On the Range Maximum-Sum Segment Query Problem

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On the Range Maximum-Sum Segment Query Problem

• Kuan-Yu Chen and Kun-Mao Chao
• Department of Computer Science and Information
  Engineering,
• National Taiwan University, Taiwan

2
Outlines

• Motivations
• Problems arising from some bioinformatics
  applications
• Defining the RMSQ problem
• Our main idea
• Reducing the RMSQ problem to the Range Minima
  Query problem (RMQ)
• Conclusions and applications
• Solving some relevant problems in O(n) time

3
Applications to Biomolecular Sequence Analysis

• Locating conserved regions or GC-rich regions in
  a DNA sequence
• Assign a real number (also called scores) to each
  residue
• Looking for the maximum-sum or maximum-average
  segments
• Add length constraints or average constraints

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An example Locating GC-rich regions (1)

• One reasonable scoring expression to measure the
  richness of a region is x-pl , where x is the
  CG count of the region, l is the length of the
  region, and p is a positive ratio constant.
• The goal is to design an algorithm to report the
  region that maximizes the expression x-pl

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An example Locating GC-rich regions (2)

• Let x be the CG count of the region, and y be
  the AT count of the region
• Hence, we have
• x-pl x-p(xy) (1-p)x - py
• Therefore, to calculate the value of x-pl, one
  can assign
• w(G) w(C)1-p
• w(A)w(T)-p

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The Maximum-Sum Segment

• Also called the maximum-sum interval or the
  maximum-scoring region
• Given a sequence of numbers, the maximum-sum
  segment is simply the contiguous subsequence
  having the greatest total sum.
• lt5, -5.1, 1, 3, -4, 2, 3, -4, 7gt

With greatest total sum 8
Zero prefix-/suffix-sums are possible.
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A Relevant Problem - RMQ

• Range Minima (Maxima) Query Problem (also called
  Discrete Range Searching)
• Given a sequence of numbers, by preprocessing the
  sequence we wish to retrieve the minimum
  (maximum) value within a given querying interval
  efficiently
• lt5, -5.1, 1, 3, -4, 2, 3, -4, 7gt

Minimum
Maximum
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Range Maximum-Sum Segment Query Problem

• Definition
• The input is a sequence lta1,a2, angt of real
  numbers which is to be preprocessed.
• A query is comprised of two intervals S and E.
• Our goal is to return the maximum-sum segment
  whose starting index lies in S and end index lies
  in E.

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A Nonoverlapping Example

• Input Sequence
• 9, -10, 4, -2, 4, -5, 4, -3, 6, -11, 8, -3, 4,
  -5, 3

Total sum 6
Starting region
End region
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An Overlapping Example

• Input Sequence
• 9, -10, 4, -2, 4, -5, 4, -3, 6, -11, 8, -3, 4,
  -5, 3

Total sum 8
Starting region
End region
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Our Results

• We propose an algorithm that runs in O(n)
  preprocessing time and O(1) query time under the
  unit-cost RAM model.
• We show that the RMSQ techniques yield
  alternative O(n) time algorithms for the
  following problems
• The maximum-sum segment with length constraints
• All maximal-sum segments

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Strategy

• Reduce the RMSQ to the RMQ problem
• Theorem. If there is a ltf(n), g(n)gt-time solution
  for the RMQ problem, then there is a ltf(n)O(n),
  g(n)O(1)gt-time solution for the RMSQ problem.

O(n)
RMSQ
RMQ
O(1)
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Cumulative Sum/ Prefix Sum
prefix-sum(i) a1a2ai
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Computing sum(i,j) in O(1) time

• prefix-sum(i) a1a2ai
• all n prefix sums are computable in O(n) time.
• sum(i, j) prefix-sum(j) prefix-sum(i-1)

j
i
prefix-sum(j)
prefix-sum(i-1)
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Case 1 Nonoverlapping
Maximize
Maximize
Minimize

• sum(i, j ) prefix-sum(j) prefix-sum(i-1)
• Prefix-sum sequence
• 9, -10, 4, -2, 4, -5, 4, -3, 6, -11, 8, -3, 4,
  -5, 3

Range Minima Query
Find the highest point here
Find the lowest point here
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Case 2 Overlapping

• Some problems may occur
• Prefix-sum sequence
• 9, -10, 4, -2, 5, -5, 4, -3, 6, -11, 8, -3, 4,
  -5, 3

Negative Sum !!
Find the highest point here
Find the lowest point here
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Case 2 Overlapping

• Divide into 3 possible cases
• Prefix-sum sequence
• 9, -10, 4, -2, 4, -5, 4, -3, 6, -11, 8, -3, 4,
  -5, 3

Range Minima Query Preprocessing time
f(n) Query time g(n)
Range Minima Query Preprocessing time
f(n) Query time g(n)
Find the highest point here
Find the highest point here
What should we do?
Find the lowest point here
Find the lowest point here
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Dealing with the Special CaseSingle Range Query

• Input Sequence
• 9, -10, 4, -2, 4, -5, 4, -3, 6, -11, 8, -3, 4,
  -5, 3
• Challenge Can this special case be reduced to
  the RMQ problem?

