Lecture note on MAS480A Matroid Theory

Donggyu Kimdonggyu@kaist.ac.kr

2019 Fall

This lecture note is based on the lecture by Prof. Oum in KAIST in 2019 Fall.

1 Week01-1, 2019.09.02. Mon

Motivations of the matroid theory are graphs and vector spaces.

Example 1.1 (Matrices). Let A = R × E matrix over a field F. Let X ⊆ E.What are the properties of A[X], the submatrix of A by taking all columns inX?

• Independency/dependency of column vectors

• (column) Basis: maximal independent set of (column) vectors

• Rank

It is fact that all bases have the same size.

Example 1.2 (Graphs). Let G = (V,E) be a graph. It is fact that all maximalsets of edges inducing no cycles have the same size. Note that such sets are edge-sets of spanning forests of G. Moreover, the size is (#vertices−#components).

Example 1.3 (Matchings in a bipartite graph). Let G be a bipartite graph with abipartition (A,B). We say that X ⊂ A is matchable to B if there is a matching(a set of edges pairwise vertex-disjoint) M covering X. It is fact that all maximalmatchable subsets of A have the same size.

Proof. Let X,Y ⊆ A be maximally matchable to B. Suppose |X| < |Y |. LetM,N be matchings covering X,Y , respectively. Note that by the maximality|X| = |M | and |Y | = |N |. Let consider H := G[M ∪N ]. Each component of His a path or a cycle. Moreover, each path of length (= #edges) ≥ 2 or cycle isalternatively labeled by M and N . A component isomorphic to K2, i.e., just anedge, is possibly labeled by both M and N . If there is an edge e ∈ N −M , thenwe can directly add it into M . It contradicts to the maximality of M . If thereis no such an edge, we can find a path P which is a component of H and its twoend-edges are contained in N since |N | > |M |. Let M ′ = M−(M∩P )+(N∩P ).It is also a matching in G, and gives X ′ ⊆ A which is matchable to B. Notethat X ( X ′, so it is a contradiction.

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If we solve problems in matroid, we can think it (matroid) as one of them(vector spaces/graphs/matchings). Then we may get idea to solve.

Definition 1.1. A matroid is a pair M = (E, I) of a finite set E and a set Iof subsets of E satisfying the following:

(I1) ∅ ∈ I,

(I2) X ∈ I, Y ⊆ X ⇒ Y ∈ I, and

(I3) X,Y ∈ I, |X| < |Y | ⇒ ∃e ∈ Y −X s.t. X ∪ {e} ∈ I.

Let us say that

• X is independent if X ∈ I,

• X is dependent if X 6∈ I,

• X is a base if X is a maximal independent set, and

• X is a circuit if X is a minimal dependent set.

The definition of matroid is possibly extended to infinite E. However, thereare no settled one. The extensions are messy and hard to handle.

See Example 1.2, an independent set is a forest of G. A base is a spanningforest. A dependent set is an edge-set containing a cycle, and a circuit is a cycle.

Theorem 1.1. For X ⊆ E, all maximal independent subset of X have the samesize.

Proof. Let A,B ∈ I with A,B ⊆ X. Suppose |A| < |B|. By the axiom (I3),there is e ∈ B−A such that A∪{e} ∈ I, which is a subset of X. It contradictsto a maximality of A.

Definition 1.2. The rank function of a matroid M is

rM (X) := (a size of a maximal independent subset of X).

It is well-deinfed by the previous theorem. The rank of a matroid is

r(M) := rM (E).

E(M) = E is the ground set of the matroid M .

Example 1.4 (Basic classes of matroids). Now we will introduce some basiccalsses of matroids.

1. Uniform matroid, Ur,n (0 ≤ r ≤ n). |E| = n. X is independent iff |X| ≤ r.It satisfies all axioms of the matroid.

2. Vector matroid (or linear matroid). A := R × E matrix over F. Xis independent iff the column vectors of A[X] are linearly independent.Here, we denote M = M(A).

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Theorem 1.2. A vector matroid is a matroid.

Proof. (I1) and (I2) are trivial. Also, (I3) is obvious by an argument of dimen-sions, but we will explain it more precisely. By a permutation of E, WMA first|X ∩ Y | columns of A correspond to X ∩ Y , next |X − Y | columns correspondto X − Y , next |Y −X| columns correspond to Y −X, and remainings corre-sponds to E − (X ∪ Y ). Note that elementary row operation does not affecton dependency of column vectors. So we can change first |X| columns to I|X|.By the independency of Y , next |Y −X| > 0 columns cannot be a zero matrix.This implies that we can find such e from these column vectors.

In this case, we say that M is representable over F or F-representable. A isa representation of M .

Similarly, we can define matroids from a graph G (Example 1.2 and 1.3).Setting E = E(G) and X ∈ I iff X is a forest, then M = (E, I) is a matroid.We call it as a graphic matroid or a cycle matroid. Assume G is bipartite witha bipartition (A,B). Setting E = A and X ∈ I iff X is matchable (to B), thenM is also a matroid. We call it as a transversal matroid or a mathcing matroid.

1.1 Whitney, 1935

Definition 1.3. Let M be a finite set. Let r be a function from 2M to Z≥0.Then we call a system (M, r) a matroid if satisfies below three axioms:

(R1) r(∅) = 0,

(R2) for any N ⊆M and e ∈M −N , r(N + e) = r(N) or = r(N) + 1, and

(R3) for any N ⊆ M and e1, e2 ∈ M − N , if r(N + e1) = r(N + e2) = r(N)then r(N + e1 + e2) = r(N).

We call r a rank function, and r(N) a rank of N .

Without doubt, it is fact that for any N ⊆ a system (N, r|2N ) is a matroid.We call it a submatroid of (M, r). When we write a matroid, we can omit arank function if it is apparent in a context.

Definition 1.4. Let define a two functions ρ, n : 2M → Z≥0 as

ρ(N) = |N |,n(N) = ρ(N)− r(N).

ρ(N) is just a cardinality of N . We call n a nullity function, and n(N) a nullityof N .

• N is independent if n(N) = 0,

• N is dependent if n(N) > 0,

• N is a base if it is a maximal independent set (M − N is called a basecomplement), and

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• N is a circuit if it is a minimal dependent set.

By the axiom (R1) and (R2), 0 ≤ r(N) ≤ ρ(N). So 0 ≤ n(N) ≤ ρ(N).

Lemma 1.3. For N ⊆ N ′(⊆M

), r(N) ≤ r(N ′) and n(N) ≤ r(N ′).

Proof. By (R2), r(N) ≤ r(N ′). Also the axiom implies that r(N ′) ≤ r(N) +ρ(N ′)− ρ(N). It is equivalent with n(N) ≤ n(N ′).

Lemma 1.4. Any subset of an independent set is independent.

Proof. Let N ⊆M be an independent set, i.e., n(N) = 0. Let N ′ ⊆ N . By theprevious theorem,

(0 ≤

)n(N ′) ≤ n(N) = 0. Therefore, n(N ′) = 0.

Theorem 1.5. N is independent iff N is contained in a base iff N does notcontain any circuit.

Proof. (1⇒2) Let N be an independent set. If N is a base, done. If not, thereis e ∈M−N such that N+e is independent. Repeat this work until get a base.It actually finishes since M is finite.

(2⇒1) By the previous lemma, done.(1⇒3) Suppose N contains a circuit C. Then 0 < n(C) ≤ n(N), which is a

contradiction.(3⇒1) We will show a contraposition. Let N be a dependent set. If N is a

circuit, done. If not, there is e ∈ N such that N − e is a dependent. Repeatthis work until get a circuit.

Theorem 1.6. A circuit is a minimal submatroid contained in no bases, i.e.,containing at least one element from each base complement. A base is a maximalsubmatroid containing no circuit. A base complement is a minimal submatroidcontaining at least one element from each circuit.

Proof. By the previous theorem, N is ‘contained in no bases’ iff it is dependent,and N is ‘containing no circuit’ iff it is independent. Then the first two state-ments are obvious. The last one is obtained by taking the complement operatorfor each set in the second statement.

We can observe the reciprocal relation ship between circuits and base com-plements. Do we derive a dual concept of a matroid from this?

Definition 1.5.∆(N,N ′) = r(N +N ′)− r(N).

Proposition 1.7.

∆(A,B + C) = ∆(A+ C,B)−∆(A+ C,C).

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Proof.

δ(A,B + C) = r(A+B + C)− r(A)

= r(A+B + C)− r(A+ C)− r(A) + r(A+ C)

= ∆(A+ C,B)−∆(A+ C,C).

Lemma 1.8.∆(N + e2, e1) ≤ ∆(N, e1).

Proof. ∆(N + e2, e1) = r(N + e1 + e2)− r(N + e2) and ∆(N, e1) = r(N + e1)−r(N). So the given statement is equivalent with

∆(N, e1 + e2) = r(N + e1 + e2)− r(N)

≤(r(N + e1)− r(N)

)+(r(N + e2)− r(N)

)= ∆(N, e1) + ∆(N, e2).

∆(N, e1 + e2) is possible 0, 1, or 2. If it is 0, nothing to do. If it is 2, we caneasily deduce that ∆(N, e1) = ∆(N, e2) = 1 since

r(N + e1)

r(N) r(N + e1 + e2)

r(N + e2)

+1

+1

+1

+1

by the axiom (R2).For ∆(N, e1 + e2) = 1, suppose that ∆(N, e1) = ∆(N, e2) = 0. Then

∆(N, e1 + e2) = 0 by (R3), which is a contradiction. Therefore, the giveninequality holds.

Before this lemma, we did not use the axiom (R3).

Lemma 1.9.∆(N +N ′, e) ≤ ∆(N, e).

Proof. Denote N ′ as e1 + e2 + . . .+ ek where k = ρ(N ′). Then inductively

∆(N +N ′, e) ≤ ∆(N + e1 + . . .+ ek−1, e) ≤ · · · ≤ ∆(N, e)

by the previous lemma.

Theorem 1.10.∆(N +N2, N1) ≤ ∆(N,N1),

or equivalently,

r(N +N1 +N2) ≤ r(N +N1) + r(N +N2)− r(N).

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Proof. Consider an induction on ρ(N1). For ρ(N1) = 1, it is a case of theprevious lemma.

∆(N +N2, N1 + e) = ∆(N +N2 + e,N1)−∆(N +N2, e)

≤ ∆(N + e,N1)−∆(N, e)

= ∆(N,N1 + e).

The equalities are guaranteed by the earlier proposition. The middle inequalityholds by the induction hypothesis.

The above theorem is equivalent with

r(N1 +N2) ≤ r(N1) + r(N2)− r(N1 ∩N2).

The interpretation of these lemmas and theorem is when N ⊆ N ′(⊆ M

),

adding elements does not increase a rank of N ′ more than a rank of N , i.e.,r(N ′ + e1 + . . . + ek) − r(N ′) ≤ r(N + e1 + . . . + ek) − r(N). This agree withsetting M as a set of column vectors for a given matrix A, and r as a dimensionof a spanning space. (Increase of a dimension after adding some vectors in abigger space is smaller than increase of a dimension after adding the vectors ina smaller space.)

2 Week01-2, 2019.09.04. Wed

Example 2.1 (Basic classes of matroids). Continue to exhibit classic examples.

3. Graphic matroid (or cycle matroid). Let G = (V,E) be a graph. LetM := M(G) be the cycle matroid of G if it is a matroid on E (i.e.,E(M) = E) where X is indep (i.e., X ∈ I) iff X contains no edge set ofa cycle of G.We say that a matroid M is graphic if M = M(G) for some graph G.

4. Transversal matroid (or matching matroid). Let G be a simple bipartitegraph with a bipartition (A,B). Letting M(E) = A, and X ∈ I iff X ⊆ Ais matchable to B.

5. Partition matroid. Let E =∐ri=1Ei, where each Ei is finite. X is inde-

pendent iff |X ∩ Ei| ≤ 1 for each i. It is obviously a matroid.One remark is that if |E1| = . . . = |Er| = 1, then it is Ur,r. Especially, wecall it a free matroid.

Remark. Not every matroid is graphic. Consider U2,4. E(U2,4) = {a, b, c, d}.Let us try to represent it as a graph. Since {a, b, c} is dependent, it is a cycle.Also, {a, b, d} is a cycle. It implies that c and d are parallel edges. So {c, d} isa cycle of length two, i.e., {c, d} 6∈ I. It is a contradiction.

Not every matroid is a binary matroid. Also, U2,4 is a counter-example.Suppose it is F2-representable. Let ea, eb, ec, ed be column vectors correspondingto a, b, c, d, respectively. Then ea + eb + ec = ea + eb + ed = 0. It deducesec + ed = 0, which is a contradiction.

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Remark. Any Ur, n is representable over a big field.

Proof. ???

Remark. There is a matroid which is not representable over any fields. V8matroid(?).

Proposition 2.1. A graphic matroid M(G) is a matroid.

Proof. (I1) and (I2) are trivial. Now consider (I3). Let X,Y ∈ I with |X| < |Y |.Let C1, C2, . . . , Cl be the components of a subgraph (V,X) of G. If Y containsan edge e joining Ci and Cj with i 6= j. Then X+ e ∈ I. So WMA there are nosuch edges. It implies that Y ⊆

∐iE(G[V (Ci)]). Since Y contains no cycles,

|Y | ≤∑i

(|V (Ei)| − 1

)= |X|.

It is a contradiction.

Proposition 2.2. A transversal matroid is a matroid

Proof. (I1) and (I3) are obvious. Showing (I3) is similar with the proof afterExample 1.3. So I will roughly describe a proof. Pick matchings M and Ncovering X and Y , respectively. WMA |M | = |X| < |Y | = |N |. ConsiderG[M∆N ]. Each component is a path or a cycle of which edges are alternativelyin M or N . Since |M | < |N |, there is a path P such that |P ∩M | < |P ∩N |.Consider M∆E(P ). It is a matching covering X ′ := X+a where a ∈ Y −X.

Proposition 2.3. Every transversal matroid is representable over a big field.

Proof. Let G be a simple bipartite graph with a bipartition (A,B). Let (A, I) bea transversal matroid satisfying X ∈ I iff X ⊆ A is matchable to B. Consider anincidence matrix of G such that rows are labeled by elements of B, and columnsare labeled by elements of A. Put 0 for each entry (b, a) where a, b are notadjacent. Put a non-zero undefined value xb,a for an entry (b, a) where a, b areadjacent. We want that X is matchable to B iff the columns of X are linearlyindependent. The last statement is equivalent that the columns of X are rank|X| iff there is a set of |X| rows inducing a |X| × |X| non-singular submatrix.Note that a square matrix is non-singular iff its determinant is non-zero.

A matching M = {a1b1, . . . , akbk} in a complete bipartite graph with abipartition (A,B) corresponds to a set of entries {(b1, a1), . . . , (bk, ak)} of aB × A incidence matrix of the complete bipartite graph. When consider anoriginal bipartite graph G, a matching M covering X with size |X| exactlycorresponds to a set of entries {(b1, a1), . . . , (bk, ak)} where each value of entryis non-zero.

Here, a non-trivial fact (we skip a proof) is that we can give non-zero xb,a’ssatisfying any two product

∏i xbi,ai and

∏j xb′j ,a′j are different over a big field F.

Moreover, we can find xb,a’s such that summations of ±∏i xbi,ai ’s are non-zero.

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This fact implies that detS =∑σ(−1)sgnσ

∏i πiσ(i) 6= 0 iff there exists a

matching covering X, where S is a square submatrix chosen in the first para-graph and πij is a value of (i, j)-entry of S. Therefore, the transversal matroid(A, I) is representable over a big field F.

What are the interesting problems in this course? Here are some motivatingquestions in the matroid theory.

• How to find a maximum weight independent set?

• How to find a largest common independent set in two matroids (on a sameground set)?

• How to find a largest set that is a union of k sets X1, . . . , Xk where Xi isindependent in the i-th matroid Mi?

• When is a matroid representable over the binary field?

• When is a matroid graphic?

• What is the relation between M(G) and M(G∗), where G is a plane graph?

The first question is related to the problem finding a maximum(or minimum)weight spanning tree in a connected graph G with a weighted edge-set. It issolved by a greedy algorithm. We will show that a matroid can be defined by astructure holding a greedy algorithm.

As mentioned in the previous lecture, we can imagine a matroid as a graphor a vector space. When we consider a matroid as a vector space, its dimensiongoes up very quickly.

Let us think about representing low-rank matroids.

Definition 2.1 (Affine geometry). Let v1, v2, . . . , vn ∈ Fm are affinely indepen-dent whenever

∑i civi = 0 with

∑i ci = 0 implies ci = 0 for all i.

Note that v1, . . . , vn ∈ Fm are affinely independent iff (1, v1), . . . , (1, vn) ∈Fm+1 are linearly independent.

Definition 2.2. Let M be an affine matroid on E if X ⊆ E is independent iffX is affinely independent.

Proposition 2.4. An affine matroid is a matroid.

Proof.

Proposition 2.5. An affine matroid over F is representable over F.

