Fundamental Structures of Computer Science Feb. 24, 2005 Ananda Guna Lempel-Ziv Compression

Recap

Huffman Trees Huffman Trees can be used to construct an optimal prefix code. What does optimal mean? Greedy algorithm to assemble a Huffman tree. locally optimal steps to global optimization Requires symbol frequencies. read the file twice – counting and encoding

Huffman Encoding Process

Adaptive Huffman or Dynamic Huffman Clearly, having to read the data twice (first for frequency count, then for actual compression) is a bit cumbersome. Perhaps data is available in blocks (streaming data) Can build an adaptive Huffman tree that adjusts itself as more frequency data become available.

Adaptive Huffman ctd..  Mapping from source messages to code words based upon a running estimate of the source message probabilities  Change the tree to remain optimal for the current estimates  adaptive Huffman codes respond to locality  Requires only a single pass of the data

Beating Huffman How about beating the compression achieved by Huffman? Impossible! It produces an optimal prefix code. Right. But who says we have to use a prefix code?

Dictionary-Based Compression

Dictionary-based methods  Here is a simple idea:  Keep track of “words” that we have seen, and replace them with a code number when we see them again.  We can maintain dictionary entries  (word, code)  and make additions to the dictionary as we read the input file.

Lempel & Ziv (1977/78)

Fred Hacker’s algorithm…  Fred now knows what to do… (, 1 ) Transmit 1, done.

Right?  Fred’s algorithm provides excellent compression, but…  …the receiver does not know what is in the dictionary!  And sending the dictionary is the same as sending the entire uncompressed file  Thus, we can’t decompress the “1”.

Hence…  …we need to build our dictionary in such a way that the receiver can rebuild the dictionary easily.

LZW Compression: The Byte Version

Byte method LZW  We start with a trie that contains a root and n children  one child for each possible character  each child labeled 0…n  When we compress as before, by walking down the trie  but, after emitting a code and growing the trie, we must start from the root’s child labeled c, where c is the character that caused us to grow the trie

LZW: Byte method example  Suppose our entire character set consists only of the four letters:  {a, b, c, d}  Let’s consider the compression of the string  baddad

Byte LZW: Compress example baddad Input: ^ a b Dictionary: Output: 1032 cd

Byte LZW: Compress example baddad Input: ^ a b Dictionary: Output: 1032 cd 1 4 a

Byte LZW: Compress example baddad Input: ^ a b Dictionary: Output: 1032 cd 10 4 a 5 d

Byte LZW: Compress example baddad Input: ^ a b Dictionary: Output: 1032 cd a 5 d 6 d

Byte LZW: Compress example baddad Input: ^ a b Dictionary: Output: 1032 cd a 5 d 6 d 7 a

Byte LZW: Compress example baddad Input: ^ a b Dictionary: Output: 1032 cd a 5 d 6 d 7 a

Byte LZW output  So, the input  baddad  compresses to   which again can be given in bit form, just like in the binary method…  …or compressed again using Huffman

Byte LZW: Uncompress example  The uncompress step for byte LZW is the most complicated part of the entire process, but is largely similar to the binary method

Byte LZW: Uncompress example Input: ^ a b Dictionary: Output: 1032 cd

Byte LZW: Uncompress example Input: ^ a b Dictionary: Output: 1032 cd b

Byte LZW: Uncompress example Input: ^ a b Dictionary: Output: 1032 cd ba 4 a

Byte LZW: Uncompress example Input: ^ a b Dictionary: Output: 1032 cd bad 4 a 5 d

Byte LZW: Uncompress example Input: ^ a b Dictionary: Output: 1032 cd badd 4 a 5 d 6 d

Byte LZW: Uncompress example Input: ^ a b Dictionary: Output: 1032 cd baddad 4 a 5 d 6 d 7 a

LZW Byte method: An alternative presentation

Getting off the ground Suppose we want to compress a file containing only letters a, b, c and d. It seems reasonable to start with a dictionary a:0 b:1 c:2 d:3 At least we can then deal with the first letter. And the receiver knows how to start.

Growing pains Now suppose the file starts like so: a b b a b b … We scan the a, look it up and output a 0. After scanning the b, we have seen the word ab. So, we add it to the dictionary a:0 b:1 c:2 d:3 ab:4

Growing pains We output a 1 for the b. Then we get another b. a b b a b b … output 1, and add bb it to the dictionary a:0 b:1 c:2 d:3 ab:4 bb:5

So? Right, so far zero compression. But now we get a followed by b, and ab is in the dictionary a b b a b b … so we output 4, and put bab into the dictionary … d:3 ab:4 bb:5 ba:6 bab:7

And so on Suppose the input continues a b b a b b b b a … We output 5, and put bbb into the dictionary … ab:4 bb:5 ba:6 bab:7 bbb:8

More Hits As our dictionary grows, we are able to replace longer and longer blocks by short code numbers. a b b a b b b b a … And we increase the dictionary at each step by adding another word.

