Program C++ untuk Enkripsi Algoritma RSA dengan CRT

#include <iostream.h>

#include <math.h>

#include <string.h>

#include <stdlib.h>

#include <conio.h>

#include <stdio.h>

long int p,q,n,r,t,i,s,flag,e[100],d[100],temp[100],j,m[100],en[100],x,y;

char kata[100];

int prime(long int);

void ce();

long int cd(long int);

void encrypt();

int hitmod(int a,int b,int n);

int phi(int n);

int GCD(int x,int y);

void main()

{cout<<":::::::::::::::::::Chinese Remainder Theorem:::::::::::::::::::"<<endl<<endl;

 cout<<"\n Masukkan bilangan prima p : ";

 cin>>p;

 flag=prime(p);

  if(flag == 0)

  {cout<<"\n inputan anda salah \n";

   exit(1); }

 cout<<"\n Masukkan bilangan prima q : ";

 cin>>q;

 flag=prime(q);

  if(flag == 0 || p == q)

  {cout<<"\n inputan anda salah \n";

   exit(1); }

 n=p*q;

 cout<<"\n Diperoleh bilangan n = "<<n<<endl;

 cout<<"\n Masukkan kata : ";

 fflush(stdin);

 cin>>kata;

    for(i=0;kata[i] != NULL;i++)

       m[i]=kata[i];

    hitmod(p,q,n);

    cout<<"\n ::::Hasil CRT:::: \n";

    cout<<"\n Nilai untuk x yang diperoleh adalah : "<<hitmod(p,q,n)<<endl;

    encrypt();

 getch();

}

int prime(long int pr)

{ int i;

 j=sqrt(pr);

 for(i=2;i<=j;i++)

 {  if(pr%i == 0)

   return 0; }

  return 1; }

long int cd(long int x)

{ long int k = 1;

 while (1)

 {  k=k+t;

  if(k%x == 0)

  return(k/x); } }

void encrypt()

{ long int pt, ct, key = hitmod(p,q,n), k, len;

 i = 0;

 len = strlen(kata);

 while (i != len)

 {  pt=m[i];

  pt=pt-96;

  k=1;

   for(j=0;j<key;j++)

   {    k=k*pt;

    k=k%n;   }

    temp[i]=k;

    ct=k+96;

    en[i]=ct;

     i++;  }

  en[i] = -1;

  cout << "\n Hasil enkripsi : ";

   for (i = 0; en[i] != -1; i++)

    printf("%c", en[i]);

}

int hitmod(int a,int b,int n)

{ int p,q,r;

   p=a%n;

   q=b%phi(n);

   r=pow(p,q);

   r=r%n;

   return r; }

int phi(int n)

{ int i,s=0;

   for (i=1;i<n;i++){

   if (GCD(i,n)==1)

    s++;   }

   return s;}

int GCD(int x,int y)

{ int t,s;

   if (x<y)

{   t=x;

      x=y;

      y=t;   }

   s=x%y;

   while(s!=0)

{    x=y;

      y=s;

      s=x%y;   }

   return y;

}

Output Program

 

                  

Cara Manual

Misal dipilih p= 13 dan q= 23, sehingga diperoleh n=299 dan ɸ(n)=264. Kita ambil e=7, dimana GCD(e, ɸ(n) )= 1.

Diambil pesan : kompleks

Kode ASCII : 107111109112108101107115

Dengan memecah plainteks menjadi 3 digit, diperoleh m_i sebagai berikut:

m₁= 107                                     m₅= 108

m₂= 111                                     m₆= 101

m₃= 109                                     m₇= 107

m₄= 112                                     m₈= 115

Dengan rumus enkripsi c=m⁷ mod 299, diperoleh :

c₁= 172                                      c₅= 225

c₂= 84                                        c₆= 257

c₃= 112                                      c₇= 172

c₄= 44                                        c₈= 184

Jadi, diperoleh cipherteks : 172 84 112 44 225 257 172 184

Program c++ untuk RSA

#include <iostream.h>

#include <conio.h>

#include <stdio.h>

#include <math.h>

#include <string.h>

#include <stdlib.h>

int a,b,q,j,goprim, m[100], e[100];

long double c[100], n;

char pesan[100];

int prima(int x);

void kuncienkrip();

void enkrip();

void main()

{  int i;

                       goprim = 0;

                       while(goprim==0)

                       { cout<<"Masukkan bilangan prima p : ";

                       cin>>a;

                       goprim = prima(a);

                       if(goprim==0)

                       cout<<"Inputan bukan bilangan prima "<<endl;            }

                       goprim = 0;

                       while(goprim==0||a==b)

                       { cout<<"Masukkan bilangan prima q : ";

                           cin>>b;

                           goprim=prima(b);

                           if(goprim==0||a==b)

                           cout<<"Inputan salah "<<endl;           }

                       cout<<"\nMasukkan pesan yang ingin dienkripsi : ";

                       fflush(stdin);

                       cin>>pesan;

                       for (i=0; pesan[i]!=0;i++)

                       { m[i]=pesan[i];  }

                       n = a*b;

                       q = (a-1)*(b-1);

                       kuncienkrip();

                       cout<<"\nNilai e (kunci enkripsi) : ";

                       for (i=0;i<1;i++)

                       { cout<<e[i+1]<<"\t";    }

                       enkrip();

                       getch();   }

void faktor()

{                     int a, b, f, fktr[100];

                       b=0;

                       for(a=2;a<=f;a++)

                       {  if((f%a)==0)

                           {  fktr[b]=a;

                                       f/=a;

                                       a--;

                                       b++;                 }          }          }

int gcd(int r, int t)

{  int s, x;

                       if(t>r)

                       {             x=r;

                           r=t;

                           t=x;    }

                       s=r%t;

                       while (s!=0)

                       {             r=t;

                           t=s;

                           s=r%t;              }

                       return t;   }

int psy(int m)

{                     int i, pi;

                       pi=0;

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

                       {  if(gcd(m,i)==1)

                           pi++;    }

                       return pi;             }

int prima(int x)

{  int i;

                       j = sqrt(x);

                       for(i=2;i<=j;i++)

                       {  if(x%i==0)

                                       return 0;           }

                       return 1;  }

void kuncienkrip()

{  int k, i;

                       k = 0;

                       for (i=2;i<q;i++)

                       {  if((q%i)==0)

                           continue;

                           goprim=prima(i);

                           if(goprim==1 && i!=a && i!=b)

                           {  e[k]=i;

                                       if(goprim>0)

                                                   k++;

                                       if(k==99)

                                                   break;              }          }          }

void enkrip()

{  int l, key=e[1];

                       cout<<endl<<"\nCipherteks : ";

                       for(l=0;m[l]!=0;l++)

                       {             c[l]=fmod((pow(m[l],key)), n);

                                       cout<<c[l]<<" ";          }}

Output Program

 

Hasil yang diperoleh pada cara manual dengan program adalah sama, yaitu

cipherteks : 172 84 112 44 225 257 172 184

*PS: Program ini masih belum sepenuhnya benar (Anissa Ittaqullah)