Proof. By the earlier note about relation between affine independency and lien-arly independency, an affine matroid on E = {v1, . . . , vn} ⊆ Fm is M(A), where

A =

(1 1 · · · 1v1 v2 . . . vn

),

an (m+ 1)× n matrix over F.

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The converse holds, i.e., every linear matroid is an affine matroid. (?)How to draw matroids of rank ≤ 3 on the plane?

• A dependent set of rank 2 corresponds to a line segment,

• a dependent set of rank 1 corresponds to a set of multiple points at thesame location, and

• a loop is outside of the picture.

Here, e ∈ E is a loop if {e} is dependent, i.e., it is an element of rank 0. Thename of a loop is originated from the graph theory.

Example 2.2. U2,n with n ≥ 3 is drawn on the plane by n colinear points. U3,n

is drawn on the plane by n points without line segments.Consider a drawing of points a, b, c, d, e, f such that a, b, c are colinear and

c, d, e, f are colinear. Then {a, b, f} is independent.

Example 2.3 (Fano matroid). Let

A =

1 0 0 1 0 1 10 1 0 1 1 0 10 0 1 0 1 1 1

over the binary field. Then we call M = M(A) a Fano matroid. Its drawing isnot embeddable on R2. (Moreover, not embeddable on any Rn.)

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Can we recover a matroid from a drawing on the plane? What rules areneeded?

3 Week02-1, 2019.09.09. Mon

Chapter 2. Cryptomorphisms.Remind that the three axioms of a matroid M = (E, I):

(I1) ∅ ∈ I,

(I2) X ∈ I, Y ⊆ X ⇒ Y ∈ I, and

(I3) X,Y ∈ I, |X| < |Y | ⇒ ∃e ∈ Y −X s.t. X ∪ {e} ∈ I.

We will explore several equivalent definitions of the matroid.

Chapter 2.1. Circuit (= minimally dependent set)

Proposition 3.1. If C is a collection of all circuits in a matroid, then

(C1) ∅ 6∈ C,

(C2) X,Y ∈ C, X ⊆ Y ⇒ X = Y , and

(C3) X,Y ∈ C, e ∈ X ∩ Y , X 6= Y ⇒ ∃D ∈ C s.t. D ⊆ X ∪ Y − e.

We call (C3) the (weak) circuit elimination axiom.

Proof. (C1) ⇐ (I1). (C2) is trivial from the definition (minimality). Now wewill show (C3). Suppose (C3) does not hold, i.e., there are X,Y, e such that(C3) fails. Then X ∪ Y − e is independent, and it is a maximal independentset in X ∪ Y . Since X,Y ∈ C and X 6= Y , X 6⊆ Y . So there is f ∈ X − Y .X − f is independent. Then there is a maximal independent set Z in X ∪ Y

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containing X − f . Since X is dependent, Z 6⊇ X, i.e., Z 63 f . Note that|Z| = |X ∪Y − e|

(= |X ∪Y |− 1

)by the Theorem 1.1: For X ⊆ E, all maximal

independent subset of X has the same size. It implies that Z = X ∪ Y − f , soZ ⊇ Y . It is a contradiction.

Proposition 3.2. If C is a collection of subsets of a finite set E satisfying (C1),(C2), (C3), and let I := {X ⊆ E : Y 6⊆ X,∀Y ∈ C}, then (E, I) is a matroidwith C as the collection of circuits.

Proof. (I1) ⇐ (C1). (I2) holds trivially from the construction of I. Now wewill show (I3). Suppose that (I3) does not holds, i.e., there are X,Y ∈ I with|X| < |Y | such that X + e 6∈ I for all e ∈ Y −X. Choose a pair (X,Y ) so that|X ∩Y | is maximized. Since X + e 6∈ I for e ∈ Y −X, there is Ce ∈ C such thate ∈ Ce ⊆ X + e. Note that such Ce is unique since if there are two differentCe, C

′e ∈ C such that e ∈ Ce, C ′e ⊆ X + e, then we can find D ∈ C such that

D ⊆ Ce ∪ C ′e − e ⊆ X by (C3), which is a contradiction. Since Y ∈ I, Ce 6⊆ Y .So there is

(e 6=

)f ∈ Ce − Y . Let X ′ := X + e− f . Then X ′ ∈ I since if not,

i.e., there is D ∈ C such that D ⊆ X ′ ⊂ X + e and e ∈ D, then by the earlierobservation D = Ce 3 f , which is a contradiction. Since |X ′ ∩ Y | > |X ∩ Y |(X ′ ∩ Y = X ∩ Y + e), (I3) holds for (X ′, Y ) by our choice of (X,Y ). There is(e, f 6=

)g ∈ Y −X ′ such that X ′+g = X+e+g−f ∈ I. Since also g ∈ Y −X,

there is a unique Cg ∈ C such that g ∈ Cg ⊆ X + g. Cg 3 f since if not,Cg ⊆ X ′ + g, which is a contradiction. Remind that Ce, Cg ∈ C, f ∈ Ce ∩ Cg,and Ce 6= Cg. Then by (C3) there is D ∈ C such that D ⊆ Ce∪Cg−f ⊆ X ′+g,which is a contradiction.

By the construction of I, any C ∈ C is not in I, i.e., dependent in thematroid M := (E, I). For any e ∈ C, C − e is in not in C by (C2), andmoreover, C − e ∈ I. It implies that C is a minimal dependent set, i.e., acircuit in M . Hence C is a collection of circuits (but we do not know it containsall circuits). Let D be a circuit in M . Since D 6∈ I, there is C ∈ C such thatC ⊆ D. By the minimality (both C,D are minimal dependent sets), D = C ∈ C.Therefore, C is the collection of circuits.

Chapter 2.2. Base (= maximally independent set)

Proposition 3.3. If B is a collection of all bases of a matroid, then

(B1) B 6= ∅, and

(B2) B1, B2 ∈ B, e ∈ B1 −B2 ⇒ ∃f ∈ B2 −B1 s.t. B1 − e+ f ∈ B.

We call (B2) the base exchange axiom.

Proof. (B1) ⇐ (I1). Now we will show (B2). B1 − e is an independent set withcardinality = |B1| − 1 < |B2| (remind all bases have the same size). By (I3)there is f ∈ B2 − (B1 − e) = B2 − B1 such that B1 − e + f is an independentset. Because of its cardinality, it is a base.

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Proposition 3.4. If B is a collection of subsets of a finite set E satisfying(B1) and (B2), and I := {X ⊆ E : X ⊆ B for some B ∈ B}, then (E, I) is amatroid with B as the collection of bases.

Proof. (I1)⇐ (B1). By the construction of I, (I2) is obvious. Now we will show(I3). Let X,Y ∈ I with |X| < |Y |. There are X ′, Y ′ ∈ B such that X ⊆ X ′

and Y ⊆ Y ′. By (B2), every set in B has the same size. (Let B1, B2 ∈ B with|B1| < |B2|. Choose a pair (B1, B2) so that |B1 ∩ B2| is maximized. By (B2),B1 ( B2 is impossible since for e ∈ B2 −B1 we cannot find f ∈ B1 −B2

(= ∅)

such that B2−e+f ∈ B. So there is e ∈ B1−B2. By (B2), there is f ∈ B2−B1

suc that B′1 = B1 − e + f ∈ B. |B′1| = |B1| < |B2| and |B′1 ∩ B2| > |B1 ∩ B2|,which contradicts to our choice of (B1, B2).) So |X ′| = |Y ′|. Choose a pair(X ′, Y ′) so that |X ′ ∩ Y ′| is maximized. Suppose (Y −X) ∩X ′ = ∅. Note thatX ′ is a disjoint union of X, (Y −X) ∩X ′, X ′ ∩ Y ′ −X ∪ Y , and X ′ − Y ′ −X.Similarly, Y ′ is a disjoint union of Y , (X − Y ) ∩ Y ′, X ′ ∩ Y ′ − X ∪ Y , andY ′ −X ′ − Y . Note that X ′ ∩ Y ′ −X ∪ Y = X ′ ∩ Y ′ −X by the assumption,(Y −X) ∩X ′ = ∅.

|X ′| = |X|+ |(Y −X) ∩X ′|+ |X ′ ∩ Y ′ −X ∪ Y |+ |X ′ − Y ′ −X|= |X|+ |X ′ ∩ Y ′ −X|+ |X ′ − Y ′ −X|,

|Y ′| = |Y |+ |(X − Y ) ∩ Y ′|+ |X ′ ∩ Y ′ −X ∪ Y |+ |Y ′ −X ′ − Y |≥ |Y |+ |X ′ ∩ Y ′ −X|+ |Y ′ −X ′ − Y |.

Since |X ′| = |Y ′| and |X| < |Y |, we can derive that |X ′−Y ′−X| ≥ |Y |− |X|+|Y ′−X ′−X| > 0. There is e ∈ Y ′−X ′−X. By (B2), there is f ∈ Y ′−X ′ suchthat X ′′ := X ′−e+f ∈ B. Then |X ′′∩Y ′| > |X ′∩Y ′|, which is a contradiction.So the choice of (X ′, Y ′) so that |X ′∩Y ′| is maximized implies (Y −X)∩X ′ 6= ∅.There is e ∈ (Y − X) ∩ X ′ such that X + e ⊆ X ′ (so X + e ∈ I). Therefore,(I3) holds.

By the construction of I, it is obvious that every B ∈ B is a base in thematroid M := (E, I). (Remind that all sets in B has the same size, so no setin B is a proper subset of another set in B.) Hence B is a collection of bases inM (but it is not sure that B contains all bases). Let A be a base in M . Sinceit is in I, there is B ∈ B such that A ⊆ B. Since B is independent and A ismaximally independent, A = B ∈ B. Therefore, B is the collection of bases inM .

The fundamental circuit of e 6∈ B with respect to B ∈ B is a unique circuitCe contained in B + e. Note that e ∈ Ce. The uniqueness is from (C3) (seethe proof of the Proposition 3.1). This terminology is from the fundamentalcycle of e 6∈ E(T ) with respect to a tree T in a graph G. Remind that T ande(6∈ E(T )

)determines a unique cycle containing e.

4 Week02-2, 2019.09.11. Wed

Chapter 2.3. Rank

12

Remind that rM (X) := max{|I| : I ⊆ X, I ∈ I} for a matroid M = (E, I)is the rank of X. By (I3), all maximal independent subsets of X have sizerM (X). We call rM the rank function of M . (We omit a subscription so writesimply r when a given matroid is clear.)

Proposition 4.1. The rank function r of a matroid satisfies

(R1) 0 ≤ r(x) ≤ |X|,

(R2) X ⊆ Y ⇒ r(X) ≤ r(Y ), and

(R3) r(X) + r(Y ) ≥ r(X ∩ Y ) + r(X ∪ Y ).

We call (R3) the submodular inequality.

Proof. (R1) and (R2) are trivial from the definition of r. Let I be an inde-pendent set contained in X ∩ Y with |I| = r(X ∩ Y ). Let J be a maximalindependent set containing I contained in X ∪ Y . Then |J | = r(X ∪ Y ).

r(X) + r(Y ) ≥ |X ∩ J |+ |Y ∩ J | = |I|+ |J | = r(X ∩ Y ) + r(X ∪ Y ),

i.e., (R3) holds.

Proposition 4.2. If r : 2E → Z satisfies (R1), (R2), (R3), and I := {X ⊆ E :r(X) = |X|}, then (E, I) is a matroid with the rank function r.

Proof. (I1) is directly from (R1) since 0 ≤ r(∅) ≤ |∅| = 0. Let X ⊆ Y andY ∈ I, i.e., r(Y ) = |Y |. By (R3),

r(X) + r(Y −X) ≥ r(∅) + r(Y ) = |Y |.

By (R1), r(X) ≤ |X| and r(Y ) ≤ |Y −X|, so r(X) + r(Y −X) ≤ |Y |. It impliesthat r(X) = |X|, i.e., X ∈ I. Hence (I2) holds.

Before showing (I3), we claim a lemma: If r(X + e) = r(X) for all e ∈ Z,then r(X ∪Z) = r(X). Its proof is following. Let W be a maximal subset of Zsuch that r(X ∪W ) = r(X). If W = Z, done. WMA W ( Z. Let e ∈ Z −W .Then by (R3),

r(X + e) + r(X ∪W ) ≥ r(X) + r(X ∪W + e).

By the assumptions, r(X + e) = r(X ∪W ) = r(X). It implies that r(X) ≥r(X ∪ (W + e)), which contradicts to our choice of W .

Now we will show (I3). Let X,Y ∈ I and |X| < |Y |. Suppose that X+e 6∈ Ifor all e ∈ Y − X. Note that |X| = r(x) ≤ r(X + e) ≤ |X + e| by (R1) and(R2). Since X + e 6∈ I, i.e., r(X + e) < |X + e|, r(X + e) = r(X). Then by theabove lemma,

|Y | = r(Y ) ≤ r(X ∪ (Y −X)) = r(X) = |X|.

It is a contradiction.We claim that r is the rank function of M := (E, I). Let I be a maximal

independent subset of X in M , for a given X ⊆ E. WTS r(X) = |I|(

= rM (X)).

By the maximality, r(I + e) < |I + e| for any e ∈ X − I. Then by the abovelemma, r(X) = r(I ∪ (X − I)) = r(I) = |I|.

13

One question that can we deduce that

0 ≥ r(X∪Y ∪Z)−r(X)−r(Y )−r(Z)+r(X∪Y )+r(X∪Z)+r(Y ∪Z)−r(X∩Y ∩Z)

from the submodular inequality and others??? Or, can we deduce more generalone???

Example 4.1. Consider a graphic matroid M = M(G) for a graph G = (V,E).Then

rM (X) = |E(a union of a spanning tree of each component of (V,X))|

=∑

C: a component of (V,X)

(|V (C)| − 1

)= |V | −#(components of (V,X)).

Here, (V,X) is a subgraph of G.

Chapter 2.4. Closure

Definition 4.1. Let M = (E, I) be a matroid. For X ⊆ E, the closure (orspan) of X is

clM (X) := {e ∈ E : rM (X ∪ {e}) = rM (X)}.

We can omit a subscription of clM (so simply write cl) if there is no confusion.

From the definition, it is obvious that X ⊆ cl(X).

Example 4.2. Consider a vector matroid M = M(A) where A is an R × Ematrix over a field. Let X ⊆ E. We can think X as a set of column vectors ofA[X]. Also, we can regard e ∈ E as a column vector in A. If e ∈ spanX, thendim(span(X + e)) = dim(spanX), and vice versa. Note that rM ≡ dim span, sofor the vector matroid M ,

clM (X) = {e ∈ E : e ∈ spanX}.

Example 4.3. Let M = M(G) be a graphic matroid, where G = (V,E) is agraph described in below. E = {a, b, c, d, e, f}. Then

clM ({a, b, h}) = {a, b, h, g, j}.

Adding g or j to {a, b, h} does not affect on the number of components.

14

Lemma 4.3. rM (clM (X)) = rM (X).

Proof. It is equivalent with the claim in the proof of the Proposition 4.2: IfrM (X + e) = rM (X) for all e ∈ Z, then rM (X ∪ Z) = rM (Z).

Proposition 4.4. The closure cl of a matroid satisfies

(CL1) X ⊆ cl(X) for all X ⊆ E,

(CL2) X ⊆ Y ⇒ cl(X) ⊆ cl(Y ),

(CL3) cl(cl(X)) = cl(X), and

(CL4) ∃f ∈ cl(X + e)− cl(X) ⇒ e ∈ cl(X + f).

Proof. (CL1) is trivial from the definition. Let X ⊆ Y . Suppose that there ise ∈ cl(X)−Y (otherwise, it is directly deduced that cl(X) ⊆ Y ⊆ cl(Y )). Then

r(X + e) + r(Y ) ≥ r(X) + r(Y + e)

by the submodular inequality, (R3). Since r(X + e) = r(X), r(Y ) ≥ r(Y + e),i.e., e ∈ cl(Y ). So (CL2) holds. Let e ∈ cl(cl(X)), i.e., r(cl(X) + e) = r(cl(X)).By the previous lemma, r(cl(X)) = r(X). Then

r(X) ≤ r(X + e) ≤ r(cl(X) + e) = r(X).

It implies r(X + e) = r(X), i.e., e ∈ cl(X). So cl(cl(X)) ⊆ cl(X). The axiom(CL3) holds.

Now WTS (CL4). Assume its sufficient condition. Then we get

r(X + e+ f) = r(X + e),

r(X + f) = r(X) + 1.

Note that by the submodular inequality, r(X) + r(e) ≥ r(X ∩{e}) + r(X+ e) ≥r(X + e), so r(X + e)− r(X) ≤ r(e) ≤ 1.

15

r(X + e)

r(X) r(X + e+ f)

r(X + f)

≤ 1

+1

≤ 1

=

?

The above diagram means that

r(X + e+ f)− r(X + f) + 1 = r(X + e+ f)− r(X + f) + r(X + f)− r(X)

= r(X + e+ f)− r(X)

= r(X + e+ f)− r(X + e) + r(X + e)− r(X)

≤ 1.

So r(X + e+ f)− r(X + f) = 0, i.e., e ∈ cl(X + f).

