Network Flow Lecturer: Shi Li Department of Computer Science and - - PowerPoint PPT Presentation

CSE 632: Analysis of Algorithms II: Combinatorial Optimization and Linear Programming (Fall 2020)

Network Flow

Lecturer: Shi Li

Department of Computer Science and Engineering University at Buffalo

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Outline

1

Network Flow

2

Ford-Fulkerson Method Correctness of Ford-Fulkerson’s Method Running Time of Ford-Fulkerson-Type Algorithm

3

Applications of Network-Flow Single Commodity Flow with Multiple Sources and Sinks s-t Edge-Disjoint Paths Problem Bipartite Matching Problem

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Flow Network

Abstraction of fluid flowing through edges Digraph G = (V, E) with source s ∈ V and sink t ∈ V

No edges enter s No edges leave t

Edge capacity c(e) ∈ R>0 for every e ∈ E

s t a b d c 12 14 9 4 7 16 13 20 4

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• Def. An s-t flow is a function f : E → R such that

for every e ∈ E: 0 ≤ f(e) ≤ c(e) (capacity conditions) for every v ∈ V \ {s, t}:

• e into v

f(e) =

• e out of v

f(e). (conservation conditions) The value of a flow f is val(f) =

• e out of s

f(e). Maximum Flow Problem Input: directed network G = (V, E), capacity function c : E → R>0, source s ∈ V and sink t ∈ V Output: an s-t flow f in G with the maximum val(f)

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Maximum Flow Problem: Example s t a b d c 12/12 11/14 0/9 0/4 7/7 12/16 11/13 19/20 4/4

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Outline

1

Network Flow

2

Ford-Fulkerson Method Correctness of Ford-Fulkerson’s Method Running Time of Ford-Fulkerson-Type Algorithm

3

Applications of Network-Flow Single Commodity Flow with Multiple Sources and Sinks s-t Edge-Disjoint Paths Problem Bipartite Matching Problem

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Greedy Algorithm Start with empty flow: f(e) = 0 for every e ∈ E Define the residual capacity of e to be c(e) − f(e) Find an augmenting path: a path from s to t, where all edges have positive residual capacity Augment flow along the path as much as possible Repeat until we got stuck

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Greedy Algorithm: Example s t a b d c 12/12 11/14 0/9 0/4 7/7 12/16 11/13 19/20 4/4

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Greedy Algorithm Does Not Always Give a Optimum Solution

0/1 1/1 1/1 0/1 a b s t 1/1 0/1 0/1

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Fix the Issue: Allowing “Undo” Flow Sent

a b s t 0/1 1/1 1/1 1/1 1/1

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Assumption (u, v) and (v, u) can not both be in E

• Def. For a s-t flow f, the residual graph Gf of G = (V, E)

w.r.t f contains: the vertex set V , for every e = (u, v) ∈ E with f(e) < c(e), a forward edge e = (u, v), with residual capacity cf(e) = c(e) − f(e), for every e = (u, v) ∈ E with f(e) > 0, a backward edge e′ = (v, u), with residual capacity cf(e′) = f(e).

0/1 1/1 1/1 0/1 a b s t 1/1 0/1 0/1

Original graph G and f

1 a b s t 1 1 1 1

Residual Graph Gf

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Residual Graph: One More Example

s t a b c d 4 12 8 5 1 4 4 4 s t a b c d 4/16 / 1 3 4/12 4 / 9 4/14 0/7 / 2 4/4 0/4 G Gf 4 13 7 20 4

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Agumenting Path

Augmenting the flow along a path P from s to t in Gf Augment(P)

1: b ← min

e∈P cf(e)

2: for every (u, v) ∈ P do 3:

if (u, v) is a forward edge then

4:

f(u, v) ← f(u, v) + b

5:

else ⊲ (u, v) is a backward edge

6:

f(v, u) ← f(v, u) − b

7: return f

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Example for Augmenting Along a Path

0/1 1 / 1 1/1 0/1 a b s t 1/1 0/1 0/1 1 a b s t 1 1 1 1

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Ford-Fulkerson’s Method

Ford-Fulkerson(G, s, t, c)

1: let f(e) ← 0 for every e in G 2: while there is a path from s to t in Gf do 3:

let P be any simple path from s to t in Gf

4:

f ←augment(f, P)

