From: brnstnd@kramden.acf.nyu.edu (Dan Bernstein)
Subject: An introduction to Y coding
Message-ID: <8667:Mar610:36:2291@kramden.acf.nyu.edu>
Date: 6 Mar 91 10:36:22 GMT

Save this article: you may want it later.

---Dan


Y coding

Daniel J. Bernstein

Copyright 1991. All rights reserved.
Draft 4, March 6, 1991.

This is a draft. That means it's supposed to be unreadable, misleading,
boring, useless, and generally wrong. Any deviations from the norm are
accidents---happy accidents, but accidents nonetheless. End of warning.


----- 1. Introduction

--- LZW coding

Fix an alphabet A, and take a string I over that alphabet. Construct a
``dictionary'' D---a one-to-one mapping from strings to integers---as
follows:

  0. Start by mapping each symbol of A to a unique integer.
  1. Find the longest prefix p of I contained in D. (Rather, contained
     in the domain of D. It is convenient to ignore this distinction.)
  2. Take the next letter c after p in I.
  3. Add pc to the dictionary, mapping to another unique integer.
  4. Strip p from the front of I, so that I begins with c.
  5. Repeat from step 1 until I is empty (i.e., until step 2 fails).

For example, say A is the set abcdefghijklmnopqrstuvwxyz, and say I is
the string yabbadabbadabbadoo. D starts with all single characters from
A. Now the longest match between I and D is I's first character, y; and
the charater after that is a. So we add ya to the dictionary and strip y
from the front of I.

Now I is abbadabbadabbadoo, and D has all single characters plus ya. The
longest match between D and I is a, so we add ab to the dictionary and
remove the a. We continue in this manner until I is empty:

  I                     match  add    new dictionary
  yabbadabbadabbadoo    y      ya     ya (plus all single characters)
   abbadabbadabbadoo    a      ab     ya ab
    bbadabbadabbadoo    b      bb     ya ab bb
     badabbadabbadoo    b      ba     ya ab bb ba
      adabbadabbadoo    a      ad     ya ab bb ba ad
       dabbadabbadoo    d      da     ya ab bb ba ad da
	abbadabbadoo    ab     abb    ya ab bb ba ad da abb
	  badabbadoo    ba     bad    ya ab bb ba ad da abb bad
	    dabbadoo    da     dab    ya ab bb ba ad da abb bad dab
	      bbadoo    bb     bba    ya ab bb ba ad da abb bad dab bba
		adoo    ad     ado    ya ab bb ba ad da abb bad dab bba ado
		  oo    o      oo     ya ab bb ba ad da abb bad dab bba ado oo
		   o    o

While we construct the dictionary we can output the value of each match
under the dictionary. This produces a sequence of numbers which, as we
will see below, is sufficient to reconstruct the original string. This
mapping from strings to sequences of numbers is called LZW coding.

Typically the numbers in the dictionary are chosen sequentially, from 0
to |A| - 1 for the initial single-character strings and then from |A| on
up. In the above example, the matches are y a b b a d ab ba da bb ad o o,
which have numbers 24 0 1 1 0 3 27 29 31 28 30 14 14. That sequence is
the result of yabbadabbadabbadoo under LZW coding.


--- LZW decoding

How do we reconstruct the original string if we're given the coded
sequence? The secret is to reconstruct the dictionary.

  0. Start by mapping each symbol of A to a unique integer. Set I to the
     empty string. Read a number from the input, and set p to the
     corresponding single-character string.
  1. Append p to I.
  2. Read a number from the input, and let c be the first character of
     the corresponding string in the dictionary.
  3. Add pc to the dictionary, mapping to the next unique integer.
  4. Set p to the dictionary string corresponding to the number read.
  5. Repeat from step 1 until end-of-input (i.e., until step 2 fails).

For example, take the input sequence 24 0 1 1 0 3 27 29 31 28 30 14 14.
D starts with all single characters from A; p starts as the single
character y; and I starts empty.