Proposition 4.5. If cl : 2E → 2E satisfies (CL1), (CL2), (CL3), (CL4), andI := {X ⊆ E : e 6∈ cl(X − e), ∀e ∈ X}, then (E, I) is a matroid whose closurefunction is cl.

Proof. (I1) is trivial. (Since 6 ∃e ∈ ∅, so the statement of I vaguely holds for ∅.It implies that ∅ ∈ I.) Let X ⊆ Y and Y ∈ I. For all e ∈ Y , e 6∈ cl(Y − e)by the construction of I. Let f ∈ X. By (CL2), cl(X − f) ⊆ cl(Y − f). Sof 6∈ cl(X − f) (since f 6∈ cl(Y − f)), and it implies that X ∈ I. (I2) holds.

Now WTS (I3). Let X,Y ∈ I and |X| < |Y |. We claim that if f ∈ Y −cl(X),then X + f ∈ I. The proof of the claim is following. Suppose e ∈ X + f . Ife = f , then e = f 6∈ cl(X) = cl((X + f) − e). If e 6= f , then we can supposee ∈ cl(X + f − e). Remind that X ∈ I, so e 6∈ cl(X − e). Then by (CL4),f ∈ cl((X − e) + e) = cl(X), which is a contradiction. So e 6∈ cl(X + f − e).Therefore, the claim is proved.

So WMA cl(X) ⊇ Y (otherwise, (I3) holds by the above claim). Let choosea counter-example (X,Y ) of (I3) with maximum |X ∩ Y |. Let e ∈ Y − X.If X ⊆ cl(Y − e), then cl(X) ⊆ cl(cl(Y − e)) = cl(Y − e) by (CL3). Sincee ∈ Y −X ⊆ cl(X) and e 6∈ cl(Y − e) (since Y ∈ I), it is a contradiction. SoX 6⊆ cl(Y − e). So WMA there is f ∈ X − Y (since if not, then X ⊂ Y , so (I3)holds) with f 6∈ cl(Y − e).

We claim that Y + f − e ∈ I. If not, then there is g ∈ Y + f − e so thatg ∈ cl(Y + f − e− g). If g = f , then g = f 6∈ cl(Y − e), so g 6∈ cl(Y − e− g). Ifg 6= f , then g 6∈ cl(Y − e− g) since Y − e ∈ I by (I2). So g 6∈ cl(Y − e− g). By(CL4), f ∈ cl((Y − e− g) + g) = cl(Y − e), which is a contradiction. Therefore,the claim holds.

Let Y ′ := Y + f − e ∈ I. Then a pair (X,Y ′) satisfies (I3) by the maximumcondition for |X ∩ Y |

(= |X ∩ Y ′| − 1

). So there is e′ ∈ Y ′ −X

(⊂ Y −X

)so

that X + e′ ∈ I. Therefore, (I3) holds.So we know that M := (E, I) is a matroid. Now WTS

clM (X) = cl(X)

16

for each X ⊆ E. Suppose e ∈ clM (X) − cl(X). Let r be the rank function ofM . Then r(X + e) = r(X) since e ∈ clM (X). Let I be a maximal independentsubset of X. cl(I) ⊆ cl(X) by (CL2). I + e is dependent (note that x 6∈ cl(X)so x 6∈ I) since

r(X) = r(X + e) ≥ r(I + e) ≥ r(I) = |I| = r(X).

So there is f ∈ I+e such that f ∈ cl(I+e−f). It is impossible that e = f sinceif not, f = e 6∈ cl(X) = cl((I + e)− f). So f ∈ I, and f 6∈ cl(I − f) since I ∈ I.Then by (CL4), e ∈ cl((I − f) + f) = cl(I) ⊆ cl(X), which is a contradiction.So clM (X) ⊆ cl(X).

Before proving the other direcion, we claim a lemma: If I ∈ I (I is definein the problem) and e ∈ E − I such that I + e 6∈ I, then e ∈ cl(I). Its proofis following. There is f ∈ I + e such that f ∈ cl(I + e − f). If f = e, thene = f ∈ cl((I + e)− f) = cl(I). If f 6= e, then f ∈ I so f 6∈ cl(I − f). By (CL4),e ∈ cl((I − f) + f) = cl(I).

For the other direction, suppose e ∈ cl(X). WMA e 6∈ X. Let I be amaximal independent subset of X. cl(I) ⊆ cl(X). I + e′ 6∈ I for all e′ ∈ X − Iby the maximality. By the above lemma, X − I ⊆ cl(I) so X ⊆ cl(I). By(CL2) and (CL4), cl(X) ⊆ cl(cl(I)) = cl(I) so cl(X) = cl(I). It implies thate ∈ cl(X) = cl(I) = cl((I + e)− e). So I + e 6∈ I, i.e., dependent. Then

|I| = r(I) ≤ r(I + e) < |I + e| = |I|+ 1.

It implies that r(I) = r(I + e), i.e., e ∈ clM (I) ⊆ clM (X). We can concludethat cl(X) ⊆ clM (X).

(The proof of the Proposition 4.5 is completed in the lecture Week03-1.)

5 Week03-1, 2019.09.16. Mon

Chapter 2.5. Flat (or Closed set)

Definition 5.1. X ⊆ E is a flat (or a clsoed set) if cl(X) = X. X is spanningif cl(X) = E(M). X is a hyperplane if it is a flat of rank r(M)− 1.

Example 5.1. Let G be a below graph with labels on edges. Let M := M(G)be a graphic (or cycle) matroid.

17

(Flats of rank 0) = cl(∅) = (the set of all loops) = {k, l},(Flats of rank 1) = cl(non-loop edge)

= (sets of all edges joining 2 adj. vertices union w/ all loops)

= {a, k, l}, {f, g, k, l}, . . . ,(Flats of rank 2) = {a, c, k, l}, {a, b, j, k, l}, {d, e, f, g, k, l}, . . . .

Let V1, V2, . . . , Vn−r be a partition of V = V (G), where n = |V | and 0 ≥ r ≥n− 1. Then F :=

∐iE(G[Vi]) is a flat of rank r in M = M(G).

Example 5.2. Let M := M(K6) be a graphic matroid. How many hyperplanes?It is (

6

1

)+

(6

2

)+

1

2

(6

3

).

First term means the number of (K1,K5) pairs, the second means the numberof (K4,K2), and the last means the number of (K3,K3). Generally, M(Kn) hasthe below number of hyperplanes,(

n

1

)+

(n

2

)+ . . .+

(n

k

)if n = 2k + 1, and (

n

1

)+

(n

2

)+ . . .+

(n

k − 1

)+

1

2

(n

k

)if n = 2k.

Definition 5.2. We say X spans Y if Y ⊆ cl(X), and X spans e if e ∈ cl(X).

Lemma 5.1. X spans e if and only if cl(X + e) = cl(X).

Proof. ‘If’ part is trivial. Now we will show ‘only if’ part. Let e ∈ cl(X). ThenX + e ⊆ cl(X). By (CL2) and (CL3), cl(X + e) ⊆ cl(cl(X)) = cl(X). Socl(X + e) = cl(X).

The above lemma still holds if we replace an element e to a set Y . The proofis almost same.

Lemma 5.2. The followings hold:

1. X is spanning ⇔ r(X) = r(M),

2. X is a hyperplane ⇔ X is a maximal set not spanning, and

3. X is a flat of rank k ⇔ X is a maximal set of rank k.

18

Proof. 1. X is spanning, i.e., cl(X) = E(M). Then r(X) = r(cl(X)) =r(E(M)) = r(M). Here the first equality holds by the Lemma 4.3. Conversely,suppose r(M) = r(X). For any e ∈ E(M)−X,

r(E(M)) ≥ r(X + e) ≥ r(X) = r(M).

It implies that r(X + e) = r(X), i.e., e ∈ cl(X). So E(M) = cl(X).2. Let X be a hyperplane. Then r(X) = r(M)− 1, so cl(X) = X ( E(M),

i.e., X is not spanning. Let Y be a set containing X and not spanning. Thenfor any e ∈ Y ,

r(X) ≤ r(X + e) < r(M) = r(X) + 1.

It implies that r(X) = r(X + e), i.e., e ∈ cl(X). Y ⊆ X. So X is a maximalset not spanning. Conversely, let X be a maximal set not spanning. Thenr(cl(X)) = r(X) < r(M). By the maximality, r(X + e) = r(M) for anye ∈ E(M) − X. It implies two facts: First, (E(M) − X) ∩ cl(X) = ∅, i.e.,cl(X) ⊆ X. Second, r(X) = r(M)− 1 since r(X + e) ≤ r(X) + 1. Therefore, Xis a flat of rank r(M)− 1, i.e., a hyperplane.

3. Let X be a flat of rank k, i.e., X = cl(X) and r(X) = k. Let Y be a setor rank k containing X. Then for any e ∈ Y ,

k = r(X) ≤ r(X + e) ≤ r(Y ) = k,

so e ∈ cl(X). It implies that Y ⊆ X. Therefore, X is a maximal set ofrank k. Conversely, let X be a maximal set of rank k. Let e ∈ cl(X), i.e.,r(X + e) = r(X). Then by the maximality, X = X + e, i.e., e ∈ X. HenceX = cl(X).

Lemma 5.3. Let e 6∈ X. e ∈ cl(X) iff there is a circuit C ⊆ X + e containinge.

Proof. Let I be a maximal independent set of X. Since e ∈ cl(X)−X,

|I| = r(I) = r(X) = r(cl(X)) ≥ r(I + e)

and |I + e| = |I| + 1. It implies that I + e is dependent. So there is a circuitC in I + e (so in X + e). It must contain e. Conversely, e ∈ cl(C − e) sincer(C) = r(C − e) (note that C − e is independent and C is dependent). Soe ∈ cl(C − e) ⊆ cl(X).

Proposition 5.4 (Strong circuit elimination axiom). Let C1 and C2 be distinctcircuits, and e ∈ C1∩C2, f ∈ C1−C2. Then there is a circuit C3 ⊆ (C1+C2)−esuch that f ∈ C3.

Proof. Since C2−e is independent and C2 is dependent, r(C2−e) = r(C2), i.e.,e ∈ cl(C2 − e). Similarly, f ∈ cl(C1 − f).

cl((C1 + C2)− e− f) ⊇ cl(C2 − e) 3 e.

19

Then cl((C1 + C2)− e− f) ⊇ (C1 + C2)− f , so by (CL2) and (CL3)

cl((C1 + C2)− e− f) = cl((C1 + C2)− f) ⊇ cl(C1 − f) 3 f.

By the previous lemma, there is a circuit C3 in ((C1 + C2) − e − f) + f =(C1 + C2)− e containing f .

6 Week03-2, 2019.09.18. Wed

Proposition 6.1. Let M = (E, I) be a matroid and F be the collection of allflats. Then

(F1) E ∈ F ,

(F2) F1, F2 ∈ F ⇒ F1 ∩ F2 ∈ F , and

(F3) F ∈ F and {F1, F2, . . . , Fk} is the collection of all minimal members ofF containing F properly ⇒ {F1 −F, F2 −F, . . . , Fk −F} is a partition ofE − F .

Proof. (F1): By (CL1), E ⊆ cl(E)(⊆ E

). Hence cl(E) = E, i.e., E ∈ F .

(F2): Suppose not, i.e., there is e ∈ cl(F1∩F2)−(F1∩F2). By symmetry, letus assume e 6∈ F1. There is a circuit C ⊆ (F1 ∩ F2) + e containing e by Lemma5.3. Then e ∈ C ⊆ F1 + e, so e ∈ cl(C − e) ⊆ cl(F1) = F1. It is a contradiction.

(F3): Let F1, F2, . . . , Fk be a collection of minimal flats containing F prop-erly. Then Fi ∩ Fj = F for i 6= j by the minimality. Note that for anye ∈ E − F , cl(F + e) is a flat minimally containing F properly. It is obvi-ous that cl(F + e) is a flat containing F properly. The minimality is guaranteedby r(cl(F + e)) = r(F + e) = r(F ) + 1. Hence cl(F + e) = Fi for some i.Therefore, {F1 − F, F2 − F, . . . , Fk − F} is a partition of E − F .

What is an intuition of (F3)? For a vector matroid M , let F be a flat ofrank 1, i.e., it is a set of vectors contained in a given line. Then each Fi is aplane containing the line corresponds to F . Then we can easily observe that{F1−F, . . . , Fk−F} is a partition of E−F . For a graph G and a graphic matroidM = M(G), a flat F of rank k is a subgraph G[F ] with r(M)−k+1 = |V (G)|−kcomponents which are induced subgraphs of G. Then Fi−F corresponds to anedge-cut between two different components.

Proposition 6.2. If F is a collection of subsets of a finite set E satisfying(F1), (F2), (F3), and let cl(X) = ∩X⊆F∈FF for all X ⊆ E, then there is amatroid on E whose closure funcion is cl.

Proof. By (F1), cl(X) is well-defined. We will show that cl satisfies the closureaxioms.

(CL1): By its construction, X ⊆ cl(X).(CL2): Also by its construction, if X ⊆ Y then cl(X) ⊆ cl(Y ).(CL3): cl(X) ∈ F by (F2). So cl(cl(X)) ⊆ cl(X).

20

(CL4): Let f ∈ cl(X + e) − cl(X). Let F := cl(X) ∈ F . By (F3), thereis {F1, F2, . . . , Fk} ⊆ F of which members minimally contain F properly, and{F1−F, F2−F, . . . , Fk −F} is a partition of E −F . WLOG e ∈ F1−F . ThenF ( cl(X + e) ⊆ F1 and cl(X + e) ∈ F , so cl(X + e) = F1 by the minimality.It implies that f ∈ F1 − F . Then F ( cl(X + f) ⊆ F1 and cl(X + f) ∈ F , socl(X + f) = F1 3 e.

Therefore, cl defines a matroid M = (E, I), where I = {X ⊆ E : e 6∈cl(X − e), ∀e ∈ X}, by Proposition 4.5. In addition, the closure function of Mis cl. Let FM be the collection of flats of M , i.e., FM = {X ⊆ E : X = cl(X)}.WTS F = FM .

If F ∈ F , then cl(F ) ⊆ F by the definition of cl. Hence cl(F ) = F , i.e.,F ∈ FM .

Let F ∈ FM . By (F2),(F =

)cl(F ) ∈ F .

Chapter 2.6. Greedy Algorithm and Matroids

Definition 6.1 (Greedy algorithm). An algorithm described in below is calledthe greedy algorithm.

Input: A finite set E, a collection F ⊆ 2E , and a weight function ω : E → R.

Goal: Find X ∈ F s.t. ω(X) =∑e∈X ω(e) is maximized.

(i.e., ω(X) ≥ ω(X ′) for all X ′ ∈ F .)

Algorithm: 1. Start with X0 = ∅.2. Sort E so that ω(e1) ≥ ω(e2) ≥ . . . ≥ ω(en), where n = |E|.3. For each i = 1, 2, . . . , n,

Xi =

{Xi−1 + ei, if there is a set Y in F so that Xi−1 + ei ⊆ Y ,Xi−1, otherwise.

Output: Xn.

Note that we know that Xn ⊆ Y for some Y ∈ F . Moreover, we can alwaysguarantee that Xn ∈ F . Suppose not, i.e., Xn is properly contained in Y forall Y ∈ F containing Xn. Choose any such Y . Let i be the maximum numberso that ei ∈ Xn. Let ej ∈ Y −Xn. Then j 6= i. It is impossible that j < i bythe third step of the algorithm. However, j > i is also impossible. Therefore,we can conclude that Xn ∈ F .

Remind that the above is just an algorithm. We do not know that the goalis really achieved or not, i.e., possibly, Xn is not optimal, i.e., possibly, there isX ′ ∈ F such that ω(Xn) < ω(X ′).

Note that we can replace Xi−1 + ei ⊆ Y with Xi−1 + ei = Y ∩{e1, . . . , ei} inthe third step in the algorithm. Obviously, the new condition, Xi−1 + ei = Y ∩{e1, . . . , ei}, deduces the old condition, Xi−1+ei ⊆ Y . In addition, we can easilycheck that the old condition implies the new condition. If not, there is Y ∈ F

21

such that Y ∩{e1, . . . , ei} 6= Xi−1+ei ⊆ Y . So there is ej ∈ (Y ∩{e1, . . . , ei−1})−Xi−1. However, it implies that Xj−1 + ej ⊆ Y , so Xj−1 + ej = Xj ⊆ Xi−1. Itis a contradiction.

One example of applying the greedy algorithm is a spanning tree of a con-nected graph G. See Modern Graph Theory by Bela Bollobas, page 11-14.

Proposition 6.3. The greedy algorithm always finds an optimum solution if Fis a set of all bases of some matroid M .

Proof. Suppose not, i.e., there is a base Y such that ω(Y ) > ω(Xn). Let i bethe minimum such that Y ∩{e1, . . . , ei} 6= Xn ∩{e1, . . . , ei} = Xi. Choose Y sothat i is maximized. Note that by our choice of i, Y ∩ {e1, . . . , ei−1} = Xi−1.(If i = 1, then Y ∩ ∅ = X0.)

