We study the problem of finding a longest common increasing subsequence (LCIS) of multiple sequences of numbers. The LCIS problem is a fundamental issue in various application areas, including the whole genome alignment. In this paper we give an efficient algorithm to find the LCIS of two sequences in O(min(r log ℓ, nℓ+r) log log n+Sort(n)) time where n is the length of each sequence and r is the number of ordered pairs of positions at which the two sequences match, ℓ is the length of the LCIS, and Sort(n) is the time to sort n numbers. For m sequences where m ≥ 3, we find the LCIS in O(min(mr 2, r log ℓ log m r)+m·Sort(n)) time where r is the total number of m-tuples of positions at which the m sequences match. The previous results find the LCIS of two sequences in O(n 2) and O(nℓ log log n+Sort(n)) time. Our algorithm is faster when r is relatively small, e.g., for r < min(n 2 /(log ℓ log log n), nℓ / log ℓ). 1

1 Introduction Algorithms that search for the longest common subsequence (LCS) of twoinput sequences or the longest increasing subsequence (LIS) of one input sequence date back several decades.

Given a string S = {a1, a2, a3, · · · , an}, the longest increasing subsequence (LIS) problem is to find a subsequence of the given string such that the subsequence is increasing and its length is maximal. In a previous result, to find the longest increasing subsequences of each sliding windows with a fixed size w of a given string can be solved in O(w log log n + OUT P UT) time, where O(w log log n) is taken for preprocessing. In this paper, we solve the problem for finding the longest increasing subsequence of every substring of S. With the straightforward implementation of the previous result, the time required for the preprocessing would be O(n 2 log log n). With the modification of the data structure used in the algorithm, our algorithm needs only O(n 2) preprocessing time. Since there are O(n 2) substrings totally in a given string with length n, our algorithm is optimal. The time required for the reporting stage is linear to the size of the output. In other words, our algorithm can find the LIS of every substring in O(n 2 +OUT P UT) time, where OUTPUT is the sum of the lengths of the outputs. Key words: longest increasing subsequence, substring, sliding window, row tower 1

www.elsevier.com/locate/ipl