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CS 6824: Network Flow

T. M. Murali

February 12, 2014

T. M. Murali February 12, 2014 CS 6824: Network Flow

Maximum Flow and Minimum Cut

I Two rich algorithmic problems.

I Fundamental problems in combinatorial optimization.

I Beautiful mathematical duality between flows and cuts.

I Numerous non-trivial applications:

I Bipartite matching.

I Data mining.

I Project selection.

I Airline scheduling.

I Baseball elimination.

I Image segmentation.

I Network connectivity.

I Open-pit mining.

I Network reliability.

I Distributed computing.

I Egalitarian stable matching.

I Security of statistical data.

I Network intrusion detection.

I Multi-camera scene reconstruction.

I Gene function prediction.


History

(Soviet Rail Network, Tolstoi, 1930; Harris and Ross, 1955; Alexander Schrijver,Math Programming, 91: 3, 2002.)


Flow NetworksI Use directed graphs to model transporation networks:

I edges carry traffic and have capacities.I nodes act as switches.I source nodes generate traffic, sink nodes absorb traffic.

I A flow network is a directed graph G (V ,E )I Each edge e ∈ E has a capacity c(e) > 0.I There is a single source node s ∈ V .I There is a single sink node t ∈ V .I Nodes other than s and t are internal.


Defining Flow

I In a flow network G (V ,E ), an s-t flow is a function f : E → R+ such that

(i) (Capacity conditions) For each e ∈ E , 0 ≤ f (e) ≤ c(e).(ii) (Conservation conditions) For each internal node v ,∑

e into v

f (e) =∑

e out of v

f (e)

I The value of a flow is ν(f ) =∑

e out of s f (e).


Maximum-Flow Problem

Maximum Flow

INSTANCE: A flow network G

SOLUTION: The flow with largest value in G , where the maximum istaking over all possible flows on G .

I Output should assign a flow value to each edge in the graph.

I The flow on each edge should satisfy the capacity condition.

I The flow into and out of each internal node should satisfy the conservationconditions.

I The value of the output flow, i.e., the total flow out of the source node in theoutput flow, must be the largest over all possible flows on G .

I Assumptions:

1. No edges enter s, no edges leave t.2. There is at least one edge incident on each node.3. All edge capacities are integers.


Examples of Flows

20

30

10

2010


Examples of Flows

20

30

10

2010

0 0

0

0 0


Examples of Flows

20

30

10

2010

10 0

10

0 10


Examples of Flows

20

30

10

2010

10 10

0

0 0


Examples of Flows

20

30

10

2010

10 10

10

0 10


Examples of Flows

20

30

10

2010

20 10

10

0 10


Examples of Flows

20

30

10

2010

20 10

10

10 20


Developing the Algorithm

I Let us take a greedy approach.

1. Start with zero flow along all edges (Figure 7.3(a)).2. Find an s-t path and push as much flow along it as possible (Figure 7.3(b)).

3. Idea to increase flow: Push flow along edges with leftover capacity and undoflow on edges already carrying flow.




1. Start with zero flow along all edges (Figure 7.3(a)).

2. Find an s-t path and push as much flow along it as possible (Figure 7.3(b)).3. Idea to increase flow: Push flow along edges with leftover capacity and undo

flow on edges already carrying flow.




1. Start with zero flow along all edges (Figure 7.3(a)).2. Find an s-t path and push as much flow along it as possible (Figure 7.3(b)).

3. Idea to increase flow: Push flow along edges with leftover capacity and undoflow on edges already carrying flow.




1. Start with zero flow along all edges (Figure 7.3(a)).2. Find an s-t path and push as much flow along it as possible (Figure 7.3(b)).3. Idea to increase flow: Push flow along edges with leftover capacity and undo

flow on edges already carrying flow.


Residual Graph

I Given a flow network G (V ,E ) and a flow f on G , the residual graph Gf of Gwith respect to f is a directed graph such that

(i) (Nodes) Gf has the same nodes as G .(ii) (Forward edges) For each edge e = (u, v) ∈ E such that f (e) < c(e), Gf

contains the edge (u, v) with a residual capacity c(e)− f (e).(iii) (Backward edges) For each edge e ∈ E such that f (e) > 0, Gf contains the

edge e′ = (v , u) with a residual capacity f (e).


Augmenting Paths in a Residual GraphI Let P be a simple s-t path in Gf .

I b = bottleneck(P, f ) is the minimumresidual capacity of any edge in P.

I The following operation augment(f ,P)yields a new flow f ′ in G :

I e is forward edge in Gf ⇒ flow increasesalong e in G .

I e = (u, v) is backward edge in Gf ⇒ flowdecreases along (v , u) in G .


