COMPUTER NETWORKS LAB

COMPUTER NETWORKS LAB




S.No.
Name of the experiment

1
Implement the data link layer framing methods such as character stuffing and bit stuffing.

2
Implement on a data set of characters the three CRC polynomials-CRC 12,CRC 16 and CRC CCIP.

3
Implement Dijkstra’s algorithm to compute the Shortest path through a graph.

4
Take an example subnet graph with weights indicating delay between nodes. Now obtain Routing table art each node using distance vector routing algorithm.

5
Take an example subnet of hosts. Obtain broadcast tree for it.

6
Take a 64 bit plain text and encrypt the same using DES algorithm.

7
Write a program to break the above DES coding.

8
Using RSA algorithm Encrypt a text data and Decrypt the same.
















/* CHARACTER STUFFING AND BIT STUFFING. */

Aim:-To implement the data link layer framing methods- character stuffing and bit stuffing.

Program:-

main()
{
  int j,n,k=0,c=0,i,t;
  int os[80],ns[100],s[]={0,1,1,1,1,1,1,0};
  clrscr();
  printf(“\n enter the number of bits:”);
  scanf(“%d”,&n);
  printf(“\n enter the bits:”);
  for(i=0;i
      scanf(“%d”,&os[i]);
  for(j=0;j<8;j++)
  {          ns[k]=s[j];
              k++;
  }
 for(j=0;j
  {       t=j;
      if(os[t]==1&&os[t++]==1&&os[t++]==1&&os[t++]==1&&os[t++]==1)
          {
             for(i=0;i<5;i++)
             {       ns[k]=0;
                      k++;
                      j=j+5;
       else
       {         ns[k]=os[j];
                  k++;
        }  }
 for(j=0;j<8;j++)
 {
                 ns[k]=s[j];
                 k++; }
 for(j=0;j
printf(“%d”,ns[j]);
getch(); }
/*1. IMPLEMENT THE DATALINK LAYER FRAMING METHODS SUCH AS CHARACTER STUFFING AND BIT STUFFING. */

main()
{
           int i, a[50],n,k=1,j=1,m=1,p=0,q=1,c=0,b[10],po,h;
           clrscr();
           printf(“\n enter total length of the frame::”);
           scanf(“%d”,&n);
           printf(“\n now enter the string::”);
           for(i=0;i
                  scanf(“%d”,&a[i]);
                  i=0;
            while(p
             {
                printf(“\n the number of characters in the %d frame is %d”,k,a[i]);
                b[q]=a[i];
               for(m=1;m
                           printf(“\n the frame data is:: %d”,a[j]);
               q++;
               c++;
               p=p+a[i];
     j++;
     i=i+a[i];
     k++;
             }
           printf(“\n enter the frame number and the frame header”);
           scanf(“%d%d”,&po,&h);
           for(q=1;q<=c;q++)
           {
              if(po==q)
              {
                if(b[q]==h)
                       printf(“\n there is no error in the frame”);
            else
                       printf(“\n there is an error in the frame”);
               }
           }
           getch();
 }
/*1. IMPLEMENT THE DATALINK LAYER FRAMING METHODS SUCH AS CHARACTER STUFFING AND BIT STUFFING. */

main()
 {
       int j,n,i,t,l,p;
       char os[40];
       clrscr();
       printf(“\n enter the text”);
             gets(os);
          l=strlen(os);
          printf(“STX DEL”);
          for(j=0;j<1;j++)
          {
             t=j;
             if(os[t++]==’d’&&os[t++]==’I’&&os[t]==’e’)
              {
                         printf(“DEL del”);
                           j=j+3;
             }
              else
                  printf(“%c”,os[j]);
          }
          printf(“ETX DEL”);
          getch();
}














/* 2.IMPLEMENT ON A DATA SET OF CHARECTERS THE CRC POLYNOMIALS-CRC12,CRC16 AND CCIP. */

Aim:-To implement CRC polynomials.

