CN LAB

//CN LAB
======================================================================================================================
1) Write a program for error detecting code using CRC-CCITT (16- bits).

//Server
#include<stdio.h>
int main()
{
    int i,j,n=24,dw[24],cw[24],p[]={1,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,1};
    printf("Enter the 8 bit data word in binary\n");
    for(i=0;i<8;i++)
        scanf("%d",&dw[i]);
    for(i=0;i<8;i++)
        cw[i]=dw[i];
    for(i=8;i<24;i++)
        cw[i]=0;
    for(i=0;i<8;i++)
        if(cw[i]==1)
            for(j=0;j<17;j++)
                cw[i+j]=cw[i+j]^p[j];
    for(i=0;i<8;i++)
        cw[i]=dw[i];
    printf("The final code word is \n");
    for(i=0;i<24;i++)
        printf("%d",cw[i]);
    return 0;
}


/*OUTPUT:
Enter the 8 bit data word in binary
1 0 0 0 0 0 0 1
The final code word is
100000011000000110101001
*/
//Client
#include<stdio.h>
int main()
{
    int i,j,r[24],cw[24],p[]={1,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,1};
    printf("Enter the reviced code word in binary\n");
    for(i=0;i<24;i++)
        scanf("%d",&cw[i]);
    for(i=0;i<8;i++)
        r[i]=cw[i];
    for(i=0;i<8;i++)
        if(cw[i]==1)
            for(j=0;j<17;j++)
                cw[i+j]=cw[i+j]^p[j];
    for(i=0;i<24;i++)
        if(cw[i]!=0)
        {
            printf("Error in recived data\n");
            exit(0);
        }
    printf("The recived data word \n");
    for(i=0;i<8;i++)
        printf("%d",r[i]);
    printf(" is correct\n");
    return 0;
}
/*OUTPUT:
1)Enter the reviced code word in binary
1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1 0 1 0 1 0 0 1
The recived data word
10000001 is correct


2)Enter the reviced code word in binary
1 0 1 0 0 0 0 1 1 0 0 0 0 0 0 1 1 0 1 0 1 0 0 1
Error in recived data

*/


==========================================================================================================================

2) Write a program for Hamming code generation for error detection and correction.

#include<stdio.h>
int main()
{
    int a0,a1,a2,a3,r0,r1,r2,s0,s1,s2;
    printf("Enter the 4 bit data word in binary\n");
    scanf("%d%d%d%d",&a3,&a2,&a1,&a0);
    r0=a2^a1^a0;
    r1=a2^a1^a3;
    r2=a3^a1^a0;
    printf("The final code word is \n");
    printf("%d %d %d %d %d %d %d\n",a3,a2,a1,a0,r2,r1,r0);
    printf("Enter the 7 bit code word in binary\n");
    scanf("%d%d%d%d%d%d%d",&a3,&a2,&a1,&a0,&r2,&r1,&r0);
    s0=a2^a1^a0^r0;
    s1=a2^a1^a3^r1;
    s2=a3^a1^a0^r2;
    if(!(s0||s1||s2))
    {
        printf("NO error\nRecived data word is \n");
        printf("%d %d %d %d",a3,a2,a1,a0);
    }
    else if((!s2)&&(s1)&&(s0))
    {
        printf("Error at a2 bit and corrected data word is ");
        printf("%d %d %d %d",a3,!a2,a1,a0);
    }
    else if((s2)&&(!s1)&&(s0))
    {
        printf("Error at a0 bit and corrected data word is ");
        printf("%d %d %d %d",a3,a2,a1,!a0);
    }
    else if((s2)&&(s1)&&(!s0))
    {
        printf("Error at a3 bit and corrected data word is ");
        printf("%d %d %d %d",!a3,a2,a1,a0);
    }
    else if(s0&&s1&&s2)
    {
        printf("Error at a1 bit and corrected data word is ");
        printf("%d %d %d %d",a3,a2,!a1,a0);
    }
    return 0;
}
/*OUTPUT:
1)Enter the 4 bit data word in binary
0 1 0 0
The final code word is
0 1 0 0 0 1 1
Enter the 7 bit code word in binary
1 1 0 0 0 1 1
Error at a3 bit and corrected data word is 0 1 0 0

