maximum flow problems
DESCRIPTION
Maximum flow problems. Maximum flow problems. Introduction of: network, max-flow problem capacity, flow Ford-Fulkerson method pseudo code, residual networks, augmenting paths cuts of networks max-flow min-cut theorem example of an execution analysis, running time, - PowerPoint PPT PresentationTRANSCRIPT
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Maximum flow problems
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Maximum flow problems- Introduction of:
network, max-flow problem
capacity, flow
- Ford-Fulkerson method
pseudo code, residual networks, augmenting paths
cuts of networks
max-flow min-cut theorem
example of an execution
analysis, running time,
variations of the max-flow problem
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Introduction – flow network
Practical examples of a flow network
- liquids flowing through pipes
- parts through assembly lines
- current through electrical network
- information through communication network
- goods transported on the road…
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Introduction – flow networkExample: oil pipelineRepresentation
Flow network: directed graph G=(V,E)
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v4source sink
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Introduction – max-flow problemRepresentation
Flow network: directed graph G=(V,E)
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v1
v2
v3
v4
S t
v1
v2
v3
v4source sink
Informal definition of the max-flow problem:
What is the greatest rate at which material can be shipped from the source to the sink without violating any capacity contraints?
Example: oil pipeline
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Introduction - capacity
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v4
RepresentationFlow network: directed graph G=(V,E)
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u v12
u v6
c(u,v)=12
c(u,v)=6
Big pipe
Small pipe
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3
6
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3
3
Example: oil pipeline
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Introduction - capacity
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RepresentationFlow network: directed graph G=(V,E)
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v4
8
3
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6
3
3
Example: oil pipeline
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If (u,v) E c(u,v) = 0
0
0
0
6
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Introduction – flow
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v4
RepresentationFlow network: directed graph G=(V,E)
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u v6/12
u v6/6
f(u,v)=6
f(u,v)=6
Flow below capacity
Maximum flow
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3
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6
3
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Example: oil pipeline
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Introduction – flow
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RepresentationFlow network: directed graph G=(V,E)
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v2
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v4
8
3
6
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3
3
Example: oil pipeline
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Introduction – flow
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RepresentationFlow network: directed graph G=(V,E)
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v4
8
3
6
6/8
6/6
6/6
3
3
Example: oil pipeline
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Introduction – flow
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v2 v4
v3
RepresentationFlow network: directed graph G=(V,E)
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v4
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3
6
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6/6
3
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Example: oil pipeline
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Introduction – flow
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v2 v4
v3
RepresentationFlow network: directed graph G=(V,E)
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v2
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v4
3/8
3/3
3/6
6/8
6/6
6/6
3
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Example: oil pipeline
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Introduction – flow
S t
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v2 v4
v3
RepresentationFlow network: directed graph G=(V,E)
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v3
v4
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3/3
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6/8
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Example: oil pipeline
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Introduction – flow
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v2 v4
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RepresentationFlow network: directed graph G=(V,E)
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Example: oil pipeline
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Introduction – cancellation
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RepresentationFlow network: directed graph G=(V,E)
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Example: oil pipeline
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Introduction – cancellation
S t
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v2 v4
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RepresentationFlow network: directed graph G=(V,E)
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v4
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3/3
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u
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10 4
u
v
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u
v
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u
v
5/10 4
Example: oil pipeline
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Flow properties
Flow in G = (V,E): f: V x V R with 3 properties:
1) Capacity constraint: For all u,v V : f(u,v) < c(u,v)
2) Skew symmetry: For all u,v V : f(u,v) = - f(v,u)
3) Flow conservation: For all u V \ {s,t} : f(u,v) = 0 v V
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Flow properties
Flow in G = (V,E): f: V x V R with 3 properties:
1) Capacity constraint: For all u,v V : f(u,v) < c(u,v)
2) Skew symmetry: For all u,v V : f(u,v) = - f(v,u)
3) Flow conservation: For all u V \ {s,t} : f(u,v) = 0 v V
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Flow network G = (V,E)Note:
by skew symmetry
f (v3,v1) = - 12
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Net flow and value of a flow
Net Flow: positive or negative value of f(u,v) Value of a Flow f: Def:
|f| = f(s,v) v V
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u
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u
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f(u,v) = 5
f(v,u) = -5
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The max-flow problem
Informal definition of the max-flow problem:
What is the greatest rate at which material can be shipped from the source to the sink without violating any capacity contraints?
Formal definition of the max-flow problem:
The max-flow problem is to find a valid flow for a given weighted directed graph G, that has the maximum value over all valid flows.
