Cn Case Tools Lab

Geethanjali College Of Engineering and TechnologyCheeryal(V),Keesara (M),Ranga Reddy Dist.

Program 1 : Character Stuffing

Program :

#include<stdio.h>#include<conio.h>#include<string.h>void main() {int n,i;long a;char str[100]=" ",temp[20]=" ",ch;FILE *fp1,*fp2,*fp3;clrscr();fp1=fopen("input1.txt","r");if(fp1==NULL){printf("not created");}fp2=fopen("z:/input2.txt","w");printf("\n\t **Character Stuffing** \t\n");printf("The Original Text is :");fseek(fp1,0,SEEK_END);a=ftell(fp1);rewind(fp1);while((ch=fgetc(fp1))!=EOF) printf("%c",ch);rewind(fp1);printf("\n\t The Encoded Text is :");while(!feof(fp1)){ if(a==ftell(fp1)) break; else { fscanf(fp1,"%s",str); fputs("stxdle ",fp2); printf("stxdle "); if(strstr(str,"dle")!=0) { fputs(str,fp2); printf("%s",str); fputs("dle",fp2); printf("dle"); }

Page 1CN CASE TOOLS LAB III yr IT


else { fputs(str,fp2); printf("%s",str);} fputs(" etxdle ",fp2); printf(" etxdle "); }}fcloseall();fp2=fopen("z:/input2.txt","r");fp3=fopen("z:/input3.txt","w");printf("\n\t The Decoded Text is :") ;while(!feof(fp2)){ fscanf(fp2,"%s",str); if(strcmp(str,"stxdle")==0) fputs("",fp3); else if(strcmp(str,"etxdle")==0) fputs(" ",fp3); else if(strstr(str,"dledle")!=0) { n=strlen(str); n=n-3; strcpy(temp,str+n); fputs(str,fp3); } else { fputs(str,fp3); }} //for(i=0;i<n;i++) //temp[i]='\0'; fcloseall(); fp3=fopen("z:/input3.txt","r"); while((ch=fgetc(fp3))!=EOF) printf("%c",ch); fcloseall(); getch();}



Result:

Input File: (Input1.txt)

This Is CN Lab.The Program dle is about idle character stuffing.

Output:

** CHARACTER STUFFING **

The Original Text is :

This Is CN Lab.The Program dle is about idle character stuffing.

The Encoded Text is:

stxdle This etxdle stxdle Is etxdle stxdle CN etxdle stxdle Lab. etxdle stxdle The etxdle stxdle Program etxdle stxdle dledle etxdle stxdle is etxdle stxdle about etxdle stxdle idledle etxdle stxdle character etxdle stxdle stuffing. etxdle

The Decoded Text is:

This Is CN Lab. The Program dle is about idle character stuffing.



Program 2: Bit Stuffing

Program:

#include<stdio.h>#include<conio.h>#include<math.h>main(){int n,r,j,a=0,b=0,l=0,i=0,arr[200],bs[200];char ch;FILE *fp1;clrscr();fp1=fopen("input1.txt","r");printf("\t\t *** Bit Stuffing ** \n\n");printf("The Original Text is:\n\n ");while((n=ch=fgetc(fp1))!=EOF){printf("%c",ch);for(i=0;n!=0;i++){r=n%2;bs[i]=r;n/=2;}for(;i<8;i++)bs[i]=0;for(j=i-1;j>=0;j--,a++)arr[a]=bs[j];b=l=a;}getch();printf("\n\n The Text after Convertion to binary is:\n\n");for(i=0;i<b;i++)printf("%d",arr[i]);getch();printf("\n\n The Encoded Text is:\n\n");for(i=0;i<=7;i++){if(i==0||i==7)bs[i]=0;elsebs[i]=1;}for(i=0,j=8;i<=l;i++){



if(arr[i]==1){n++;bs[j]=arr[i];if(n==5){j++;bs[j]=0;}j++;}else{n=0;bs[j]=arr[i];j++;}}for(i=0;i<=7;i++){if(i==0||i==7)bs[i+j]=0;elsebs[i+j]=1;}j=j+7;for(i=0;i<=j;i++)printf("%d",bs[i]);getch();printf("\n\n The Decoded Text is:\n ");for(i=8,r=0;i<j-7;i++,r++)bs[r]=bs[i];for(i=r=l=n=0;i<=j-16;i++){if(bs[i]==1)n++;elsen=0;r=i;if(n==5){r++;l++;for(a=r+1;a<=j-16;a++,r++)bs[r]=bs[a];}}



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

printf("%d",bs[i]);getch();printf("\n The Text aft Conversion from binary text is:\n\n");for(i=0;i<b; ){for(a=0,n=7;n>=0;n--,i++)a=a+bs[i]*pow(2,n);printf("%c",a);

}fclose(fp1);getch();return (0);}



Result:

Input File: (Input1.txt)

This is Computer_networks Lab.