Program:-

int n,a[50],I,co,p,d[40],b[40],e,f;
main()
{
            clrscr();
            printf(“\n enter the generator length::”);
                   scanf(“%d”,&e);
            printf(“\n enter the generator ::”);
            for(i=0;i
                   scanf(“%d”,&b[i]);
           printf(“\n enter the length of polynomial ::”);
                   scanf(“%d”,&f);
           printf(“\n the polynomial ::”);
           for(i=0;i
             {        scanf(“%d”,&a[i]);
                       d[i]=a[i];
             }
           printf(“\n the polynomial at sender side is ::”);
func();
func();
if(co>0)
         printf(“\n error occurred in transmission”);
else
         printf(“\n no error occurred in transmission”);
getch();
}
     int func()
     {               int k,c=0,q=0,j,m,l,x;
                      if(p==0)
                      {            n=f+e-1;
                                    for(i=0;i
                                    printf(“%d”,a[i]);
                                    p++;
                        printf(“\n crc division on sender side”);                      }
                     else
                     {   printf(“\n enter the remainder ::”);
for(i=f;i
         scanf(“%d”,&d[i]);
printf(“\n now append this remainder to polynomial”);
printf(“\n now the polynomial at receiver ::”);
printf(“\n crc division on receiver side”);
for(i=0;i
{      a[i]=d[i];
      printf(“%d”,a[i]);
 }     }
 for(q=0;q
{
    for(k=0;j=c;j
    {
        if(a[j]!=b[k])
             a[j]=1;
        else
             a[j]=0;
    }
     j=c;
    for(j=c;j
     {
       if(a[j]==0)
            c++;
       else
            break;
      }
        q=c;
   }
        printf(“\n the remainder is ::”);
        co=0;
        for(i=f;f
         {
              printf(“%d,a[i]”);
              if(a[i]==1)
              co++;
         }
          return co;
}
/* 3. IMPLEMENT DIJKSTRA’S ALGORITHM TO COMPUTE A SHORTEST PATH THROUGH GRAPH. */

Aim:-To implement Dijkstra’s algorithm to compute shortest path through graph.

Program:-

#include
#define INFINITY 5000
int n,dist[10][10],path[10];
struct state
{
            int predecessor;
            int length;
            enum {permanent,tentative} lable;
}state[10];
struct state *p;
main()
{
    int s,t,i,j;
    clrscr();
    printf(“Enter the no of vertices”);
             scanf(“%d”,&n);
    printf(“Enter adjacency matrix”);
    for(i=0;i
    for(j=0;j
             scanf(“%d”,&dist[i][j]);
    printf(“enter the source vertex and destination vertex”);
              scanf(“%d%d”,&s,&t);
    shortest_path(int s,int t)
    {
        int i,k,min,j;
        for(p=&state[0];p<&state[n];p++)
        {
          p->predecessor=-1;
          p->length=INFINITY;
          p->lable=tentative;
       }
        state[t].length=0;
        state[t].lable=permanent;
   k=t;
  do
  {
    for(i=0;i
   if(dist[k][i]!=0&&state[i].lable==tentative)
   {
    if(state[k].length+dist[k][i]
     {
       state[i].predecessor=k;
       state[i].length=state[k].length+dist[k][i];
    }
  }
k=0;
min=INFINITY;
for(i=0;i
if(state[i].lable==tentative&&state[i].length
{
min=state[i].length;
k=i;
}
state[k].lable=permanent;
}while(k!=s);
i=0;
k=s;
do{
path[i++]=k;
k=state[k].predecessor;
}while(k.=0);
printf(“the shortest path is :”);
for(j=i-1;j>=0;j--)
printf(“\t %d\t”,path[j]);
printf(“\n the shortest distance is %d”,state[s].length);
}






4. TAKE AN EXAMPLE SUBNET GRAPH WITH WEIGHTS INCLUDING DELAY BETWEEN NODES.

Aim:-To obtain Routing table for a given subnet graph using distance vector routing algorithm.