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The Ford-Fulkerson methoda way how to find the max-flow
This method contains 3 important ideas:
1) residual networks
2) augmenting paths
3) cuts of flow networks
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Ford-Fulkerson – pseudo code
1 initialize flow f to 0
2 while there exits an augmenting path p
3 do augment flow f along p
4 return f
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Ford Fulkerson – residual networks
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The residual network Gf of a given flow network G with a valid flow f consists of the same vertices v V as in G which are linked with residual edges (u,v) Ef that can admit more strictly positive net flow.
The residual capacity cf represents the weight of each edge Ef and is the amount of additional net flow f(u,v) before exceeding the capacity c(u,v)
Flow network G = (V,E)
cf(u,v) = c(u,v) – f(u,v)
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residual network Gf = (V,Ef)
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Ford Fulkerson – residual networks
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Flow network G = (V,E)
cf(u,v) = c(u,v) – f(u,v)
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residual network Gf = (V,Ef)
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The residual network Gf of a given flow network G with a valid flow f consists of the same vertices v V as in G which are linked with residual edges (u,v) Ef that can admit more strictly positive net flow.
The residual capacity cf represents the weight of each edge Ef and is the amount of additional net flow f(u,v) before exceeding the capacity c(u,v)
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Ford Fulkerson – augmenting paths
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cf(p) = min{cf (u,v): (u,v) is on p}
Flow network G = (V,E) residual network Gf = (V,Ef)
Definition: An augmenting path p is a simple (free of any cycle) path from s to t in the residual network Gf
Residual capacity of p
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154
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Ford Fulkerson – augmenting paths
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cf(p) = min{cf (u,v): (u,v) is on p}
Flow network G = (V,E)
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v4
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8
12
5
3
43
5
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5
residual network Gf = (V,Ef)
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5
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154
Definition: An augmenting path p is a simple (free of any cycle) path from s to t in the residual network Gf
Residual capacity of p
Augmenting path
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Ford Fulkerson – augmenting paths
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Flow network G = (V,E)
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residual network Gf = (V,Ef)
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154
We define a flow: fp: V x V R such as:
cf(p) if (u,v) is on p
fp(u,v) = - cf(p) if (v,u) is on p
0 otherwise
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Ford Fulkerson – augmenting paths
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Flow network G = (V,E)
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11
-4/8
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-4/5
residual network Gf = (V,Ef)
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-4/154/4
Our virtual flow fp along the augmenting path p in Gf
cf(p) if (u,v) is on p
fp(u,v) = - cf(p) if (v,u) is on p
0 otherwise
We define a flow: fp: V x V R such as:
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Ford Fulkerson – augmenting the flow
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Flow network G = (V,E)
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4/5
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residual network Gf = (V,Ef)
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Our virtual flow fp along the augmenting path p in Gf
New flow: f´: V x V R : f´=f + fp
cf(p) if (u,v) is on p
fp(u,v) = - cf(p) if (v,u) is on p
0 otherwise
We define a flow: fp: V x V R such as:
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Ford Fulkerson – new valid flowproof of capacity constraint
Proof:
fp (u ,v) < cf (u ,v) = c (u ,v) – f (u ,v)
(f + fp) (u ,v) = f (u ,v) + fp (u ,v) < c (u ,v)
Lemma:
f´ : V x V R : f´ = f + fp in G
Capacity constraint:
For all u,v V, we require f(u,v) < c(u,v)
cf(p) if (u,v) is on p
fp(u,v) = - cf(p) if (v,u) is on p
0 otherwise
cf(p) = min{cf (u,v): (u,v) is on p}
cf(u,v) = c(u,v) – f(u,v)
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Ford Fulkerson – new valid flowproof of Skew symmetry
Proof:
(f + fp)(u ,v) = f (u ,v) + fp (u ,v) = - f (v ,u) – fp (v ,u)
= - (f (v ,u) + fp (v ,u)) = - (f + fp) (v ,u)
Skew symmetry:
For all u,v V, we require f(u,v) = - f(v,u)
Lemma:
f´ : V x V R : f´ = f + fp in G
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Ford Fulkerson – new valid flowproof of flow conservation
Proof:
u V – {s ,t} (f + fp) (u ,v) = (f(u ,v) + fp (u ,v))
= f (u ,v) + fp (u ,v) = 0 + 0 = 0
Flow conservation:
For all u V \ {s,t} : f(u,v) = 0 v V
v V v V
v V v V
Lemma:
f´ : V x V R : f´ = f + fp in G
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Ford Fulkerson – new valid flow
Proof:
| (f + fp) | = (f + fp) (s ,v) = (f (s ,v) + fp (s ,v))
= f (s ,v) + fp (s ,v) = | f | + | fp |
Lemma:
| (f + fp) | = | f | + | fp |
v V v V
v V v V
Value of a Flow f:
Def: |f| = f(s,v)
v V
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Ford Fulkerson – new valid flow
Lemma:
| (f + fp) | = | f | + | fp | > | f |
f´ : V x V R : f´ = f + fp in G
Lemma shows:
if an augmenting path can be found then the above flow augmentation will result in a flow improvement.