Now p is appended to I, so that I contains y. The 0 in the input means
that the next character of I is an a; and ya is added to the dictionary.
p is then set to a.

Next, p is appended to I, so that I contains ya. The 1 in the input
means that the next character of I is b; and ab is added to the
dictionary. p is then set to b. We continue this way through the entire
input:

  input  c   added    new p   I
  24                  y       y
  0      a   (ya,26)  a       ya
  1      b   (ab,27)  b       yab
  1      b   (bb,28)  b       yabb
  0      a   (ba,29)  a       yabba
  3      d   (ad,30)  d       yabbad
  27    *a*  (da,31)  ab      yabbadab      27 is ab, so *a* is the 1st char
  29    _b_  (abb,32) ba      yabbadabba    29 is ba, so _b_ is the 1st char
  31     d   (bad,33) da      yabbadabbada
  28     b   (dab,34) bb      yabbadabbadabb
  30     a   (bba,35) ad      yabbadabbadabbad
  14     o   (ado,36) o       yabbadabbadabbado
  14     o   (oo,37)  o       yabbadabbadabbadoo

Notice the slightly twisted statement of steps 2 through 4. p is set to
the dictionary string corresponding to the number read from the input,
but first c is set to the first character of that string, and the old pc
is added to the dictionary. This roundabout presentation is necessary
because the number on the input may be the number of the very last
dictionary string added---and that last string depends on the first
character of the current string. This overlap is always safe but is one
of the first problems to look out for in a new implementation.


--- So who cares about this LZW stuff anyway?

What's the point of LZW coding? When the input string has lots of the
same characters or sequences of characters, the dictionary will rapidly
grow to include those sequences. Then a single number in the output
sequence might represent several characters of input, and will use less
space.

For example, take the simplest natural encoding of a string over our
lowercase alphabet A. There are 26 symbols, so we can represent each
with 5 bits. Our string yabbadabbadabbadoo has 18 characters and takes
90 bits under this encoding.

On the other hand, the string after encoding has just 13 numbers:
24 0 1 1 0 3 27 29 31 28 30 14 14. For these values there are 26 27 28
29 30 31 32 33 34 35 36 37 38 choices respectively, so they take 5 5 5 5
5 5 5 6 6 6 6 6 6 bits in the natural way, for a total of just 71 bits.

Most strings that come up in practice have similar redundancy, and LZW
coding will produce an output that takes less space than its input. This
is why LZW coding is usually called LZW compression. It often reduces
longer strings to 50% or even 30% of their original size, so that they
take a fraction as much disk space and communication time as they would
in their uncompressed form.


--- A bit of jargon

Many audio and video compression methods throw away some information.
They don't reconstruct exactly the original pictures and sounds, but
nobody really notices. These are called inexact, or sometimes lossy,
compressors. In contrast, LZW always lets you get back exactly the
original string. It is an exact, or lossless, compressor.

Any compressor that splits its input into substrings and codes the
strings with a dictionary is called (logically enough) a dictionary
compressor.

A substring of an input string---particularly a substring that is coded
as a single output number---is called a ``phrase.'' In LZW, one string
is added to the dictionary per phrase.

Some compression methods use fixed tables: ``the'' becomes a special,
single character, ``of'' becomes another, etc. They're called static, or
nonadaptive, compressors. Others, like LZW, build their dictionaries
dynamically to adapt to any input; LZW is an example of an adaptive
compressor. Somewhere in between are compressors that make two passes
over the input, building a dictionary the first time through and then
producing output the second time through. They're called semiadaptive.

One useless theoretical property of LZW is that it is universal. This
means that on a certain wide class of inputs it will give as close as
you want to the best possible compression of any method. The longer your
strings are, the closer it comes to optimal. Universality doesn't make
one whit of difference in practice---to get within 1% of optimal with
LZW you'd have to start compressing the planet---but it generally
indicates that a method is both simple and trustworthy. Non-universal
compressors are usually much more complicated, and will fail miserably
on inputs they weren't designed specifically to handle.



