The Greedy Algorithm
5.2 Characterizations of Matroids
We begin with two characterizations of matroids which show that these struc-tures can be viewed as generalizations of the notion of linear independence.
Theorem 5.2.1 Let M = (E,S) be an independence system. Then the fol-lowing conditions are equivalent:
linear and algebraic independence, respectively. In the next section, we give some other possible definitions. The generalization of the algorithm of Kruskal to matroids was found independently by [Gal68], [Wel68] and—actually a bit earlier—by Edmonds; see [Edm71].
Early forms of the underlying ideas go back even to [Bor26a] and [Rad57].
2The construction of graphic matroids admits an obvious generalization to multigraphs.
While this observation may, at first glance, not seem very interesting, it will be important in the context of duality.
3Thetail-partition matroid is defined analogously.
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(1) M is a matroid.
(2) ForJ, K∈Swith|J|=|K|+ 1,there always exists somea∈J\K such that K∪ {a} is also inS.
(3) For every subset A of E,all maximal independent subsets of Ahave the same cardinality.
Proof Suppose first thatM is a matroid for which (2) is not satisfied. Then there areJ, K∈Swith |J|=|K|+ 1 such that, for every a∈J\K, the set
Note that K is not the solution of the associated optimization problem:
w(K) =k(k+ 2)<(k+ 1)2≤w(J). On the other hand, the greedy algo-rithm first chooses all elements of K, because they have maximal weight.
Afterwards, the weight of the solution cannot be increased any more: all re-maining elementseeither have w(e) = 0 or are in J\K, so thatK∪ {e} is not in S, according to our assumption above. Thus M is not a matroid, a contradiction. Hence (1) implies (2).
Now let A be an arbitrary subset of E and J and K two maximal in-dependent subsets contained in A; thus there is no independent subset of A containing J or K, except J or K itself, respectively. Suppose we have
|K|<|J|. As S is closed under inclusion, there is a subset J of J with
|J|=|K|+ 1. By (2), there exists an elementa∈J\K such that K∪ {a} is independent, contradicting the maximality ofK. Thus (2) implies (3).
Finally, suppose thatM is not a matroid, but satisfies condition (3). Then the greedy algorithm does not work for the corresponding optimization prob-lem. Thus we may choose a weight function w for which Algorithm 5.1.1 constructs an independent setK={e1, . . . , ek}, even though there exists an independent setJ ={e1, . . . , eh} of larger weight. We may assume that the elements of J and K are ordered according to decreasing weight and that J is a maximal independent subset of E. By construction, K is maximal too. Then (3), withA=E, implies h=k. We use induction onm to estab-lish the inequality w(ei)≥w(ei) for i= 1, . . . , m; the instance m=k then gives a contradiction to our assumptionw(K)< w(J). Now the greedy algo-rithm choosese1as an element of maximal weight; thus the desired inequality holds form= 1. Now suppose that the assertion holds form≥1 and assume w(em+1)< w(em+1). Consider the set
w(e)≤w(em+1)< w(em+1), since the greedy algorithm chose the element em+1 after having chosenem; hencee /∈Aso thatS is indeed a maximal sub-set ofA. But{e1, . . . , em+1}is also an independent subset ofA, contradicting
condition (3). Thus (3) implies (1).
Note that condition (2) of Theorem 5.2.1 is analogous to a well-known result from linear algebra, the Steinitz exchange theorem; therefore (2) is usually called the exchange axiom. Similarly, condition (3) is analogous to the fact that all bases of a linear subspace have the same cardinality. In fact, Theorem5.2.1immediately gives the following result.
Theorem 5.2.2 Let E be a finite subset of a vector space V, and let Sbe the set of all linearly independent subsets ofE. Then(E,S) is a matroid.
A matroid constructed as in Theorem5.2.2 is called a vectorial matroid or amatric matroid. The second name comes from the fact that a subsetE of a vector space V can be identified with the set of columns of a suitable matrix (after choosing a basis for V); then the independent sets are the linearly independent subsets of this set of columns. An abstract matroid is called representable over F, whereF is a given field, if it is isomorphic to a vectorial matroid in a vector space V over F. (We leave it to the reader to give a formal definition of the termisomorphic.)
Exercise 5.2.3 Prove that every graphic matroid is representable over F for every field F. Hint: Use the incidence matrix of an arbitrary orientation of the underlying graph.
Exercise 5.2.4 LetG= (V, E) be a graph. A set A⊆E is called ak-forest ofGif it splits into a forestF and at mostkedges not inF. Prove that the set of all k-forests of G forms a matroid Mk(G). Hint: Use Theorem 5.2.1 and reduce the assertion to the casek= 0, where the matroid in question is just the graphic matroid.
