Topics in Combinatorial Optimization Orlando Lee Unicamp 15 de - - PowerPoint PPT Presentation

Topics in Combinatorial Optimization

Orlando Lee – Unicamp 15 de abril de 2014

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Agradecimentos

Este conjunto de slides foram preparados originalmente para o curso T´

• picos de Otimiza¸

c˜ ao Combinat´

• ria no primeiro

semestre de 2014 no Instituto de Computa¸ c˜ ao da Unicamp. Preparei os slides em inglˆ es simplesmente porque me deu vontade, mas as aulas ser˜ ao em portuguˆ es (do Brasil)! Agradecimentos especiais ao Prof. M´ ario Leston Rey. Sem sua ajuda, certamente estes slides nunca ficariam prontos a tempo. Qualquer erro encontrado nestes slide ´ e de minha inteira responsabilidade (Orlando Lee, 2014).

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Partially ordered sets (posets)

A partially ordered set or poset is a pair (S, ) where S is a set and is partial order, that is, a binary relation satisfying: (i) s s, (reflexive) (ii) if s t and t s, then s = t, (anti-symmetric) (iii) if s t and t u, then s u, (transitive) for every s, t, u ∈ S. We also write t s to denote s t. We restrict ourselves to finite posets, that is, with S finite.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Chains and antichains

A subset C of S is a chain if s t or t s for every s, t ∈ C, that is, any two elements of C are comparable. A subset A of S is an antichain if s t and t s for every s, t ∈ S with s = t, that is, any two elements of A are incomparable. Lemma 1. If C is a chain and A is an antichain, then |C ∩ A| 1.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Posets and digraphs

Given a poset (S, ) we can define a digraph D = (S, A) where (s, t) ∈ A if s t. Note that D is a transitive digraph whose only cycles are loops at each vertex (so we could just delete them). Similarly, if we have a digraph with these properties, we have a poset. A chain in S corresponds to a path (clique) in D. An antichain in S corresponds to an independent set in D (if we have deleted all the loops). We will describe two results in terms of posets, but it is useful to remember this equivalence.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Mirsky’s theorem

Theorem 1. (Mirsky, 1971) Let (S, ) be a poset. Then the minimum size of a collection of aintichains covering S is equal to the maximum size of a chain.

• Proof. It follows from Lemma 1 that the maximum cannot be

larger than the minimum. Let us show that these numbers must be equal. For an element s ∈ S define the height of s as the maximum size of a chain in S with maximum s. Let Ai denote the set of elements of S with height i. Clearly Ai is an antichain. Let k be the maximum height (= maximum size of a chain). Then A1, . . . , Ak is a collection of antichains covering S.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Some remarks

Since every subset of an antichain is also an antichain, we can actually require that the antichains in the collection are pairwise

• disjoint. So the theorem says that we can decompose S into k

antichains, where k is the maximum size of a chain of S. Notice that in the proof, we constructed a covering by disjoint antichains. In general, a covering result does not necessarily imply a decomposition result. Here, this holds because of monoticity of antichains.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Some consequences

Let D = (V , A) be a digraph. A cut δout(X) (∅ = X = V ) is a dicut if δin(X) = ∅.

• Corollary. (Vidyasankar and Younger, 1975) Let D = (V , A) be an

acyclic digraph. Then the minimum size of a collection of dicuts covering A is equal to the length of a longest path.

• Proof. Let denote the binary relation on A such that a a′ if

there exists a path in D traversing arcs a and a′ in this order. Clearly, (A, ) is a partial order.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Some consequences

Moreover, (a) if C is a maximal chain in (A, ) then C is the arc-set of a path in D (why?), and (b) if B is an antichain in (A, ) then B is contained in some dicut of D. To see that (b) holds, let T be set of tails of the arcs in B. Let X be the set of vertices that can reach T. Then δout(X) is a dicut containing B. Applying Mirsky’s theorem we obtain the desired result.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Some consequences

We also have a weighted version of Mirsky’s theorem.