2)Enter the 4 bit data word in binary
0 1 0 0
The final code word is
0 1 0 0 0 1 1
Enter the 7 bit code word in binary
0 0 0 0 0 1 1
Error at a2 bit and corrected data word is 0 1 0 0

3)Enter the 4 bit data word in binary
0 1 0 0
The final code word is
0 1 0 0 0 1 1
Enter the 7 bit code word in binary
0 1 1 0 0 1 1
Error at a1 bit and corrected data word is 0 1 0 0

4)Enter the 4 bit data word in binary
0 1 0 0
The final code word is
0 1 0 0 0 1 1
Enter the 7 bit code word in binary
0 1 0 1 0 1 1
Error at a0 bit and corrected data word is 0 1 0 0

5)Enter the 4 bit data word in binary
0 1 0 0
The final code word is
0 1 0 0 0 1 1
Enter the 7 bit code word in binary
0 1 0 0 0 1 1
NO error
Recived data word is
0 1 0 0
*/






==========================================================================================================================





3) Write a program for even / odd parity checking on binary data.
#include<stdio.h>
int main()
{
    int dw[5],cw[5],i,p,s,ch;
    printf("Sender side\n");
    printf("Enter data word of 4 bits in binary\n");
    for(i=0;i<4;i++)
        scanf("%d",&dw[i]);
    printf("Enter your choice\n1.Even dataword\n2.Odd dataword\n");
    scanf("%d",&ch);
    switch(ch)
    {
        case 1:p=dw[0]^dw[1]^dw[2]^dw[3];
                dw[4]=p;
                printf("Code word is\t");
                for(i=0;i<5;i++)
                    printf("%d\t",dw[i]);
                printf("\n\nReceiver side\n");
                printf("Enter code word of 5 bits in binary\n");
                for(i=0;i<5;i++)
                    scanf("%d",&cw[i]);
                s=cw[0]^cw[1]^cw[2]^cw[3]^cw[4];
                if(s==0)
                {
                    printf("Code word is received correctly\n");
                    printf("Data word is\t");
                    for(i=0;i<4;i++)
                        printf("%d\t",cw[i]);
                    printf("\n");
                }
                else
                    printf("Code word is incorrect\n");
                                    break;
        case 2:p=dw[0]^dw[1]^dw[2]^dw[3];
                dw[4]=!p;
                printf("Code word is\t");
                for(i=0;i<5;i++)
                    printf("%d\t",dw[i]);
                printf("\n\nReceiver side\n");
                printf("Enter code word of 5 bits in binary\n");
                for(i=0;i<5;i++)
                    scanf("%d",&cw[i]);
                s=!(cw[0]^cw[1]^cw[2]^cw[3]^cw[4]);
                if(s==0)
                {
                    printf("Code word is received correctly\n");
                    printf("Data word is\t");
                    for(i=0;i<4;i++)
                        printf("%d\t",cw[i]);
                    printf("\n");
                }
                else
                    printf("Code word is incorrect\n");
                                    break;
        default:printf("Invalid choice\n");
                    return 0;
    }
    return 0;
}
/* 1) Sender side
Enter data word of 4 bits in binary
1 0 1 0
Enter your choice
1.Even dataword
2.Odd dataword
1
Code word is    1       0       1       0       0

Receiver side
Enter code word of 5 bits in binary
1 0 1 0 0
Code word is received correctly
Data word is    1       0       1       0

2)Sender side
Enter data word of 4 bits in binary
1 1 0 1
Enter your choice
1.Even dataword
2.Odd dataword
2
Code word is    1       1       0       1       0

Receiver side
Enter code word of 5 bits in binary
1 1 0 1 0
Code word is received correctly
Data word is    1       1       0       1

3)Sender side
Enter data word of 4 bits in binary
1 1 0 0
Enter your choice
1.Even dataword
2.Odd dataword
1
Code word is    1       1       0       0       0