5: return f

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Ford-Fulkerson: Example

s t a b c d s t a b c d 1 1 / 1 3 11/14 7/7 1 5 / 2 G Gf 8/16 8/12 / 9 4/4 0/4 8 8 4 4 8 4 9 2 11 5 15 7 3 1 1

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Outline

1

Network Flow

2

Ford-Fulkerson Method Correctness of Ford-Fulkerson’s Method Running Time of Ford-Fulkerson-Type Algorithm

3

Applications of Network-Flow Single Commodity Flow with Multiple Sources and Sinks s-t Edge-Disjoint Paths Problem Bipartite Matching Problem

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Correctness of Ford-Fulkerson’s Method

1

The procedure augment(f, P) maintains the two conditions:

for every e ∈ E: 0 ≤ f(e) ≤ c(e) (capacity conditions) for every v ∈ V \ {s, t}:

• e into v

f(e) =

• e out of v

f(e). (conservation conditions)

2

When Ford-Fulkerson’s Method terminates, val(f) is maximized

3

Ford-Fulkerson’s Method will terminate

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Correctness of Ford-Fulkerson’s Method

1

The procedure augment(f, P) maintains the two conditions:

for every e ∈ E: 0 ≤ f(e) ≤ c(e) (capacity conditions) for every v ∈ V \ {s, t}:

• e into v

f(e) =

• e out of v

f(e). (conservation conditions)

2

When Ford-Fulkerson’s Method terminates, val(f) is maximized

3

Ford-Fulkerson’s Method will terminate

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for every e ∈ E: 0 ≤ f(e) ≤ c(e) (capacity conditions) for every v ∈ V \ {s, t}:

• e into v

f(e) =

• e out of v

f(e). (conservation conditions)

s t +b +b −b −b +b

for an edge e correspondent to a forward edge : b ≤ c(e) − f(e) = ⇒ f(e) + b ≤ c(e) for an edge e correspondent to a backward edge : b ≤ f(e) = ⇒ f(e) − b ≥ 0

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Correctness of Ford-Fulkerson’s Method

1

The procedure augment(f, P) maintains the two conditions:

for every e ∈ E: 0 ≤ f(e) ≤ c(e) (capacity conditions) for every v ∈ V \ {s, t}:

• e into v

f(e) =

• e out of v

f(e). (conservation conditions)

2

When Ford-Fulkerson’s Method terminates, val(f) is maximized

3

Ford-Fulkerson’s Method will terminate

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• Def. An s-t cut of G = (V, E) is a pair (S ⊆ V, T = V \ S)

such that s ∈ S and t ∈ T.

• Def. The cut value of an s-t cut is

c(S, T) :=

• e=(u,v)∈E:u∈S,v∈T

c(e).

• Def. Given an s-t flow f and an s-t cut (S, T), the net flow

sent from S to T is f(S, T) :=

• e=(u,v)∈E:u∈S,v∈T

f(e) −

• e=(u,v)∈E:u∈T,v∈S

f(e).

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s t a b c d 6/16 5 / 1 3 6/12 4 / 9 9/14 5 / 7 7/20 4/4 / 4 G

c(S, T) = 14 + 12 = 26 f(S, T) = 9 + 6 − 4 = 11

S T

• Obs. f(S, T) ≤ c(S, T) s-t cut (S, T).
• Obs. f(S, T) = val(f) for any s-t flow f and any s-t cut

(S, T). Coro. val(f) ≤ min

s-t cut (S,T) c(S, T) for every s-t flowf.

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Coro. val(f) ≤ min

s-t cut (S,T) c(S, T) for every s-t flowf.

We will prove Main Lemma The flow f found by the Ford-Fulkerson’s Method satisfies val(f) = c(S, T) for some s-t cut (S, T). Corollary and Main Lemma implies Maximum Flow Minimum Cut Theorem sup

s-t flow f

val(f) = min

s-t cut (S,T) c(S, T).

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Maximum Flow Minimum Cut Theorem sup

s-t flow f

val(f) = min

s-t cut (S,T) c(S, T).

s t a b d c 12/12 11/14 0/9 0/4 7 / 7 12/16 11/13 19/20 4/4

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Main Lemma The flow f found by the Ford-Fulkerson’s Method satisfies val(f) = c(S, T) for some s-t cut (S, T). Proof of Main Lemma. When algorithm terminates, no path from s to t in Gf, What can we say about Gf? There is a s-t cut (S, T), such that there are no edges from S to T For every e = (u, v) ∈ E, u ∈ S, v ∈ T, we have f(e) = c(e) For every e = (u, v) ∈ E, u ∈ T, v ∈ S, we have f(e) = 0 Thus, val(f) = f(S, T) =

• e=(u,v)∈E,u∈S,v∈T

f(e) −

• e=(u,v)∈E,u∈T,v∈S

f(e) =

• e=(u,v)∈E,u∈S,v∈T

c(e) = c(S, T).

