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Fusion Trees. Advanced Data Structures Aris Tentes. 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

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fusion trees

Fusion Trees

Advanced Data Structures

Aris Tentes

slide2
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
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
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
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.
slide6
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!!
slide7
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.
slide8
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 ( )
slide9
If we find such an m, then we compute

which is long.

  • Note that k sketches fit in a word.
slide10
Can we find such an 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
slide11
The set of the sketched keys of a node is denoted by S(K)
  • Def.: We define the sketch of an entire node as follows:
lemma
Lemma
  • 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.
slide13
Using the previous lemma, we can reduce the computation of 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 rank sketch y in s k
Finding the rank(Sketch(y)) in S(K)
  • Firstly, we compute
  • Then the substraction
  • And finally
  • Observing that

.

slide15
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).
lemma1
Lemma
  • 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.
slide17

msb(x)

  • 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.
slide18

First Phase

  • 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).
slide19
Let be the set of the first 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.
slide20
The position of the most significant one in lead(x) is f=sb’
  • To extract the desired block we multiply by and right justify the significant portion.
second phase
Second Phase
  • 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
Conclusions
  • 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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