Receiver side
Enter code word of 5 bits in binary
1 1 1 0 0
Code word is incorrect

4)Sender side
Enter data word of 4 bits in binary
1 1 1 0
Enter your choice
1.Even dataword
2.Odd dataword
2
Code word is    1       1       1       0       0

Receiver side
Enter code word of 5 bits in binary
1 1 0 0 0
Code word is incorrect
5) Sender side
Enter data word of 4 bits in binary
1 1 0 1
Enter your choice
1.Even dataword
2.Odd dataword
3
Invalid choice*/

==========================================================================================================================

4) Write a program to perform stuffing and destuffing on given information.
#include<stdio.h>
#include<string.h>
int main()
{
    char a[50];
    int ch,i,j=0,k;
    printf("Enter your choice\n1.Stuffing\n2.Destuffing\n");
    scanf("%d",&ch);
    switch(ch)
    {
        case 1: printf("Enter data for stuffing\n");
                scanf("%s",a);
                for(i=0;i<strlen(a);i++)
                {
                    if(a[i]=='0')
                        j=0;
                    if(a[i]=='1')
                        j++;
                    if(j==5)
                    {   for(k=strlen(a);k>i;k--)
                            a[k+1]=a[k];
                        a[i+1]='0';
                    }

                }
                printf("The stuffed data is = %s\n",a);
                break;
        case 2:printf("Enter data for destuffing\n");
                scanf("%s",a);
                for(i=0;i<strlen(a);i++)
                {
                    if(a[i]=='0')
                        j=0;
                    if(a[i]=='1')
                        j++;
                    if(j==5)
                    {   for(k=i+1;k<=strlen(a);k++)
                            a[k]=a[k+1];
                        j=0;
                    }
                }
                printf("The destuffed data is = %s\n",a);
                break;
        default:printf("Invalid choice\n");
    }
    return 0;
}


/*OUTPUT:
1) Enter your choice
1.Stuffing
2.Destuffing
1
Enter data for stuffing
11110111111011111
The stuffed data is = 1111011111010111110

2)Enter your choice
1.Stuffing
2.Destuffing
2
Enter data for destuffing
0011111010111110111
The destuffed data is = 00111111011111111
*/




==========================================================================================================================



5) Write a program for distance vector algorithm to find suitable path for transmission.
#include<stdio.h>
struct node
{
    int dist[20],from[20];
}rt[10];
int main()
{
    int c[20][20],s,d,arr[20],n,i,j,s1,k,count=0;
    printf("Enter the number of nodes\n");
    scanf("%d",&n);
    printf("Enter the cost of each node\n");
    for(i=0;i<n;i++)
    for(j=0;j<n;j++)
    {
        scanf("%d",&c[i][j]);
        rt[i].dist[j]=c[i][j];
        rt[i].from[j]=j;
    }
    for(i=0;i<n;i++)
    {
        printf("Routing information of routing %d\n",i);
        printf("Source\tDestination\tVia\tCost\n");
        for(j=0;j<n;j++)
            printf("\n%d\t%d\t\t%d\t%d\n",i,j,rt[i].from[j],rt[i].dist[j]);
    }

    do
    {
        count=0;
        for(i=0;i<n;i++)
        for(j=0;j<n;j++)
        for(k=0;k<n;k++)
        if(rt[i].dist[j]>(c[i][k]+rt[k].dist[j]))
        {
            rt[i].dist[j]=rt[i].dist[k]+rt[k].dist[j];
            rt[i].from[j]=k;
            count++;
        }
    }while(count!=0);
    printf("\nAfter updating\n");
    for(i=0;i<n;i++)
    {
        printf("Routing information of router %d\n",i);
        printf("Source\tDestination\tVia\tCost\n");
        for(j=0;j<n;j++)
            printf("\n%d\t%d\t\t%d\t%d\n",i,j,rt[i].from[j],rt[i].dist[j]);
    }
    printf("Enter the source node\n");
    scanf("%d",&s);
    s1=s;
    printf("Enter the desination node\n");
    scanf("%d",&d);
    printf("The main cost from router %d to router %d = %d",s,d,rt[s].dist[d]);
    arr[0]=s;
    i=1;
    while(s!=d)
    {
        arr[i]=rt[s].from[d];
        s=rt[s].from[d];
        i++;
    }
    printf("The shortest path from router %d to router %d \n",s1,d);
    for(j=0;j<i-1;j++)
        printf("%d -->",arr[j]);
    printf("%d\n",arr[j]);
    return 0;
}
/*OOUTPUT:
 Enter the number of nodes
4
Enter the cost of each node
0 2 4 9999
2 0 4 3
4 4 0 5
9999 3 5 0
Routing information of routing 0
Source  Destination     Via     Cost