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Correctness of Ford-Fulkerson’s Method

1

The procedure augment(f, P) maintains the two conditions:

for every e ∈ E: 0 ≤ f(e) ≤ c(e) (capacity conditions) for every v ∈ V \ {s, t}:

• e into v

f(e) =

• e out of v

f(e). (conservation conditions)

2

When Ford-Fulkerson’s Method terminates, val(f) is maximized

3

Ford-Fulkerson’s Method will terminate

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Ford-Fulkerson’s Method will Terminate

Intuition: In every iteration, we increase the flow value by some amount There is a maximum flow value So the algorithm will finally reach the maximum value However, the algorithm may not terminate if some capacities are irrational numbers. (“Pathological cases”)

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Lemma Ford-Fulkerson’s Method will terminate if all capacities are integers. Proof. The maximum flow value is finite (not ∞). In every iteration, we increase the flow value by at least 1. So the algorithm will terminate. Integers can be replaced by rational numbers.

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Correctness of Ford-Fulkerson’s Method

1

The procedure augment(f, P) maintains the two conditions:

for every e ∈ E: 0 ≤ f(e) ≤ c(e) (capacity conditions) for every v ∈ V \ {s, t}:

• e into v

f(e) =

• e out of v

f(e). (conservation conditions)

2

When Ford-Fulkerson’s Method terminates, val(f) is maximized

3

Ford-Fulkerson’s Method will terminate

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Outline

1

Network Flow

2

Ford-Fulkerson Method Correctness of Ford-Fulkerson’s Method Running Time of Ford-Fulkerson-Type Algorithm

3

Applications of Network-Flow Single Commodity Flow with Multiple Sources and Sinks s-t Edge-Disjoint Paths Problem Bipartite Matching Problem

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Running time of the Generic Ford-Fulkerson’s Algorithm

Ford-Fulkerson(G, s, t, c)

1: let f(e) ← 0 for every e in G 2: while there is a path from s to t in Gf do 3:

let P be any simple path from s to t in Gf

4:

f ←augment(f, P)

5: return f

O(m)-time for 3 and 4 in each iteration Total time = O(m)× number of iterations Assume all capacities are integers, then algorithm may run up to val(f ∗) iterations, where f ∗ is the optimum flow Total time = O(mval(f ∗)) Running time is “Pseudo-polynomial”

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The Upper Bound on Running Time Is Tight!

s t a b

2 / 1000 2 / 1000 2 / 1 2 / 1000 0 / 1 Better choices for choosing augmentation paths: Choose the shortest augmentation path Choose the augmentation path with the largest bottleneck capacity

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Choosing Augmenting Path with Largest Bottleneck Capacity

Idea: find the augmenting path from s to t with the largest bottleneck capacity Assumption: Capacities are integers between 1 and C capacity-scaling(G, s, t, c)

1: let f(e) ← 0 for every e in G 2: ∆ ← largest power of 2 which is at most C 3: while ∆ ≥ 1 do 4:

while while there exists an augmenting path P with bottleneck capacity at least ∆ do

5:

f ←augment(f, P)

6:

∆ ← ∆/2

7: return f

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• Obs. The outer while loop repeats 1 + ⌊log2 C⌋ times.

Lemma At the beginning of ∆-scale phase, the value of the max-flow is at most val(f) + 2m∆. Each augmentation increases the flow value by at least ∆ Thus, there are at most 2m augmentations for ∆-scale phase. Theorem The number of augmentations in the capacity-scaling max-flow algorithm is at most O(m log C).

• Coro. Running time of capacity-scaling max-flow algorithm is

polynomial.