Case 1, ei 6∈ Xi. Then Xi−1 + ei is dependent. It implies ei 6∈ Y since ifnot, Xi−1 + ei ⊆ Y , a contradiction. Hence Y ∩ {e1, . . . , ei} = Xi, which is acontradiction.

Case 2, ei ∈ Xi. Then ei 6∈ Y . There is the fundamental circuit C of eiwith respect to Y . Since Xi is independent, C 6⊆ Xi = (Y ∩{e1, . . . , ei−1}) + ei.There is f ∈ C − Xi = C − {e1, . . . , ei}. Then f = ej for some j > i, soω(f) ≤ ω(ei). In addition, Y − f + ei is independent since it does not containa circuit. Y − f + ei is a base. Let Y ′ := Y − f + ei. Then ω(Y ′) ≥ ω(Y ), andi′ > i where i′ be the smallest number such that Y ′ ∩ {e1, . . . , ei′} 6= Xi′ . Itcontradicts to our choice of Y .

Proposition 6.4. Let E be a finite set and I be a set of subset of E satisfying(I1), (I2), and (G): The greedy algorithm finds an optimum solution for allω : E → R when F is the set of all maximal members of I. Then (E, I) is amatroid.

Proof. WTS (I3) holds. Let X,Y ∈ I with |X| < |Y |. Suppose X + e 6∈ I forall e ∈ Y −X. It implies X ( Y , i.e., X − Y 6= ∅. Let ε > 0 be arbitrary. Letus define a weight function ω : E → R as

ω(e) =

1 + ε, if e ∈ X,1, if e ∈ Y −X,0, otherwise.

Let Xn ∈ F be an output of the greedy algorithm with the above inputs. Xi’sare defined also in a procedure of the algorithm. Here, it is okay that we choosean arbitrary sorting of E in the second step of the algorithm. By (G), Xn isoptimal, i.e., ω(Xn) ≥ ω(X) for all X ′ ∈ F . Then X = X|X| ⊆ Xn since X ∈ I.In addition, Xn ⊆ X ∪ (E−Y ) since X+ e 6∈ I for all e ∈ Y −X. We can checkthat

ω(Xn) = (1 + ε)|X|,ω(Y ) = |Y −X|+ (1 + ε)|X ∩ Y | = |Y |+ ε|X ∩ Y |.

22

Since ω(Xn) ≥ ω(Y ′) ≥ ω(Y ) where Y ⊆ Y ′ ∈ F ,

ε|X − Y | ≥ |Y | − |X| > 0.

Remind that ε > 0 is arbitrary. Choose small ε < |Y |−|X||X−Y | . Then it contradicts

to the above inequality. Therefore, we can conclude that I satisfies (I3), so(E, I) is a matroid.

6.1 Homework 2.3, Hyperplane Axioms

Proposition 6.5. Let E be a finite set and let H be a collection of subsets ofE. Prove that H is the set of all hyperplanes of a matroid if and only if thefollowing three properties hold.

(H1) E 6∈ H.

(H2) If H1, H2 ∈ H and H1 ⊆ H2, then H1 = H2.

(H3) For all distinct H1 and H2 in H and for all x ∈ E, there exists H ∈ Hwith

(H1 ∩H2) + x ⊆ H.

Proof. (⇒) Let M be a matroid with a ground set E. Let H be the collectionof all hyperplanes of M . Let r be the rank function of M . We will show Hsatisfies (H1), (H2), and (H3).

Since r(E) = r(M), E is not a hyperplane, i.e., E 6∈ H. So (H1) holds.Let H1, H2 ∈ H and H1 ⊆ H2. Suppose there is e ∈ H2 −H1. Then

r(M)− 1 = r(H1) ≤ r(H1 + e) ≤ r(H2) = r(M)− 1.

It implies that e ∈ cl(H1) = H1, which is a contradiction. Hence H1 = H2.(H2) holds.

Let H1, H2 be a distinct hyperplanes, and x ∈ E. Denote A := (H1∩H2)+x.By the submodular inequality,

r(H1 ∩H2) + r(H1 ∪H2) ≤ r(H1) + r(H2) = 2r(M)− 2.

Since H1 ∪ H2 ) H1, r(H1 ∪ H2) = r(M). So r(H1 ∩ H2) ≤ r(M) − 2. Notethat r(X + e) ≤ r(X + e) + r(X ∩ {e}) ≤ r(X) + r({e}) ≤ r(X) + 1 for anyX ⊆ E and e ∈ E. So r(A) ≤ r(M) − 1. It also implies that through weadd elements of E − A into A one-by-one, we can find a set B ⊆ E − A suchthat r(A + B) = r(M) − 1. Choose H := cl(A + B). Then it is a flat of rankr(cl(A+B)) = r(A+B) = r(M)− 1, which contains A. So (H3) holds.

(⇐) We will show the opposite direction. Let H be a collection of subsetsof E satisfying (H1), (H2), and (H3). Let δ : 2E → 2E be a function defined by

δ(X) = ∩H∈HX⊆H

H.

23

If there is no such H ∈ H satisfying X ⊆ H, then we define δ(X) = E. First,we claim that δ satisfies the closure axioms.

By its construction, δ satisfies (CL1) and (CL2), i.e., X ⊆ δ(X), and ifX ⊆ Y then δ(X) ⊆ δ(Y ).

Let X ⊆ E. If H ∈ H contains X, then it must contain δ(X) by itsconstruction. It implies that δ(X) ⊇ δ(δ(X)) for δ(X) ( E. If δ(X) = E, i.e.,there is no H ∈ H containing X, then δ(δ(X)) = δ(E) = E = δ(X). Hence(CL3) holds.

Now we will show that δ satisfies (CL4). Let f ∈ δ(X + e)− δ(X). Supposee 6∈ δ(X + f). Then

1. [f ∈ δ(X + e)] implies that [if there is H1 ∈ H with H1 ⊇ X + e, thenH1 3 f ].

2. [f 6∈ δ(X)] implies that [there is H3 ∈ H s.t. H3 ⊇ X and H3 63 f ].

3. [e 6∈ δ(X+f)] implies that [there is H2 ∈ H s.t. H2 ⊇ X+f and H2 63 e].

If there is no H3 ∈ H containing X, then δ(X) = E 3 f . So there is H3 ∈ Hcontaining X. If all such H3 contains f , then f ∈ δ(X). Therefore, (2) holds.Similarly, (3) holds. Note that f ∈ H2 and f 6∈ H3, so H2 and H3 are distinct.

By (H3), there is H1 ∈ H such that H1 ⊇ (H2 ∩H3) + e (so H1 ⊇ X + e).By the above argument e, f ∈ H1. f ∈ H2 but e 6∈ H2. f 6∈ H3 so e 6∈ H3 by(1). Hence H1, H2, H3 are distinct.

Let us say that a triple (G1, G2, G3) is good if

• G1, G2, G3 ∈ H,

• X + e+ f ⊆ G1 (equivalently, X + e ⊆ G1 by (1)),

• X + f ⊆ G2 and e 6∈ G2, and

• X ⊆ G3 and e, f 6∈ G3.

The existence of a good triple is already shown. Now, choose a good triple(G1, G2, G3) so that |G1∩G2∩G3| is maximized. Among such candidates, choose(G1, G2, G3) so that |G1 ∩ G2| is maximized. Also, among such candidates,choose (G1, G2, G3) so that |G1 ∩G3| is maximized. Let us denote a such goodtriple as (I1, I2, I3). In other words, (I1, I2, I3) is a good tripe satisfying

• |I1 ∩ I2 ∩ I3| ≥ |G1 ∩G2 ∩G3| for any good triple (G1, G2, G3),

• |I1∩I2| ≥ |G1∩G2| for any good tripe (G1, G2, G3) with |G1∩G2∩G3| =|I1 ∩ I2 ∩ I3|, and

• |I1∩I3| ≥ |G1∩G3| for any good tripe (G1, G2, G3) with |G1∩G2∩G3| =|I1 ∩ I2 ∩ I3| and |G1 ∩G2| = |I1 ∩ I2|.

24

WTS (I2 ∩ I3) − I1 = ∅. By (H3), there is I ′1 ∈ H containing (I2 ∩ I3) + e.(Possibly, I ′1 = I1.) Then X + e ⊆ I ′1, so (I ′1, I2, I3) is a good triple. Suppose(I2∩ I3)− I1 6= ∅. Then |I ′1∩ I2∩ I3| > |I1∩ I2∩ I3|. It contradicts to our choiceof (I1, I2, I3).

By (H2), there is g ∈ I2 − I1(

= I2 − (I1 ∪ I3)). (Otherwise, I2 ( I1.) Also,

by (H2), there is h ∈ I3 − I1(

= I3 − (I1 ∪ I2)). (Otherwise, I3 ( I1.) By (H3),

there is H ∈ H containing (I1 ∩ I2) + h. Note that X + f ⊆ H.Case 1: e 6∈ H. Let I ′2 := H. By (H3), there is I ′1 ∈ H containing (I ′2∩I3)+e.

Then X + e ⊆ I ′1, X + f ⊆ I ′2, and e 6∈ I ′2. So (I ′1, I′2, I3) is a good triple. And

I ′1 ∩ I ′2 ∩ I3 ⊇ (I1 ∩ I2 ∩ I3) + h. It contradicts to our choice of (I1, I2, I3).Case 2: e ∈ H. Let I ′1 := H. Note that (I ′1, I2, I3) is a good triple satisfying

I ′1 ∩ I2 ∩ I3 ⊇ I1 ∩ I2 ∩ I3, and I ′1 ∩ I2 ⊇ I1 ∩ I2.Sub-case 2.1: I ′1 ⊇ I1 ∩ I3. Then I ′1 ∩ I3 ⊇ (I1 ∩ I3) + h. It contradicts to

our choice of (I1, I2, I3).Sub-case 2.2: ∃i ∈ (I1 ∩ I3) − I ′1. Then I1 and I ′1 are distinct. By (H3),

there is I ′′1 ∈ H containing (I1 ∩ I ′1) + g. Remind that (I1 ∩ I2) + e ⊆ I1 ∩ I ′1, soX + e+ f ⊆ I1. A tripe (I ′′1 , I2, I3) is good. However, I ′′1 ∩ I2 ∩ I3 ⊇ I1 ∩ I2 ∩ I3,and I ′′1 ∩ I2 ⊇ (I1 ∩ I2) + g. It contradicts to our choice of (I1, I2, I3).

Hence we can conclude that the basic assumption, e 6∈ δ(X + f), fails viaabove case analysis. So δ satisfies (CL4): f ∈ δ(X + e) − δ(X) implies thate ∈ δ(X + f).

Let M be a matroid on a ground set M defined by setting its closure functionas δ. Let HM be the collection of hyperplanes on M . WTS HM = H.

For any H ∈ H, δ(H) ⊆ H by the definition of δ. By (CL1), δ(H) = H. SoH is a flat on M . By (H1), H 6= E. Let e ∈ E−H. By (H2), there is no H ′ ∈ Hcontaining H + e. It implies that δ(H + e) = E. It holds for any e ∈ E−H. SoH is a maximal set not spanning, i.e., H is a hyperplanes on M , i.e., H ∈ HM .

Let G ∈ HM , i.e., G is a maximal set not spanning, i.e., δ(G) = G ( E andδ(G + e) = E for any e ∈ E − G. Since δ(G) ( E, there is H ∈ H such thatG ⊆ H. Suppose there is f ∈ H −G. Then G+ f ⊆ H, so E = δ(G+ f) ⊆ H.By (H1), H 6= E. This is a contradiction. Hence G = H ∈ H.

Therefore, we can conclude that H = HM .

7 Week05-1, 2019.09.30. Mon

Chapter. Symmetric Greedy Algorithm and Delta-matroids

Definition 7.1 (Symmetric greedy algorithm). An algorithm described in be-low is called the symmetric greedy algorithm.

Input: A finite set E, a collection F ⊆ 2E , and a weight function ω : E → R.

Goal: Find X ∈ F s.t. ω(X) =∑e∈X ω(e) is maximized.

(i.e., ω(X) ≥ ω(X ′) for all X ′ ∈ F .)

Algorithm: 1. Sort E so that |ω(e1)| ≥ |ω(e2)| ≥ . . . ≥ |ω(en)|, where n = |E|.

25

2. Start with J0 = ∅. For each i = 1, 2, . . . , n,

Ji =

{Ji−1 + ei, if (∗),Ji−1, otherwise.

(∗): For ω(ei) ≥ 0, Ji−1 + ei can be extended to a member of F , i.e.,∃B ∈ F so that B ∩ {e1, e2, . . . , ei} = Ji−1 + ei. For ω(ei) < 0, Ji−1cannot be extended to a member of F not containing ei, i.e., ???

Output: Jn.

Definition 7.2. We say (E,F) is a delta-matroid if E is a finite set, and F isa collection of subsets of E satisfying

(D1) F 6= ∅, and

(D2) X,Y ∈ F and e ∈ X4Y ⇒ ∃f ∈ X4Y s.t. X4{e, f} ∈ F .

We call (D2) the symmetric greedy axiom.

Note that X4Y := (X − Y ) ∪ (Y −X).The symmetric greedy axiom, (D2), is similar with the base exchange axiom,

(B2), but not equivalent.

Remark.

Example 7.1. One trivial example of delta-matroids, which is not a matroid, is(E, 2E) for a finite set E.

8 Week05-2, 2019.10.02. Wed

Chapter. Dual MatroidsLet B be the collection of bases a given matroid M . Remind that

(B2) X,Y ∈ B, e ∈ X − Y ⇒ ∃f ∈ Y −X so that X − e+ f ∈ B, and

(B2’) X,Y ∈ B, e ∈ Y −X ⇒ ∃f ∈ X − Y so that X − f + e ∈ B.

Let B∗ := {E−B : B ∈ B}(

= c(B)), the collection of complements of members

of B. Then it is trivial that B∗ satisfies (B1). Consider (B2) once again. X,Y ∈B and e ∈ X − Y . There is f ∈ Y −X such that

X − e+ f ∈ B.

We observe that(E −X)− f + e ∈ B∗.

In addition, E−X,E−Y ∈ B∗, e ∈ (E−Y )−(E−X), and f ∈ (E−X)−(E−Y ).Hence (B2) for B corresponds to (B2’) for B∗. It implies that B∗ defines amatroid.

26

Definition 8.1. The dual matroid M∗ of a matroid M is a matroid on E(M)having B∗ as the collection of bases of M∗ when B is the collection of bases ofM .

By the definition, it is directly deduced that (M∗)∗ = M .

Example 8.1. There are easy examples of dual matroids.

1. (Ur,n)∗ = Un−r,n.

2. M(K3)∗ = M(K32 ). Here K3

2 is a graph on two vertices, which has threeparallel edges. Note that K3

2 is a dual graph of K3 = ∆. It is a reason whyWhitney developed the concept of matroids. He wanted to generalized theconcept of dual on planar graphs.

Definition 8.2. We say that

• X is conindependent in M if X is independent in M∗,

• X is a cobase in M if X is a base in M∗,

• X is a cocircuit in M if X is a circuit in M∗,

• X is a cohyperplane in M if X is a hyperplane in M∗, and

• e is a coloop in M if e is a loop in M∗.

Proposition 8.1. X is a cocircuit of M iff E −X is a hyperplane of M .

Proof. The followings are equivalent.

X is a cocircuit of M

⇔ X is a circuit of M∗

⇔ X is a minimal set which is not a subset of a base of M∗

⇔ X is a minimal set intersecting with every base of M

⇔ E −X is a maximal set containing no base of M

⇔ E −X is a maximal non-spanning set

⇔ E −X is a hyperplane.

Example 8.2. Let G be a connected plane graph. Let M := M(G). Let E −Xbe a hyperplane of M . It is an edge-set of a subsgraph G[E −X] of G, whichconsists of two component, and each component is an induced subgraph of G.By the previous proposition, X is a cocircuit of M . In the graph sense, X isa cycle of G∗, or equivalently X is an edge-cut of G. More precisely, it is anedge-cut between two components of G[E −X].

27

Proposition 8.2 (Rank of M∗).

rM∗(X) = |X| − r(M) + rM (E −X).

Proof.

rM∗(X) = maxB∗∈B∗

|X ∩B∗|

= maxB∈B|X −B|

= maxB∈B

(|X| − |X ∩B|

)= |X| −min

B∈B|X ∩B|

= |X| −minB∈B

(|B| − |B ∩ (E −X)|

)= |X| − r(M) + max

B∈B|B ∩ (E −X)|

= |X| − r(M) + rM (E −X).

In particular, if X ∈ B, i.e., E −X ∈ B∗, then

rM∗(X) = rM (E −X).

Note that r(M) + r(M∗) = |E|.We can also define dual matroids in the sense of the rank function.

Proposition 8.3. Let M be a matroid on a finite set E, and r be the rankfunction of M . Let define r∗ : 2E → Z as

r∗(X) = |X| − r(M) + r(E −X).

Then r∗ satisfies the rank axioms, and a matroid defined by r∗ is same with thedual matroid M∗ of M .

Proof. ???

One important question in dual matroids is following: Suppose that M isrepresentable over F. Can we argue that M∗ is representable over F? Theanswer is positive.

