June 28, 2011

Wavelet Trees

In my last post I introduced a data structure called RRR, which is used to quickly answer rank queries on binary sequences, and provide implicit compression.

Today I will talk about an elegant way of answering rank queries on sequences over larger alphabets. The structure is called the Wavelet Tree, which organises a string into a hierarchy of bit vectors. A rank query has time complexity is O(log_2 A), where A is the size of the alphabet. It was introduced by Grossi, Gupta and Vitter in their 2003 paper High-order entropy-compressed text indexes [4] (see the Further Reading section for more papers). It has since been featured in many papers [1, 2, 3, 5, 6].

If you store the bit vectors in RRR sequences, it may take less space than the original sequence. Alternatively, you could store the bit vectors in the rank indexes proposed by Sadakane and Okonohara [7]. It has a different approach to compression. I will talk about it another time ;) – fortunately, I will be studying under Sadakane-san at a later date.

In a different future post, I will show how Suffix Arrays can be used to find arbitrary patterns of length P, by issuing 2P rank queries. If using a Wavelet Tree, this means a pattern search has O(P log_2 A) time complexity, that is, the size of size of the ‘haystack’ doesn’t matter, it instead depends on the size of the ‘needle’ and size of the alphabet.

Construction

A Wavelet Tree converts a string into a balanced binary-tree of bit vectors, where a 0 replaces half of the symbols, and a 1 replaces the other half. This creates ambiguity, but at each level this alphabet is filtered and re-encoded, so the ambiguity lessens, until there is no ambiguity at all.

The tree is defined recursively as follows:

Take the alphabet of the string, and encode the first half as 0, the 2nd half as 1: { a, b, c, d } would become { 0, 0, 1, 1 }
Group each 0-encoded symbol, { a, b }, as a sub-tree;
Group each 1-encoded symbol, { c, d }, as a sub-tree;
Reapply this to each subtree recursively until there is only one or two symbols left (when a 0 or 1 can only mean one thing).

For the string “Peter Piper picked a peck of pickled peppers” (spaces and a string terminator have been represented as _ and $ respectively, due to convention in the literature) the Wavelet Tree would look like this:

note: the strings aren’t actually stored, but are shown here for convenience

It has the alphabet { $, P, _, a, c, d, e, f, i, k, l, o, p, r, s, t }, which would be mapped to { 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1 }. So, for example, $ would map to 0, and r would map to 1.

The left subtree is created by taking just the 0-encoded symbols { $, P, _, a, c, d, e, f } and then re-encoding them by dividing this new alphabet: { 0, 0, 0, 0, 1, 1, 1, 1 }. Note that on the first level an e would be encoded as a 0, but now it is encoded as a 1 (it becomes a 0 again at a leaf node).

We can store the bit vectors in RRR structures for fast binary rank queries (which are needed, as described below), and compression :) In fact, since it is a balanced tree, we can concatenate each of the levels and store it as one single bit vector.

Querying

Recall from my last post that a rank query is the count of 1-bits up to a specified position. Rank queries over larger alphabets are analogous – instead of a 1, it may be any other symbol:

After the tree is constructed, a rank query can be done with log A (A = alphabet size) binary rank queries on the bit vectors – O(1) if you store them in RRR or another binary rank index. The encoding at each internal node may be ambiguous, but of course it isn’t useless – we use the ambiguous encoding to guide us to the appropriate sub-tree, and keep doing so until we have our answer.

For example, if we wanted to know rank(5, e), we use the following procedure which is illustrated below. We know that e is encoded as 0 at this level, so we take the binary rank query of 0 at position 5:

Which is 4, which we then use to indicate where to rank in the 0-child: the 4th bit (or the bit at position 3, due to 0-basing). We know to query the 0-child, since that is what e was encoded as at the parent level. We then repeat this recursively:

At a leaf node we have our answer. I would love to explain why this works, but it is fun and rewarding to think about it yourself ;)

There are also ways to provide fast select queries, but once again I will leave that up to you to research. The curious among you might also be interested in the Huffman-Shaped Wavelet Tree described by Mäkinen and Navarro [5].

