Post on 06-Apr-2018
transcript
8/3/2019 Maxflow Problem
http://slidepdf.com/reader/full/maxflow-problem 1/73
Network Flow, ChinesePostman,Topological Sorting and
Planarity Detection and Embedding
Firas M Abu Hassan20911833
CS 530: Design and Analysis of Algorithms
Fall 2010 1
8/3/2019 Maxflow Problem
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Network flows Problem
Flow network: a flow network G = (V,E) is a directed graphin which each edge (u,v) ∈ E has a nonnegative capacityc(u,v) >= 0. if (u,v) ∉ E , we assume that c(u,v) = 0.
We distinguish two vertices in flow network , source s andsink t.
The source produces the material at a steady rate . The sink consumes the material at a steady rat
Objective : how much of material can be imposed throughnetwork that network can handle. (Find a maximal flow over adirected graph)
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Definition of Flow
Each edge can carry an entity called “flow”.
For flow on an edge to be valid, it must satisfythree conditions: Capacity constraint: The flow can not be greater than
the capacity of the edge f ( u,v ) <= c ( u,v )
Skew symmetry: The flow on the edge (u,v) is equal tothe negation of the flow on the edge (v,u)
Flow conservation: The total flow entering a vertex(excluding the source or sink) must equal the totalpositive flow leaving that vertex.
V vV v
t v f vs f f ),(),(||3
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Definition of Flow
S t
v1
v2
v3
v4
Flow network: directed graph G=(V,E)
S t
v1
v2
v3
v4
u v6/12
u v6/6
f(u,v)=6
f(u,v)=6
Flow below capacity
Maximum flow
8
3
6
8
6
6
3
3
Example: oilpipeline
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Example of a Flow:
f(v2, v1) = 1, c(v2, v1) = 4. f(v1, v2) = -1, c(v1, v2) = 10.
f(v3, s) + f(v3, v1) + f(v3, v2) + f(v3, v4) + f(v3, t) =
0 + (-12) + 4 + (-7) + 15 =
0
flowcapacity
capacity
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Residual Networks
The residual capacity of an edge (u,v) in a network with a
flow f is given by:
),(),(),( vu f vucvuc f
• The residual network of a graph G induced by a flow f is the graph including only the edges with positive
residual capacity, i.e., ( , ), wheref f
G V E {( , ) : ( , ) 0}f f
E u v V V c u v
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The basic Ford Fulkerson method
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 theresidual network G f
5 do cf (p) = min {cf (u, v) | (u, v) is in p}
6 for each edge (u, v) in p
7do
f [u, v] = f [u, v] + cf (p)8 f [v, u] = - f [u, v]
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The basic Ford Fulkersonaugmenting paths Algorithm
S t
v1
v2
v3
v4
12
14
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 G f 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 Fulkersonaugmenting paths Algorithm
S t
v1
v2
v3
v4
0 /12
0 /14
(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 G f 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 Fulkersonaugmenting paths Algorithm
S t
v1
v2
v3
v4
12
14
(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 G f 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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8/3/2019 Maxflow Problem
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The basic Ford Fulkersonaugmenting paths Algorithm
S t
v1
v2
v3
v4
12
14
(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 G f 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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8/3/2019 Maxflow Problem
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The basic Ford Fulkersonaugmenting paths Algorithm
S t
v1
v2
v3
v4
12
14
(residual) network Gf
temporary variable:cf (p) = 12
S t
v1
v2
v3
v4
12/12
14
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 G f 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 Fulkersonaugmenting paths Algorithm
S t
v1
v2
v3
v4
12
14
(residual) networkGf
S t
v1
v2
v3
v4
12/12
14
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 G f 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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8/3/2019 Maxflow Problem
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The basic Ford Fulkersonaugmenting paths Algorithm
S t
v1
v2
v3
v4
12
14
(residual) network Gf
S t
v1
v2
v3
v4
12/12
14
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 G f 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 Fulkersonaugmenting paths Algorithm
S t
v1
v2
v3
v4
12
14
(residual) network Gf
S t
v1
v2
v3
v4
12/12
14
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 G f 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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8/3/2019 Maxflow Problem
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The basic Ford Fulkersonaugmenting paths Algorithm
(residual) networkGf
S t
v1
v2
v3
v4
12/12
14
new flow network
G
temporary variable:cf (p) = 4
S t
v1
v2
v3
v4
12
14
