I am not asking how Huffman coding is working, but instead, I want to know why it is good.
I have the following two questions:
Q1
I understand the ultimate purpose of Huffman coding is to give certain char a less bit number, so space is saved. What I don't understand is that why the decision of number of bits for a char can be related to the char's frequency?
Huffman Encoding Trees says
It is sometimes advantageous to use variable-length codes, in which
different symbols may be represented by different numbers of bits. For
example, Morse code does not use the same number of dots and dashes
for each letter of the alphabet. In particular, E, the most frequent
letter, is represented by a single dot.
So in Morse code, E can be represented by a single dot because it is the most frequent letter. But why? Why can it be a dot just because it is most frequent?
Q2
Why the probability / statistics of the chars are so important to Huffman coding?
What happen if the statistics table is wrong?
If you assign less number or bits or shorter code words for most frequently used symbols you will be saving a lot of storage space.
Suppose you want to assign 26 unique codes to English alphabet and want to store an english novel ( only letters ) in term of these code you will require less memory if you assign short length codes to most frequently occurring characters.
You might have observed that postal code and STD codes for important cities are usually shorter ( as they are used very often ). This is very fundamental concept in Information theory.
Huffman encoding gives prefix codes.
Construction of Huffman tree:
A greedy approach to construct Huffman tree for n characters is as follows:
places n characters in n sub-trees.
Starts by combining the two least weight nodes into a tree which is assigned the sum of the two leaf node weights as the weight for its root node.
Do this until you get a single tree.
For example consider below binary tree where E and T have high weights ( as very high occurrence )
It is a prefix tree. To get the Huffman code for any character, start from the node corresponding to the the character and backtrack till you get the root node.
Indeed, an E could be, say, three dashes followed by two dots. When you make your own encoding, you get to decide. If your goal is to encode a certain text so that the result is as short as possible, you should choose short codes for the most frequent characters. The Huffman algorithm ensures that we get the optimal codes for a specific text.
If the frequency table is somehow wrong, the Huffman algorithm will still give you a valid encoding, but the encoded text would be longer than it could have been if you had used a correct frequency table. This is usually not a problem, because we usually create the frequency table based on the actual text that is to be encoded, so the frequency table will be "perfect" for the text that we are going to encode.
well.. you want assign shorter codes to the symbols which appear more frequently... huffman encoding works just by this simple assumption.. :-)
you compute the frequency of all symbols, sort them all, and start assigning bit codes to each one.. the more frequent a symbol is, the shorter the code you'll assign to it.. simple as this.
the big question is: how large the window in which we compute such frequencies should be? should it be as large as the entire file? or should it be smaller? and if the latter apply, how large? Most huffman encoding have some sort of "test-run" in which they estimate the best window size a little bit like TCP/IP do with its windows frame sizes.
Huffman codes provide two benefits:
they are space efficient given some corpus
they are prefix codes
Given some set of documents for instance, encoding those documents as Huffman codes is the most space efficient way of encoding them, thus saving space. This however only applies to that set of documents as the codes you end up are dependent on the probability of the tokens/symbols in the original set of documents. The statistics are important because the symbols with the highest probability (frequency) are given the shortest codes. Thus the symbols most likely to be in your data use the least amount of bits in the encoding, making the coding efficient.
The prefix code part is useful because it means that no code is the prefix of another. In morse code for instance A = dot dash and J = dot dash dash dash, how do you know where to break reading the code. This increases the inefficiency of transmitting data using morse as you need a special symbol (pause) to signify the end of transmission of one code. Compare that to Huffman codes where each code is unique, as soon as you discover the encoding for a symbol in the input, you know that that is the transmitted symbol because it is guaranteed not to be the prefix of some other symbol.
It's the dual effect of having the most frequent characters using the shortest bit sequences that gives you the savings.
For a concrete example, let's say you have a piece of text that consists of 1024 e characters and 1024 of all other characters combined.