Theorem 8.4. If M is reprsentable over F, then M∗ is reprsentable over F.Of course, the converse holds.

Proof. Let A be a matrix over F satisfying M = M(A). Note that the rowoperation does not affect on the independence/dependence of column vectors.Also changing of ordering of column vectors does not affect. WMA

A =(Ir A0

),

28

where r = r(M), n = |E(M)|, and A0 is an r × (n − r)-matrix. In this case,we say that A is of the standard form (or the standard representation) of thevector matroid M . Let define

A∗ :=(AT0 In−r

),

an (n− r)× n-matrix. We claim that M∗ = M(A∗).Denote E := E(M) = E(M∗). Let F ⊆ E corresponding to Ir of A, and

G ⊆ E corresponding to A0 of A. Then F corresponds to AT0 of A∗, and G ⊆ Ecorresponds to In−r of A∗.

Let X ⊆ E. Then

X is a base of M = M(A)

⇔ A[X], a submatrix of A consisting of column vectors corresponding to X,is an invertible r × r-matrix.

Let s = |F ∩X|. Note that

A[X] =

(0r−s,s BIs C

).

where B is an (r−s)× (r−s)-matrix, and C is a s× (r−s)-matrix. In addition,the above is equivalent that

B is an invertible matrix.

Similarly, we observe that

E −X is a base of M(A∗)

⇔ A∗[E−X], a submatrix of A∗ consisting of column vectors correspondingto E −X, is an invertible (n− r)× (n− r)-matrix

⇔ B′ is an invertible (r − s)× (r − s)-matrix,

where B′ is an (r− s)× (r− s)-matrix, and D′ is a (n− 2r+ s)× (r− s)-matrixwith

A∗[E −X] =

(B′ 0r−s,n−2r+sD′ In−2r+s

).

For example, by taking X as last s elements from F and first r− s elementsfrom G, we can consider A and A∗ as like below

A =

(In−r 0 B D

0 Ir C ∗

)and

A∗ =

(B′ C ′ Ir−s 0D′ ∗ 0 In−2r+s

).

We can confirm that B′ = BT . In addition, it holds for any choice of X.Therefore, we can conclude that

29

E −X is a base of M∗

⇔ X is a base of M = M(A)

⇔ B is invertible

⇔ E −X is a base of M(A∗),

i.e., the collection of bases of M(A∗) is same with B∗, the collection of bases ofM∗. It implies that M(A∗) = M∗, so M∗ is representable over F .

Note that for a plane graph G, if C is a cycle of G and D is an edge-cut ofD, i.e., D is a cycle of G∗, then |C ∩D| is always even. We can prove a weakerversion of this fact for matroids.

Proposition 8.5. If C is a circuit and D is a cocircuit, then |C ∩D| 6= 1.

Proof. Suppose |C∩D| = 1, and denote {e} = C∩D. By the earlier proposition,E −D is a hyperplane. So E −D does not span e ∈ D.

By the assumption, |(E−D)∩C| = |C|−1. Since C is a circuit, C−e = C−Dspans e ∈ C, i.e., C ⊆ cl(C −D). It implies that E −D

(⊇ C −D

)spans e by

(CL2). It is a contradiction.

What is an example that |C ∩D| = 2k + 1 for some k ≥ 1???

Definition 8.3. Let H be a hypergraph if it is a pair (V,E) with E ⊆ 2V and∅ 6∈ E. We call V (H) := V the vertex set, and E(H) := E the (hyper)edge setof H.

Obviously, we can understand (E, I−{∅}) as a hypergraph when M = (E, I)is a matroid.

Definition 8.4. Let H = (V,E) be a hypergraph. We say that H is a clutterif no edge is a proper subset of another edge. In other words, E is an antichainunder the inclusion, ⊆. Usually, we say that E is a clutter on V .

Example 8.3. Let M be a matroid. The followings are clutters on E(M).

• B(M), the collection of bases of M ,

• C(M), the collection of circuits of M ,

• H(M), the collection of hyperplanes of M ,

• B∗(M), the collection of cobases of M ,

• C∗(M), the collection of cocircuits of M , and

• H∗(M), the collection of cohyperplanes of M .

When the given matroid M is clear, we can omit M , and simply write them asB, C, H, B∗, C∗, and H∗.

30

Definition 8.5. Let H = (V,E) be a hypergraph. Let b(H) be a hypergraphon V , whose edges are minimal sets intersecting every edge of H. We call it theblocker of H.

If V is clear, then we can write it as b(E), and also we can understand it asa set-system on V , i.e., understand the hypergraph b(E) as its edge set.

Since we choose minimal sets, b(H) is a clutter on V (H).For a convenience, we do not consider a unfruitful case that I = B = {∅}

for a matroid M .

Proposition 8.6. b(B(M)) = C∗(M).

Proof. See the first four equivalent statements of the proof of Proposition 8.1.

Theorem 8.7 (Edmonds-Fulkerson, ’70). Let H be a clutter. Then

b(b(H)) = H.

Proof. In this proof, we understand each hypergraphs as its edge set. Sincethe vertex set of hypergraphs H, b(H), b(b(H)) is V (H), it does not make aconfusion.

I referred the below proof in chapter 9, Extremal Combinatorics with Appli-cations in Computer Science by Stasys Jukna.

Let us say that X ⊆ V (H) is a blocking set of H if X intersects with everyY ∈ H. Then b(H) is the collection of minimal blocking sets of H. Note thateach member of H is a blocking set of b(H).

Let X ∈ b(b(H)). Since it is a minimal blocking set of b(H), there is nomember of H properly contained in X. Suppose X 6∈ H. Then by the previousobservation, for any Y ∈ H, Y − X 6= ∅. So we can choose xY ∈ Y − X foreach Y ∈ H. Let Z := {xY : Y ∈ H}. Obviously, Z is a blocking set of H,and there is Z ′ ⊆ Z such that Z ′ ∈ b(H). Then Z ′ ∩X = ∅, it contradicts toX ∈ b(b(H)). Therefore, X ∈ H, so b(b(H)) ⊆ H.

Let X ∈ H. Remind that it is a blocking set of b(H). Then there isX ′ ⊆ X such that X ′ ∈ b(b(H)). Since H is a clutter, X = X ′ ∈ H. HenceH ⊆ b(b(H)).

It implies that B(M) = b(C∗(M)).Note that for a general hypergraph H, the above theorem does not holds.

However, it is always true that b(b(H)) ⊆ H.

Definition 8.6. Let H = (V,E) be a hypergraph. Let c(H) be a hypergraphon V , of which edge set is

{V −X : X ∈ E}.

We call it the complement of H.If V is clear, then we can write it as c(E), and also we can understand it as

a set-system on V , i.e., understand c(E) as its edge set.

31

By the definition of the dual matroid, Proposition 8.1, 8.6, and Theorem8.7, we can summarized relations between clutters induced by a matroid M .

B B∗

C C∗

H H∗

c

b

c

9 Week06-1, 2019.10.07. Mon

Chapter. Duals of a cycle matroid

Definition 9.1. A matroid M is cographic if M∗ is graphic.

Proposition 9.1. M(K5) is not cographic.

Proof. Suppose M(K5) is cographic, i.e., there is a graph G so that(M(K5)

)∗=

M(G). The collection of circuits of M(G) (cycles of G) does not change eventhough we identify vertices for each component of G. So WMA G is connected.

r(M(G)) = |E(M(K5))| − r(M(K5)) =

(5

2

)− 4 = 6.

It implies that |E(G)| = |E(M(G))| = 10 and |V (G)| = |r(M(G))| + 1 = 7.Then the average degree d(G) of G is 2×10

7 < 3. There is a vertex v of degree1 or 2. Note that deleting incident edges of a vertex induces a hyperplane ofM(G). Since a cocircuit is a complement of a hyperplane, the set of incidentedges of v is a cocircuit of M(G). So it is a circuit of M(K5), i.e., there is acycle of length 1 or 2 in G. It is a contradiction.

Proposition 9.2. M(K3,3) is not cographic.

Proof. Suppose that M(K3,3) is cographic, i.e., there is a graph G so that(M(K3,3)

)∗= M(G). WMA G is connected.

r(M(G)) = |E(M(K3,3))| − r(M(K3,3)) = 9− 5 = 4.

It implies that |E(G)| = |E(M(G))| = 9 and |V (G)| = |r(M(G))|+1 = 5. Thenthe average degree d(G) of G is 2×9

5 < 4. There is a vertex v of degree ≤ 3. Theset of incident edges of v is a cocircuit of M(G). So there is a cycle of length≤ 3 in G, which is a contradiction.

32

What is a cocircuit of M(G)? Remind that

rM (X) = |V (G)| −#(components of the subgraph (V (G), X) of G)

for X ⊆ E(G) = E(M(G)). (It holds whether G is connected or not.) SupposeG is connected. Let F be a flat of rank r(M(G))− k = |V (G)| − 1− k, and Hbe a hyperplane of M(G). Then F induces k + 1 components, and H induces2 components, i.e., subgraphs (V (G), F ) and (V (G), H) of G have k + 1 and 2components, respectively. From the fact that a cocircuit is a complement of ahyperplane, i.e., C∗ = c(H), E(G)−H is a cocircuit, and E(G)− F is a unionof cocircuits of M(G). Moreover,

X is a cocircuit of M(G)

⇔ E(G)−X is a hyperplane of M(G)

⇔ E(G)−X is a maximal edge-set inducing two components of G

⇔ X is a minimal edge-cut of G.

Remind that C∗ = b(B).

X is a cocircuit of M(G), i.e., X is a minimal edge-cut of G

⇔ X is a minimal set intersecting with every base of M(G)

⇔ X is a minimal set intersecting with every spanning tree of G.

Definition 9.2. A bond of a graph G is a minimal nonempty edge-cut of G.

Definition 9.3. The bond matroid M∗(G) of a graph G is a matroid on E(G)whose circuits are bonds of G.

Note that the collection of circuits of M∗(G) (bonds of G) does not changeeven though we identify vertices for each component of G. So WMA G isconnected when we define a bond matroid.

By the above observation, C(M∗(G)) = C∗(M(G))(

= C(M(G)∗)). In addi-

tion, E(M∗(G)) = E(M(G)) = E(G). It implies that

(M(G))∗ = M∗(G),

i.e., the dual of the cycle matroid is a bond matroid, and vice versa. So everybond matroid is cographic.

If G is a planar graph then M(G) is a both graphic and cographic. It is fromsome facts of graph theory. Let G be a plane graph, and G∗ be its geometricdual. (There is the natural one-to-one corresponding between E(G) and E(G∗).)Then

X is an edge-set of a cycle of G

⇔ X∗ is a minimal nonempty edge-cut of G∗.

33

It implies that C(M(G)) = C(M∗(G∗)). In addition, if G is connected, then(G∗)∗ = G. Therefore,

M(G) = M∗(G∗),

M∗(G) = M(G∗).

(The second holds by M∗(G) = M(G)∗ =(M∗(G∗)

)∗=(M(G∗)

)∗∗= M(G∗).)

We can interpret them as, first, cycles of G correspond to bonds of G∗, andsecond, bonds of G correspond to cycles of G∗. Moreover, the converse holds.

Proposition 9.3. If M(G) is both graphic and cographic then G is planar.

It will be proved in later. See Proposition 10.3.

Chapter. Matroid MinorsRemind graph minors. There are three operations: vertex deletion G\v,

edge deletion G\e and edge contraction G/e. We say that H is a minor of G ifH is obtained from G by contracting edges and deleting edges/vertices, i.e., afinite sequence of the three operations.

Definition 9.4. For a set T ⊆ E = E(M), we define M\T as a matroid onE − T such that

I(⊆ E − T

)is independent in M\T if and only if it is independent in M .

Simply, when M = (E, I),

M\T = (E − T, {I ⊆ E − T : I ∈ I}).

We call it a deletion. When T = {e}, we can write M\e = M\{e}.We define

M/T = (M∗\T )∗.

It is a contraction. Here, E(M/T ) = E(M∗\T ) = E(M∗)− T = E − T .

In graph sense, for a connected plane graphG, a contraction inG correspondsto a deletion in G∗. Hence the concept of minors in graph theory agrees withthe concept of minors in matroid theory.

Example 9.1. Why we do not define a matroid minor by using circuits? Let Gbe a below first graph. C is a cycle emphasized by blue dotted line, and e is anedge emphasized by red bold line. Then G/e is a second figure. We can checkthat C is no longer a cycle in G/e, but it is a union of cycles. This exampleshows why defining a matroid contraction as the way of describing circuits ishard.

34

We will discuss more about circuits of a matroid contraction in later.

Proposition 9.4 (Ranks of M\T and M/T ). For X ⊆ E − T ,

rM\T (X) = rM (X),

rM/T (X) = rM (X ∪ T )− rM (T ).

Proof. Denote E = E(M). The first equation is obvious by the definition ofM\T . In addition, it implies that

r(M\T ) = rM (E − T ).

Remind Proposition 8.2: rM∗(X) = |X| − r(M) + rM (E(M)−X).

rM/T (X) = |X| − r(M∗\T ) + rM∗\T ((E − T )−X)

= |X| − rM∗(E − T ) + rM∗(E − T −X)

= |X| −(|E − T | − r(M) + rM (T )

)+(|E − T −X| − r(M) + rM (X ∪ T )

)= rM (X ∪ T )− rM (T ).

Proposition 9.5. Let BT be a maximal independent subset of T in M . Then

X is independent in M/T ⇔ X ∪BT is independent in M .

Proof. (⇐) If X ∪BT is independent in M , then

rM (X ∪BT ) = |X ∪BT | = |X|+ |BT | = |X|+ rM (T ).

Hence

rM/T (X) = rM (X ∪ T )− rM (T ) ≥ rM (X ∪BT )− rM (T ) = |X|.

35

It implies that X is independent in M/T .(⇒) If X is independent in M/T , then

rM (X ∪ T ) = rM/T (X) + rM (T ) = |X|+ |BT |.

Let Y be a maximal independent subset of X ∪ T containing BT . By themaximality of BT , Y ∩ T = BT . In addition,

|Y −BT |+ |BT | = |Y | = rM (Y ) = rM (X ∪ T ) = |X|+ |BT |.

It implies that |Y − BT | = |X|, so Y − BT = X, i.e., Y = X ∪ BT . HenceX ∪BT is independent in M .

In a graph sense, we can interpret the previous proposition as follow: Let Gbe a graph, and E = E(G). Let T ⊆ E. Then BT is a maximal tree in T . Xis independent in G/T if and only if X is a tree in G/T . Possibly, a vertex ofthe tree X can be identified by T/T . Then we can observe that X ∪BT is alsoa tree in M . The converse is similarly done. In addition, we can check that themaximality of BT needs to guarantee the converse.

Sub-chapter. Circuits of M/T

Proposition 9.6. The collection of circuits of M/T is equal to the collectionof minimal nonempty sets in {C − T : C is a circuit of M}.

We checked in Example 9.1 that a member of {C − T : C is a circuit of M}is possibly not a circuit of M/T .

Lemma 9.7. Let A and B be hypergraphs on a same vertex set. Suppose thatthe followings hold:

1. For each A ∈ E(A), there is B ∈ E(B) contained in A.

2. For each B ∈ E(B), there is A ∈ E(A) contained in B.

Then m(A) = m(B).

Here, m(H) = (V,m(E)) for a hypergraph H = (V,E), where m(E) be thecollection of minimal sets of E. Obviously, m(H) is a clutter.

Proof. Let A ∈ E(m(A)) = m(E(A)). By the first condition, there is B ∈ E(B)such that B ⊆ A. WMA B ∈ m(E(B)) by taking a minimal set contained in theold B ∈ B. By the second condition, there is A′ ∈ E(A) such that A′ ⊆ B. ThenA′ ⊆ B ⊆ A. By the minimality of A, A′ = A, so A = B ∈ m(E(B)). It impliesthat m(A) ⊆ m(B), i.e., E(m(A)) ⊆ E(m(B)). Similarly, m(B) ⊆ m(A).

Similarly, we can prove that if A and B are two hypergraphs on a samevertex set satisfying

1′. for each A ∈ E(A), there is B ∈ E(B) containing A, and

36

2′. for each B ∈ E(B), there is A ∈ E(A) containing B,

then M(A) = M(B). Here, M(H) = (V,M(E)) for a hypergraph H = (V,E),where M(E) be the collection of maximal sets of E. Obviously, M(H) is aclutter.

Proof of Proposition 9.6. Let C(M/T ) be the collection circuits of M/T . Let

A = {C − T : C is a circuit of M, C − T 6= ∅}.

Note that m(C(M/T )) = C(M/T ) since it is a clutter. Note that m(A) is equalto the collection of minimal nonempty sets in {C − T : C is a circuit of M}. Inaddition, both C(M/T ) and A are edge sets of hypergraphs on E(M) − T =E(M/T ).

By the lemma, ETS

1. if C − T ∈ A is nonempty then C − T contains a circuit of M/T , and

2. if C ′ is a circuit of M/T then it contains D − T ∈ A.

