Question

#include <string.h>
#include <conio.h>
#include <math.h>

int mult(unsigned int x, unsigned int y, unsigned int n) {
    unsigned long int k = 1;
    int j;
    for (j = 1; j <= y; j++)
        k = (k * x) % n;
    return (unsigned int) k;
}

void main() {
    char msg[100];
    unsigned int pt[100], ct[100], n, d, e, p, q, i;
    printf("Enter message : ");
    gets(msg);
    //strcpy(pt, msg);
    for (i = 0; i < strlen(msg); i++)
        pt[i] = msg[i];
    n = 253;
    d = 17;
    e = 13;
    printf("\nCT = ");
    for (i = 0; i < strlen(msg); i++)
        ct[i] = mult(pt[i], e, n);
    for (i = 0; i < strlen(msg); i++)
        printf("%d ", ct[i]);
    printf("\nPT = ");
    for (i = 0; i < strlen(msg); i++)
        printf("%c", pt[i]);
    for (i = 0; i < strlen(msg); i++)
        pt[i] = mult(ct[i], d, n);
}

This is simple program for implementing RSA algorithm can anybody explain me what’s happening here and why particularly n=253, d=17 , e=13.

Help would be much appreciated.

Answer 1

http://en.wikipedia.org/wiki/RSA_(algorithm).

253=11*23, // n=253 is product of two “large” primes p,q

phi(n)=10*22=220. // Eulers totient function is calculated from p,q
e= 13 // a small number, that is coprime with phi(n).
Then d is calculated so, that e*d=1 mod 220 –> d=17
-> d*e=221 === 1 mod 220.

Throw p,q away.