cvw/examples/crypto/gfmul/gfmul.c
2024-05-15 19:29:42 -07:00

73 lines
2.0 KiB
C

// gfmul.c - Galois Field multiplication
// James Stine and David Harris 16 May 2024
#include <stdio.h>
/* return ab mod m(x) - long multiplication in GF(2^n) with polynomial m */
int gfmul(int a, int b, int n, int m) {
int result = 0;
while (b) {
if (b & 1) result = result ^ a; /* if bit of b is set add a */
a = a << 1; /* multiply a by x */
if (a & 1 << n)
a = a ^ m; /* reduce/sub modulo AES m(x) = 100011011 */
//printf("a = %x, b = %x, result = %x\n", a, b, result);
b = b >> 1; /* get next bit of b */
}
return result;
}
void inverses(void) {
int i, j, k, num;
printf("\nTable of inverses in GF(2^8) with polynomial m(x) = 100011011\n");
for (i=0; i<16; i++) {
for (j=0; j<16; j++) {
num = i*16+j;
if (num ==0) printf ("00 ");
else for (k=1; k<256; k++) {
if (gfmul(num, k, 8, 0b100011011) == 1) {
printf("%02x ", k);
break;
}
}
}
printf("\n");
}
}
void inverses3(void) {
int k, num;
printf("\nTable of inverses in GF(2^8) with polynomial m(x) = 100011011\n");
for (num=0; num<8; num++) {
if (num == 0) printf ("0 ");
else for (k=1; k<8; k++) {
if (gfmul(num, k, 3, 0b1011) == 1) {
printf("%d ", k);
break;
}
}
}
printf("\n");
}
int main() {
int a = 0xC5;
int b = 0xA1;
printf("The GF(2^8) result is %x\n", gfmul(a,b, 8, 0b100011011));
printf("The GF(2^8) result is %x\n", gfmul(0xC1, 0x28, 8, 0b100011011));
inverses();
// tabulate inverses for GF(2^3)
inverses3();
// check worked examples
printf("The GF(2^3) result is %d\n", gfmul(0b101,0b011, 3, 0b1011));
printf("The GF(2^3) result is %d\n", gfmul(0b101,0b010, 3, 0b1011));
printf("The GF(2^3) result is %d\n", gfmul(0b101,0b100, 3, 0b1011));
printf("The GF(2^3) result is %d\n", gfmul(0b101,0b011, 3, 0b1011));
}