With 8 bits for code, that's a full 2048 bytes used in uncompressed form.
Now let's say we represent e as a single 1-bit and every other letter as a 0-bit followed by its original 8 bits (a very primitive form of Huffman).
You can see that half the characters have been expanded from 8 bits to 9, giving 9216 bits, or 1152 bytes. However, the e characters have been reduced from 8 bits to 1, meaning they take up 1024 bits, or 128 bytes.
The total bytes used is therefore 1152 + 128, or 1280 bytes, representing a compression ratio of 62.5%.
You can use a fixed encoding scheme based on the likely frequencies of characters (such as English text), or you can use adaptive Huffman encoding which changes the encoding scheme as characters are processed and frequencies are adjusted. While the former may be okay for input which has high probability of matching frequencies, the latter can adapt to any input.
Statistic table can't be wrong, because in general Huffman algorithm, analyze hole text at the beginning, and builds frequent-statistics of the given text, while Morse has a static symbol -code map.
Huffman algorithm uses the advantage of a given text. As an example, if E is most frequent letter in English in general, that doesn't mean that E is most frequent in a given text for a given author.
Another advantage of Huffman algorithm is that you can use it for any alphabet starting from [0, 1] finished Chinese hieroglyphs, while Morse is defined only for English letters
So in Morse code, "E" can be represented by a single dot, because it is the most frequent letter. But why? Why is it a dot because of its frequency?
"E" can be encoded to any unique code for a specific code dictionary, so it can be "0", we choose it to be short to save memory, so the average bytes used after encode is minimized.
Why is the probability / statistics of the chars so important to Huffman coding? What happens if the statistics table is wrong?
why do we encode? save space right? Space used after encode is freq(wordi)*Length(wordi), it is what we should try to minimize, so we choose to assign words with high prob short code greedly to save space.
If the statistics table is wrong, then the encoding is not the best way to save space.
Problem:
I want to compress an array of non-negative integers of non-fixed length (but it should be 300 to 400), containing mostly 0's, some 1's, a few 2's. Although unlikely, it is also possible to have bigger numbers.
For example, here is an array of 360 elements:
0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,2,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,0,
0,0,4,0,0,0,0,0,0,3,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,5,2,0,0,0,
0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,1,2,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.
Goal:
The goal is to compress an array like this, into a shortest possible encoding using letters and numbers. Ideally, something like: sd58x7y
What I've tried:
I tried to use "delta encoding", and use zeroes to denote any value higher than 1. For example: {0,0,1,0,0,0,2,0,1} would be denoted as: 2,3,0,1. To decode it, one would read from left to right, and write down "2 zeroes, one, 3 zeroes, one, 0 zeroes, one (this would add to the previous one, and thus have a two), 1 zero, one".
To eliminate the need of delimiters (commas) and thus saves more space, I tried to use only one alphanumerical character to denote delta values of 0 to 35 (using 0 to y), while leaving letter z as "35 PLUS the next character". I think this is called "variable bit" or something like that. For example, if there are 40 zeroes in a row, I'd encode it as "z5".
That's as far as I got... the resultant string is still very long (it would be about 20 characters long in the above example). I would ideally want something like, 8 characters or even shorter. Thanks for your time; any help or inspiration would be greatly appreciated!
Since your example contains long runs of zeroes, your first step (which it appears you have already taken) could be to use run-lenth encoding (RLE) to compress them. The output from this step would be a list of integers, starting with a run-length count of zeroes, then alternating between that and the non-zero values. (a zero-run-length of 0 will indicate successive non-zero values...)
Second, you can encode your integers in a small number of bits, using a class of methods called universal codes. These methods generally compress small integers using a smaller number of bits than larger integers, and also provide the ability to encode integers of any size (which is pretty spiffy...). You can tune the encoding to improve compression based on the exact distribution you expect.
You may also want to look into how JPEG-style encoding works. After DCT and quantization, the JPEG entropy encoding problem seems similar to yours.
