Algorithm Design and Complexity - Course 12

Algorithm Design and Complexity

Course 12

Overview
 Flow Networks
 Positive Flow and Net Flow
 Flow Properties
 Maximum Flow
 Minimum Cut
 Ford-Fulkerson Method
 Edmonds-Karp Algorithm
 Sample Applications

Flow Networks
 Graphs that model materials that are transported
through conduits
 Each edge represents a conduit: it has a capacity
that is an upper bound for the flow rate
 Some vertices may produce materials: sources
 Other vertices may consume materials: sinks

Flow Networks (2)
 The simplest version of flow networks:
 Think of edges as pipes of different sizes (capacities)
 The flow is the effective quantity of liquid that is transported
through the pipes

 Flow Networks are also called transportation networks
 But you can transport a lot of things beside liquids, like
commodities, information, electricity, traffic

 There are some interesting and difficult problems for flow
networks:
 Maximum flow problem
 Maximum flow with minimum cost problem
 Circulation problem
 Multi-commodity problem

Flow Networks (3)
 G(V, E) – directed graph
 The capacity function c: E → R+
 c(u, v) = the capacity of the edge (u, v)
  (u,v)  E  c(u,v) = 0

 Assume we have a single source vertex: s
 Assume we have a single sink vertex: t
 Assume we have a path s .. v .. t  vV

Positive Flow
 We can define a function called positive flow
p: V x V  R+

 Properties of positive flow:
 Capacity constraint:
 u, vV: p(u, v) <= c(u, v)
 Flow conservation:
 uV {s, t}: flow_in(u) = flow_out(u)
 uV {s, t}: vV(p(v, u)) = vV(p(u, v))

Positive Flow – Example
 Notation on each edge: positive flow/capacity
 Note that the two properties are respected
 The flow is always lower than the capacity of each edge
 The flow is conserved in each vertex
 The source produces flow
 The sink consumes flow

Cancellation for Positive Flow
 Note that in the previous example we have:
 1 unit of flow from x  z
 2 units of flow from z  x
 We can cancel 1 unit of flow in both directions
 0 units of flow from x  z
 1 unit of flow from z  x
 Flow conservation is still respected (we cancel the same
amount of flow in each direction)
 Capacity constraint is still respected (the flow is decreased)
 In both cases, the “net flow” is 1 from z  x

 Without loss of generality, we may always define a
positive flow in a flow network such that for each (u, v),
u,vV, we have positive flow either from u  v or from v
 u, but not in both directions!

Net Flow
 We can define a function called net flow
f: V x V  R

 Properties of net flow:
 u, vV: f(u, v) <= c(u, v)
 uV {s, t}: total_net_flow_in(u) = 0
 uV {s, t}: vV(f(u, v)) = 0
 Skew symmetry:
 u, vV: f(u, v) = - f(v, u)

Net Flow (2)
 Conservation of net flow can be rewritten:

vV (f(u, v)) = 0

As f(u, v) might be either positive or negative
 uV {s, t}: vV and f(u,v)<0(f(u, v)) + vV and f(u,v)>0 (f(u, v))
=0
 uV {s, t}: vV and f(v,u)>0(f(v, u)) = vV and f(u,v)>0 (f(u, v))
“total positive flow entering u” = “total positive flow
leaving u”
“flow in” = “flow out”

Net Flow – Example
 Only positive flow is illustrated in graphical
representations
 Nevertheless, this does not mean that the (negative)
net flow does not exist in the other direction

Positive Flow and Net Flow
 Differences

 Net flow satisfies skew symmetry

 p(u, v) >= 0, while
f(u, v) >= 0 or f(u, v) <= 0

Positive Flow and Net Flow (2)
 Equivalence
 We can define net flow using only positive flow and vice-
versa

 f(u, v) = p(u, v) – p(v, u)  u, vV
 Need to show that using this definition, the net flow still
satisfies the three properties: capacity constraint,
conservation and skew symmetry

 p(u, v) = f(u, v) if f(u, v) > 0
 p(u, v) = 0 otherwise
 Need to show that this definition satisfies the two
properties of positive flow: capacity constraint and
conservation

Positive Flow and Net Flow (3)
 f(u, v) = p(u, v) – p(v, u)  u, vV
p(u, v) <= c(u, v) and p(v, u) >= 0
=> f(u, v) = p(u, v) – p(v, u) <= p(u, v) <= c(u, v)
vV (f(u, v)) = vV (p(u, v) – p(v, u)) = vV (p(u, v)) – vV
(p(v, u)) = (from positive flow conservation) = 0
 Skew symmetry:
f(u, v) = p(u, v) – p(v, u) = –(p(v, u) – p(u, v)) = –f(v,u)

Similar for the transformation from f(u, v)  p(u, v). Prove
them at home!

