J. Fischer, V.Heun: Improvements on the RMQ-Problem

Improvements on theRange-Minimum-Query-

Problem

Johannes FischerVolker Heun

Universität München, Institut für Informatik

J. Fischer, V.Heun: Improvements on the RMQ-Problem

Introduction

3J. Fischer, V.Heun: Improvements on the RMQ-Problem

Introduction►given: array A of size n►Task: preprocess A such that

RMQA(l,r) = argminl≤i≤r A[i]

can be answered efficientlyl r

min ⇒ return 5

1 2 3 4 5 6 7 8 9 10 11

A = 2 5 2 4 1 7 5 8 3 8 6

► Break ties to the left

1 2 3 4 5 6 7 8 9 10 11

A = 2 5 2 4 1 7 5 8 3 8 61 2 3 4 5 6 7 8 9 10 11

A = 2 5 2 4 1 7 5 8 3 8 6

l r

min ⇒ return 1

4J. Fischer, V.Heun: Improvements on the RMQ-Problem

Applications► Lowest Common Ancestors (LCA)

A

B C

I

D E

KJ

G HF

A = I D B E A J F K C G C H

0

1

2

3

H = 3 2 1 2 0 3 2 3 1 2 1 2

J

G

H = 3 2 1 2 0 3 2 3 1 2 1 2A = I D B E A J F K C G C H

C

5J. Fischer, V.Heun: Improvements on the RMQ-Problem

Applications►Longest common extensions of

strings (LCE)

t=i jabba abbax z

• RMQs on the LCP-table of suffix array

►Other applications• Document Retrieval (Muthukrishnan SODA’02)• Suffix links in ESA (Abouldhoda et al. WABI’02)• Maximum-Sum Queries (Chen/Chao ISAAC‘04)• … ⇒ basic ingredient!

6J. Fischer, V.Heun: Improvements on the RMQ-Problem

Previous Results for RMQ►Berkman/Vishkin FOCS‘89:

• Preprocessing O(n)• Query time O(1)

►Rediscovered & simplified by Bender/Farach-Colton (LATIN’00)

►Reduction Chain:RMQ ➾ LCA ➾ ±1RMQ

Cartesian Tree Euler Tour 4-Russians Trick

4-Russians Trick

►cf. suffix array vs. suffix treetext ➾ suffix tree ➾ suffix array

7J. Fischer, V.Heun: Improvements on the RMQ-Problem

Cartesian Tree►Cartesian Tree for A[1,n]:

• Root: minimal element of A[1,n] at pos i• Left Child: Cartesian Tree for A[1,i-1]• Right Child: Cartesian Tree for A[i+1,n]

1 2 3 4 5 6 7 8 9 10 11

A = 2 5 2 4 1 7 5 8 3 8 6

51

3

2 4

9117

1086

O(n2)

J. Fischer, V.Heun: Improvements on the RMQ-Problem

The New Algorithm

9J. Fischer, V.Heun: Improvements on the RMQ-Problem

Overview

► Divide A into blocks B1,…,Bn/s of sizes = log(n)/4

► Answer queries seperately• Long queries than span several blocks (O(1))• Short in-block-queries (O(1))

► return position where overall minimum occurs (O(1))

A=

s

l rB1 Bn/s

10J. Fischer, V.Heun: Improvements on the RMQ-Problem

Answering Long Queries (B/F-C’00)

►Precompute all RMQs that span 2k blocks• M[i][k] = position of min in Bi,…,Bi+2^k-1

• Filled in optimal time with Dyn. Prog.

►Query: select 2 blocks covering interval

A=B1 Bn/sBi

M[i,1]M[i,2]M[i,3]

Size of M:n/s · log(n/s)

=O(n/logn·log(n/logn)) =O(n)

M[i,0]

11J. Fischer, V.Heun: Improvements on the RMQ-Problem

Answering In-block-queries►Computing the in-block-queries for all

n/s occurring blocks is too much►Really necessary?

3 4 3 2 4 3 4 8 11 10 -5 1 -4 0

41

32

►Fact: B and B‘ have the same answers to all RMQs iff they have the same Cartesian Tree.

65 7

41

32

65 7

n/s·s2

=O(n logn)

12J. Fischer, V.Heun: Improvements on the RMQ-Problem

Answering In-block-queries►Number of unlabelled bin. trees with

n nodes: n’th Catalan number Cn

►Cn=O(4n/n3/2)

►Theorem: We can store answers to all in-block-queries in space O(n)

►Proof: O(4s/s3/2)·s2

= O(22s·s1/2) = O(2log(n)/2·log1/2n) = O(n1/2·log1/2n)

13J. Fischer, V.Heun: Improvements on the RMQ-Problem

Answering In-block-queries►One problem remains:

• For each block Bi we need to know its type in time O(s)

►Type: bijection t from arrays of size s to {0,…,Cs-1} with t(B)= t(B’)

iffB and B’ have same Cartesian Tree

►build Cartesian Tree for each block Bj

►give tree a number 0 ≤ t(Bj) < Cs

14J. Fischer, V.Heun: Improvements on the RMQ-Problem

O(n)-Algo for Cartesian Tree►Let Ti be the Cartesian Tree for B[1,i]

►Ti obtained from Ti-1 as follows:

x

y

B[x] ≤ B[i]

> B[i]

x

i

y⇒

15J. Fischer, V.Heun: Improvements on the RMQ-Problem

Computing the block type►Don’t have to calculate tree!

• just keep “rightmost path” p on stack

• compute sequence of numbers l1,…,ls:

li=# nodes deleted from p in step i

• l1,…,ls satisfies “prefix property”

0 ≤ ∑1≤k≤i lk<i

• ...because one cannot delete more elements than have been inserted…

• … and each element is removed from p at most once!

16J. Fischer, V.Heun: Improvements on the RMQ-Problem

Computing the block type►l1,…,ls with ∑1≤k≤i lk<i corresponds to

path from to in

0 0

0 1

0 2

0 3

1 1

1 2

1 3

2 2

2 3 3 3

s s 0 0

►# paths from to given byCp,q= Cp-1,q + Cp,q-1 (“ballot numbers”)

p q 0 0

Cn,n= Cn

In step i:Go up li cells, go one to the left

17J. Fischer, V.Heun: Improvements on the RMQ-Problem

Computing the block type

►Paths with greater numbers than path q: at some point above q

►⇒ add # paths from current cell before going upwards

q

18J. Fischer, V.Heun: Improvements on the RMQ-Problem

Computing the block type► Precompute ballot numbers up to

s=logn/4. For all blocks Bj:

► let S be an empty stack, push(S,-∞)► q ← s, N ← 0► for i ← 1,…, s

• while top(S)>Bj[i]- pop(S)

- N ← N + C(s-i) q

- q ← q - 1

• push(S, Bj[i])

► return N

19J. Fischer, V.Heun: Improvements on the RMQ-Problem

Summary and Outlook►Direct construction algorithm for RMQ

• no dynamic data structures• never uses more space than in the end

►not the first… see Alstrup et al. SPAA’02

►Our method can be augmented with techniques from Sadakane SODA’02 to give a succinct data structure (2n+o(n) bits) with direct construction algorithm

J. Fischer, V.Heun: Improvements on the RMQ-Problem

Any Questions?