Friday, November 25, 2016
Simple Encryption and Decryption program in C
#include<iostream.h>
#include<stdio.h>
#include<conio.h>
#include<string.h>
void decrypt(char a[],int n)
{
int i;
for(i=0;i<strlen(a);i++)
{
a[i]=a[i]-n;
}
}
void encrypt(char a[],int n)
{
int i;
for(i=0;i<strlen(a);i++)
{
a[i]=a[i]+n;
}
}
void main()
{
char name[100];
clrscr();
cout<<"enter a string\n";
// cin>>name;
gets(name);
cout<<"original string:"<<name<<"\n";
encrypt(name,4);
cout<<"encrypted string:"<<name<<"\n";
decrypt(name,4);
cout<<"decrypted string:"<<name<<"\n";
getch();
}
#include<stdio.h>
#include<conio.h>
#include<string.h>
void decrypt(char a[],int n)
{
int i;
for(i=0;i<strlen(a);i++)
{
a[i]=a[i]-n;
}
}
void encrypt(char a[],int n)
{
int i;
for(i=0;i<strlen(a);i++)
{
a[i]=a[i]+n;
}
}
void main()
{
char name[100];
clrscr();
cout<<"enter a string\n";
// cin>>name;
gets(name);
cout<<"original string:"<<name<<"\n";
encrypt(name,4);
cout<<"encrypted string:"<<name<<"\n";
decrypt(name,4);
cout<<"decrypted string:"<<name<<"\n";
getch();
}
C Program to find given IP Address belongs to which Class(A,B,C,D,E)
#include<iostream.h>
#include<conio.h>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int
check_digit_dot(char ipstr[])
{
for(int i=0;i<strlen(ipstr);i++)
{
if((ipstr[i]>=48 &&
ipstr[i]<=57)||(ipstr[i]=='.'))
{
//i++;
}
else
{ return 0; }
}
return 1;
}
int
check_ip(char ipstr[])
{
char temp[20];
int a,num,dot=0,j=0,numcount=0;
a=check_digit_dot(ipstr);
temp[0]='\0';
if(a==0)
return 0;
else
{
for(int i=0;i<=strlen(ipstr);i++)
{
if(ipstr[i]!='.')
{
temp[j]=ipstr[i];
j++;
temp[j]='\0';
}
else
{
num=atoi(temp);
//
cout<<"num="<<num<<"\n";
dot++;
if((num>=0 && num<=255)
&& strlen(temp)>0)
{
numcount++; }
//cout<<"numcount="<<numcount<<"\n";
j=0;
temp[j]='\0';
}
}
// after last dot
num=atoi(temp);
//
cout<<"num="<<num<<"\n";
if ((num>=0 && num<=255)
&& strlen(temp)>0)
{ numcount++; }
//
cout<<"numcount="<<numcount<<"\n";
}
//cout<<"dots="<<dot<<"
numcount="<<numcount<<"\n";
if(dot==3 && numcount==4)
return 1;
else
return 0;
}
void
main()
{
clrscr();
char a[50],temp[50];
int check,j=0,num;
cout<<" enter ip address\n";
cin>>a;
check=check_ip(a);
if(check==0)
{
cout<<"not valid ip
address\n";
}
else
{
cout<<" valid ip address\n";
temp[0]='\0';
for(int i=0;i<strlen(a);i++)
{
if(a[i]=='.')
{ break; }
else
{
temp[j]=a[i];
j++;
temp[j]='\0';
}
}
num=atoi(temp);
cout<<"num:"<<num<<"\n";
if((num>=0)&&(num<128))
cout<<"given ip address is class
A\n";
else
if((num>=128)&&(num<192))
cout<<"given ip address is class
B";
else
if((num>=192)&&(num<224))
cout<<"given ip address is class
C";
else
if((num>=224)&&(num<240))
cout<<"given ip address is class
D";
else
cout<<"given ip address is class
E";
}
getch();
}
C Program to find given IP Address is Valid or not
#include<iostream.h>
#include<conio.h>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int
check_digit_dot(char ipstr[])
{
for(int i=0;i<strlen(ipstr);i++)
{
if((ipstr[i]>=48 &&
ipstr[i]<=57)||(ipstr[i]=='.'))
{
//i++;
}
else
{
return 0;
}
}
return 1;
}
int
check_ip(char ipstr[])
{
char temp[20];
int a,num,dot=0,j=0,numcount=0;
a=check_digit_dot(ipstr);
temp[0]='\0';
if(a==0)
return 0;
else
{
for(int i=0;i<=strlen(ipstr);i++)
{
if(ipstr[i]!='.')