To prove the first statement, ETS C − T ∈ A is dependent in M/T .

rM/T (C − T ) = rM ((C − T ) ∪ T )− rM (T )

= rM (C ∪ T )− rM (T )

≤ rM (C)− rM (C ∩ T )

= |C| − 1− |C ∩ T |= |C − T | − 1.

The inequality holds by (R3), the submodular inequality. Note that since C −T 6= ∅, C ∩ T is independent in M . From the above inequality, we can deducethat C − T is dependent in M/T .

Now we will prove the second statement. Let C ′ be a circuit of M/T . LetBT be a maximal independent subset of T in M . By the earlier proposition,X is independent in M/T if and only if X ∪ BT is independent in M . HenceC ′ ∪ BT is dependent in M . There is a circuit D of M such that C ′ ∪ BT .Obviously, D − T = D − BT ⊆ C ′. Since BT is independent in M , D 6⊆ BT ,i.e., D − T is nonempty, i.e., D − T ∈ A.

Some people may define a contraction of a matroid as this way. In otherwords, it might be possible defining M/T by giving its circuit as the collectionof minimal nonempty sets in {C − T : C is a circuit of M}.

What is the collection C(M\T ) of circuits of M\T?

C(M\T ) = {C ∈ C(M) : C ∩ T = ∅}.

Let C ⊆ E(M)− T . Then

C ∈ C(M\T )

37

⇔ C ∩(E(M\T )− I

)for any I ∈ I(M\T )

⇔ C ∩(E(M)− I

)for any I ∈ I(M)

⇔ C ∈ C(M).

(The remaining proof of Proposition 9.6 was done in October 14th.)

10 Week07-1, 2019.10.14. Mon

Sub-chapter. Closure in M/T

Proposition 10.1. clM/T (X) = clM (X ∪ T )− T.

Proof. Let e ∈ E(M)− T . Then

rM/T (X + e)− rM/T (X)

=(rM ((X + e) ∪ T )− rM (T )

)−(rM (X ∪ T )− rM (T )

)= rM ((X ∪ T ) + e)− rM (X ∪ T ).

It implies that e ∈ clM/T (X) if and only if e ∈ clM (X ∪ T )− T .

Without doubt, the closure function of M\T satisfies that

clM\T (X) = clM (X)− T.

(For e ∈ E(M)− T , rM\T (X + e)− rM\T (X) = rM (X + e)− rM (X).)

Sub-chapter: Minor

Definition 10.1. Let M,N be matroids. N is a minor of M if

N = M\T1/T2

for some T1, T2 ⊆ E(M) with T1 ∩ T2 = ∅.

The above definition of a matroid-minor is enough because of a below propo-sition.

Proposition 10.2. Let M be a matroid. Let T1, T2 be disjoint subsets of E(M).Then

M\T1\T2 = M\(T1 ∪ T2),

M/T1/T2 = M/(T1 ∪ T2),

M/T1\T2 = M\T2/T1.

38

Proof. The first one is obvious from the definition of the deletion. The secondone is a dualized problem of the first. More precisely,

M/T1/T2 = (M∗\T1)∗/T2

= (M∗\T1\T2)∗

= (M∗\(T1 ∪ T2))∗

= M/(T1 ∪ T2).

The last equality is proved by comparing rank functions. For any X ⊆E(M)− (T1 ∪ T2),

rM/T1\T2(X) = rM/T1

(X)

= rM (X ∪ T1)− rM (T1)

= rM\T2(X ∪ T1)− rM\T2

(T1)

= rM\T2/T1(X).

Sub-chpater. Minors of graphic matroidsLet G be a graph, and e be an edge of G. Then we can easily check that

M(G\e) = M(G)\e,M(G/e) = M(G)/e

by comparing circuits. First,

C ∈ C(M(G\e))

⇔ C is a cycle of G\e

⇔ C is a cycle of G not containing e

⇔ C ∈ C(M(G)) and e 6∈ G

⇔ C ∈ C(M(G))\e.

Second, ... ???Hence simply we can write M(G\T1/T2) = M(G)\T1/T2 for disjoint T1, T2 ⊆

E(G).This is the motivation why we define minors of matroids in this way. More-

over, it implies that

every minor of a graphic matroid is graphic.

Note that if N is a minor of M , then N∗ is a minor of M∗ since

N = M\T1/T2⇒ N∗ = (M\T1/T2)∗ = (M\T1)∗\T2 = M∗/T1\T2.

Hence we can deduce that

39

every minor of a cographic matroid is cographic.

In fancy words, we say that the matroid propreties, graphic and cogrphic, areclosed under the matroid-minor.

Proposition 10.3. M(G) is cogrpahic if and only if G is planar.

Proof. (⇐) Remind that M(G)∗ = M(G∗) whenever G is planar.(⇒) Remind that M(K5) and M(K3,3) are not cographic. Since the co-

graphic property is hereditary, a cographic matroid M(G) has no minor iso-morphic to M(K5) nor M(K3,3). By the previous observation, M(G\T1/T2) =M(G)\T1/T2 for disjoint T1, T2 ⊆ E(G). Note that deleting or adding isolatedvertices in G does not affect on M(G). Hence G has no minor isomorphic toK5 nor K3,3. By the Kuratowski’s theorem (or Kuratowski-Wagner theorem),G is planar.

Theorem 10.4 (Tutte). M is graphic iff M has no minors isomorphic to U2,4,M∗(K5), M∗(K3,3), F7 nor F ∗7 .

Here, F7 is the Fano matroid. Remind that it is a vector matroid inducedby 1 0 0 1 0 1 1

0 1 0 1 1 0 10 0 1 0 1 1 1

over the binary field.

We will not cover the proof of the Tutte’s theorem in this lecture.

Sub-chapter. Minors of representable matroids

Proposition 10.5. Every minor of F-representable matroids is F-representable.

Proof. Let M = M(A) be a vector matroid, where A is a matrix over the fieldF. ETS M\e and M/e are F-representable for each e ∈ E := E(M). WMA Ais standard form, see Theorem 8.4.

A =(Ir A0

),

where r = r(M), n = |E|, and A0 is an r × (n − r)-matrix. Moreover we willshow that we can conserve the structure of standard form A in M\e and M/e.

For a convenience, let k be an integer such that k-th column of A correspondsto e. Denote i-th row of A0 as βi for 1 ≤ i ≤ r. Let A0(i) be a submatrix of A0

deleting the i-th row βi.First, consider M\e. Without doubt, M\e = M(A\e) where A\e is a sub-

matrix of A deleting a column corresponding to e. Remaining works are aboutconserving a standard form. If k > r, A\e is still a standard form. When k ≤ r,we can check that A\e is not a standard form. However, we can obtain a new

40

standard form A′ of M by the row operations and swapping columns (which donot affect on the column space) such that

A′ =(Ir A′0

),

and e corresponds to a column in A′0, unless βk =(0 0 · · · 0

). In a case

of k ≤ r and βk = 01×(n−r) (actually, it is equivalent that e is a coloop),M(A)\e = M(A\e) = M(A/e), where A/e is a submatrix of A deleting bothk-th row and column. In addition, A/e is a standard form.

Second, consider M/e. Remind that M/e = (M∗\e)∗, and M∗ = M(A∗)where

A∗ =(AT0 In−r

)=(βT1 βT2 · · · βTr In−r

).

Moreover, column indexing of both A and A∗ is E with same ordering. Whenk ≤ r, by the earlier observation, M∗\e = M(A∗\e) where A∗\e is a submatrixof A∗ deleting βTk , i.e., A∗\e =

(A0(k)T In−r

). In addition, we can check that

M/e = (M∗\e)∗ = M((A∗\e)∗) where

(A∗\e)∗ :=(Ir−1 A0(k)

)is a standard form. When k > r, as like above, we can obtain a new standardform (A∗)′ of M∗ by the row operations and swapping columns (which do notaffect on the column space) such that

(A∗)′ =((AT0 )′ In−r

),

and e corresponds to a column in (AT0 )′, unless (k − r)-row of AT0 is zero (i.e.,(k − r)-column of A0 is zero, i.e., e is a loop in M). In a case of k > r and(k− r)-row of AT0 is zero, we can check that M(A∗)\e = M(A∗\e) = M(A∗/e),where A∗/e is a submatrix of A deleting k-row and (k − r)-column. HenceM/e = M(A∗/e)∗ = M((A∗/e)∗) where

(A∗/e)∗ =(Ir A0[k − r]

),

and A0[k − r] is a submatrix of A0 deleting (k − r)-th column.

Remark. If e is a loop or a coloop then

M\e = M/e.

First, when e is a loop, it is simply proved by comparing circuits,

C(M\e) = C(M)− {e} = C(M/e).

Second, when e is a coloop in M , i.e., e is a loop in M∗, M∗\e = M∗/e. Hence

M/e = (M∗\e)∗ = (M∗/e)∗ = (M∗∗\e)∗∗ = M\e.

41

Remark. Minors of transversal matroids are not necessarily transversal. Forexample, let G = (V = [4], E = {a, a′, b, b′, c, c′, d}) where edges a, a′ are inci-dent with vertices 1,2, b, b′ are incident 2,3, c, c′ are incident with 3, 4, and dis incident with 4,1. Then we can easily check that M(G) is transversal, butM(G/d) is not transversal.

Chapter. Matroid IntersectionOne basic question on matroid theory is following: Let M1 and M2 be two

matroids on E. Find a set X independent in both M1 and M2 with maximizing|X|. Why we consider this? By the greedy algorithm, we can maximize a givenweight for one matroid. When we think two matroid (or more), can we do likethat?

Example 10.1 (Maximum matching in a bipartite graph). Let G be a bipartitegraph with a bipartition (A,B). Let M1 be a partition matroid on E whosepartition is given by the vertices of A, i.e., e, f ∈ E are in a same partition setif e, f are incident with a same vertex in A. Let M2 be a partition matroid onE whose partition is given by the vertices of B. Then X ⊆ E is independentin both M1 and M2 if and only if X is a matching in G. In addition, it isa well-known problem finding a maximum matching in a bipartite graph. (Iremember that we can do it in poly-time. (?))

Example 10.2 (Intersection of three matroids is hard). Let G be a directedgraph. Let M1 be a matroid on E(G) such that X is independent if everyvertex has ≤ 1 incoming edge in X. Let M2 be a matroid on E(G) such that Xis independent if every vertex has ≤ 1 outgoing edge in X. Let M3 be a cyclematroid of G as a undirected graph. Then X is independent in M1,M2,M3 and|X| = |V (G)| − 1 if and only if it is a directed path passing through all verticesif and only if it is a (directed) Hamiltonian path. (Note that X is independentin M1,M2 if and only if it is directed path or directed cycle. Note that X isindependent with |X| = |V (G)| − 1 in M3 if and only if it is a spanning tree inthe underlying graph of G.) Note that an algorithm finding a Hamiltonian pathis NP-complete.

Theorem 10.6 (Matroid intersection theorem). Let M1 = (E, I1) and M2 =(E, I2) be matroids with rank functions r1 and r2, respectively. Then

maxI∈I1∩I2

|I| = minX⊆E

(r1(X) + r2(E −X)

).

(Moreover, there is an efficient algorithm finding I ∈ I1∩I2 achieving the aboveequation. Here, the oracle is about rank function.)

Proof. (Unfortunately, this proof gives an exponential-time algorithm.)(≤) If I is a common independent set of M1 and M2, then

r1(X) ≥ |X ∩ I|,r2(E −X) ≥ |(E −X) ∩ I|.

42

Hence r1(X) + r2(E −X) ≥ |X ∩ I|+ |(E −X) ∩ I| = |I|.(≥) Let k = minX⊆E

(r1(X) + r2(E −X)

). We will prove this problem by

an induction on |E|. Suppose that there is no e so that {e} is independent inboth M1 and M2. Then ∅ is the largest independent set in both M1 and M2,i.e., maxI∈I1∩I2 |I| = 0. In addition, we can easily check that k = 0. Let Xbe the set of loops in M1. Then E −X is the set of loops in M2 by the givenassumption (there is no e so that {e} ∈ I1∩I2). Hence r1(X) = r2(E−X) = 0.

Now WMA there is e so that {e} is independent in both M1 and M2. Ifthere is a common independent set of size k in M1\e and M2\e, then it is also acommon independent set in M1 and M2, so the proof is done. So WMA everycommon independent set in M1\e and M2\e is of size < k, i.e., by the inductionhypothesis, there is S ⊆ E − e so that

r1(S) + r2((E − e)− S) = rM1(S) + rM2

((E − e)− S) ≤ k − 1.

Let M be a matroid, and e ∈ E(M) not be a loop. If J is an independentset in M/e, then J + e is an independent set in M . Suppose not, i.e., J + eis dependent in M , i.e., there is a circuit C in M contained in J + e. Then∅ 6= C − e ⊆ J since e is not a loop (note that possibly e 6∈ C). It impliesthat C − e contains a circuit of M/e, so J is a dependent set in M/e. It is acontradiction, so we can conclude that J + e is an independent set in M .

If there is a common independent J set of size k − 1 in M1/e and M2/e,then J + e is a common independent set of size k in M1 and M2 by the aboveobservation, so the proof is done. So WMA every common independent set inM1/e and M2/2 is of size < k − 1, i.e., by the induction hypothesis, there isT ⊆ E − e so that

r1(T + e) + r2(E − T )− 2

= r1(T + e)− r1({e}) + r2(((E − e)− T ) + e)− r2({e})= rM1/e(T ) + rM2/e((E − e)− T )

≤ k − 2.

Then

2k − 1

≥ r1(S) + r2(E − e− S) + r1(T + e) + r2(E − T )

=(r1(S) + r1(T + e)

)+(r2(E − e− S) + r2(E − T )

)≥ r1(S ∩ T ) + r1(S + T + e) + r2(E − (S + T + e)) + r2(E − (S ∩ T ))

=(r1(S ∩ T ) + r2(E − (S ∩ T ))

)+(r1(S + T + e) + r2(E − (S + T + e))

)≥ k + k = 2k.

Here, the second inequality holds by applying the submodular inequality twice.It is a contradiction. Therefore, we can conclude that there is a common inde-pendent set of size k in both M1\e and M2\e, or a common independent set ofsize k − 1 in both M1/e and M2/e. It makes the proof complete.

43

Remark (Konig’s theorem). Remind that Konig’s theorem in graph theory: Fora bipartite graph G, ν(G) = τ(G) where ν(G) is the maximum size of a matchingin G, and τ(G) is the minimum size of a set of vertices hitting every edge (weusually call this set a vertex cover in G). We can prove this theorem usingTheorem 10.6, the matroid intersection theorem.

Note that ν(G) ≤ τ(G) is obvious. (Give a minimum vertex cover C in G,and a maximum matching M . Then a vertex in C is incident with at most oneedge in M since M is the matching. It implies that ν(G) = |M | ≤ |C| = τ(G).)

Let G be a bipartite graph with a bipartition (A,B). Let M1 = (E(G), I1)and M2 = (E(G), I2) be defined as like in Example 10.1. Let ri be the rankfunction of Mi for i = 1, 2. Then by the matroid intersection theorem,

ν(G) = maxI∈I1∩I2

|I|

= minX⊆E

(r1(X) + r2(E −X)

)= minX⊆E

(#(vertices in A hitting X) + #(vertices in B hitting E −X)

)≥ τ(G).

Check the last line....................

11 Week07-2, 2019.10.16. Wed

Lemma 11.1. Let M be a matroid on E = {e1, e2, . . . , en}. Let (m1,m2, . . . ,mn)be a finite sequence of positive integers, and

E′ = {e11, e21, . . . , em11 , e12, . . . , e

m22 , . . . , e1n, . . . , e

mnn }.

Let M ′ be a pair (E′, I ′) such that I ′ ∈ I ′ if and only if (i) I ′ contains at mostone of eki ’s for varying 1 ≤ k ≤ mi, and (ii) the underlying set I of I ′ is anindependent set in M . Here, the underlying set I of I ′ is a set of ei’s satisfyingeki ∈ I ′ for some k. Then M ′ is a matroid.

Proof. Obviously, I ′ satisfies (I1) and (I2). Let X ′, Y ′ ∈ I ′ with |X ′| < |Y ′|. LetX and Y be the underlying sets of X ′ and Y ′, respectively. By (i), |X| < |Y |, sothere is ei ∈ Y −X such that X+ei is independent in M . There is eki ∈ Y ′−X ′.In addition, the underlying set of Xi + eki is X + ei, so Xi + eki ∈ I. Therefore,I ′ satisfies (I3).

By considering a matroid-deletion, we can consider that (m1,m2, . . . ,mn)be a finite sequence of non-negative integers in the above lemma.

Theorem 11.2 (Rado’s theorem, 1962). Let G be a bipartite graph with abipartition (A,B). Let N be a matroid on B with the rank function r. Then Ghas a matching M covering A such that V (M) ∩B is independent in N if andonly if r(nG(X)) ≥ |X| for all X ⊆ A.

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Here, nG(X) =(∪x∈X nG(x)

)−X is the open neighbor set of X.

The Rado’s theorem is a generalization of the Hall’s theorem: G has amatching M covering A if and only if |nG(X)| ≥ |X| for all X ⊆ A. Set amatroid N in the Rado’s theorem as (B, 2B). It implies the Hall’s theorem.