Output:

** BIT STUFFING **

The Original Text is:


The Text after Conversion to binary is:

010101000110100001101001011100110010000001101001011100110010000001100011011010110110101110000011101010111010001100101011100100101111101101110011001010111001110111011011110111001001101011011100110010000001101100011000010110001000101

The Encoded Text is:

01111110010101000110100001101001011100110010000001101001011100110010000001100011011110110110101110000011101010111010001100101011100100101111100110111001101011101000111011101101111011100100110101101110011001000000110110001100001011000010111001111110

The Decoded Text is:

010101000110100001101001011100110010000001101001011100110010000001100011011011110110110101110000011101010111010001100101011100100010111110110111001100100111010001110111011011110111001001101011011100110010000001101100011000010110001000101110

The Text after Conversion from binary to Text is:




Program 3: Cyclic Redundancy Check,CRC-12

Program:

#include<stdio.h>#include<conio.h>#include<string.h>#include<dos.h>#include<stdlib.h>int copy();int check();int i,j,k,t,count=0,num=0;int gen[10],frame[30],rem[30],temp[30];void main(){char c,plym[50];char ch[]={'0','1','2','3','4','5','6','7','8','9'};clrscr();printf("\t\t\t***CYCLIC REDUNDANCY CHECK-12***\n\n\n");for(i=0;i<50;i++)plym[i]='\0';for(i=0;i<30;i++)temp[i]=rem[i]=frame[i]='\0';for(i=0;i<10;i++)gen[i]='\0';printf("enter the polynomial: ");gets(plym);plym[strlen(plym)]='+';for(i=0;i<strlen(plym);i++){if(plym[i]=='x'){i++;for(;plym[i]!='+';count++,i++){}if(count==3){for(i=i-1,j=3;j<=9;j++)if(plym[i]==ch[j]){printf("\Enter the polynomial's");printf("degree is high");getch(); exit(0);}for(j=0,num=10;j<=2;j++)if(plym[i]==ch[j])



{num=num+j;frame[num]=1;}}if(count==2){for(i=i-1,j=1,num=0;j<=9;j++)if(plym[i]==ch[j])num=j;frame[num]=1;}if(count==0)frame[1]=1;count=0;}else if(plym[i]=='1')frame[0]=1;}printf("FRAME is: ");for(i=12,j=0;i>=0;i--,j++){temp[j]=frame[i];printf("%d",frame[i]);}printf("\n\n\n>>>>both high & low orders");printf("bits of GENERATOR must be 1<<<<");printf("\n enter the generator: ");for(num=i=0;(c=getchar())!='\n';i++,num++){if(c=='1')gen[i]=1;else if(c=='0')gen[i]=0;else{printf("\n Enter the GENERATOR");printf("is other then 0 or 1");getch(); exit(0);}}for(j=13,i=i-1;i>0;i--,j++)temp[j]=0;printf("\n\n FRAME after appending 0's: ");copy(); check();printf("\n The REMAINDER is: ");for(i=13;i<j;i++)



{temp[i]=rem[i];printf("%d",rem[i]);}printf("\n\n\n Transmitting FRAME......");delay(10000);printf("\n\n\n Transmitted FRAME is: ");copy(); check();printf("\n frame recieved");printf("\n\n\n checking for errors......");delay(10000);printf("\n\n\n recieved frame is: ");copy(); check();printf("\n the remainder is: ");for(i=13;i<j;i++)printf("%d",rem[i]);printf("\n DATA SENT SUCCESSFULLY");getch();}check(){for(i=0;i<=12;i++){if(rem[i]==0)continue;else{for(k=0,t=i;k<num;k++,t++){if(rem[t]==1 && gen[k]==1)rem[t]=0;else if(rem[t]==0 && gen[k]==0)rem[t]=0;else if(rem[t]==1 && gen[k]==0)rem[t]=1;else if(rem[t]==0 && gen[k]==1)rem[t]=1;}}}return 0;}copy(){for(i=0;i<j;i++){printf("%d",temp[i]);



rem[i]=temp[i];}return 0;}

Result:

Output :

*** CYCLIC REDUNDANCY CHECK-12 ***

Enter the Polynomial: x^12+x^6+x+1FRAME is: 1000001000011

>>>> Both High & Low order bits of GENERATOR must be 1 <<<<Enter the GENERATOR: 10010

FRAME after appending 0’s: 10000010000110000The REMAINDER is: 0110

Transmitting FRAME ………

The Transmitted FRAME is: 10000010000110110FRAME RECEIVED.

Checking for ERRORS………

Received FRAME is: 10000010000110110The REMAINDER is: 0000

DATA SENT SUCCESSFULLY.



