1

Maximum flow

sender

receiver

Capacity constraintLecture 6: Jan 25

2

Network transmission

Given a directed graph G A source node s A sink node t

Goal: To send as much information from s to t

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Flows

An s-t flow is a function f which satisfies:

(capacity constraint)

(conservation of flows)

An s-t flow is a function f which satisfies:

(capacity constraint)

(conservation of flows (at intermediate vertices)

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Value of the flow

s t 10

10

9

8

4

10

10 6 2

10

3

9

9 9 10

7

0

G:6

Value = 19

Maximum flow problem: maximize this value

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Flow decomposition

Any flow can be decomposed into at most m flow paths.

The same idea applies to the Chinese postman problem

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An upper bound

sender

receiver

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Cuts

An s-t cut is a set of edges whose removal

disconnect s and t

The capacity of a cut is defined as the sum of the

capacity of the edges in the cut

Minimum s-t cut problem:

minimize this capacity of a s-t cut

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Flows ≤ cuts

Let C be a cut and S be the connected component of

G-C containing s. Then:

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Main result

Value of max s-t flow ≤ capacity of min s-t cut

(Ford Fulkerson 1956)

Max flow = Min cut

A polynomial time algorithm

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Greedy method?

Find an s-t path where every edge has f(e) < c(e)

Add this path to the flow

Repeat until no such path can be found.

Does it work?

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A counterexample

Hint: Find an augmenting path

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Residual graph

Key idea: allow flows to push back

c(e) = 10

f(e) = 2

c(e) = 8

c(e) = 2

Advantage of this representation

is not to distinguish send forward

or push back (which are irrelevant)

Can send 8 units forward

or push 2 units back.

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Finding an augmenting path

Find an s-t path in the residual graph

Add it to the current flow to obtain a larger flow.

Why?

1. Flow conservations

2. More flow going out from s

Key: don’t think about flow paths!

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Ford-Fulkerson Algorithm

1. Start from an empty flow f

2. While there is an s-t path P in G

update f along P

3. Return f

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Max-flow min-cut theorem

Consider the set S of all vertices reachable from s

So, s is in S, but t is not in S

No incoming flow coming in S (otherwise push back)

Achieve full capacity from S to T

Min cut!

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Integrality theorem

If every edge has integer capacity,

then there is a flow of integer value.

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Complexity

Assume edge capacity between 1 to C

At most nC iterations

Finding an s-t path can be done in O(m) time

Total running time O(nmC)

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Speeding up

Capacity scaling (find paths with large capacity)

Find a shortest s-t path time

Preflow-push

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Faster Algorithms

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Even Faster Algorithms

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Applications

of the algorithm

of the min-max theorem

of the integrality theorem

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Multi-source multi-sink

A set of sources S = {s1,…,sk} A set of sinks T = {t1,…,tm} Maximum flow from S to T

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Bipartite matching

Bipartite matching <= Maximum flow

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Disjoint paths

Find the maximum number of disjoint s-t paths

directed edge => directed vertex (vertex splitting)

directed vertex => undirected vertex (bidirecting)

undirected vertex => undirected edge (line graph)

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Minimum Path Cover

Given a directed graph, find a minimum number of paths to cover all vertices

Directed graph => Bipartite graph

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Matrix Rounding

Round the entries to “keep” the row sums and column sums

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League winner

See if your favorite team can still win the leaque

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Bonus Question 3

(25%) In a soccer tournament of n  teams, every pair of teams plays one match.

The winner gets 3 points, the loser gets 0, while both teams receive 1 point in a draw.

Is there a polynomial algorithm to decide whether a given score sequence

(a score for each team) can be the score sequence at the end of a valid championship?

League winner version is NP-hard.