altos: Rename 'core' to 'kernel'

core remains a bad name to use -- dirvish skips files (and
directories, it seems) with that name.

Signed-off-by: Keith Packard <keithp@keithp.com>

1 files changed, 318 insertions, 0 deletions

diff --git a/src/kernel/ao_fec_rx.c b/src/kernel/ao_fec_rx.c
new file mode 100644
index 00000000..c4f5559a
--- /dev/null
+++ b/src/kernel/ao_fec_rx.c
@@ -0,0 +1,318 @@
+/*
+ * 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>
+
+#ifdef TELEMEGA
+#include <ao.h>
+#endif
+
+#if AO_PROFILE
+#include <ao_profile.h>
+
+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;
+}