Program 4: Cyclic Redundancy Check,CRC-16

Program:

#include<stdio.h>#include<conio.h>#include<string.h>#include<dos.h>#include<stdlib.h>int copy();int check();int i,j,k,t,count=0,num=0;int gen[10],frame[30],rem[30],temp[30];void main(){char c,plym[50];char ch[]={'0','1','2','3','4','5','6','7','8','9'};clrscr();printf("\t\t\t***CYCLIC REDUNDANCY CHECk-16***\n\n\n");for(i=0;i<50;i++)plym[i]='\0';for(i=0;i<30;i++)temp[i]=rem[i]=frame[i]='\0';for(i=0;i<10;i++)gen[i]='\0';printf("Enter the polynomial: ");gets(plym);plym[strlen(plym)]='+';for(i=0;i<strlen(plym);i++){if(plym[i]=='x'){i++;for(;plym[i]!='+';count++,i++){}if(count==3){for(i=i-1,j=7;j<=9;j++)if(plym[i]==ch[j]){printf("\entered polynomial's");printf("degree is high");getch(); exit(0);}for(j=0,num=10;j<=6;j++)



if(plym[i]==ch[j]){num=num+j;frame[num]=1;}}if(count==2){for(i=i-1,j=1,num=0;j<=9;j++)if(plym[i]==ch[j])num=j;frame[num]=1;}if(count==0)frame[1]=1;count=0;}else if(plym[i]=='1')frame[0]=1;}printf("FRAME is: ");for(i=16,j=0;i>=0;i--,j++){temp[j]=frame[i];printf("%d",frame[i]);}printf("\n\n\n>>>>both high & low orders");printf("bits of GENERATOR must be 1<<<<");printf("\n Enter the generator: ");for(num=i=0;(c=getchar())!='\n';i++,num++){if(c=='1')gen[i]=1;else if(c=='0')gen[i]=0;else{printf("\n the Enter the generator");printf("is other then 0 or 1");getch(); exit(0);}}for(j=17,i=i-1;i>0;i--,j++)temp[j]=0;printf("\n\n FRAME after appending 0's: ");copy(); check();printf("\n The REMAINDER is: ");



for(i=17;i<j;i++){temp[i]=rem[i];printf("%d",rem[i]);}printf("\n\n\n Transmitting frame......");delay(10000);printf("\n\n\n TRANSMITTED FRAME is: ");copy(); check();printf("\n frame RECEIVED");printf("\n\n\n Checking for ERRORS......");delay(10000);printf("\n\n\n Recieved Frame is: ");copy(); check();printf("\n The REMAINDER is: ");for(i=17;i<j;i++)printf("%d",rem[i]);printf("\n DATA SENT SUCCESSFULLY");getch();}check(){for(i=0;i<=16;i++){if(rem[i]==0)continue;else{for(k=0,t=i;k<num;k++,t++){if(rem[t]==1 && gen[k]==1)rem[t]=0;else if(rem[t]==0 && gen[k]==0)rem[t]=0;else if(rem[t]==1 && gen[k]==0)rem[t]=1;else if(rem[t]==0 && gen[k]==1)rem[t]=1;}}}return 0;}copy(){for(i=0;i<j;i++){



printf("%d",temp[i]);rem[i]=temp[i];}return 0;}

Result:

Output:

*** CYCLIC REDUNDANCY CHECK-16 ***

Enter the Polynomial: x^16+x^12+x^10+x^6+x^5+x^4+x^3+x^2+x^1+1FRAME is: 10001010001111111

>>>> Both High & Low order bits of GENERATOR must be 1 <<<<Enter the GENERATOR: 100101

FRAME after appending 0’s: 1000101000111111100000The REMAINDER is: 01110

Transmitting FRAME ………

The Transmitted FRAME is: 1000101000111111101110FRAME RECEIVED.

Checking for ERRORS………

Received FRAME is: 1000101000111111101110The REMAINDER is: 0000

DATA SENT SUCCESSFULLY.



Program- 5: Shortest Path Routing.

Program:

#include<stdio.h>#include<conio.h>main(){ int z,fe[10],fc[10],fg[10],hi[10],hf[10]; int a[10]={0,5,2,3,0,0,0,0,0,0}; int b[10]={0,0,0,0,3,0,0,0,0,0}; int c[10]={0,0,0,0,0,4,0,0,0,0}; int d[10]={0,0,0,0,0,0,4,0,0,0}; int e[10]={0,0,0,0,0,2,0,0,2,0}; int f[10]={0,0,0,0,0,0,0,5,0,0}; int g[10]={0,0,0,0,0,6,0,0,0,0}; int h[10]={0,0,0,0,0,0,0,0,0,1}; int i[10]={0,0,0,0,0,0,0,2,0,0}; int j[10]={0,0,0,0,0,0,0,0,0,0}; clrscr(); printf("\t\t***shortest path routing***\n\n"); printf("\n the shortest path routing table \n\n"); for(z=0;z<10;z++) { if(b[z]!=0) b[z]=a[1]+b[z]; else b[z]=0; if(c[z]!=0) c[z]=a[2]+c[z]; else c[z]=0; if(d[z]!=0) d[z]=a[1]+d[z]; else d[z]=0; } for(z=0;z<10;z++) { if(e[z]!=0) e[z]=e[z]+b[4]; else e[z]=0; if(g[z]!=0) g[z]=g[z]+d[6]; else g[z]=0;



