15-211 fundamental data structures and algorithms aleks nanevski february 10, 2004 based on a...
TRANSCRIPT
15-211Fundamental Data Structures and Algorithms
Aleks NanevskiFebruary 10, 2004
based on a lecture by Peter Lee
LZW Compression
Last Time…
Problem: data compression
Convert a string into a shorter string.Lossless – represents exactly
the same information.Lossy – approximates the
original information. Uses of compression:
Images over the web: JPEGMusic: MP3General-purpose: ZIP, GZIP, JAR, …
Huffman trees
Huffman’s algorithm
Huffman’s algorithm gives the optimal prefix code.
For a nice online demo, see http://ciips.ee.uwa.edu.au/~morris/Year2/PLDS210/huffman.html
Huffman compression
Huffman trees provide a straightforward method for file compression.1. Scan the file and compute frequencies2. Build the code tree3. Write code tree to the output file as a
header4. Scan input, encode, and write into the
output file
Huffman decompression
Read the header in the compressed file, and build the code tree
Read the rest of the file, decode using the tree
Write to output
Beating Huffman
How about doing better than Huffman!
Impossible!Huffman’s algorithm gives the optimal
prefix code!
Right.But who says we have to use a prefix
code?
Example
Suppose we have a file containingabcdabcdabcdabcdabcdabcd…
abcdabcd
This could be expressed very compactly asabcd^1000
Dictionary-BasedCompression
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.
The code is typically shorter than the word
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…
Create the dictionary:
( <the-whole-file>, 1 )
Transmit 1, done.
Right?
Fred’s algorithm provides excellent compression, but…
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 Binary Version
LZW=variant of Lempel-Ziv Compression, by Terry Welch (1984)
Maintaining a dictionary
We need a way of incrementally building up a dictionary during compression in such a way that…
…someone who wants to uncompress can “rediscover” the very same dictionary
And we already know that a convenient way to build a dictionary incrementally is to use a trie
Binary LZW
In this method, we build up binary tries In a binary trie, each node has two
children In addition, we will add the following:
each left edge is marked 0each right edge is marked 1each leaf has a label from the set {0,…,n}
A binary trie
0 1
0
3
1 2
4 5
0 0
0
0
11
1
1
Binary LZW: Compression
1. We start with a binary trie consisting of a root node and two children
left child labeled 0, and right labeled 1
2. We read the bits of the input file, and follow the trie
3. When a leaf is reached, we emit the label at the leaf
4. Then, add two new children to that leaf (converting it into an internal node)
Binary LZW: Compression, pt.2
5. The new left child takes the old label
6. The new right child takes a new label value that is one greater than the current maximum label value
Binary LZW: Compression example
10010110011Input:^
0 1
0 1
Dictionary:
Output:
Binary LZW: Compression example
10010110011Input:^
0 1
0
Dictionary:
Output: 1
1 2
0 1
Binary LZW: Compression example
10010110011Input:^
0 1
3
Dictionary:
Output: 10
1 2
0 1
0
0 1
Binary LZW: Compression example
10010110011Input:^
0 1
4
Dictionary:
Output: 103
1 2
0 1
0
0 1
3
0 1
Binary LZW: Compression example
10010110011Input:^
0 1
Dictionary:
Output: 1034
1 2
0 1
0
0 1
3
0 1
4 5
0 1
Binary LZW: Compression example
10010110011Input:^
0 1
Dictionary:
Output: 10340
1 2
0 10 1
3
0 1
4 5
0 1
0 6
Binary LZW: Compression example
10010110011Input:^
0 1
Dictionary:
Output: 103402
1
0 10 1
3
0 1
4 5
0 1
0 6 2
0 1
7
Binary LZW output
So from the input10010110011
we get output103402
To represent this output we can keep track of the number of labels n each time we emit a codeand use log(n) bits for that code
Binary LZW output
We started with input
10010110011
Encoded it as 103402, for which we get the bit sequence 001 000 011 100 000 010
This looks like an expansion instead of a compression
But what if we have a larger input, with more repeating sequences?
Try it!
