1. Lectures on Network Flows
COMP 523: Advanced Algorithmic Techniques
Lecturer: Dariusz Kowalski
Lectures on Network Flows

2. 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

3. Lectures on Network Flows
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

4. Lectures on Network Flows
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 v f(e) = ∑e out v f(e)
there is a node t (target) with no outgoing edges
Property: ∑e in t f(e) = ∑e out s f(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

5. Lectures on Network Flows
Useful definitions
Given s-t flow f in G = (V,E) and any subset of
nodes S
f in(S) = ∑e in S f(e)
f out(S) = ∑e out S f(e)
Property: f in(t) = f out(s)
Example: f in(u,v) = f out(u,v) = 30 u u
Network: 20 10
Flow: 20 10 30 t s 10 t s 10 20 10 20
Lectures on Network Flows v v

6. Lectures on Network Flows
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 u u
Network: 20 10
Flow: 20 10 30 t s 10 t s 10 20 10 20
Lectures on Network Flows v v

7. Lectures on Network Flows
Residual graph
Assume that we are given a flow f in graph 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) u
Network: u
Residual
Graph: 20 u
Flow: 20 10 20 10 20 t s 30 t 20 10 t s s 20 10 20 10 20 v
Lectures on Network Flows v v

8. Augmenting path & augmentation
Assume that we are given a flow f in graph G, and the 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: u
New
flow:
Network: u 10 20 u 20 10 10 10 20 10 t 20 s 30 t 10 10 20 t s 20 s v 10 10 20 20
Lectures on Network Flows v v

9. u 20 t s 10 v
Residual
Graph: u 20 t s v
Flow: 20 t s 10 v
New residual graph: u
New
flow: 20 t s 10 v u

10. 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
Augment f through path P :
Find minimum residual capacity on the path
Go through the path and modify residual capacities u
Network:
New
flow: u
New
residual
graph: u 20 10 10 10 20 20 30 t 10 20 10 10 20 t s s t s 10 10 10 20 20 20 v v v
Lectures on Network Flows

11. 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 u
Network:
New
flow: u
New
residual
graph: u 20 10 10 10 20 20 30 t 10 20 10 10 20 t s s t s 10 10 10 20 20 20 v v v
Lectures on Network Flows

12. 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

13. Lectures on Network Flows
Implementing DFS
Structures:
Adjacency list
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

14. 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 A c(e) = ∑e in B c(e) is the capacity of this cut
Property:
Minimum cut is equal to the maximum flow
Example:
c(A,B) = 50 u u A 20 10 20 10 B 30 t 30 t s s 10 20 10 20 v v
Lectures on Network Flows

15. Lectures on Network Flows
Max Flow vs. Min Cut
For any set A containing s we proceed in 3 steps:
value(f) = ∑v in A ∑e out v f(e) - ∑v in A ∑e in v f(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 and f out(A) = 30 u u A 20 10 20
10/10 B 30 t
30/10 t s s 10 20
10/10 20 v v
Lectures on Network Flows

16. 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 that f is maximum u u 20 10
Residual
graph: 20 10 30 t 10 20 t s s
DFS 10 20 10 20
Lectures on Network Flows v
Min-cut v

17. Lectures on Network Flows
Conclusions
Network flow algorithms:
Ford-Fulkerson algorithm in time O(mC)
Correspondence between max-flows and min-cuts
Lectures on Network Flows

18. Lectures on Network Flows
Exercises
READING:
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

19. 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

20. Lectures on Network Flows
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

21. Lectures on Network Flows
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 v f(e) = ∑e out v f(e)
there is a node t (target) with no outgoing edges
Property: ∑e in t f(e) = ∑e out s f(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

22. Lectures on Network Flows
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: 20 10
Flow: 20 10 30 t s 10 t s 10 20 10 20
Lectures on Network Flows v v

23. 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
Augment f through path P :
Find minimum residual capacity on the path
Go through the path and modify residual capacities u
Network:
New
flow: u
New
residual
graph: u 20 10 10 10 20 20 30 t 10 20 10 10 20 t s s t s 10 10 10 20 20 20 v v v
Lectures on Network Flows

24. 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 u
New
flow: u
New
residual
graph: u
Graph: 10 10 10 20 20 20 t 30 t 10 20 10 10 20 s s t s 10 10 10 20 20 20 v v v
Lectures on Network Flows

25. 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 A c(e) = ∑e in B c(e) is a capacity of the cut
Property:
Minimum cut is equal to the maximum flow
Example:
c(A,B) = 50 u u A 20 10 20 10 B 30 t 30 t s s 10 20 10 20 v v
Lectures on Network Flows

26. 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 t s
Lectures on Network Flows

27. 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 t s
Lectures on Network Flows

28. 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 t s
Lectures on Network Flows

29. 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 t s
Lectures on Network Flows

30. 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 t s
Lectures on Network Flows

31. 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

32. Textbook and Exercises
READING:
Chapter 7, Sections 7.5, 7.6 and 7.7
EXERCISES:
Find a network with real capacities for which FF-algorithm works very slowly
Prove formally that FF-algorithm gives the largest matching in the last application
Design the algorithm for finding the largest number of edge-disjoint paths from s to t in undirected network
Design the algorithm for finding the largest number of node-disjoint paths from s to t in both directed and undirected networks
Lectures on Network Flows