Understanding Fusion Trees?

Question

I stumbled across the Wikipedia page for them:

Fusion tree

And I read the class notes pdfs linked at the bottom, but it gets hand-wavy about the data structure itself and goes into a lot of detail about the sketch(x) function. I think part of my confusion is that the papers are trying to be very general, and I would like a specific example to visualize.

Is this data structure appropriate for storing data based on arbitrary 32 or 64 bit integer keys? How does it differ from a B-tree? There is one section that says it's basically a B-tree with a branching factor B = (lg n)^(1/5). For a fully populated tree with 32 bit keys, B would be 2. Does this just become a binary tree? Is this data structure intended to use much longer bit-strings as keys?

My Googling didn't turn up anything terribly useful, but I would welcome any good links on the topic. This is really just a passing curiosity, so I haven't been willing to pay for the PDFs at portal.acm.org yet.

Answer 1

You've asked a number of great questions here:

1. Is a fusion tree a good data structure for storing 32-bit or 64-bit numbers? Or is it designed to store longer bitstrings?
2. How does a fusion tree differ from a B-tree?
3. A fusion tree picks b = w^1/5, where w is the machine word size. Does this mean that b = 2 on a 32-bit machine, and does that make it just a binary tree?
4. Why is so much of the discussion of a fusion tree focused on sketching?
5. Is there a visualization of a fusion tree available to help understand how the structure works?

I'd like to address each of these questions in turn.

Q1: What do you store in a fusion tree? Are they good for 32-bit integers?

Your first question was about what fusion trees are designed to store. The fusion tree data structure is specifically designed to store integers that fit into a single machine word. As a result, on a 32-bit machine, you'd use the fusion tree to store integers of up to 32 bits, and on a 64-bit machine you'd use a fusion tree to store integers of up to 64 bits.

Fusion trees are not designed to handle arbitrarily long bitstrings. The design of fusion trees, which we'll get to in a little bit, is based on a technique called word-level parallelism, in which individual operations on machine words (multiplications, shifts, subtractions, etc.) are performed to implicitly operate on a large collection of numbers in parallel. In order for these techniques to work correctly, the numbers being stored need to fit into individual machine words. (It is technically possible to adapt the techniques here to work for numbers that fit into a constant number of machine words, though.)

But before we go any further, I need to include a major caveat: fusion trees are of theoretical interest only. Although fusion trees at face value seem to have excellent runtime guarantees (O(log_w n) time per operation, where w is the size of the machine word), the actual implementation details are such that the hidden constant factors are enormous and a major barrier to practical adoption. The original paper on fusion trees was mostly geared toward proving that it was possible to surpass the Ω(log n) lower bound on BST operations by using word-level parallelism and without regard to wall-clock runtime costs. So in that sense, if your goal in understanding fusion trees is to use one in practice, I would recommend stopping here and searching for another data structure. On the other hand, if you're interested in seeing just how much latent power is available in humble machine words, then please read on!

Q2: How does a fusion tree differ from a regular B-tree?

At a high level, you can think of a fusion tree as a regular B-tree with some extra magic thrown in to speed up searches.

As a reminder, a B-tree of order b is a multiway search tree where, intuitively, each node stores (roughly) b keys. The B-tree is a multiway search tree, meaning that the keys in each node are stored in sorted order, and the child trees store elements that are ordered relative to those keys. For example, consider this B-tree node:

             +-----+-----+-----+-----+
             | 103 | 161 | 166 | 261 |
             +-----+-----+-----+-----+
            /      |     |     |      \
           /       |     |     |       \
          A        B     C     D        E

Here, A, B, C, D, and E are subtrees of the root node. The subtree A consists of keys strictly less than 103, since it's to the left of 103. Subtree B consists of keys between 103 and 161, since subtree B is sandwiched between 103 and 161. Similarly, subtree C consists of keys between 161 and 166, subtree D consists of keys between 166 and 261, and subtree E consists of keys greater than 261.

To perform a search in a B-tree, you begin at the root node and repeatedly ask which subtree you need to descend into to continue the search. For example, if I wanted to look up 137 in the above tree, I'd need to somehow determine that 137 resides in subtree B. There are two "natural" ways that we could do this search:

• Run a linear search over the keys to find the spot where we need to go. Time: O(b), where b is the number of keys in the node.
• Run a binary search over the keys to find the spot where we need to go. Time: O(log b), where b is the number of keys in the node.

Because each node in a B-tree has a branching factor of b or greater, the height of a B-tree of order b is O(log_b n). Therefore, if we use the first strategy (linear search) to find what tree to descend into, the worst-case work required for a search is O(b log_b n), since we do O(b) work per level across O(log_b n) levels. Fun fact: the quantity b log_b n is minimized when b = e, and gets progressively worse as we increase b beyond this limit.

On the other hand, if we use a binary search to find the tree to descend into, the runtime ends up being O(log b · log_b n). Using the change of base formula for logarithms, notice that

log b · log_b n = log b · (log n / log b) = log n,

so the runtime of doing lookups this way is O(log n), independent of b. This matches the time bounds of searching a regular balanced BST.