Program:-

#include
int A[12]={0,12,25,40,14,23,18,17,21,9,24,29};
int I[12]={24,36,18,27,7,20,31,20,0,11,22,33};
int H[12]={20,31,19,8,30,19,6,0,14,7,21,9};
int k[12]={21,28,36,24,22,40,31,19,22,10,0,9};
char s[12]={‘A’,’B’,’C’,’D’,’E’,’F’,’G’,’H’,’I’,’J’,’K’,’L’};
char s1[4]={‘A’,’I’,’H’,’K’};
int ejd[4]={8,10,12,6},i,j,k,l,temp,b[4],a[4],f;
main()
{
        clrscr();
        printf(“\n the estimated hops for the nodes A,I,H,K are:”);
        for(i=0;i<12;i++)
                  printf(“\n%d\t%d\t%d\t%d”,A[i],I[i],H[i],K[i]);
                  printf(“\n the delays from j are :”);
        for(i=0;i<4;i++)
                 printf(“\t%d”,ejd[i]);
                 printf(“\n the new routing table for j is:”);
       for(i=0;i<12;i++)
       {
         if(i==9)
         printf(“\n jàj\t0\t-“);
      else
     {
         k=0;
         j=0;
         b[j++]=A[i]+ejd[k];
         b[j++]=I[i]+ejd[k+1];
         b[j++]=H[i]+ejd[k+1];
         b[j]=k[i]+ejd[k+1];
        for(k=0;k<4;k++)
        for(j=k+1;j<4;j++)
        {
           if(b[k]
            f=0;
          else
         {
            temp=b[j];
            b[j]=b[k];
            b[k]=temp;
          }
     }
                printf(“\n”);
               for(j=0;j<4;j++)
               if(b[0]==a[j])
               printf(“Jà%c\t%d\t%c”,s[i],b[0],s1[j]);
          }
     }
        getch();
}






















5. TAKE AN EXAMPLE SUBNET OF HOSTS.OBTAIN BROABCAST TREE FOR IT.

Aim:-To obtain broadcast tree for a given sunet.

Program:-

main()
{
       int a[4][4]={{99,3,5,99},{3,99,4,6},{5,4,99,6},{99,6,7,99}};
       int i,j,b[4],k,n,flag,temp,l;
       clrscr();
       n=4;
       printf(“\n broadcast routing edges are: \n”);
      for(i=0;i
      {       k=0;
              for(j=0;j
              {   b[k]=a[i][j];
                   k++;
              }   }
         for(k=0;k
        for(j=k+1;j
         {
             if(b[k]
               flag=0;
         else
         {
            temp=b[j];
            b[j]=b[k];
           b[k]=temp;
        }     }
     for(j=0;j
    {       if(b[0]==a[i][j])
      {        if(i
                 {      printf(“\n %dà%d\t%d”,i,j,a[i][j]);
                         a[i][j]=a[j][i]=99;      }
                 else
                       l=1;
                  }            }        }
       getch();           }
6. TAKE A 64-BIT PLAIN TEXT AND ENCRYPT THE SAME USING DES ALGORITHM.


Aim:-To encrypt a 64-bit plain text using DES algorithm.