0       0               0       0

0       1               1       2

0       2               2       4

0       3               3       9999
Routing information of routing 1
Source  Destination     Via     Cost

1       0               0       2

1       1               1       0

1       2               2       4

1       3               3       3
Routing information of routing 2
Source  Destination     Via     Cost

2       0               0       4

2       1               1       4

2       2               2       0

2       3               3       5
Routing information of routing 3
Source  Destination     Via     Cost

3       0               0       9999

3       1               1       3

3       2               2       5

3       3               3       0

After updating
Routing information of router 0
Source  Destination     Via     Cost

0       0               0       0

0       1               1       2

0       2               2       4

0       3               1       5
Routing information of router 1
Source  Destination     Via     Cost

1       0               0       2

1       1               1       0

1       2               2       4

1       3               3       3
Routing information of router 2
Source  Destination     Via     Cost

2       0               0       4

2       1               1       4

2       2               2       0

2       3               3       5
Routing information of router 3
Source  Destination     Via     Cost

3       0               1       5

3       1               1       3

3       2               2       5

3       3               3       0
Enter the source node
0
Enter the desination node
3
The main cost from router 0 to router 3 = 5The shortest path from router 0 to ro
uter 3
0 -->1 -->3
*/



==========================================================================================================================





6) Write a program for congestion control using leaky bucket algorithm
#include<stdio.h>
int main()
{
    int i,t,c,b[50],op,sum=0,s=0,m;
    printf("Enter the time in sec:");
    scanf("%d",&t);
    printf("\nEnter the capacity of bucket:");
    scanf("%d",&c);
    printf("\nEnter the incoming bandwidth:\nTime in sec\tBandwith in MBPS\n");
    for(i=0;i<t;i++)
    {
        printf("\n%d\t\t",i+1);
        scanf("%d",&b[i]);
        if((sum+b[i])<=c)
            sum+=b[i];
        else
            printf("\t packet %d is rejected",i+1);
    }
    printf("Enter the operate:");
    scanf("%d",&op);
    printf("\nOutput transfer of leaky bucket\nTime in sec\tBandwith in MBPS\n");
    m=sum/op;
    i=0;
    for(i=0;i<t;i++)
    {
        if(i<m)
        {
            printf("%d\t\t%d\n",i+1,op);
            s+=op;
        }
        else if(s<sum)
        {
            printf("%d\t\t%d\n",i+1,sum%op);
            s+=sum%op;
        }
        else
            printf("%d\t\t%d\n",i+1,0);
    }
    return 0;
}
/*OUTPUT:
1)Enter the time in sec:10
Enter the capacity of bucket:30
Enter the incoming bandwidth:
Time in sec     Bandwith in MBPS

1               12

2               3

3               2

4               0

5               50
         packet 5 is rejected
6               0

7               0

8               0

9               0

10              0
Enter the operate:2

Output transfer of leaky bucket
Time in sec     Bandwith in MBPS
1               2
2               2
3               2
4               2
5               2
6               2
7               2
8               2
9               1
10              0

2) Enter the time in sec:10
Enter the capacity of bucket:40
Enter the incoming bandwidth:
Time in sec     Bandwith in MBPS

1               15

2               10

3               0

4               0

5               5

6               1

7               0

8               0

9               0

10              2
Enter the operate:3

Output transfer of leaky bucket
Time in sec     Bandwith in MBPS
1               3
2               3
3               3
4               3
5               3
6               3
7               3
8               3
9               3
10              3


