Abstract

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 $$O({\rm min}(r {\rm log} \ell, n \ell +r) {\rm log} {\rm log} n + Sort(n))$$ time where n is the length of each sequence andr 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 wherem ≥ 3, we find the LCIS in $$O({\rm min}(mr^2, r {\rm log}\ell {\rm log}^m r)+m\cdot $$ 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 $$O(n\ell {\rm log} {\rm log} n+$$ Sort(n)) time. Our algorithm is faster when r is relatively small, e.g., for $$r < {\rm min}(n^2/({\rm log} \ell {\rm log}{\rm log} n), n\ell/{\rm log}\ell)$$ .

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