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QuestionNeed to improve performance Multi-Threading code Pin
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Questionecc diffie hellman Pin
Member 1152168512-Mar-15 22:45
Member 1152168512-Mar-15 22:45 
plz give me decryption code for following progrm.....


import java.io.*;
import java.util.Scanner;
import java.util.Random;
import java.lang.Math.*;
class Diffie_Hellman{
public static void main(String[] args) {
Random rand = new Random();
Scanner in = new Scanner(System.in);
System.out.println("Enter any prime number");
int n = in.nextInt();
int g1, g2;
int pb1,pb2,pa1,pa2;
int e1, e2;
int i;
int bk1, bk2, ak1, ak2;

System.out.println("Enter valu of g");
g1= in.nextInt();
g2= in.nextInt();

System.out.println("Enter valu of e");
e1= in.nextInt();
e2= in.nextInt();

int na = rand.nextInt(5)+2;
if(na%2!=0)
na=na+1;
int nb = rand.nextInt(5)+2;
if(nb%2!=0)
nb=nb+1;
System.out.println("Private key of Alice: "+ na + " \n Private key Bob " +nb + " ");
//int pa = (na * g)
// int pb = (nb * g)

int p = na/2;
int g11= g1;
int g22= g2;
for(i=1; i<= p ; i++)
{
double m1=(3*g11*g11+ e1);
double m2= 2*g22;
double inverse=0;
inverse = (Math.pow(m2,-1))%n;
double m =(na*(m1*inverse))%n;
double x = (m*m +(n-(2*g11)))%n;
double y= (m*(x+(n-g11))+g22)%n;
g11=(int)x;
g22=(int)y;
}
if(na%2 == 0)
{
pa1=g11;
pa2=g22;
System.out.println("Alice's Public key is :" +pa1+ " " +pa2 );
}
else
{

double m = (g2-g22)/(g1-g11);
double x3 = (m*m+(n-g1)+(n-g11))%n;
double y3 = (m*(x3+(n-g11)) + g22)%n;
pa1= g11+(int)x3;
pa2= g22+(int)y3;
System.out.println("Alice's Public key is :" +pa1+ " " +pa2 );
}


p = nb/2;
g11= g1;
g22= g2;
for(i=1; i<= p ; i++)
{
double m1=(3*g11*g11+ e1);
double m2= 2*g22;
double inverse=0;
inverse = (Math.pow(m2,-1))%n;
double m =(na*(m1*inverse))%n;
double x = (m*m +(n-(2*g11)))%n;
double y= (m*(x+(n-g11))+g22)%n;
g11=(int)x;
g22=(int)y;
}
if(nb%2 == 0)
{
pb1=g11;
pb2=g22;
System.out.println("Bob's Public key is :" +pb1+ " " +pb2 );
}
else
{
double m = (g2-g22)/(g1-g11);
double x3 = (m*m+(n-g1)+(n-g11))%n;
double y3 = (m*(x3+(n-g11)) + g22)%n;
pb1= g11+(int)x3;
pb2= g22+(int)y3;
System.out.println("Bob's Public key is :" +pb1+ " " +pb2 );
}


System.out.println("******Generating shared key for Alice *******");
p = na/2;
g11= pb1;
g22= pb2;
for(i=1; i<= p ; i++)
{
double m1=(3*g11*g11+ e1);
double m2= 2*g22;
double inverse=0;
inverse = (Math.pow(m2,-1))%n;
double m =(na*(m1*inverse))%n;
double x = (m*m +(n-(2*g11)))%n;
double y= (m*(x+(n-g11))+g22)%n;
g11=(int)x;
g22=(int)y;
}
if(na%2 == 0)
{
ak1=g11;
ak2=g22;
System.out.println("Alice's shared key is :" +ak1+ " " +ak2 );
}
else
{
double m = (g2-g22)/(g1-g11);
double x3 = (m*m+(n-g1)+(n-g11))%n;
double y3 = (m*(x3+(n-g11)) + g22)%n;
ak1= g11+(int)x3;
ak2= g22+(int)y3;
System.out.println("Alice's shared key is :" +ak1+ " " +ak2 );
}


System.out.println("*****Generation of shared key for Bob*****");
p = nb/2;
g11= pa1;
g22= pa2;
for(i=1; i<= p ; i++)
{
double m1=(3*g11*g11+ e1);
double m2= 2*g22;
double inverse=0;
inverse = (Math.pow(m2,-1))%n;
double m =(na*(m1*inverse))%n;
double x = (m*m +(n-(2*g11)))%n;
double y= (m*(x+(n-g11))+g22)%n;
g11=(int)x;
g22=(int)y;
}
if(nb%2 == 0)
{
bk1=g11;
bk2=g22;
System.out.println("Bob's shared key is :" +bk1+ " " +bk2 );
}
else
{
double m = (g2-g22)/(g1-g11);
double x3 = (m*m+(n-g1)+(n-g11))%n;
double y3 = (m*(x3+(n-g11)) + g22)%n;
bk1= g11+(int)x3;
bk2= g22+(int)y3;
System.out.println("Bob's shared key is :" +bk1+ " " +bk2 );
}

System.out.println("Enter Your Messgae");
int msg1 = in.nextInt();
int msg2= in.nextInt();
int k= (bk1*bk2)%n;
System.out.println("Value of K is : " +k);

p = k/2;
if(p<0)
p=p*(-1);
g11= msg1;
g22= msg2;
for(i=1; i<=p ; i++)
{
double m1=(3*g11*g11+ e1);
double m2= 2*g22;
double inverse=0;
inverse = (Math.pow(m2,-1))%n;
double m =(na*(m1*inverse))%n;
double x = (m*m +(n-(2*g11)))%n;
double y= (m*(x+(n-g11))+g22)%n;
g11=(int)x;
g22=(int)y;
}
if(na%2 == 0)
{
pa1=g11;
pa2=g22;
System.out.println("C :" +pa1+ " " +pa2 );
}
else
{

double m = (g2-g22)/(g1-g11);
double x3 = (m*m+(n-g1)+(n-g11))%n;
double y3 = (m*(x3+(n-g11)) + g22)%n;
int c1= g11+(int)x3;
int c2=g22+(int)y3;
pa1= (int)c1;
pa2= (int)c2;
System.out.println("C:" +pa1+ " " +pa2 );
}
//decryption code

}
}
AnswerRe: ecc diffie hellman Pin
Richard MacCutchan12-Mar-15 23:53
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