

}

}

}}


Output:

Enter frame 111111111

Enter generator 1101

Frame received correctly


AIM:- To implement dijkstra’s algorithm to compute the shortest path.

PROGRAM:-

#include<stdio.h>

main()

{

char ch='y',sp[10],nb[10];

int ds[10],j=0,k,sm,i,n;

clrscr();

printf("enter the source node:");

sp[0]=getche();

j++;

do

{

printf("\nhow many neighbours:");

scanf("%d",&n);

for(i=0;i<n;i++)

{

printf("enter the neighbour name:");

nb[i]=getche();

printf("\n enter distance:");

scanf("%d",&ds[i]);

}

sm=ds[0];


for(i=1;i<n;i++)

{

if(sm>ds[i])

{

sm=ds[i];k=i;

}

}

sp[j]=nb[k];

j++;

sp[j]='\0';

printf("\n shortest path is:");

for(i=0;sp[i]!='\0';i++)

printf("%c",sp[i]);

printf("\n do u want more neighbours?y/n");

ch=getche();

}while(ch=='y');

getch();

}


OUTPUT:-

enter source node:A

enter how many neighbours:4

enter neighbor name:B

enter distance:2

enter neighbor name:C

enter distance:3

enter neighbor name:D

enter distance:4

enter neighbor name:E

enter distance:1

shortrest path is:AE

do you want more neighbours?y/n:n


AIM:-Take A 64 Bit Playing Text And Encrypt The Same Using DES Algorithm

PROGRAM:

#include<stdio.h>

#include<conio.h>

#include<string.h>

void main()

{

char cipher[50],plain[50],key[50];

int cipher I,lp,lk;

clrscr();

while(1)

printf(“\n menu”);

printf(“\n 1.data encryption”);

printf(“\n2.decryption;”);

printf(“\n 3.exit”);

printf(“\n enter your choice;”);

scanf(“%d”,&choice);

switch(choice)

case1:

printf(“\n data encryption:”);

printf(“\n encryption text ”);

fflush(stdin);

gets(plain);

printf(“\n enter the encryption key;”);


gets(key);

lp=strlen(plain);

lk=strlen(key);

while(lp==lk)

{

printf(“\n invalid key”);

printf(“\n encryption key”);

gets(key);

lk=strlen(key);

}

for(i=0;plaini!=’10’;i++)

cipher[i]=plain[i]^key[i];

printf(“\n encryptyed text is :”);

capture[i]=’10’;

puts (capture);

break;

case2:

printf(“\n decryption:”);

fflush(stdin);

printf(“\n enter encrypted key:”);

gets(key);

lp=strlen(plain);

lk=strlen(key);

while(lp===lk)


{

printf(“\n invalid key”);

printf(“\n enter encryption key:”);

gets(key);

lk=strlen(key);

}

for(i=0;capture[i]!=’10’;i++)

plain[i]=capture[i]^key[i];

printf(“\n decrypted text is:”);

puts(plain);

case3:

exit(0);

default:printf(“invalid option”);

}

}

}


Output:

Menu:1.data encryption

2.decryption

3.exit

Enter your choice : 1

data encryption

enter plain text : kits

enter the encryption key:cse

encrypted text is:

Menu

1.data encryption

2.decryption

3.exit

Enter your choice :2

Decryption

Enter the encrypted key:cse

The decrypted text is:kits

Menu

1.data encryption

2.decryption

3.exit

Enter your choice :3


+

AIM:-Take An Example Subnet Graph With Weights Indicating Delay Between Nodes.Now Obtain Routing Table Ar Each Node Using Diastance Vector Routing Algorithm

PROGRAM:

#include<stdio.h>

#include<conio.h>

struct vertex

{

int num,delay;

char name;

};

void main()

{

int I,j,ch,n,adj,min,del[20][20];

struct vertex su,varchar[20];

clrcsr();

printf(“enter no of vertex;”);

scanf(“%d”,&n);

printf(“\n enter source vertex name:”);

fflush(stdin);

scanf(“%c”,&su.name);

printf(“enter source vertex number:”);


scanf(“%d”,&su.number);

fflush(stdin);

printf(“enter adjacent lines for su:”);

scanf(“%d”,&adj);

printf(“enter all line adjacent delay.[vetex_name,delay]:”);

for(i=1;i<=adj;i++)

{

flushall();

scanf(“%c”,”%d”,&var[i].name,&var[i].delay);

}

printf(“enter all routing delays for adjacent nodes for others:”);

for(i=1;i<=adj;i++)

{

printf(“enter routing delay for router number #%d”,i);

for(j=1;j<=n;j++)

scanf(“%d”,&del[I],[j]);

}

for(i=1;i<=n;i++)

{

if(i==su.number)

{

min=0;

}

else


{

min=3000;

ch=0;

for(j=1;j<=adj;j++)

{

if(min >ver[j].delay+del[i][j])

{

min=var[j].delay+del[j][i];

ch=j;

