Topics in Combinatorial Optimization Orlando Lee Unicamp 28 de - PowerPoint PPT Presentation

Topics in Combinatorial Optimization Orlando Lee – Unicamp 28 de maio 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

Dependent sets A matroid is a pair M = ( E , I ) in which I ⊆ 2 E that satisfies the following properties: (I1) ∅ ∈ I . (I2) If I ∈ I and I ′ ⊆ I , then I ′ ∈ I . (I3) If I 1 , I 2 ∈ I and | I 1 | < | I 2 | , then there exists e ∈ I 2 − I 1 such that I 1 ∪ { e } ∈ I . (Independence augmenting axiom) We say that the members of I are the independent sets of M . We say that a subset of E which is not in I is dependent. Orlando Lee – Unicamp Topics in Combinatorial Optimization

Circuits A minimal dependent set of a matroid is called circuit. Question: what are the circuits in the uniform matroid U n , k ? Question: what are the circuits in a graphic matroid? (This one is easy!) Question: what are the circuits in a linear matroid? And affine matroids? Orlando Lee – Unicamp Topics in Combinatorial Optimization

Circuits Let C := C ( M ) denote the collection of circuits of a matroid M := ( E , I ). Note that if I is known, so is C . The converse is true as well: I := { I ⊆ E : there exists no C ∈ C s.t. C ⊆ I } . We want to derive a set of necessary and sufficient conditions (axioms) that a collection C must satisfy in order to be the collection of circuits of some matroid. The following conditions are obvious. (C1) ∅ �∈ C . (C2) If C 1 , C 2 ∈ C then C 1 �⊆ C 2 . Orlando Lee – Unicamp Topics in Combinatorial Optimization

Circuits Theorem. Let C be the collection of circuits of a matroid. Let C 1 , C 2 be two distinct members of C and suppose e ∈ C 1 ∩ C 2 . Then there exists a member C ∈ C such that C ⊆ ( C 1 ∪ C 2 ) − { e } . Proof. Suppose for a contradiction that ( C 1 ∪ C 2 ) − { e } does not contain a circuit, that is, is independent. By (C2), C 1 − C 2 � = ∅ . Let f ∈ C 1 − C 2 . So C 1 − f is independent. Extend C 1 − f to a maximal independent set I in C 1 ∪ C 2 . Note that f �∈ I . Since C 2 is a circuit, some element g of C 2 − C 1 is not in I . Hence | I | � | ( C 1 ∪ C 2 ) − { f , g }| = | C 1 ∪ C 2 | − 2 < | ( C 1 ∪ C 2 ) − e | . By (I3) some element of ( C 1 ∪ C 2 ) − e can be used to extend I to a larger independent set, contradciting the choice of I . Thus, ( C 1 ∪ C 2 ) − { e } contains a circuit. Orlando Lee – Unicamp Topics in Combinatorial Optimization

Circuit axioms Consider the following axioms. (C1) ∅ �∈ C . (C2) If C 1 , C 2 ∈ C then C 1 �⊆ C 2 . (C3) If C 1 , C 2 are distinct members of C and e ∈ C 1 ∩ C 2 , then there exists a member C ∈ C such that C ⊆ ( C 1 ∪ C 2 ) − { e } . Condition (C3) is known as the weak circuit (elimination) axiom. We will show that if C satisfies (C1)(C2)(C3) then C is the collection of circuits of a matroid. Orlando Lee – Unicamp Topics in Combinatorial Optimization

Circuit axioms Theorem. Let C ⊆ 2 E a collection satisfying (C1)(C2)(C3). Let I := { I ⊆ E : there exists no C ∈ C s.t. C ⊆ I } . Then M := ( E , I ) is a matroid. Proof. For convenience, call each member of C of circuit. Clearly by (C1) we have that ∅ ∈ I , so (I1) holds. If I contains no member of C and I ′ ⊆ I , then I ′ contains no member of C , so (I2) holds. Orlando Lee – Unicamp Topics in Combinatorial Optimization

Circuit axioms Claim. Let I ∈ I and let e ∈ E . Then I + e contains at most one circuit. Proof. Suppose for a contradiction that I + e contains two distinct circuits C 1 and C 2 . Clearly, e belongs to both circuits. By (C3) there exists a circuit C ⊆ ( C 1 ∪ C 2 ) − { e } . But then C ⊆ I , which is a contradiction. In particular, note that if I + e contains a circuit, then removing any element of the circuit containing e we obtain a member of I . Remark: the result above actually holds for a matroid M = ( E , I ). Orlando Lee – Unicamp Topics in Combinatorial Optimization

Circuit axioms Let us show that (I3) holds. Suppose for a contradiction that there exist I 1 , I 2 ∈ I with | I 1 | < | I 2 | such that I 1 cannot be extended to a larger member of I by adding any element of I 2 − I 1 . Choose such pair with | I 1 ∩ I 2 | maximum. There must exist e ∈ I 1 − I 2 , otherwise I 1 ⊂ I 2 and I 1 can be extended. If I 2 + e ∈ I then let f be any arbitrary element in I 2 − I 1 , otherwise let f be an element in the unique circuit of I 2 + e . In both cases, we have that I := I 2 + e − f belongs to I . Since | I 1 ∩ I | > | I 1 ∩ I 2 | , there must exist g ∈ I − I 1 such that I 1 + g ∈ I . But g ∈ I 2 − I 1 , which is a contradiction. Thus I is the independence set system of a matroid. Orlando Lee – Unicamp Topics in Combinatorial Optimization