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Outline

1

Network Flow

2

Ford-Fulkerson Method Correctness of Ford-Fulkerson’s Method Running Time of Ford-Fulkerson-Type Algorithm

3

Applications of Network-Flow Single Commodity Flow with Multiple Sources and Sinks s-t Edge-Disjoint Paths Problem Bipartite Matching Problem

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Applications of Network-Flow

Many problems can be reduced to the network-flow problem. single commodity flow with multiple sources and sinks s-t disjoint paths bipartite graph matching preemptive-scheduling of jobs graph orientation and densest subgraph load balance For each problem, we have two aspects: The problem can be solved efficiently. maximum-flow-minimum-cut theorem: dual structure that gives a witness on the optimality of a solution.

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Outline

1

Network Flow

2

Ford-Fulkerson Method Correctness of Ford-Fulkerson’s Method Running Time of Ford-Fulkerson-Type Algorithm

3

Applications of Network-Flow Single Commodity Flow with Multiple Sources and Sinks s-t Edge-Disjoint Paths Problem Bipartite Matching Problem

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Single Commodity Flow with Multiple Sources and Sinks Input: digraph G = (V, E) with many sources S ⊆ V and many sinks T ⊆ V

no edges enter S, no edges leave T

edge capacity c(e) ∈ R>0 for every e ∈ E Output: a maximum value flow f from S to T

s3 s2 s1 t2 t3 t4 t5 t1 3 5 8 S T

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S-T Flow

An S-T flow is a function f : E → R such that for every e ∈ E: 0 ≤ f(e) ≤ c(e) (capacity conditions) for every v ∈ V \ (S ∪ T):

• e into v

f(e) =

• e out of v

f(e). (conservation conditions) The value of a flow f is val(f) =

• e out of S

f(e).

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S-T Flow

s3 s2 s1 t2 t3 t4 t5 t1 3 5 8 S T

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“Single Commodity Flow with Multiple Sources and Sinks” Reason for “Single Commodity” there is only one commodity sending a flow from any source to any sink is acceptable

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Reduction of Problem to s-t Maximum Flow

s3 s2 s1 t2 t3 t4 t5 t1 3 5 8 S T ∞ ∞ ∞ super source s t super sink ∞ ∞ ∞ ∞ ∞

create a super-source s and super-sink t add edges of capacity ∞ from s to S add edges of capacity ∞ from T to t not hard to see that the reduction works

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Dual Structure for Problem

s3 s2 s1 t2 t3 t4 t5 t1 3 5 8 S T A B

MFMC theorem: maximum flow = value of minimum cut (A, B), s ∈ A, t ∈ B (A, B) can not cut new edges, since they have capacity ∞ So, S ⊆ A, T ⊆ B

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Dual Structure for Problem

s3 s2 s1 t2 t3 t4 t5 t1 3 5 8 S T A B

Lemma The value of the maximum flow is equal to the minimum value of (A, B), over all cuts (A, B) such that S ⊆ A and T ⊆ B.

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Unlimited Supplies and Demands

s3 s2 s1 t2 t3 t4 t5 t1 S T

each source si ∈ S is a supply node each sink ti ∈ T is a demand node every supply node has unlimited amount of commodity, every demand node has unlimited demand for the commodity.

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Variant: Limited Supplies and Demands

s3 s2 s1 t2 t3 t4 t5 t1 S T u1 = 10 u2 = 40 u3 = 30 d1 = 20 d2 = 5 d3 = 15 d4 = 30 d5 = 10

every si can supply ui ∈ R>0 units of commodity (ui is called the supply for si) every sink ti di ∈ R>0 units of commodity (di is called the demand for ti) assume

si∈S ui = ti∈T di

Goal: decide if all the demands can be satisfied

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Reduction to s-t Network Flow

s3 s2 s1 t2 t3 t4 t5 t1 S T s t 10 40 30 20 5 15 30 10

add a super-source s and a super-sink t add edges from s to S and from T to t the capacity of edge (s, si) is ui the capacity of edge (ti, t) is di decide if there is a flow that saturates all the new (red) edges

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Dual structure to the Problem

s3 s2 s1 t2 t3 t4 t5 t1 S T s t 10 40 30 20 5 15 30 10

MFMC: the instance is infeasible if and only if there is a s-t cut (A, B) of value strictly less than

i si = i ti

Now we do not have S ⊆ A, T ⊆ B any more.

