computer networks algorithms in c language

Computer Networks Algorithms in C language Contents: 1. Bit Stuffing and Character Stuffing 2. CRC,CRC-12,CRC-16,CRC-CCIT 3. Dijkstra's Shotest path routing algorithm 4. Distance vector routing algorithm 5. Link state routing algorithm 6. Floding 7.Broadcasting 1.IMPLEMENTATION OF BIT STUFFING AND CHARACTER STUFFING #include #include #include main() { int ch; clrscr(); do { printf("\n********************:"); printf("\n1.BIT STUFFING:"); printf("\n2.CHARACTER STUFFING:"); printf("\n3.EXIT:"); printf("\n*********************"); printf("\nEnter your choice:"); scanf("%d",&ch); switch(ch) { case 1:bit(); break; case 2:character(); break; case 3:exit(0); } }while(ch!=3); getch(); } bit() { FILE *fp,*fp1; char ch; int i; if((fp=fopen("source.txt","w"))==NULL) { printf("\nError in opening the file"); exit(0); } printf("\nEnter data to send press 'e' to end :\n"); i=0; while(1) { scanf("%c",&ch); if(ch=='e') break; if(ch=='1') i++; else i=0;

putc(ch,fp); if(i==5) { putc('0',fp); i=0; } } fclose(fp); i=0; fp=fopen("source.txt","r"); fp1=fopen("dest.txt","w"); printf("\nDATA AFTER STUFFING\n"); while((ch=getc(fp))!=EOF) { putc(ch,stdout); } fseek(fp,0L,0); printf("\nDATA AFTER UNSTUFFING\n"); while((ch=getc(fp))!=EOF) { if(ch=='1') i++; else i=0; if(i==5) { putc(ch,stdout); putc(ch,fp1); ch=getc(fp); i=0; } else { putc(ch,stdout); putc(ch,fp1); } } fclose(fp); fclose(fp1); } character() { FILE *fp,*fp1; char ch,c[2],k[4],j[9]; long beg,end; if((fp=fopen("Input.txt","w"))==NULL) { printf("\n Input.txt file opening problem..."); exit(0); } printf("\nEnter data to send at end put '}': \n\n"); while(1) { scanf("%c",&ch); if(ch=='}') break; putc(ch,fp); }

fclose(fp); if((fp=fopen("Input.txt","r"))==NULL) { printf("\nInput.txt file opening problem..."); exit(0); } if((fp1=fopen("csource.txt","w"))==NULL) { printf("\ncsource.txt file opening problem..."); exit(0); } fputs(" DLE STX ",fp1); while((ch=getc(fp))!=EOF) {

if(ch=='D') {

c[0]=getc(fp); c[1]=getc(fp); if(c[0]=='L'&&c[1]=='E') fputs("DLE",fp1); putc(ch,fp1); putc(c[0],fp1); putc(c[1],fp1); } else putc(ch,fp1); } fputs(" DLE ETX ",fp1); fclose(fp); fclose(fp1); if((fp=fopen("csource.txt","r"))==NULL) { printf("\ncsource.txt file opening problem..."); exit(0); } if((fp1=fopen("cdest.txt","w"))==NULL) { printf("\ncdest.txt file opening problem..."); exit(0); } beg=ftell(fp); beg+=9; fseek(fp,-0L,2); end=ftell(fp); end-=9; fclose(fp); printf("\nData after stuffing "); fp=fopen("csource.txt","r"); while((ch=getc(fp))!=EOF) putc(ch,stdout); fclose(fp); printf("\n"); printf("\nThe data after destuffing"); fp=fopen("csource.txt","r"); fgets(j,9,fp); while(beg<=end)

{ ch=getc(fp); if(ch=='D') { c[0]=getc(fp); c[1]=getc(fp); if(c[0]=='L'&&c[1]=='E') { fgets(k,4,fp); beg+=4; } else { putc(ch,fp1); putc(c[0],fp1); putc(c[1],fp1); putc(ch,stdout); putc(c[0],stdout); putc(c[1],stdout); } } else { putc(ch,fp1); putc(ch,stdout); } beg++; } fclose(fp); fclose(fp1); }

