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´ opicos de Otimiza¸ c˜ ao Combinat´ oria 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 A i denote the set of elements of S with height i . Clearly A i is an antichain. Let k be the maximum height (= maximum size of a chain). Then A 1 , . . . , A k 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 δ out ( X ) ( ∅ � = X � = V ) is a dicut if δ in ( 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 δ out ( 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 obtain 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
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