# 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 ${\displaystyle P=(X,\leq )}$ is an interval order if and only if there exists a bijection from ${\displaystyle X}$ to a set of real intervals, so ${\displaystyle x_{i}\mapsto (\ell _{i},r_{i})}$, such that for any ${\displaystyle x_{i},x_{j}\in X}$ we have ${\displaystyle x_{i} in ${\displaystyle P}$ exactly when ${\displaystyle r_{i}<\ell _{j}}$. Such posets may be equivalently characterized as those with no induced subposet isomorphic to the pair of two element chains, the ${\displaystyle (2+2)}$ free posets .[1]

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

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

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

 Unsolved problem in mathematics: What is the complexity of determining the order dimension of an interval order? (more unsolved problems in mathematics)

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-hard, the complexity of determining the order dimension of an interval order is unknown.[3]

## Combinatorics

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

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

${\displaystyle F(t)=\sum _{n\geq 0}\prod _{i=1}^{n}(1-(1-t)^{i})}$.

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

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