} for(z=0;z<10;z++) { if(f[z]!=0) { fe[z]=e[5]+f[z]; fc[z]=c[5]+f[z]; fg[z]=g[5]+f[z]; } else { fe[z]=0; fc[z]=0; fg[z]=0; } } for(z=0;z<10;z++) { if(fc[z]<fe[z]&&fc[z]<fg[z]) f[z]=fc[z]; else if(fe[z]<fg[z]) f[z]=fe[z]; else f[z]=fg[z]; } for(z=0;z<10;z++) { if(i[z]!=0) i[z]=i[z]+e[8]; else i[z]=0; if(h[z]!=0) { hi[z]=h[z]+i[7]; hf[z]=h[z]+f[7]; } else { hi[z]=0; hf[z]=0; } if(hi[z]<hf[z]) h[z]=hi[z]; else h[z]=hf[z]; } printf("\tA\tB\tC\tD\tE\n");



printf("A\t"); for(z=0;z<5;z++) printf("%d \t",a[z]); printf("\nB\t"); for(z=0;z<5;z++) printf("%d \t",b[z]); printf("\nC\t"); for(z=0;z<5;z++) printf("%d \t",c[z]); printf("\nD\t"); for(z=0;z<5;z++) printf("%d \t",d[z]); printf("\nE\t"); for(z=0;z<5;z++) printf("%d \t",e[z]); printf("\nF\t"); for(z=0;z<5;z++) printf("%d \t",f[z]); printf("\nG\t"); for(z=0;z<5;z++) printf("%d \t",g[z]); printf("\nI\t"); for(z=0;z<5;z++) printf("%d \t",i[z]); printf("\nH\t"); for(z=0;z<5;z++) printf("%d \t",h[z]); printf("\n\n"); printf("\tF\tG\tH\tI\tJ\n"); printf("\nA\t"); for(z=5;z<10;z++) printf("%d \t",a[z]); printf("\nB\t"); for(z=5;z<10;z++) printf("%d \t",b[z]); printf("\nC\t"); for(z=5;z<10;z++) printf("%d \t",c[z]); printf("\nD\t"); for(z=5;z<10;z++) printf("%d \t",d[z]); printf("\nE\t"); for(z=5;z<10;z++) printf("%d \t",e[z]); printf("\nF\t"); for(z=5;z<10;z++) printf("%d \t",f[z]);



printf("\nG\t"); for(z=5;z<10;z++) printf("%d \t",g[z]); printf("\nI\t"); for(z=5;z<10;z++) printf("%d \t",i[z]); printf("\nH\t"); for(z=5;z<10;z++) printf("%d \t",h[z]); printf("\nthe shortest path from source 'a' to destination 'j' is: %d",h[9]); getch();}



Result:

Output:

*** SHORTEST PATH ROUTING ***

THE SHORTEST PATH ROUTING TABLE

A B C D E A 0 5 2 3 0B 0 0 0 0 8C 0 0 0 0 0D 0 0 0 0 0E 0 0 0 0 0F 0 0 0 0 0G 0 0 0 0 0H 0 0 0 0 0I 0 0 0 0 0

F G H I JA 0 0 0 0 0B 0 0 0 0 0C 6 0 0 0 0D 0 7 0 0 0E 10 0 0 10 0F 0 0 11 0 0G 13 0 0 0 0I 0 0 12 0 0H 0 0 0 0 12

The Shortest Path from SOURCE ‘A’ to DESTINATION’J’ is : 12



Program-6: Distance Vector Routing Algorithm

Program:

#include<stdio.h>#include<conio.h>Main(){

int i=0,a[5],b[5],c[5],d[5],e[5],ab[5];int ac[5],ad[5],ao[5],ba[5],bc[5],be[5];int bo[5],ca[5],cb[5],ce[5],co[5],da[5];int ec[5],eb[5],eo[5],dop[5];char node[5]={‘A’,’B’,’C’,’D’,’E’};clrscr();printf(“\t\t\t\ *** DISTANCE VECTOR ROUTING *** \n\n\n\”);printf(“\n Enter values for ‘a’:\n”);for(i=0;i<5;i++)

scanf(“%d”,&a[i]);printf(“\n Enter the values for ‘b’:\n “);for(i=0;i<5;i++)

scanf(“%d”,&b[i]);printf(“\n Enter the values for ‘c’:\n “);for(i=0;i<5;i++)

scanf(“%d”,&c[i]);printf(“\n Enter the values for ‘d’:\n “);for(i=0;i<5;i++)

scanf(“%d”,&d[i]);printf(“\n Enter the values for ‘e’:\n “);for(i=0;i<5;i++)

scanf(“%d”,&e[i]);printf(“\n\n THE ROUTERS TABLE WITH OPTIMUM VALUES:\n”);printf(“ \t ------------------------------------ \n”);printf(“\t A \t B \t C \t D\t E\n”);printf(“ \t ------------------------------------ \n”);printf(“%c\t”,node[0]);for(i=0;i<5;i++){

Ab[i]=a[1]+b[i];Ac[i]=a[2]+c[i];Ad[i]=a[3]+d[i];

}for(i=0;i<5;i++){

if(a[i]<ab[i] && a[i]<ac[i] &&a[i]<ad[i])Ao[i]=a[i];

else if(ab[i]<ac[i] && ab[i]<ad[i])



Ao[i]=ab[i];else if(ac[i]<ad[i])

Ao[i]=ac[i];else

Ao[i]=ad[i];printf(“%d\t”,ao[i]);}printf(“\n %c \t “,node[1]);for(i=0;i<5;i++){

ba[i]=b[0]+ao[i];bc[i]=b[2]+c[i];be[i]=b[4]+e[i];