The magic of the fusion tree is in finding a way to determine which subtree to descend into in time O(1). Let that sink in for a minute - we can have multiple children per node in our B-tree, stored in sorted order, and yet we can find which two keys our element is between in time O(1)! Doing so is decidedly nontrivial and is the bulk of the magic of the fusion tree. But for now, assuming that we can do this, notice that the runtime of searching the fusion tree would be O(log_b n), since we do O(1) work times O(log_b layers) in the tree!

The question now is how to do this.

Q3: A fusion tree picks b = w^1/5, where w is the machine word size. Does this mean that b = 2 on a 32-bit machine, and does that make it just a binary tree?

For technical reasons that will become clearer later on, a fusion tree works by choosing, as the branching parameter for the B-tree, the value b = w^1/5, where w is the machine word size. On a 32-bit machine, that means that we'd pick

b = w^1/5 = (2⁵)^1/5 = 2,

and on a 64-bit machine we'd pick

b = w^1/5 = (2⁶)^1/5 = 2^6/5 ≈ 2.29,

which we'd likely round down to 2. So does that mean that a fusion tree is just a binary tree?

The answer is "not quite." In a B-tree, each node stores between b - 1 and 2b - 1 total keys. With b = 2, that means that each node stores between 1 and 3 total keys. (In other words, our B-tree would be a 2-3-4 tree, if you're familiar with that lovely data structure). This means that we'll be branching slightly more than a regular binary search tree, but not much more.

Returning to our earlier point, fusion trees are primarily of theoretical interest. The fact that we'd pick b = 2 on a real machine and barely do better than a regular binary search tree is one of the many reasons why this is the case.

On the other hand, if we were working on, say, a machine whose word size was 32,768 bits (I'm not holding my breath on seeing one of these in my lifetime), then we'd get a branching factor of b = 8, and we might actually start seeing something that beats a regular BST.

Q4: Why is so much of the discussion of a fusion tree focused on sketching?

As mentioned above, the "secret sauce" of the fusion tree is the ability to augment each node in the B-tree with some auxiliary information that makes it possible to efficiently (in time O(1)) determine which subtree of the B-tree to descend into. Once you have the ability to get this step working, the remainder of the data structure is basically just a regular B-tree. Consequently, it makes sense to focus extensively (exclusively?) on how this step works.

This is also, by far, the most complicated step in the process. Getting this step working requires the development of several highly nontrivial subroutines that, collectively, give the overall behavior.

The first technique that we'll need is a parallel rank operation. Let's return to the key question about our B-tree search: how do we determine which subtree to descend into? Let's look back to our B-tree node, as shown here:

             +-----+-----+-----+-----+
             | 103 | 161 | 166 | 261 |
             +-----+-----+-----+-----+
            /      |     |     |      \
           /       |     |     |       \
         T0       T1    T2    T3       T4

This is the same drawing as before, but instead of labeling the subtrees A, B, C, D, and E, I've labeled them T0, T1, T2, T3, and T4.

Let's imagine I want to search for 162. That should put me into subtree T2. One way to see this is that 162 is bigger than 161 and less than 166. But there's another perspective we can take here: we want to search T2 because 162 is greater than both 103 and 161, the two keys that come before it. Interesting - we want tree index 2, and we're bigger than two of the keys in the node. Hmmm.

Now, search for 196. That puts us in tree T3, and 196 happens to be bigger than 103, 161, and 166, a total of three keys. Interesting. What about 17? That would be in tree T0, and 17 is greater than zero of the keys.

This hints at a key strategy we're going to use to get the fusion tree to work:

To determine which subtree to descend into, we need to count how many keys our search key is greater than. (This number is called the rank of the search key.)

The key insight in fusion tree is how to do this in time O(1).

Before jumping into sketching, let's build out a key primitive that we'll need for later on. The idea is the following: suppose that you have a collection of small integers, where, here, "small" means "so small that lots of them can be packed into a single machine word." Through some very clever techniques, if you can pack multiple small integers into a machine word, you can solve the following problem in time O(1):

Parallel rank: Given a key k, which is a small integer, and a fixed collection of small integers x₁, ..., x_b, determine how many of the x_i's are less than or equal to k.

For example, we might have a bunch of 6-bit numbers, for example, 31, 41, 59, 26, and 53, and we could then execute queries like "how many of these numbers are less than or equal to 37?"

To give a brief glimpse of how this technique works, the idea is to pack all of the small integers into a single machine word, separated by zero bits. That number might look like this:

00111110101001011101100110100110101
0  31  0  41  0  59  0  26  0  53

Now, suppose we want to see how many of these numbers are less than or equal to 37. To do so, we begin by forming an integer that consists of several replicated copies of the number 37, each of which is preceded by a 1 bit. That would look like this:

11001011100101110010111001011100101
1  37  1  37  1  37  1  37  1  37  

Something very cool happens if we subtract the first number from this second number. Watch this:

  11001011100101110010111001011100101    1  37  1  37  1  37  1  37  1  37
- 00111110101001011101100110100110101  - 0  31  0  41  0  59  0  26  0  53
  -----------------------------------    ---------------------------------
  10001100111100010101010010110110000    1   6  0  -4  0  -12 1   9  0 -16
  ^      ^      ^      ^      ^          ^      ^      ^      ^      ^

The bits that I've highlighted here are the extra bits that we added in to the front of each number Notice that

• if the top number is greater than or equal to the bottom number, then the bit in front of the subtraction result will be 1, and
• if the top number is smaller than the bottom number, then the bit in front of the subtraction result will be 0.