Program:-

int s00[4][4]={{1,0,3,2},{3,2,1,0},{0,2,1,3},{3,1,3,2}},s0,s1;
int s11[4][4]={{0,1,2,3},{2,0,1,3},{3,0,1,0},{2,1,0,3}};
int f1[4],f2[4],k1[8]={1,0,1,0,0,1,0,0},k2[8]={0,1,0,0,0,0,1,1};
int ip[8],ipi[8]={1,5,2,0,3,7,4,6},cp[8],epi[8]={3,0,1,2,1,2,3,0},ep[8];
int p[4],r1,c1,r2,c2,p4[4],p44[4]={1,3,2,0},i,j,k[10],r[4],c=0,rs[8];
int p[8]={1,0,0,0,1,1,0,1},ipii[8]={3,0,2,4,6,1,7,5};
main()
{
 clrscr();
 printf(“\n Enter the 8 bits of plain text:”);
 for(i=0;i<8;i++)
        scanf(“%d”,&p1[i]);
printf(“\n enter the 10-bit key:”);
        scanf(“%d”, &k[i]);
 for(i=0;i<8;i++)
      ip[i]=p1[ipi[i]];
fn();
c++;
fn();
for(i=0;i<4;i++)
      ip[i]=r[i];
for(i=0,j=4;i<4;i++,j++)
      ip[j]=f2[i];
for(i=0;i<8;i++)
      cp[i]=ip[ipii[i]];
printf(“\n The cipher text is \n”);
for(i=0;i<8;i++)
         printf(“%d,\t”,cp[i]);
getch();
}

int fn()
{
  for(i=0,j=0;i<4;i++,j++)
          f1[j]=ip[i];
  for(i=4,j=0;i<8;i++,j++)
          f2[j]=ip[i];
 for(i=0;i<8;i++)
ep[i]=f2[epi[i]];
if(c==0)
{
       for(i=0;i<8;i++)
        {
               if(k1[i]==ep[i])
                          rs[i]=0;
               else
                         rs[i]=1;
        }
 }
 else
  {
    for(i=0;i<8;i++)
        {
                  if(k2[i]==ep[i])
                    rs[i]=0;
                 else
                    rs[i]=1;
         }
  }
      r1=rs[0]*2+rs[3]*1;
      c1=rs[1]*2+rs[2]*1;
      r2=rs[4]*2+rs[7]*1;
      c2=rs[5]*2+rs[6]*1;
      s0=s00[r1][c1];
      s1=s11[r2][c2];
      i=0;
      p[i]=s0/2;i++;
      p[i]=s0%2;i++;
      p[i]=s1/2;i++;
      p[i]=s1%2;

    for(i=0;i<4;i++)
     p4[i]=p[p44[i]];
    for(i=0;i<4;i++)
     {
         if(p4[i]==f1[i])
           r[i]=0;
        else
           r[i]=1;
    }
      for(i=0;i<4;i++)
          ip[i]=f2[i];
      for(i=0,j=4;i<4;i++,j++)
          ip[j]=r[i];
}


























7.TAKE A 64-BIT PLAIN TEXT AND ENCRYPT THE SAME USING DES ALGORITHM.

Aim:- To encrypt a 64-bit plain text using DES algorithm.


Program:-

int s00[4][4]={{1,0,3,2},{3,2,1,0},{0,2,1,3},{3,1,3,2}},s0,s1;
int s11[4][4]={{0,1,2,3},{2,0,1,3},{3,0,1,0},{2,1,0,3}};
int f1[4],f2[4],k1[8]={1,0,1,0,0,1,0,0},k2[8]={0,1,0,0,0,0,1,1};
int ip[8],ipi[8]={1,5,2,0,3,7,4,6},cp[8],epi[8]={3,0,1,2,1,2,3,0},ep[8];
int p[4],r1,c1,r2,c2,p4[4],p44[4]={1,3,2,0},i,j,k[10],r[4],c=0,rs[8];
int p[8]={1,0,0,0,1,1,0,1},ipii[8]={3,0,2,4,6,1,7,5};
main()
{
 clrscr();
 printf(“\n Enter the 8 bits of plain text:”);
 for(i=0;i<8;i++)
        scanf(“%d”,&p1[i]);
printf(“\n enter the 10-bit key:”);
        scanf(“%d”, &k[i]);
 for(i=0;i<8;i++)
      ip[i]=p1[ipi[i]];
fn();
c++;
fn();
for(i=0;i<4;i++)
      ip[i]=r[i];
for(i=0,j=4;i<4;i++,j++)
      ip[j]=f2[i];
for(i=0;i<8;i++)
      cp[i]=ip[ipii[i]];
printf(“\n The cipher text is \n”);
for(i=0;i<8;i++)
         printf(“%d,\t”,cp[i]);
getch();
}


