COMPSCI 611: Advanced Algorithms
Lecture 5: Matroid
Hung Le
University of Massachusetts Amherst
01 Sept 2022
Recall: Minimum Spanning Tree
I Kruskal: Keep adding the lightest edge as long as it does not
creat a cycle.
I Proof of correctness: Exchange argument.
I Switching an edge of greedy algorithm and an edge of the
optimal solution.
I Today lecture: generalzing the idea of Kruskal’s algorithm to a
more general structure called matroid.
Matroid: What is?
Matroid
A matroid is a pair (S, J ) where S is a set of n elements and
J is a collection of subsets of S such that:
1. Non-emptiness: ∅ ∈ J .
2. Heredity: if X ∈ J , then every subset of X belongs to
J.
3. Exchange: Any two sets X and Y in J such that
|X| < |Y |, then there exists y ∈ Y \ X such that
X ∪ {y} ∈ J .
Y
X
y
Y
X∪ {y}
y
Examples
Matroid
A matroid is a pair (S, J ) where S is a set of n elements and
J is a collection of subsets of S such that:
1. Non-emptiness: ∅ ∈ J .
2. Heredity: if X ∈ J , then every subset of X belongs to
J.
3. Exchange: Any two sets X and Y in J such that
|X| < |Y |, then there exists y ∈ Y \ X such that
X ∪ {y} ∈ J .
I (S, {∅}) is a matroid.
I (S, 2S ) is a matroid. 2S is the power set of S.
More Examples
I ({1, 2, 3}, {∅, {1}, {2}, {3}}) is a matroid.
I ({1, 2, 3}, {∅, {2}, {1, 3}}) is NOT a matroid.
Independent Set and Basis
Let (S, J ) be a matroid
I Every subset of S in J is called an independent set.
I A subset I ∈ J is a basis if it is NOT a proper subset of
another independent set.
Example: matroid ({1, 2, 3}, {∅, {1}, {2}, {3}}).
I {1} is a basis, ∅ is not a basis.
Example: matroid (S, {∅}).
I ∅ is a basis.
Size of Bases
Lemma
Any two bases have the same size.
Proof: Exchange property
I Suppose that there are two bases X and Y s.t. |Y | > |X|.
I By the exchange property, there exists y ∈ Y \ X such that
X ∪ {y} is an independent set.
I As X ⊆ X ∪ {y}, we get a contradiction. (X is a basis if it is
NOT a proper subset of another independent set.)
Linear Matroid
Linear Matroid
Let A be an n × n matrix. S = {1, 2, . . . , n} and I ⊆
{1, 2, . . . n} is independent if and only if the coorresponding
subset of columns is linearly independent.

a11
a

A =  21
. . .
an1
...
...
....
...

a1n
a2n 


. . .
ann
(1)
Uniform Matroid
Uniform Matroid
Let k ∈ [0, n] be an integer. S = {1, 2, . . . , n} and I ⊆
{1, 2, . . . n} is independent if and only if |I| ≤ k.
Example:
I S = {1, 2, 3} and k = 2, then:
I Any subset of size at most 2, for example {1} or {1, 2}, is
independent.
I Bases: {1, 2}, {2, 3}, {1, 3}.
Graphic Matroid
Graphic Matroid
Let G = (V, E) be an undirected graph. S = E and I ⊆ E
is independent if (V, I) has no cycle.
Verifying matroid properties of Graphic Matroid:
I Non-emptiness: ∅ is independent as (V, ∅) has no cycle.
I Heredity: if (V, I) is acyclic, so is (V, I 0 ) ∀ I 0 ⊆ I.
I Exchange property is more difficult to verify.
Exchange Property of Graphic Matroid
Graphic Matroid
Let G = (V, E) be an undirected graph. S = E and I ⊆ E
is independent if (V, I) has no cycle.
I Let X, Y be two independent sets such that |X| < |Y |.
I (V, X) and (V, Y ) have no cycle.
1
1
2
3
4
6
5
X
e
2
6
3
4
Y
5
I Need to show: ∃e ∈ Y \ X such that adding e to (V, X)
creates no cycle.
Proof of Exchange Property (1)
1
1
2
3
4
6
5
X
e
2
6
3
4
Y
5
Lemma
∃ e ∈ Y \ X connecting two different connected components
of X.
Proof: Let C be any connected component of (V, X), then
|Y ∩ E(C)| ≤ |E(C) ∩ X| since E(C) ∩ X induces a spanning tree
of vertices in C.
I The total number of edges of Y insides all connected
components of X is at most |X|.
I Since |Y | > |X|, there must be an edge connecting two
different connected components of X.
Proof of Exchange Property (2)
1
1
2
3
4
6
5
X
e
2
6
3
4
Y
5
Lemma
∃ e ∈ Y \ X connecting two different connected components
of X.
I Let e ∈ Y \ X be an edge in the lemma.
I By the lemma, adding e to (V, X) creates no cycle.
I Thus, X ∪ {e} is an independent set.
Basis of Graphic Matroid
Graphic Matroid
Let G = (V, E) be an undirected graph. S = E and I ⊆ E
is independent if (V, I) has no cycle.
Question: What is a basis of a graphic matroid?
Matroid Optimization
Matroid Optimization Problem
Let (S, J ) be a matroid, such that for every e ∈ S has a
weight w(e) ≥ 0. Find a basis of maximum total weight of
the matroid (S, J ).
Example:
I (S, J ) is a graphic matroid, then the matroid optimization
problemm is to find a spanning tree of maximum weight.
Greedy Algorithm
Idea: repeatedly add to the solution the maximum weight element if
it does not destroy independence.
GreedyBasis((S, J ), w)
1. sort S in decreasing weights
2. A ← ∅
3. for each e ∈ S in the sorted order
4. if A ∪ {e} ∈ J
5.
A ← A ∪ {e}
6. return A
I Let F (n) be the running time to test whether a set I belongs
to J or not. Here n = |S|.
I Running time of GreedyBasis is O(n log n + F (n)n)
Correctness: Exchange Argument
I A = {a1 , . . . , ak } greedy solution, sortedy decreasingly.
I O = {o1 , . . . , ok } optimal solution, sorted decreasingly.
Since O is optimal:
w(o1 ) + . . . + w(ok ) > w(a1 ) + . . . + w(ak )
(2)
Let i be smallest index such that w(oi ) > w(ai ). Let
Ai−1 = {a1 , . . . , ai−1 }
Oi = {o1 , . . . , oi−1 , oi }
(3)
As |Oi | > |Ai−1 |, there exists o ∈ Oi \ Ai−1 such that
{Ai−1 } ∪ {o} is independent by the exchange property.
I But w(o) ≥ w(oi ) > w(ai ), the greedy algorithm will choose o
instead of ai to add to A, a contradiction.
Greedy for non-matroid
If (S, J ) is not a matroid, then the greedy algorithm does not
produce an optimal solution.
A counter example:
I Let X, Y be subsets of S s.t. |X| < |Y | and that ∀ y ∈ Y ,
X ∪ {y} 6∈ J .
I Let |X| = m, and assign w(x) = m + 2 for every x ∈ X.
I For every y ∈ Y \ X, assign w(y) = m + 1.
I For all z ∈ S \ (X ∪ Y ) has w(z) = 0.
The greedy algorithm will produce X, but
w(X) = m(m + 2) = m2 + 2m.
The optimal solution is Y that has
w(Y ) ≥ (m + 1)(m + 1) = m2 + 2m + 1 > w(X).