Understanding Cryptography 2.11

lfsr6x5.c

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define STREAM (256 * 8)

/* short program to understand linear feed shift register
 * and its use in stream ciphers.
 *
 * modified to solve problem 2.11 in Understanding Cryptography by C. Paar
 * stream cipher has 6 elements and letters are represented as 5 bit values.
 * plain/cipher text input: a-z = 0-25, 0-5 = 26-31 represented as 00000-11111 binary.
 */


int
main(int argc, char argv[]) {
    
        unsigned char in_s, cs, cp, p, nbit, s[STREAM];
        unsigned char input[STREAM], output[STREAM];
        int i, j, k=0, stl;
    

        in_s = 0x3f;     /* initial value 111111 */
        p = 0x03;       /* polynomial x^6+x^5+1 */

        cs = in_s;      /* copy initial state */

        scanf("%s",&input);
        stl = strlen(input) * 5;

        /* first solve key bits s[k] */
        while (k < stl) {
                for (j = 0;j < 6;j++,k++) {
                        cp = nbit = cs & p;
                        
                        for (i = 1;i < 6; i++) { /* xor all bits together */
                                nbit ^= (cp >> i);
                        }
                        s[k] = cs & 0x01;
                        nbit &= 1;
                        cs = (cs >> 1) | (nbit << 5); /*  rotate in new bit */
                }

        }

        printf("\ncipher-text: ");
        unsigned char out = 0;
        k = 4;
        int c=0;

        /* then xor cipher/plain text with key bits and print */
        for (i = 0;i < stl;i++) {
                out |= s[i]<<k;
                k--;
                if ((i+1) % 5 == 0) {
                        if (input[c] < 54)      /* adjust for 0-5 = 26-31 in cipher */
                                input[c] += 75;
                        output[c] = ((input[c] - 0x61) ^ out) + 0x61;
                        if (output[c] > 0x7a)
                                output[c] -= 75;
                        printf("%c", output[c]);
                        k = 4;out = 0;c++;
                }
        }
        printf("\n");
}