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Experiment for finding a short BLC encoding for binary data
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#include <stdio.h> | |
#include <stdlib.h> | |
#include <math.h> | |
static void to_digits(unsigned char *bytes, int n_bytes, unsigned int *digits, | |
int n_digits, unsigned int base) | |
{ | |
for (int i_byte = 0; i_byte < n_bytes; i_byte++) { | |
unsigned char byte = bytes[i_byte]; | |
for (int mask = 0x80; mask; mask >>= 1) { | |
int bit = (byte & mask) != 0; | |
digits[0] = digits[0] * 2 + bit; | |
for (int i_digit = 1; i_digit < n_digits; i_digit++) { | |
digits[i_digit] *= 2; | |
if (digits[i_digit - 1] > (base - 1)) { | |
digits[i_digit - 1] -= base; | |
digits[i_digit]++; | |
} | |
} | |
} | |
} | |
} | |
// returns the maximum digits needed for encoding n bytes in a base | |
static size_t digits_needed(int n_bytes, int base) | |
{ | |
return (size_t)ceil(log(pow(256, n_bytes)) / log(base)); | |
} | |
static void write_bit(char val, FILE *file, char *byte, int *bit) | |
{ | |
if (*bit > 7) { // flush byte | |
fwrite(byte, 1, 1, file); | |
*byte = 0; | |
*bit = 0; | |
} | |
if (val) | |
*byte |= 1UL << (7 - *bit); | |
(*bit)++; | |
} | |
int main(int argc, char *argv[]) | |
{ | |
if (argc != 2) { | |
fprintf(stderr, "Usage: %s <file>\n", argv[0]); | |
return 1; | |
} | |
FILE *in = fopen(argv[1], "rb"); | |
if (!in) { | |
fprintf(stderr, "Could not open file\n"); | |
return 1; | |
} | |
fseek(in, 0, SEEK_END); | |
size_t n_bytes = ftell(in); | |
fseek(in, 0, SEEK_SET); | |
fprintf(stderr, "n_bytes: %zu\n", n_bytes); | |
unsigned char *bytes = calloc(n_bytes, sizeof(*bytes)); | |
fread(bytes, 1, n_bytes, in); | |
fclose(in); | |
int base = 8; | |
size_t n_digits = digits_needed(n_bytes, base); | |
fprintf(stderr, "n_digits: %zu\n", n_digits); | |
unsigned int *digits = calloc(n_digits, sizeof(*digits)); | |
to_digits(bytes, n_bytes, digits, n_digits, base); | |
for (; n_digits > 0 && !digits[n_digits - 1]; n_digits--) | |
; | |
FILE *out = stdout; | |
char byte = 0; | |
int bit = 0; | |
for (size_t i = 0; i < n_digits; i++) { | |
write_bit(0, out, &byte, &bit); | |
for (unsigned int j = 0; j < digits[i] + 1; j++) { | |
write_bit(1, out, &byte, &bit); | |
} | |
} | |
if (bit) // flush final | |
fwrite(&byte, 1, 1, out); | |
return 0; | |
} |
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#!/bin/env python3 | |
from random import randint | |
def to_base(n, b): | |
if n == 0: | |
return [0] | |
digits = [] | |
while n: | |
digits.append(int(n % b)) | |
n //= b | |
return digits[::-1] | |
def random_with_n_digits(n): | |
range_start = 10 ** (n - 1) | |
range_end = (10**n) - 1 | |
return randint(range_start, range_end) | |
def gen_for_increasing(): | |
N = 10**6 | |
B = range(2, 10) | |
for b in B: | |
avg = 0 | |
for n in range(1, N + 1): | |
out = 2 * b # abstractions (00) | |
base = to_base(n, b) | |
for e in base[:-1]: | |
out += 2 # application (01) | |
out += e + 2 | |
out += base[-1] + 2 | |
avg += out | |
print(b, avg / N) | |
def gen_for_fixed_big(): | |
N = 1000000 | |
I = 5 | |
B = range(2, 10) | |
for b in B: | |
avg = 0 | |
for i in range(I): | |
n = random_with_n_digits(N) | |
out = 2 * b # abstractions (00) | |
base = to_base(n, b) | |
for e in base[:-1]: | |
out += 2 # application (01) | |
out += e + 2 | |
out += base[-1] + 2 | |
avg += out | |
print(b, float(avg) / I) |
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This program is part of my article here.