Question: If we cannot find any more an augmenting path is our flow then maximum?
Idea: The flow in G is maximum the residual Gf contains no augmenting path.
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Ford Fulkerson – cuts of flow networksNew notion: cut (S,T) of a flow network
A cut (S,T) of a flow network G=(V,E) is a partiton of V into S and T = V \ S such that s S and t T.
Implicit summation notation: f (S, T) = f (u, v)
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TS
In the example:
S = {s,v1,v2) , T = {v3,v4,t}
Net flow f(S ,T) = f(v1,v3) + f(v2,v4) + f(v2,v3)
= 12 + 11 + (-0) = 23
Capacity c(S,T) = c(v1,v3) + c(v2,v4)
= 12 + 14 = 26
Practical example
u S
v T
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Ford Fulkerson – cuts of flow networks
Lemma:
the value of a flow in a network is the net flow across any cut of the network
f (S ,T) = | f |
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proof
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Ford Fulkerson – cuts of flow networksAssumption:
The value of any flow f in a flow network G is bounded from above by the capacity of any cut of G
Lemma: | f | < c (S, T)
| f | = f (S, T)
= f (u, v)
< c (u, v)
= c (S, T) v T
u S
v T
u S
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F. Fulkerson: Max-flow min-cut theorem
If f is a flow in a flow network G = (V,E) with source s and sink t, then the following conditions are equivalent:
1. f is a maximum flow in G.
2. The residual network Gf contains no augmenting paths.
3. | f | = c (S, T) for some cut (S, T) of G.proof:
(1) (2):
We assume for the sake of contradiction that f is a maximum flow in G but that there still exists an augmenting path p in Gf.
Then as we know from above, we can augment the flow in G according to the formula: f´= f + fp. That would create a flow f´that is strictly greater than the former flow f which is in contradiction to our assumption that f is a maximum flow.
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F. Fulkerson: Max-flow min-cut theorem
If f is a flow in a flow network G = (V,E) with source s and sink t, then the following conditions are equivalent:
1. f is a maximum flow in G.
2. The residual network Gf contains no augmenting paths.
3. | f | = c (S, T) for some cut (S, T) of G.proof:
(2) (3):
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1/3S t
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residual network Gf original flow network G
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F. Fulkerson: Max-flow min-cut theorem
If f is a flow in a flow network G = (V,E) with source s and sink t, then the following conditions are equivalent:
1. f is a maximum flow in G.
2. The residual network Gf contains no augmenting paths.
3. | f | = c (S, T) for some cut (S, T) of G.proof:
(2) (3): Define
S = {v V | path p from s to v in Gf }
T = V \ S (note t S according to (2)) S t
v1
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v3
v4
2
3
2
8
1
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6 14
residual network Gf
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F. Fulkerson: Max-flow min-cut theorem
If f is a flow in a flow network G = (V,E) with source s and sink t, then the following conditions are equivalent:
1. f is a maximum flow in G.
2. The residual network Gf contains no augmenting paths.
3. | f | = c (S, T) for some cut (S, T) of G.proof:
(2) (3): Define
S = {v V | path p from s to v in Gf }
T = V \ S (note t S according to (2))
for u S, v T: f (u, v) = c (u, v) (otherwise (u, v) Ef and v S)
| f | = f (S, T) = c (S, T)
1
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original network G
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F. Fulkerson: Max-flow min-cut theorem
If f is a fow in a flow network G = (V,E) with source s and sink t, then the following conditions are equivalent:
1. f is a maximum flow in G.
