Maximum flow

1
Maximum flow
receiver
Capacity constraint
Lecture 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

3
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)

4
Value of the flow

Maximum flow problem maximize this value
3
4
G
9
10
7
8
6
6
10
10
2
0
9
10
9
s
t
10
10
9
Value 19
5
Flow decomposition

• Any flow can be decomposed into at most m flow
  paths.
• The same idea applies to the Chinese postman
  problem

6
An upper bound
receiver
7
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
8
Flows cuts

• Let C be a cut and S be the connected component
  of G-C containing s. Then

9
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

10
Greedy method?

• Find an s-t path where every edge has f(e) lt c(e)
• Add this path to the flow
• Repeat until no such path can be found.
• Does it work?

11
A counterexample
Hint Find an augmenting path
12
Residual graph

• Key idea allow flows to push back

f(e) 2
Can send 8 units forward or push 2 units back.
c(e) 10
c(e) 8
Advantage of this representation is not to
distinguish send forward or push back (which are
irrelevant)
c(e) 2
13
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?

Key dont think about flow paths!

• Flow conservations
• More flow going out from s

14
Ford-Fulkerson Algorithm

• Start from an empty flow f
• While there is an s-t path P in G
  
  update f along P
• Return f

15
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!
16
Integrality theorem

• If every edge has integer capacity,
• then there is a flow of integer value.

17
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)

18
Speeding up

• Capacity scaling (find paths with large capacity)
• Find a shortest s-t path ? time
• Preflow-push

19
Faster Algorithms
20
Even Faster Algorithms
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Applications

• of the algorithm
• of the min-max theorem
• of the integrality theorem

22
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 lt Maximum flow

24
Disjoint paths

• Find the maximum number of disjoint s-t paths

directed edge gt directed vertex (vertex
splitting)
directed vertex gt undirected vertex (bidirecting)
undirected vertex gt undirected edge (line graph)
25
Minimum Path Cover

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

Directed graph gt Bipartite graph
26
Matrix Rounding

• Round the entries to keep the row sums and
  column sums

27
League winner

• See if your favorite team can still win the leaque

28
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.