To see why this is, if the top number is greater than or equal to the bottom number, then when we perform the subtraction, we'll never need to "borrow" from that extra 1 bit we put in front of the top number, so that bit will stay a 1. Otherwise, the top number is smaller, so to make the subtraction work out we have to borrow from that 1 bit, marking it as a zero. In other words, this single subtraction operation can be thought of as doing a parallel comparison between the original key and each of the small numbers. We're doing one subtraction, but, logically, it's five comparisons!

If we can count up how many of the marked bits are 1s, then we have the answer we want. This turns out to require some additional creativity to work in time O(1), but it is indeed possible.

This parallel rank operation shows that if we have a lot of really small keys - so small that we can pack them into a machine word - we could indeed go and compute the rank of our search key in time O(1), which would tell us which subtree we need to descend into. However, there's a catch - this strategy assumes that our keys are really small, but in general, we have no reason to assume this. If we're storing full 32-bit or 64-bit machine words as keys, we can't pack lots of them into a single machine word. We can fit exactly one key into a machine word!

To address this, fusion trees use another insight. Let's imagine that we pick the branching factor of our B-tree to be very small compared to the number of bits in a machine word (say, b = w^1/5). If you have a small number of machine words, the main insight you need is that only a few of the bits in those machine words are actually relevant for determining the ordering. For example, suppose I have the following 32-bit numbers:

A: 00110101000101000101000100000101
B: 11001000010000001000000000000000
C: 11011100101110111100010011010101
D: 11110100100001000000001000000000

Now, imagine I wanted to sort these numbers. To do so, I only really need to look at a few of the bits. For example, some of the numbers differ in their first bit (the top number A has a 0 there, and the rest have a 1). So I'll write down that I need to look at the first bit of the number. The second bit of these numbers doesn't actually help sort things - anything that differs at the second bit already differs at the first bit (do you see why?). The third bit of the number similarly does help us rank them, because numbers B, C, and D, which have the same first bit, diverge at the third bit into the groups (B, C) and D. I also would need to look at the fourth bit, which splits (B, C) apart into B and C.

In other words, to compare these numbers against one another, we'd only need to store these marked bits. If we process these bits, in order, we'd never need to look at any others:

A: 00110101000101000101000100000101
B: 11001000010000001000000000000000
C: 11011100101110111100010011010101
D: 11110100100001000000001000000000
   ^ ^^

This is the sketching step you were referring to in your question, and it's used to take a small number of large numbers and turn them into a small number of small numbers. Once we have a small number of small numbers, we can then use our parallel rank step from earlier on to do rank operations in time O(1), which is what we needed to do.

Of course, there are a lot of steps that I'm skipping over here. How do you determine which bits are "interesting" bits that we need to look at? How do you extract those bits from the numbers? If you're given a number that isn't in the group, how do you figure out how it compares against the numbers in the group, given that it might differ in other bit positions? These aren't trivial questions to answer, and they're what give rise to most of the complexity of the fusion tree.

Q5: Is there a visualization of a fusion tree available to help understand how the structure works?

Yes, and no. I'll say "yes" because there are resources out there that show how the different steps work. However, I'll say "no" because I don't believe there's any one picture you can look at that will cause the whole data structure to suddenly click into focus.

I teach a course in advanced data structures and spent two 80-minute lectures building up to the fusion tree by using techniques from word-level parallelism. The discussion here is based on those lectures, which go into more depth about each step and include visualizations of the different substeps (how to compute rank in constant time, how the sketching step works, etc.), and each of those steps individually might give you a better sense for how the whole structure works. Those materials are linked here:

• Part One discusses word-level parallelism, computing ranks in time O(1), building a variant of the fusion tree that works for very small integers, and computing most-significant bits in time O(1).

• Part Two explores the full version of the fusion tree, introducing the basics behind the sketching step (which I call "Patricia codes" based on the connection to the Patricia trie).

To Summarize

In summary:

• A fusion tree is a modification of a B-tree. The basic structure matches that of a regular B-tree, except that each node has some auxiliary information to speed up searching.

• Fusion trees are purely of theoretical interest at this point. The hidden constant factors are too high and the branching factor too low to meaningfully compete with binary search trees.

• Fusion trees use word-level parallelism to speed up searches, commonly by packing multiple numbers into a single machine word and using individual operations to simulate parallel processing.

• The sketching step is used to reduce the number of bits in the input numbers to a point where parallel processing with a machine word is possible.

• There are lecture slides detailing this in a lot more depth.

Hope this helps!

Answer 2

I've read (just a quick pass) the seminal paper and seems interesting. It also answers most of your questions in the first page.

You may download the paper from here

HTH!