mirror of
https://github.com/henrydcase/pqc.git
synced 2024-11-24 00:11:27 +00:00
238 lines
8.2 KiB
C
238 lines
8.2 KiB
C
#include "parameters.h"
|
|
#include "reed_muller.h"
|
|
#include <stdint.h>
|
|
#include <string.h>
|
|
/**
|
|
* @file reed_muller.c
|
|
* Constant time implementation of Reed-Muller code RM(1,7)
|
|
*/
|
|
|
|
|
|
|
|
// number of repeated code words
|
|
#define MULTIPLICITY CEIL_DIVIDE(PARAM_N2, 128)
|
|
|
|
// copy bit 0 into all bits of a 32 bit value
|
|
#define BIT0MASK(x) (-((x) & 1))
|
|
|
|
|
|
static void encode(uint8_t *word, uint8_t message);
|
|
static void hadamard(uint16_t src[128], uint16_t dst[128]);
|
|
static void expand_and_sum(uint16_t dest[128], const uint8_t src[16 * MULTIPLICITY]);
|
|
static uint8_t find_peaks(const uint16_t transform[128]);
|
|
|
|
|
|
|
|
/**
|
|
* @brief Encode a single byte into a single codeword using RM(1,7)
|
|
*
|
|
* Encoding matrix of this code:
|
|
* bit pattern (note that bits are numbered big endian)
|
|
* 0 aaaaaaaa aaaaaaaa aaaaaaaa aaaaaaaa
|
|
* 1 cccccccc cccccccc cccccccc cccccccc
|
|
* 2 f0f0f0f0 f0f0f0f0 f0f0f0f0 f0f0f0f0
|
|
* 3 ff00ff00 ff00ff00 ff00ff00 ff00ff00
|
|
* 4 ffff0000 ffff0000 ffff0000 ffff0000
|
|
* 5 ffffffff 00000000 ffffffff 00000000
|
|
* 6 ffffffff ffffffff 00000000 00000000
|
|
* 7 ffffffff ffffffff ffffffff ffffffff
|
|
*
|
|
* @param[out] word An RM(1,7) codeword
|
|
* @param[in] message A message
|
|
*/
|
|
static void encode(uint8_t *word, uint8_t message) {
|
|
uint32_t e;
|
|
// bit 7 flips all the bits, do that first to save work
|
|
e = BIT0MASK(message >> 7);
|
|
// bits 0, 1, 2, 3, 4 are the same for all four longs
|
|
// (Warning: in the bit matrix above, low bits are at the left!)
|
|
e ^= BIT0MASK(message >> 0) & 0xaaaaaaaa;
|
|
e ^= BIT0MASK(message >> 1) & 0xcccccccc;
|
|
e ^= BIT0MASK(message >> 2) & 0xf0f0f0f0;
|
|
e ^= BIT0MASK(message >> 3) & 0xff00ff00;
|
|
e ^= BIT0MASK(message >> 4) & 0xffff0000;
|
|
// we can store this in the first quarter
|
|
word[0 + 0] = (e >> 0x00) & 0xff;
|
|
word[0 + 1] = (e >> 0x08) & 0xff;
|
|
word[0 + 2] = (e >> 0x10) & 0xff;
|
|
word[0 + 3] = (e >> 0x18) & 0xff;
|
|
// bit 5 flips entries 1 and 3; bit 6 flips 2 and 3
|
|
e ^= BIT0MASK(message >> 5);
|
|
word[4 + 0] = (e >> 0x00) & 0xff;
|
|
word[4 + 1] = (e >> 0x08) & 0xff;
|
|
word[4 + 2] = (e >> 0x10) & 0xff;
|
|
word[4 + 3] = (e >> 0x18) & 0xff;
|
|
e ^= BIT0MASK(message >> 6);
|
|