• Theorem. Let (S, ) be a poset and let ω : S → Z+ be a weight
• function. Then the minimum size of of a collection of antichains

covering each s ∈ S exactly ω(s) times is equal to the maximum weight of a chain.

• Proof. Remove all elements of weight zero. Replace each

remaining element s of S by a chain of ω(s) new elements so that an element s′ of this chain is smaller than an element t′ of the chain corresponding to t ∈ S if and only if s t. This results in a poset (S′, ′).

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Some consequences

The maximum weight ω(C) of an chain C in S is equal to maximum size |C ′| of a chain C ′ in S′. By Mirsky’s theorem, S′ can be covered by a collection of |C ′|(= ω(C)) (disjoint) antichains. Replacing the elements of each antichain by their originals in S, we

• btain ω(C) antichains covering each s ∈ S exactly ω(s)

times.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Polarity

We describe next a polar (dual) version of Mirsky’s theorem in which we switch the roles of chains and antichains. It is remarkable that this phenomena occurs very often in

• combinatorics. We already have seen the following polar results.
• Theorem. (Exercise) Let D = (V , A) be a digraph and let s, t ∈ V

with s = t. Then the size of shortest st-path is equal to the maximum size of a packing of st-cuts in D.

• Theorem. (Menger) Let D = (V , A) be a digraph and let s, t ∈ V

with s = t. Then the size of a minimum st-cut is equal to the maximum size of a packing of st-paths in D.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Dilworth’s decomposition theorem

• Theorem. (Dilworth, 1950) Let (S, ) be a poset. Then the

minimum size of a collection of chains covering S is equal to the maximum size of an antichain.

• Proof. It follows from Lemma 1 that the maximum cannot be

larger than the minimum. To show that these numbers are equal we use induction on |S|. The basis case |S| = 0 is trivial. So assume that |S| > 0. Let A be a maximum antichain and let k := |A|. Define A↓ := {s ∈ S : there exists t ∈ A such that s t}, A↑ := {s ∈ S : there exists t ∈ A such that s t}. Then A↓ ∩ A↑ = A and A↓ ∪ A↑ = S. (Why?)

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Dilworth’s decomposition theorem

First suppose that A↓ = S and A↑ = S. Then A↓ and A↑ are nonempty posets whose partial orders are induced by . Morever, A is a maximum antichain in both posets. By induction, A↓ can be covered by k = |A| chains. Moreover, each of these chains contains (exactly) one element of A. For each s ∈ A, let C s denote the chain containing s. Similarly, there exist k chains C ′

s (for s ∈ A) covering A↑, where

C ′

s contains s.

Then for each s ∈ A, C s ∪ C ′

s is a chain in S. Moreover, these

chains cover S.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Dilworth’s decomposition theorem

Assume now that A↓ = S or A↑ = S for every antichain A of size

• k. This means that every antichain of size k is either the set of

minimal elements of S or the set of maximal elements of S. Choose a minimal element s and a maximal element t of S with s ≤ t. Then the maximum size of an antichain in the poset S − {s, t} is k − 1 (because every antichain of size k contains either s or t). By induction, S − {s, t} can be covered by k − 1 chains. Adding the chain {s, t} yields a covering of S by k chains.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Some consequences

• Remark. Since every subset of a chain is also a chain, we can

actually require that the chains in the cover are pairwise disjoint. So Dilworth’s (decomposition) theorem says that we can decompose S into k chains, where k is the maximum size of an antichain of S. The next result is equivalent to Dilworth’s decomposition theorem. A subset X of vertices of a digraph is path-independent if no two vertices in X belong to the same path.

• Corollary. (Gallai and Milgram) Let D = (V , A) be an acyclic
• digraph. Then the minimum size of a family of paths covering V is

equal to the maximum size of a path-independent set of vertices.

• Proof. Let (V , ) be the poset where u v if and only if v is

reachable from u in D. The result follows from Dilworth’s decomposition theorem.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Some consequences

The following is an analogue for covering the arc set by paths.