OUTPUT

********************: 1.BIT STUFFING: 2.CHARACTER STUFFING: 3.EXIT: ********************* Enter your choice:1

Enter data to send press 'e' to end : 11100000111111111101e

DATA AFTER STUFFING

1110000011111011111001 DATA AFTER UNSTUFFING

11100000111111111101 ********************: 1.BIT STUFFING: 2.CHARACTER STUFFING: 3.EXIT: ********************* Enter your choice:2

Enter data to send at end put '}':

This is DLE idle DLE program}

Data after stuffing DLE STX This is DLEDLE idle DLEDLE program DLE ETX

The data after destuffing This is idle program ********************: 1.BIT STUFFING: 2.CHARACTER STUFFING: 3.EXIT: ********************* Enter your choice:3

2. IMPLEMENTATION OF DIJKSTRA’S ALGORITHM*/ #include #include #define maxnode 10 #define perm 1 #define tent 2 #define infinity INT_MAX typedef struct nodelabel { int predecessor; int length; int label; }nodelabel; int shortpath(a,n,s,t,path,dist) int a[maxnode][maxnode],n,s,t,path[maxnode],*dist; { nodelabel state[maxnode]; int i,k,min,count,rpath[maxnode]; *dist=0; for(i=1;i<=n;i++) { state[i].predecessor=0; state[i].length=infinity; state[i].label=tent; } state[s].predecessor=0; state[s].length=0; state[s].label=perm; k=s; do { for(i=1;i<=n;i++) { if(a[k][i]>0 && state[i].label==tent) { if(state[k].length+a[k][i]{ state[i].predecessor=k; state[i].length=state[k].length+a[k][i]; } }

} min=infinity; k=0; for(i=1;i<=n;i++) { if(state[i].label==tent&&state[i].length{ min=state[i].length; k=i; } } if(k==0) return 0; state[k].label=perm; }while(k!=t); k=t; count=0; do { count++; rpath[count]=k; k=state[k].predecessor; }while(k!=0); for(i=1;i<=count;i++) path[i]=rpath[count-i+1]; for(i=1;i*dist+=a[path[i]][path[i+1]]; return count; } void main() { int a[maxnode][maxnode],i,j; int path[maxnode]; int from,to,dist,count,n; clrscr(); printf("Enter no. of nodes:"); scanf("%d",&n); for(i=1;i<=n;i++) { printf("Enter node %d",i); for(j=1;j<=n;j++) scanf("%d",&a[i][j]); } printf("\n From to"); scanf("%d %d",&from,&to); count=shortpath(a,n,from,to,path,&dist); if(dist) { printf("\n Short path"); printf("%d",path[1]); for(i=2;i<=count;i++) printf("-->%d",path[i]); printf("\nMinimum distance is %d \n",dist); } else printf("\n Path does not exists");

getch();

}

OUTPUT

Enter no. of nodes:4 Enter node 1 0 3 4 0 Enter node 2 0 0 6 20 Enter node 3 0 0 0 9 Enter node 4 15 0 0 0

From to 2 4

Short path2-->3-->4 Minimum distance is 15

3.DISTANCE VECTOR ROUTING ALGORITHM #include #include #define LIMIT 10 #define INFINITY 10000 int n,m,i,j,mi,dist[LIMIT][LIMIT],a[LIMIT],adj[LIMIT],sn,dn,c=0,x[LIMIT],min; main() { clrscr(); printf("Enter no.of nodes::"); scanf("%d",&n); for(i=1;i<=n;i++) for(j=1;j<=n;j++) dist[i][j]=-1; for(i=1;i<=n;i++) { printf("\nEnter distance for %d node:",i); for(j=1;j<=n;j++) if(dist[i][j]==-1) { if(i==j) { dist[i][j]=0; printf("\n 0"); } else { scanf("%d",&m); dist[i][j]=dist[j][i]=m; } } else printf("\n%d",dist[i][j]); } printf("\ndistance matrix is:"); for(i=1;i<=n;i++) { printf("\n"); for(j=1;j<=n;j++)

