1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
|
/*
* Copyright © 2012 Keith Packard <keithp@keithp.com>
*
* 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 <ao_fec.h>
#include <stdio.h>
/*
* 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;
}
|
