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0 Simon Gog:Advanced Data Structures

Institute of Theoretical InformaticsAlgorithmics

Institute of Theoretical Informatics - Algorithmics

Advanced Data StructuresSimon Gog – [email protected]

KIT – University of the State of Baden-Wuerttemberg andNational Research Center of the Helmholtz Association www.kit.edu

Predecessor data structures



We want to support the following operations on a set of integers from thedomain U = [u].

insert(x) Add x to S. I.e. S′ = S ∪ x.delete(x) Delete x from S. I.e. S′ = S \ x.member(x) = |x ∈ S|predecessor(x) = maxy |y ≤ x ∧ y ∈ Ssuccessor(x) = miny |y ≥ x ∧ y ∈ S

where x is an integer in U and S the set of integers of size n stored in thedata structure.

minS = successor(0)maxS = predecessor(u − 1)

Solution know from „Algo I”: Balanced search trees. E.g. red-black trees.In all comparison based approaches at least one operation takesΩ(log n) time. Why?

Predecessor data structures



Ω(log n) bound can be beaten in the word RAM model:Memory is organized in words of b ∈ O(log u) bitsA word can be accessed in constant timeWe can address all data using one wordStandard arithmetic operations take constant time on words (i.e.addition, subtraction, division, shifts . . .)

We first concentrate on the static case: The set S is fixed. I.e. no insertand delete operations.

x-fast trie (Willard, 1982)



Conceptional: Complete binary tree of hight w = dlog ue

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Operations member/successor/prodecessor can be answered inO(w) time by traversing the tree





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x-fast trieFrom O(log u) to O(log log u)...



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Generate a perfect hash table hi for each level i (keys are the presentnodes)Note: Node numbers (represented in binary) are prefixes of keys in S




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Generate a perfect hash table hi for each level i (keys are the presentnodes)Note: Node numbers (represented in binary) are prefixes of keys in S

x-fast trieQuery time from O(log u) to O(log log u)...



Member queries can be answered by a binary search using prefixes ofthe searched key.There are w prefixes, i.e. binary search takes O(log w) or O(log log u)timeSpace: w ·O(n) words, i.e. O(n log u) bits

Predecessor/Successor queries

For each node store a pointer to the maximal/minimal leaf in itssubtreeUse a double linked list to represent leaf nodes.Solving predecessor: Search for a node v which represents thelongest prefix of x with any key in S. Two cases:

Minimum in subtree of v is larger x : Return element to the left of the leaf.Maximum in the subtree of v is smaller than x . Return maximum.

y-fast trie (Willard, 1982)Space from O(n log u) to O(n) words...



Split S into nw blocks of O(log u) elements B0,B1, . . . ,Bd n

w e−1.

maxBi < minBi+1 for 0 ≤ i < d nw e − 1

Let ri = maxBi be a representative of block Bi .Build x-fast trie over representatives.

B0 B1 . . .

Total space: nw ·O(w) + O(n) = O(n) words

y-fast trieSpace from O(n log u) to O(n) words...



Use sorted array to represent BiA member query is answered as follows

Search for successor of x in x-trie of ri ’sLet Bk be the block of the successor of xSearch in O(log w) = O(log log u) time for x in Bk

How does predecessor/successor work?

y-fast trie



Changes to make structure dynamicuse cuckoo hashing for x-fast trieuse balanced search trees of size between 1

2w and 2w for Bisrepresentative is not the maximum, but any element separating twoconsecutive groups

Summary:

Operation static y-fast trie dynamic y-fast triepred(x)/succ(x) O(log log u) w.c. O(log log u)insert(x)/delete(x) O(log log u) exp. & am.construction O(n) exp.

Van Emde Boas Trees



Conceptual bitvector B of length u with B[i ] = 1 for all i ∈ SSplit B into u/

√u blocks (blue blocks) B0,B1, ...

Set bit in R[i ] if there is at least one bit set in Bi

Also store the minimum/maximum of S

0 0 0 0Summary R

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0B= 1 1 1 1 1 1 1

1 1 1

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

min = 1max = 14

Here: u = 16

Van Emde Boas Trees



Define van Emde Boas tree (vEB tree) recursivelyI.e. use vEB to represent Bi ’s and RvEB(u) denotes the vEB tree on a universe of size uBase case: u = 2. Only one node and variables min, max.

Technicalities↑√u = 2d(log u)/2e

↓√u = 2b(log u)/2c

high(x) =⌊

x↓√u

⌋(block that contains x)

low(x) = x mod ↓√u (relative position of x in Bhigh(x))

Van Emde Boas Trees



Predecessor(x , B) (first attempt)

Let y = hight(x) and z = Predecessor (low(x),By )

If z 6= ⊥ return z + y · ↓√

uLet b = Predecessor (high(x),R)

If b 6= ⊥ return max(Bb) + b · ↓√

uReturn ⊥

Problem

Recurrence for time complexity: T (u) = 2T (√

u) + O(1)Solution (Master Theorem or drawing recursion tree): T (u) = Θ(log u)Now: avoid one of the recursive calls

Van Emde Boas Trees



Predecessor(x , B) (second attempt)

If x > max return maxLet y = high(x), if min(By ) < x return Predecessor (low(x),By )

Let b = Predecessor (high(x),R)

If b 6= ⊥ return max(Bb) + b · ↓√

uReturn ⊥

Recurrence for time complexity: T (u) = T (√

u) + O(1)Solution (Master Theorem or drawing recursion tree):T (u) = Θ(log log u)

advanced data structures - kitalgo2.iti.kit.edu/gog/3_ads.pdfvan emde boas trees 11 simon gog:...

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