Correctness of augment(f ,P)

I A simple s-t path in the residual graph is an augmenting path.I Let f ′ be the flow returned by augment(f ,P).I Claim: f ′ is a flow. Verify capacity and conservation conditions.

I Only need to check edges and internal nodes in P.I Capacity condition on e = (u, v) ∈ Gf : Note that b = bottleneck(P, f ) ≤

residual capacity of (u, v).

I e is a forward edge:0 ≤ f (e) ≤ f ′(e) = f (e) + b ≤ f (e) +

(c(e)− f (e)

)= c(e).

I e is a backward edge: c(e) ≥ f (e) ≥ f ′(e) = f (e)− b ≥ f (e)− f (e) = 0.

I Conservation condition on internal node v ∈ P.

Four cases to work out.

After augmentation

Residual graph

Forward Forward

Flow into v increases by b

Flow out of v increases by b

After augmentation

Residual graph

Backward Backward

Flow into v decreases by b

Flow out of v decreases by b

After augmentation

Residual graph

Forward Backward

Flow into v increases by b and decreases by b

Flow out of v does not change

After augmentation

Residual graph

ForwardBackward

Flow into v does not change

Flow out of v increases by b and decreases by b




I Only need to check edges and internal nodes in P.

I Capacity condition on e = (u, v) ∈ Gf : Note that b = bottleneck(P, f ) ≤residual capacity of (u, v).


(c(e)− f (e)

)= c(e).




After augmentation

Residual graph

Forward Forward



After augmentation

Residual graph

Backward Backward



After augmentation

Residual graph

Forward Backward



After augmentation

Residual graph

ForwardBackward







residual capacity of (u, v).


(c(e)− f (e)

)= c(e).

I e is a backward edge: c(e) ≥ f (e) ≥ f ′(e) = f (e)− b ≥ f (e)− f (e) = 0.I Conservation condition on internal node v ∈ P.


After augmentation

Residual graph

Forward Forward



After augmentation

Residual graph

Backward Backward



After augmentation

Residual graph

Forward Backward



After augmentation

Residual graph

ForwardBackward







residual capacity of (u, v).I e is a forward edge:

0 ≤ f (e) ≤ f ′(e) = f (e) + b ≤ f (e) +(c(e)− f (e)

)= c(e).



Before augmentation

Forward edge

After augmentation

Residual graph

After augmentation

Residual graph

Forward Forward



After augmentation

Residual graph

Backward Backward



After augmentation

Residual graph

Forward Backward



After augmentation

Residual graph

ForwardBackward








0 ≤ f (e) ≤ f ′(e) = f (e) + b ≤ f (e) +(c(e)− f (e)

)= c(e).




Before augmentation

Forward edge

After augmentation

Residual graph

After augmentation

Residual graph

Forward Forward



After augmentation

Residual graph

Backward Backward



After augmentation

Residual graph

Forward Backward



After augmentation

Residual graph

ForwardBackward








0 ≤ f (e) ≤ f ′(e) = f (e) + b ≤ f (e) +(c(e)− f (e)

)= c(e).



After augmentation

Residual graph

Forward Forward



After augmentation

Residual graph

Backward Backward



After augmentation

Residual graph

Forward Backward



After augmentation

Residual graph

ForwardBackward








0 ≤ f (e) ≤ f ′(e) = f (e) + b ≤ f (e) +(c(e)− f (e)

)= c(e).

I e is a backward edge: c(e) ≥ f (e) ≥ f ′(e) = f (e)− b ≥ f (e)− f (e) = 0.I Conservation condition on internal node v ∈ P. Four cases to work out.

After augmentation

Residual graph

Forward Forward



After augmentation

Residual graph

Backward Backward



After augmentation

Residual graph

Forward Backward



After augmentation

Residual graph

ForwardBackward




Ford-Fulkerson Algorithm


Analysis of the Ford-Fulkerson Algorithm

I Running timeI Does the algorithm terminate?I If so, how many loops does the algorithm take?

I Correctness: if the algorithm terminates, why does it output a maximumflow?


Termination of the Ford-Fulkerson Algorithm

I Claim: at each stage, flow values and residual capacities are integers.

Proveby induction.

I Claim: Flow value strictly increases when we apply augment(f ,P).v(f ′) = v(f ) + bottleneck(P, f ) > v(f ).

I Claim: Maximum value of any flow is C =∑

e out of s c(e).

I Claim: Algorithm terminates in at most C iterations.

I Claim: Algorithm runs in O(mC ) time.



I Claim: at each stage, flow values and residual capacities are integers. Proveby induction.



e out of s c(e).






