Applications of Network Flow

🎯 Learning Objectives

• Model real-world problems as flow networks
• Apply max-flow min-cut to bipartite matching
• Solve scheduling with flow networks
• Analyze baseball elimination using min-cut
• Formulate project selection as flow

1. Bipartite Matching

1.1 Problem

1.2 Flow Formulation

1.3 Tracing

2. Scheduling with Release Times and Deadlines

2.1 Problem

2.2 Flow Construction

• Time slots: discretize into unit slots between min $r_i$ and max $d_i$
• Source → each job: capacity $p_i$
• Each job → eligible slots: capacity 1
• Each slot → sink: capacity 1
• Feasible schedule exists iff max flow = $\sum p_i$

3. Baseball Elimination

3.1 Problem

3.2 Tracing Example

4. Project Selection

4.1 Problem

4.2 Reduction to Min-Cut

• Source → profitable projects: capacity = revenue
• Costly projects → sink: capacity = |cost|
• Prerequisite edges: capacity = ∞
• Selected projects = reachable from source in min-cut

5. Common Pitfalls

Pitfall 1: Wrong Capacity Assignment

Pitfall 2: Missing Intermediate Nodes

Pitfall 3: Infinite Capacity on Wrong Edges

6. Key Concepts Reference

7. 📝 Practice Questions

Q1: A bipartite graph has U={a,b,c}, V={1,2,3}. Edges: a-1, a-2, b-2, b-3, c-1, c-3. Find maximum matching.

Answer: Max matching = 3. Possible matching: (a,1), (b,2), (c,3). All three vertices on both sides matched (perfect matching). Flow formulation would find this in 3 augmentations. Q2: What does the min-cut in the baseball elimination graph represent?

Answer: The min-cut represents a set of teams that must exceed team x's wins. If Team A and B are on the source side of the cut, ALL remaining games between A and B must go to A or B (cannot go to x's side). The cut capacity = wins A/B must get + remaining games involving them. If this capacity < total remaining games, some wins must go to x's opponents, meaning x is eliminated. Q3: For project selection with profits [5, -3, -2, 4] and prerequisites: 1→3, 2→3, 3→4, find optimal selection.

Answer: Flow model: s→1 (5), s→2 (0 since 2 not profitable... wait, 2 has 0 profit? No -2 is a cost). Actually: P1=5, P2=-3, P3=-2, P4=4. Source connects to P1(5) and P4(4). P2 connects to sink(3), P3 connects to sink(2). Prereq: P1→P3, P2→P3, P3→P4. Min cut determines optimal set. Projects reachable from source in min cut are selected. Q4: How does max flow find the min cut?

Answer: After max flow is found, BFS from source along edges with residual capacity > 0. The reachable set = S side of min cut. T side = all other vertices. The min-cut capacity equals max flow value (max-flow min-cut theorem). This also finds the bottleneck edges. Q5: Can all NP problems be solved with network flow?

Answer: No — flow is in P (solvable in polynomial time). Only problems that can be formulated as flow are efficiently solvable. NP-complete problems (like general TSP, 3SAT) cannot be expressed as pure flow networks without exponential blowup. However, flow is a subroutine in approximation algorithms for some NP-hard problems.

8. 🔗 Cross-References

• Week 4 - Network Flow: Ford-Fulkerson, max-flow min-cut
• Week 6 - NP-Completeness: Limits of flow
• BSCS4020 (DSA): Graph algorithms Join Discord PreviousNetwork FlowNextString Algorithms