Proof. (⇒) It is trivial. (Let MX be a matching whose edges are in M andincident with at least one vertex of X. Remind that M covers A, and V (M)∩Bis independent in N . Then r(nG(X)) ≥ r(V (MX) ∩B) = |X|.)

(⇐) Denote E := E(G). Let M1 be a matroid on E such that X is indepen-dent in M1 if and only if no 2 edges share a vertex in A. In other words, M1 isa partition matroid on E whose partition is given by the vertices of A. Let M2

be a matroid on E such that X is independent in M2 if and only if no 2 edgesshare a vertex in B, and the vertices of B incident with edges in X for a inde-pendent set of N . We can consider M2 as a matroid defined as like the previouslemma with (m1, . . . ,m|B|) is a sequence of degrees of corresponding vertices inB. Hence we can observe that M2 is actually a matroid by the previous lemma.

Let r1 and r2 be the rank functions of M1 and M2, respectively. Then forX ⊆ E,

r1(X) = |Z|,r2(E −X) ≥ r(nG(A− Z)) ≥ |A− Z|,

where Z(

= V (X) ∩ A)

is the set of vertices in A hitting edges in X. Thefirst inequality in the second line since E −X contains all edges incident withvertices of A − Z (so V (E −X) ∩ B ⊇ nG(A − Z)). The second inequality inthe second line holds by the given condition. Then

r1(X) + r2(E −X) ≥ |Z|+ |A− Z| = |A|.

By the matroid intersection theorem, M1 and M2 have a common independentset M of size |A|. We can check that M is a matching covering A such thatV (M) ∩B is independent in N , from the construction of M1 and M2.

Theorem 11.3 (Generalized version of Rado’s theorem). Let G be a bipartitegraph with a bipartition (A,B). Let N be a matroid on B with the rank functionr. Let d ≥ 0 be fixed. Then G has a matching M covering ≥ |A|−d vertices in A(or B) such that V (M)∩B is independent in N if and only if r(nG(X)) ≥ |X|−dfor all |X| ⊆ A.

Proof. The proof is almost same. Here, we notice some different parts.(⇒) Let MX be a matching whose edges are in M and incident with at least

one vertex of X. Then r(nG(X)) ≥ r(V (MX) ∩B)≥|X|−d.(⇐)

r2(E −X) ≥ r(nG(A− Z)) ≥ |A− Z|−d,r1(X) + r2(E −X) ≥ |A|−d.

By the matroid intersection theorem, M1 and M2 have a common independentset M of size |A|−d.

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By setting N = (B, 2B) in the above theorem, we can deduce a variation ofHall’s theorem: Let G be a bipartite graph with bipartition (A,B). Then Ghas a matching covering |A| − d vertices of A if and only if |nG(X)| ≥ |X| − dfor all X ⊆ A. From this,

ν(G) = max{|X| : X ⊆ A matchable to B}= |A| −min{d ∈ Z≥0 : |nG(X)| ≥ |X| − d for all X}= |A| − max

X⊆A

(|X| − |nG(X)|

)= minX⊆A

(|A| − |X|+ |nG(X)|

).

(Here, note that maxX⊆A(|X| − |nG(X)|

)≥ 0 by setting X = ∅.)

Remark (The rank function of transversal matroid). Let G be a bipartite graphwith a bipartition (A,B) Let M be a transversal matroid on A with respect toG. Remind that X ⊆ A is an independent set in M if and only if it is matchableto B. From the previous observation, we can deduce that

rM (Y ) = minX⊆Y

(|Y | − |X|+ |nG(X)|

).

for any Y ⊆ A. (Consider the above variation of the Hall’s theorem for aninduced subgraph G[Y ∪B]).

Theorem 11.4 (Matroid union theorem). Let M1,M2, . . . ,Mn be matroids onE. Let Ii be the collection of independent sets of Mi, and ri be the rank functionof Mi. Let I = {I1 ∪ · · · ∪ In : Ii ∈ I}. Then M = (E, I) is a matroid, and itsrank function satisfies that

rM (X) = minY⊆X

(r1(Y ) + r2(Y ) + . . .+ rn(Y ) + |X − Y |

).

Here, remind that rM (X) = maxZ⊆X,Z∈I |Z|. So the matrix union theoremis also regarded as a min-max type theorem.

Remark (LP duality). Why we consider these kinds of min-max problems? No-tice the duality of the linear programming (LP). LP is a problem having a formof the following:

Objective: max cTx,

Restrictions: Ax ≤ b, x ≥ 0 and x ∈ Rn.

Its dual is

Objective: max bT y,

Restrictions: AT y ≥ c, y ≥ 0 and y ∈ Rm.

In addition, the integer programming (IP) is same with LP except x ∈ Zn.Similarly, IP dual is same with LP dual except y ∈ Zm.

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One surprising observation is the optimal solutions of LP and LP dual areexactly same. However, it does not generally hold for IP and IP dual.

... img ...Researchers in combinatorial optimization work to find a condition that op-

timal solutions of IP and IP dual are same. Here, the matroid intersectiontheorem and the matroid union theorem (and their corollaries) are good exam-ples of these kinds. Konig’s theorem (ν(G) = τ(G) for a bipartite graph G) isone of the most popular example.

Proof of Theorem 11.4. First, WTS I satisfies the independent set axioms. Ob-viously, I satisfies that (I1) and (I2). Let X,Y ∈ I with |X| < |Y |. ThenX = ∪ni=1Ii and Y = ∪nj=1Ji where Ii, Ji ∈ Ii for each i. WMA Ii ∩ Ij = ∅ and

Ji ∩ Jj = ∅ for all distinct i, j. (Set new I ′ as I ′i = Ii −∪i−1t=1It, and do similarlyfor Ji.) Choose Ii’s and Ji’s so that

∑i |Ii ∩ Ji| is maximized. There is k such

that |Ik| < |Jk| since∑|Ii| = |X| < |Y | =

∑|Ji|, so there is e ∈ Jk − Ik such

that Ik + e is independent in Mk. If e 6∈ X, then X + e ∈ I makes (I3) hold.So WMA e ∈ X. It implies that e ∈ It for some t 6= k. Let us define

I ′i =

It − e, if i = t,

Ik + e, if i = k,

Ii, o.w..

Then I ′i ∈Mi for each i, I ′i’s are disjoint, and ∪I ′i = X. In addition,∑|I ′i∩Ji| =∑

|Ii∩Ji|+1. It contradicts to our choice of Ii’s and Ji’s. Hence we can concludethat I satisfies (I3), so M is a matroid on E.

Second, WTS

rM (X) = minY⊆X

(r1(Y ) + r2(Y ) + . . .+ rn(Y ) + |X − Y |

).

WMA X = E. (........ why?? ETS for X = E?) Let E1, E2, . . . , En be ndisjoint copies of E. Let πi : Ei → E be the natrual bijection. Let N1 be amatroid on ∪ni=1Ei such that X is independent in N1 if and only if πi(X ∩Ei) isindependent in Mi for all i. (We can easily show that N1 is actually a matroidby checking the three independent axioms. Combining matroids on disjointgrounds sets always makes a matroid.) Let N2 be a matroid on ∪Ei such thatX is independent in N2 if and only if no two copies of some element of E are inX, i.e., πi(X ∩ Ei) ∩ πj(X ∩ Ej) = ∅ for any distinct i, j. N1 is a nothing buta partition matroid (of which partition is the collection of the set of copies in∪Ei of each element of E).

Let I is a common independent set of N1 and N2. The the correspondingset I ′ = ∪πi(I ∩ Ei) of elements of E is independent in M , and |I ′| = |I|. Itimplies that

rM (E) ≥ maxI∈I(N1)∩I(N2)

|I|.

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Let J ⊆ I such that rM (E) = rM (J). J = ∪Ji for Ji ∈ Ii. WMA Ji’s aredisjoint. Let I = ∪π−1i (Ji). Then I is a common independent set of N1 andN2, so

rM (E) ≤ maxI∈I(N1)∩I(N2)

|I|.

By Theorem 10.6, the matroid intersection theorem,

rM (E) = minY⊆∪Ei

(rN1

(Y ) + rN2

((∪Ei)− Y

))= min∀i, Yi⊆Ei

(rN1

(∪ Yi

)+ rN2

((∪Ei)− (∪Yi)

))= min∀i, Yi⊆Ei

((∑ri(πi(Yi)

))+∣∣ ∪ (E − πi(Yi))∣∣)

= min∀i, Yi⊆Ei

((∑ri(πi(Yi)

))+∣∣E − ∩πi(Yi)∣∣).

Take Yi’s so that the minimum achieved. Set Z = ∩πi(Yi). It implies that

min∀i, Yi⊆Ei

((∑ri(πi(Yi)

))+∣∣E − ∩πi(Yi)∣∣)

≥ minZ⊆E

((∑ri(Z)

)+ |E − Z|

).

In addition, by taking Yi = π−1i (Z) for each i, we can deduce that the aboveinequality is actually the equality. In conclusion,

rM (E) = minZ⊆E

((∑ri(Z)

)+ |E − Z|

).

Let M be a matroid on E. Its rank function is r. Let us write M1 ∨M2 todenote the union of two matroids M1 and M2 on E.

Corollary 11.4.1. M has k bases whose union is E if and only if k ·r(X) ≥ |X|for all X ⊆ E.

Proof. Let Mi’s be copies of M (on the same ground set). Then by Theorem11.4, the matroid union theorem, (check first line....)

rM1∨M2∨···∨Mk(E) = |E|

⇔ minX⊆E(k · r(X) + |E −X|

)≥ |E|

⇔ For all X ⊆ E, k · r(X) ≥ |X|.

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Corollary 11.4.2. M has k disjoint bases if and only if k · r(X) + |E −X| ≥k · r(M) for all X ⊆ E.

Proof.

We can interpret them in the graph sense. Let G be a graph.

Corollary 11.4.3. G has k forests covering all edges if and only if k(|T |−1) ≥|E(G[T ])| for all T ⊆ V (G).

Proof.

12 Week09-1, 2019.10.28. Mon

Corollary 12.0.1 (Nash-Williams 1961; Tutte 1961). Let G be connected. ThenG has k edge-disjoint spanning trees if and only if for any partition of V (G) intoP1, P2, . . . , Ps for some s (Pi 6= ∅ and ∪Pi = V (G))

#(edges having ends in distinct Pi’s) ≥ (s− 1)k.

Proof. (⇒) Trivial.(⇐) Our goal is showing that k · r(Y ) + |E − Y | ≥ k · r(E) for any Y ⊆ E,

where E = E(G) and r is the rank function of M(G). (Then by a corollary ofthe matroid union theorem, the proof is done.) WMA F is a flat (otherwisewe adding edges not to increase its rank, and to decrease |E − Y |.) It impliesthat each component of G[Y ] = (V (G), Y ) is an induced subgraph of G. Let usdenote that P1, P2, . . . , Ps are the vertex-sets of the components of G[Y ]. Then

|E − Y | = #(edges having ends in distinct Pi’s) ≥ (s− 1)k,

k · r(Y ) = k(|V (G)| − s).

Adding above two, we obtain

k · r(Y ) + |E − Y | ≥ k(|V (G)| − 1) = k · r(M(G)).

Corollary 12.0.2. If G is 2k-edge-connected, then G has k edge-disjoint span-ning trees.

Proof. Let P1, . . . , Ps be a partition of V (G). By 2k-edge-connectedness, thereare at least 2k edges having ends in Pi and V (G)\Pi for each i. Then bydouble-counting,

#(edges having ends in distinct Pi’s) ≥ 2k · s · 1

2= ks.

By the previous corollary, the proof is done.

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Does the above corollary give us a tight bound? How to prove it without amatroid argument (i.e., only graph sense proof)?

Note that ifG has k edge-disjoint spanning trees, thenG is k-edge-connected.For two distinct vertices of G, we get k paths in k edge-disjoint spanning trees.Then paths are edge-disjoint. By the Menger’s theorem, the argument is com-pleted.

Application on mad(G), the maximum average degree of G.

mad(G) := maxH⊂G, |V (H)|≥1

∑v∈V (H) degH(v)

|V (H)|= max

2|E(H)||V (H)|

.

How can we decide mad(G) ≤ α for some α ∈ Q. Is this decision problem solvedin poly-time? Yes, it is by assuming that the algorithm of the matroid uniontheorem is poly-time.

Definition 12.1 (Bicircular matroid B(G)). Let define a matroid B(G) onE(G) satisfying one of below two: X ⊆ E(G).

• X is a circuit iff X is a theta graph, two edge-disjoint cycles with exactlyone common vertex, or two (vertex-)disjoint cycles and a path from avertex in a cycle to a vertex in the other cycle.

• X is independent iff each component of the subgraph (V (G), X) has atmost 1 cycle.

We call B(G) a Bicircular matroid.

We can easily check that the above two conditions are equivalent, and actu-ally a bicircular matroid is a matroid.

Proposition 12.1. B(G) is a matroid.

Proof. Our goal is showing that B(G) satisfies the independet axioms. (I1)and (I2) holds obviously. Now WTS (I3). Suppose that X,Y is independent,|X| < |Y |, and X ∪ {e} is not independent for all e ∈ Y −X. WMA E(G) =X∪Y . WMA G is connected (choose a component C of G so that |X∩E(C)| <|Y ∩E(C)|). WMA V (Y ) ⊆ V (X) (otherwise there is an edge e of Y so that oneof its ends is in V (Y ), and we can add it into X to make a larger independentset). Here V (Y ) is the set of vertices incident with Y .

Let X1 be the union of the edge-set of the components of (V,X) havinga cycle. (We call a graph having exactly one cycle a unicyclic graph.) LetX2 = X−X1, i.e., the union of edge-set of acyclic components of (V,X). WMAthere are no edges of Y −X, of which ends vertices are in X1 and X2, or bothare in X2 (of course there are two sub-cases: two ends are in a same component,or not). (Otherwise a edge of a previous case can be added into X to makea larger independent set.) It implies that every edge of Y , of which ends arein V (X2), is in X2. In addition, the number of edges of Y , of which endsare in V (X1), is less or equal than |V (X1)| (= |X1|) since each components of(V (G), Y )[V (X1)] is acyclic (so |E| = |V |−1) or unicyclic (so |E| = |V |). Hence|Y | ≤ |X1|+ |X2| = |X|, which is a contradiction.

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Note that the rank of B(G) is equal to |V (G)| −#(acyclic components).

Proposition 12.2. Let k > 0 and l > 0 be natrual numbers. Let Gl be a graphobtained from G by replacing each edge with l parallel edges. Then the followingare equivalent:

(1) mad(G) ≤ 2kl

(i.e., |E(H)|/|V (H)| ≤ k/l for every subgraph H of G with V (H) 6= ∅),

(2) |F | ≤ k · rB(Gl)(F ) for all F ⊆ E(Gl), and

(3) B(Gl) has k bases whose union is E(Gl).

Proof. (2) ⇔ (3) by a corollary of matroid union theorem.(1) ⇒ (2). Let F ⊆ E(Gl). WMA F is a flat of B(Gl). WMA F induces a

connected graph. If F has at least 1 cycle, then F contains all l parallel edgeswhenever it contains one of them. Let H be a subgraph of G consisting of edgesin F . Then

|F | = l|E(H)|,rB(Gl)(F ) = |V (F )| = |V (H)|.

Since |E(H)|/|V (H)| ≤ k/l, we can conclude that |F | ≤ k · rB(Gl).If F has no cycles, then rB(Gl)(F ) = |F |. So k · rB(Gl) ≥ |F |.(2) ⇐ (1). Let H be an induced subgraph of G. Let F = E(Hl) be the set

of edges of Gl corresponding to edges in H. Then

rB(Gl)(F ) = |V (H)| −#(acyclic components of E(Hl)) ≤ |V (H)|,|F | = l|E(H)|.

Then |F | ≤ k · rB(Gl)(F ) ≤ k|V (H)|, so |E(H)|/|V (H)| ≤ k/l.

By the above proposition, a poly-time alogrithm for the matroid union the-orem implies that there is a poly-time algorithm for the decision problem: de-ciding whether mad(G) ≤ α or not for a fixed α ∈ Q.

Note that a cycle matroid M(G) is represented by a signed vertex-edgeincident matrix of G over Q or R, of which (v, e)-entry is 1 if v is an end of e,−1 if v is the other ends of e, and 0 otherwise. A bicircular matroid can berepresented by a matrix replacing −1 to some other non-negative value in somefield. There is a survey on this topic...

Remark (Equivalence of the matroid intersection/union theorems). Let M1 andM2 be matroids. How do we get a common independent set from the matroidunion theorem?

A common independent set of size k exists

⇔ |B1 ∩B2| ≥ k for some Bi ∈ B(Mi), i = 1, 2,

⇔ |B1∪ (E−B2)| = |E|− |B2−B1| = |E|− |B2|+ |B1∩B2| ≥ |E|− |B2|+k

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⇔ M1 ∨M∗2 has rank ≥ |E| − r(M2) + k.

From this, the matroid union theorem implies that the matroid intersectiontheorem. (???) We proved that the matroid union theorem from the matroidintersection theorem. Hence both theorems are equivalent.