I Claim: Flow value strictly increases when we apply augment(f ,P).

v(f ′) = v(f ) + bottleneck(P, f ) > v(f ).


e out of s c(e).








e out of s c(e).




Correctness of the Ford-Fulkerson Algorithm

I How large can the flow be?

I Can we characterise the magnitude of the flow in terms of the structure ofthe graph? For example, for every flow f , ν(f ) ≤ C =

∑e out of s c(e).

I Is there a better bound?

I Idea: An s-t cut is a partition of V into sets A and B such that s ∈ A andt ∈ B.

I Capacity of the cut (A,B) is c(A,B) =∑

e out of A c(e).I Intuition: For every flow f , ν(f ) ≤ c(A,B).



I How large can the flow be?I Can we characterise the magnitude of the flow in terms of the structure of

the graph? For example, for every flow f , ν(f ) ≤ C =∑

e out of s c(e).I Is there a better bound?I Idea: An s-t cut is a partition of V into sets A and B such that s ∈ A and

t ∈ B.

I Capacity of the cut (A,B) is c(A,B) =∑

e out of A c(e).I Intuition: For every flow f , ν(f ) ≤ c(A,B).

A B



I How large can the flow be?I Can we characterise the magnitude of the flow in terms of the structure of

the graph? For example, for every flow f , ν(f ) ≤ C =∑

e out of s c(e).I Is there a better bound?I Idea: An s-t cut is a partition of V into sets A and B such that s ∈ A and

t ∈ B.I Capacity of the cut (A,B) is c(A,B) =

∑e out of A c(e).

I Intuition: For every flow f , ν(f ) ≤ c(A,B).

A B


Some Useful Notation

A B

f out(v) =∑

e out of v

f (e) f in(v) =∑

e into v

f (e)

For S ⊆ V ,

f out(S) =∑

e out of S

f (e) f in(S) =∑

e into S

f (e)


Fun Facts about Cuts

A B

I Let f be any s-t flow and (A,B) any s-t cut.

I Claim: ν(f ) = f out(A)− f in(A).

I ν(f ) = f out(s) and f in(s) = 0⇒ ν(f ) = f out(s)− f in(s).I For every other node v ∈ A, 0 = f out(v)− f in(v).I Summing up all these equations, ν(f ) =

∑v∈A

(f out(v)− f in(v)

).

I An edge e that has both ends in A or both ends out of A does not contribute.I An edge e that has its tail in A contributes f (e).I An edge e that has its head in A contributes −f (e).

I∑

v∈A(f out(v)− f in(v)

)=∑

e out of A f (e)−∑

e into A f (e) = f out(A)− f in(A).

I Corollary: ν(f ) = f in(B)− f out(B).



A B

I Let f be any s-t flow and (A,B) any s-t cut.I Claim: ν(f ) = f out(A)− f in(A).


∑v∈A


).


I∑


)=∑






A B


I ν(f ) = f out(s) and f in(s) = 0⇒ ν(f ) = f out(s)− f in(s).

I For every other node v ∈ A, 0 = f out(v)− f in(v).I Summing up all these equations, ν(f ) =

∑v∈A


).


I∑


)=∑






A B


I ν(f ) = f out(s) and f in(s) = 0⇒ ν(f ) = f out(s)− f in(s).I For every other node v ∈ A, 0 = f out(v)− f in(v).

I Summing up all these equations, ν(f ) =∑


).


I∑


)=∑






A B



∑v∈A


).


I∑


)=∑






A B



∑v∈A


).


I∑


)=∑


e into A f (e) = f out(A)− f in(A).I Corollary: ν(f ) = f in(B)− f out(B).


Important Fact about Cuts

A B

I ν(f ) ≤ c(A,B).

ν(f ) = f out(A)− f in(A)

≤ f out(A) =∑

e out of A

f (e)

≤∑

e out of A

c(e) = c(A,B).


Max-Flows and Min-Cuts

I Let f be any s-t flow and (A,B) any s-t cut. We proved ν(f ) ≤ c(A,B).

I Very strong statement: The value of every flow is ≤ capacity of any cut.

I Corollary: The maximum flow is at most the smallest capacity of a cut.

I Question: Is the reverse true? Is the smallest capacity of a cut at most themaximum flow?

I Answer: Yes, and the Ford-Fulkerson algorithm computes this cut!


Flows and Cuts

I Let f̄ denote the flow computed by the Ford-Fulkerson algorithm.

I Enough to show ∃ s-t cut (A∗,B∗) such that ν(f̄ ) = c(A∗,B∗).

I When the algorithm terminates, the residual graph has no s-t path.

I Claim: If f is an s-t flow such that Gf has no s-t path, then there is an s-tcut (A∗,B∗) such that ν(f ) = c(A∗,B∗).