Exercise 5.2.5 Let G= (V, E) be a connected graph which is not a tree.
Prove that the subsets ofE which contain at most one cycle form a matroid on E. Does the analogous result hold if we choose as independent sets all subsets ofE containing at most two cycles?
Let us introduce some more terminology chosen in analogy to that used in linear algebra. The maximal independent sets of a matroid M = (E,S) are called itsbases. Therank ρ(A) of a subset A of E is the cardinality of a maximal independent subset of A. Any subset ofE not contained inSis calleddependent.
Exercise 5.2.6 Let ρbe the rank function of a matroid M= (E,S). Show that ρhas the following properties.
140 5 The Greedy Algorithm
(1) ρ(A)≤ |A|for allA⊂E;
(2) ρismonotonic, that is,A⊂B impliesρ(A)≤ρ(B) (for all A, B⊂E);
(3) ρ is submodular, that is, ρ(A∪B) +ρ(A∩B)≤ρ(A) +ρ(B) for all A, B⊂E.
Conversely, matroids can be defined using their rank function. LetE be a set andρa function from the power set ofE toN0satisfying conditions (1), (2), and (3) above. Then the subsetsX of E satisfyingρ(X) =|X| are the independent sets of a matroid onE; for example, see [Wel76]. Submodular functions are important in combinatorial optimization and matroid theory;
see, for instance, [PymPe70], [Edm70], [FraTa88], [Qi88], and the monograph by [Fuj91].
To solve Exercise5.2.6, we need a result worth noting explicitly, although it is a direct consequence of condition (2) of Theorem5.2.1.
Theorem 5.2.7(Basis completion theorem) LetJ be an independent set of the matroidM= (E,S).ThenJ is contained in a basis of M.
We will now use the rank function to introduce another important concept;
this rests on the following simple observation.
Lemma 5.2.8 Let J be an independent set of the matroid (E,S), and let X, Y ⊆E. IfJ is a maximal independent set ofX as well as of Y,thenJ is also a maximal independent set of X∪Y.
Theorem 5.2.9 Let M = (E,S) be a matroid and A a subset of E. Then there is a unique maximal setB containingAsuch thatρ(A) =ρ(B),namely
B=
e∈E:ρ
A∪ {e}
=ρ(A) .
Proof First letCbe an arbitrary superset ofAsatisfyingρ(A) =ρ(C). Then ρ(A∪ {e}) =ρ(A) holds for each e∈C: otherwise we would have ρ(C)≥ ρ(A∪ {e})> ρ(A). Thus we only need to show that the setB defined in the assertion satisfies the conditionρ(A) =ρ(B). LetJ be a maximal independent subset ofA; thenJ is also a maximal independent subset ofA∪ {e}for each e∈B. By Lemma5.2.8,J is also a maximal independent subset ofB.
The setB defined in Theorem5.2.9is called thespan ofAand is denoted byσ(A). By analogy with the terminology of linear algebra, agenerating set ofM is a setAwithE=σ(A). A setAsatisfyingσ(A) =Ais called aclosed set, and a hyperplane is a maximal closed proper subset of E. Matroids may be characterized by systems of axioms using the notion of span or of hyperplane; we refer again to [Wel76]. Let us pose some exercises concerning the concepts just introduced.
Exercise 5.2.10 Let M = (E,S) be a matroid. Then the span operator σ has the following properties:
(1) X⊂σ(X) for allX⊂E;
(2) Y ⊂X⇒σ(Y)⊂σ(X) for allX, Y ⊂E;
(3) σ(σ(X)) =σ(X) for allX⊂E;
(4) Ify /∈σ(X) andy∈σ(X∪ {x}), thenx∈σ(X∪ {y}).
Property (3) explains why the sets σ(A) are called closed; property (4) is again called the exchange axiom because it is basically the same as condi-tion (2) of Theorem5.2.1. Conversely, the conditions given above can be used for an axiomatic characterization of matroids by the span operator.
Exercise 5.2.11 Show that the bases of a matroid are precisely the minimal generating sets.
Exercise 5.2.12 Let (E,S) be a matroid. Prove the following assertions:
(a) The intersection of closed sets is closed.
(b) σ(X) is the intersection of all closed sets containing X.
(c) X is closed if and only if ρ(X∪ {x}) =ρ(X) + 1 for somex∈E\X. Exercise 5.2.13 Let (E,S) be a matroid of rankr, that is,ρ(E) =r). Show that (E,S) contains at least 2r closed subsets.
Let us introduce one further notion, this time generalizing a concept from graph theory. Acircuit in a matroid is a minimal dependent set—by analogy with a cycle in a graph. We have the following result; the special case of a graphic matroid should be clear from the preceding discussion.
