Message-ID: <015501cc56b5$30816f00$91844d00$@systemfabricworks.com>
Date:	Tue, 9 Aug 2011 11:55:47 -0500
From:	"Bob Pearson" <rpearson@...temfabricworks.com>
To:	"'George Spelvin'" <linux@...izon.com>,
	<akpm@...ux-foundation.org>, <fzago@...temfabricworks.com>,
	<joakim.tjernlund@...nsmode.se>, <linux-kernel@...r.kernel.org>,
	<rdunlap@...otime.net>
Subject: RE: [patch v4 1/7] crc32: move-to-documentation.diff



> -----Original Message-----
> From: George Spelvin [mailto:linux@...izon.com]
> Sent: Tuesday, August 09, 2011 6:45 AM
> To: akpm@...ux-foundation.org; fzago@...temfabricworks.com;
> joakim.tjernlund@...nsmode.se; linux-kernel@...r.kernel.org;
> linux@...izon.com; rdunlap@...otime.net;
> rpearson@...temfabricworks.com
> Subject: Re: [patch v4 1/7] crc32: move-to-documentation.diff
> 
> Here's a hopefully-improved Documentation file, which explains the slicing
> algorithm as well.  As long as you have a big diff, it seems worth
tweaking.
> 
> You also might want to leave a pointer in lib/crc32.c to the relocated
docs.
> 
> (I'm just inclding the whole changed file because I assume it's easier
> to review that way; do you prefer a diff?)
> 
> === Cut here ===
> A brief CRC tutorial.
> 
> A CRC is a long-division remainder.  You add the CRC to the message,
> and the whole thing (message+CRC) is a multiple of the given
> CRC polynomial.  To check the CRC, you can either check that the
> CRC matches the recomputed value, *or* you can check that the
> remainder computed on the message+CRC is 0.  This latter approach
> is used by a lot of hardware implementations, and is why so many
> protocols put the end-of-frame flag after the CRC.
> 
> It's actually the same long division you learned in school, except that
> - We're working in binary, so the digits are only 0 and 1, and
> - When dividing polynomials, there are no carries.  Rather than add and
>   subtract, we just xor.  Thus, we tend to get a bit sloppy about
>   the difference between adding and subtracting.
> 
> Like all division, the remainder is always smaller than the divisor.
> To produce a 32-bit CRC, the divisor is actually a 33-bit CRC polynomial.
> Since it's 33 bits long, bit 32 is always going to be set, so usually the
> CRC is written in hex with the most significant bit omitted.  (If you're
> familiar with the IEEE 754 floating-point format, it's the same idea.)
> 
> Note that a CRC is computed over a string of *bits*, so you have
> to decide on the endianness of the bits within each byte.  To get
> the best error-detecting properties, this should correspond to the
> order they're actually sent.  For example, standard RS-232 serial is
> little-endian; the most significant bit (sometimes used for parity)
> is sent last.  And when appending a CRC word to a message, you should
> do it in the right order, matching the endianness.
> 
> 
> Just like with ordinary division, you proceed one digit (bit) at a time.
> Each step of the division, division, you take one more digit (bit) of the
> dividend and append it to the current remainder.  Then you figure out the
> appropriate multiple of the divisor to subtract to being the remainder
> back into range.  In binary, this is easy - it has to be either 0 or 1,
> and to make the XOR cancel, it's just a copy of bit 32 of the remainder.
> 
> When computing a CRC, we don't care about the quotient, so we can
> throw the quotient bit away, but subtract the appropriate multiple of
> the polynomial from the remainder and we're back to where we started,
> ready to process the next bit.
> 
> A big-endian CRC written this way would be coded like:
> for (i = 0; i < input_bits; i++) {
> 	multiple = remainder & 0x80000000 ? CRCPOLY : 0;
> 	remainder = (remainder << 1 | next_input_bit()) ^ multiple;
> }
> 
> Notice how, to get at bit 32 of the shifted remainder, we look
> at bit 31 of the remainder *before* shifting it.
> 
> But also notice how the next_input_bit() bits we're shifting into
> the remainder don't actually affect any decision-making until
> 32 bits later.  Thus, the first 32 cycles of this are pretty boring.
> Also, to add the CRC to a message, we need a 32-bit-long hole for it at
> the end, so we have to add 32 extra cycles shifting in zeros at the
> end of every message,
> 
> These details lead to a standard trick: rearrange merging in the
> next_input_bit() until the moment it's needed.  Then the first 32 cycles
> can be precomputed, and merging in the final 32 zero bits to make room
> for the CRC can be skipped entirely.  This changes the code to:
> 
> for (i = 0; i < input_bits; i++) {
> 	remainder ^= next_input_bit() << 31;
> 	multiple = (remainder & 0x80000000) ? CRCPOLY : 0;
> 	remainder = (remainder << 1) ^ multiple;
> }
> 
> With this optimization, the little-endian code is particularly simple:
> for (i = 0; i < input_bits; i++) {
> 	remainder ^= next_input_bit();
> 	multiple = (remainder & 1) ? CRCPOLY : 0;
> 	remainder = (remainder >> 1) ^ multiple;
> }
> 
> The most significant coefficient of the remainder polynomial is stored
> in the least significant bit of the binary "remainder" variable.
> The other details of endianness have been hidden in CRCPOLY (which must
> be bit-reversed) and next_input_bit().
> 