Finally, if you want to go for maximum compression, you might want to look up arithmetic encoding, which can compress your data arbitrarily close to the statistical minimum entropy.
The above links explain how to compress to a stream of raw bits. In order to convert them to a string of letters and numbers, you will need to add another encoding step, which converts the raw bits to such a string. As one commenter points out, you may want to look into base64 representation; or (for maximum efficiency with whatever alphabet is available) you could try using arithmetic compression "in reverse".
Additional notes on compression in general: the "shortest possible encoding" depends greatly on the exact properties of your data source. Effectively, any given compression technique describes a statistical model of the kind of data it compresses best.
Also, once you set up an encoding based on the kind of data you expect, if you try to use it on data unlike the kind you expect, the result may be an expansion, rather than a compression. You can limit this expansion by providing an alternative, uncompressed format, to be used in such cases...
In your data you have:
14 1s (3.89% of data)
4 2s (1.11%)
1 3s, 4s and 5s (0.28%)
339 0s (94.17%)
Assuming that your numbers are not independent of each other and you do not have any other information, the total entropy of your data is 0.407 bits per number, that is 146.4212 bits overall (18.3 bytes). So it is impossible to encode in 8 bytes.
Noticing that byte-pair encoding (BPE) is sorely lacking from the large text compression benchmark, I very quickly made a trivial literal implementation of it.
The compression ratio - considering that there is no further processing, e.g. no Huffman or arithmetic encoding - is surprisingly good.
The runtime of my trivial implementation was less than stellar, however.
How can this be optimized? Is it possible to do it in a single pass?
This is a summary of my progress so far:
Googling found this little report that links to the original code and cites the source:
Philip Gage, titled 'A New Algorithm
for Data Compression', that appeared
in 'The C Users Journal' - February
1994 edition.
The links to the code on Dr Dobbs site are broken, but that webpage mirrors them.
That code uses a hash table to track the the used digraphs and their counts each pass over the buffer, so as to avoid recomputing fresh each pass.
My test data is enwik8 from the Hutter Prize.
|----------------|-----------------|
| Implementation | Time (min.secs) |
|----------------|-----------------|
| bpev2 | 1.24 | //The current version in the large text benchmark
| bpe_c | 1.07 | //The original version by Gage, using a hashtable
| bpev3 | 0.25 | //Uses a list, custom sort, less memcpy
|----------------|-----------------|
bpev3 creates a list of all digraphs; the blocks are 10KB in size, and there are typically 200 or so digraphs above the threshold (of 4, which is the smallest we can gain a byte by compressing); this list is sorted and the first subsitution is made.
As the substitutions are made, the statistics are updated; typically each pass there is only around 10 or 20 digraphs changed; these are 'painted' and sorted, and then merged with the digraph list; this is substantially faster than just always sorting the whole digraph list each pass, since the list is nearly sorted.
The original code moved between a 'tmp' and 'buf' byte buffers; bpev3 just swaps buffer pointers, which is worth about 10 seconds runtime alone.
Given the buffer swapping fix to bpev2 would bring the exhaustive search in line with the hashtable version; I think the hashtable is arguable value, and that a list is a better structure for this problem.
Its sill multi-pass though. And so its not a generally competitive algorithm.
If you look at the Large Text Compression Benchmark, the original bpe has been added. Because of it's larger blocksizes, it performs better than my bpe on on enwik9. Also, the performance gap between the hash-tables and my lists is much closer - I put that down to the march=PentiumPro that the LTCB uses.
There are of course occasions where it is suitable and used; Symbian use it for compressing pages in ROM images. I speculate that the 16-bit nature of Thumb binaries makes this a straightforward and rewarding approach; compression is done on a PC, and decompression is done on the device.
I've done work with optimizing a LZF compression implementation, and some of the same principles I used to improve performance are usable here.
