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Dilworth theorem and Duality in graph
Zhao-H. Yin
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Review : Set Order A set π is partially ordered over βΌ if
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Chain A chain is a subset πΆβπ , in which each pair of π,π is comparable. (i.e. πβΌπ or πβΌπ or π=π holds) An antichain is a subset π΄βπ , in which each pair of π,π is non-comparable.
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Division Any nonempty set π can be divided into chains πΆ 1 , πΆ 2 ,β¦ πΆ π Any nonempty set π always contains antichain.
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Observation Size of the largest antichain in the following sets ?
The smallest number of chains dividing the set? π΄ πππ₯ =2 π΄ πππ₯ =1 π· πππ =2 π· πππ =1
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Theorem π΄ πππ₯ = π· πππ (Dilworth,1950)
For any finite partially ordered set π π΄ πππ₯ = π· πππ
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Observation Having learnt basic proving skills graph theory,
one can try to figure out the proof by himself. Do some observation first. 1. What does π΄ πππ₯ imply ? We can not add any new point π’ into the antichain π΄. β π’βπ βπ΄,β π£βπ΄, there exists a path connecting π’,π£ (Local Maximal)
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Observation 2. Easy to verify π· β₯|π΄|
By Contradiction. If βπ΄,π·,|π·|<|π΄| Applying Pigeonhole principle, β πβπ·,π,πβπ΄ s.t. π,πβπ β chain πΆβπ, π,πβπΆ
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Observation A π overlappable path cover implies a π chain division and vice versa. Inspiration: path cover β vertex cover Build a bipartite graph G to illustrate the possibilities.
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Proof Consider a vertex cover π of G, exclude the element covered by π, the remained elements forms an antichain π΄. i.e. π΄ β₯πβ π β₯πβπΌ(πΊ)
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Proof π· πππ =πβπ½ πΊ =πβπΌ πΊ β€ π΄ πππ₯ π· β₯|π΄| π· πππ = π΄ πππ₯ Q.E.D.
On the other hand, the graph is a DAG , then the minimal disjoint path cover can be calculated by π· πππ =πβπ½ πΊ =πβπΌ πΊ β€ π΄ πππ₯ Since π· β₯|π΄| it defines π· πππ = π΄ πππ₯ Q.E.D.
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Primal-Dual Schema Dual LP(ILP) ? Use LP to solve minimal vertex cover
Minimize π’βπΊ π₯ π’ Subject to π₯ π’ + π₯ π£ β₯1 (π’,π£)βπΈ(πΊ) π₯ π’ β 0,1 Dual LP(ILP) ?
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Primal-Dual Schema Use LP to solve minimal vertex cover
Minimize π’βπΊ π₯ π’ Subject to π₯ π’ + π₯ π£ β₯1 (π’,π£)βπΈ(πΊ) π₯ π’ β{0,1} Define a factor πΌ π’π£ β₯0 πΌ π’π£ π₯ π’ + π₯ π£ β₯ πΌ π’π£ (π’,π£)βπΈ(πΊ) Subject to Maximize πΌ π’π£
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Primal-Dual Schema πΌ π’π£ Maximize Subject to
If we only define edges once, i.e. πΌ π’π£ = πΌ π£π’ Maximize πΌ π’π£ Subject to Thus, we have shown that matching problem is a duality of vertex cover problem
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Primal-Dual Schema
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Primal-Dual Schema More interestingly for this problem, if we can prove Strong duality defines The SOL of the LP-Relax program is an integer Then we will induce πΌ πΊ =π½(πΊ)
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