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 G f 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 Fulkersonaugmenting paths Algorithm
S t
v1
v2
v3
v4
12
14
(residual) networkGf
S t
v1
v2
v3
v4
12/12
new flow network
G 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 G f 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 Fulkersonaugmenting paths Algorithm
S t
v1
v2
v3
v4
12
10
(residual) network Gf
S t
v1
v2
v3
v4
12/12
new flow network
G
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 G f 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 Fulkersonaugmenting paths Algorithm
S t
v1
v2
v3
v4
12
10
(residual) network Gf
S t
v1
v2
v3
v4
12/12
new flow network
G
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 G f 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 Fulkersonaugmenting paths Algorithm
S t
v1
v2
v3
v4
12
10
(residual) networkGf
S t
v1
v2
v3
v4
12/12
new flow network
G
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 G f 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]
20
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The basic Ford Fulkersonaugmenting paths Algorithm
S t
v1
v2
v3
v4
12
10
(residual) networkGf
S t
v1
v2
v3
v4
12/12
new flow network
G
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 G f 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]
21
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The basic Ford Fulkersonaugmenting paths Algorithm
S t
v1
v2
v3
v4
12
10
(residual) networkGf
S t
v1
v2
v3
v4
12/12
new flow network
G
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 G f 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]
22
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The basic Ford Fulkersonaugmenting paths Algorithm
(residual) networkGf
S t
v1
v2
v3
v4
12/12
new flow network
G
S t
v1
v2
v3
v4
12
3
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 G f 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]
23
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The basic Ford Fulkersonaugmenting paths Algorithm
(residual) network Gf
S t
v1
v2
v3
v4
12/12new flow network
G
S t
v1
v2
v3
v4
12
3
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 G f
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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Analysis of the Ford Fulkerson algorithm
The running time depends on how the augmenting path p in line 4 isdetermined.
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 G f 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]
25
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Analysis of the Ford Fulkerson algorithm
O(|E|)
O(|E|) O(|E||fmax|)
running time: O ( |E| |fmax|)with fmax as maximumflow
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 G f 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|)
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Ford Fulkerson – cuts of flow networks
A cut (S,T) of a flow network G=(V,E) is a partiton of V into Sand T = V \ S such that s S and t T.
S t
v1
v2
v3
v4
12/12
11/14
TS
For 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
Max Flow = Min Cut = 23
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Edmonds-Karp Algorithm
Edmonds-Karp = Ford-Fulkerson using breadth-first search inline 4 to find an augmenting path,i.e., the augmenting path found is shortest in terms of thenumber of edges.
Edmonds-Karp runs in O(VE2) time… 28
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Analysis of Edmonds-Karp
If the Edmonds-Karp algorithm is run on a flow networkG = (V, E) with source s and sink t, then the total numberof flow augmentations performed by the algorithm is
O(VE).
Since, breadth-first search can be implemented in O(E)time, the running time of Edmonds-Karp is O(VE * E) =
O(VE2).
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A BFS Example
g
s
d
b
f
e
c
a
L = s 0
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s
d
b
g
f
e
c
a
L = s 0
L = a, c 1
A BFS Example
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s
d
b
g
f
e
c
a
L = s 0
L = a, c 1
L = d, e, f 2
A BFS Example
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s
d
b
g
f
e
c
a
L = s 0
L = a, c 1
L = d, e, f 2
L = b , g 3
A BFS Example
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s
d
b
g
f
e
c
a
L = s 0
L = a, c 1
L = d, e, f 2
L = b , g 3
A BFS Example
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s
d
b
g
f
e
c
a
0
1
1
2
2
2
3
3
d (b ): shortest distance from s to b
A BFS Example
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History
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Discoverer Method Big-Oh Year Ford, Fulkerson Augmenting path mF*1955
Edmonds-Karp Shortest path m2n1970
Dinitz Shortest path mn2 1970
Edmonds-Karp, Dinitz Capacity scaling m2 logn1972
Dinitz-Gabow Capacity scaling mn logn1973
Karzanov Preflow-push n3 1974
Sleator-Tarjan Dynamic trees mn log n1983
Goldberg-Tarjan FIFO preflow-push mn log (n2 / m)1986
. . . . . . . . .. . .
