# Interval order

(Redirected from Interval dimension)

In mathematics, especially order theory, the interval order for a collection of intervals on the real line is the partial order corresponding to their left-to-right precedence relation—one interval, I1, being considered less than another, I2, if I1 is completely to the left of I2. More formally, a poset $P = (X, \leq)$ is an interval order if and only if there exists a bijection from $X$ to a set of real intervals, so $x_i \mapsto (\ell_i, r_i)$, such that for any $x_i, x_j \in X$ we have $x_i < x_j$ in $P$ exactly when $r_i < \ell_j$. Such posets may be equivalently characterized as those with no induced subposet isomorphic to the pair of two element chains, the $(2+2)$ free posets .[1]

The subclass of interval orders obtained by restricting the intervals to those of unit length, so they all have the form $(\ell_i, \ell_i + 1)$, is precisely the semiorders.

The complement of the comparability graph of an interval order ($X$, ≤) is the interval graph $(X, \cap)$.

Interval orders should not be confused with the interval-containment orders, which are the containment orders on intervals on the real line (equivalently, the orders of dimension ≤ 2).

## Interval dimension

The interval dimension of a partial order can be defined as the minimal number of interval order extensions realizing this order, in a similar way to the definition of the order dimension which uses linear extensions. The interval dimension of an order is always less than its order dimension,[2] but interval orders with high dimensions are known to exist. While the problem of determining the order dimension of general partial orders is known to be NP-complete, the complexity of determining the order dimension of an interval order is unknown.[3]

## Combinatorics

In addition to being isomorphic to $(2+2)$ free posets, unlabeled interval orders on $[n]$ are also in bijection with a subset of fixed point free involutions on ordered sets with cardinality $2n$ .[4] These are the involutions with no left or right neighbor nestings where, for $f$ an involution on $[2n]$, a left nesting is an $i \in [2n]$ such that $i < i+1 < f(i+1) < f(i)$ and a right nesting is an $i \in [2n]$ such that $f(i) < f(i+1) < i < i+1$.

Such involutions, according to semi-length, have ordinary generating function [5]

$F(t) = \sum_{n \geq 0} \prod_{i = 1}^n (1-(1-t)^i)$.

Hence the number of unlabeled interval orders of size $n$ is given by the coefficient of $t^n$ in the expansion of $F(t)$.

1, 2, 5, 15, 53, 217, 1014, 5335, 31240, 201608, 1422074, 10886503, 89903100, 796713190, 7541889195, 75955177642, … (sequence A022493 in OEIS)