Circuits A one-element circuit is called loop. If { e , f } is a circuit we say that e and f are parallel. A matroid which has no loop or parallel elements is called simple. Lemma. If e , f are parallel and f , g are parallel, then e , g are parallel. Proof. Note that none of e , f , g are loops because no circuit is contained in another one. Suppose for a contradiction that { e , g } is independent. Then { e , f , g } contains two circuits, which is a contradiction. Easy one: what are loops and parallel elements in a linear matroid? Orlando Lee – Unicamp Topics in Combinatorial Optimization

Bonds Let G = ( V , E ) be a connected graph. A subset B ⊆ E is a bond of G if there exists S ⊂ V , S � = ∅ , such that B = δ ( S ) and both G [ S ] and G [¯ S ] are connected. Equivalenty, a bond is a minimal cut (it is not properly contained in another cut). Let C denote the collection of bonds of G . Proposition. C is the collection of circuits of a matroid. Exercise. Prove that C satisfies the circuit axioms. A matroid constructed this way is called cographic matroid. What are the independent sets of this matroid? What are the bases? Orlando Lee – Unicamp Topics in Combinatorial Optimization

Equivalent axioms Consider the axiom. (C3’) If C 1 , C 2 are distinct members of C , e ∈ C 1 ∩ C 2 and f ∈ C 1 − C 2 , then there exists a member C ∈ C such that f ∈ C ⊆ ( C 1 ∪ C 2 ) − { e } . Condition (C3’) is called strong circuit (elimination) axiom. We claim that (C1)(C2)(C3) are equivalent to (C1)(C2)(C3’) Obviously, (C1)(C2)(C3’) ⇒ (C1)(C2)(C3) Orlando Lee – Unicamp Topics in Combinatorial Optimization

Equivalent axioms (C1)(C2)(C3) ⇒ (C1)(C2)(C3’) Suppose for a contradiction that there exist circuits C 1 , C 2 of C and elements e , f violating (C3’). Choose them so that | C 1 ∪ C 2 | is minimum. By (C3) there exists a circuit C 3 ∈ C such that C 3 ⊆ ( C 1 ∪ C 2 ) − { e } . By our choice, we have that f �∈ C 3 . Since C 3 �⊆ C 1 , there exists g ∈ C 3 − C 1 which is in C 2 . Since | C 3 ∪ C 2 | < | C 1 ∪ C 2 | , (C3) holds for C 2 , C 3 , g , e and hence, there exists a circuit C ′ ⊆ ( C 2 ∪ C 3 ) − { g } containing e . Note that f ∈ C 1 − C ′ and hence C 1 � = C ′ . Now | C 1 ∪ C ′ | < | C 1 ∪ C 2 | , e ∈ C 1 ∩ C ′ and e ∈ C 1 ∩ C ′ . So by (C3) there exists a circuit C ⊆ ( C 1 ∪ C ′ ) − { e } ⊆ ( C 1 ∪ C 2 ) − { e } containing f , which is a contradiction. Orlando Lee – Unicamp Topics in Combinatorial Optimization

Bases and fundamental circuits Let I be an independent set and let e an element such that I + e is dependent. Let C ( I , e ) denote the unique circuit of I + e . Let B denote the collection of bases of M . Recall the exchange basis axiom. (B2) If B 1 , B 2 ∈ B and x ∈ B 1 − B 2 , then there exists y ∈ B 2 − B 1 such that ( B 1 − x + y ) ∈ B . Proposition. If B 1 , B 2 ∈ B and x 2 ∈ B 2 − B 1 , then there exists x 1 ∈ B 1 − B 2 such that B 1 − x 1 + x 2 ∈ B . Proof. Consider the fundamental circuit C ( B 1 , x 2 ). This circuit cannot be contained in B 2 . So there exists some element x 1 ∈ C ( I , x 2 ) which is in B 1 . Thus B 1 − x 1 + x 2 ∈ B . Orlando Lee – Unicamp Topics in Combinatorial Optimization

Bases and fundamental circuits Theorem. (symmetric basis exchange) If B 1 , B 2 ∈ B and x 1 B 1 − B 2 , then there exists x 2 ∈ B 2 − B 1 such that B 1 − x 1 + x 2 ∈ B and B 2 − x 2 + x 1 ∈ B . Proof. Let C 2 := C ( B 2 , x 1 ). Let C be a circuit such that x 1 ∈ C ⊆ B 1 ∪ B 2 and C − B 1 ⊆ C 2 − B 1 ( ∗ ) and | C − B 1 | is minimum. (Such circuit must exist since C 2 satisfies ( ∗ ).) Note that | C − B 1 | � 1 because B 1 contains no circuit. Orlando Lee – Unicamp Topics in Combinatorial Optimization