}for(i=0;i<5;i++){

if(b[i]<ba[i] && b[i]<bc[i] && b[i]<be[i])bo[i]=b[i];

else if(ba[i]<bc[i] && ba[i]<be[i])bo[i]=ba[i];

else if(bc[i]<be[i])bo[i]=bc[i];

elsebo[i]=be[i];

printf(“%d\t”,bo[i]);}printf(“\n%c\t”,node[2]);For(i=0;i<5;i++){

ca[i]=c[0]+ao[i];cb[i]=c[1]+bo[i];ce[i]=c[4]+e[i];

}for(i=0;i<5;i++){

if(c[i]<ca[i] && c[i]<cb[i] && c[i]<ce[i])co[i]=c[i];

else if(ca[i]<cb[i] && ca[i]<ce[i])co[i]=ca[i];

else if(cb[i]<ce[i])co[i]=cb[i];

elseco[i]=ce[i];

printf(“%d\t”,co[i]);}printf(“\n %c \t”,node[3]);For(i=0;i<5;i++)

da[i]=d[0]+ao[i];



for(i=0;i<5;i++){

if(d[i]<da[i])dop[i]=d[i];

elsedop[i]=da[i];

printf(“%d\t”,dop[i]);}printf(“\n%c\t”,node[4]);for(i=0;i<5;i++){

ec[i]=e[2]+co[i];eb[i]=e[1]+bo[i];

}for(i=0;i<5;i++){

if(e[i]<ec[i] && e[i]<eb[i])eo[i]=e[i];

else if(ec[i]<eb[i])eo[i]=ec[i];

elseeo[i]=eb[i];

printf(“%d\t”,eo[i]);}printf(“\n\t--------------------------- \n”);printf(“\n OPTIMUM COST FROM ROUTER-A TO ROUTER-E:”);printf(“%d”,ao[4]);getch();return();

}



Result:

Output:

*** DISTANCE VECTOR ROUTING ***

Enter the values for ‘a’:0 5 2 3 6

Enter the values for ‘b’:5 0 4 8 3

Enter the values for ‘c’:2 4 0 5 4

Enter the values for ‘d’:3 8 5 0 9

Enter the values for ‘e’:6 3 4 9 0

THE ROUTERS TABLE WITH OPTIMUM VALUES

------------------------------A B C D E------------------------------

A 0 5 2 3 6B 5 0 4 8 3C 2 4 0 5 4D 3 8 5 0 9E 6 3 4 9 0

------------------------------

THE OPTIMUM COST FROM ROUTER-A TO ROUTER-E: 6



Program-7: Simplified Decryption Encryption Standard

Program:#include<stdio.h>#include<conio.h>#include<math.h>int p,q,i=0,j=0,k=0,a=0,r=0,x=0,c=0.t=0,y=0,b[2];int binary[10],binmsg[500],temp[8],e[8],t4[4],tp[4];int p[10],key[10],newkey[10],key1[8],key2[8],p8[8];int ip[8]={2,6,3,1,4,8,5,7}ip_1[8]={4,1,3,5,7,2,8,6};int ep[8]={4,1,2,3,2,3,4,1},p4[4]={2,4,3,1},binmsg1[500];int s0[4][4]={1,0,3,2,3,2,1,0,0,2,1,3,3,1,3,2};int s1[4][4]={0,1,2,3,2,0,1,3,3,0,1,0,2,1,0,3};char ch;main(){

FILE *fp;clrscr();fp=fopen(“input1.txt”,”r”);printf(“\t\t\t ***SIMPLIFIED-DES***\n”);printf(“\n Original Message:\n”);

//Coverting Text to Binarywhile((a=ch=fgetc(fp))!=EOF){

printf(“%c”,ch);for(j=0;a!=0;j++){

r=a%2;a/=2;binary[j]=r;

}for(i=8-j;i>0;i__,j++)

binary[j]=0;for(i=j-1;i>=0;i--,x++)

binmsg[x]=binary[i];}printf(“\n\n Enter the 10 bit key: “);for(i=0;i<10;i++)

scanf(“%d”,&key[i]);//P10 User Defined

printf(“\n Enter the P10 Sequence” “);for(i=0;i<10;i++)

scanf(“%d”,&p10[i]);//Generating Key according to P10

for(i=0;i<10;i++){

j=p10[i]; j--;



newkey[i]=key[j];}

//LS-1 operationj=newkey[4];for(i=4,k=3;i>0;i--,k--)

newkey[i]=newkey[k];newkey[0]=j;J=newkey[9];for(i=9,k=8;i>5;i--,k--)

newkey[i]=newkey[k];newkey[5]=j;printf(“\n Enter the p8 sequence: “);for(i=0;i<8;i++)

scanf(“%d”,&p8[i]);//Generating KEY-1

printf(“\n key-1:”);for(i=0;i<8;i++){

j=p8[i]; j--;key1[i]=newkey[j];printf(“%d”,key1[i]);

