/* * Copyright © 2012 Keith Packard * * This program is free software; you can redistribute it and/or modify * it under the terms of the GNU General Public License as published by * the Free Software Foundation; version 2 of the License. * * This program is distributed in the hope that it will be useful, but * WITHOUT ANY WARRANTY; without even the implied warranty of * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU * General Public License for more details. * * You should have received a copy of the GNU General Public License along * with this program; if not, write to the Free Software Foundation, Inc., * 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA. */ #include #include /* * byte order repeats through 3 2 1 0 * * bit-pair order repeats through * * 1/0 3/2 5/4 7/6 * * So, the over all order is: * * 3,1/0 2,1/0 1,1/0 0,1/0 * 3,3/2 2,3/2 1,3/2 0,3/2 * 3,5/4 2,5/4 1,5/4 0,5/4 * 3,7/6 2,7/6 1,7/6 0,7/6 * * The raw bit order is thus * * 1e/1f 16/17 0e/0f 06/07 * 1c/1d 14/15 0c/0d 04/05 * 1a/1b 12/13 0a/0b 02/03 * 18/19 10/11 08/09 00/01 */ static inline uint16_t ao_interleave_index(uint16_t i) { uint8_t l = i & 0x1e; uint16_t h = i & ~0x1e; uint8_t o = 0x1e ^ (((l >> 2) & 0x6) | ((l << 2) & 0x18)); return h | o; } struct ao_soft_sym { uint8_t a, b; }; #define NUM_STATE 8 #define NUM_HIST 8 #define MOD_HIST(b) ((b) & 7) #define V_0 0xc0 #define V_1 0x40 static const struct ao_soft_sym ao_fec_decode_table[NUM_STATE][2] = { /* next 0 1 state */ { { V_0, V_0 }, { V_1, V_1 } } , /* 000 */ { { V_0, V_1 }, { V_1, V_0 } }, /* 001 */ { { V_1, V_1 }, { V_0, V_0 } }, /* 010 */ { { V_1, V_0 }, { V_0, V_1 } }, /* 011 */ { { V_1, V_1 }, { V_0, V_0 } }, /* 100 */ { { V_1, V_0 }, { V_0, V_1 } }, /* 101 */ { { V_0, V_0 }, { V_1, V_1 } }, /* 110 */ { { V_0, V_1 }, { V_1, V_0 } } /* 111 */ }; static inline uint8_t ao_next_state(uint8_t state, uint8_t bit) { return ((state << 1) | bit) & 0x7; } static inline uint16_t ao_abs(int16_t x) { return x < 0 ? -x : x; } static inline uint16_t ao_cost(struct ao_soft_sym a, struct ao_soft_sym b) { return ao_abs(a.a - b.a) + ao_abs(a.b - b.b); } /* * 'in' is 8-bits per symbol soft decision data * 'len' is input byte length. 'out' must be * 'len'/16 bytes long */ uint8_t ao_fec_decode(uint8_t *in, uint16_t len, uint8_t *out, uint8_t out_len, uint16_t (*callback)()) { static uint16_t cost[2][NUM_STATE]; /* path cost */ static uint16_t bits[2][NUM_STATE]; /* save bits to quickly output them */ uint16_t i; /* input byte index */ uint16_t b; /* encoded symbol index (bytes/2) */ uint16_t o; /* output bit index */ uint8_t p; /* previous cost/bits index */ uint8_t n; /* next cost/bits index */ uint8_t state; /* state index */ uint8_t bit; /* original encoded bit index */ const uint8_t *whiten = ao_fec_whiten_table; uint16_t interleave; /* input byte array index */ struct ao_soft_sym s; /* input symbol pair */ uint16_t avail; p = 0; for (state = 0; state < NUM_STATE; state++) { cost[0][state] = 0xffff; bits[0][state] = 0; } cost[0][0] = 0; if (callback) avail = 0; else avail = len; o = 0; for (i = 0; i < len; i += 2) { b = i/2; n = p ^ 1; if (!avail) { avail = callback(); if (!avail) break; } /* Fetch one pair of input bytes, de-interleaving * the input. */ interleave = ao_interleave_index(i); s.a = in[interleave]; s.b = in[interleave+1]; /* Reset next costs to 'impossibly high' values so that * the first path through this state is cheaper than this */ for (state = 0; state < NUM_STATE; state++) cost[n][state] = 0xffff; /* Compute path costs and accumulate output bit path * for each state and encoded bit value */ for (state = 0; state < NUM_STATE; state++) { for (bit = 0; bit < 2; bit++) { int bit_cost = cost[p][state] + ao_cost(s, ao_fec_decode_table[state][bit]); uint8_t bit_state = ao_next_state(state, bit); /* Only track the minimal cost to reach * this state; the best path can never * go through the higher cost paths as * total path cost is cumulative */ if (bit_cost < cost[n][bit_state]) { cost[n][bit_state] = bit_cost; bits[n][bit_state] = (bits[p][state] << 1) | (state & 1); } } } #if 0 printf ("bit %3d symbol %2x %2x:", i/2, s.a, s.b); for (state = 0; state < NUM_STATE; state++) { printf (" %5d(%04x)", cost[n][state], bits[n][state]); } printf ("\n"); #endif p = n; /* A loop is needed to handle the last output byte. It * won't have any bits of future data to perform full * error correction, but we might as well give the * best possible answer anyways. */ while ((b - o) >= (8 + NUM_HIST) || (i + 2 >= len && b > o)) { /* Compute number of bits to the end of the * last full byte of data. This is generally * NUM_HIST, unless we've reached * the end of the input, in which case * it will be seven. */ int8_t dist = b - (o + 8); /* distance to last ready-for-writing bit */ uint16_t min_cost; /* lowest cost */ uint8_t min_state; /* lowest cost state */ /* Find the best fit at the current point * of the decode. */ min_cost = cost[p][0]; min_state = 0; for (state = 1; state < NUM_STATE; state++) { if (cost[p][state] < min_cost) { min_cost = cost[p][state]; min_state = state; } } /* The very last byte of data has the very last bit * of data left in the state value; just smash the * bits value in place and reset the 'dist' from * -1 to 0 so that the full byte is read out */ if (dist < 0) { bits[p][min_state] = (bits[p][min_state] << 1) | (min_state & 1); dist = 0; } #if 0 printf ("\tbit %3d min_cost %5d old bit %3d old_state %x bits %02x whiten %0x\n", i/2, min_cost, o + 8, min_state, (bits[p][min_state] >> dist) & 0xff, *whiten); #endif if (out_len) { *out++ = (bits[p][min_state] >> dist) ^ *whiten++; --out_len; } o += 8; } } return len/16; }