printf("%d\t",dist[i][j]); } printf("\nnEnter the sorce node :"); scanf("%d",&sn); printf("Enter the distance node:"); scanf("%d",&dn); j=1;m=0; for(i=1;i<=n;i++) { if(dist[sn][i]>0) { m++; adj[j++]=i; } } min=0; mi=INFINITY; for(i=1;i<=m;i++) { printf("\nVector table for %d to all nodes:",adj[i]); for(j=1;j<=n;j++) { if(adj[i]==j) x[j]=0; else x[j]=spath(adj[i],j,min); if(j==dn) if(x[j]{ mi=x[j]; c=adj[i]; } printf("\n%d",x[j]); } } for(i=1;i<=n;i++) { if(i==sn) a[i]=0; else if(i==dn) a[i]=dist[sn][c]+mi; else a[i]=spath(sn,i,min); } printf("\nNew vector table for %d node:",sn); for(i=1;i<=n;i++) printf("\n%d",a[i]); getch(); }

int spath(int s,int t,int min) { struct state { int pred; int len; enum{permanent,tentative}label;

}state[LIMIT]; int i=1,k; struct state *p; for(p=&state[i];p<=&state[n];p++) { p->pred=0; p->len=INFINITY; p->label=tentative; } state[t].len=0; state[t].label=permanent; k=t; do { for(i=1;i<=n;i++) if(dist[k][i]!=0 && state[i].label==tentative) { if(state[k].len+dist[k][i]{ state[i].pred=k; state[i].len=state[k].len+dist[k][i]; } } k=0; min=INFINITY; for(i=1;i<=n;i++) if(state[i].label==tentative && state[i].len{ min=state[i].len; k=i; } state[k].label=permanent; } while(k!=s); return(min); }

output Enter no.of nodes::4

Enter distance for 1 node: 0 1 2 0



distance matrix is: 0 1 2 0 1 0 0 4 2 0 0 8 0 4 8 0 Enter the source node :3 Enter the distance node:2

Vector table for 1 to all nodes: 0 1 2 5 Vector table for 4 to all nodes: 5 4 7 0 New vector table for 3 node: 2 3 0 7

4.LINK STATE ROUTING ALGORITHM #include #include #define LIMIT 10 #define INFINITY 10000 int m,n,k,i,j,dist[LIMIT][LIMIT],sn,dn,min=0; struct node { int hello[LIMIT],from,adj[LIMIT]; }node[LIMIT]; main() { clrscr(); printf("Enter how many nodes do u want::"); scanf("%d",&n); for(i=1;i<=n;i++) for(j=1;j<=n;j++) dist[i][j]=-1; for(i=1;i<=n;i++) { printf("\nEnter distance for %d node:",i); for(j=1;j<=n;j++) if(dist[i][j]==-1) if(i==j) { dist[i][j]=0; printf(" 0"); } else { scanf("%d",&m); dist[i][j]=dist[j][i]=m; }

else printf(" %d",dist[i][j]); } printf("\nDistance matrix is:"); for(i=1;i<=n;i++) { printf("\n"); node[i].from=0; for(j=1;j<=n;j++) printf("\t%d",dist[i][j]); } printf("\nSENDING HELLO PACKETS"); for(i=1;i<=n;i++) { k=1; for(j=1;j<=n;j++) { if(dist[i][j]>0) node[i].from++; if(dist[j][i]>0) node[i].hello[k++]=j; } } for(i=1;i<=n;i++) { printf("\nHello packets to node %d:",i); for(j=1;j<=node[i].from;j++) printf("\n%d",node[i].hello[j]); } printf("\nSENDING ECHO PACKETS"); for(i=1;i<=n;i++) { for(j=1;j<=n;j++) if(dist[i][j]>0) printf("\nfrom %d to %d, the distance is:%d",i,j,dist[i][j]); } printf("\n CONSTRUCTION LINKSTATE PACKETS"); for(i=1;i<=n;i++) { printf("\nLINK STATE PACKET FOR %d",i); printf("\n|-------------|"); printf("\n| %d |",i); printf("\n|-------------|"); printf("\n| seq |"); printf("\n|-------------|"); printf("\n| Age |"); printf("\n|-------------|"); for(j=1;j<=n;j++) if(dist[i][j]>0) { printf("\n| %d | %d |",j,dist[i][j]); printf("\n|-------------|"); }

