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

Independent 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.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Rank function of a matroid

In linear algebra we have important concepts such as dimension of a vector space (in particular, rank of a matrix), span of a set of vectors and bases. They all can be presented in matroid terminology. Let M = (E, I) be a matroid.

• Definition. The rank of a subset X ⊆ E, denoted r(X), is the size
• f a maximal independent set contained in X.

Thus r(X) |X| for every X ⊆ E and X is independent if and

• nly if |X| = r(X). Moreover, if X ⊆ Y then r(X) r(Y ).

Let r(M) := r(E) be the rank of the matroid M.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Rank

• Lemma. The rank function r is submodular, that is,

r(X) + r(Y ) r(X ∩ Y ) + r(X ∪ Y ), for every X, Y ⊆ E.

• Proof. Let I a maximal independent set of X ∩ Y . Thus

|I| = r(X ∩ Y ). Extend I to a maximal independent set J of X ∪ Y . Then |J| = r(X ∪ Y ). The choice of I implies that J ∩ X ∩ Y = I and so |J ∩ X| + |J ∩ Y | = |I| + |J|. Since J ∩ X is an independent of X, we have r(X) |J ∩ X|. Similarly, r(Y ) |J ∩ Y |. Then r(X)+r(Y ) |J∩X|+|J∩Y | = |I|+|J| = r(X ∩Y )+r(X ∪Y ).

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Dimension

Compare this result with the classic theorem from linear algebra relating the dimensions of two linear subspaces.

• Theorem. If V and W are linear (sub)spaces, then

dim(V ) + dim(W ) = dim(V ∩ W ) + dim(V + W ), where V + W := {v + w : v ∈ V , w ∈ W }. Also, if X and Y are subsets of columns of a matrix, then rank(X) + rank(Y ) = rank(X ∩ Y ) + rank(X ∪ Y ). and the rank function of a linear matroid is exactly the rank of matrices.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Graphs

Consider now a graph G = (V , E) and its associated graphic matroid M. Suppose for the moment that G is connected. What is the value of r(M)? |V | − 1. Now let c(G) denote the number of components of G. What is the value of r(M)? |V | − c(G). In general, for X ⊆ E we have r(X) = |V (G[X])| − c(G[X]).

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Rank axioms

Let us try to determine which properties a set-function r : 2E → Z+ must satisfy to be a rank function of a matroid. (R1) r(∅) = 0 (zero on the empty set), (R2) r(X) r(Y ) for X ⊆ Y ⊆ E (non-decreasing), (R3) r(X) |X| (sub-cardinal), (R4) r(X) + r(Y ) r(X ∩ Y ) + r(X ∪ Y ) for every X, Y ⊆ E (submodular). We have seen that a rank function of a matroid satisfies all these conditions.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Rank axioms

• Theorem. A set-function r : 2E → Z+ is the rank function of a

matroid if and only if it satisfies (R1)(R2)(R3)(R4).

• Proof. We have shown that the conditions are necessary. Let us

prove the sufficiency. Let I := {X ⊆ E : r(X) = |X|}. We will show that M := (E, I) is a matroid. First we prove the following: (R3’) r(A + e) r(A) + 1 for every A ⊆ E and every e ∈ E − A. This follows from submodularity: r(A) + 1 r(A) + r(e) r(A ∩ {e}) + r(A ∪ {e}) r(A + e).

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Rank axioms

• Lemma. Let A ⊆ E and e1, . . . , ek ∈ E − A. If r(A + ei) = r(A)

for i = 1, . . . , k, then r(A ∪ {e1, . . . , ek}) = r(A). (In other words, if the addition of elements cannot individually increase the rank of A, then neither can their simultaneous addition.)

