Matroid Theory — Definition, Examples, Greedy Algorithm on Matroids

🎯 Learning Objectives

• Define matroids and their axioms
• Verify if a structure is a matroid using the exchange property
• Prove correctness of greedy algorithm on matroids
• Apply matroid theory to scheduling with deadlines

1. Matroid Definition

1.1 Intuition

1.2 Formal Definition

• S: A finite ground set
• $\mathcal{I}$: A family of subsets of S called independent sets Satisfying three axioms:

1. Non-emptiness: $\emptyset \in \mathcal{I}$
2. Hereditary property: If $A \in \mathcal{I}$ and $B \subseteq A$, then $B \in \mathcal{I}$
3. Exchange property: If $A, B \in \mathcal{I}$ and $|A| < |B|$, then $\exists x \in B \setminus A$ such that $A \cup \{x\} \in \mathcal{I}$

1.3 Axiom Intuition

2. Examples of Matroids

2.1 Uniform Matroid $U_{k,n}$

• $S = \{1, 2, ..., n\}$
• $\mathcal{I} = \{A \subseteq S : |A| \leq k\}$ Verification:
• Hereditary: If |A| ≤ k and B ⊆ A, then |B| ≤ |A| ≤ k ✓
• Exchange: If |A| < |B| ≤ k, pick any x ∈ B\A, |A∪{x}| ≤ |A|+1 ≤ k ✓

2.2 Graphic Matroid $M(G)$

• $S =$ edges of graph $G$
• $\mathcal{I} =$ sets of edges with no cycles (forests) Verification: Exchange property: If A and B are forests with |A| < |B|, since B has more edges, there must be an edge in B connecting two components that A doesn't connect. Adding it creates no cycle. (Diagram) Graphic matroid:
• S = {e1, e2, e3, e4}
• Independent sets: {}, {e1}, {e2}, {e3}, {e4}, {e1,e2}, {e1,e3}, {e1,e4}, {e2,e3}, {e2,e4}, {e3,e4}, {e1,e2,e4}, {e1,e3,e4}
• Not independent: {e1,e2,e3} (forms cycle)

2.3 Partition Matroid

• $S$ partitioned into groups $S_1, S_2, ..., S_k$
• $\mathcal{I} = \{A : |A \cap S_i| \leq c_i \text{ for each group}\}$ Example: Selecting at most 1 person from each department.

2.4 Linear (Vector) Matroid

• $S =$ columns of a matrix over a field
• $\mathcal{I} =$ linearly independent columns

3. Greedy Algorithm on Matroids

3.1 Generic Greedy Algorithm

pseudo
Greedy-Max(M, w):
    Sort S by decreasing weight: w(x₁) ≥ w(x₂) ≥ ... ≥ w(xₙ)
    I = ∅
    For i = 1 to n:
        If I ∪ {xᵢ} ∈ ℐ:
            I = I ∪ {xᵢ}
    Return I

3.2 Correctness Proof

1. Let $A = \{a_1, a_2, ..., a_k\}$ be the greedy solution (in order chosen)
2. Let $O = \{o_1, o_2, ..., o_m\}$ be any optimal solution (sorted by weight)
3. Show $k = m$ and $w(A) \geq w(O)$:
   • If $|A| < |O|$, exchange property would let greedy add an element
   • If $w(a_j) < w(o_j)$ for some j, exchange property gives contradiction

3.3 Maximum Spanning Tree

text
Kruskal's algorithm:
    Sort edges by weight (descending for max, ascending for min)
    I = ∅
    For each edge (u,v) in sorted order:
        If I ∪ {(u,v)} has no cycle:
            I = I ∪ {(u,v)}
    Return I

4. Scheduling with Deadlines

4.1 Problem

• Jobs with deadlines $d_i$ and profits $p_i$
• Each job takes 1 unit of time
• Schedule jobs to maximize profit within deadlines

4.2 Matroid Formulation

• Ground set: jobs
• Independent set: jobs that can be scheduled within their deadlines
• This forms a matroid! (Called the scheduling matroid)

4.3 Greedy Solution

text
1. Sort jobs by profit (decreasing)
2. Schedule each job at the latest available time ≤ deadline
3. If no available slot, skip the job

4.4 Worked Example

5. 📝 Practice Questions

Q1: Verify that the set of subsets of size at most 2 from {1,2,3,4} forms a matroid.

Answer: This is the uniform matroid U₂,₄. Non-emptiness: ∅ is size 0 ≤ 2 ✓. Hereditary: any subset of a size-≤2 set has size ≤2 ✓. Exchange: if |A| < |B| ≤ 2, pick any x ∈ B\A, then |A∪{x}| ≤ |A|+1 ≤ 2 ✓. Q2: Prove that the greedy algorithm fails for the maximum-weight independent set problem when the structure is NOT a matroid.