} printf("\nDISTRIBUTING THE LINK STATE PACKETS"); for(i=1;i<=n;i++) {

printf("\nNeighbours for %d node are:",i); for(j=1;j<=n;j++) { if(dist[i][j]>0) printf("%d",j); } for(j=1;j<=node[i].from;j++) { printf("\n the distance from %d to ",i); printf("%d is %d",node[i].hello[j],dist[i][node[i].hello[j]]); } } printf("\nCOMPUTING THE SHORTEST PATH"); printf("\nEnter source and destination:"); scanf("%d%d",&sn,&dn); min=spath(sn,dn,min); printf("\n minimum distance:%d",min); getch(); } int spath(int s,int t,int min) { struct state { int pred; int len; enum{permanent,tentative}label; }state[LIMIT]; int i=1,k; struct state *p; for(p=&state[i];p<=&state[n];p++) { p->pred=0; p->len=INFINITY; p->label=tentative; } state[t].len=0; state[t].label=permanent; k=t; do { for(i=1;i<=n;i++) if(dist[k][i]!=0 && state[i].label==tentative) { if(state[k].len+dist[k][i]{ state[i].pred=k; state[i].len=state[k].len+dist[k][i]; } } k=0; min=INFINITY; for(i=1;i<=n;i++) if(state[i].label==tentative && state[i].len{ min=state[i].len; k=i; } state[k].label=permanent;

} while(k!=s); return(min); }

output Enter how many nodes do u want::4




Enter distance for 4 node: 4 8 6 0 Distance matrix is: 0 2 3 4 2 0 7 8 3 7 0 6 4 8 6 0

SENDING HELLO PACKETS Hello packets to node 1: 2 3 4 Hello packets to node 2: 1 3 4 Hello packets to node 3: 1 2 4 Hello packets to node 4: 1 2 3 SENDING ECHO PACKETS from 1 to 2, the distance is:2 from 1 to 3, the distance is:3 from 1 to 4, the distance is:4 from 2 to 1, the distance is:2 from 2 to 3, the distance is:7 from 2 to 4, the distance is:8 from 3 to 1, the distance is:3 from 3 to 2, the distance is:7 from 3 to 4, the distance is:6 from 4 to 1, the distance is:4 from 4 to 2, the distance is:8 from 4 to 3, the distance is:6

CONSTRUCTION LINKSTATE PACKETS LINK STATE PACKET FOR 1 |-------------| | 1 | |-------------| | seq | |-------------| | Age | |-------------| | 2 | 2 | |-------------| | 3 | 3 | |-------------| | 4 | 4 |

LINK STATE PACKET FOR 2 |-------------| | seq | |-------------| | Age | |-------------| | 1 | 2 | |-------------| | 3 | 7 | |-------------| | 4 | 8 | |-------------| LINK STATE PACKET FOR 3 |-------------| | 3 | |-------------| | seq | |-------------| | Age | |-------------| | 1 | 3 | |-------------| | 2 | 7 | |-------------| | 4 | 6 | LINK STATE PACKET FOR 4 |-------------| | 4 | |-------------| | seq | |-------------| | Age | |-------------| | 1 | 4 | |-------------| | 2 | 8 | |-------------| | 3 | 6 | |---------------| | 2 | 8 | |-------------| | 3 | 6 | |-------------| DISTRIBUTING THE LINK STATE PACKETS Neighbours for 1 node are:234 the distance from 1 to 2 is 2 the distance from 1 to 3 is 3 the distance from 1 to 4 is 4 Neighbours for 2 node are:134 the distance from 2 to 1 is 2 the distance from 2 to 3 is 7 the distance from 2 to 4 is 8 Neighbours for 3 node are:124 the distance from 3 to 1 is 3 the distance from 3 to 2 is 7 the distance from 3 to 4 is 6 Neighbours for 4 node are:123 the distance from 4 to 1 is 4 the distance from 4 to 2 is 8