Binary LZW output
One can also use Huffman compression on the output…
Binary LZW termination
Note that binary LZW has a serious problem, in that the input might end while we are in the middle of the trie (instead of at a leaf node)
This is a nasty problemwhich is why we won’t use this binary
methodBut this is still good for illustration
purposes…
Binary LZW: Uncompress
To uncompress, we need to read the compressed file and rebuild the same trie as we go along
To do this, we need to maintain the trie and also the maximum label value
Binary LZW: Uncompress example
103402Input:^
0 1
Dictionary:
Output:
10
Binary LZW: Uncompress example
103402Input:^
0 1
Dictionary:
Output: 1
1
0 1
2
0
Binary LZW: Uncompress example
103402Input:^
0 1
Dictionary:
Output: 10
1
0 10 1
203
Binary LZW: Uncompress example
103402Input:^
0 1
Dictionary:
Output: 1001
1
0 10 1
3
0 1
20
4
Binary LZW: Uncompress example
103402Input:^
0 1
Dictionary:
Output: 1001011
1
0 10 1
3
0 1
4 5
0 1
0 2
Binary LZW: Uncompress example
103402Input:^
0 1
Dictionary:
Output: 100101100
1
0 10 1
3
0 1
4 5
0 1
0 6
2
Binary LZW: Uncompress example
103402Input:^
0 1
Dictionary:
Output: 10010110011
1
0 10 1
3
0 1
4 5
0 1
0 6 2
0 1
7
LZW Compression:The Byte Version
Byte method
The binary LZW method doesn’t really workwe show it for illustrative purposes
Instead, we use a slightly more complicated version that works on bytes or charactersWe can think of each byte as a
“character” in the range {0…255}
Byte method trie
Instead of a binary trie, we use a more general trie in whicheach node can have up to n children
(where n is the size of the alphabet), one for each byte/character
every node (not just the leaves) has an integer label from the set {0…m}, for some m• except the root node, which has no label
Byte method LZW
We start with a trie that contains a root and n childrenone child for each possible charactereach child labeled 0…n
When we compress as before, by walking down the triebut, 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 stringbaddad
Byte LZW: Compress example
baddadInput:^
a bDictionary:
Output:
10 32
c d
Byte LZW: Compress example
baddadInput:^
a bDictionary:
Output:
10 32
c d
1
4
a
Byte LZW: Compress example
baddadInput:^
a bDictionary:
Output:
10 32
c d
10
4
a
5
d
Byte LZW: Compress example
baddadInput:^
a bDictionary:
Output:
10 32
c d
103
4
a
5
d
6
d
Byte LZW: Compress example
baddadInput:^
a bDictionary:
Output:
10 32
c d
1033
4
a
5
d
6
d
7
a
Byte LZW: Compress example
baddadInput:^
a bDictionary:
Output:
10 32
c d
10335
4
a
5
d
6
d
7
a
Byte LZW output
So, the inputbaddad
compresses to10335
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
10335Input:^
a bDictionary:
Output:
10 32
c d
Byte LZW: Uncompress example
10335Input:^
a bDictionary:
Output:
10 32
c d
b
Byte LZW: Uncompress example
10335Input:^
a bDictionary:
Output:
10 32
c d
ba
4
a
Byte LZW: Uncompress example
10335Input:^
a bDictionary:
Output:
10 32
c d
bad
4
a
5
d
Byte LZW: Uncompress example
10335Input:^
a bDictionary:
Output:
10 32
c d
badd
4
a
5
d
6
d
Byte LZW: Uncompress example
10335Input:^
a bDictionary:
Output:
10 32
c d
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 already scanned the first b.
a b b a b b …
Then we get another b.
We output a 1 for the first b, and add bb to the dictionary
a:0 b:1 c:2 d:3 ab:4 bb:5
So?
Right, so far zero compression.
We already scanned the second b.
a b b a b b …
After scanning a, we output 1 for the b, and put ba in the dictionary
… d:3 ab:4 bb:5 ba:6
Still zero compression.
But now…
We already scanned a.
a b b a b b …
We scan the next b, and ab : 4 is in the dictionary.
We scan the next b, output 4, and put abb into the dictionary.
… d:3 ab:4 bb:5 ba:6 abb:7
We got compression, because 4 is shorter than ab.
