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Dilworth's Chain Covering Theorem and its Dual

In this section, we prove the following theorem of R.P. Dilworth, which is truly one of the classic results of combinatorial mathematics.

Dilworth's Chain Covering Theorem and its Dual

In this section, we prove the following theorem of R.P. Dilworth, which is truly one of the classic results of combinatorial mathematics.

Before proceeding with the proof of Dilworth's theorem later in this section, we pause to discuss the dual version for partitions into antichains, as it is even easier to prove.

Theorem: Dual of Dilworth's Theorem

If \(\PXP\) is a poset and \(\height(P)=h\), then there exists a partition \(X=A_1\cup A_2\cup\dots\cup A_h\), where \(A_i\) is an antichain for \(i=1,2,\dots,h\). Furthermore, there is no partition using fewer antichains.

Proof

For each \(x\in X\), let \(\height(x)\) be the largest integer \(t\) for which there exists a chain \[\begin{aligned}\end{aligned}\] with \(x=x_t\). Evidently, \(\height(x)\le h\) for all \(x\in X\). Then for each \(i=1,2,\dots,h\), let \(A_i=\{x\in X:\height(x)=i\}\). It is easy to see that each \(A_i\) is an antichain, as if \(x,y\in A_i\) are such that \(x\lt y\), then there is a chain \(x_1\lt x_2\lt \cdots\lt x_i=x \lt x_{i+i}=y\), so \(\height(y)\ge i+1\). Since \(\height(P)=h\), there is a maximum chain \(C=\{x_1,x_2,\dots,x_h\}\). If it were possible to partition \(\bfP\) into \(t\lt h\) antichains, then by the Pigeon Hole Principle, one of the antichains would contain two points from \(C\), but this is not possible.

When \(\PXP\) is a poset, a point \(x\in X\) with \(\height(x)=1\) is called a minimal point of \(\bfP\). We denote the set of all minimal points of a poset \(\PXP\) by \(\min(X,P)\).Since we use the notation \(\bfP= (X,P)\) for a poset, the set of minimal elements can be denoted by \(\min(\bfP)\) or \(\min(X,P)\). This convention will be used for all set valued and integer valued functions of posets.

The argument given for the proof of yields an efficient algorithm, one that is defined recursively. Set \(\bfP_0= \bfP\). If \(\bfP_i\) has been defined and \(\bfP_i\neq \emptyset\), let \(A_i=\min(\bfP_i)\) and then let \(\bfP_{i+1}\) denote the subposet remaining when \(A_i\) is removed from \(\bfP_i\).

In , we illustrate the antichain partition provided by this algorithm for the \(17\) point poset from . The darkened points form a chain of size\(5\).

Maximal antichains and maximum antichains are defined analogously.

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

Nou jy Geen sakrekenaar los hierdie een nie, maar die stukke daarvan is kon nie wees kondifiseerbaar. Probeer een hieronder, of tik jou eie.

Artikel (1)

Probeer elkeen eers op papier. Onthul die antwoord om te toets; bevestigde persone kan oopgemaak word in die oplosder vir elke stap.

Consider implementing an algorithm from on the poset of . First, verify that you agree with the labeling of the points with their heights, which assigns the points to antichains. Explain why this poset is a good illustration of why finding the chain of five points is best done by starting with a point labeled \(5\) and working down instead of starting with a point labeled \(1\) and working up.

Hou jou eie werk

'n Vry rekening voeg notas by oor elke les, 'n verslag van wat jy klaar gemaak het, jou opgelos probleme op een plek, en' n tugmeester wat jy kan vra oor hierdie bladsy. Die wiskunde self is oop vir almal, geteken in of nie.

Meld op Meld aan

Simbools wat hier gebruik word

Maak enige simbool van die volledige definisie,'n prent en wat elke letter daarin beteken.

Vrae wat mense vra

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.

Dele van hierdie bladsy word aangepas Keller & Trotter, Applied Combinatorics (CC BY-SA 4.0). Verklaar en weer opklaar hier, foute is ons s'n.

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