*/

==========================================================================================================================
7) Using TCP/IP sockets, write a client – server program to make the client send the file name and to make the server send back the contents of the requested file if present.
//server.c
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<errno.h>
#include<string.h>
#include<fcntl.h>
#include<sys/types.h>
#include<sys/ioctl.h>
#include<net/if_arp.h>
#include<sys/socket.h>
#include<unistd.h>
#include<netinet/in.h>
#include<netdb.h>
#include<arpa/inet.h>
#define MAX 25
#define MAXBUFF 100

int main()
{
    int fd,n,b;
    char buf1[MAXBUFF], buf[MAX];
    int sersock,clisock;
    struct sockaddr_in seraddr,cli_info;
    socklen_t addr_size;
    sersock=socket(AF_INET,SOCK_STREAM,0);
    if(sersock!=-1)
        printf("socket created\n");
    else
        printf("socket not created\n");
    seraddr.sin_family=AF_INET;
    seraddr.sin_port=htons(2500);
    b=bind(sersock,(struct sockaddr*)&seraddr,sizeof(seraddr));
    if(b==0)
        printf("binded successfully\n");
    else
        printf("binding failed\n");
    listen(sersock,5);
    addr_size=sizeof(cli_info);
    clisock=accept(sersock,(struct sockaddr*)&cli_info ,&addr_size);
    n=read(clisock,buf,MAX);
    buf[n]='\0';
    printf("the filereceived by client is %s=\n",buf);
    fd=open(buf,0,O_RDONLY);
    if(fd<0)
    {
        printf("file open error\n");
        strcpy(buf,"file open error");
        n=strlen(buf);
        buf[n]='\0';
        write(clisock,buf,n);
    }
    while((n=read(fd,buf1,MAXBUFF))>0)
    {
        write(clisock,buf1,n);
    }
    printf("\n");
    close(clisock);
    return 0;
}
//client
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<errno.h>
#include<string.h>
#include<fcntl.h>
#include<sys/types.h>
#include<sys/ioctl.h>
#include<net/if_arp.h>
#include<sys/socket.h>
#include<unistd.h>
#include<netinet/in.h>
#include<netdb.h>
#define MAXBUFF 1000
#define MAX 25
int main()
{
    int n,clisock,c;
    char buf1[MAXBUFF];
    char buffer[1024];
    socklen_t addr_size;
    struct sockaddr_in seraddr;
    clisock=socket(AF_INET,SOCK_STREAM,0);
    seraddr.sin_family=AF_INET;
    seraddr.sin_port=htons(2500);
    addr_size=sizeof(seraddr);
    c=connect(clisock,(struct sockaddr*)&seraddr,addr_size);
    printf("c=%d",c);
    if(c==0)
        printf("connected to server\n");
    else
        printf("connection failed\n");
    printf("enter the file name\n");
    scanf("%s",buffer);
    write(clisock,buffer,strlen(buffer));
    printf("file contents are\n");
    n=read(clisock,buf1,MAXBUFF);
    write(1,buf1,n);
    printf("\n");
    return 0;
}
/*OUTPUT:
gcc ser.c
./a.out
socket created
binded successfully
the filereceived by client is abc.txt=

gcc cli.c
./a.out
c=0connected to server
enter the file name
abc.txt
file contents are
hi hello

*/



==========================================================================================================================