Using Your New Powers for Good

Feel free to implement this yourself, but if you want to get your hands dirty right away, all-around-clever-guy Francisco Claude has made an implementation available in his Compressed Data Structure Library (libcds). If you create something neat with it be sure to report back ;)

And if you read this far, consider following me on Twitter: @alexbowe.

YaRRR Me Hearties - a post about a succinct data structure (not pirates, sorry)

This blog post will give an overview of a static bitsequence data structure known as RRR, which answers arbitrary length rank queries in O(1) time, and provides implicit compression.

As my blog is informal, I give an introduction to this structure from a birds eye view. If you want, read my thesis for a version with better markup, and follow the citations for proofs by people smarter than myself :)

My intended future posts will cover the other aspects of my thesis, including generalising RRR (for sequences over small alphabets), Wavelet Trees (which answer rank queries over bigger alphabets), and Suffix Arrays (a text index which – when combined with the above structures – can answer queries in O(P log A) time, when P is the length of the search pattern, and A is the alphabet size).

Update: I have now posted about Wavelet Trees! Check it out here.

Example Problem

Cracking the Oyster, the first column of Programming Pearls, opens with a programmer asking for advice when sorting around ten million unique seven-digit integers – phone numbers.

After some discussion, the author concludes that a bitmap should be used. If we wanted to store ten million integers, we could use an array of 32-bit integers, consuming 38 MB, or we could represent our numbers as positions on a number line.

All of these phone numbers will be within the range [0000000, 9999999]. To represent the presence of these numbers, we only need a bitmap 10⁷ bits long, about 1 MB, which would represent our number line. Then, for a bitmap M, if we want to store phone number p, we set the bit M[p] to 1. Sorting would involve setting the numbers that are present to 1, then iterating over the bitmap, printing the positions of the 1-bits – O(N) time.

In the following sections, I will detail operations that can be done on bitmaps, named rank and select, and explain how to answer rank queries in O(1) time, and implicitly compress the bitmap. Using rank and select, a compressed bitmap can be a very powerful way to store sets. This isn’t limited to just sets of numbers, all sorts of things, such as sets of graph nodes for example; A friend of mine is using succinct bitmaps to represent De Bruijn graphs, which are used in genome assembly.

Extension: Rank

Allow me to extend the problem. I want to query our simple phone number database to see how many phone numbers are allocated within the range $[0005000, 0080000]$. I could iterate over that range and update a counter whenever I encounter a 1-bit. Actually, this operation is what is known as a rank operation.

The operation rank(i) is defined as the number of set bits (1s) in the range [0, i] (or [0, i) in some papers). In the bitstring above, the answer to rank(5) is 3… This is a generalisation of the popcount operation which counts all set bits, which I have discussed before. rank(i) can be implemented by left-shifting L - i bits (where L is the length of the datatype you are using, int, long, etc) to remove the unwanted bits, then calling popcount on the resulting value. This could be done iteratively over an array if you want, but I will discuss a much faster way below.

Then, the above question can be answered as: rank(0080000) - rank(0005000 - 1). This will give us just the number of 1s between 0005000 and 0080000.

This isn’t the only place we would use a popcounts; it happens that popcounts are common enough that we want to optimise them. Check out this blog post at valuedlessons.com for a discussion and empirical comparison of several fast approaches.

RRR

As it happens, we can build a data structure for static bitmaps that answers rank queries in O(1) time, and provides implicit compression. It is what is known as a succinct data structure, which means that even though it is compressed, we don’t need to decompress the whole thing t operate on it efficiently. Sadakane (a big name in succinct data structures) gives a nice analogy in his introduction of the field, likening it to forcing dehydrated noodles apart with your chopsticks (decompression) as you are rehydrating them, but before the whole thing is fully cooked and separated. This allows you to keep some of the noodles compressed while you eat the decompressed fragment.

