Abstract

The monotone minimal perfect hash function (MMPHF) problem is the following indexing problem. Given a set S={s1,…,sn}S= \{s_1,\ldots,s_n\} of nn distinct keys from a universe UU of size uu, create a data structure DSDS that answers the following query: RankOp(q)=rank of q in S for all q∈S  and arbitrary answer otherwise. RankOp(q) = \text{rank of } q \text{ in } S \text{ for all } q\in S ~\text{ and arbitrary answer otherwise.} Solutions to the MMPHF problem are in widespread use in both theory and practice. The best upper bound known for the problem encodes DSDS in O(nlog⁡log⁡log⁡u)O(n\log\log\log u) bits and performs queries in O(log⁡u)O(\log u) time. It has been an open problem to either improve the space upper bound or to show that this somewhat odd looking bound is tight. In this paper, we show the latter: specifically that any data structure (deterministic or randomized) for monotone minimal perfect hashing of any collection of nn elements from a universe of size uu requires Ω(n⋅log⁡log⁡log⁡u)Ω(n \cdot \log\log\log{u}) expected bits to answer every query correctly. We achieve our lower bound by defining a graph G\mathbf{G} where the nodes are the possible (un){u \choose n} inputs and where two nodes are adjacent if they cannot share the same DSDS. The size of DSDS is then lower bounded by the log of the chromatic number of G\mathbf{G}. Finally, we show that the fractional chromatic number (and hence the chromatic number) of G\mathbf{G} is lower bounded by 2Ω(nlog⁡log⁡log⁡u)2^{Ω(n \log\log\log u)}.

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