Value of the Flow
 From now on, we shall only use net flow

 Value of the flow = |f| = vV (f(s, v)) = vV (f(v, t))
 In the example below: |f| = 3

Cancellation of (Net) Flow
 Let’s say that f(u, v) = 5 and c(u, v) = 8
 We can “ship” at most 3 units of flow from u  v
 => f(u, v) = 8 = c(u, v)
 But we can also “ship” up to 5 units of flow back from v 
u
 => f(u, v) = 3 if we cancel 2 units of flow!
 Called flow cancellation
 If (v, u) E => f(v, u) <= c(v, u) = 0
 We are not allowed to have a positive flow from v  u
 But we are allowed to have negative flow!
 This negative flow is equal in absolute value with the
positive value from u  v

Cancellation of (Net) Flow (2)
 Due to cancellation, a flow only gives us this so-
called “net” effect. We cannot reconstruct actual
“shipments” of products from a net flow.
 A flow f(u, v) = 5 and f(v, u) = -5 can mean either:

5 units u  v 6 units u  v 8 units u  v
0 units v  u 1 units v  u 3 units v  u

Flow Operations
 Extend the definition of the flow to sets of vertices
 X,Y  V
 f(X,Y) = ΣxXΣyY f(x,y) = total flow between X and Y

 Possible operations:
 XV f(X,X) = 0;
 X,YV f(X,Y) = -f(Y,X);
 X,Y,ZV si Y⊆X
 f(X Y, Z) = f(X,Z) - f(Y,Z);
 f(Z, X Y) = f(Z,X) - f(Z,Y);
 X,Y,ZV si XY=
 f(X Y, Z) = f(X,Z) + f(Y,Z);
 f(Z, X Y) = f(Z,X) + f(Z,Y)
 |f| = f(s,V) = f(V,t)

Examples (1)

X 1/2
a b
1/8 2/2

4/4 3/3
s t

3/9 Y
c d
3/5

f(X,Y) = ΣxXΣyY f(x,y)

f(X,X) = f(s,a) + f(a,s) + f(s,b) + f(b,s) + f(a,b) + f(b,a) = 0
f(X,Y) = f(b,c) + f(b,t) = -f(c,b) - f(t,b) = -f(Y,X)
Proiectarea Algoritmilor 2010

Examples (2)

X 1/2
a b
1/8 2/2
Y
s 4/4 3/3 t

c 3/9 Z
3/5 d

X, Y, Z  V si Y ⊆ X
f(X Y, Z) = f(X, Z) - f(Y, Z)
f(Z, X Y) = f(Z, X) - f(Z, Y)

f(X Y, Z) = 0 = f(b,t) + f(c,d) - f(b,t) - f(c,d) = f(X,Z) - f(Y,Z)
f(Z, X Y) = 0 = f(t,b) + f(d,c) - f(t,b) - f(d,c) = f(Z,X) - f(Z,Y)


Examples (3)
Z
X 1/2
a b
1/8 2/2

s 4/4 3/3 t

3/9 Y
c d
3/5

X, Y, Z  V si X  Y = 
f(XY, Z) = f(X,Z) + f(Y,Z)
f(Z, XY) = f(Z,X) + f(Z,Y)

f(X Y, Z) = f(s,b) + f(a,b) + f(t,b) + f(d,c) = f(X,Z) + f(Y,Z)
f(Z, X Y) = f(b,a) + f(b,s) + f(b,t) + f(c,d) = f(Z,X) + f(Z,Y)


Examples (4)

1/2
a b
1/8 2/2
4/4 3/3
s t

3/9
c d
3/5

f(s, V) = f(V, t)

f(s, V) = f(s,a) + f(s,b) = 5 = f(d,t) + f(b,t) = f(V, t)


Maximum Flow Problem
 Given G(V, E), c: E  R+, sV and tV
 Find the maximum flow that can be “transported” from s
to t

 We start with an initial void flow on all edges
 We want to compute the net flow between each pair of
vertices that determines a maximum flow!