----- 2. Implementing LZW

--- Encoding

The most natural way to find the longest match between I and strings in
D is one character at a time. Start with p as the first character of I
(or as the null string if that is more convenient). Read the next
character, c. If pc is in the dictionary, set p to pc and repeat.
Otherwise p and c are set.

So the dictionary has to support two basic operations: searching for pc
if p is known to be there already, and adding pc if it is not there.
Most implementors use some form of trie---array trie, list trie, hash
trie, huptrie---for this job. The array trie, as in Woods' ``hyper-LZ''
[], offers extreme speed at the expense of |A| (typically 256) words of
memory for each string in the trie. The huptrie and straight hash trie
use only a small amount of memory per string but still allow reasonably
fast searches in the average case. We will not consider these structures
in further detail.

In any case at most a fixed number of operations happen for every input
character, so LZW coding takes linear time no matter how long the input
is. Unfortunately, computers don't have infinite memory. Implementations
typically limit the dictionary to some size, usually between 1024 and
65536 strings. When the dictionary reaches that size it can either
remain fixed for the rest of the input, or it can be cleared and start
from single characters again. The second strategy effectively breaks the
input into ``blocks'' with independent dictionaries. If the input
``feels'' different in different blocks, blocking will produce good
results, because it will weed useless strings out of the dictionary.

The ``compress'' program [] introduced an important blocking variant.
Instead of clearing the dictionary as soon as it reaches its maximum
size, compress periodically checks how well it is doing. It will only
clear the dictionary when its compression ratio deteriorates. This way
the blocks adapt to the input. Note that all of these blocking
techniques require space for a special output code to clear the
dictionary.

Another variant is LRU replacement: the least-recently-used strings are
removed from the dictionary at some opportune time. This provides a
smoother transition between blocks than clearing the dictionary.
Unfortunately, it is difficult to find good heuristics for when to
remove what strings. Blocking is still more of an art than a science.


--- Preparing for encryption

It is very useful to compress a message before encrypting it, as
compression removes much of the statistical regularity of straight text.
However, the usual coding leaves a lot of redundancy. If the dictionary
has 33 strings, for example, and 6 bits are used to represent a choice
among those strings, then the high bit will almost always be 0. These
high bits come in predictable locations, so an attacker can almost
always figure out the key given a long enough stretch of compressed
text.

To prevent this, the compressor should introduce random bits as follows:
Given a choice between 0 and 40, 0 through 22 are encoded as either
themselves or from 41 through 63. The choice should be made with a
high-delay random number generator such as a subtractive generator.
23 through 40 are always encoded as themselves. This method greatly
reduces the amount of extra information available per bit without
appreciably slowing down the coding. Because random bits are used at
unpredictable times, an attacker cannot easily take advantage of the
determinism of the pseudo-random generator.

Most implementations also add a recognizable header to compressed text.
This header should be removed before encryption.


--- Decoding

LZW decoding is even easier than encoding: the dictionary does not have
to support searching. The easiest (and generally fastest) method is to
keep I in memory as it is reconstructed, and to keep track of which
substring of I a given dictionary number corresponds to. To add pc to
the dictionary means to add the pair (pointer to start of pc within I,
length of pc) at the right spot in an array indexed by dictionary
numbers. There are methods which take slightly less memory than this,
but they are slower.



----- 3. MW and AP coding

--- MW: Adapting better

LZW adapts to its input relatively slowly: strings in the dictionary
only get one character longer at a time. If LZW is fed a million
consecutive x's, its longest dictionary string will only have around
1400 x's, and it will only achieve around 1000:1 compression. Is there
any better way for it to notice the redundancy?

