1 =================================
2 Brief tutorial on CRC computation
3 =================================
4
5 A CRC is a long-division remainder. You add the CRC to the message,
6 and the whole thing (message+CRC) is a multiple of the given
7 CRC polynomial. To check the CRC, you can either check that the
8 CRC matches the recomputed value, *or* you can check that the
9 remainder computed on the message+CRC is 0. This latter approach
10 is used by a lot of hardware implementations, and is why so many
11 protocols put the end-of-frame flag after the CRC.
12
13 It's actually the same long division you learned in school, except that:
14
15 - We're working in binary, so the digits are only 0 and 1, and
16 - When dividing polynomials, there are no carries. Rather than add and
17 subtract, we just xor. Thus, we tend to get a bit sloppy about
18 the difference between adding and subtracting.
19
20 Like all division, the remainder is always smaller than the divisor.
21 To produce a 32-bit CRC, the divisor is actually a 33-bit CRC polynomial.
22 Since it's 33 bits long, bit 32 is always going to be set, so usually the
23 CRC is written in hex with the most significant bit omitted. (If you're
24 familiar with the IEEE 754 floating-point format, it's the same idea.)
25
26 Note that a CRC is computed over a string of *bits*, so you have
27 to decide on the endianness of the bits within each byte. To get
28 the best error-detecting properties, this should correspond to the
29 order they're actually sent. For example, standard RS-232 serial is
30 little-endian; the most significant bit (sometimes used for parity)
31 is sent last. And when appending a CRC word to a message, you should
32 do it in the right order, matching the endianness.
33
34 Just like with ordinary division, you proceed one digit (bit) at a time.
35 Each step of the division you take one more digit (bit) of the dividend
36 and append it to the current remainder. Then you figure out the
37 appropriate multiple of the divisor to subtract to being the remainder
38 back into range. In binary, this is easy - it has to be either 0 or 1,
39 and to make the XOR cancel, it's just a copy of bit 32 of the remainder.
40
41 When computing a CRC, we don't care about the quotient, so we can
42 throw the quotient bit away, but subtract the appropriate multiple of
43 the polynomial from the remainder and we're back to where we started,
44 ready to process the next bit.
45
46 A big-endian CRC written this way would be coded like::
47
48 for (i = 0; i < input_bits; i++) {
49 multiple = remainder & 0x80000000 ? CRCPOLY : 0;
50 remainder = (remainder << 1 | next_input_bit()) ^ multiple;
51 }
52
53 Notice how, to get at bit 32 of the shifted remainder, we look
54 at bit 31 of the remainder *before* shifting it.
55
56 But also notice how the next_input_bit() bits we're shifting into
57 the remainder don't actually affect any decision-making until
58 32 bits later. Thus, the first 32 cycles of this are pretty boring.
59 Also, to add the CRC to a message, we need a 32-bit-long hole for it at
60 the end, so we have to add 32 extra cycles shifting in zeros at the
61 end of every message.
62
63 These details lead to a standard trick: rearrange merging in the
64 next_input_bit() until the moment it's needed. Then the first 32 cycles
65 can be precomputed, and merging in the final 32 zero bits to make room
66 for the CRC can be skipped entirely. This changes the code to::
67
68 for (i = 0; i < input_bits; i++) {
69 remainder ^= next_input_bit() << 31;
70 multiple = (remainder & 0x80000000) ? CRCPOLY : 0;
71 remainder = (remainder << 1) ^ multiple;
72 }
73
74 With this optimization, the little-endian code is particularly simple::
75
76 for (i = 0; i < input_bits; i++) {
77 remainder ^= next_input_bit();
78 multiple = (remainder & 1) ? CRCPOLY : 0;
79 remainder = (remainder >> 1) ^ multiple;
80 }
81
82 The most significant coefficient of the remainder polynomial is stored
83 in the least significant bit of the binary "remainder" variable.
84 The other details of endianness have been hidden in CRCPOLY (which must
85 be bit-reversed) and next_input_bit().
86
87 As long as next_input_bit is returning the bits in a sensible order, we don't
88 *have* to wait until the last possible moment to merge in additional bits.