Total sum 6
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Reduction Procedure

• Step 1. Find a partner for each index.
• Step 2. Record the sum of each pair in an array
• Step 3. Retrieve the maximum-sum pair by applying
  the RMQ techniques

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Our First Attempt (1)

• Step 1 For each index i, we define the lowest
  point preceding i as its partner
• Prefix-sum sequence

i
Lowest point
Find a partner within this region
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Our First Attempt (2)

• Step 2 Record sum(partner(i), i) in an array

i
Lowest point
sum(partner(i), i)
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Our First Attempt (3)

• Step 3 Apply the RMQ techniques to the array

i
The maximum-sum pair can be retrieved
Applying RMQ to this sequence
Querying this interval
Lowest point
sum(partner(i), i)
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Bump into Difficulties

• What if its partners go beyond the querying
  interval?

i
We might have to update every pair!
Needs to be updated
partner(i)
sum(partner(i), i)
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A Better Partner

• Prefix-sum sequence

Find the nearest point at least as large as i
i
Left_bound(i)
Find the lowest point
New partner(i)
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Why Is It Better? (1)

• It remains the best choice.
• It saves lots of update steps.
• It turns out that zero or one point needs to be
  updated.

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Why Is It Better? (2)-- Remains the Best
Find the nearest higher point
i
Left_bound(i)
Find the lowest point
partner(i)
Impossible region
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Why Is It Better? (3)-- Minimal-Maximal Property

• Height(partner(i))lt Height(j) lt Height(i), for
  all partner(i)lt jlt i

Next higher point
Maximal point
i
Minimal point
partner(i)
No one higher than i
No one lower than partner(i)
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Why Is It Better? (4)-- Save Some Updates

• Prefix-sum sequence

Next higher point
Can not be the right end of the maximum-sum
segment
Querying interval
i
partner(i)
No one higher than i
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Why Is It Better? (5)-- Nesting Property

• For two indices i lt j, it cannot be the case that
  partner(i)ltpartner(j) ?iltj

Maximal point
j
i
Minimal point
Minimal point
Maximal point
partner(j)
partner(i)
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Why Is It Better? (6)-- An example

• No overlapping is allowed
• 9, -10, 4, -2, 4, -5, 4, -3, 6, -11, 8, -3, 4,
  -5, 3
• Nesting Property
• 9, -10, 4, -2, 4, -5, 4, -3, 6, -11, 8, -3, 4,
  -5, 3

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When a Query Comes-- Case 1 No Exceeding

• The maximum pair (partner(i), i) lies in the
  querying interval

Retrieve the maximum pair
Querying interval
i
partner(i)
We are done. Output (partner(i), i).
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When a Query Comes-- Case 2 Exceeding

• The maximum pair (partner(i), i) goes beyond the
  querying interval

Retrieve the maximum pair
Retrieve the maximum pair
Querying interval
j
i
Maximal
Minimal
partner(i)
Update partner(i)
partner(j)
(Partner(i), i) is the maximum pair.
Nesting property
Can not be the right end of the maximum-sum
segment.
Compare (new_partner(i), i) and (partner(j), j)
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Time Complexity

• RMSQ can be reduced to the RMQ problem in O(n)
  time
• Since under the unit-cost RAM model, there is a
  ltO(n), O(1)gt-time solution for the RMQ problem,
  there is a ltO(n), O(1)gt-time solution for the
  RMSQ problem.
• On the other hand, RMQ can be reduced to the RMSQ
  problem in O(n) time, too. (Range Maxima Query
  For each two adjacent elements, we augment a
  negative number whose absolute value is larger
  than them.)

O(n)
RMQ
RMSQ
O(1)
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Use RMSQ Techniques to Solve Two Relevant
Problems

• 1. Finding the Maximum-Sum Segment with length
  constraints in O(n) time.
• - Y.-L. Lin, T. Jiang, K.-M. Chao, 2002
• - T.-H Fan et al., 2003
• 2. Finding all maximal scoring subsequences in
  O(n) time.
• - W. L. Ruzzo M. Tompa, 1999

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Problem 1The Maximum-Sum Segment with Length
Constraints

• Lin, Jiang, and Chao JCSS 2002 and Fan et al.
  CIAA 2003 gave O(n)-time algorithms for this
  problem.
• Length at least L, and at most U

L
U
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Problem 1 Finding the Maximum-Sum Segment with
Length Constraints

• Length at least L, at most U
• For each index i, find the maximum-sum segment
  whose starting point lies in i-U1, i-L1 and
  end point is i

i
RMSQ query
L
U
Runs in O(n) time since each query costs O(1) time
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Problem 2 All Maximal-Sum Segments

• Ruzzo and Tompa ISMB 1999 gave a O(n)-time
  algorithm for this problem.
• Recursive definition.

R(S)
L(S)
S
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Problem 2 Finding All Maximal Scoring
Subsequences

• Recursive calls.
• Input sequence

R(S)
L(S)
S
RMSQ query
Runs in O(n) time since each query costs O(1) time