 Various methods are used
 Only look at one of the simplest:
 Ford-Fulkerson and Edmonds-Karp
 Uses the notion of residual network

Residual Network
 Given G(V, E), c: E  R+, sV and tV
 Knowing a given flow in the graph, f: E  R

 We can compute the residual capacity for any pair of
vertices:
 cf(u, v) = c(u, v) – f(u, v) >= 0 (due to capacity constraint)
 Residual means: how much flow we can still transport from u 
v
 We can compute the residual network of the graph:
 It is a graph Gf = (V, Ef)
 Ef = {(u, v)  V x V | cf(u, v) > 0}
 It is a graph that contains edges that have residual capacity
(that can still transport flow!)
 How many edges does Gf have? Find the minimum and the
maximum value!

Augmenting Path
 Any path from the residual graph admits more flow
from the start vertex to the end vertex
 Therefore, we want to find a path from the source s
to the sink t in the residual graph: augmenting path

 Look for p = s..t in Gf
 If we can find this augmenting path, compute how
much flow we can send along it!
 We are constrained by the minimum residual capacity of
any edge along the augmenting path
 cf(p) = min(cf(u, v) | (u, v) is on p)

Augmenting Path – Example
 p = <(s, w), (w, y), (y, z), (z, t)>
 Is it unique? Can you find another?

Summing Flows
 If the graph G has a flow f
 And we can find another flow f’
 We can sum up the two flows to find the new flow:
f(u,v) + f’(u,v)
 The value of the new flow is |f| + |f’|

 Therefore, if we can find an augmenting path, p, in the residual
network of a graph, we can also find another flow, fp: V x V 
R
 fp(u,v) = cf(p) if (u,v) is on p
 fp(u,v) = -cf(p) if (v,u) is on p
 fp(u,v) = 0

 We can thus define a new flow in G: |f’| = |f| + |fp| > |f|
 Thus, we can improve the existing flow!

Graph Cuts
 A cut (S, T) of a flow network G(V, E) is a partitionof
the vertices into two sets S (sS) and T = V S (t
T)
 Already know cuts for minimum spanning trees
 The flow across the cut is f(S, T)
 The capacity of the cut is c(S, T) = summing up the
capacities of all the edges starting from a vertex in S
and ending in a vertex in T
 A minimum cut in G is a cut whose capacity is
minimum over all the possible cuts in G

Graph Cuts – Example
 S = {s, w, y}, T = {x, z, t}
 f(S, T) = f(w, x) + f (y, x) + f(y, z) = 2 + (-1) + 2 = 3
 Includes positive flow that goes either direction across the
cut
 But only takes into account net flow that goes from any
vertex in S to any vertex in T
 c(S, T) = c(w, x) + c(y, z) = 2 + 3 = 5

Max-Flow Min-Cut Theorem
 The following statements are equivalent:
 f is a maximum flow in G
 There is no augmenting path in Gf
 |f| = c(S, T) for all minimum cuts (S, T)

 This theorem gives us a method to compute the
maximum flow: push flow along augmenting paths
until you cannot find any
 This theorem also tells us how to compute a
minimum cut once the maximum flow is known
 When f is a maximum flow, all the vertices reachable from
s in Gf are in S, the others are in T => (S, T) is a minimum
cut

Ford-Fulkerson Method
1. Start with a void flow
2. Compute the residual graph for the given flow
3. If there is an augmenting path in this residual graph
a) Increase the flow along it
b) Go to step 2
4. The maximum flow has been reached!

 The disadvantage of this method is that it does not
specify how to look for the augmenting path in step 3
 Any method can be used
 Pick a random augmenting path
 Use any search algorithms to compute the path: DFS, BFS
 Be greedy: look for the fattest pipe each time! How?