The answer is yes. Instead of adding p plus one character of the next
phrase to the dictionary, we can add p plus the entire next phrase to
the dictionary. This defines MW coding. For example:

  I                     match  added to dictionary
  yabbadabbadabbadoo    y        
   abbadabbadabbadoo    a      ya (concatenation of last two matches)
    bbadabbadabbadoo    b      ab
     badabbadabbadoo    b      bb
      adabbadabbadoo    a      ba
       dabbadabbadoo    d      ad
	abbadabbadoo    ab     dab
	  badabbadoo    ba     abba
	    dabbadoo    dab    badab
	       badoo    ba     dabba
		 doo    d      bad
		  oo    o      do
		   o    o

Even such a short example shows how quickly MW begins to adapt. By the
fifteenth character ``dabba'' is already established as a dictionary
phrase. On the other hand, MW will miss shorter phrases where there is
less redundancy, so it generally performs only somewhat better than LZW
in practice. In this example it outputs 13 numbers, just like LZW. But
given a million x's it will end up with a dictionary string of length
around half a million, and it will output just 30 bytes.




--- Problems of MW

MW lacks some properties of LZW that we don't realize are so helpful
until they are taken away. Most importantly, not every prefix of a
dictionary string is in the dictionary. This means that the
character-at-a-time search method we saw for LZW doesn't work. Instead,
every prefix of a string in the dictionary must be added to the trie,
and every node in the trie must be given a tag saying whether it is in
the dictionary or not. Furthermore, finding the longest string may
require backtracking: if the dictionary contains xxxx and xxxxxxxx, we
don't know until we've read to the eighth character of xxxxxxxy that we
have to choose the shorter string. This seems to imply that MW coding is
fundamentally slower than LZW coding. (Decoding can be made reasonably
fast, though.)

Another problem is that a string may be added to the dictionary twice.
This rules out certain implementations and requires extra care in
others. It also makes MW a little less safe than LZW before encryption.


--- AP: Adapting faster

There is a natural way to preserve the good adaptation of MW while
eliminating its need for backtracking: instead of just concatenating the
last two phrases and putting the result into the dictionary, put all
prefixes of the concatenation into the dictionary. (More precisely, if S
and T are the last two matches, add St to the dictionary for every
nonempty prefix t of T, including T itself.) This defines AP coding. For
example:

  I                     match  added to dictionary
  yabbadabbadabbadoo    y        
   abbadabbadabbadoo    a      ya
    bbadabbadabbadoo    b      ab
     badabbadabbadoo    b      bb
      adabbadabbadoo    a      ba
       dabbadabbadoo    d      ad
	abbadabbadoo    ab     da, dab        (d + all prefixes of ab)
	  badabbadoo    ba     abb, abba      (ab + all prefixes of ba)
	    dabbadoo    dab    bad, bada, badab
	       badoo    ba     dabb, dabba
		 doo    d      bad
		  oo    o      do
		   o    o

Since AP adds more strings to the dictionary, it takes more bits to
represent a choice. However, it does provide a fuller range of matches
for the input string, and in practice it achieves slightly better
compression than MW in quite a bit less time.


----- 4. Y coding

--- Completing the square

LZW adds one dictionary string per phrase and increments strings by one
character at a time. MW adds one dictionary string per phrase and
increments strings by several characters at a time. AP adds one
dictionary string per input character and increments strings by several
characters at a time. These properties define three broad classes of
methods and point naturally to a fourth: coders that add one dictionary
string per input character and increment strings by one character at a
time. An example of such a method is Y coding. (It is worth noting at
this point that LZ77 variants are characterized by adding several
dictionary strings per input character.)


--- The incomprehensible definition

Y coding is defined as follows: The dictionary starts with all single
characters. One string pc starting at each input position is added to
the dictionary, where p is in the dictionary before that character and
pc is not.

To put it differently, ``yowie'' appears in the dictionary as soon as
the regular expression yo.*yow.*yowi.*yowie matches the text. It is
added at the final e. So yabbadabbadabbadoo leads to the dictionary ya,
ab, bb, ba, ad, da, abb, bba, ada, dab, abba, bbad, bado, ado, oo.