}//LS-2 Operation

j=newkey[3]; a=newkey[4];for(i=4,k=2;i>=2;i--,k--)

newkey[i]=newkey[k];newkey[0]=j; a=newkey[9];for(i=9,k=7;i>=7;i--,k--)

newkey[i]=newkey[k];newkey[5]=j; newkey[6]=a;

//Generating KEY-2printf(“\n Key-2: “);for(i=0;i<8;i++){

j=p8[i];key2[i]=newkey[j];printf(“%d”,key2[i]);

}printf(“\n\n Cipher Message: “);for(i=0;i<x;){// considering 8-bit plain text

for(j=0;j<8;j++,i++)temp[j]=binmsg[i];

initp(); extp(); xor 1();fk(); switching();extp(); xor2(); fk(); ipinv();for(j=7,a=0,k=0;k<8;j--,k++,y++)



{binmsg1[y]=temp[k];a=a+temp[k]*pow(2,j);

}printf(“%c”,a);

}//Decryption

printf(“\n\n Decrypted Message is:”);for(i=0;i<y;){

for(j=0;j<8;j++,i++)temp[j]=binmsg1[i];

initp(); extp();xor2();fk(); switching();extp(); xor1(); fk(); ipinv();for(j=7,a=0,k=0;k<8;j--,k++)

a=a+temp[k]*pow(2,j);print(“%c”,a);

}getch();return 0;

}INITP(){//permutating using IP Function

For(j=0;j<8;j++){

a=ip[j]; a--;binary[j]=tamp[a];

}return 0;

}ipinv(){//Inverse Initial Permutation

for(j=0;j<8;j++){

a=ip_1[j]; a--;temp[j]=binary[a];

}return 0;

}switching(){//Switching

for(k=0,j=4;k<=3;k++,j++){



a=binary[k];binary[k]=binary[j];binary[j]=a;

}return 0;

}extp(){//Applying E/P Operation

for(j=0;j<8;j++){

a=ep[j]; a+=3;e[j]=binary[a];

}return 0;

}fk(){//Converting Binary to Decimal

p=e[0];q=e[3];r=p*2+q;p=e[1];q=e[2];c=p*2+q;

//Finding value in S-0 boxt=s0[r][c]a=0;compare();p=e[4];q=e[7];r=p*2+q;t=s1[r][c];compare();

//EX-OR operation between p4 & first 4bits of IPfor(j=0;j<4;j++){

a=p4[j];a--;tp[j]=t4[a];if(tp[j]==0&& binary[j]==0)

binary[j]=0;else if(tp[j]==1 && binary[j]==1)

binary[j]=0;else if(tp[j]==0 && binary[j]==1)

binary[j]=1;else if (tp[j]==1 && binary[j]==0)

binary[j]=1;}return 0;



}compare(){

if(t==0){

t4[a]=0; a++;t4[a]=0; a++;

}else{

for(j=0;t!=0;j++){

k=t%2;t/=2;b[j]=k;}

for(j=1;j>=0;j--,a++)T4[a]=b[j];

}return 0;

}Xor1(){//EX-OR operation b/w ep & key-2

for(j=0;j<8;j++){

if(e[j]==0 &&key2[j]==0)e[j]=0;

else if(e[j]==1 && key2[j]==1)e[j]=0;

else if(e[j]==1&&key2[j]==0)e[j]=1;

else if(e[j]==0&&key2[j]==1)e[j]=1;

}return 0;

}



Result:

Input File (input1.txt):

Surendra and Pradeep

Output:

*** SIMPLIFIED –DES ***

Original Message:Suerndra and pradeep

Enter the 10 BIT KEY: 1 0 1 01 0 1 0 1 0

Enter the P10 Sequence: 4 2 1 7 8 3 9 5 10 6

Enter the P8 Sequence: 2 1 7 8 3 9 5 10

Key-1: 00110110Key-2: 11000101

Cipher Message:

Decrypted Message is: surendra and pradeep



Program-8: Caeser Cipher Susbstitution Technique.

Program:#include<stdio.h>#include<conio.h>#include<string.h>main(){int c,i,n=0,key;char ch,cipher[100];FILE *fp;fp=fopen("input1.txt","r");clrscr();for(i=0;i<100;i++)cipher[i]='\0';printf("\t\t\t*** ceaser cipher substitution tech***\n\n\n");printf("the original text is:");while((ch=fgetc(fp))!=EOF)printf("%c",ch);fseek(fp,0,0);y:printf("\n\nenter the key");scanf("%d",&key);if(key>25){printf("\n\n***minimum value allowed for key is 25***\n");getch(); goto y;}printf("\n cipher text is :");/*encryption*/for(i=0;(n=fgetc(fp))!=EOF;){if(n==32){cipher[i]=' ';i++;printf(" ");}else{if(n>64&&n<91){c=(n+key)%91;if(c<65)



c=c+65;cipher[i]=c;i++;printf("%c",c);}else if(n>96 &&n<123){c=(n+key)%123;if(c<97)c=c+97;cipher[i]=c;i++;printf("%c",c);}}}printf("\n\nthe decrypted text is :");for(i=0;i<strlen(cipher);i++){n=cipher[i];if(n==32)printf(" ");else{if(n>64&&n<91){c=(n-key)%91;if(c<64)c=90-64+c;printf("%c",c);}else if(n>96&&n<123){c=(n-key)%123;if(c<97)c=122-96+c;printf("%c",c);}}}getch();return 0;}