word[12 + 0] = (e >> 0x00) & 0xff;
|
|
word[12 + 1] = (e >> 0x08) & 0xff;
|
|
word[12 + 2] = (e >> 0x10) & 0xff;
|
|
word[12 + 3] = (e >> 0x18) & 0xff;
|
|
e ^= BIT0MASK(message >> 5);
|
|
word[8 + 0] = (e >> 0x00) & 0xff;
|
|
word[8 + 1] = (e >> 0x08) & 0xff;
|
|
word[8 + 2] = (e >> 0x10) & 0xff;
|
|
word[8 + 3] = (e >> 0x18) & 0xff;
|
|
}
|
|
|
|
|
|
|
|
/**
|
|
* @brief Hadamard transform
|
|
*
|
|
* Perform hadamard transform of src and store result in dst
|
|
* src is overwritten: it is also used as intermediate buffer
|
|
* Method is best explained if we use H(3) instead of H(7):
|
|
*
|
|
* The routine multiplies by the matrix H(3):
|
|
* [1 1 1 1 1 1 1 1]
|
|
* [1 -1 1 -1 1 -1 1 -1]
|
|
* [1 1 -1 -1 1 1 -1 -1]
|
|
* [a b c d e f g h] * [1 -1 -1 1 1 -1 -1 1] = result of routine
|
|
* [1 1 1 1 -1 -1 -1 -1]
|
|
* [1 -1 1 -1 -1 1 -1 1]
|
|
* [1 1 -1 -1 -1 -1 1 1]
|
|
* [1 -1 -1 1 -1 1 1 -1]
|
|
* You can do this in three passes, where each pass does this:
|
|
* set lower half of buffer to pairwise sums,
|
|
* and upper half to differences
|
|
* index 0 1 2 3 4 5 6 7
|
|
* input: a, b, c, d, e, f, g, h
|
|
* pass 1: a+b, c+d, e+f, g+h, a-b, c-d, e-f, g-h
|
|
* pass 2: a+b+c+d, e+f+g+h, a-b+c-d, e-f+g-h, a+b-c-d, e+f-g-h, a-b-c+d, e-f-g+h
|
|
* pass 3: a+b+c+d+e+f+g+h a+b-c-d+e+f-g-h a+b+c+d-e-f-g-h a+b-c-d-e+-f+g+h
|
|
* a-b+c-d+e-f+g-h a-b-c+d+e-f-g+h a-b+c-d-e+f-g+h a-b-c+d-e+f+g-h
|
|
* This order of computation is chosen because it vectorises well.
|
|
* Likewise, this routine multiplies by H(7) in seven passes.
|
|
*
|
|
* @param[out] src Structure that contain the expanded codeword
|
|
* @param[out] dst Structure that contain the expanded codeword
|
|
*/
|
|
static void hadamard(uint16_t src[128], uint16_t dst[128]) {
|
|
// the passes move data:
|
|
// src -> dst -> src -> dst -> src -> dst -> src -> dst
|
|
// using p1 and p2 alternately
|
|
uint16_t *p1 = src;
|
|
uint16_t *p2 = dst;
|
|
uint16_t *p3;
|
|
for (uint32_t pass = 0; pass < 7; pass++) {
|
|
for (uint32_t i = 0; i < 64; i++) {
|
|
p2[i] = p1[2 * i] + p1[2 * i + 1];
|
|
p2[i + 64] = p1[2 * i] - p1[2 * i + 1];
|
|
}
|
|
// swap p1, p2 for next round
|
|
p3 = p1;
|
|
p1 = p2;
|
|
p2 = p3;
|
|
}
|
|
}
|
|
|
|
|
|
|
|
/**
|
|
* @brief Add multiple codewords into expanded codeword
|
|
*
|
|
* Accesses memory in order
|
|
* Note: this does not write the codewords as -1 or +1 as the green machine does
|
|
* instead, just 0 and 1 is used.