the distance from 4 to 3 is 6 COMPUTING THE SHORTEST PATH Enter source and destination:2 3

minimum distance:5

5. PROGRAM TO IMPLEMENT BROADCAST ROUTING ALGORITHM #include int a[10][10],n; main() { int i,j,root; clrscr(); printf("Enter no.of nodes:"); scanf("%d",&n); printf("Enter adjacent matrix\n"); for(i=1;i<=n;i++) for(j=1;j<=n;j++) { printf("Enter connecting of %d-->%d::",i,j); scanf("%d",&a[i][j]); } printf("Enter root node:"); scanf("%d",&root); adj(root); } adj(int k) { int i,j; printf("Adjacent node of root node::\n"); printf("%d\n\n",k); for(j=1;j<=n;j++) { if(a[k][j]==1 || a[j][k]==1) printf("%d\t",j); } printf("\n"); for(i=1;i<=n;i++) { if((a[k][j]==0) && (a[i][k]==0) && (i!=k)) printf("%d",i); } }

OUTPUT

Enter no.of nodes:5 Enter adjacent matrix Enter connecting of 1-->1::0 Enter connecting of 1-->2::1 Enter connecting of 1-->3::1 Enter connecting of 1-->4::0 Enter connecting of 1-->5::0 Enter connecting of 2-->1::1 Enter connecting of 2-->2::0 Enter connecting of 2-->3::1 Enter connecting of 2-->4::1 Enter connecting of 2-->5::0

Enter connecting of 3-->1::1 Enter connecting of 3-->2::1 Enter connecting of 3-->3::0 Enter connecting of 3-->4::0 Enter connecting of 3-->5::0 Enter connecting of 4-->1::0 Enter connecting of 4-->2::1 Enter connecting of 4-->3::0 Enter connecting of 4-->4::0 Enter connecting of 4-->5::1 Enter connecting of 5-->1::0 Enter connecting of 5-->2::0 Enter connecting of 5-->3::0 Enter connecting of 5-->4::1 Enter connecting of 5-->5::0 Enter root node:2 Adjacent node of root node:: 2

1 3 4 5

6. PROGRAM TO SIMULATE ROUTING USING FLOODING #include #include int a[30][30],i,j,k=0,l=0,b[20],cnt=0,n,seq[30],ch; char c1='x'; main() { clrscr(); b[1]=1; printf("Enter no.of nodes:"); scanf("%d",&n); printf("\nConstruct graph"); for(i=1;i<=n;i++) { printf("\nEnter edges for %d:",i); for(j=1;j<=n;j++) scanf("%d",&a[i][j]); } printf("\n\t\tMENU"); printf("\n\t\t*********************"); printf("\n\t\t1. FLODDING WITH OUT SOLUTION"); printf("\n\t\t1. FLODDING WITH SOLUTION"); printf("\n\t\t**********************"); printf("\nEnter your choice:"); scanf("%d",&ch); switch(ch) { case 1: for(i=1;i<=n;i++) { l=0; for(j=1;j<=n;j++) if(a[i][j]!=0) l++; if(icnt++; b[i+1]=l*cnt;

} flood(); break; case 2: for(i=1;i<=n;i++) for(j=1;j<=n;j++) if(a[i][j]!=0 && iif(seq[j]!=1) { printf("\n%d-->%d",i,j); printf("%c",c1); seq[j]=1; } break; default: exit(0); } getch(); }

flood() { printf("\n RECEIVED PACKETS ARE"); for(i=1;i<=n;i++) { for(j=2;j<=n;j++) if(a[i][j]!=0 && i{ printf("\n%d-->%d",i,j); for(k=1;k<=b[i];k++) printf("%c",c1); } } }