9) Implement the  program 7 using the FIFO  IPC channels:
//fifoserver.c
#include<stdio.h>
#include<sys/types.h>
#include<sys/stat.h>
#include<fcntl.h>
#include<errno.h>
#include<string.h>
#include<stdlib.h>
#include<unistd.h>
#define FIFO1  "fifo1"
#define FIFO2  "fifo2"
#define PERMS 0666
char fname[256];
int main()
{
    int readfd,writefd,fd;
    char buff[512];
    ssize_t n;
    if(mkfifo(FIFO1,PERMS)<0||mkfifo(FIFO2,PERMS)<0)
    {
        printf("cant create file\n");
        exit(0);
    }
    printf("waiting for connection request\n");
    readfd=open(FIFO1,O_RDONLY,0);
    writefd=open(FIFO2,O_WRONLY,0);
    printf("Connection Established.....\n");
    read(readfd,fname,255);
    printf("Client has requested the file %s\n",fname);
    if((fd=open(fname,O_RDWR))<0)
    {
        strcpy(buff,"File does not exist..\n");
        write(writefd,buff,strlen(buff));
    }
    else
    {
        while((n=read(fd,buff,512))>0)
            write(writefd,buff,n);
    }
    close(readfd);
    close(writefd);
    unlink(FIFO1);
    unlink(FIFO2);
    return 0;
}
//client
#include<stdio.h>
#include<sys/types.h>
#include<sys/stat.h>
#include<fcntl.h>
#include<errno.h>
#include<string.h>
#include<stdlib.h>
#include<unistd.h>
#define FIFO1  "fifo1"
#define FIFO2  "fifo2"
#define PERMS 0666
char fname[256];
int main()
{
    int readfd,writefd,fd;
    char buff[512];
    ssize_t n;
    if(mkfifo(FIFO1,PERMS)<0||mkfifo(FIFO2,PERMS)<0)
    {
        printf("cant create file\n");
        exit(0);
    }
    printf("waiting for connection request\n");
    readfd=open(FIFO1,O_RDONLY,0);
    writefd=open(FIFO2,O_WRONLY,0);
    printf("Connection Established.....\n");
    read(readfd,fname,255);
    printf("Client has requested the file %s\n",fname);
    if((fd=open(fname,O_RDWR))<0)
    {
        strcpy(buff,"File does not exist..\n");
        write(writefd,buff,strlen(buff));
    }
    else
    {
        while((n=read(fd,buff,512))>0)
            write(writefd,buff,n);
    }
    close(readfd);
    close(writefd);
    unlink(FIFO1);
    unlink(FIFO2);
    return 0;
}
/*OUTPUT:
gcc fifoserver.c
./a.out
waiting for connection request
Connection Established.....
Client has requested the file abc.txt

gcc fifoclient.c
./a.out
Trying to connect to server...
Connection Established.....
Enter the file name to request from server
abc.txt
Waiting for server to reply..
Contents of the file are...
hi hello

*/













==========================================================================================================================





10) Write a program for simple RSA algorithm to encrypt and decrypt the data.

#include<stdio.h>
#include<math.h>
int phi,d,c,m,e,n;
void encrypt()
{
    int i,c=1;
    for(i=0;i<e;i++)
        c=c*m%n;
    c=c%n;
    printf("Encrypted data is = %d\n",c);
}
void decrypt()
{
    int i,m=1;
    for(i=0;i<d;i++)
        m=m*c%n;
    m=m%n;
    printf("Decrypted data is = %d\n",m);
}
int main()
{
    int p,q,s;
    printf("Enter two large numbers\n");
    scanf("%d%d",&p,&q);
    n=p*q;
    phi=(p-1)*(q-1);
    printf("phi(%d)=%d\n",n,phi);
    printf("Enter the exponent value\n");
    scanf("%d",&e);
    d=e;
    do
    {
        s=(d*e)%phi;
        d++;
    }while(s!=1);
    d=d-1;
    printf("Public key is {%d,%d}\n",e,n);
    printf("Private key is {%d,%d}\n",d,n);
    printf("Enter plain text\n");
    scanf("%d",&m);
    encrypt();
    printf("Enter the cipher text\n");
    scanf("%d",&c);
    decrypt();
    return(0);
}



/*OUTPUT:
1)Enter two large numbers
53
67
phi(3551)=3432
Enter the exponent value
47
Public key is {47,3551}
Private key is {3359,3551}
Enter plain text
99
Encrypted data is = 1793
Enter the cipher text
1793
Decrypted data is = 99

2) Enter two large numbers
67
71
phi(4757)=4620
Enter the exponent value
23
Public key is {23,4757}
Private key is {1607,4757}
Enter plain text
69
Encrypted data is = 610
Enter the cipher text
610
Decrypted data is = 69
*/