To speed up performance on byte-pair encoding:
Limit the block size to less than 65kB (probably 8-16 kB will be optimal). This guarantees not all bytes will be used, and allows you to hold intermediate processing info in RAM.
Use a hashtable or simple lookup table by short integer (more RAM, but faster) to hold counts for a byte pairs. There are 65,656 2-byte pairs, and BlockSize instances possible (max blocksize 64k). This gives you a table of 128k possible outputs.
Allocate and reuse data structures capable of holding a full compression block, replacement table, byte-pair counts, and output bytes in memory. This sounds wasteful of RAM, but when you consider that your block size is small, it's worth it. Your data should be able to sit entirely in CPU L2 or (worst case) L3 cache. This gives a BIG speed boost.
Do one fast pass over the data to collect counts, THEN worry about creating your replacement table.
Pack bytes into integers or short ints whenever possible (applicable mostly to C/C++). A single entry in the counting table can be represented by an integer (16-bit count, plus byte pair).
Code in JustBasic can be found here complete with input text file.
Just BASIC Files Archive – forum post
EBPE by TomC 02/2014 – Ehanced Byte Pair Encoding
EBPE features two post processes to Byte Pair Encoding
1. Is compressing the dictionary (believed to be a novelty)
A dictionary entry is composed of 3 bytes:
AA – the two char to be replaced by (byte pair)
1 – this single token (tokens are unused symbols)
So "AA1" tells us when decoding that every time we see a "1" in the
data file, replace it with "AA".
While long runs of sequential tokens are possible, let’s look at this
8 token example:
AA1BB3CC4DD5EE6FF7GG8HH9
It is 24 bytes long (8 * 3)
The token 2 is not in the file indicating that it was not an open token to
use, or another way to say it: the 2 was in the original data.
We can see the last 7 tokens 3,4,5,6,7,8,9 are sequential so any time we
see a sequential run of 4 tokens or more, let’s modify our dictionary to be:
AA1BB3<255>CCDDEEFFGGHH<255>
Where the <255> tells us that the tokens for the byte pairs are implied and
are incremented by 1 more than the last token we saw (3). We increment
by one until we see the next <255> indicating an end of run.
The original dictionary was 24 bytes,
The enhanced dictionary is 20 bytes.
I saved 175 bytes using this enhancement on a text file where tokens
128 to 254 would be in sequence as well as others in general, to include
the run created by lowercase pre-processing.
2. Is compressing the data file
Re-using rarely used characters as tokens is nothing new.
After using all of the symbols for compression (except for <255>),
we scan the file and find a single "j" in the file. Let this char do double
duty by:
"<255>j" means this is a literal "j"
"j" is now used as a token for re-compression,
If the j occurred 1 time in the data file, we would need to add 1 <255>
and a 3 byte dictionary entry, so we need to save more than 4 bytes in BPE
for this to be worth it.
If the j occurred 6 times we would need 6 <255> and a 3 byte dictionary
entry so we need to save more than 9 bytes in BPE for this to be worth it.
Depending on if further compression is possible and how many byte pairs remain
in the file, this post process has saved in excess of 100 bytes on test runs.
Note: When decompressing make sure not to decompress every "j".
One needs to look at the prior character to make sure it is not a <255> in order
to decompress. Finally, after all decompression, go ahead and remove the <255>'s
to recreate your original file.
3. What’s next in EBPE?
Unknown at this time
I don't believe this can be done in a single pass unless you find a way to predict, given a byte-pair replacement, if the new byte-pair (after-replacement) will be good for replacement too or not.
Here are my thoughts at first sight. Maybe you already do or have already thought all this.
I would try the following.
Two adjustable parameters:
Number of byte-pair occurrences in chunk of data before to consider replacing it. (So that the dictionary doesn't grow faster than the chunk shrinks.)
Number of replacements by pass before it's probably not worth replacing anymore. (So that the algorithm stops wasting time when there's maybe only 1 or 2 % left to gain.)