We haven't defined the coding yet. While we build the dictionary, we
keep track of a current longest match (initially empty). Right after
reading an input character and before integrating it into the
dictionary, we see whether the longest match plus that character is
still in the dictionary *constructed before the start of that longest
match*. If so, we use that as the new longest match, and proceed.
Otherwise, we output the number of the longest match, and set the
longest match to just that new character.

To decode, we read these matches, one by one. We take each one as a
sequence of characters as defined by the dictionary, and output that
sequence. Meanwhile we take each character and feed it as input to the
dictionary-building routine.


--- A comprehensible example

We can run through the string keeping track of dictionary matches, as
follows:

  I                        add    current matches
  abcabcabcabcabcabcabcx          a
   bcabcabcabcabcabcabcx   ab     b
    cabcabcabcabcabcabcx   bc     c
     abcabcabcabcabcabcx   ca     a
      bcabcabcabcabcabcx          ab b
       cabcabcabcabcabcx   abc       bc c
        abcabcabcabcabcx   bca          ca a
         bcabcabcabcabcx   cab             ab  b
          cabcabcabcabcx            _____  abc bc  c
           abcabcabcabcx   abca    / \   \___  bca ca a
            bcabcabcabcx   bcab   cab ab   b \____  \_/
             cabcabcabcx   cabc       abc  bc   c \__/
              abcabcabcx              abca bca  ca   a
               bcabcabcx   abcab           bcab cab  ab    b
                cabcabcx   bcabc                cabc abc   bc   c
                 abcabcx   cabca                     abca  bca  ca  a
                  bcabcx           ,-----------------abcab bcab cab ab b
                   cabcx   abcabc  `--bcabc cabc  abc    bc    c
		    abcx   bcabca           cabca abca   bca   ca   a
		     bcx   cabcab                 abcab  bcab  cab  ab  b
		      cx                          abcabc bcabc cabc abc bc c
		       x   abcabcx x
			    bcabcx
			     cabcx
			      abcx
			       bcx
				cx

The strings on the right side are the current matches-so-far after each
input character. We start with no matches. When we read a character, we
append it to each match so far; if the results are in the dictionary, we
make them the new matches, and add the character as a match by itself.
If any of them are not in the dictionary we take them out of the list
and add them to the dictionary.

Before reading the fifth c, for example, the matches are bcab, cab, ab,
and b. We append c to each one: bcabc doesn't match, so we add it to the
dictionary. The rest are still in the dictionary, so our new list is
cabc, abc, bc, and a lone c. And so on.

When the x is read, the current list is abcab bcab cab ab b. Now none of
abcabx bcabx cabx abx bx are in the dictionary, so we add all of them,
and the list becomes just a single x.


--- The crucial observation

It is not clear so far that Y is a linear-time algorithm: each input
character demands additions to several matches. However, *every
substring of a dictionary string is in the dictionary*. This is clear
from the regular-expression characterization of dictionary strings: if
yo.*yow.*yowi.*yowie matches the text, then ow.*owi.*owie (for example)
certainly does as well.

Say the current match list is wonka onka nka ka a, and we read the
character x. The next list has to be one of the following:

  wonkax onkax nkax kax ax x
         onkax nkax kax ax x
               nkax kax ax x
                    kax ax x
                        ax x
                           x

If onkax matches, for example, then nkax must match as well, and so must
kax and ax and x.

So the match list will always consist of one string and its suffixes.
Hence we can keep track of just the longest string in the match list.
For example, with the input string oompaoompapaoompaoompapa:

  input  test   add    match
  o                    o
  o      oo     oo     o
  m      om     om     m
  p      mp     mp     p
  a      pa     pa     a
  o      ao     ao     o
  o      oo            oo
  m      oom    oom    om
  p      omp    omp    mp
  a      mpa    mpa    pa
  p      pap    pap
                ap     p
  a      pa            pa
  o      pao    pao    ao
  o      aoo    aoo    oo
  m      oom           oom
  p      oomp   oomp   omp
  a      ompa   ompa   mpa
  o      mpao   mpao
	        pao    ao
  o      aoo           aoo
  m      aoom   aoom   oom
  p      oomp          oomp
  a      oompa  oompa  ompa
  p      ompap  ompap
	        mpap   pap
  a      papa   papa
                apa    pa


--- A comprehensible definition

Here is another definition of how to construct the Y dictionary, based
on the chart above.