{
temp[j]=ipstr[i];
j++;
temp[j]='\0';
}
else
{
num=atoi(temp);
cout<<"num="<<num<<"\n";
dot++;
if((num>=0 && num<=255)
&& strlen(temp)>0)
{
numcount++; }
cout<<"numcount="<<numcount<<"\n";
j=0;
temp[j]='\0';
}
}
// after last dot
num=atoi(temp);
cout<<"num="<<num<<"\n";
if ((num>=0 && num<=255)
&& strlen(temp)>0)
{ numcount++; }
cout<<"numcount="<<numcount<<"\n";
}
cout<<"dots="<<dot<<"
numcount="<<numcount<<"\n";
if(dot==3 && numcount==4)
return 1;
else
return 0;
}
void
main()
{
clrscr();
char a[50];
int check;
cout<<" enter ip address\n";
cin>>a;
check=check_ip(a);
if(check==1)
{
cout<<"valid ip address";
}
else
{
cout<<"not valid address";
}
getch();
}
RSA Encryption and Decryption with Numerical Example and C Program on RSA Encryption and Decryption
To generate the encryption and
decryption keys, we can proceed as follows.
1. Generate randomly two “large”
primes p and q.
2. Compute n = pq and Ï•= (p − 1)(q − 1).
3. Choose a number e so that gcd(e, Ï•)
= 1.
4. Find the multiplicative inverse of
e modulo Ï•, i.e., find d so that ed ≡ 1 (mod Ï•).
This can be done efficiently using
Euclid’s Extended Algorithm.
The encryption public key is KE
= (n, e) and the decryption private key is KD = (n, d).
The encryption function is E(M) = Me mod n.
The decryption function is D(M) = Md mod n.
These functions satisfy D(E(M)) = M and E(D(M)) = M for any 0 ≤ M ≤
n.
*****
Let’s look at a numerical example.1. Let p = 7 and q = 13 be the two primes.
2. n = pq = 91 and Ï• = (p − 1)(q − 1)
= 72.
3. Choose e. Let’s look among the
primes.
•
Try e = 2; gcd(2, 72) = 2 (does not
work)
•
Try e = 3; gcd(3, 72) = 3 (does not
work)
•
Try e = 5; gcd(5, 72) = 1 (it works)
We
choose e = 5.
4. Let’s find d. We want to find d
such that ed ≡ 1
(mod Ï•) which is equivalent to find d such that ed + Ï• k = 1 for some integer k. Recall that
gcd(e, Ï•) = 1.
We
can use the Extended Euclid’s Algorithm to find integers x and y such that
ex + Ï•y = gcd(e, Ï•).
If
e = 5 and Ï• = 72, we
find x = 29 and y = −2.
Indeed, 5(29) +
72(−2) = gcd(5, 72) = 1. Then, d = 29.
In
general, we use d = x mod Ï•.
5. The encryption function is E(M) = Me mod n = M5
mod 91.
The decryption function is D(M) = Md mod n = M29
mod 91.
6. Suppose the message is M = 10.
E(M)
= E(10) = 105 mod 91 = 82
Here 82 is called as CipherText
D(E(M))
= D(82) = 8229 mod 91 = 10
7. Let’s see how to compute
efficiently 8229 mod 91 using the square-and-multiply algorithm.
(82)1
≡
82 (mod 91)
(82)2
≡ 81
(mod 91)
(82)4
≡ (81)2 ≡ 9 (mod 91)
(82)8
≡ (9)2 ≡ 81
(mod 91)
(82)16
≡ (81)2 ≡ 9 (mod 91
Since
29 = 16 + 8 + 4 + 1 (in binary 29 is 11101),
we deduce that
8229
≡
(82)16(82)8(82)4(82)1
(mod 91)
≡
(9)(81)(9)(82) (mod 91)
≡
10 (mod 91)
We
conclude that 8229 mod 91 = 10 (Original Message).
Consider an another example
1. Let p = 19 and q = 23 be the two
primes.
2. n = p*q = 437 and Ï• = (p − 1)*(q −
1) = 396.
3. Choose e. Let’s look among the
primes.
•
Try e = 2; gcd(2, 396) = 0 (does not
work because it has factors)
•
Try e = 3; gcd(3, 396) = 0 (does not
work because it has factors)
•
Try e = 5; gcd(5, 396) = 1 (it works)
We
choose e = 5.
4. Let’s find d. We want to find d
such that ed ≡ 1
(mod Ï•) which is equivalent to find d such that ed + Ï• k = 1 for some integer k. Recall that
gcd(e, Ï•) = 1.