Theorem 5.2.14 LetM = (E,S)be a matroid,J an independent set ofM, ande any element of E\J. Then eitherJ∪ {e} is independent, or J∪ {e} contains a unique circuit.
Proof Suppose thatJ∪ {e} is dependent, and put C=
c∈E:
J∪ {e}
\ {c} ∈S .
NoteC=∅, sincee∈Cby definition. Also,Cis dependent, because otherwise it could be completed to a maximal independent subsetK ofJ∪ {e}. AsJ is independent itself, we would have |K|=|J|, so thatK= (J∪ {e})\ {d}for some elementd. But thendwould have to be an element ofC, a contradiction.
It is easy to see thatC is even a circuit: if we remove any elementc, we get a subset of (J∪ {e})\ {c}which is, by definition ofC, an independent set. It remains to show thatCis the only circuit contained inJ∪{e}. Thus letDbe any circuit contained inJ∪ {e}. Suppose there exists an elementc∈C\D.
142 5 The Greedy Algorithm
ThenDis a subset of (J∪ {e})\ {c}which is an independent set. Therefore
C⊂D, and henceC=D.
We conclude this section by characterizing matroids in terms of their cir-cuits. We begin with a simple observation.
Lemma 5.2.15 Let (E,S) be a matroid. A subset A of E is dependent if and only ifρ(A)<|A|.Moreover,ρ(A) =|A| −1 for every circuitA.
Theorem 5.2.16 Let M= (E,S) be a matroid, and let C be the set of all circuits ofM.ThenC has the following properties:
(1) If C⊂D, thenC=D for allC, D∈C;
(2) For allC, D∈CwithC=D and for eachx∈C∩D,there always exists someF∈C withF⊂(C∪D)\ {x}.
Conversely,assume that a set system(E,C)satisfies the preceding twocircuit axioms.Then there is a unique matroid(E,S)havingCas its set of circuits.
Proof First, let C be the set of circuits of M. As circuits are minimal de-pendent sets, condition (1) is trivial. The submodularity ofρyields, together with Lemma5.2.15,
ρ(C∪D) +ρ(C∩D)≤ρ(C) +ρ(D) =|C|+|D| −2 =|C∩D|+|C∪D| −2.
As C and D are minimal dependent sets, C∩D is independent; therefore ρ(C∩D) =|C∩D|, and hence
ρ
(C∪D)\ {x}
≤ρ(C∪D)≤ |C∪D| −2<(C∪D)\ {x}. By Lemma5.2.15, (C∪D)\ {x} is dependent and hence contains a circuit.
Conversely, supposeC satisfies the conditions (1) and (2). If there exists a matroid (E,S) with set of circuitsC, its independent sets are given by
S={J⊂E:J does not contain any element ofC}.
Obviously, S is closed under inclusion, and it suffices to show that (E,S) satisfies condition (2) of Theorem5.2.1. Suppose that this condition is not satisfied, and choose a counterexample (J, K) such that|J∪K| is minimal.
LetJ\K={x1, . . . , xk}. Note k= 1, because otherwise|J|=|K|+ 1 would imply thatKis a subset ofJ, and henceJ=K∪{x1}would be independent.
Our assumption meansK∪{xi}∈/Sfori= 1, . . . , k. In particular, there exists C∈C with C⊂K∪ {x1}; as K is independent, x1 must be in C. As J is independent, there is an elementy∈K\J which is contained inC. Consider the set Z= (K\ {y})∪ {x1}. If Z is not in S, then there exists D∈C withD⊂Z and x1∈D, and the circuit axiom (2) yields a set F ∈C with F⊂(C∪D)\ {x1} ⊂K, contradictingK∈S. HenceZ must be independent.
Note that|Z∪J|<|K∪J|. As we chose our counterexample (J, K) to be minimal, (J, Z) has to satisfy condition (2) of Theorem 5.2.1. Thus there exists some xi, say x2, such that Z∪ {x2} ∈S. But K∪ {x2}∈/S, so that there is a circuitC∈CwithC⊂K∪ {x2}. We must havex2∈C, because K is independent; and (K\ {y})∪ {x1, x2} ∈Syields y∈C. ThusC=C, andy∈C∩C. Using the circuit axiom (2) again, there exists a setF∈C with F⊂(C∪C)\ {y} ⊂(K\ {y})∪ {x1, x2} ∈S. This contradicts the definition of S. Therefore M= (E,S) is indeed a matroid, and clearly C is
the set of circuits ofM.
Exercise 5.2.17 Show that the set C of circuits of a matroid (E,S)