• Corollary. Let D = (V , A) be an acyclic digraph. Then the

minimum size of a collection of paths covering A is equal to the maximum size of a dicut.

• Proof. Let (A, ) be the poset where a a′ if and only if there

exists a path in D traversing a and a′ in this order. The result follows from Dilworth’s decomposition theorem.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Some consequences

We have a similar result for st-paths.

• Corollary. Let D = (V , A) be an acyclic digraph with exactly one

source s and exactly one sink t. Then the minimum size of a collection of of st-paths covering A is equal to the maximum size

• f an st-dicut.
• Proof. Let (A, ) be the poset where a a′ if and only if there

exists an st-path in D traversing a and a′ in this order. The result follows from Dilworth’s decomposition theorem.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Some consequences

If we require to cover only a subset of arcs, we have the following general result.

• Corollary. Let D = (V , A) be an acyclic digraph and let B ⊆ A.

Then the minimum number of paths covering B is equal to the maximum of |C ∩ B| where C is a dicut of D.

• Proof. Let (B, ) be the poset where a a′ if and only if there

exists a path in D traversing a and a′ in this order. For each chain K in (B, ) there exists a path in D containing K, and for each antichain L in (B, ) there exists a dicut C in D with L ⊆ C ∩ B. The result follows from Dilworth’s decomposition theorem.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

The weighted version

• Theorem. Let (S, ) be a poset and let ω : S → Z+ be a weight
• function. Then the maximum weight ω(A) of an antichain A is

equal to minimum size of a collection of chains covering each s ∈ S exactly ω(s) times.

• Proof. Remove all elements of weight zero. Replace each

remaining element s of S by an antichain with ω(s) elements so that an element s′ in this antichain is smaller than an element t′ of the antichain corresponding to t ∈ S if and only if s t. This results in a poset (S′, ′).

Orlando Lee – Unicamp Topics in Combinatorial Optimization

The weighted version

The maximum weight ω(A) of an antichain A in S is equal to maximum size |A′| of an antichain A′ in S′. By Dilworth’s decomposition theorem, S′ can be covered by a collection of |A′|(= ω(A)) (disjoint) chains. Replacing the elements of each chain by their original in S, we

• btain ω(A) chains covering each s ∈ S exactly ω(s) times.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Some consequences

We have the equivalent result for digraphs.

• Corollary. (Gallai, 1958) Let D = (V , A) be an acyclic digraph and

let S, T be subsets of V such that every vertex belongs to some ST-path. Let c : V → Z+ be a weight function. Then the minimum size of a collection of ST-paths covering each v ∈ V times is equal to the maximum weight c(X) of an ST-separator X (that is, D − X has no ST-path).

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Algorithmic aspects

The proof given here for Mirsky’s theorem can be turned easily into an efficient greedy algorithm. You could try to find it yourself. Neither Schrijver nor Frank says this result was proved by Mirsky in their books. . . I found out Mirsky’s name on Google and in some papers. The proof given here for the weighted version does not yield a polynomial algorithm, but we will describe later a simple greedy polynomial algorithm.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Algorithmic aspects

The proof of Dilworth’s decomposition theorem given here does not yield a polynomial algorithm (since it requires to find a maximum antichain). In Frank’s book, he describes a (polynomial) reduction to the bipartite matching problem (page 77). For the weighted version of the weighted Dilworth’s decomposition theorem, the proof given here does not yield an efficient algorithm. However, there exist polynomial algorithms for this problem. One possible way (I think!) is to reduce it to a MinFlow MaxCut problem (Exercise).

Orlando Lee – Unicamp Topics in Combinatorial Optimization

References

• A. Frank, Connections in Combinatorial Optimization, Oxford.

(pages 77–79)

• A. Schrijver, Combinatorial Optimization, Vol. A, Springer.

(pages 217–221)

Orlando Lee – Unicamp Topics in Combinatorial Optimization