}

}

}

printf(“in shortest distance \n:”);

if(min!=0)

printf(“%d\t%c\n”,min,var[ch].name);

else

printf(“”%d”\t%c\n”,min,”__”);

}

}

}


Input:

Enter no of vertex :12

Enter source vertex name:j

Enter source vertex number:10

Enter adjacent lines for source vertex=4

Enter all lines adjacent delay

A 8

I 10

H 12

K 6

Enter all routing delay from adjacent nodes from others

Enter routing delay for router number 1

0 12 25 40 14 23 18 17 21 9 24 29

Enter routing delay for router numer 2

24 26 18 27 7 20 31 20 0 11 22 33 ‘


20 31 19 8 30 19 6 0 14 7 9 22


21 28 36 24 22 40 31 19 22 10 0 9


Output:

Shortest distance

17 i

Shortest distance

30 i

Shortest distance

18 H

Shortest distance

12 H

Shortest distance

10 I

Shortest distance

0 T

Shortest distance

6 K

Shortest distance

15 K


AIM:- To encrypt and decrypt a text data using RSA algorithm.

PROGRAM:-

#include<stdio.h>

main()

{

int p,q,n,msg,encrypt=1,decrypt=1,phi,e,d,i;

clrscr();

printf("enter two prime numbers");

scanf("%d %d",&p,&q);

n=p*q;

printf("n:%d\n",n);

phi=(p-1)*(q-1);

printf("Phi:%d\n",phi);

e=e(phi);

d=d(e,phi);

printf("Public key e:%d\n",e);

printf("Private key d:%d\n",d);

printf("enter msg");

scanf("%d",&msg);

for(i=0;i<e;i++)

{

encrypt*=(msg%n);

encrypt%=n;


}

for(i=0;i<d;i++)

{

decrypt*=encrypt%n;

decrypt%=n;

}

printf("message:%d\n",msg);

printf("encrypted message:%d\n",encrypt);

printf("decrypted message:%d\n",decrypt);

if(msg==decrypt)

printf("Satisfied");

getch();

}

int E(int phi)

{

int i,d;

for(i=2;i<phi;i++)

if(gcd(phi,i) == 1)

printf("\td = %d",i);

printf("\nSelect e value:");

scanf("%d",&d); return d;

}

int d(int e,int phi)

{


int d=1;

while((e*d)%phi!=1)

d++;return d;

}

int gcd(int a,int b)

{

int c;

while(1)

{

c = a%b;

if(c==0) return b;

a = b;b = c;

}

}


OUTPUT :

enter two prime numbers 17 11

n:187

phi:160

d=3 d=7 d=9 d=11 d=13 d=17 d=19 d=21 d=23

d=27 d=29 d=31 d=33 d=37 d=39 d=41 d=43 d=47 d=49

d=51 d=53 d=57 d=59 d=61 d=63 d=67 d=69 d=71 d=73

d=77 d=79 d=81 d=83 d=87 d=89 d=91 d=93 d=97 d=99

d=101 d=103 d=107 d=109 d=111 d=113 d=117 d=119 d=121 d=123

d=127 d=129 d=131 d=133 d=137 d=139 d=141 d=143 d=147 d=149

d=151 d=153 d=157 d=159

Select e value:7

Public key e:7

Private key d:23

enter msg 88

message:88

encrypted message:11

decrypted message:88

Satisfied


AIM:-Take An Example Of Subnet Hosts. Obtain Broadcast Tree for It.

PROGRAM:

#include<stdio.h>

#include<conio.h>

#include<graphics.h>

void main()

{

int i,j,n,radius=5,temp[5];

int s[10];

int link[10][10];

int l[5][2]={{150,200},{200,150},{200,200},{200,250},{250,200}};

int gdriver=detect,gmode,errorcode;

clrscr();

printf(“\n enter no of nodes:”);

scanf(“%d”,&n);

for(i=0;i<n;i++)

{

for(j=0;j<n;j++)

{

printf(“%d &di:”,i,j);

scanf(“%d”,&link[I,j]);

}

}

initgraph (&gdriver,&gmode,” “);


printf(“\n\n\n\n\t\t subnet \t\t\t corresponding true \n;”);

printf(“\t\t\t\t -------\t\t --------\n\n”);

for(i=0;i<n<i++)

circle(l[i][0],l[i][1],radius);

for(i=0;i<n;i++)

circle(l[i][0]+250,l[i][1],radius)

for(i=n-1;i>=0;i--)

line(l[i][0],l[i][1],l[j][0],l[j][1]);

for(i=n-1;i>=0;i--)

{

for(j=n-1;j>=0;j--)

{

if(link[i][j]==1&&temp[i]!=1)

{

line(l[i][0]+250,l[i][1],l[j][0]+250,l[j][1])

temp[i]=1

}

}

}

getch();

}


Input:

Enter No of nodes:3

0 0 1

0 1 1

0 2 1

1 0 1

1 1 1

1 2 1

2 0 1

2 1 1

2 2 1

Output:

Subnet corresponding tree

l l

l l

l l


cn lab manual

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