We present an algorithm for constructing a tree to satisfy a set of lineage constraints on common ancestors. We then apply this algorithm to synthesize a relational algebra expression from a simple tableau, a problem arising in the theory of relational databases.

Previously published algorithms for finding the longest common subsequence of two sequences of length n have had a best-case running time of O(n 2 ). An algorithm for this problem is presented which has a running time of O((r + n) log n), where r is the total number of ordered pairs of positions at which the two sequences match. Thus in the worst case the algorithm has a running time of O(n 2 log n). However, for those applications where most positions of one sequence match relatively few positions in the other sequence, a running time of O(n log n) can be expected.

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