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Dual structure to the Problem

s3 s2 s1 t2 t3 t4 t5 t1 S T s t 10 40 30 20 5 15 30 10

A B

cut value =

si∈S∩B ui + ti∈T∩A di + Q.

cut value <

si∈S ui = ti∈T di iff

• si∈S∩B

ui +

• ti∈T∩A

+Q <

• si∈S

ui

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Dual structure to the Problem

s3 s2 s1 t2 t3 t4 t5 t1 S T

A B

• si∈S∩B

ui +

• ti∈T∩A

di + Q <

• si∈S

ui ⇔

• ti∈T∩A

di + Q <

• si∈S∩A

ui ⇔ Q <

• si∈S∩A

ui −

• ti∈T∩A

di

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Dual structure to the Problem

s3 s2 s1 t2 t3 t4 t5 t1 S T

A B

Q <

• si∈S∩A

ui −

• ti∈T∩A

di. Meaning: “the capacity going out of A is smaller than the net supply inside A”

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Dual structure to the Problem

s3 s2 s1 t2 t3 t4 t5 t1 S T

A B

Lemma The instance is infeasible, if and only if there exists a cut (A, B) of the graph, such that the total capacity of edges going out of A is strictly less than the total supply in A, minus the total demand in A.

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The reduction only works for single-commodity flow problem. Multi-Commodity Flow We have many pairs (s1, t1), (s2, t2), ..., (sk, tk). Each pair corresponds to one commodity. si needs to send di units of commodity i to ti. Goal : decide if we can satisfy all the demands. We can not reduce the problem to s-t network flow. Q: Why the reduction we used before (i.e, creating a super-source s and a super-sink t) fails? A: In the resulting flow, we may send a flow from si to some tj for j = i.

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Outline

1

Network Flow

2

Ford-Fulkerson Method Correctness of Ford-Fulkerson’s Method Running Time of Ford-Fulkerson-Type Algorithm

3

Applications of Network-Flow Single Commodity Flow with Multiple Sources and Sinks s-t Edge-Disjoint Paths Problem Bipartite Matching Problem

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s-t Edge Disjoint Paths Input: a directed (or undirected) graph G = (V, E) and s, t ∈ V Output: the maximum number of edge-disjoint paths from s to t in G

s t

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Solving the maximum flow problem, where all capacities are 1 All flow values are integral (i.e, either 0 or 1) From flow to disjoint paths find an arbitrary s → t path where all edges have flow value 1 change the flow values of the path to 0 and repeat

s t

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Theorem The maximum number of edge disjoint paths from s to t equals the minimum value of an s-t cut (S, T).

s t

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s-t Edge Disjoint Paths in Undirected Graphs

s t

an undirected edge → two anti-parallel directed edges. Solving the s-t maximum flow problem in the directed graph Convert the flow to paths Issue: both e = (u, v) and e′ = (v, u) are used Fix: if this happens we change f(e) = f(e′) = 0

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Menger’s Theorem

Menger’s Theorem In an undirected graph, the maximum number of edge-disjoint paths between s to t is equal to the minimum number of edges whose removal disconnects s and t.

s t

s-t connectivity measures how well s and t are connected.

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Global Min-Cut Problem Input: a connected graph G = (V, E) Output: the minimum number of edges whose removal will disconnect G

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Solving Global Min-Cut Using Maximum Flow

1: let G′ be the directed graph obtained from G by replacing

every edge with two anti-parallel edges

2: for every pair s = t of vertices do 3:

• btain the minimum cut separating s and t in G, by

solving the maximum flow instance with graph G′,source s and sink t

4: output the smallest minimum cut we found

Need to solve Θ(n2) maximum flow instances Can we do better? Hint: only need to enumerate t

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Outline

1

Network Flow

2

Ford-Fulkerson Method Correctness of Ford-Fulkerson’s Method Running Time of Ford-Fulkerson-Type Algorithm

3

Applications of Network-Flow Single Commodity Flow with Multiple Sources and Sinks s-t Edge-Disjoint Paths Problem Bipartite Matching Problem

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Bipartite Graphs

• Def. A graph G = (V, E) is bipartite if the vertices V can be

partitioned into two subsets L and R such that every edge in E is between a vertex in L and a vertex in R.