We already scanned the last b
a b b a b b …
Suppose the input continues
a b b a b b b b a …
We scan the next b, and bb:5 is in the dictionary
We scan the next b, output 5, and put bbb into the dictionary
… ab:4 bb:5 ba:6 abb:7 bbb:8
And so on
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 …
0 1 1 4 5 6
And we increase the dictionary at each step by adding another word.
Summary
where each prefix is in the dictionary.
We stop when we fall out of the dictionary:
a1 a2 a3 …. ak b
We scan a sequence of symbols
a1 a2 a3 …. ak
Summary (cont’d)
We output the code for a1 a2 a3 …. ak and
put a1 a2 a3 …. ak b into the dictionary.
Then we set
a1 = b
And start all over.
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.
Sort of
Let's take a closer look at an example.
Assume alphabet {a,b,c}.
The code for aabbaabb is 0 0 1 1 3 5.
The decoding starts with dictionary D:
0:a, 1:b, 2:c
Moving along
The first 4 code words are already in D.
0 0 1 1 3 5
and produce output a a b b.
As we go along, we extend D:
0:a, 1:b, 2:c, 3:aa, 4:ab, 5:bb
For the code numbers 3 5, get
a a b b a a b b
Done
We have also added to D:
6:ba, 7:aab
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 ==> 0 0 1 5 3
After reading 0 0 1, we output
a a b
and extend D with codes for aa and ab
0:a, 1:b, 2:c, 3:aa, 4:ab
Disaster
We have read 0 0 1 from the input
0 0 1 5 3
The dictionary is
0:a, 1:b, 2:c, 3:aa, 4:ab
The next code number to read is 5, but it’s not in D.
How could this have happened?
Can we recover?
… narrowly averted
This problem only arises when on the compressor end:
• the input contains a substring
…s s s …
• compressor read s , output code c for s , and added c+1: s s to the dictionary.
• Here s is a single symbol, but a (possibly empty) word.
… narrowly averted (pt. 2)
On the decompressor end, D contains
c: s
• but does not contain c+1: s s
• the decompressor has already output
x = s
and is now looking at unknown code number c+1.
… narrowly averted (pt. 3)
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.
Because x=s was already output, we get the required
s s s
We also update the dictionary to contain the new entry x+first(x) = s s.
In our example we have read 0 0 1 from the input
0 0 1 5 3
The last decompressed word is b, and the next code number to read is 5. Thus
• s = b
• = empty
•The next word to output and add to D is
s s = bb
Example
Summary
Let x be the last added word.
Ordinarily, D contains a word y matching to the input code number.
We output y and extend D with
x+ first (y)
But sometimes we immediately use x.
Then it must be x = s and we output
x + first(x) = s s
Example (extended)
0 0 1 5 3 6 7 9 5 aabbbaabbaaabaababb
Input Output add to D
0 a
0 + a 3:aa
1 + b 4:ab
5 - bb 5:bb
3 + aa 6:bba
6 + bba 7:aab
7 + aab 8:bbaa
9 - aaba 9:aaba
5 + bb 10:aabab
Pseudo Code: Compression
Initialize dictionary D to all words of length 1.
Read all input characters:
output code numbers 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 numbers 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 number
x = word(pc); // corresponding word
output x;
First code number is easy: codes only a single symbol.
Remember as pc (previous code) and x (previous word).
More Less Pseudo
while ( c = nextcode ) {
if ( c is in D ) {
y = word(c);
ww = x + first(y);
insert ww in D;
output y;
}
else {
The hard case
else {
y = x + first(x);
insert y in D;
output y;
}
pc = c;
x = y;
}
One more detail…
One detail remains: how to build the dictionary for compression (decompression is easy).
We need to be able to scan through a sequence of symbols and check if they form a prefix of a word already in the dictionary.
We use tries for dictionaries.
Tries!
a b
10 32
c d
4
a
5
d
6
d
a:0 b:1 c:2 d:3 ba:4 ad:4 dd:6
Corresponds to dictionary
Tries
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 (i.e. proceed to compress the rest of the input with an empty dictionary).
• 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.