int fn()
{
  for(i=0,j=0;i<4;i++,j++)
          f1[j]=ip[i];
  for(i=4,j=0;i<8;i++,j++)
          f2[j]=ip[i];
 for(i=0;i<8;i++)
ep[i]=f2[epi[i]];
if(c==0)
{
       for(i=0;i<8;i++)
        {
               if(k1[i]==ep[i])
                          rs[i]=0;
               else
                         rs[i]=1;
        }
 }
 else
  {
    for(i=0;i<8;i++)
        {
                  if(k2[i]==ep[i])
                    rs[i]=0;
                 else
                    rs[i]=1;
         }
  }
      r1=rs[0]*2+rs[3]*1;
      c1=rs[1]*2+rs[2]*1;
      r2=rs[4]*2+rs[7]*1;
      c2=rs[5]*2+rs[6]*1;
      s0=s00[r1][c1];
      s1=s11[r2][c2];
      i=0;
      p[i]=s0/2;i++;
      p[i]=s0%2;i++;
      p[i]=s1/2;i++;
      p[i]=s1%2;

     for(i=0;i<4;i++)
     p4[i]=p[p44[i]];
    for(i=0;i<4;i++)
     {
         if(p4[i]==f1[i])
           r[i]=0;
        else
           r[i]=1;
    }
      for(i=0;i<4;i++)
          ip[i]=f2[i];
      for(i=0,j=4;i<4;i++,j++)
          ip[j]=r[i];
}


























8. USING RSA ALGORITHM ENCRYPT A TEXT DATA AND DECRYPT THE SAME.

Aim:- To encrypt and decrypt a text RSA algorithm.


Program:-

#include
#include
char *r,en[10],de[10],b[10],g;
char a[26]={‘a’,’b’,’c’,’d’,’e’,’f’,’g’,’h’,’I’,’j’,’k’,’l’,’m’,’n’,’o’,’p’,’q’,’r’,’s’,’t’,’u’,’v’,’w’,’x’,’y’,’z’};
main()
{
    int q,l=0,j,cop[50],flag=0,f;
    int temp[10],p,e,d,I,n,z,m,g;
    unsigned long c;
    clrscr();
     printf(“\n Enter the plaintext at the end type”);
             while((b[i]=getchar())!=’@’)
        i++;
        r=b;
     printf(“\n Enter two prime numbers”);
              scanf(“%d%d”,&p&q);
     n=p*q;
     z=(p-1)*(q-1);
     i=0;j=0;
     for(f=3;f
     {
           if(f%2==0)
               flag=0;
           else
           {
               cop[i]=f;
                i++;j++;
            }
      }
      printf(“\n The coprimes for %d are”,z);
      for(i=0;i
      printf(“%d\t”,cop[i]);
      printf(“\n enter one coprime”);
       scanf(“%d”,&e);
      j=0;
      while(*r!=’@’)
      {
        for(i=0;i!=25&&flag==0;i++)
       {
           if(*r==a[i])
           {
             c=pow(I,e)
             g=c%n;
             temp[j]=g;
             flag=1;
             en[j]=toupper(a[g]);
              j++;
         }    }
        r++;
        flag=0; }
printf(“\n The cipher text is:”);
for(i=0;i
printf(“%c”,en[i]);
for(d=1;;d++)
{
   if((e*d)%z==1)
   break;
}
printf(“\n The decryption key is %d”,d);
printf(“\n Enter decryption key”);
scanf(“%d”,&d);
printf(“\n the decrypted text is:”);
for(i=0;i
{
   c=pow(temp[i],d);
         m=c%n;
printf(“%c”,a[m]);
}
getch();
}