/* * 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; either version 2 of the License, or * (at your option) any later version. * * 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 #ifdef TELEMEGA #include #endif #if AO_PROFILE #include uint32_t ao_fec_decode_start, ao_fec_decode_end; #endif /* * 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 const uint8_t ao_interleave_order[] = { 0x1e, 0x16, 0x0e, 0x06, 0x1c, 0x14, 0x0c, 0x04, 0x1a, 0x12, 0x0a, 0x02, 0x18, 0x10, 0x08, 0x00 }; static inline uint16_t ao_interleave_index(uint16_t i) { return (i & ~0x1e) | ao_interleave_order[(i & 0x1e) >> 1]; } #define NUM_STATE 8 #define NUM_HIST 24 typedef uint32_t bits_t; #define V_0 0xff #define V_1 0x00 /* * These are just the 'zero' states; the 'one' states mirror them */ static const uint8_t ao_fec_decode_table[NUM_STATE*2] = { V_0, V_0, /* 000 */ V_0, V_1, /* 001 */ V_1, V_1, /* 010 */ V_1, V_0, /* 011 */ V_1, V_1, /* 100 */ V_1, V_0, /* 101 */ V_0, V_0, /* 110 */ V_0, V_1 /* 111 */ }; static inline uint8_t ao_next_state(uint8_t state, uint8_t bit) { return ((state << 1) | bit) & 0x7; } /* * '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(const uint8_t *in, uint16_t len, uint8_t *out, uint8_t out_len, uint16_t (*callback)(void)) { static uint32_t cost[2][NUM_STATE]; /* path cost */ static bits_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 */ const uint8_t *whiten = ao_fec_whiten_table; uint16_t interleave; /* input byte array index */ uint8_t s0, s1; uint16_t avail; uint16_t crc = AO_FEC_CRC_INIT; #if AO_PROFILE uint32_t start_tick; #endif p = 0; for (state = 0; state < NUM_STATE; state++) { cost[0][state] = 0x7fffffff; bits[0][state] = 0; } cost[0][0] = 0; if (callback) avail = 0; else avail = len; #if AO_PROFILE if (!avail) { avail = callback(); if (!avail) return 0; } start_tick = ao_profile_tick(); #endif o = 0; for (i = 0; i < len; i += 2) { b = i/2; n = p ^ 1; if (!avail) { avail = callback(); if (!avail) return 0; } /* Fetch one pair of input bytes, de-interleaving * the input. */ interleave = ao_interleave_index(i); s0 = in[interleave]; s1 = in[interleave+1]; avail -= 2; /* Compute path costs and accumulate output bit path * for each state and encoded bit value. Unrolling * this loop is worth about > 30% performance boost. * Decoding 76-byte remote access packets is reduced * from 14.700ms to 9.3ms. Redoing the loop to * directly compare the two pasts for each future state * reduces this down to 5.7ms */ /* Ok, of course this is tricky, it's optimized. * * First, it's important to realize that we have 8 * states representing the combinations of the three * most recent bits from the encoder. Flipping any * of these three bits flips both output bits. * * 'state<<1' represents the target state for a new * bit value of 0. '(state<<1)+1' represents the * target state for a new bit value of 1. * * 'state' is the previous state with an oldest bit * value of 0. 'state + 4' is the previous state with * an oldest bit value of 1. These two states will * either lead to 'state<<1' or '(state<<1)+1', depending * on whether the next encoded bit was a zero or a one. * * m0 and m1 are the cost of coming to 'state<<1' from * one of the two possible previous states 'state' and * 'state + 4'. * * Because we know the expected values of each * received bit are flipped between these two previous * states: * * bitcost(state+4) = 510 - bitcost(state) * * With those two total costs in hand, we then pick * the lower as the cost of the 'state<<1', and compute * the path of bits leading to that state. * * Then, do the same for '(state<<1) + 1'. This time, * instead of computing the m0 and m1 values from * scratch, because the only difference is that we're * expecting a one bit instead of a zero bit, we just * flip the bitcost values around to match the * expected transmitted bits with some tricky * arithmetic which is equivalent to: * * m0 = cost[p][state] + (510 - bitcost); * m1 = cost[p][state+4] + bitcost * * Then, the lowest cost and bit trace of the new state * is saved. */ #define DO_STATE(state) { \ uint32_t bitcost; \ \ uint32_t m0; \ uint32_t m1; \ uint32_t bit; \ \ bitcost = ((uint32_t) (s0 ^ ao_fec_decode_table[(state<<1)]) + \ (uint32_t) (s1 ^ ao_fec_decode_table[(state<<1)|1])); \ \ m0 = cost[p][state] + bitcost; \ m1 = cost[p][state+4] + (510 - bitcost); \ bit = m0 > m1; \ cost[n][state<<1] = bit ? m1 : m0; \ bits[n][state<<1] = (bits[p][state + (bit<<2)] << 1) | (state&1); \ \ m0 -= (bitcost+bitcost-510); \ m1 += (bitcost+bitcost-510); \ bit = m0 > m1; \ cost[n][(state<<1)+1] = bit ? m1 : m0; \ bits[n][(state<<1)+1] = (bits[p][state + (bit<<2)] << 1) | (state&1); \ } DO_STATE(0); DO_STATE(1); DO_STATE(2); DO_STATE(3); #if 0 printf ("bit %3d symbol %2x %2x:", i/2, s0, s1); for (state = 0; state < NUM_STATE; state++) { printf (" %8u(%08x)", 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 */ uint32_t min_cost; /* lowest cost */ uint8_t min_state; /* lowest cost state */ uint8_t byte; /* 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 byte = (bits[p][min_state] >> dist) ^ *whiten++; *out++ = byte; if (out_len > 2) crc = ao_fec_crc_byte(byte, crc); if (!--out_len) { if ((out[-2] == (uint8_t) (crc >> 8)) && out[-1] == (uint8_t) crc) out[-1] = AO_FEC_DECODE_CRC_OK; else out[-1] = 0; out[-2] = 0; goto done; } o += 8; } } done: #if AO_PROFILE ao_fec_decode_start = start_tick; ao_fec_decode_end = ao_profile_tick(); #endif return 1; }