|
|
* The resulting hadamard transform has:
|
|
* all values are halved
|
|
* the first entry is 64 too high
|
|
*
|
|
* @param[out] dest Structure that contain the expanded codeword
|
|
* @param[in] src Structure that contain the codeword
|
|
*/
|
|
static void expand_and_sum(uint16_t dest[128], const uint8_t src[16 * MULTIPLICITY]) {
|
|
size_t part, bit, copy;
|
|
// start with the first copy
|
|
for (part = 0; part < 16; part++) {
|
|
for (bit = 0; bit < 8; bit++) {
|
|
dest[part * 8 + bit] = (uint16_t) ((src[part] >> bit) & 1);
|
|
}
|
|
}
|
|
// sum the rest of the copies
|
|
for (copy = 1; copy < MULTIPLICITY; copy++) {
|
|
for (part = 0; part < 16; part++) {
|
|
for (bit = 0; bit < 8; bit++) {
|
|
dest[part * 8 + bit] += (uint16_t) ((src[16 * copy + part] >> bit) & 1);
|
|
}
|
|
}
|
|
}
|
|
}
|
|
|
|
|
|
|
|
/**
|
|
* @brief Finding the location of the highest value
|
|
*
|
|
* This is the final step of the green machine: find the location of the highest value,
|
|
* and add 128 if the peak is positive
|
|
* if there are two identical peaks, the peak with smallest value
|
|
* in the lowest 7 bits it taken
|
|
* @param[in] transform Structure that contain the expanded codeword
|
|
*/
|
|
static uint8_t find_peaks(const uint16_t transform[128]) {
|
|
uint16_t peak_abs = 0;
|
|
uint16_t peak = 0;
|
|
uint16_t pos = 0;
|
|
uint16_t t, abs, mask;
|
|
for (uint16_t i = 0; i < 128; i++) {
|
|
t = transform[i];
|
|
abs = t ^ ((-(t >> 15)) & (t ^ -t)); // t = abs(t)
|
|
mask = -(((uint16_t)(peak_abs - abs)) >> 15);
|
|
peak ^= mask & (peak ^ t);
|
|
pos ^= mask & (pos ^ i);
|
|
peak_abs ^= mask & (peak_abs ^ abs);
|
|
}
|
|
pos |= 128 & ((peak >> 15) - 1);
|
|
return (uint8_t) pos;
|
|
}
|
|
|
|
|
|
|
|
|
|
/**
|
|
* @brief Encodes the received word
|
|
*
|
|
* The message consists of N1 bytes each byte is encoded into PARAM_N2 bits,
|
|
* or MULTIPLICITY repeats of 128 bits
|
|
*
|
|
* @param[out] cdw Array of size VEC_N1N2_SIZE_64 receiving the encoded message
|
|
* @param[in] msg Array of size VEC_N1_SIZE_64 storing the message
|
|
*/
|
|
void PQCLEAN_HQCRMRS128_CLEAN_reed_muller_encode(uint8_t *cdw, const uint8_t *msg) {
|
|
for (size_t i = 0; i < VEC_N1_SIZE_BYTES; i++) {
|
|
// encode first word
|
|
encode(&cdw[16 * i * MULTIPLICITY], msg[i]);
|
|
// copy to other identical codewords
|
|
for (size_t copy = 1; copy < MULTIPLICITY; copy++) {
|
|
memcpy(&cdw[16 * i * MULTIPLICITY + 16 * copy], &cdw[16 * i * MULTIPLICITY], 16);
|
|
}
|
|
}
|
|
}
|
|
|
|
|
|
|
|
/**
|
|
* @brief Decodes the received word
|
|
*
|
|
* Decoding uses fast hadamard transform, for a more complete picture on Reed-Muller decoding, see MacWilliams, Florence Jessie, and Neil James Alexander Sloane.
|
|
* The theory of error-correcting codes codes @cite macwilliams1977theory
|
|
*
|
|
* @param[out] msg Array of size VEC_N1_SIZE_64 receiving the decoded message
|
|
* @param[in] cdw Array of size VEC_N1N2_SIZE_64 storing the received word
|
|
*/
|
|
void PQCLEAN_HQCRMRS128_CLEAN_reed_muller_decode(uint8_t *msg, const uint8_t *cdw) {
|
|
uint16_t expanded[128];
|
|
uint16_t transform[128];
|
|
for (size_t i = 0; i < VEC_N1_SIZE_BYTES; i++) {
|
|
// collect the codewords
|
|
expand_and_sum(expanded, &cdw[16 * i * MULTIPLICITY]);
|
|
// apply hadamard transform
|
|
hadamard(expanded, transform);
|
|
// fix the first entry to get the half Hadamard transform
|
|
transform[0] -= 64 * MULTIPLICITY;
|
|
// finish the decoding
|
|
msg[i] = find_peaks(transform);
|
|
}
|
|
}
|