The Complexity of the Partial Order Dimension Problem - Closing the Gap

TL;DR

This paper proves that determining whether a height-2 partial order has dimension 3 is NP-complete, resolving a long-standing open problem in the complexity of partial order dimension.

Contribution

It establishes the NP-completeness of the dimension 3 problem for height-2 partial orders, closing a major gap in the complexity classification.

Findings

Proves the NP-completeness of the dimension 3 problem for height-2 partial orders.

Shows the equivalence of the problem to bipartite triangle containment representations.

Reduces from a known NP-hard planar satisfiability problem to establish complexity.

Abstract

The dimension of a partial order $P$ is the minimum number of linear orders whose intersection is $P$. There are efficient algorithms to test if a partial order has dimension at most $2$. In 1982 Yannakakis showed that for $k\geq 3$ to test if a partial order has dimension $\leq k$ is NP-complete. The height of a partial order $P$ is the maximum size of a chain in $P$. Yannakakis also showed that for $k\geq 4$ to test if a partial order of height $2$ has dimension $\leq k$ is NP-complete. The complexity of deciding whether an order of height $2$ has dimension $3$ was left open. This question became one of the best known open problems in dimension theory for partial orders. We show that the problem is NP-complete. Technically we show that the decision problem (3DH2) for dimension is equivalent to deciding for the existence of bipartite triangle containment representations (BTCon). This problem then allows a reduction from a class of planar satisfiability problems (P-3-CON-3-SAT(4)) which is known to be NP-hard.