Result:

Output:

***CAESER CIPHER SUBSTITUTION ***

The Original text is : This is Anitha

Enter the key: 2

Cipher text is: vjku ku cpkvjc

The Decrypted text is : This is Anitha



LiBrary management system

SYSTEM REQUIREMENT SPECIFICATION

To model the library management system using ‘Object Oriented Analysis and Designing’ concept and the unified modelling language (UML) objective behind the development of this software is to model an object oriented system which helps the librarians to easily performs their activities. It must as a best interface between the librarian and the computer so as to perform the required activities effectively. It must cover the activites of maintaining the records of issued and received books and

Other things available in the library separately for both faculty and student when by taking the secario as a college. Uml diagrams drew below helps us to easily understand how the software is developed. The diagrams are class, usecase, sequence, collaborations, component and deployment diagrams.



CLASS DIAGRAMClass diagram shows structure of the software system. The class diagram shows a set of

class, interfaces and their relationships. The components are:a) Classb) Relationship

The forms of relationship are:1. Associations 2. Aggregations3. Generalization4. Compositions5. Dependency

Data base

student id : intfaculty id : int

update data()

book details

book no : intauthor : stringcost : float

add book()del book()

mags & jornels

mid : intjid : int

add()del()

publisher

pid : intpname : stringpaddras : string

add publisher()del publisher()

faculty

name : stringid : int

issue()renewal()return()cancel()enquiry()

issued books

<<->> book name : string<<->> book id : num<<->> issued date : date

<<+>> add books()

librarian

name : stringid : int

check details()check availability()check database()

Reservation

student idbook nameauthor name

priority()

book bank

studentname : stringid : int

issuing()returning()

library

coll name : stringcoll id : int

digital library

id : stringpassword : stringsys no : int

accessing()download()

student

name : srtingid : intlib card

issue of book bank()issue()renewal()returning()cancel()enquiry()



USE CSE DIAGRAMUse case diagram is created to visualize the interaction of our system with the outside

world. The components of use case diagrams are:

Use Case: Scenarios of the system

Actors: some or something who is interacting with the system

Relationship: semantic link between use case and actor. The forms of relationship are:

a) Associationb) Dependencyc) Generalization

Database

impossing fines

updates

librarian

publisher

reservation

recieve books

digital library

student

issue books

return books

faculty

reading books



INTERACTION DIAGRAM

An interaction diagram models the dynamic aspects of the system by showing the relationship among the objects and message they may dispatch. There are two types of interaction diagrams:

SEQUENCE DIAGRAM

Sequence diagram shows the steps to steps what much happen to accomplish a piece of functionality provided by the system. The components are:

a) Actors b) Objectsc) Messaged) Lifelinee) Focus of Control

COLLABORATION DIAGRAM

Collaboration diagrams displays object interactions organized around object and their links to one another. The components are:

a) Actorb) Objectc) Link



Sequence diagram for issue and return of books

:student :books :Librarian :Faculty

refers

refer books

issue books

issue books

return books

update books



Collaboration diagram for issue and return of books

:books:student

:Librarian

:Faculty

1: refers

2: refer books

3: issue books

4: issue books

5: return books

6: update books



Sequence diagram for book bank

:student:librarian :books

refers

request for book bank

update records

issues book bank

Collaboration diagram for book bank

:librarian :books

:student

4: update records

2: request for book bank3: issues book bank1: refers



Sequence diagram for database

:librarian :books details :student details :faculty details :database

checks

checks

checks

update all details



Collaboration diagram for database

1: checks

:librarian

:books details

:student details

:faculty details

:database

3: checks

4: update all details

2: checks



Sequence diagram for reservation

:librarian:books:student

refers

request for reservation

ask for book details

tells book details

issue book when available



:books:student

:librarian

1: refers

2: request for reservation

3: sk for book details

4: tells book details

5: issue book when available

ACTIVITY DIAGRAM

Activity diagram shows the flow of events within our systems. The components are:

a) Start stateb) End statec) Transactiond) Decision Boxe) Synchronization Barf) Swim lane



show id card

select operation

show lib card

issue renewal return

start operation

close operation

valid student

update database

validation/sucess

databaselibrarianstudent



STATE CHART State chart diagrams shows a life cycle of a single class. The state is a condition where the object may be in. The components are:

a) Start stateb) End statec) Stated) Transition

verify student details

make/remove reservation

pay dues(if any)

take/return book



calculate dues

issue book

update catalog

request for stock

manages

supply

Component diagramComponent diagrams provide a physical view of the current model. A component

diagram shows the organizations and dependencies among software components, including source code components, binary code components, and executable components.

Component diagram for library system

library

book stock details.jar

update application.dll



library

book stock details.jar

update application.dll



Deployment DiagramDeployment diagram shows the physical architecture of the hardware and software in the system.

