Lectures on Network Flows

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Lectures on Network Flows. COMP 523: Advanced Algorithmic Techniques Lecturer: Dariusz Kowalski. Overview. Previous lectures: Dynamic programming Weighted interval scheduling Sequence alignment These lectures: Network flows Applications: largest matching in bipartite graphs. Network.

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### Lectures on Network Flows

Lecturer: Dariusz Kowalski

Lectures on Network Flows

Overview

Previous lectures:

• Dynamic programming
• Weighted interval scheduling
• Sequence alignment

These lectures:

• Network flows
• Applications: largest matching in bipartite graphs

Lectures on Network Flows

Network

A directed graph G = (V,E) such that

• each directed edge e has its nonnegative capacity denoted by ce
• there is a node s (source) with no incoming edges
• there is a node t (target) with no outgoing edges

u

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u,v - internal

nodes

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Lectures on Network Flows

Flow

s-t flow in G = (V,E) is a function f from E to R+

• capacity condition: for each e, 0  f(e) ce
• conservation condition: for each internal node v, ∑e in vf(e) =∑e out vf(e)
• there is a node t (target) with no outgoing edges

Property: ∑e in tf(e) =∑e out sf(e)

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Network:

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Flow:

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Lectures on Network Flows

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Useful definitions

Given s-t flow f in G = (V,E) and any subset of

nodes S

• f in(S) = ∑e in Sf(e)
• f out(S) = ∑e out Sf(e)

Property: f in(t) = f out(s)

Example: f in(u,v) = f out(u,v) = 30

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Network:

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Flow:

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Lectures on Network Flows

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Problem(s)
• What is the maximum value of f in(t) (flow) for a given graph G = (V,E) ?
• How to compute it efficiently?

Assumption: capacities are positive integers.

Example: f in(t) = f out(s) = 30

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Network:

Flow:

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Lectures on Network Flows

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Residual graph

Assume that we are given a flow f ingraph G.

Residual graph Gf

• The same nodes, internal and s,t
• For each edge e in E with ce > f(e) we put weight ce - f(e) (residual capacity)
• For each edge e = (u,v) in E we put weight f(e) to the backward edge (v,u) (residual capacity)

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Residual

Graph:

Network:

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0

Flow:

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Lectures on Network Flows

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Augmenting path & augmentation

Assume that we are given a flow f ingraph G, andthe corresponding residual graph Gf

• Find a new flow in residual graph - through a path with no repeating nodes, and value equal to the minimum capacity on the path (augmenting path)
• Update residual graph along the path

New

residual

graph:

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flow:

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Network:

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Lectures on Network Flows

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Ford-Fulkerson Algorithm
• Initialize f(e) = 0 for all e
• While there is s-t path P in residual graph
• Augment f through path P and get new f and new residual graph

Augmentf through path P:

• Find minimum capacity on the path
• Go through the path and modify weights

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New

residual

graph:

Network:

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flow:

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Lectures on Network Flows

Analysis

Correctness:

maximum flow - proof later

termination - each time the flow is an integer and advances by at least 1 (assumption about integer capacities)

Time: O(mC)

• at most C iterations, where C is the value of the maximum flow, m is the number of edges
• each iteration takes O(m+n) steps - use DFS to find path P

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residual

graph:

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Network:

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flow:

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Lectures on Network Flows

Reminder: Depth-First Search (DFS)

Given a directed graph G = (V,E) of diameter D and the root node

Goal: find a directed rooted spanning tree such that each edge in graph G corresponds to the ancestor relation in the tree

Recursive idea of the algorithm:

Repeat until no neighbor of the root is free

• Select a free out-neighbor v of the root and add the edge from root to v to partial DFS
• Recursively find a DFS’ in graph G restricted to free nodes and node v as the root’ and add it to partial DFS

root’

root

root’’

Lectures on Network Flows

Implementing DFS

Structures:

• List (stack) S
• Array Discovered[1…n]

Algorithm:

• Set S= {root}
• Consider the top element v in S
• For each out-neighbor w of node v, if w is not Discovered then put w into the stack S and start the next iteration for w as the top element
• Otherwise remove v from the stack, add edge (z,v) to partial DFS, where z is the current top element, and start the next iteration for z as the top element

Remark: after considering the out-neighbor of node v we remove this neighbor from adjacency list to avoid considering it many times!

root’

root

root’’

Lectures on Network Flows

Flows vs. Cuts

(A,B) - cut in graph G:

• A,B is a partition of nodes, s in A, t in B

c(A,B)= ∑e out Ac(e) = ∑e in Bc(e) is the capacity of this cut

Property:

Minimum cut is equal to the maximum flow

Example:

c(A,B)= 50

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Lectures on Network Flows

Max Flow vs. Min Cut

For any setAcontaining s we proceed in 3 steps:

• value(f)= ∑v in A ∑e out vf(e) - ∑v in A ∑e in vf(e)
• value(f)= f out(A) - f in(A)
• c(A,B)  f out(A) - f in(A) = value(f)

Conclusion:

Min-cut  value(f)

Example:c(A,B)= 50 andf out(A) = 30

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Lectures on Network Flows

FF-algorithm gives Max-flow

Suppose FF-algorithm stopped with flow f :