2. The residual network Gf contains no augmenting paths.
3. | f | = c (S, T) for some cut (S, T) of G.proof:
(3) (1): as proofed before | f | = f (S, T) < c (S, T)
the statement of (3) : | f | = c (S, T) implies that f is a maximum flow
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1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t in the
residual network Gf 5 do cf (p) = min {cf (u, v) | (u, v) is in p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf (p) 8 f [v, u] = - f [u, v]
The basic Ford Fulkerson algorithm
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The basic Ford Fulkerson algorithmexample of an execution
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1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t
in the residual network Gf 5 do cf(p) = min{cf(u, v) | (u, v) p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf(p) 8 f [v, u] = - f [u, v]
(residual) network Gf
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The basic Ford Fulkerson algorithmexample of an execution
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(residual) network Gf 1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t
in the residual network Gf 5 do cf(p) = min{cf(u, v) | (u, v) p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf(p) 8 f [v, u] = - f [u, v]
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The basic Ford Fulkerson algorithmexample of an execution
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(residual) network Gf 1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t
in the residual network Gf 5 do cf(p) = min{cf(u, v) | (u, v) p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf(p) 8 f [v, u] = - f [u, v]
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The basic Ford Fulkerson algorithmexample of an execution
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(residual) network Gf
temporary variable:
cf (p) = 12
1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t
in the residual network Gf 5 do cf(p) = min{cf(u, v) | (u, v) p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf(p) 8 f [v, u] = - f [u, v]
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The basic Ford Fulkerson algorithmexample of an execution
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(residual) network Gf
temporary variable:
cf (p) = 12S t
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new flow network G
1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t
in the residual network Gf 5 do cf(p) = min{cf(u, v) | (u, v) p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf(p) 8 f [v, u] = - f [u, v]
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The basic Ford Fulkerson algorithmexample of an execution
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(residual) network Gf
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new flow network G
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1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t
in the residual network Gf 5 do cf(p) = min{cf(u, v) | (u, v) p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf(p) 8 f [v, u] = - f [u, v]
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The basic Ford Fulkerson algorithmexample of an execution
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(residual) network Gf
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new flow network G
1212
1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t
in the residual network Gf 5 do cf(p) = min{cf(u, v) | (u, v) p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf(p) 8 f [v, u] = - f [u, v]
51
The basic Ford Fulkerson algorithmexample of an execution
S t
v1
v2
v3
v4
10
13
12
4
4
4
14
8
7
9
(residual) network Gf
S t
v1
v2
v3
v4
10
13
12/12
12/16
4
4
14
12/20
7
9
new flow network G
1212
1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t
in the residual network Gf 5 do cf(p) = min{cf(u, v) | (u, v) p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf(p) 8 f [v, u] = - f [u, v]
52
The basic Ford Fulkerson algorithmexample of an execution
(residual) network Gf
S t
v1
v2
v3
v4
10
13
12/12
12/16
4
4
14
12/20
7
9
new flow network G
temporary variable:
cf (p) = 4
S t
v1
v2
v3
v4
10
13
12
4
4
4
14
8
7
9
1212
1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t
in the residual network Gf 5 do cf(p) = min{cf(u, v) | (u, v) p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf(p) 8 f [v, u] = - f [u, v]
53
The basic Ford Fulkerson algorithmexample of an execution
S t
v1
v2
v3
v4
10
13
12
4
4
4
14
8
7
9
(residual) network Gf
S t
v1
v2
v3
v4
10
4/13
12/12
12/16
4
4/4
4/14
12/20
7
9
new flow network G
1212
temporary variable:
cf (p) = 4
1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t
in the residual network Gf 5 do cf(p) = min{cf(u, v) | (u, v) p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf(p) 8 f [v, u] = - f [u, v]
54
The basic Ford Fulkerson algorithmexample of an execution
S t
v1
v2
v3
v4
10
9
12
4
4
4
10
8
7
9
(residual) network Gf
S t
v1
v2
v3
v4
10
4/13
12/12
12/16
4
4/4
4/14
12/20
7
9
new flow network G
1212
4
4
1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t