89 We can do it 8 bits at a time rather than 1 bit at a time::
90
91 for (i = 0; i < input_bytes; i++) {
92 remainder ^= next_input_byte() << 24;
93 for (j = 0; j < 8; j++) {
94 multiple = (remainder & 0x80000000) ? CRCPOLY : 0;
95 remainder = (remainder << 1) ^ multiple;
96 }
97 }
98
99 Or in little-endian::
100
101 for (i = 0; i < input_bytes; i++) {
102 remainder ^= next_input_byte();
103 for (j = 0; j < 8; j++) {
104 multiple = (remainder & 1) ? CRCPOLY : 0;
105 remainder = (remainder >> 1) ^ multiple;
106 }
107 }
108
109 If the input is a multiple of 32 bits, you can even XOR in a 32-bit
110 word at a time and increase the inner loop count to 32.
111
112 You can also mix and match the two loop styles, for example doing the
113 bulk of a message byte-at-a-time and adding bit-at-a-time processing
114 for any fractional bytes at the end.
115
116 To reduce the number of conditional branches, software commonly uses
117 the byte-at-a-time table method, popularized by Dilip V. Sarwate,
118 "Computation of Cyclic Redundancy Checks via Table Look-Up", Comm. ACM
119 v.31 no.8 (August 1998) p. 1008-1013.
120
121 Here, rather than just shifting one bit of the remainder to decide
122 in the correct multiple to subtract, we can shift a byte at a time.
123 This produces a 40-bit (rather than a 33-bit) intermediate remainder,
124 and the correct multiple of the polynomial to subtract is found using
125 a 256-entry lookup table indexed by the high 8 bits.
126
127 (The table entries are simply the CRC-32 of the given one-byte messages.)
128
129 When space is more constrained, smaller tables can be used, e.g. two
130 4-bit shifts followed by a lookup in a 16-entry table.
131
132 It is not practical to process much more than 8 bits at a time using this
133 technique, because tables larger than 256 entries use too much memory and,
134 more importantly, too much of the L1 cache.
135
136 To get higher software performance, a "slicing" technique can be used.
137 See "High Octane CRC Generation with the Intel Slicing-by-8 Algorithm",
138 ftp://download.intel.com/technology/comms/perfnet/download/slicing-by-8.pdf
139
140 This does not change the number of table lookups, but does increase
141 the parallelism. With the classic Sarwate algorithm, each table lookup
142 must be completed before the index of the next can be computed.
143
144 A "slicing by 2" technique would shift the remainder 16 bits at a time,
145 producing a 48-bit intermediate remainder. Rather than doing a single
146 lookup in a 65536-entry table, the two high bytes are looked up in
147 two different 256-entry tables. Each contains the remainder required
148 to cancel out the corresponding byte. The tables are different because the
149 polynomials to cancel are different. One has non-zero coefficients from
150 x^32 to x^39, while the other goes from x^40 to x^47.
151
152 Since modern processors can handle many parallel memory operations, this
153 takes barely longer than a single table look-up and thus performs almost
154 twice as fast as the basic Sarwate algorithm.
155
156 This can be extended to "slicing by 4" using 4 256-entry tables.
157 Each step, 32 bits of data is fetched, XORed with the CRC, and the result
158 broken into bytes and looked up in the tables. Because the 32-bit shift
159 leaves the low-order bits of the intermediate remainder zero, the
160 final CRC is simply the XOR of the 4 table look-ups.
161
162 But this still enforces sequential execution: a second group of table
163 look-ups cannot begin until the previous groups 4 table look-ups have all
164 been completed. Thus, the processor's load/store unit is sometimes idle.
165
166 To make maximum use of the processor, "slicing by 8" performs 8 look-ups
167 in parallel. Each step, the 32-bit CRC is shifted 64 bits and XORed
168 with 64 bits of input data. What is important to note is that 4 of
169 those 8 bytes are simply copies of the input data; they do not depend
170 on the previous CRC at all. Thus, those 4 table look-ups may commence
171 immediately, without waiting for the previous loop iteration.
172
173 By always having 4 loads in flight, a modern superscalar processor can
174 be kept busy and make full use of its L1 cache.
175
176 Two more details about CRC implementation in the real world:
177
178 Normally, appending zero bits to a message which is already a multiple
179 of a polynomial produces a larger multiple of that polynomial. Thus,
180 a basic CRC will not detect appended zero bits (or bytes). To enable
181 a CRC to detect this condition, it's common to invert the CRC before
182 appending it. This makes the remainder of the message+crc come out not
183 as zero, but some fixed non-zero value. (The CRC of the inversion
184 pattern, 0xffffffff.)
185
186 The same problem applies to zero bits prepended to the message, and a
187 similar solution is used. Instead of starting the CRC computation with
188 a remainder of 0, an initial remainder of all ones is used. As long as
189 you start the same way on decoding, it doesn't make a difference.
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