Ford-Fulkerson Method (2)
 If the augmenting path is not picked smartly, then the
worst case running time for the algorithm is O(E*|fmax|)
 Why?
 Each augmenting path has at most O(E) edges that will
cause a change in the value of the flow along them at
each step!
 Maximum number of steps is fmax if a each step cf(p)=1
 See the example on the next slides!
 It is not good that the running time depends on fmax
 What happens if the capacities of the edges are irrational
number? => so is fmax

Exemplu Ford – Fulkerson (1)
1000 B 1000
0/1000 B 0/1000
1
G
0/1 Gf A D
A D
1000 C 1000
0/1000 C 0/1000
Residual Path: A-B-C-D; Cf = 1
1
1/1000 B 0/1000 B 1000
999 G
1/1 Gf A 1 D
A D 999

0/1000 C 1/1000 1000 C 1

Residual Path : A-B-C-D; Cf = 1
1/1000 B 1/1000 1 B 999 1
Gf 999 G
A 0/1 D A 1 D
999
1/1000 C 1/1000 999 C 1
1
Residual Path : A-B-C-D; Cf = 1

Exemplu Ford – Fulkerson (2)

2
2/1000 B 1/1000 B 999 1
998 G
1/1 Gf 1
A D A D
998
999 C
1/1000 C 2/1000 1 2

… Residual Path: A-C-B-D; Cf = 1

1000/1000 B 1000/1000 B
1000 1000
0/1
Gf
A D A 1 D
1000/1000 C 1000/1000 1000 C 1000

Residual Path: Ø
How many steps are needed to reach the maximum flow?


Edmonds-Karp Algorithm
 Is a variant of the Ford-Fulkerson algorithm that
guarantees to have a running time that is
independent of fmax
 Always picks the augmenting path using BFS on Gf
 Source vertex for BFS is s
 You may stop when t is reached
 This way, the length of the augmenting paths is
increasing when running the algorithm
 We always pick an augmenting path with an equal or
greater length than the previous one

Edmonds-Karp Algorithm (2)
Edmonds–Karp(G, s, t)
FOREACH (u,v) in E
f(u,v) = f(v,u) = 0 // O(E)
WHILE (1) // O(E*V)
p = find-augmenting-path-using-bfs(G, f, s, t) // O(E)
IF (p == null)
BREAK
cf(p) = min{cf(u,v) | (u,v) in p} // O(E)
FOREACH (u,v) in p // O(E)
f(u,v) = f(u,v) + cf(p)
f(v,u) = -f(u,v)
RETURN f

Complexity: O(V*E2)

Edmonds-Karp – Example
1000 B 1000
0/1000 B 0/1000
1
G
0/1 Gf A D
A D
1000 C 1000
0/1000 C 0/1000
Residual Path: A-B-D; Cf = 1000
1000/1000 B 1000/1000 1000 B
1000
0/1 Gf 1
G
A D A D
0/1000 C 1/1000 1000 C 1000

Residual Path: A-C-D; Cf = 1000
1000/1000 B 1000/1000
1000 B
1000
0/1 Gf
A D 1
A D
1000/1000 C 1000/1000 1000 1000
C
Residual Path: Ø

Application: Maximum Bipartite Matching
 Bipartite graph: we can partition the graph into two sets
of vertices L, R such as L U R = V and for all (u, v)E,
uL, vR
 Matching: a subset of edges, M, such that for all vV at
most an edge in M is incident in v
 Need to transform this problem into a maximum flow

Application: The Escape Problem
 http://webcourse.cs.technion.ac.il/234247/Spring2006/ho/
WCFiles/The%20Escape%20Problem.pdf

Conclusions
 Flow networks are met very often in real life
 A lot of applications and interesting problems to be solved:
maximum flow is only one of them

 It is important to know the main concepts of flow networks:
positive flow, net flow, cuts, augmenting networks, etc.

 Ford-Fulkerson gives us a generic method to compute the
maximum flow in a graph
 Edmonds-Karp is a variant of FF that runs in O(V*E2)
 There are other algorithms that work better

 A lot of problems can be solved using maximum flow, but you
need to be able to “transform” them into a flow problem!

References
 CLRS – Chapter 26

 http://www.cs.princeton.edu/courses/archive/fall04/cos22
6/lectures/maxflow.4up.pdf

 www.brics.dk/~sskyum/dSoegOpt/public_html/ek.pdf

Algorithm Design and Complexity - Course 12

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