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´

• 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

Dependent sets

A matroid is a pair M = (E, I) in which I ⊆ 2E 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 Un,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 ⊆ 2E 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

• f 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 C3 ⊆ 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 B1, B2 ∈ B and x ∈ B1 − B2, then there exists y ∈ B2 − B1 such that (B1 − x + y) ∈ B.

• Proposition. If B1, B2 ∈ B and x2 ∈ B2 − B1, then there exists

x1 ∈ B1 − B2 such that B1 − x1 + x2 ∈ B.

• Proof. Consider the fundamental circuit C(B1, x2). This circuit

cannot be contained in B2. So there exists some element x1 ∈ C(I, x2) which is in B1. Thus B1 − x1 + x2 ∈ B.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Bases and fundamental circuits

• Theorem. (symmetric basis exchange) If B1, B2 ∈ B and

x1B1 − B2, then there exists x2 ∈ B2 − B1 such that B1 − x1 + x2 ∈ B and B2 − x2 + x1 ∈ B.

• Proof. Let C 2 := C(B2, x1). Let C be a circuit such that

x1 ∈ C ⊆ B1 ∪ B2 and C − B1 ⊆ C 2 − B1 (∗) and |C − B1| is minimum. (Such circuit must exist since C 2 satisfies (∗).) Note that |C − B1| 1 because B1 contains no circuit.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Bases and fundamental circuits

We claim that |C − B1| = 1. Suppose for a contradiction that |C − B1| > 1. Let x ∈ C − B1 and let C 1 := C(B1, x). By the choice of C, we have that x1 ∈ C 1. By the strong circuit axiom there exists a circuit C ′ ⊆ (C 1 ∪ C) − {x} containing x1. This contradicts the choice of C. So |C − B1| = 1. Let x2 be the unique element in C − B1. Thus C = C(B1, x2) and x1 ∈ C. Furthermore, x2 ∈ C(B2, x1). Therefore, x1 and x2 are the desired elements.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

References

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

J.G. Oxley, Matroid Theory, Oxford.

Orlando Lee – Unicamp Topics in Combinatorial Optimization