• Proof. We use induction on k. If k = 1 the result is obvious. So

suppose that k 2 and the result holds for k − 1. Let A′ := A ∪ {e1, . . . , ek−1}. By the induction hypothesis, we have r(A′) = r(A). From (R2) and (R4) we obtain r(A) + r(A) = r(A + ek) + r(A′) r((A + ek) ∩ A′) + r((A + ek) ∪ A′) = r(A) + r(A ∪ {e1, . . . , ek}) r(A) + r(A), and the result follows.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Rank axioms

(R1) implies that (I1) holds. To prove that (I2) holds, let X ⊆ Y ∈ I with r(Y ) = |Y | (that is, Y ∈ I). By repeated applications of (R3’) we have r(Y ) r(X) + |Y − X| and hence r(X) |X|. By (R3) it follows that r(X) = |X| and so X ∈ I. This proves (I2). Let us show that (I3’) holds, that is, every maximal subset of a set X which is in I has the same size. Let X ⊆ E. Take a maximal subset I of X which is in I (so r(I) = |I|). We claim that |I| = r(X). Indeed, the maximality of I implies that I + e ∈ I for any e ∈ X − I. So r(I) r(I + e) |I| = r(I). So equality holds throughout and by the previous lemma we have |I| = r(I) = r(X). Thus every maximal subset of X and is in I has the same size r(X). So (I3’) holds.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Other concepts

A subset X ⊆ E is closed if r(X + e) > r(X) for every e ∈ E − X. A closed set is also called flat. The complement of a closed set is

• pen. The ground set E is closed and ∅ is open. A closed set with

rank r(M) − 1 is called hyperplane. The closure cl(X) of a subset X consists of the elements whose addition to X does not increase its rank. We say that cl(X) is spanned or generated by X. Equivalently, cl(X) is the unique largest superset of X that has the same rank of X. Or, cl(X) is the intersection of all closed sets containing X.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Other concepts

For a graph G = (V , E) let M := M(G) denote the graphic matroid associated with G. What are the closed sets of M? What are the hyperplanes of M? What is a closure of a subset X ⊆ E?

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Exercises

1) Show that closed sets are closed under intersection. 2) Let I a maximal independent set of X. Prove that X is closed if and only if I + e is independent for every e ∈ E − X. Let G = (V , E) be a graph and let P a partition of V . The border

• f P is the set of edges that connect vertices in distinct members
• f P.

3) Consider a graphic matroid M := M(G) where G = (V , E). Show that a subset F ⊆ E is open if and only if it is the border of some partition of V .

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Maximal and minimal

Let F a collection of sets. Let max F denote the collection of the maximal sets in F and let min F denote the collection of minimal sets in F. For a matroid M let I(M), B(M), C(M) and D(M) denote the collections of independent sets, bases, circuits and dependent sets

• f M. Then

B(M) = max I(M) and C(M) = min D(M) − {∅}.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Deletion

• Deletion. Let M := (E, I) be a matroid and S ⊆ E. The matroid

M − S is the matroid with ground set E − S and independence set system: I′ := {I − S : I ∈ I}. We say that M \ e is the matroid obtained from M by deleting S. (it is easy to see that I′ satisfies the independence axioms.) When S = {e} we denote M \ e. The restriction of M to S is the matroid M \ (E − S).

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Deletion

What does it mean to delete an element/set in a graphic matroid? And in a linear matroid? Consider the uniform matroid Un,k and let S be a S-element set of the ground set. Then Un,k \ T ≡ Un−s,n−s if n − k s n Un−s,k if s < n − k.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Deletion

• Theorem. Let M = (E, I) a matroid and let S ⊆ E. Then

(a) I(M \ S) = {I − S : I ∈ I(M)} (definition), (b) B(M \ S) = max{B − S : B ∈ B(M)}, (c) C(M \ S) = {C ∈ E − S : C ∈ C(M)}, (d) r M\S(X) = r M(X) for every X ⊆ E − S, (e) clM\S(X) = clM(X) − S for every X ⊆ E − S.

• Exercise. Prove the theorem.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Contraction

The contraction operation for matroids is a little more complicated to understand. Let us begin with a graphic matroid. Let G = (V , E) be a graph and let S ⊆ E. Let G ′ := G/S obtained from G by contracting the edges in S. Let us describe the circuits (rather than independent sets = forests) of G ′ in terms of circuits

• f G.

