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Dilworth's Theorem for Interval Orders

As remarked previously, we do not yet have an efficient process for determining the width of a poset and a minimum partition into chains. For interval orders, there is indeed a simple way to find both.

Dilworth's Theorem for Interval Orders

As remarked previously, we do not yet have an efficient process for determining the width of a poset and a minimum partition into chains. For interval orders, there is indeed a simple way to find both. The explanation is just to establish a connection with coloring of interval graphs as discussed in .

Let \(\PXP\) be an interval order and let \(\{[a_x,b_x]:x\in X\}\) be intervals of the real line so that \(x\lt y\) in \(\bfP\) if and only \(b_x\lt a_y\). Then let \(\bfG\) be the interval graph determined by this family of intervals. Note that if \(x\) and \(y\) are distinct elements of \(X\), then \(x\) and \(y\) are incomparable in \(\bfP\) if and only if \(xy\) is an edge in \(\bfG\). In other words, \(\bfG\) is just the incomparability graph of \(\bfP\).

Recall from that interval graphs are perfect, , \(\chi(\bfG)=\omega(\bfG)\) for every interval graph \(\bfG\). Furthermore, you can find an optimal coloring of an interval graph by applying first fit to the vertices in a linear order that respects left end points. Such a coloring concurrently determines a partition of \(\bfP\) into chains.

In fact, if you want to skip the part about interval representations, take any linear ordering of the elements as \(x_1\), \(x_2,\dots,x_n\) so that \(i\lt j\) whenever \(D(x)\) is a proper subset of \(D(y)\). Then apply First Fit with respect to chains. For example, using the \(10\) point interval order illustrated in , here is such a labeling: \[\begin{aligned}x_1 \amp=g \amp x_2 \amp=f \amp x_3 \amp=c \amp x_4 \amp=d \amp x_5 \amp=h \\ x_6 \amp=a \amp x_7 \amp=j \amp x_8 \amp=b \amp x_9 \amp=i \amp x_{10} \amp=e\end{aligned}\]

Now apply the First Fit algorithm to the points of \(\bfP\), in this order, to assign them to chains \(C_1\), \(C_2,\dots\). In other words, assign \(x_1\) to chain\(C_1\). Thereafter if you have assigned points \(x_1\), \(x_2,\dots,x_i\) to chains, then assign \(x_{i+1}\) to chain \(C_j\) where \(j\) is the least positive integer for which \(x_{i+1}\) is comparable to \(x_k\) whenever \(1\le k\le i\) and \(x_k\) has already been assigned to \(C_j\). For example, this rule results in the following chains for the interval order \(\bfP\) shown in . \[\begin{aligned}C_1 \amp = \{g,h,b\} \\ C_2 \amp = \{f,a,e\} \\ C_3 \amp = \{c,d\} \\ C_4 \amp = \{j\} \\ C_5 \amp = \{i\}\end{aligned}\] In this case, it is easy to see that the chain partition is optimal since the width of \(\bfP\) is \(5\) and \(A=\{a,b,d,i,j\}\) is a \(5\)-element antichain.

Condensed — the full section is in Keller & Trotter, Applied Combinatorics.

Practice (2)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

  1. Start applying the First Fit algorithm for chain partitioning to the interval order shown below, ordering the intervals by left endpoint (and then right endpoint if you have a tie to break). Only apply the algorithm until you assign interval \(q\) to a chain! What chain is interval \(q\) assigned to? Why?

  2. Consider the interval order with representation shown below. Can you see how to visually inspect the interval representation to find its width without needing to apply the First Fit algorithm? If so, what is the width of this interval order and explain how you came up with this. If you're not able to see how to do this, briefly describe some things you've thought about and ask a question that might help get you unstuck.

Symbols used here

x \in A,\ A \subseteq B
element of, subset
x belongs to A; every element of A is in B.
i
imaginary unit
i² = −1.
\leq,\ \geq
less/greater than or equal
Inequalities that allow equality; < and > exclude it.
n!
factorial
n × (n−1) × … × 1; the number of orderings of n things. 0! = 1.
\binom{n}{k}
binomial coefficient, "n choose k"
Number of k-element subsets of n things: n!/(k!(n−k)!).
\sum_{k=1}^{n} a_k
summation
Add a_k for k = 1 up to n.
\prod_{k=1}^{n} a_k
product
Multiply a_k for k = 1 up to n.
\emptyset,\ |A|
empty set, cardinality
The set with no elements; the number of elements of A.

Questions people ask

Permutation or combination?

Ask whether order matters. A lock code is a permutation (order matters); a hand of cards is a combination (it does not).

What is a graph in this sense?

Dots (vertices) joined by lines (edges) — not a plot. Road maps, social networks and molecules are graphs; questions like "is there a route" and "how few colours" are graph theory.

Thử đi.

Parts of this page are adapted from Keller & Trotter, Applied Combinatorics (CC BY-SA 4.0). Condensed and re-explained here; errors are ours.

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