Since it is static it isn’t well suited for a bitmap which you want to update (although work has been done toward this), it is still really cool :)

The structure I’m referring to is named RRR. It sounds like a radio station, but it is named after its creators: Raman, Raman, Rao, from their 2002 paper Succinct indexable dictionaries with applications to encoding k-ary trees and multisets. Its a data structure I had to become intimately involved with for my honours thesis, where I extended it for sequences of larger (but still small) alphabets. If you want to answer rank queries on large alphabets, a wavelet tree might be what you are after, but that will be covered in a different blog post (or you could read my thesis!).

In my last post (Generating Binary Permutations in Popcount Order) I discussed how to compress a bitstring by replacing blocks of a certain blocksize with their corresponding pop number, and (variable length) offset into a lookup table. I briefly mentioned building an index over it to improve lookup as well.

RRR: Construction

To construct a RRR sequence we divide our bitmap into blocks, as I mentioned in my previous blog post. These are grouped in superblocks, too, which allows us to construct an index to enable O(1) rank queries. In the following image, I have fragmented the bitmap using a blocksize of b = 5, and grouped them with a superblock factor of f = 3 – so each superblock is three blocks.

First we replace the blocks with a pair of values, a class value C and offset value O, which are used together as a lookup key into a table of precomputed (small – for each possible block only) ranks – this is demonstrated in the figure below. This is the same as the previous blog post, although in that I called the “class” P. This is because the class of a block is defined as the popcount – the number of set bits – in the block: class(B) = popcount(B) for block B.

The table is shared among all your RRR sequences, and is in fact a table of tables, where C points to the first element for the ranks of a given popcount:

For this table (let’s call it G), for a given class C, the sub-table at G[C] has b Choose C entries, which correspond to all possible permutations that have a popcount of C. This means that while our C values will always be log(b + 1) (the number of bits to represent values 0, 1, 2… b – these are all possible popcount values for the blocksize), but our O values will vary in size, requiring log(b Choose C) bits (oh yeah, and of course I’m using log_2 here :)). During a query, we can use our C values to work out how many bits will follow for the O values.

Using this approach alone we get the compression, but not O(1) ranks. C is fixed width, the compression comes from O being varied width.

In order to get the O(1) ranks we use a method discussed by Munro in Tables, 1996. This is where the superblocks come in to play:

For each superblock boundary we store the global rank up to that position. We also store a prefix sum of the bits, which gives us the address to the first block in the next superblock (since it is variable length!). This allows us to not require iterating over the whole RRR sequence, but instead going straight to the required superblock. We will only need to iterate over the blocks within a superblock, so it is now bound by whatever your superblock factor is.

RRR: Querying

To calculate rank(i):

Calculate which block our index is in as i_b = i/b. (i_b is the global index of the block)
Calculate which superblock our block resides in as i_s = i_b/f. (i_s is the index of the superblock)
Set result to the sum of previous ranks at is boundary (which is pre- calculated).
Using each blocks class-offset pair (c,o) after the boundary at is, add the rank for that entire block to result. 5. Repeat previous step until we reach i_b. We then add rank(j,c) (from i_b, not the global rank) to our result,where j = i mod b, and is the position we are querying local to i_b. Our final answer is result.

Select

Select is the inverse operation to rank; it answers the question “at which position is the ith set bit?”. To tie this in with the phone numbers example, maybe we want to find out the fiftieth phone number in the set (excluding unassigned numbers). This is a way we can index just the present elements of a bitmap. It turns out select can be answered in O(1) time as well. I won’t cover select here, as my future posts (and thesis) will mainly use rank. You can read about it in the RRR paper.

Go Forth and…

Feel free to implement this (somewhat complicated) data structure yourself, or you can use a pre-rolled one by my friend Francisco Claude in his LIBCDS – Compressed Data Structure Library.

If you read this far, consider adding me to twitter :) or you may enjoy reading my post on Wavelet Trees.

Alex Bowe

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