Let C ′ be a circuit of of G ′. If we uncontract the edges of S then we see that there exists some circuit of G such that C ′ = C − S. The converse is not necessarily true: if C is a circuit in G then C − S might not correspond to a circuit in G ′ (for example, if C has a chord which is in S), but it contains a circuit of G ′. So a set X is dependent (contains a circuit) in G ′ if and only if there exists a circuit C in G such that C − S ⊆ X. So C(M(G ′)) = min{C − S : C ∈ C(M)} − {∅}.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Contraction

• Contraction. Let M := (E, I) be a matroid and S ⊆ E. The

matroid M/S is the matroid with ground set E − S and circuit set system: C′ := min{C − S : C ∈ C} − {∅}. We say that M/e is the matroid obtained from M by contracting

• S. When S = {e} we denote M/e.
• Exercise. Prove that C′ satisfies the circuit axioms.

Probably this is not the easiest way to define contraction, but I find it a little more intuitive because of the example of graphic matroids.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

A simple lemma

We say that a maximal independent subset of S is a basis of S.

• Lemma. Let M = (E, I) be a matroid. Let S ⊆ E and I ⊆ E − S

be an independent set. Suppose that S has a basis B such that B ∪ I ∈ I. Then for every basis B′ of S, we have that B′ ∪ I ∈ I.

• Proof. Clearly B ∪ I is a basis of S ∪ I. Let B′ be an arbitrary

basis of S. Extend B′ to a basis I ′ of S ∪ I. This set must have size |B ∪ I| and cannot have any element of S − B′ because B′ is a basis of S. So B′ ∪ I is independent.

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Contraction

• Lemma. Let M := (E, I) be a matroid and S ⊆ E. Then

I(M/S) = {I ⊆ E − S : S has a basis B s.t. B ∪ I ∈ I}.

• Proof. Let I′ denote the set on the right-side. Let I ∈ I′. So there

exists a basis B of S such that B ∪ I ∈ I. Suppose for a contradiction that there exists C ′ ∈ C(M/S) such that C ′ ⊆ I. So there exists a circuit C of M such that C ′ = C − S. Extend S ∩ C to a basis B′ of S. By the previous lemma, B′ ∪ I ∈ I, that is, it is independent in M. But C is contained in this set which is a

• contradiction. So I is independent in M/S and hence I′ ⊆ I(M).

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Contraction

Now let I ∈ I(M/S). Let B a basis of S. If B ∪ I is not independent in M, then there exists a circuit C contained in B ∪ I. But then C − S is a circuit of M/S contained in I, which is a contradiction. Consider the uniform matroid Un,k and let S be a S-element set of the ground set. Then Un,k/T ≡ Un−s,0 if k s n Un−s,k−s if s < k. What does it mean to contract an element/set in a linear matroid?

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Contraction

• Lemma. Let M := (E, I) be a matroid and S ⊆ E. Then

r M/S(X) = r M(X ∪ S) − r M(S) for every X ⊆ E − S.

• Proof. Let r denote the rank in M. Let I be a maximal

independent set of X in M/S. By the previous result, S has a basis B such that B ∪ I is independent in M. We claim that B ∪ I is a basis of X ∪ S. Since B is a basis of S, it is not possible extend B ∪ I by adding an element of S − B. On the other end, it is not possible extend B ∪ I by adding an element x of X − I since I + x would be an independent set of X in M/S, contradicting the choice of I. So r(X ∪ S) = |B ∪ I| and r(S) = |B|. Therefore r M/S(X) = |I| = |B ∪ I| − |B| = r(X ∪ S) − r(S).

Orlando Lee – Unicamp Topics in Combinatorial Optimization

Contraction

• Theorem. Let M = (E, I) a matroid and let S ⊆ E. Then

(a) I(M/S) = {I ⊆ E − S : S has a basis B s.t. B ∪ I ∈ I}, (b) B(M/S) = max{I ⊆ E − S : S has a basis B s.t. B ∪ I ∈ I}, (c) C(M/S) = min{C − S : C ∈ C} − {∅}, (d) r M/S(X) = r M(X ∪ S) − r(S) for every X ⊆ E − S, (e) clM/S(X) = clM(X ∪ S) − S for every X ⊆ E − S.

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