I Claim applies to any flow f such that Gf has no s-t path, and not just to theflow f̄ computed by the Ford-Fulkerson algorithm.


Proof of Claim Relating Flows to Cuts

I Claim: f is an s-t flow and Gf has no s-t path ⇒ ∃ s-t cut (A∗,B∗),ν(f ) = c(A∗,B∗).

I A∗ = set of nodes reachable from s in Gf , B∗ = V − A∗.

I Claim: (A∗,B∗) is an s-t cut in G .

I Claim: If e = (u, v) such thatu ∈ A∗, v ∈ B∗, then

f (e) = c(e).

I Claim: If e′ = (u′, v ′) such thatu′ ∈ B∗, v ′ ∈ A∗, then

f (e′) = 0.

I Claim: ν(f ) = c(A∗,B∗).


=∑

e out of A

f (e)−∑

e into A

f (e)

=∑

e out of A

c(e)−∑

e into A

0 = c(A,B).






I Claim: If e = (u, v) such thatu ∈ A∗, v ∈ B∗, then f (e) = c(e).

I Claim: If e′ = (u′, v ′) such thatu′ ∈ B∗, v ′ ∈ A∗, then

f (e′) = 0.



=∑

e out of A

f (e)−∑

e into A

f (e)

=∑

e out of A

c(e)−∑

e into A

0 = c(A,B).






I Claim: If e = (u, v) such thatu ∈ A∗, v ∈ B∗, then f (e) = c(e).

I Claim: If e′ = (u′, v ′) such thatu′ ∈ B∗, v ′ ∈ A∗, then f (e′) = 0.



=∑

e out of A

f (e)−∑

e into A

f (e)

=∑

e out of A

c(e)−∑

e into A

0 = c(A,B).


Max-Flow Min-Cut Theorem

I The flow f̄ computed by the Ford-Fulkerson algorithm is a maximum flow.

I Given a flow of maximum value, we can compute a minimum s-t cut in O(m)time.

I In every flow network, there is a flow f and a cut (A,B) such thatν(f ) = c(A,B).

I Max-Flow Min-Cut Theorem: in every flow network, the maximum value ofan s-t flow is equal to the minimum capacity of an s-t cut.

I Corollary: If all capacities in a flow network are integers, then there is amaximum flow f where f (e), the value of the flow on edge e, is an integerfor every edge e in G .


Real-Valued Capacities

I If capacities are real-valued, Ford-Fulkerson algorithm may not terminate!

I But Max-Flow Min-Cut theorem is still true. Why?


Bad Augmenting Paths


Improving Ford-Fulkerson Algorithm

I Bad case for Ford-Fulkerson algorithm is when the bottleneck edge is theaugmenting path has a low capacity.

I Idea: decrease number of iterations by picking s-t path with bottleneck edgeof largest capacity.

Computing this path can slow down each iterationconsiderably.

I Modified idea: Maintain a scaling parameter ∆ and choose only augmentingpaths with bottleneck capacity at least ∆.

I Gf (∆): residual network restricted to edges with residual capacities ≥ ∆.


Improving Ford-Fulkerson Algorithm

I Bad case for Ford-Fulkerson algorithm is when the bottleneck edge is theaugmenting path has a low capacity.

I Idea: decrease number of iterations by picking s-t path with bottleneck edgeof largest capacity. Computing this path can slow down each iterationconsiderably.

I Modified idea: Maintain a scaling parameter ∆ and choose only augmentingpaths with bottleneck capacity at least ∆.

I Gf (∆): residual network restricted to edges with residual capacities ≥ ∆.


Scaling Max-Flow Algorithm


Correctness of the Scaling Max-Flow Algorithm

I Flow and residual capacities are integer valued throughout.

I When ∆ = 1, Gf (∆) and Gf are identical.

I Therefore, when the scaling algorithm terminates, the flow is a maximumflow.


Running time of the Scaling Max-Flow Algorithm I

I ∆-scaling phase: one iteration of the algorithm’s outer loop, with ∆ fixed.

I Claim: the number of ∆-scaling phases is at most

1 + dlog2 Ce.I Need to bound the number of iterations in each ∆-scaling phase.

I Claim: During a ∆-scaling phase, each iteration increases the flow by ≥ ∆.




I Claim: the number of ∆-scaling phases is at most 1 + dlog2 Ce.

I Need to bound the number of iterations in each ∆-scaling phase.





I Claim: the number of ∆-scaling phases is at most 1 + dlog2 Ce.I Need to bound the number of iterations in each ∆-scaling phase.



