1. 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
2019/2/21
Chen and Chao

2. An example – Locating GC-rich regions (1)
One reasonable scoring expression to measure the richness of a region is x-p×l , where x is the C+G 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-p×l
2019/2/21
Chen and Chao

3. An example – Locating GC-rich regions (2)
Let x be the C+G count of the region, and y be the A+T count of the region
Hence, we have
x-p×l = x-p×(x+y) = (1-p)×x - p×y
Therefore, to calculate the value of x-p×l, one can assign
w(G)= w(C)=1-p
w(A)=w(T)=-p
2019/2/21
Chen and Chao

4. 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.
<5, -5.1, 1, 3, -4, 2, 3, -4, 7>
With greatest total sum = 8
Zero prefix-/suffix-sums are possible.
2019/2/21
Chen and Chao

5. 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
<5, -5.1, 1, 3, -4, 2, 3, -4, 7>
Minimum
Maximum
2019/2/21
Chen and Chao

6. Range Maximum-Sum Segment Query Problem
Definition:
The input is a sequence <a1,a2, …… an> 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.
2019/2/21
Chen and Chao

7. 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
2019/2/21
Chen and Chao

8. 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
2019/2/21
Chen and Chao

9. Our Results
We propose an algorithm that runs in O(n) preprocessing time and O(1) query time under the unit-cost RAM model.
In fact, we show that RMSQ and RMQ are computationally linearly equivalent.
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
2019/2/21
Chen and Chao

10. Strategy Reduce the RMSQ to the RMQ problem
Theorem. If there is a <f(n), g(n)>-time solution for the RMQ problem, then there is a <f(n)+O(n), g(n)+O(1)>-time solution for the RMSQ problem.
O(n)
RMSQ
RMQ
O(1)
2019/2/21
Chen and Chao

11. Computing sum(i,j) in O(1) time
prefix-sum(i) = a1+a2+…+ai
all n prefix sums are computable in O(n) time.
sum(i, j) = prefix-sum(j) – prefix-sum(i-1) i j
prefix-sum(j)
prefix-sum(i-1)
2019/2/21
Chen and Chao

12. Find the highest point here Find the lowest point here
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
2019/2/21
Chen and Chao

13. Find the highest point here Find the lowest point here
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
2019/2/21
Chen and Chao

14. 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
2019/2/21
Chen and Chao

15. Dealing with the Special Case: Single 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
2019/2/21
Chen and Chao

16. 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
2019/2/21
Chen and Chao

17. Find a partner within this region
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
2019/2/21
Chen and Chao

18. Our First Attempt (2) Step 2: Record sum(partner(i), i) in an array i
Lowest point
sum(partner(i), i)
2019/2/21
Chen and Chao

19. Applying RMQ to this sequence The maximum-sum pair can be retrieved
Our First Attempt (3)
Step 3: Apply the RMQ techniques to the array i
Applying RMQ to this sequence
Querying this interval
The maximum-sum pair can be retrieved
Lowest point
sum(partner(i), i)
2019/2/21
Chen and Chao

20. 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)
2019/2/21
Chen and Chao

21. Find the nearest point at least as large as prefix-sum(i)
A Better Partner
How?
Prefix-sum sequence
Find the nearest point at least as large as prefix-sum(i) i
Left_bound(i)
Find the lowest point
New partner(i)
2019/2/21
Chen and Chao

22. 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.
2019/2/21
Chen and Chao

23. Why Is It Better? (2) -- Remains the Best
Find the nearest point at least as large as prefix-sum(i) i
Left_bound(i)
Find the lowest point
partner(i)
2019/2/21
Impossible region
Chen and Chao

24. Why Is It Better? (3) -- Minimal-Maximal Property
Height(partner(i))< Height(j) < Height(i), for all partner(i)< j< i
Next higher point
Maximal point
Minimal point i
partner(i)
No one higher than i
No one lower than partner(i)
2019/2/21
Chen and Chao

25. 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
2019/2/21
Chen and Chao

26. Why Is It Better? (5) -- Nesting Property
For two indices i < j, it cannot be the case that partner(i)<partner(j) ≦i<j
Maximal point i j
Minimal point
Minimal point
Maximal point
partner(j)
partner(i)
2019/2/21
Chen and Chao

27. 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
2019/2/21
Chen and Chao

28. 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).
2019/2/21
Chen and Chao

29. 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.
Compare (new_partner(i), i) and (partner(j), j)
Can not be the right end of the maximum-sum segment.
Nesting property
2019/2/21
Chen and Chao

30. Time Complexity RMSQ can be reduced to the RMQ problem in O(n) time
Since under the unit-cost RAM model, there is a <O(n), O(1)>-time solution for the RMQ problem, there is a <O(n), O(1)>-time solution for the RMSQ problem.
Preprocessing: O(n)
RMQ
RMSQ
Query: O(1)
2019/2/21
Chen and Chao

31. RMQ  RMSQ
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 or simply a negative number larger than the maximum number of the sequence.)
RMQ Instance: 
RMSQ Instance: or
RMSQ Instance:
2019/2/21
Chen and Chao

32. 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
2019/2/21
Chen and Chao

33. Problem 1:The 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
2019/2/21
Chen and Chao

34. 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-U+1, i-L+1] and end point is i i
RMSQ query L U
Runs in O(n) time since each query costs O(1) time
2019/2/21
Chen and Chao

35. Problem 2: All Maximal-Sum Segments
Ruzzo and Tompa [ISMB 1999] gave a O(n)-time algorithm for this problem.
Recursive definition.
L(S)
R(S) S
2019/2/21
Chen and Chao

36. Problem 2: Finding All Maximal Scoring Subsequences
Recursive calls.
Input sequence:
L(S)
R(S) S
RMSQ query
Runs in O(n) time since each query costs O(1) time
2019/2/21
Chen and Chao