in the residual network Gf 5 do cf(p) = min{cf(u, v) | (u, v) p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf(p) 8 f [v, u] = - f [u, v]
55
The basic Ford Fulkerson algorithmexample of an execution
S t
v1
v2
v3
v4
10
9
12
4
4
4
10
8
7
9
(residual) network Gf
S t
v1
v2
v3
v4
10
4/13
12/12
12/16
4
4/4
4/14
12/20
7
9
new flow network G
1212
4
4
1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t
in the residual network Gf 5 do cf(p) = min{cf(u, v) | (u, v) p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf(p) 8 f [v, u] = - f [u, v]
56
The basic Ford Fulkerson algorithmexample of an execution
S t
v1
v2
v3
v4
10
9
12
4
4
4
10
8
7
9
(residual) network Gf
S t
v1
v2
v3
v4
10
4/13
12/12
12/16
4
4/4
4/14
12/20
7
9
new flow network G
1212
4
4
1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t
in the residual network Gf 5 do cf(p) = min{cf(u, v) | (u, v) p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf(p) 8 f [v, u] = - f [u, v]
57
The basic Ford Fulkerson algorithmexample of an execution
S t
v1
v2
v3
v4
10
9
12
4
4
4
10
8
7
9
(residual) network Gf
S t
v1
v2
v3
v4
10
4/13
12/12
12/16
4
4/4
4/14
12/20
7
9
new flow network G
1212
4
4
temporary variable:
cf (p) = 7
1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t
in the residual network Gf 5 do cf(p) = min{cf(u, v) | (u, v) p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf(p) 8 f [v, u] = - f [u, v]
58
The basic Ford Fulkerson algorithmexample of an execution
S t
v1
v2
v3
v4
10
9
12
4
4
4
10
8
7
9
(residual) network Gf
S t
v1
v2
v3
v4
10
11/13
12/12
12/16
4
4/4
11/14
19/20
7/7
9
new flow network G
1212
4
4
temporary variable:
cf (p) = 7
1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t
in the residual network Gf 5 do cf(p) = min{cf(u, v) | (u, v) p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf(p) 8 f [v, u] = - f [u, v]
59
The basic Ford Fulkerson algorithmexample of an execution
(residual) network Gf
S t
v1
v2
v3
v4
10
11/13
12/12
12/16
4
4/4
11/14
19/20
7/7
9
new flow network G
S t
v1
v2
v3
v4
10
2
12
4
4
4
3
1
7
9
1219
11
11
1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t
in the residual network Gf 5 do cf(p) = min{cf(u, v) | (u, v) p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf(p) 8 f [v, u] = - f [u, v]
60
The basic Ford Fulkerson algorithmexample of an execution
(residual) network Gf
S t
v1
v2
v3
v4
10
11/13
12/12
12/16
4
4/4
11/14
19/20
7/7
9
new flow network G
S t
v1
v2
v3
v4
10
2
12
4
4
4
3
1
7
9
1219
11
11
Finally we have:
| f | = f (s, V) = 23
1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t
in the residual network Gf 5 do cf(p) = min{cf(u, v) | (u, v) p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf(p) 8 f [v, u] = - f [u, v]
61
Analysis of the Ford Fulkerson algorithmRunning time
The running time depends on how the augmenting path p in line 4 is determined.
1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t
in the residual network Gf 5 do cf(p) = min{cf(u, v) | (u, v) p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf(p) 8 f [v, u] = - f [u, v]
62
Analysis of the Ford Fulkerson algorithmRunning time (arbitrary choice of p)
O(|E|)
O(|E|)
O(|E| |fmax|)
running time: O ( |E| |fmax| )
with fmax as maximum flow
(1) The augmenting path is chosen arbitrarily and all capacities are integers
1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t
in the residual network Gf 5 do cf(p) = min{cf(u, v) | (u, v) p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf(p) 8 f [v, u] = - f [u, v]
O(|E|)
63
Analysis of the Ford Fulkerson algorithmRunning time (arbitrary choice of p)
(1) The augmenting path is chosen arbitrarily and all capacities are integers
Consequencies of an arbitrarily choice:
Example if |f*| is large:
t
v2
v1
s
1,000,000 1,000,000
1,000,0001,000,000
1
running time: O ( |E| |fmax| )
with fmax as maximum flow
64
Analysis of the Ford Fulkerson algorithmRunning time (arbitrary choice of p)
(1) The augmenting path is chosen arbitrarily and all capacities are integers
Consequencies of an arbitrarily choice:
Example if |f*| is large:
t
v2
v1
s
1 / 1,000,000 1,000,000
1,000,0001 / 1
,000,000
1 / 1
t
v2
v1
s
999,999 1,000,000
1,000,000999,999
1
residual network Gf
1
1
running time: O ( |E| |fmax| )
with fmax as maximum flow
65
Analysis of the Ford Fulkerson algorithmRunning time (arbitrary choice of p)
(1) The augmenting path is chosen arbitrarily and all capacities are integers
Consequencies of an arbitrarily choice:
Example if |f*| is large:
t
v2
v1
s
1 / 1,000,000 1 / 1,000,000
1 / 1,000,0001 / 1
,000,000
1
t
v2
v1
s
999,999 999,999
999,999999,999
1
residual network Gf
1
1
1
1
running time: O ( |E| |fmax| )
with fmax as maximum flow
66
Analysis of the Ford Fulkerson algorithmRunning time with Edmonds-Karp algorithm
running time: O ( |V| |E|² )
S t
u1
u2
v3
v44
4
6
4
4
6
2
2
Informal idea of the proof:
(1) for all vertices v V\{s,t}:
the shortest path distance f(s,v) in Gf increases monotonically with each flow augmentation
. f(s,v) < f´(s,v) f(s,u2) = 1
(2) Edmonds-Karp algorithm
augmenting path is found by breath-first search and has to be a shortest path from p to t