More importantly  Since we extend our dictionary in such a simple way, it can be easily reconstructed on the other end.  Start with the same initialization, then  Read one code number after the other, look up the each one in the dictionary, and extend the dictionary as you go along.

Again: Extending where each prefix is in the dictionary. We stop when we fall out of the dictionary: a 1 a 2 a 3 …. a k b We scan a sequence of symbols a 1 a 2 a 3 …. a k

Again: Extending We output the code for a 1 a 2 a 3 …. a k and put a 1 a 2 a 3 …. a k b into the dictionary. Then we set a 1 = b And start all over.

Sort of Let's take a closer look at an example. Assume alphabet {a,b,c}. The code for aabbaabb is The decoding starts with dictionary a:0, b:1, c:2

Moving along The first 4 code words are already in D and produce output a a b b. As we go along, we extend D: a:0, b:1, c:2, aa:3, ab:4, bb:5 For the rest we get a a b b

Done We have also added to D: ba:6, aab:7 But these entries are never used. Everything is easy, since there is already an entry in D for each code number when we encounter it.

Is this it? Unfortunately, no. It may happen that we run into a code word without having an appropriate entry in D. But, it can only happen in very special circumstances, and we can manufacture the missing entry.

A Bad Run Consider input a a b b b a a ==> After reading 0 0 1, D looks like this: a:0, b:1, c:2, aa:3, ab:4

Disaster The next code is 5, but it’s not in D. a:0, b:1, c:2, aa:3, ab:4 How could this have happened? Can we recover?

… narrowly averted This problem only arises when the input contains a substring …s  s … s  was just added to the dictionary. Here s is a single symbol, but  a (possibly empty) word.

… narrowly averted But then the fix is to output x + first(x) where x is the last decompressed word, and first(x) the first symbol of x. And, we also update the dictionary to contain this new entry.

Example In our example we had s = b w = empty The output and new dictionary word is bb.

Another Example aabbbaabbaaabaababb ==> Decoding (dictionary size: initial 3, final 11) a 0 a+0aa b+1ab bb-5bb aa+3bba bba+6aab aab+7bbaa aaba-9aaba bb+5aabab

The problem cases code position in D a 0 a+0aa 3 b+1ab 4 bb-5bb 5 aa+3bba 6 bba+6aab 7 aab+7bbaa 8 aaba-9aaba 9 bb+5aabab 10

Old vs. new Ordinarily, we use an old dictionary word for the next code word. But sometimes we immediately use what was last added to the dictionary. But then it must be of the form s  s and we can still decompress.

Pseudo Code: Compression Initialize dictionary D to all words of length 1. Read all input characters: output code words from D, extend D whenever a new word appears. New code words: just an integer counter.

Less Pseudo initialize D; c = nextchar; // next input character W = c; // a string while( c = nextchar ) { if( W+c is in D ) // dictionary W = W + c; else output code(W); add W+c to D; W = c; } output code(W)

Pseudo Code: Decompression Initialize dictionary D with all words of length 1. Read all code words and - output corresponding words from D, - extend D at each step. This time the dictionary is of the form ( integer, word ) Keys are integers, values words.

Less Pseudo initialize D; pc = nextcode; // first code word pw = word(pc); // corresponding word output pw; First code word is easy: codes only a single symbol. Remember as pc (previous code) and pw (previous word).

More Less Pseudo while( c = nextcode ) { if( c is in D ) { cw = word(c); pw = word(pc); ww = pw + first(cw); insert ww in D; output cw; } else {

The hard case else { pw = word(pc); cw = pw + first(pw); insert cw in D; output cw; } pc = c; }

Implementation - Tries Tries are the best way to implement LZW in the LZW situation, we can add the new word to the trie dictionary in O(1) steps after discovering that the string is no longer a prefix of a dictionary word. Just add a new leaf to the last node touched.

LZW details In reality, one usually restricts the code words to be 12 or 16 bit integers. Hence, one may have to flush the dictionary ever so often. But we won’t bother with this.

LZW details Lastly, LZW generates as output a stream of integers. It makes perfect sense to try to compress these further, e.g., by Huffman.

Summary of LZW LZW is an adaptive, dictionary based compression method. Encoding is easy in LZW, but uses a special data structure (trie). Decoding is slightly complicated, requires no special data structures.