Value of Flow at the End of a ∆-Scaling PhaseI Let f be the flow at the end of a ∆-scaling phase.I Claim: Then there is an s-t cut in (A,B) in G such that

ν(f ) ≤ ν(f̄ ) ≤ c(A,B) ≤ ν(f ) + m∆

I There is no s-t path in Gf (∆).I Let A∗ be the set of nodes reachable from s in Gf (∆); B∗ = V − A∗.I Claim: (A∗,B∗) is an s-t cut in G .I Claim: If e = (u, v) such that u ∈ A∗, v ∈ B∗, then

c(e)− f (e) < ∆.

I Claim: If e′ = (u′, v ′) such that u′ ∈ B∗, v ′ ∈ A∗, then

f (e′) < ∆.

I Claim: ν(f ) ≥ c(A∗,B∗)−m∆.

ν(f ) = f out(A∗)− f in(A∗)

=∑

e out of A∗

f (e)−∑

e into A∗

f (e)

≥∑

e out of A∗

(c(e)−∆)−∑

e into A∗

∆

≥∑

e out of A∗

c(e)−∑e in G

∆

≥ c(A∗,B∗)−m∆.



ν(f ) ≤ ν(f̄ ) ≤ c(A,B) ≤ ν(f ) + m∆

I There is no s-t path in Gf (∆).

I Let A∗ be the set of nodes reachable from s in Gf (∆); B∗ = V − A∗.I Claim: (A∗,B∗) is an s-t cut in G .I Claim: If e = (u, v) such that u ∈ A∗, v ∈ B∗, then

c(e)− f (e) < ∆.


f (e′) < ∆.



=∑

e out of A∗

f (e)−∑

e into A∗

f (e)

≥∑

e out of A∗

(c(e)−∆)−∑

e into A∗

∆

≥∑

e out of A∗

c(e)−∑e in G

∆

≥ c(A∗,B∗)−m∆.



ν(f ) ≤ ν(f̄ ) ≤ c(A,B) ≤ ν(f ) + m∆

I There is no s-t path in Gf (∆).I Let A∗ be the set of nodes reachable from s in Gf (∆); B∗ = V − A∗.

I Claim: (A∗,B∗) is an s-t cut in G .I Claim: If e = (u, v) such that u ∈ A∗, v ∈ B∗, then

c(e)− f (e) < ∆.


f (e′) < ∆.



=∑

e out of A∗

f (e)−∑

e into A∗

f (e)

≥∑

e out of A∗

(c(e)−∆)−∑

e into A∗

∆

≥∑

e out of A∗

c(e)−∑e in G

∆

≥ c(A∗,B∗)−m∆.



ν(f ) ≤ ν(f̄ ) ≤ c(A,B) ≤ ν(f ) + m∆

I There is no s-t path in Gf (∆).I Let A∗ be the set of nodes reachable from s in Gf (∆); B∗ = V − A∗.I Claim: (A∗,B∗) is an s-t cut in G .

I Claim: If e = (u, v) such that u ∈ A∗, v ∈ B∗, then

c(e)− f (e) < ∆.


f (e′) < ∆.



=∑

e out of A∗

f (e)−∑

e into A∗

f (e)

≥∑

e out of A∗

(c(e)−∆)−∑

e into A∗

∆

≥∑

e out of A∗

c(e)−∑e in G

∆

≥ c(A∗,B∗)−m∆.



ν(f ) ≤ ν(f̄ ) ≤ c(A,B) ≤ ν(f ) + m∆

I There is no s-t path in Gf (∆).I Let A∗ be the set of nodes reachable from s in Gf (∆); B∗ = V − A∗.I Claim: (A∗,B∗) is an s-t cut in G .I Claim: If e = (u, v) such that u ∈ A∗, v ∈ B∗, then

c(e)− f (e) < ∆.I Claim: If e′ = (u′, v ′) such that u′ ∈ B∗, v ′ ∈ A∗, then

f (e′) < ∆.



=∑

e out of A∗

f (e)−∑

e into A∗

f (e)

≥∑

e out of A∗

(c(e)−∆)−∑

e into A∗

∆

≥∑

e out of A∗

c(e)−∑e in G

∆

≥ c(A∗,B∗)−m∆.



ν(f ) ≤ ν(f̄ ) ≤ c(A,B) ≤ ν(f ) + m∆

I There is no s-t path in Gf (∆).I Let A∗ be the set of nodes reachable from s in Gf (∆); B∗ = V − A∗.I Claim: (A∗,B∗) is an s-t cut in G .I Claim: If e = (u, v) such that u ∈ A∗, v ∈ B∗, then c(e)− f (e) < ∆.


f (e′) < ∆.



