Lempel-Ziv Factorization May Be Harder Than Computing All Runs

Abstract

The complexity of computing the Lempel-Ziv factorization and the set of all runs (= maximal repetitions) is studied in the decision tree model of computation over ordered alphabet. It is known that both these problems can be solved by RAM algorithms in $O(n\log\sigma)$ time, where $n$ is the length of the input string and $\sigma$ is the number of distinct letters in it. We prove an $\Omega(n\log\sigma)$ lower bound on the number of comparisons required to construct the Lempel-Ziv factorization and thereby conclude that a popular technique of computation of runs using the Lempel-Ziv factorization cannot achieve an $o(n\log\sigma)$ time bound. In contrast with this, we exhibit an $O(n)$ decision tree algorithm finding all runs in a string. Therefore, in the decision tree model the runs problem is easier than the Lempel-Ziv factorization. Thus we support the conjecture that there is a linear RAM algorithm finding all runs.Comment: 12 pages, 3 figures, submitte