> As long as next_input_bit is returning the bits in a sensible order, we
don't
> *have* to wait until the last possible moment to merge in additional bits.
> We can do it 8 bits at a time rather than 1 bit at a time:
> for (i = 0; i < input_bytes; i++) {
> 	remainder ^= next_input_byte() << 24;
> 	for (j = 0; j < 8; j++) {
> 		multiple = (remainder & 0x80000000) ? CRCPOLY : 0;
> 		remainder = (remainder << 1) ^ multiple;
> 	}
> }
> Or in little-endian:
> for (i = 0; i < input_bytes; i++) {
> 	remainder ^= next_input_byte();
> 	for (j = 0; j < 8; j++) {
> 		multiple = (remainder & 1) ? CRCPOLY : 0;
> 		remainder = (remainder << 1) ^ multiple;
> 	}
> }
> 
> If the input is a multiple of 32 bits, you can even XOR in a 32-bit
> word at a time and increase the inner loop count to 32.
> 
> You can also mix and match the two loop styles, for example doing the
> bulk of a message byte-at-a-time and adding bit-at-a-time processing
> for any fractional bytes at the end.
> 
> 
> To reduce the number of conditional branches, software commonly uses
> the byte-at-a-time table method, popularized by Dilip V. Sarwate,
> "Computation of Cyclic Redundancy Checks via Table Look-Up", Comm. ACM
> v.31 no.8 (August 1998) p. 1008-1013.
> 
> Here, rather than just shifting one bit of the remainder to decide
> in the correct multiple to subtract, we can shift a byte at a time.
> This produces a 40-bit (rather than a 33-bit) intermediate remainder,
> and the correct multiple of the polynomial to subtract is found using
> a 256-entry lookup table indexed by the high 8 bits.
> 
> (The table entries are simply the CRC-32 of the given one-byte messages.)
> 
> When space is more constrained, smaller tables can be used, e.g. two
> 4-bit shifts followed by a lookup in a 16-entry table.
> 
> 
> It is not practical to process much more than 8 bits at a time using this
> technique, because tables larger than 256 entries use too much memory and,
> more importantly, too much of the L1 cache.
> 
> To get higher software performance, a "slicing" technique can be used.
> See "High Octane CRC Generation with the Intel Slicing-by-8 Algorithm",
> ftp://download.intel.com/technology/comms/perfnet/download/slicing-by-
> 8.pdf
> 
> This does not change the number of table lookups, but does increase
> the parallelism.  With the classic Sarwate algorithm, each table lookup
> must be completed before the index of the next can be computed.
> 
> A "slicing by 2" technique would shift the remainder 16 bits at a time,
> producing a 48-bit intermediate remainder.  Rather than doing a single
> lookup in a 65536-entry table, the two high bytes are looked up in
> two different 256-entry tables.  Each contains the remainder required
> to cancel out the corresponding byte.  The tables are different because
the
> polynomials to cancel are different.  One has non-zero coefficients from
> x^32 to x^39, while the other goes from x^40 to x^47.
> 
> Since modern processors can handle many parallel memory operations, this
> takes barely longer than a single table look-up and thus performs almost
> twice as fast as the basic Sarwate algorithm.
> 
> This can be extended to "slicing by 4" using 4 256-entry tables.
> Each step, 32 bits of data is fetched, XORed with the CRC, and the result
> broken into bytes and looked up in the tables.  Because the 32-bit shift
> leaves the low-order bits of the intermediate remainder zero, the
> final CRC is simply the XOR of the 4 table look-ups.
> 
> But this still enforces sequential execution: a second group of table
> look-ups cannot begin until the previous groups 4 table look-ups have all
> been completed.  Thus, the processor's load/store unit is sometimes idle.
> 
> To make maximum use of the processor, "slicing by 8" performs 8 look-ups
> in parallel.  Each step, the 32-bit CRC is shifted 64 bits and XORed
> with 64 bits of input data.  What is important to note is that 4 of
> those 8 bytes are simply copies of the input data; they do not depend
> on the previous CRC at all.  Thus, those 4 table look-ups may commence
> immediately, without waiting for the previous loop iteration.
> 
> By always having 4 loads in flight, a modern superscalar processor can
> be kept busy and make full use of its L1 cache.
> 
> 
> Two more details about CRC implementation in the real world:
> 
> Normally, appending zero bits to a message which is already a multiple
> of a polynomial produces a larger multiple of that polynomial.  Thus,
> a basic CRC will not detect appended zero bits (or bytes).  To enable
> a CRC to detect this condition, it's common to invert the CRC before
> appending it.  This makes the remainder of the message+crc come out not
> as zero, but some fixed non-zero value.  (The CRC of the inversion
> pattern, 0xffffffff.)
> 
> The same problem applies to zero bits prepended to the message, and a
> similar solution is used.  Instead of starting the CRC computation with
> a remainder of 0, an initial remainder of all ones is used.  As long as
> you start the same way on decoding, it doesn't make a difference.

Thanks George! I'll replace the document patch with this. Can I add a signed
off line from you?

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