Data Compression With The Burrows Wheeler Transform - Online Article

Introduction

In mathematics, difficult problems can often be simplified by performing a transformation on a data set. For example, digital signal processing programs use the FFT to convert sets of sampled audio data points to equivalent sets of frequency data. Pumping up the bass or applying a notch filter is then just a matter of multiplying some frequency data points by a scaling factor. Perform an inverse FFT on the resulting points, and voila, you have a new audio waveform that has been transformed according to your specifications.

The BWT is an algorithm that takes a block of data and rearranges it using a sorting algorithm. The resulting output block contains exactly the same data elements that it started with, differing only in their ordering. The transformation is reversible, meaning the original ordering of the data elements can be restored with no loss of fidelity.

The BWT is performed on an entire block of data at once. Most of today's familiar lossless compression algorithms operate in streaming mode, reading a single byte or a few bytes at a time. But with this new transform, we want to operate on the largest chunks of data possible. Since the BWT operates on data in memory, you may encounter files too big to process in one fell swoop. In these cases, the file must be split up and processed a block at a time. The demo programs that accompany this article work comfortably with block sizes of 50Kbytes up to 250 Kbytes.

For purposes of illustration, I'll start with a much smaller data set. This string contains seven bytes of data. In order to perform the BWT, the first thing we do is treat a string S, of length N, as if it actually contains N different strings, with each character in the original string being the start of a specific string that is N bytes long. (In this case, the word string doesn't have the C/C++ semantics of being a null terminated set of characters. A string is just a collection of bytes.) We also treat the buffer as if the last character wraps around back to the first.

It's important to remember at this point that we don't actually make N -1 rotated copies of the input string. In the demonstration program, we just represent each of the strings by a pointer or an index into a memory buffer.

The next step in the BWT is to perform a lexicographical sort on the set of input strings. That is, we want to order the strings using a fixed comparison function. This means we should compare using a function similar to the C functions strcmp() or memcmp(). In this high level view of the algorithm the comparison function has to be able to wrap around when it reaches the end of the buffer, so a slightly modified comparison function would be needed.

After sorting, the set of strings is arranged as shown in Figure 3. There are two important points to note in this picture. First, the strings have been sorted, but we've kept track of which string occupied which position in the original set. So, we know that the String 0, the original unsorted string, has now moved down to row 4 in the array.

Second, tag the first and last columns in the matrix with the special designations F and L, for the first and last columns of the array. Column F contains all the characters in the original string in sorted order. So our original string "DRDOBBS" is represented in F as "BBDDORS".

The characters in column L don't appear to be in any particular order, but in fact they have an interesting property. Each of the characters in L is the prefix character to the string that starts in the same row in column F.

The actual output of the BWT, oddly enough, consists of two things: a copy of column L, and the primary index, an integer indicating which row contains the original first character of the buffer B. So performing the BWT on our original string generates the output string L which contains "OBRSDDB", and a primary index of 5.

The integer 5 is found easily enough since the original first character of the buffer will always be found in column L in the row that contains S1. Since S1 is simply S0 rotated left by a single character position, the very first character of the buffer is rotated into the last column of the matrix. Therefore, locating S1 is equivalent to locating the buffer's first character position in L.

Two Obvious Questions

At this point in the exposition, there are two obvious questions. First, it doesn't seem possible that this is a reversible transformation. Generally, a sort() function doesn't come with an unsort() partner that can restore your original ordering. In fact, it isn't likely that you've ever even considered this as something you might like. And second, what possible good does this strange transformation do you ?

I'll defer the answer to the second question while I explain the reversibility of this transform. Unsorting column L requires the use of something called the transformation vector. The transformation vector is an array that defines the order in which the rotated strings are scattered throughout the rows of the matrix.

The transformation vector, T, is an array with one index for each row in column F. For a given row i, T[ i ] is defined as the row where S[ i + 1 ] is found. In Figure 3, row 3 contains S0, the original input string, and row 5 contains S1, the string rotated one character to the left. Thus, T[ 3 ] contains the value 5. S2 is found in row 2, so T[ 5 ] contains a 2. For this particular matrix, the transformation vector can be calculated to be { 1, 6, 4, 5, 0, 2, 3 }.

How the transformation vector is used to walk through the various rows. For any row that contains S[ i ], the vector provides the value of the row where S[ i + 1 ] is found.

So now we come to the core premise of the Burrows-Wheeler transform: given a copy of L, you can calculate the transformation vector for the original input matrix. And consequently, given the primary index, you can recreate S0, or the input string.

The key that makes this possible is that calculating the transformation vector requires only that you know the contents of the first and last columns of the matrix. And believe it or not, simply having a copy of L means that you do, in fact, have a copy of F as well.

Given just the copy of L, we don't know much about the state of the matrix. L, which I've moved into the first column for purposes of illustration. In this figure, F is going to be in the next column. And fortunately for us, F has an important characteristic: it contains all of the characters from the input string in sorted order. Since L also contains all the same characters, we can determine the contents of F by simply sorting L!

Now we can start working on calculating T. The character 'O' in row 0 clearly moves to row 4 in column F, which means T[ 4 ] = 0. But what about row 1? The 'B' could match up with either the 'B' in row 0 or row 1. Which do we select ?

Fortunately, the choice here is not ambiguous, although the decision making process may not be intuitive. Remember that by definition, column F is sorted. This means that all the strings beginning with 'B' in column L also appear in sorted order. Why ? They all start with the same character, and they are sorted on their second character, by virtue of their second characters appearing in column F.

Since by definition the strings in F must appear in sorted order, it means that all the strings that start with a common character in L appear in the same order in F, although not necessarily in the same rows. Because of this, we know that the 'B' in row 1 of L is going to move up to row 0 in F. The 'B' in row 6 of L moves to row 1 of F.

Once that difficulty is cleared, it's a simple matter to recover the transformation matrix from the two columns. And once that is done, recovering the original input string is short work as well. Simply applying the C++ code shown earlier does the job.

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