OUTPUT

Enter no.of nodes:4

Construct graph Enter edges for 1:0 1 0 1 Enter edges for 2:1 0 1 0 Enter edges for 3:0 1 0 1 Enter edges for 4:1 0 0 0

MENU ********************* 1. FLODDING WITH OUT SOLUTION 1. FLODDING WITH SOLUTION ********************** Enter your choice:1

RECEIVED PACKETS ARE 1-->2x 1-->4x 2-->3xx 3-->4xxxx Enter no.of nodes:4

Construct graph Enter edges for 1:0 1 0 1 Enter edges for 2:1 0 1 0 Enter edges for 3:0 1 0 1 Enter edges for 4:1 0 0 0

MENU ********************* 1. FLODDING WITH OUT SOLUTION 1. FLODDING WITH SOLUTION ********************** Enter your choice:2

1-->2x 1-->4x 2-->3x

7. CRC PROGRAM #include #include int g1[]={1,0,0,1,1}; int g2[]={1,1,0,0,0,0,0,0,0,1,1,1,1}; int g3[]={1,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,1}; int g4[]={1,1,0,0,0,0,0,0,0,0,0,0,0,0,1,0,1}; int f[100],g[20],r[2],df[100]; int ch,gl,n,i,j,k; void main() { clrscr(); do{ printf("\n 1.CRC"); printf("\n 2.CRC-12"); printf("\n 3.CRC-16"); printf("\n 4.CRC-CCIT"); printf("\n 5 .DEFAULT"); printf("\n EXIT"); printf("\n ENTER YOUR CHOICE"); scanf("%d",&ch); switch(ch) { case 1 :for(i=0;i<5;i++) g[i]=g1[i]; gl=5; break; case 2 :for(i=0;i<13;i++) g[i]=g2[i]; gl=13; break; case 3 :for(i=0;i<17;i++) g[i]=g3[i]; gl=17; break; case 4 :for(i=0;i<17;i++) g[i]=g4[i]; gl=17; break; case 5 :printf("\n Enter the size of generator"); scanf("%d",&gl);

printf("\n Enter the generator bits"); for(i=o;iscanf("%d",&g[i]); } printf("\n Enter the frame size"); scanf("%d",&n); if(n>=gl) { printf("\n enter the frame bits"); for(i=0;i{ scanf("%d",&f[i]); df[i]=f[i]; } clrscr(); printf("\n\n\t The entered frame is:"); for(i=0;iprintf("%d",f[i]); printf("\n\n\t The Polynomial is:"); for(i=0;iprintf("%d",g[i]); for(i=0;if[n+i]=0; printf("\n\n\t\t\t ******STUFF*****"); printf("\n\n\t After adding zeros the frame is:"); for(i=0;i<(n+gl-1);i++) printf("%d",f[i]); division(f,g); for(i=0;i<(n+gl-1);i++) df[n+i]=r[i+1]; printf("\n\n\t The Transmitted frame is:"); for(i=0; i<(n+gl-1); i++) printf("%d",df[i]); printf("\n\n\t\t *****DESTUFF*****"); printf("\n\n\t\t The Receiving frame is :"); for(i=0;i<(n+gl-1);i++) printf("%d",df[i]); division(df,g); printf("\n\n\t There are no errors in the Received frame"); } else printf("\n\n\t CRC is not possible"); }while(ch<=5); getch(); } division(int f[], int g[]) { for(i=0;i<=(n-1);i++) { if((f[0]==1) && (g[0]==1)) { for(j=0;j{ if(f[j]==g[j]) f[j]=r[j]=0; else f[j]=r[j]=1; }

} else { for(k=0;kr[k]=f[k]; } for(k=0;k<(n+gl-1);k++) f[k]=f[k+1]; } printf("\n\n\t\t The Remainder is:"); for(i=0;iprintf("%d",r[i]); }

computer networks algorithms in c language

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