=∑

e out of A∗

f (e)−∑

e into A∗

f (e)

≥∑

e out of A∗

(c(e)−∆)−∑

e into A∗

∆

≥∑

e out of A∗

c(e)−∑e in G

∆

≥ c(A∗,B∗)−m∆.



ν(f ) ≤ ν(f̄ ) ≤ c(A,B) ≤ ν(f ) + m∆

I There is no s-t path in Gf (∆).I Let A∗ be the set of nodes reachable from s in Gf (∆); B∗ = V − A∗.I Claim: (A∗,B∗) is an s-t cut in G .I Claim: If e = (u, v) such that u ∈ A∗, v ∈ B∗, then c(e)− f (e) < ∆.I Claim: If e′ = (u′, v ′) such that u′ ∈ B∗, v ′ ∈ A∗, then

f (e′) < ∆.I Claim: ν(f ) ≥ c(A∗,B∗)−m∆.


=∑

e out of A∗

f (e)−∑

e into A∗

f (e)

≥∑

e out of A∗

(c(e)−∆)−∑

e into A∗

∆

≥∑

e out of A∗

c(e)−∑e in G

∆

≥ c(A∗,B∗)−m∆.



ν(f ) ≤ ν(f̄ ) ≤ c(A,B) ≤ ν(f ) + m∆

I There is no s-t path in Gf (∆).I Let A∗ be the set of nodes reachable from s in Gf (∆); B∗ = V − A∗.I Claim: (A∗,B∗) is an s-t cut in G .I Claim: If e = (u, v) such that u ∈ A∗, v ∈ B∗, then c(e)− f (e) < ∆.I Claim: If e′ = (u′, v ′) such that u′ ∈ B∗, v ′ ∈ A∗, then f (e′) < ∆.



=∑

e out of A∗

f (e)−∑

e into A∗

f (e)

≥∑

e out of A∗

(c(e)−∆)−∑

e into A∗

∆

≥∑

e out of A∗

c(e)−∑e in G

∆

≥ c(A∗,B∗)−m∆.



ν(f ) ≤ ν(f̄ ) ≤ c(A,B) ≤ ν(f ) + m∆

I There is no s-t path in Gf (∆).I Let A∗ be the set of nodes reachable from s in Gf (∆); B∗ = V − A∗.I Claim: (A∗,B∗) is an s-t cut in G .I Claim: If e = (u, v) such that u ∈ A∗, v ∈ B∗, then c(e)− f (e) < ∆.I Claim: If e′ = (u′, v ′) such that u′ ∈ B∗, v ′ ∈ A∗, then f (e′) < ∆.I Claim: ν(f ) ≥ c(A∗,B∗)−m∆.


=∑

e out of A∗

f (e)−∑

e into A∗

f (e)

≥∑

e out of A∗

(c(e)−∆)−∑

e into A∗

∆

≥∑

e out of A∗

c(e)−∑e in G

∆

≥ c(A∗,B∗)−m∆.


Running time of the Scaling Max-Flow Algorithm II

I Claim: the number of augmentations in a ∆-scaling phase is ≤ 2m.I Base case: In the first ∆-scaling phase, each edge incident on s can be used in

at most one augmenting path.I Induction: At the end of the some ∆-scaling phase, let value of ∆ be Γ and let

f ′ be the flow: ν(f ′) ≥ ν(f̄ )−mΓ.

I In the next ∆-scaling phase, the value of ∆ is Γ/2. Let f be the flow at theend of this phase.

I Since each iteration increases the flow by ≥ Γ/2, if the current ∆-scaling phasecontinues for more than 2m iterations, then ν(f ) ≥ ν(f ′) + 2mΓ/2 ≥ ν(f̄ ).

I Claim: the running time of the scaling max-flow algorithm is O(m2 log C ).


Running time of the Scaling Max-Flow Algorithm II

I Claim: the number of augmentations in a ∆-scaling phase is ≤ 2m.I Base case: In the first ∆-scaling phase, each edge incident on s can be used in

at most one augmenting path.I Induction: At the end of the some ∆-scaling phase, let value of ∆ be Γ and let

f ′ be the flow: ν(f ′) ≥ ν(f̄ )−mΓ.I In the next ∆-scaling phase, the value of ∆ is Γ/2. Let f be the flow at the

end of this phase.I Since each iteration increases the flow by ≥ Γ/2, if the current ∆-scaling phase

continues for more than 2m iterations, then ν(f ) ≥ ν(f ′) + 2mΓ/2 ≥ ν(f̄ ).I Claim: the running time of the scaling max-flow algorithm is O(m2 log C ).


Other Maximum Flow Algorithms

I Running time of the Ford-Fulkerson algorithm is O(mC ), which ispseudo-polynomial: polynomial in the magnitudes of the numbers in theinput.