You can show the actual computers and devices, along with connections they have to each other and it can also show dependencies between components.

Deployment diagram for Library System

backup

library server

client-1 client-2

>> >><<<<Tcp/Ip>>



Placement Section

SOFTWARE REQUIREMENT SPECIFICATION

Introduction

This product is developed for the purpose of Managing and maintaining details & its work of the placement section.

DOCUMENT CONVENTIONS

AGG : AggregateACCR : AccredationPO : Placement OfficerADMIN : Administration

INTENDED AUDIENCE

Company HR Placement Officer Developers Principal

PROJECT SCOPE

The scope of the project is to maintain the entire student database of the college ie student ID, Branch, 10th &12th marks, Aggregate details.

REFERENCES

UML 2 ToolKit by WLLEY publications.

DESCRIPTION

PRODUCT PERSPECTIVE

The origin of the product is derived from the examination section of the college. The examination section of the college contains only the academic details of the student during his course of study in the college.



In this product we maintain academic details, co-circular and extra circularactivites of every particular student

PRODUCT FEATURES

The main feature of this product is that it provides the entire details of the student in the format the company requires.

System should keep a track of students placed in organisation (ATLEAST ONE) and yet to be placed including the non placed student lacking areas.

BENEFITS

A well organised view of placements reducing confusion and steps to be taken for best results and reduce paper work and time of processing and saving turn over caused by them.

CLASS DIAGRAMClass diagram shows structure of the software system. The class diagram shows a set of

class, interfaces and their relationships. The components are:a) Classb) Relationship

The forms of relationship are:1. Associations 2. Aggregations3. Generalization4. Compositions5. Dependency



placement section

placement officer : string

collecting()approaching()

college

college code : stringaccr : stringname : string

approaching hr()counselling student()

student

name : stringid : intbranch : stringresume

fill application()appoinment letter()

company

name

return test()gd()hr interview()select student()

hr

namecompany name

shortlist students()select students()reject students()

technical interview

shortlisted student list

conduct interview()

USE CASE DIAGRAMUse case diagram is created to visualize the interaction of our system with the outside

world. The components of use case diagrams are:

Use Case: Scenarios of the system

Actors: some or something who is interacting with the system

Relationship: semantic link between use case and actor. The forms of relationship are:

a) Associationb) Dependencyc) Generalization



Use-Case Diagram For Placement Section:

college management

company

visit recuriment

aptitude

gd

technical interview

personal interview

approaching hr

confermation letter

hr

counselling

student



INTERACTION DIAGRAM

An interaction diagram models the dynamic aspects of the system by showing the relationship among the objects and message they may dispatch. There are two types of interaction diagrams:

SEQUENCE DIAGRAM

Sequence diagram shows the steps to steps what much happen to accomplish a piece of functionality provided by the system. The components are:

f) Actors g) Objectsh) Messagei) Lifelinej) Focus of Control

COLLABORATION DIAGRAM

Collaboration diagrams displays object interactions organized around object and their links to one another. The components are:

d) Actore) Objectf) Link



Sequence Diagram of Notification:

placement college company student

request

approach

accept

confermation

notify students

Collaboration Diagram of Notification:

placement

college

company

student

1: request

2: accept

3: approach4: confermation5: notify students



Sequence Diagram of Interview:

placement section

company studenttudentss

approach

fill application

eligible students

written test

shortlisted students

technical round

Collaboration Diagram Of Interview:

placement section

company

studenttudentss

1: approach

2: fill application3: eligible students4: written test

5: shortlisted students6: technical round



Sequence Diagram Of Appointment:

company student hr

hr round

shortlist student

appoinment letter

join

Collaboration Diagram of Appointment:

company

student hr

1: hr round2: shortlist student

3: appoinment letter

4: join



ACTIVITY DIAGRAM

Activity diagram shows the flow of events within our systems. The components are:

a) Start stateb) End statec) Transactiond) Decision Boxe) Synchronization Barf) Swim lane



Activity Diagram For Placement Section:

request

student

fill application

company

written test G.DH.R. Interview

Technical Round

Short Listed Students

Conduct Interview

Select/Reject Students



STATE CHART State chart diagrams shows a life cycle of a single class. The state is a condition where the object may be in. The components are:

a) Start stateb) End statec) Stated) Transition

State-Chart Diagram For College:

Conformation Letter

Approach Companies

Issue Applications



State-Chart Diagram For Company:

G.D

Written Test

H.R.Interview

Select Students

State-Chart Diagram For Student:

Fill Application

Attend Interview

Appointment Letter



State-Chart Diagram For H.R.Manager

Shortlist Students

Select Students

Reject Students

Component diagram

Component diagrams provide a physical view of the current model. A component diagram shows the organizations and dependencies among software components, including source code components, binary code components, and executable components.



Component Diagram For Placement Section:

placement.java

college.java

company.java

student.java

written test.dll

gd.dll

hr interview.dll

Deployment DiagramDeployment diagram shows the physical architecture of the hardware and

software in the system. You can show the actual computers and devices, along with connections they have to each other and it can also show dependencies between components



Deployment diagram for Placement Section:

placement

college

company

student


Date post:	01-Dec-2014
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