We
can use the Extended Euclid’s Algorithm to find integers x and y such that
ex + Ï•y = gcd(e, Ï•). i.e.
5x+396y=1x= (1-396y)/5
If
e = 5 and Ï• = 396,
we
find x = 317 and y = −4. Indeed, 5(317)
+ 396(−4) = gcd(5, 396) = 1. Then, d = 317.
In
general, we use d = x mod Ï•.
5. The encryption function is E(M) = Me mod n = M5
mod 437.
The decryption function is D(M) = Md mod n = M317 mod 437.
6. Suppose the message is M = 7.
E(M)
= E(10) = 75 mod 437 = 201
Here 201
is called as CipherText
D(E(M))
= D(201) = 201317 mod 437 = 7 (Original Message)
C program on RSA Encryption and Decryption
#include<iostream.h>
#include<conio.h>
#include<math.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
int
prime( int n)
{
int i,r;
for(i=2;i<n-1;i++)
{
r=n%i;
if(r==0)
{
return 0; }
}
return 1;
}
int
gcd(int a, int b)
{
int r=0;
while(b>a)
{
r=b%a;
if(r==0){ break;}
b=a;
a=r;
}
return a;
}
void main()
{
clrscr();
int n,pi,x,y,privatekey,
publickey,p,q,primecheck,gcdvalue;
cout<<"enter two prime
numbers"<<"\n";
cin>>p>>q;
if (prime(p)==0 || prime(q)==0 ||p==q)
{
cout<<" one of the number is
not prime or both numbers are same"<<"\n";
}
else
{
n=p*q;
pi= (p-1)*(q-1);
for(int e=3; e<pi-1; e++)
{
if (prime(e)==1)
{
if (gcd(e,pi)==1)
{
publickey= e;
break;
}
}
}
cout<<"Public
Key="<<publickey<<"\n";
y=1;
while(1)
{
x= (1-(pi*(-y)))/publickey ;
if (publickey*x+pi*(-y)==1) { break; }
y++;
}
privatekey=x;
cout<<"private key="<<
privatekey<<"\n";
cout<<"enter
message"<<"\n";
int msg;
cin>>msg;
long ctext=1,i;
for (i=0;i<publickey; i++)
ctext=(ctext*msg)%n;
ctext=ctext%n;
cout<<"ciphertext or
encrypt="<<ctext<<"\n";
long ptext=1,j;
for(j=0;j<privatekey;j++)
ptext=(ptext*ctext)%n;
ptext=ptext%n;
cout<<"plain
Text="<<ptext<<"\n";
}
getch();
}
Thursday, April 18, 2013
Radix Sort Sorting Technique and C Program
Radix
Sort
Consider an array A= { 54, 33, 25, 17, 09, 37, 13 }.
In radix sort traverse
array from left to right. Check maximum no. of digits in the array. In the above case, it is 2 so it will have
2-pass i.e, in the second pass information will be in sorted.
Traverse array
from left to right, check unit place
of each number in the array. Unit place of 54 is 4, so put 54 in bucket
4. Similarly, unit place of 33 is 3, so
put 33 in buctet 3. Similarly, check all
the numbers in the array and put those numbers in their respective buckets
Bucket
|
Pass-1
|
Pass-2
|
0
|
||
1
|
||
2
|
||
3
|
33,13
|
|
4
|
54
|
|
5
|
25
|
|
6
|
||
7
|
17,37
|
|
8
|
||
9
|
09
|
Now Merge the Output
of pass-1. Result is as follows A= {33,13,54,
25, 17, 37, 09}
Now tranverse output of Pass-1 from
left to right. Now check 10th
place of each number in the array. 10th
place of 33 is 3, put 33 in bucket 3. Similarly,
10th place of 13 is 1, put 13 in bucket 1. Similarly, check all the numbers in the array
and put those numbers in their respective buckets.
Bucket
|
Pass-1
|
Pass-2
|
0
|
09
|
|
1
|
13,17
|
|
2
|
25
|
|
3
|
33,13
|
33,37
|
4
|
54
|
|
5
|
25
|
54
|
6
|
||
7
|
17,37
|
|
8
|
||
9
|
9
|
Now Merge the output of pass-2. Result is as follows A= {09, 13, 17, 25, 33, 37,
54}
Note: if Maximum no. of digits is 3 then there will be 3
passes. If 4-digits, 4 passes etc.
The same can be done using the following calculation.
Consider an array A= { 54, 33, 25, 17, 09, 37, 13 }.