13 Week09-1, 2019.10.30. Wed

Sub-chapter. Algorithm for the matroid unionLet M1 = (E, I∞),M2 = (E, I∈), . . . ,Mn = (E, I\) be matroids on E. Our

goal is finidng a maximal independent set in M1 ∨ . . . ∨Mn. It is enough tofind disjoint n sets X1, X2, . . . , Xn where Xi ∈ Ii. First, find s 6∈ ∪Xi such that∪Xi + s is independent in ∨Mi if it exists.

For each i, let us define a directed graph DMi(Xi) on E, i.e., its vertex set

is E, so that

x→ y if x ∈ Xi, y 6∈ Xi, and Xi − x+ y ∈ Ii.

Let D = DM1(X1) ∪ . . . ∪ DMn(Xn), i.e., the graph induced by the union of(directed) edge-sets. Let Fi = {x 6∈ Xi : Xi + x ∈ Ii}. Denote X := ∪Xi andF := ∪Fi.

Lemma 13.1. Let s ∈ E −X. Then X + s is independent in ∨Mi if and onlyif D has a directed path from F to s.

Proof.

The previous lemma gives us an algorithm to find a maximal independentset in ∨Mi. Start from X0 = ∅, and get independent Xk+1 = Xk + sk+1 forsome sk+1 ∈ E −Xk satisfying the necessity condition of the lemma (can finda directed path from F to sk+1). Unfortunately, this algorithm has a bad time-efficiency since we check all elements of E −Xk for each k ≤ r(∨Mi) in a worstcase. (Is it good? Time r(∨Mi) × |E|?) (Note that there is a good algorithmto find a shortest path in a graph. Poly-time?)

Remark. In 1986, Cunningham found an algorithm of time O((r3/2 + n)mQ+

r1/2nm), where n is the number of matroids, m = |E|, r = r(∨Mi) ≤ m, and Q

is a time to test whether a set is independent (i.e., a running time of an oracle).

Chapter. ConnectednessNote that the disjoint union of 2 matroids is a matroid. More precisely, if

M1 = (E1, I1) and M2 = (E2, I2) are matroids with disjoint ground sets, then anew structure M = (E, I) with E = E1∪E2 and I = {X∪Y : X ∈ I1, Y ∈ I2}is a matroid. We can check easily that M satisfies the independent axioms.Denote it as M1 ⊕M2.

In a graph sense, a disconnected graph can be represent to the vertex-disjointunion of connected component. Moreover, the connectedness is defined as likethis in a usual sense. We also define the connectedness of matroids following the

52

usual sense. However, it is little bit different with the connectedness in graphtheory.

Definition 13.1. A matroid is connected if it cannot be written as the disjointunion of two non-trivial matroids.

Here a matroid is non-trivial means that its ground set is non-empty.

Proposition 13.2. A matroid M on E is connected if and only if r(X)+r(E−X) > r(E) for all ∅ 6= X ( E, where r is the rank function of M .

Proof. (⇐) Suppose M is disconnected, i.e., M = M1 ⊕M2 for some matroidsM1, M2 on non-empty disjoint ground sets. Denote the rank function of Mi asri, and Ei = E(Mi) for i = 1, 2. Then we can easily check that

r(Z) = r1(Z ∩ E1) + r2(Z ∩ E2).

Especially, r(E) = r1(E1) + r2(E2). By taking X = E1 6∈ {∅, E}, we canconclude that r(E) = r1(X) + r2(E −X) = r(X) + r(E −X).

(⇒) Suppose r(X) + r(E −X) = r(E) for some X 6= ∅, E. We claim that

M =(M\(E −X)

)⊕(M\X

).

ETS if P ⊆ X and Q ⊆ E − X are independent in M (i.e., independent inM\(E −X) and M\X, respectively), then P ∪Q is independent. By the sub-modular inequality,

r(P ∪Q) + r(X) ≥ r(P ) + r(X ∪Q),

r(X ∪Q) + r(E −X) ≥ r(Q) + r(E).

Therefore, we can conclude that r(P ∪ Q) ≥ r(P ) + r(Q) = |P | + |Q| = |P ∪Q|.

Proposition 13.3. A matroid M is connected if and only if its dual M∗ isconnected.

Proof. Let r and r∗ be the rank functions of M and M∗, respectively. Remindthat r∗(X) = |X|+ r(E −X)− r(M).

r∗(X) + r∗(E −X)− r∗(M)

=(|X|+ r(E −X)− r(M)

)+(|E −X|+ r(X)− r(M)

)−(|E| − r(M)

)= r(E −X) + r(X)− r(M).

Therefore, the given statements holds by the previous proposition.

Theorem 13.4 (Tutte). If M is connected, |E(M)| ≥ 2, and e ∈ E(M), thenM\e or M/e is connected.

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Proof. Suppose not, i.e., both M\e and M/e are not connected. Then by anearlier proposition, there is a partition (non-empty disjoint subsets coveringwhole set) (X1, Y1) and (X2, Y2) of E(M)− e such that rM\e(X1) + rM\e(Y1) =r(M\e) and rM/e(X2) + rM/e(Y2) = r(M/e). Equivalently,

r(X1) + r(Y1) = r(E − e),r(X2 + e) + r(Y2 + e) = r(E) + r({e}),

where E = E(M) and r is the rank function of M .Note that a connected matroid has no loops ( 6 ∃f ∈ E s.t. r(f) = 1)

and no coloops ( 6 ∃f ∈ E s.t. r(E − f) = r(E)). If M has a loop f , thenM = M\f ⊕ M\(E − f), so M is not connected. If M has a coloop f iffM∗ has a loop f , then M∗ is not connected iff M is not connected. Hencer(E − e) = r(M) and r({e}) = 1.

For a case that one of X1 ∩X2, X1 ∩ Y2, Y1 ∩X2, Y1 ∩ Y2 is empty, WMAX1 ∩ Y2 is empty by swapping labels. It implies that X1 ∩ X2 = X1 6= ∅ andY1 ∩ Y2 = Y2 6= ∅. Hence WMA X1 ∩X2 6= ∅ and Y1 ∩ Y2 6= ∅.

From these observations,

2 · r(M)− 1

=(r(X1) + r(Y1)

)+(r(X2 + e) + r(Y2 + e)

)=(r(X1) + r(X2 + e)

)+(r(Y1) + r(Y2 + e)

)≥(r(X1 ∩X2) + r((X1 ∪X2) + e)

)+(r(Y1 ∩ Y2) + r((Y1 ∪ Y2) + e)

)=(r(X1 ∩X2) + r((Y1 ∪ Y2) + e)

)+(r(X1 ∩X2) + r((Y1 ∪ Y2) + e)

)≥ 2 ·

(r(M) + 1

).

Here the first inequality is from the submodular inequality, and the second in-equality holds since (X1 ∩ X2, (Y1 ∪ Y2) + e) and (Y1 ∩ Y2, (X1 ∪ X2) + e) arepartitions of E, and M is connected. The above inequality derives a contradic-tion, so the proof is done.

(The remaining proof of Theorem 13.4 was done in November 06th.)

14 Week10-2, 2019.11.06. Wed

We say x ∼ y if x = y, or M has a circuit containing x and y. Note that ∼is transitive, i.e., if x ∼ y and y ∼ z then x ∼ z. (Check the Howework 2.1.)The reflexivity and symmetry are obvious. Hence (E(M),∼) is an equivalencerelation.

Definition 14.1. A component of a matroid M is an equivalence class of(E(M),∼).

Proposition 14.1. A matroid M is connected if and only if for all distinctx, y ∈ E(M), M has a circuit containing x and y, i.e., M has only one compo-nent.

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Proof. (⇒) Suppose there are x 6= y in E(M) such that M has no circuitscontaining both x and y. Let Z := [x], an equivalence class containing x. Thereis no circuit C of M such that C ∩ Z 6= ∅ and C − Z 6= ∅. It implies thatX is independent iff X ∩ Z and X − Z are independent. (Forward direction isobvious. Reverse direction: Suppose X is dependent, then there is a circuit Ccontained in X. By the given condition, either X ∩ Z or X − Z contains C.)Hence we can conclude that M is a disjoint union of two matroids M\Z andM\(E(M)− Z).

(⇐) Suppose r(X) + r(Y ) = r(M) for some partition (X,Y ) of E(M), i.e.,M is a disjoint union of two matroids M1 and M2 such that ∅ 6= X = E(M1)and ∅ 6= Y = E(M2). Then there is no circuit C such that C ∩ X 6= ∅ andC ∩ Y 6= ∅. (If there is such circuit C, then C ∩X and C ∩ Y are independentin M since they are proper subset of C.

r(C) + r(X) + r(Y ) ≥ r(C ∩X) + r(C ∪X) + r(Y )

≥ r(C ∩X) + r((C ∪X) ∩ Y ) + r(C ∪X ∪ Y )

= r(C ∩X) + r(C ∩ Y ) + r(M).

It implies that r(C) ≥ r(C ∩X) + r(C ∩ Y ), so r(C) = r(C ∩X) + r(C ∩ Y ) =|C ∩ X| + |C ∩ Y | = |C|, i.e., C is independent. It is a contradiction.) Hencethere is no circuit containing x ∈ X and y ∈ Y .

Corollary 14.1.1. Let G be a graph on > 2 vertices. Then TFAE:

• M(G) is connected.

• For any two distinct edges, G has a cycle containing both edges.

• G is 2-connected, and has no loops.

Proof. The second equivalence is easily checked. (It is totally graph theory, sowe skip. Remind that G is 2-connected if |V (G)| > 2, G is connected, and hasno cut-vertex.)

The first equivalence is nothing, but a special case of the previous proposi-tion.

Corollary 14.1.2. M is connected if and only if for all distinct x, y ∈ E(M),M has a cocircuit containing x and y.

Proof. Remind that M is connected iff M∗ is connected, and apply the previousproposition.

Proposition 14.2. M is connected if and only if for all distinct x, y ∈ E(M),M has a circuit or a cocircuit containing x and y.

Proof. Actually, the ealier proposition and corollary imply that if M is con-nected, then for all distinct x, y ∈ E(M), M has a circuit and a cocircuitcontaining x and y.

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Now WTS the converse. Suppose M = M1 ⊕M2 with E(M1), E(M2) 6= ∅.Then we can easily check that M∗ = M∗1 ⊕M∗2 . (E(M)−X ⊆ E(M) is a baseof M∗ iff X ⊆ E(M) is a base of M iff X ∩ E(M1) and X ∩ E(M2) are basesof M1 and M2 respectively iff (E(M)−X)∩E(M1) and (E(M)−X)∩E(M2)are bases of M∗1 and M∗2 respectively. Furthermore, check ranks also provesthe claim. r(M1) + r(M2) = r(M) iff r(M1∗) + r(M∗2 ) = r(M∗).) Choosex ∈ E(M1) = E(M∗1 ) and y ∈ E(M2) = E(M∗2 ). Then there are no circuits norcocircuits containing both x and y.

Proposition 14.3. Let M be a connected matroid with |E(M)| ≥ 3. Then forall |X| = 3, M has a circuit or a co circuit containing X.

It is not true that M has a circuit and a cocircuit containing X. For example,consider a theta graph (consider three paths, and identifying each starting andending vertex) with joining three P3 (a path of length 3, three edges and fourvertices). Take X as three edges joining with a starting vertex. Then there isno circuit containing X, but there is a cocircuit containing X (X itself).

Proof. Induction on |E(M)|.It is trivial for |E(M)| = 3. If r(M) = 0, then all elements are loops, so

M is not connected. If r(M) = 3, then all elements are coloops, so M is notconencted. For r(M) = 1 or 2, WMA r(M) = 1 by the duality. Note thatr(M) = 1, no loops, and no coloops implies that M = U1,3. Then E(M) is acocircuit. (When r(M) = 2, a dual case of rank = 1, M = U2,3. It implies thatE(M) is circuit.)

Now let us assume |E(M)| > 3. Choose e 6∈ X. By Theorem 13.4, M\e orM/e is connected. By symmetry (duality), WMA M/e is connected. If M/e hascocircuit D ⊇ X then D is a cocircuit of M (D is a circuit of M∗\e, so in M∗).If M/e has a circuit C ⊇ X then C or C + e is a circuit of M . (By Proposition9.5, C is dependent in M/e iff C + e is dependent in M . In addition, C − f isindependent in M/e iff C − f + e is independent in M , so C − f is independentin M for all f ∈ C. If C is dependent, then C is a circuit. If C = (C + e)− e isindependent, then C + e is a circuit.)

Proposition 14.4. Let M, N be connected matroids. Let e ∈ E(M) − E(N).If N is a minor of M , then N is a minor of M\e or M/e which is connected.

This proposition implies that we can choose a sequence e1, e2, . . . , el ofE(M)−E(N) such that contracting or deleting ei’s sequentially preserves con-nectedness.

It is possible that M/e (also M\e) contains N , but not connected. Forexample, ... graph...

Proof. ...

Definition 14.2. Let M be a matroid, and B be a base of M . Let us define abipartite graph GB on E(M) such that x and y is ajacent if x ∈ B, y 6∈ B, andB − x + y is a base. It is a bipartite graph with a bipartiton (B,E(M) − B).We call it the fundamental graph with respect to a base B.

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Theorem 14.5. M is connected if and only if GB is connected.

Proof. (⇐) By the definition ofGB , x ∈ B and y 6∈ B are adjacent inGB iff B+yhas a circuit containing x, so x ∼ y. If there is a path (a0 = a, a1, a2, . . . , al =b) in Gb, then a = a0 ∼ a1 ∼ a2 ∼ . . . ∼ al = b, so a ∼ b. Hence if GB isconnected, then M has only one component, i.e, M is connected.

(⇒) Suppose GB is not connected. Let H be a component of GB . LetB1 = V (H) ∩ B, Z1 = V (H) − B, B2 = (E(M) − V (H)) ∩ B and Z2 =E(M) − V (H) − B. They by Lemma 5.3, cl(B1) ⊇ Z1 and cl(B2) ⊇ Z2. Itimplies that r(M) = r(B1) + r(B2) = r(B1 ∪ Z1) + r(B2 ∪ Z2). Since (V (H) =B1 ∪ Z1, E(M) − V (H) = B2 ∪ Z2) is a partition of E(M), we can concludethat M is not connected.

By this theorem, we can reduce a problem checking connectedness of a ma-troid to a problem checking connectedness of graph. Note that BFS or DFS forneighborhoods of vertex gives a poly-time(?) algorithm for graph connectedness.Hence there is a good algorithm for checking that matroid is connected.

14.1 Homework 2.1

Proposition 14.6. Let x, y, z be distinct elements in the ground set of a ma-troid. Prove that if there are a circuit containing x and y and a circuit containingy and z, then there is a circuit containing x and z.

Proof. Let C and D be a circuit containing x and z, respectively, such thatC ∩ D 6= ∅. The existence of a such pair (C,D) is guaranteed by the givencondition (more precisely, the existence of y). Choose (C,D) so that |C ∪ D|is minimized. Suppose that C and D are distinct (so z 6∈ C and x 6∈ D).Let y′ ∈ C ∩ D. By the strong circuit elimination axiom, there is a circuitC ′ ⊆ (C∪D)−y′ containing x. Then it satisfies two properties: (i) C ′−C 6= ∅ (soC ′∩D ⊇ (C ′−C)∩D = C ′−C 6= ∅), and (ii) C ′−D = C−D. If (i) does not hold,then C ′ ⊆ C−y′ which contradicts to the minimality of circuits. If (ii) does nothold, then |C ′∪D| = |C ′−D|+ |D| < |C−D|+ |D| = |C∪D|, which contradictsto our choice of (C,D). Also, we can obtain a circuit D′ ⊆ (C ∪ D) − y′

containing z by the strong circuit elimination axiom. By the symmetry, itsatisfies (i’) D′ −D 6= ∅ (so C ∩D′ 6= ∅), and (ii) D′ − C = D − C. Then wecan observe that C ′ and D′ are circuits containing x and z, respectively, suchthat C ′ ∩ D′ ⊇ C ′ ∩ (D′ − C) = C ′ ∩ (D − C) = C ′ − C 6= ∅ by (ii’) and (i).However, C ′ ∪D′ ⊆ (C ∪D)− y′ (so |C ′ ∪D′| < |C ∪D|), which contradicts toour choice of (C,D). Therefore, we can conclude that C = D, so it is a circuitcontaining both x and z.

References

[1] Lecture by Prof. Oum

[2] James Oxley. Matroid Theory. Oxford, 2011.

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[3] Hassler Whitney. On the Abstract Properties of Linear Dependence. Amer-ican Journal of Mathematics, Vol. 57, No. 3 (Jul., 1935), pp. 509-533.

[4] Bela Bollobas. Modern Graph Theory.

[5] Stasys Jukna. Extremal Combinatorics with Applications in Computer Sci-ence.

[6] Andre Bouchet. Greedy Algorithm and Symmetric Matroids. MathematicalProgramming, 38 (1987), pp. 147-159.

[7] James F. Geelen. Matchings, Matroids and Unimodular Matrices. Universityof Waterloo, (1995).

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