• Directed DFS tree rooted in s does not contain t
• It means that cut-capacity between nodes in DFS and the remaining ones is 0 in residual graph
• It follows that each edge in this cut has been reversed by augmenting flow, which means that

c(DFS,DFS’) = value(f)

• Using Min-cut  value(f) we get thatfis maximum

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Residual

graph:

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DFS

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Lectures on Network Flows

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Min-cut

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Conclusions

Network flow algorithms:

• Ford-Fulkerson algorithm in time O(mC)
• Correspondence between max-flows and min-cuts

Lectures on Network Flows

Exercises

• Chapter 7, Sections 7.1, 7.2, and 7.3

EXERCISES:

• Modify FF-algorithm to work in time

O(m2 log C)

• Find an augmentation scheme to guarantee time O(mn) in FF-algorithm independently from integer C

Lectures on Network Flows

Overview

Previous lecture:

• Network flows
• Ford-Fulkerson algorithm
• Max-flows versus Min-cuts

This lecture:

• Rational and real capacities in network
• Applications: largest matching in bipartite graphs
• Application to disjoint paths problem

Lectures on Network Flows

Network

Given a directed graph G = (V,E) such that

• each directed edge e has its nonnegative capacity denoted by ce
• there is a node s (source) with no incoming edges
• there is a node t (target) with no outgoing edges

u

20

10

u,v - internal

nodes

30

t

s

10

20

v

Lectures on Network Flows

Flow

s-t flow in G = (V,E) is a function f from E to R+

• capacity condition: for each e, 0  f(e) ce
• conservation condition: for each internal node v, ∑e in vf(e) =∑e out vf(e)
• there is a node t (target) with no outgoing edges

Property: ∑e in tf(e) =∑e out sf(e)

u

u

Network:

20

10

Flow:

20

10

30

t

10

t

s

s

10

20

10

20

Lectures on Network Flows

v

v

Problem

What is a maximum value f in(t) = f out(s) (flow) for

a given graph G = (V,E) ?

How to compute it efficiently?

Assumption: capacities are positive integers.

Example: f in(t) = f out(s) = 30

u

u

Network:

Flow:

20

10

20

10

30

t

s

10

t

s

10

20

10

20

Lectures on Network Flows

v

v

Ford-Fulkerson Algorithm
• Initialize f(e) = 0 for all e
• While there is s-t path P in residual graph
• Augment f through path P and get new f and new residual graph

Augmentf through path P:

• Find minimum capacity on the path
• Go through the path and modify weights

u

New

residual

graph:

Network:

u

u

New

flow:

20

10

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30

t

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Lectures on Network Flows

Non-integer capacities
• Rational capacities - rescale them to integers by multiplying by the common multiply
• Real capacities - difficult to handle:
• Min-cut = Max-flow
• FF-algorithm may work very slowly

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residual

graph:

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flow:

Graph:

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Lectures on Network Flows

Flows vs. Cuts

(A,B) - cut in graph G:

• A,B is a partition of nodes, s in A, t in B

c(A,B)= ∑e out Ac(e) = ∑e in Bc(e) is a capacity of the cut

Property:

Minimum cut is equal to the maximum flow

Example:

c(A,B)= 50

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B

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Lectures on Network Flows

Applications - largest matching

Input: bipartite graph G=(V,W,E)

Goal: largest set of non-incident edges (with different ends)

Solution using flow algorithms:

• Lets direct edges from V to W, introduce s connected to all nodes in V, t connected from all nodes in W
• Capacities are 1 for all edges
• Max-flow is the largest matching

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Lectures on Network Flows

Applications - disjoint paths

Input: network graph G=(V,E)

Goal: largest set of edge-disjoint paths from s to t

Solution: Using flow algorithms, where each edge has capacity 1

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Lectures on Network Flows

Disjoint paths cont.

Suppose that k is the largest number of edge-disjoint paths from s to t.

• It is also a flow:
• capacity condition is clear since we push flow with value 1 through each path, and
• conservation is satisfied since if a path comes into a node it also goes out

Conclusion: the largest number of edge-disjoint paths from s to t is not bigger than Max-flow

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Lectures on Network Flows

Disjoint paths cont.

Suppose that x is the value of Max-flow produced by FF-algorithm.

How to construct x edge-disjoint paths from s to t ?

By induction on the number of edges in the flow.

For 0 edges trivial (nothing to do, no even a path)

Assume j edges in the flow. Take one of them (s,v) and continue going using edges in the flow:

• We go to t - done, since we have path, remove it from the graph and continue by induction
• We make a cycle - reduce the flow along the cycle to zero, and the obtained is a flow having the same value but smaller number of edges - next continue by induction

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Lectures on Network Flows

Complexity

Time of FF-algorithm: O(mn) ( since C = O(n) )

Time of extracting paths: each edge is considered once while extracting one path, hence total time O(mn)

Total time: O(mn)

Additional memory: O(m+n) for keeping paths

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Lectures on Network Flows

Conclusions

Network flow algorithms:

• Rational capacities are easy to deal with
• Real capacities are difficult to compute - although we can alternatively check Min-cut
• Application to the largest matching problem in time O(mn) (n = C since capacities are 1)
• Application to the edge-disjoint paths problem in time O(mn) (n = C since capacities are 1)

Lectures on Network Flows

Textbook and Exercises