For Pass-1 Calculation (unit place)
54/100 = 54/1
= 54 =
54%10 = 4 Put 54 in bucket 4
33/100 = 33/1
= 33 =
33%10 = 3 Put 33 in bucket 3
25/100
=
25/1 = 25 = 25%10 = 5
Put 25 in bucket 5
17/100 = 17/1
= 17 =
17%10 = 7 Put
17 in bucket 7
09/100 = 09/1
= 09 =
09%10 = 9 Put 09 in bucket 9
37/100 = 37/1
= 37 =
37%10 = 7 Put 37 in bucket 7
13/100 = 13/1
= 13 =
13%10 = 3 Put 13 in bucket 3
Bucket
|
Pass-1
|
Pass-2
|
0
|
||
1
|
||
2
|
||
3
|
33,13
|
|
4
|
54
|
|
5
|
25
|
|
6
|
||
7
|
17,37
|
|
8
|
||
9
|
09
|
Merge output of pass-1 i.e., A= {33,13,54, 25, 17, 37, 09}
For Pass-2 Calculation (10th place).
33/101 = 33/10
= 3 =
3%10 = 3 Put 33 in bucket 3
13/101 = 13/10
= 1 =
1%10 = 1 Put 13 in bucket 1
54/101
=
54/10 = 5 = 5%10 = 5
Put 54 in bucket 5
25/101 = 25/10
= 2 =
2%10 = 2 Put
25 in bucket 2
17/101 = 17/10
= 1 =
1%10 = 1 Put 17 in bucket 1
37/101 = 37/10
= 3 =
3%10 = 3 Put 37 in bucket 3
09/101 = 09/10
= 0 =
0%10 = 0 Put 09 in bucket 0
Bucket
|
Pass-1
|
Pass-2
|
0
|
09
|
|
1
|
13,17
|
|
2
|
25
|
|
3
|
33,13
|
33,37
|
4
|
54
|
|
5
|
25
|
54
|
6
|
||
7
|
17,37
|
|
8
|
||
9
|
09
|
Now Merge the output of pass-2. Result is as follows A= {09, 13, 17, 25, 33, 37,
54}
C Program for radix sort.
#include<stdio.h>
#include<math.h>
struct node
{
int data;
struct node *link;
};
struct node * createnodes(int n)
{
int i;
struct node
*p,*first,*temp;
printf("Enter
elements\n");
for (i=0; i<n;
i++)
{
p= (struct node *)
malloc(sizeof(struct node));
scanf("%d", &p->data);
p->link = NULL;
if (i==0)
{
first=p;
}
else
{
temp=first;
while
(temp->link!=NULL)
{
temp=temp->link;
}
temp->link=p;
}
}
return first;
}
void displaynodes(struct node *f)
{
struct node *temp;
temp=f;
while (temp!=NULL)
{
printf("%d-->",temp->data);
temp=temp->link;
}
printf("Null\n");
}
int MaxDigits(struct node *f)
{
struct node *temp;
int max,ct,n;
temp=f;
max=0;
while (temp!=NULL)
{
n=temp->data;
ct=0;
while(n>0)
{
ct++;
n=n/10;
}
if (ct>max)
{ max=ct; }
temp=temp->link;
}
return max;
}
void RadixSort(struct node *f1)
{
struct node
*first,*f[10],*r[10],*temp;
int max,i,j,k,p;
max=MaxDigits(f1);
first=f1;
for(i=0;i<max;i++)
{
for(j=0;
j<10;j++)
{
f[j]=NULL;
r[j]=NULL;
}
while(first!=NULL)
{
temp=first;
first=first->link;
p=(temp->data/pow(10.0,(float)i));
p=p%10;
if(f[p]==NULL
&& r[p]==NULL)
{
f[p]=temp;
r[p]=temp;
temp->link=NULL;
}
else
{
r[p]->link=temp;
temp->link=NULL;
r[p]=temp;
}
}
first=f[0];
for(k=1;k<10;k++)
{
if (first==NULL
&& f[k]==NULL)
first=NULL;
else if
(first==NULL && f[k]!=NULL)
first=f[k];
else
{
temp=first;
while(temp->link!=NULL)
{ temp=temp->link; }
temp->link=f[k];
}
}
printf("Result of Pass - %d
: ",i+1);
displaynodes(first);
}
}
void main()
{
int n;
struct node *first;
clrscr();
printf ("Enter
size of array/Linked list \n");
scanf("%d",&n);
first =
createnodes(n);
printf("Printing the elements from LL before sorting\n");
displaynodes(first);
RadixSort(first);
getch();
}
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