Efficient Algorithms for Locating Maximum Average Consecutive Substrings

1. Efficient Algorithms for Locating Maximum Average Consecutive Substrings Jie Zheng Department of Computer Science UC, Riverside

2. Outline • Problem Definition • Applications to Molecular Biology • Two Existing Algorithms • Open Problems

3. Definition of Problem • Given a sequence of real numbers, A = <a1, a2, …, an>, and a positive integer L≤ n, the goal is to find a consecutive substring of A of length at least L such that the average of the numbers in the subsequence is maximized.

4. Applications in Biology • Locating GC-rich Regions • Post-Processing Sequence Alignments • Annotating Multiple Sequence Alignments • Computing Ungapped Local Alignments with Length Constraints

5. Two Existing Algorithms • An O(nlogL)-time Algorithm (Yaw Ling Lin, Tao Jiang, Kun-Mao Chao, 2001) • A Linear Time Algorithm for Binary Strings (Hsueh-I Lu, 2002)

6. The O(nlogL)-time Algorithm(Yaw-Ling Lin, Tao Jiang, Kun-Mao Chao,2001) Basic Scheme: • Finding good partner of each element, i.e. for element ai, locate aj, such that the segment <ai,…, aj> has maximum average among all substrings starting from ai. • Choose the <ai,…, aj> with the maximum average among the n candidates.

7. Important Concepts • Right-Skew Sequence A sequence A = <a1, a2, …, an> is right-skew if and only if the average of any prefix < a1, a2, …, aj> is always less than or equal to the average of the remaining suffix subsequence <aj+1, aj+2,…, an>.

8. Important Concept • Decreasingly Right-Skew Partition A partition A=A1A2…Ak is decreasingly right-skew if each segment Ai of the partition is right-skew and μ(Ai) > μ(Aj) for any i < j.

9. Big Picture of Right-Skew Partition A B C D • Intuition: • If A is chosen, B must also be • If C is not chosen, D can not be, either.

10. Lemma 7(Huang): The Maximum Average Substring Can not be longer than 2L-1 • Proof If C is the maximum average substring with length ≥2L, let C= AB, where |A|≥L and |B|≥L, then the average of A or B is no less than that of C. Say μ(A) > μ(B), then μ(A) > μ(AB)

11. Main Idea of the O(nlogL)Algorithm • Compute the decreasingly right-skew partition in O(n) time. • Finding the good partner for each index costs O(logL) time.

12. Compute the decreasingly right-skew partition • Lemma 5: Every real sequence A=<a1,a2,…,an> has a unique decreasingly right-skew partition. • Lemma 6: All right-skew pointers for a length n sequence can be computed in O(n) amortized time.

13. Compute the right-skew pointers 4 9 5 30 4 9 5 30 4 9 5 30 4 9 5 30

14. Find good partner in O(logL) • Lemma 9(Bitonic): Let P be a real sequence, and A1A2…Am the decreasingly right-skew partition of a sequence A. Suppose that μ(PA1…Ak) = max{μ(PA1…Ai)|0≤i≤m} Then μ(PA1…Ai) > μ(Ai+1) if and only if i≥k.

15. What does Lemma 9 tell us? • Locating good partner can be done with binary search! To find k so that μ(PA1…Ak) = max{μ(PA1…Ai)|0 ≤ i ≤ m} We guess i and make it closer to k: • μ(PA1…Ai) >μ(Ai+1) implies i ≥ k • μ(PA1…Ai) ≤μ(Ai+1) implies i < k

16. Big Picture of Locating Good Partners L 1 L 2 1 L 2 1 3

17. Date Structure for Binary Search logL Pointer-Jumping Tables • j (k) denotes the right end-point of the kth right-skew segment . • p(0)[i] = p[i], where p[i] is right-skew pointer for i, p(k+1) [i] = min{p(k) [p(k) [i]+1], n}. 1 k  logL • The precomputation of the jumping tables takes at most O(nlogL) time.

18. The Time Complexity • Totally n phases • Each phase costs O(logL) • Overall: O(nlogL)-time

19. Crying Out for A Linear Time Algorithm!!

20. A Linear-Time Algorithm for Binary Strings(Hsueh-I Lu, 2002) • Build upon the previous algorithm • Improvements: - Considering an upper bound on the number of right-skew segments - Working simultaneously on the right-skew partitions of forward and reverse strings - Utilizing Properties of Binary Strings

21. Basic Scheme Let B = log3n and b = (loglogn)3 • Choose O(n/ logn) indices i of S such that if g(i)-i  B holds for any of such i, then g(i) can be found in O(logn) time. • Choose O(n/ loglogn) indices i of S such that if B  g(i) – i  b holds for any of such i, then g(i) can be found in O(loglogn) time. • Find g(i) for all indices i such that g(i) – i  b.

22. Denotations • A right-skew decompostion of any substring S [p, q] is a nonempty set of i indices i1,i2,…, ilso that S[i1,i2], S[i2, i3],…, S[il-1, il] are decreasingly right-skew partition of S. • Let DS (i, j) denote the right-skew decomposition of S[i, j] • If P = {p1, p2 ,…, pk, pk+1}, where p1< p2<…<pk+1, then

23. An Intuitive Observation • Right-skew pointers cannot cross A B C • By definition of right-skew segment: μ(A)  μ(B) μ(C) • Thus μ(A+B)  μ(C). • 2. By definition of decreasingly right-skew partition: • μ(A+B) > μ(C). Contradiction.

24. The Big Picture of Right-Skew Decomposition

25. Lemma 3: from the big picture • If j DS (P), then DS (j, n)  DS (P). Lemma 3 tells us that if j belongs to the right-skew decomposition of some set of indices, then its good partner will also be. Thus, we only need to search for its good partner among a limited number of indices.

26. Lemma 4: |DS(i, j)| = O((j - i)2/3)(It holds for binary strings only) • Define: A right-skew substring determined by DS(i, j) is the undividable right-skew segment. • A right-skew substring S[p, q] is long (short) if q - p l1/3 (q - p < l1/3) • Prove lemma 4 by showing that the number of long and short right-skew substrings for a binary string is O((j - i)2/3).

27. Phase 1: g(i) - i B; gR(j) - j B Define: Pshort = {p | p mod B  0 and 0  p < n}  {n} • We have |DS (Pshort)| = O (n/logn) • In this phase, we take care of index i such that i and g(i), i.e. good partner of i, are both in DS (Pshort)  DR (Pshort)

28. Phase 2: L + b < g(i) – i L+B L + b < gR(j) – j L+B Define: Ptiny = {p | p mod b  0 and 0  p < n}  {n} • We have |DS (Ptiny)| = O (n/loglogn) • In this phase, we take care of index i such that i and g(i), i.e. good partner of i, are both in DS (Ptiny)  DR (Ptiny)

29. Phase 3: g(i)-i L+b, gR(j)-j  L+b • We set up a table M whose (x,y) entry contains the index z, such that: If C is a binary string of L+b bits, x is the number of ‘1’ in the first L bits of C; y is the binary string consisting of the last b bits of C; z is the good partner of index 0 in C. Because b is relatively small, the number of possible value for x and y is linear • Looking up the table M, we can cope with the left-over case in O(n)-time.

30. Open Problems: How to extend the linear time algorithm for binary strings to arbitrary strings.

31. INTERESTED? Contact: Jie Zheng Department of Computer Science Surge Building # 350 UC, Riverside E-mail: zjie@cs.ucr.edu