University of Auckland · FACULTY OF COMPUTER SCIENCE

COMPSCI320 Chap.12 Faster Flow Algorithms and Flow Modelling

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Chapter 12 of 13 · COMPSCI320

Faster Flow Algorithms and Flow Modelling

This chapter fixes the running time of Ford-Fulkerson and turns flow into a modelling tool. Capacity scaling augments only along large-bottleneck paths, the shortest augmenting path rule bounds the number of augmentations by the number of edges times vertices, and Dinitz works one level graph at a time.

The second half builds assignment networks, splits vertices to cap their throughput and states each decision rule.

The figures show capacity scaling's phases on the network from the previous chapter and a layered assignment model, and the practice cards build complete feasibility models with their decision rules.

The practice cards cover vertex limits and disjoint paths, the topics of the fifth programming assignment.

In this chapter

What this chapter covers

  • 01

    Capacity scaling and its phases

  • 02

    Shortest augmenting paths and level graphs

  • 03

    Dinitz's algorithm and unit capacities

  • 04

    Assignment problems as layered networks

  • 05

    Vertex splitting for throughput limits

  • 06

    Feasibility from the flow value

Worked example · free

Capacity scaling phase by phase

Q [4 marks]. Run capacity scaling on the network with edges s→a 9, s→b 7, a→c 6, b→c 4, a→t 4, c→t 9 and b→t 2. Give the augmentations in each phase as Delta halves from 8. The marks are an AskSia practice weighting, not a university marking scheme.
Augmentations per phase when capacity scaling runs on the network of Chapter 11.Augmentations per phase when capacity scaling runs on the network of Chapter 11. Delta starts at 8, the largest power of two not above the top capacity 9, and halves each phase down to 1.0Δ = 83Δ = 41Δ = 20Δ = 1
Augmentations per phase when capacity scaling runs on the network of Chapter 11. Delta starts at 8, the largest power of two not above the top capacity 9, and halves each phase down to 1.
  • 1Delta = 8: no path uses only residual edges of capacity at least 8, so the phase makes no augmentation.
  • 1Delta = 4: augment s, a, t by 4, s, a, c, t by 5 and s, b, c, t by 4, reaching 13.
  • 1Delta = 2: augment s, b, t by 2, reaching 15.
  • 1Delta = 1: no augmenting path remains, so 15 is the maximum flow.
This four-mark allocation is an AskSia practice weighting and not an official university scheme. The phases make 0, 3, 1 and 0 augmentations and end with a maximum flow of 15.
Sia tip — In each phase ignore every residual edge whose capacity is below Delta, then halve Delta only when no path remains.
Glossary

Key terms

Capacity Scaling
A maximum flow method that augments only along paths whose bottleneck is at least a threshold that halves each phase.
Level Graph
The subgraph keeping only edges that lead from one breadth-first level to the next.
Vertex Splitting
Replacing a vertex by an in-copy and an out-copy joined by one edge whose capacity limits the vertex's throughput.
Dinitz Algorithm
A maximum flow method that saturates paths in one breadth-first level graph at a time.
Shortest Augmenting Path
A maximum flow rule that always augments along a residual path with the fewest edges.
FAQ

Faster Flow Algorithms and Flow Modelling FAQ

How many phases does capacity scaling use?

The threshold starts at the largest power of two not exceeding the largest capacity C and halves until it reaches 1, so there are one plus the floor of log base two of C phases. The lecture shows each phase uses at most 2m augmentations, giving a total running time of m squared times log C.

Why split a vertex in a flow model?

Flow networks put capacities on edges, not vertices. When a rule limits how much can pass through an item in total, such as a depot or a cow that can be fed only once, replace the vertex by an in-copy and an out-copy joined by a single edge with that limit. All flow through the item must then cross that edge.

What makes a flow model complete in an exam answer?

Four things: the vertex set, every edge with its capacity, the decision rule linking the maximum flow to the question, and a running time for the chosen algorithm on the network's size. Without the decision rule the model answers nothing, because a flow value on its own does not say whether the assignment exists.

How does maximum flow count disjoint paths?

Give every edge capacity one. An integral flow of value k then splits into k paths that share no edge, and any k edge-disjoint paths give a flow of value k. For paths that may not share vertices, split every intermediate vertex into an in-copy and an out-copy joined by an edge of capacity one first.

Study strategy

Exam move

Trace capacity scaling on the Chapter 11 network phase by phase, then compare its phase count with the plain augmentation count. For modelling, rewrite three story problems as networks, drawing a tiny instance of each first, and finish each with its decision rule and running time. For the exam, prepare one model of each kind, an assignment, a vertex limit and a disjoint-path count, so you can adapt a known structure quickly.

Compare the bounds of the methods in a single table and say which one depends on the capacities.

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