I would do passes, as long as it is still worth compressing one more level (according to parameter 2). During each pass, I would keep a count of byte-pairs as I go.
I would play with the two parameters a little and see how it influences compression ratio and speed. Probably that they should change dynamically, according to the length of the chunk to compress (and maybe one or two other things).
Another thing to consider is the data structure used to store the count of each byte-pair during the pass. There very likely is a way to write a custom one which would be faster than generic data structures.
Keep us posted if you try something and get interesting results!
Yes, keep us posted.
guarantee?
BobMcGee gives good advice.
However, I suspect that "Limit the block size to less than 65kB ... . This guarantees not all bytes will be used" is not always true.
I can generate a (highly artificial) binary file less than 1kB long that has a byte pair that repeats 10 times, but cannot be compressed at all with BPE because it uses all 256 bytes -- there are no free bytes that BPE can use to represent the frequent byte pair.
If we limit ourselves to 7 bit ASCII text, we have over 127 free bytes available, so all files that repeat a byte pair enough times can be compressed at least a little by BPE.
However, even then I can (artificially) generate a file that uses only the isgraph() ASCII characters and is less than 30kB long that eventually hits the "no free bytes" limit of BPE, even though there is still a byte pair remaining with over 4 repeats.
single pass
It seems like this algorithm can be slightly tweaked in order to do it in one pass.
Assuming 7 bit ASCII plaintext:
Scan over input text, remembering all pairs of bytes that we have seen in some sort of internal data structure, somehow counting the number of unique byte pairs we have seen so far, and copying each byte to the output (with high bit zero).
Whenever we encounter a repeat, emit a special byte that represents a byte pair (with high bit 1, so we don't confuse literal bytes with byte pairs).
Include in the internal list of byte "pairs" that special byte, so that the compressor can later emit some other special byte that represents this special byte plus a literal byte -- so the net effect of that other special byte is to represent a triplet.
As phkahler pointed out, that sounds practically the same as LZW.
EDIT:
Apparently the "no free bytes" limitation I mentioned above is not, after all, an inherent limitation of all byte pair compressors, since there exists at least one byte pair compressor without that limitation.
Have you seen
"SCZ - Simple Compression Utilities and Library"?
SCZ appears to be a kind of byte pair encoder.
SCZ apparently gives better compression than other byte pair compressors I've seen, because
SCZ doesn't have the "no free bytes" limitation I mentioned above.
If any byte pair BP repeats enough times in the plaintext (or, after a few rounds of iteration, the partially-compressed text),
SCZ can do byte-pair compression, even when the text already includes all 256 bytes.
(SCZ uses a special escape byte E in the compressed text, which indicates that the following byte is intended to represent itself literally, rather than expanded as a byte pair.
This allows some byte M in the compressed text to do double-duty:
The two bytes EM in the compressed text represent M in the plain text.
The byte M (without a preceeding escape byte) in the compressed text represents some byte pair BP in the plain text.
If some byte pair BP occurs many more times than M in the plaintext, then the space saved by representing each BP byte pair as the single byte M in the compressed data is more than the space "lost" by representing each M as the two bytes EM.)
You can also optimize the dictionary so that:
AA1BB2CC3DD4EE5FF6GG7HH8 is a sequential run of 8 token.
Rewrite that as:
AA1<255>BBCCDDEEFFGGHH<255> where the <255> tells the program that each of the following byte pairs (up to the next <255>) are sequential and incremented by one. Works great for text
files and any where there are at least 4 sequential tokens.
save 175 bytes on recent test.
Here is a new BPE(http://encode.ru/threads/1874-Alba).
Example for compile,
gcc -O1 alba.c -o alba.exe
It's faster than default.
There is an O(n) version of byte-pair encoding which I describe here. I am getting a compression speed of ~200kB/second in Java.
the easiest efficient structure is a 2 dimensional array like byte_pair(255,255). Drop the counts in there and modify as the file compresses.