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October 29, 2019 15:19
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/* | |
ECE4802 - Cryptography | |
Homework 1 | |
Part 2 | |
Nam Tran Ngoc | |
*/ | |
#include <stdio.h> | |
#include <stdint.h> | |
#define INIT_STATE 0b000110100 | |
#define LFSR_MASK 0b100011101 | |
unsigned lfsr(void); | |
void printbits(size_t const size, void const * const ptr); | |
int main() { | |
printf("%d", lfsr()); | |
return 0; | |
} | |
unsigned int lfsr(void) { | |
uint16_t bit; | |
uint16_t lfsr = INIT_STATE; | |
unsigned period = 0; | |
char s[17]; | |
do { | |
// for(int i = 0; i < 9; i++) { | |
// s[8 - i] = (lfsr & (1 << i)) ? '1' : '0'; | |
// } | |
// s[10]= '\0'; | |
// printf("%10d: %s\n", period, s); | |
printf("%10d: ", period); | |
printbits(sizeof lfsr, &lfsr); | |
// Taps: 8 4 3 2 feedback polynomial x^8 + x^4 + x^3 + x^2 + 1 | |
unsigned lsb = lfsr & 1; | |
// bit = ((lfsr >> 8) ^ (lfsr >> 4) ^ (lfsr >> 3) ^ (lfsr >> 2)) & 1; | |
//=============MESS AF | |
uint16_t bit8 = lfsr >> 8; | |
uint16_t bit4 = lfsr >> 4; | |
uint16_t bit3 = lfsr >> 3; | |
uint16_t bit2 = lfsr >> 2; | |
// printbits(sizeof lfsr, &lfsr); | |
// printbits(sizeof lfsr, &bit4); | |
bit = ((lfsr >> 8) ^ (lfsr >> 4) ^ (lfsr >> 3) ^ (lfsr >> 2)) & 1; | |
lfsr >>= 1; | |
lfsr ^= (-bit ^ lfsr) & (1 << 8); | |
// lfsr = (lfsr >> 1) | (bit << 8); | |
++period; | |
// printf("%d\n", lfsr); | |
// } while(lfsr != INIT_STATE); | |
} while(period <= 15); | |
return period; | |
} | |
// Print binary, size is the number of bits | |
void printbits(size_t const size, void const * const ptr) { | |
unsigned char *b = (unsigned char*) ptr; | |
unsigned char byte; | |
int i, j; | |
for(i=size-1;i>=0;i--) { | |
for(j=7;j>=0;j--) { | |
byte = (b[i] >> j) & 1; | |
if(j == 3 || j == 7) { | |
printf(" "); | |
} | |
printf("%u", byte); | |
} | |
} | |
puts(""); | |
} |
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