I Scaling algorithm runs in time polynomial in the size of the input (the graphand the number of bits needed to represent the capacities).

I Desire a strongly polynomial algorithm: running time is depends only on thesize of the graph and is independent of the numerical values of the capacities(as long as numerical operations take O(1) time).

I Edmonds-Karp, Dinitz: choose augmenting path to be the shortest path inGf (use breadth-first search). Algorithm runs in O(mn) iterations.

I Improved algorithms take time O(mn log n), O(n3), etc.

I Chapter 7.4: Preflow-push max-flow algorithm that is not based onaugmenting paths. Runs in O(n2m) or O(n3) time.


Global Min CutI Given an undirected graph G = (V ,E ), a cut is a partition of V into two

non-empty sets A and B.I Note distinction from s-t cut, where we had a directed graph, two

distinguished nodes s and t, and sought to separate s from t.

I Size of a cut (A,B) is the number of edges with one endpoint in A and theother endpoint in B.

I A global minimum cut is a cut of minimum size over all cuts.


Algorithm for Global Min Cut

1. Pick an arbitrary vertex s ∈ V . The global min cut must separate s from atleast one other vertex t in V .

2. Try every other vertex in V as a candidate for t and compute the minimums-t cut.

3. Pick smallest overall cut.

I Algorithm has a polynomial running time: involves n − 1 computations ofminimum s-t cut.

I Appears that finding global min cut is “harder” than finding minimum s-tcut.

I Many algorithms can compute global min-cut just as fast as the minimum s-tcut and do not even require a notion of flow.


Edge Contraction

I Due to Karger and Stein in 1992.

I Randomised algorithm.

I Based on the notion of edge contraction:I Pick an edge (u, v).I Unify u and v into a new “supernode” w ; delete u and v .I Delete edges connecting u to v .I Make every edge of the form (x , u) be incident on x and w .I Make every edge of the form (x , v) be incident on x and w .I An edge may be incident on w multiple times, i.e., the graph is a multigraph.

I Key idea of the algorithm is to pick edges at random and keep contracting tillthe graph has two nodes.


Randomised Contraction Algorithm

I For each node v , record the set S(v) of nodes that we have contracted into v .

I Initially, S(v) = v for every node v .

Contraction(G):

If G has two nodes v1 and v2, return the cut (S(v1),S(v2)).else

Choose an edge e = (u, v) in G uniformly at random.

Contract e using the new node w to replace u and v.Let G ′ be the resulting graph.

Set S(w) = S(u) ∪ S(v)Contraction(G ′).


Correctness of the Contraction Algorithm: I

I Let (A,B) be a global min cut of size k ; let F be the set of edges in the cut.

I How many edges must be incident on each node?

≥ k.

I Therefore, G has at least kn/2 edges.

I What can go “wrong” in the first step of the algorithm? The algorithm mayselect an edge in F for contraction. Then it cannot return the cut (A,B).

I What is the probability of this bad event? ≤ k/(kn/2) = 2/n.

I Consider the graph after j contractions.I How many supernodes? ≥ n − j .I How many edges if no edge in F contracted yet? ≥ k(n − j)/2.I What is the probability that the next contraction will be along an edge in F?≤ 2/(n − j).




I How many edges must be incident on each node? ≥ k.










I What can go “wrong” in the first step of the algorithm?

The algorithm mayselect an edge in F for contraction. Then it cannot return the cut (A,B).









I What is the probability of this bad event?

≤ k/(kn/2) = 2/n.









I Consider the graph after j contractions.

I How many supernodes? ≥ n − j .I How many edges if no edge in F contracted yet? ≥ k(n − j)/2.I What is the probability that the next contraction will be along an edge in F?≤ 2/(n − j).








I Consider the graph after j contractions.I How many supernodes?

≥ n − j .I How many edges if no edge in F contracted yet? ≥ k(n − j)/2.I What is the probability that the next contraction will be along an edge in F?≤ 2/(n − j).








I Consider the graph after j contractions.I How many supernodes? ≥ n − j .

I How many edges if no edge in F contracted yet? ≥ k(n − j)/2.I What is the probability that the next contraction will be along an edge in F?≤ 2/(n − j).








I Consider the graph after j contractions.I How many supernodes? ≥ n − j .I How many edges if no edge in F contracted yet?

≥ k(n − j)/2.I What is the probability that the next contraction will be along an edge in F?≤ 2/(n − j).








I Consider the graph after j contractions.I How many supernodes? ≥ n − j .I How many edges if no edge in F contracted yet? ≥ k(n − j)/2.

I What is the probability that the next contraction will be along an edge in F?≤ 2/(n − j).