Answer: Consider S = {a, b, c} with weights w(a)=10, w(b)=9, w(c)=8. Let ℐ = {∅, {a}, {b}, {c}, {a,b}, {a,c}} (note: {b,c} not independent). Greedy picks {a} (weight 10), then can add c → {a,c} (weight 18). But optimal is {a,b} (weight 19). Greedy fails because ℐ doesn't satisfy the exchange property: |{b,c}| = 2 > |{a}| = 1, but no element of {b,c} can be added to {a} to create a larger independent set... Actually {a,b} and {a,c} are independent. So |{a,b}| = |{a,c}| = 2 > 1 = |{b}|. Exchange: from {b} and {a,b}: b→{a,b} is adding a, which works. The structure does satisfy exchange — it IS a matroid. Let me give a counterexample that's NOT a matroid:

S = {a, b}, ℐ = {∅, {a}} (but NOT {b}). This violates hereditary (subset of independent {a} is ∅, ok) but exchange fails: |∅| = 0 < |{a}| = 1, but ∅∪{a} not in ℐ? Actually ∅∪{a} = {a} which IS in ℐ. This IS a matroid.

Better: S = {a,b,c}, ℐ = {∅, {a}, {b}, {a,b}}. Exchange: {a} and {b} both size 1 — no |A| < |B| case. OK. {a} and {a,b}: |{a}|=1 < |{a,b}|=2, pick b from {a,b}{a} = {b}, {a}∪{b} = {a,b} ∈ ℐ ✓. This IS a matroid.

For a non-matroid: S = {a,b,c}, ℐ = {∅, {a}, {b}, {c}, {a,b}, {a,c}, {b,c}}. This IS a matroid (uniform U₂₃).

Try: S = {a,b}, ℐ = {∅, {a}, {b}}. This is not a matroid because exchange: |{a}|=1, |{b}|=1, but if |A| < |B|, we need A smaller. |∅|=0 < |{a}|=1, need x from {a}\∅ = {a}, ∅∪{a} = {a} which is independent ✓. So this IS a matroid.

The point is: greed works on matroids. For non-matroid structures (e.g., independent set in general graphs), greedy fails. Q3: Schedule jobs with profits 60, 40, 30, 20, 10 and deadlines 2, 1, 2, 1, 2.

Answer: Sort by profit: J1(60,d=2), J2(40,d=1), J3(30,d=2), J4(20,d=1), J5(10,d=2).

• J1: schedule at slot 2. {slot2=J1}
• J2: schedule at slot 1. {slot1=J2, slot2=J1}
• J3: deadline 2, slots 1,2 taken → skip
• J4: deadline 1, slot 1 taken → skip
• J5: deadline 2, slots 1,2 taken → skip Profit = 60 + 40 = 100. Q4: What is the exchange property and why is it essential for greedy correctness?

Answer: The exchange property says: if A and B are independent and |A| < |B|, there exists an element x in B\A such that A∪{x} is also independent. This property ensures that the greedy algorithm never makes a decision that blocks future optimal choices. It's the key lemma in the exchange argument proving greedy optimality. Q5: Give an example of a subset system that satisfies hereditary but NOT the exchange property.

Answer: Let S = {a, b, c} and ℐ = {∅, {a}, {b}, {a,b}, {a,c}}. Hereditary holds. But exchange fails: |{b}|=1 and |{a,c}|=2, but the only element in {a,c}{b} = {a,c}. {b}∪{a} = {a,b} ∈ ℐ, so exchange actually holds here.

Consider: S = {a,b}, ℐ = {∅, {a}}. This satisfies hereditary. |∅|=0 < |{a}|=1, and ∅∪{a}={a}∈ℐ. Exchange holds.

Consider S = {a,b,c}, ℐ = {∅, {a}, {b}, {a,b}, {b,c}}. Exchange: |{a}|=1 < |{b,c}|=2. Need x∈{b,c}{a} = {b,c}. {a}∪{b}={a,b}∈ℐ ✓. |{a,b}|=2, |{b,c}|=2 — no |A|<|B| case. This IS a matroid.

True non-matroid: S={a,b,c}, ℐ={∅, {a}, {b}, {a,b}}. This satisfies exchange (check all pairs). It's a matroid (graphic matroid of two edges sharing one vertex). Most natural structures that have hereditary actually are matroids — that's why matroids are so useful!

6. 🔗 Cross-References

• Week 1 - Greedy Algorithms: Foundation for matroid understanding
• Week 4 - Network Flow: Matroid intersection for bipartite matching
• BSCS3021 (ToC): Independent sets, spanning trees Join Discord PreviousGreedy AlgorithmsNextDynamic Programming