  0. Start with the dictionary mapping every character of A to a unique
     integer. Set m to the empty string.
  1. Append the next character c of I to m.
  2. If m is not in the dictionary, then add m to the dictionary, remove
     the first character of m, and repeat this step.
  3. Repeat from step 1 until the input is exhausted.

We must be careful in defining output: the output is not synchronized
with the dictionary additions, and we must be careful not to have the
longest output match overlap itself. To this end we split the dictionary
into two parts, the first part safe to use for the output, the second
part not safe.

  0. Start with S mapping each single character to a unique integer;
     T empty; m empty; and o empty. (S and T are the two parts of the
     dictionary.)
  1. Read a character c. If oc is in S, set o to oc; otherwise output
     S(o), set o to c, add T to S, and remove everything from T.
  2. While mc is not in S or T, add mc to T (mapping to the next
     available integer), and chop off the first character of m.
  3. After m is short enough that mc is in the dictionary, set m to mc.
  4. Repeat as long as there are enough input characters (i.e., until
     step 1 fails).
  5. Output S(o) and quit.

Here's how to decode:

  0. Initialize (D,m) as above.
  1. Read D(o) from the input and take the inverse under D to find o.
  2. As long as o is not the empty string, find the first character c of
     o, and update (D,m) as above. Also output c and chop it off from
     the front of o.
  3. Repeat from step 1 until the input is exhausted.

The coding only requires two fast operations on strings in the
dictionary: testing whether sc is there if s's position is known, and
finding s's position given cs's position. The decoding only requires the
same operations plus finding the first character of a string given its
spot in the dictionary.


--- Some properties of Y coding

We propose ``per-phrase dictionary'' to describe the dictionaries
constructed by LZW and MW, ``per-character dictionary'' for AP and Y.
We also propose ``phrase-increment'' to describe MW and AP,
``character-increment'' for LZW and Y. According to this terminology,
Y is an exact character-increment per-character dictionary compressor.

Y has a similar feel to LZJ, which has a dictionary consisting of all
unique substrings up to a certain length in a block of the input.
However, Y can adapt much more effectively to redundant text.



----- 5. Results

The author has implemented Y coding [] along with, for instruction and
amusement, AP coding. In this section we survey various implementation
issues and compare the effectiveness of the coding methods explained
here.

Here are results of these methods on the Bell-Witten-Cleary text corpus,
along with a highly redundant 180489-byte ``makelog'' added by the
author to show an extreme case:

   Z12    Z14    Z16     MW    AP2    AP6    AP3     Y2     Y6     Y3
 54094  46817  46528  41040  47056  40770  40311  46882  40874  40456  bib
394419 357387 332056 336793 389702 338046 322178 363339 320622 306813  book1
325147 281354 250759 230862 297205 261270 228978 287110 256578 229851  book2
 78026  77696  77777  80296  84582  79471  80106  80817  76275  76695  geo
 34757  25647  25647   8299  20925  14762   8821  28159  21220  14411  makelog
230765 202594 182121 168652 219665 190502 167896 212617 185097 168287  news
 16528  14048  14048  13273  13824  13824  13825  13858  13858  13859  obj1
160099 138521 128659 109266 134547 123323 113296 141783 125900 114323  obj2
 29433  25077  25077  22749  26937  22413  22414  26131  22452  22453  paper1
 40881  37196  36161  34234  39415  34637  33320  38037  33671  32733  paper2
 23567  22163  22163  21495  22293  20869  20870  21609  20355  20356  paper3
  7091   6957   6957   6697   6595   6595   6596   6443   6443   6444  paper4
  6670   6580   6580   6169   6146   6146   6147   6033   6033   6034  paper5
 22333  18695  18695  16899  19770  16786  16787  19418  16677  16678  paper6
 66236  63277  62215  65102  74061  68349  67980  71952  66117  65377  pic
 21825  19143  19143  16976  18868  16691  16692  18897  17063  17064  progc
 31987  27116  27148  22223  27191  22716  22451  27607  23625  23512  progl
 22937  19209  19209  15095  17962  15138  15139  19429  16616  16617  progp
 46185  39605  38240  27742  38781  30415  28056  40444  33026  31300  trans