I Consider the graph after j contractions.I How many supernodes? ≥ n − j .I How many edges if no edge in F contracted yet? ≥ k(n − j)/2.I What is the probability that the next contraction will be along an edge in F?

≤ 2/(n − j).


Correctness of the Contraction Algorithm: II

I We want to estimate the probability that the algorithm will return the cut(A,B), i.e., the algorithm does not select any edge in F in any of theiterations 1 ≤ j ≤ n − 2.

I Let ξj denote the event that no edge in F is contracted in iteration j of thealgorithm.

I What do we want to compute? A lower bound on Pr[ξ1 ∩ ξ2 . . . ∩ ξn−2]I What have we shown so far?

Pr[ξ1] ≥ (1− 2/n) = (n − 2)/nPr[ξj+1|ξ1 ∩ ξ2 . . . ∩ ξj ] ≥ (n − j − 2)/(n − j)

Pr[ξ1 ∩ ξ2 . . . ∩ ξn−2] =

Pr[ξ1] Pr[ξ2|ξ1] Pr[ξ3|ξ1 ∩ ξ2] . . .Pr[ξj+1|ξ1 ∩ ξ2 . . . ∩ ξj ] . . .Pr[ξn−2|ξ1 ∩ ξ2 . . . ∩ ξn−3]

≥(

n − 2

n

)(n − 3

n − 1

). . .

(n − j − 2

n − j

). . .

(2

4

)(1

3

)=

2

n(n − 1)=

(n

2

)−1





I What do we want to compute?

A lower bound on Pr[ξ1 ∩ ξ2 . . . ∩ ξn−2]I What have we shown so far?


Pr[ξ1 ∩ ξ2 . . . ∩ ξn−2] =


≥(

n − 2

n

)(n − 3

n − 1

). . .

(n − j − 2

n − j

). . .

(2

4

)(1

3

)=

2

n(n − 1)=

(n

2

)−1





I What do we want to compute? A lower bound on Pr[ξ1 ∩ ξ2 . . . ∩ ξn−2]

I What have we shown so far?Pr[ξ1] ≥ (1− 2/n) = (n − 2)/nPr[ξj+1|ξ1 ∩ ξ2 . . . ∩ ξj ] ≥ (n − j − 2)/(n − j)

Pr[ξ1 ∩ ξ2 . . . ∩ ξn−2] =


≥(

n − 2

n

)(n − 3

n − 1

). . .

(n − j − 2

n − j

). . .

(2

4

)(1

3

)=

2

n(n − 1)=

(n

2

)−1






Pr[ξ1] ≥ (1− 2/n) = (n − 2)/n

Pr[ξj+1|ξ1 ∩ ξ2 . . . ∩ ξj ] ≥ (n − j − 2)/(n − j)

Pr[ξ1 ∩ ξ2 . . . ∩ ξn−2] =


≥(

n − 2

n

)(n − 3

n − 1

). . .

(n − j − 2

n − j

). . .

(2

4

)(1

3

)=

2

n(n − 1)=

(n

2

)−1







Pr[ξ1 ∩ ξ2 . . . ∩ ξn−2] =


≥(

n − 2

n

)(n − 3

n − 1

). . .

(n − j − 2

n − j

). . .

(2

4

)(1

3

)=

2

n(n − 1)=

(n

2

)−1







Pr[ξ1 ∩ ξ2 . . . ∩ ξn−2] =


≥(

n − 2

n

)(n − 3

n − 1

). . .

(n − j − 2

n − j

). . .

(2

4

)(1

3

)=

2

n(n − 1)=

(n

2

)−1T. M. Murali February 12, 2014 CS 6824: Network Flow

Number of Runs of the Contraction Algorithm

I One run of the algorithm produces the global min cut with a probability at

least(n2

)−1.

I How can we amplify this probability?

Run the algorithm many times andoutput the largest cut we find.

I If we run the algorithm(n2

)times, then the probability that we do not find

the global min cut in any run is at(1− 1/

(n

2

))(n2)≤ 1/e.

I Can we drive the probability of failure further down? If we run the algorithm(n2

)ln n times, then the probability of failure is at most 1/n.

I Several other optimisations can improve overall running time considerably.




least(n2

)−1.

I How can we amplify this probability? Run the algorithm many times andoutput the largest cut we find.




(n

2

))(n2)≤ 1/e.







least(n2

)−1.





(n

2

))(n2)

≤ 1/e.







least(n2

)−1.





(n

2

))(n2)≤ 1/e.

I Can we drive the probability of failure further down?

If we run the algorithm(n2






least(n2

)−1.





(n

2

))(n2)≤ 1/e.





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