Interval order

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 countable 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, in other words as the $$(2+2)$$-free posets . Fully written out, this means that for any two pairs of elements $$a > b$$ and $$c > d$$ one must have $$a > d$$ or $$c > b$$.

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 inclusion orders on intervals on the real line (equivalently, the orders of dimension ≤ 2).

Interval orders and dimension
An important parameter of partial orders is order dimension: the dimension of a partial order $$P$$ is the least number of linear orders whose intersection is $$P$$. For interval orders, dimension can be arbitrarily large. And while the problem of determining the dimension of general partial orders is known to be NP-hard, determining the dimension of an interval order remains a problem of unknown computational complexity.

A related parameter is interval dimension, which is defined analogously, but in terms of interval orders instead of linear orders. Thus, the interval dimension of a partially ordered set $$P = (X, \leq)$$ is the least integer $$k$$ for which there exist interval orders $$\preceq_1, \ldots, \preceq_k$$ on $$X$$ with $$x \leq y$$ exactly when $$x \preceq_1 y, \ldots,$$ and $$x \preceq_k y$$. The interval dimension of an order is never greater than its order dimension.

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 $$2n$$ . These are the involutions with no so-called left- or right-neighbor nestings where, for any involution $$f$$ 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


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

The coefficient of $$t^n$$ in the expansion of $$F(t)$$ gives the number of unlabeled interval orders of size $$n$$. The sequence of these numbers begins


 * 1, 2, 5, 15, 53, 217, 1014, 5335, 31240, 201608, 1422074, 10886503, 89903100, 796713190, 7541889195, 75955177642, …