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## PowerPoint Slideshow about ' Fusion Trees' - willem

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Goal

Fixed Universe Successor Problem

- We have a set of n numbers
- Each number has a length of at most log u bits (u=size of the fixed Universe)
- We want to perform the following actions:
- Predecessor/Successor
- Insertion/Deletion
in time better than O(log n)

Model

Transdichotomous RAM

- Memory is composed of words
- Each word has a length of w=logu
- Each item we store must fit in a word
- The following operations require constant time:
- Addition, Subtraction
- Multiplication, Division
- AND, OR, XOR
- left/right Shift
- Comparison

Main Idea

A fusion tree is a B-tree with fan-out and, therefore, has a height of

If we find a way to determine, where a query fits among the B keys of a node in constant time, then we have an solution to our problem

In the Nodes

- Suppose that the keys (K) in a node are
- If we view them in a binary tree then we have the following picture:
- The black nodes are the branching nodes.
- For k keys, there are exactly k-1 branching nodes.
- However, some of them may be in the same level.
- Thus, less than k bits are required to distinguish the ‘s.

- We construct the set B(K) with the branching levels (namely the bit positions required to distinguish the keys)
- Let with and
- Def. :PerfectSketch(x)= the extracted bits according to B(K) of x. Namely, the bits of x, which correspond to the positions
- If we collect the perfect sketches of all k keys, then we are able to reduce the node representation to k r-bit strings.
- That means that bits would be efficient. Less than a word!!

- However, computing PerfectSketch(x) is difficult. Therefore, we compute an approximation, called Sketch(x).
- Sketch(x) contains the samebits with PerfectSketch(x), in the same order with some extra 0’s in between, but in consistent positions.
- This is done by multiplying x by a number m, which we will see later how we choose it.

- Firstly, we compute leaving only the bits which correspond to B(K).
- If then we observe that
- All we need is to find an m such that:
- All are distinct (no collisions)
- (to preserve order)
- are concentrated in a small range ( )

- If we find such an m, then we compute bits which correspond to
which is long.

- Note that k sketches fit in a word.

- Can we find such an bits which correspond to m?
- Firstly, we show how to find such that whenever
- Suppose we have found with the desired property.
- We observe that implies
- Thus we can choose to be the least residue not represented among the fewer than residues of the form
- Then, by adding suitable values of we obtain the final values of mi

- The set of the sketched keys of a node is denoted by S(K) bits which correspond to
- Def.: We define the sketch of an entire node as follows:

Lemma bits which correspond to

- Suppose y is an arbitrary number and xi an element of S (the set of keys). Let be the elements of B(S) and m-1 the most significant bit position in which PerfectSketch(y) and PerfectSketch(xi) differ.
- Assume that p>bm is the most significant position in which y and xi differ.
- Then the rank(y) in S is uniquely determined by the interval containing p and the relative order between y and xi.

- Using the previous lemma, we can reduce the computation of bits which correspond to rank(y) in K to computing rank(Sketch(y)) in K(S).
- Having computed rank(Sketch(y)), we have determined the predecessor and successor Sketch(xi) and Sketch(xi+1) of Sketch(y) in K(S).
- If xi≤y≤xi+1, then we are done.
- Else we pick the one (from the sketched ones) with the longest prefix of significant bits with Sketch(y) and apply the previous lemma.
- Use of a look up table.

Finding the bits which correspond to rank(Sketch(y)) in S(K)

- Firstly, we compute
- Then the substraction
- And finally
- Observing that
.

- Suitable multiplication sums these ones and gives the desired rank.
- What remains is to find a way to compute in constant time, the most significant bit, in which two numbers u,v differ.
- We can easily see that this problem is reduced to the problem of finding the most significant bit of u XOR v.
- We want to compute msb(x).

Lemma desired rank.

- We call a number x d-sparse if the positions of its one bits belong to a set of the form Not all these positions have to be occupied by ones.
- If x is d-sparse, then there exist constants y,y’, such that for z=(yx)ANDy’ the i’th bit of z equals the bit in the position of a+di of x. Namely, z is a perfect compression of x.

msb(x) desired rank.

- At first consider a partitioning of the w bits of our word x into consecutive blocks of bits. The computation is divided into two phases.
- We find the leftmost block containing a one and we extract this block
- We find the leftmost one in this extracted block.

First Phase desired rank.

- Let be the number, which has ones precisely in the leftmost position of each block, namely and
- We compute lead(x)= the leftmost bit of each block is one iff x contains a one in this block. It is given by
- We observe that lead(x) is d-sparse, so we can apply the previous lemma and obtain compress(x).

- Let be the set of the first desired rank.b/s powers of two.
- We compute b’=rank(compress(x)) in P, in the same way as before.
- Note that b’ identifies the block number (counting from the right ) of the leftmost block of x containing a one.

- The position of the most significant one in desired rank.lead(x) is f=sb’
- To extract the desired block we multiply by and right justify the significant portion.

Second Phase desired rank.

- We want to find the position of the leftmost one in the extracted block.
- As before, we do a rank computation of these s bits with the first s powers of two.
- Now we have all the information needed to compute msb(x)

Conclusions desired rank.

- In the static case, the problem of successor and predecessor, is clear to be solvable in time, since this is the height of our B-tree and the computation in each node requires constant time (the data we need is precomputed)
- In the dynamic case, the total time to update a node is
- The amortized time for insertion/deletion in a B-tree is constant.Therefore, sorting requires

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