L R

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• Def. Given a bipartite graph G = (L ∪ R, E), a matching in G

is a set M ⊆ E of edges such that every vertex in V is an endpoint of at most one edge in M. Maximum Bipartite Matching Problem Input: bipartite graph G = (L ∪ R, E) Output: a matching M in G of the maximum size L R

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Reduce Maximum Bipartite Matching to Maximum Flow Problem

t s

1 ∞

L R

1 1 1 1 1 1 1 1 1 1 1 ∞ ∞ ∞

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Reduce Maximum Bipartite Matching to Maximum Flow Problem

Create a digraph G′ = (L ∪ R ∪ {s, t}, E′) with capacity c : E′ → R≥0:

Add a source s and a sink t Add an edge from s to each vertex u ∈ L of capacity 1 Add an edge from each vertex v ∈ R to t of capacity 1 Direct all edges in E from L to R, and assign ∞ capacity (or capacity 1) to them

Compute the maximum flow from s to t in G′ The maximum flow gives a matching Running time:

Ford-Fulkerson: O(m × max flow value) = O(mn). Hopcroft-Karp: O(mn1/2) time

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Lemma Size of max matching = value of max flow in G′ Proof. ≤. Given a maximum matching M ⊆ E, send a flow along each edge e ∈ M and thus we have a flow of value |M|.

L R

t s

1 ∞

L R

1 1 1 1 1 1 1 1 1 1 1 ∞ ∞ ∞

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Lemma Size of max matching = value of max flow in G′ Proof. ≥. The maximum flow f in G′ is integral since all capacities are integral Let M to be the set of edges e from L to R with f(e) = 1 M is a matching of size that equals to the flow value

t s

1 ∞

L R

1 1 1 1 1 1 1 1 1 1 1 ∞ ∞ ∞

L R

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Perfect Matching

• Def. Given a bipartite graph G = (L ∪ R, E) with |L| = |R|, a

perfect matching M of G is a matching such that every vertex v ∈ L ∪ R participates in exactly one edge in M. Assuming |L| = |R| = n, when does G = (L ∪ R, E) not have a perfect matching?

L R L R N(X) X

For X ⊆ L, define N(X) = {v ∈ R : ∃u ∈ X, (u, v) ∈ E} |N(X)| < X for some X ⊆ L = ⇒ no perfect matching

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Hall’s Theorem Let G = (L ∪ R, E) be a bipartite graph with |L| = |R|. Then G has a perfect matching if and only if |N(X)| ≥ |X| for every X ⊆ L. Proof. = ⇒. If G has a perfect matching, then vertices matched to X ⊆ N(X); thus |N(X)| ≥ |X|.

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Hall’s Theorem Let G = (L ∪ R, E) be a bipartite graph with |L| = |R|. Then G has a perfect matching if and only if |N(X)| ≥ |X| for every X ⊆ L. Proof. ⇐ =. Contrapositive: if no perfect matching, then ∃X ⊆ L, |N(X)| < |X| Consider the network flow instance There is a s-t cut (S, T) of value at most n − 1 Define Ls, Lt, Rs, Rt as in figure

S T s t

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Hall’s Theorem Let G = (L ∪ R, E) be a bipartite graph with |L| = |R|. Then G has a perfect matching if and only if |N(X)| ≥ |X| for every X ⊆ L. Proof. ⇐ =. Contrapositive: if no perfect matching, then ∃X ⊆ L, |N(X)| < |X| No edges from Ls to Rt, since their capacities are ∞ c(S, T) = |Lt| + |Rs| < n |N(Ls)| ≤ |Rs| < n − |Lt| = |Ls|.

S T s t

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Hall’s Theorem Let G = (L ∪ R, E) be a bipartite graph with |L| = |R|. Then G has a perfect matching if and only if |N(X)| ≥ |X| for every X ⊆ L. Exercise: Let G = (L ∪ R, E) be a bipartite graph with |L| ≤ |R|. Then G has a matching that matches all vertices in L, if and only if |N(X)| ≥ |X| for every X ⊆ L.

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Ford-Fulkerson’s Method Restricted to Bipartite Matching

• Def. Given a matching M in

G = (L ∪ R, E), an augmenting path in G is an path of edges alternating between M and E \ M, where both end vertices are unmatched.

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Ford-Fulkerson’s Method Restricted to Bipartite Matching

Bipartite-Matching Algorithm

1: M ← ∅ 2: while ∃ an augmenting path P do 3:

for every edge e ∈ P do

4:

if e ∈ M then

5:

remove e from m

6:

else

7:

add e to M

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Other Variants of Bipartite Matching b-matching where an edge can be chosen multiple times b-matching where an edge can be chosen at most once Applications from Bipartite Matching and b-Matching Graph orientation, densest sub-graph Load balancing Preemptive scheduling on a single machine, and on multiple machines