Z12 is LZW with 12 bits (i.e., a dictionary of at most 4096 strings),
using compress -b12. MW is MW using the author's squeeze program [],
with a dictionary of at most 300000 strings (roughly equivalent to
1500000 input characters). AP2 is AP with an input block size of 21000,
using whap -m21000; AP6 has a block size of 65533, and AP3 has a block
size of 300000. Y is like AP but using yabba.

Y is notably effective upon book1 and news.



XXX talk about implementation issues!

XXX what else do people want to see in this section?



----- 6. Conclusion

--- Life goes on

Y coding is not much more complex than LZW coding and produces generally
better results. It is one of the most effective non-Huffman-based
LZ78-style compressors.


--- How to achieve fame and fortune

It may be possible to implement Y so that the decoding does not have to
do all the work of the coding. In particular, three-fourths of the
dictionary strings are typically not used at all; they should not have
to be handled during decoding. Even if Y cannot be sped up, there is
probably some character-increment per-character dictionary compressor
that achieves similar results to Y but runs as quickly as LZW or AP.
This is a area for future research.

We have not discussed different ways to code the output numbers. Huffman
coding and arithmetic coding both produce quite respectable improvements
over and above the compression of the basic Y method. Little is known
about the best way to code a sequence from a dynamically expanding
alphabet; it is a bit counterintuitive that semiadaptive Huffman coding
upon an output Y sequence can produce worse results than adaptive
Huffman coding.

It would be interesting to compare a straight modeller with Y plus a
Huffman coder to see which produces better results.


--- Acknowledgments

The author would like to thank James Woods for his support and
encouragement, as well as for many copies of papers; Richard Stallman,
for motivating the author to find Y coding; and Herbert J. Bernstein for
general comments and for suggesting the order of presentation of this
material.


--- So who are these LZWMWAPY people anyway?

LZW stands for Lempel, Ziv, Welch. In 1977 Ziv and Lempel discovered the
first in what are now called the LZ family of dictionary compressors.
The original LZ algorithms were like LZW, but transmitted both p and c
and then skipped past pc. They also started with a dictionary of just
the null string. (For detailed surveys of various LZ methods, the reader
should consult [], [], [].) Welch popularized the LZ variant, now called
LZW, in which the extra character was not transmitted but became the
first character of the next phrase.

Miller and Wegman independently discovered several LZ variants in 1984,
including LZW and MW. In fact, Seery and Ziv had in 1978 proposed an LZ
variant where two adjacent phrases were concatenated; this is related to
MW in approximately the same way that LZ is related to LZW. (Seery and
Ziv also introduced an important improvement for binary alphabets: if
01101 and 01100 are in the dictionary, there is no way that 0110 can be
a longest match, so it can effectively be removed from the dictionary.)

AP stands for ``all prefixes.'' It was originally discovered by Storer
in 1988 (?) and independently rediscovered by this author in 1990.

Y actually does stand for yabbadabbadoo. The author discovered Y coding
on December 26, 1991; he used yabbadabbadabbadoo as an example in
explaining the method to Woods the next day. Woods replied [] ``I'll
have to look at your yabbadabbadoo code again to see how it differs from
LZR.'' The author promptly adopted ``Y coding'' as the official name.
(Ross Williams has suggested [] that ``LZDB coding'' might be more
appropriate.)


--- References

Yeah, yeah, I'm still writing up proper citations. XXX