Goldberg-Rao Length functionm3/2 log (n2 / m) log n
mn2/3 log (n2 / m) log n1997
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CHINESE POSTMAN ALGORITHM
• This is used when we are trying to find the shortest
route round a network such all the arcs are travelled
along at least once.
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Edmond’s Algorithm
procedure FloydWarshallWithPathReconstruction ()
for k := 1 to n
for i := 1 to n
for j := 1 to nif path[i][k] + path[k][j] < path[i][j] then
path[i][j] := path[i][k]+path[k][j];
next[i][j] := k;
Run time = O(n 3 )
38
Add thi t th t k
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A postman starts at the depot T and has to deliver mailalong all the streets shown in the network below. Find aroute that means he travels the shortest possible distance.
T
HG
FE
DCB
A
0.8
1.7
1.01.0
2.0 0.750.2
0.4 0.6
0.5 1.5
1.70.6
0.9
1.6
0.9
0.3
Odd vertices are B, G, H & T Possible pairings:
BG & HT
Route: BEG & HECT
Length: 1.7 + 1.8 = 3.5
BH & GT Route: BEH & GECT
Length: 1.5 + 2.0 = 3.5
BT & GH
Route: BCT & GH
Length: 0.8 + 0.9 = 1.7
The shortest extra route is:
Add this to the network
Possible route is:
TABEAGHEGHFDFECDTCBCT
39
T
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A
B C
D
EF
GH
Route is:
T A B E A G H E G H F D F E C D T C B C T
Total length of alloriginal paths is16.45 km
Add on the extraroutes BCT and GH= 1.7 km
Shortest length =16.45 + 1.7 = 18.15 km
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Topological Sort
it should be possible to perform all the tasks insuch an order as to satisfy all of thedependencies
Such as order is called a topological sort of the
vertices
The requirement is to list all of the vertices suchthat in that list, vi precedes v j whenever a path
exists from vi to v j
If this is not possible, this would imply the existenceof a cycle.
Theorem:41
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Topological Sort
int topologicalOrderTraversal( ){
int numVisitedVertices = 0;
while(there are more vertices to be visited){
if(there is no vertex with in-degree 0)
break;else{
select a vertex v that has in-degree 0;
visit v;
numVisitedVertices++;
delete v and all its emanating edges;
}}
return numVisitedVertices;
} Running time: O(V + E) =O(n)
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Example
Let us define the table of in-degrees
VertexIn-
degree
1 02 1
3 3
4 3
5 2
6 0
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Example
And a queue into which we can insert vertices 1and 6
1 6
Queue
VertexIn-
degree
1 02 1
3 3
4 3
5 2
6 0
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Example
We dequeue the head (1), decrement the in-degree of all adjacent vertices: 2 and 4
2 is decremented to zero: enqueue 2
6 2
QueueSort1
VertexIn-
degree
1 02 0
3 3
4 2
5 2
6 0
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Example
We dequeue 6 and decrement the in-degree of alladjacent vertices
Neither is decremented to zero
2
QueueSort1, 6
VertexIn-
degree
1 02 0
3 2
4 2
5 1
6 0
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Example
We dequeue 2, decrement, and enqueue vertex 5
5
QueueSort1, 6, 2
VertexIn-
degree
1 02 0
3 1
4 1
5 0
6 0
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Example
We dequeue 5, decrement, and enqueue vertex 3
3
QueueSort1, 6, 2, 5
VertexIn-
degree
1 02 0
3 0
4 1
5 0
6 0
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Example
We dequeue 3, decrement 4, and add 4 to thequeue
4
QueueSort1, 6, 2, 5, 3
VertexIn-
degree
1 02 0
3 0
4 0
5 0
6 0
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Example
We dequeue 4, there are no adjacent vertices todecrement the in degree
QueueSort1, 6, 2, 5, 3, 4
VertexIn-
degree
1 02 0
3 0
4 0
5 0
6 0
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Example
The queue is now empty, so a topological sort is1, 6, 2, 5, 3, 4
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Planar Graphs
Can be drawn on the plane without crossings
We will :
Testing if a graph is planarFinding a nice drawing of a planar graph
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Testing if a graph is planar
Euler’s Formula
- region : is a part of the plane completelydisconnected off from other parts of the planeby the edges of the graph.
r = |E | - |V | + 2
c = r - |E | + |V | c must equal 2 for planar graphs.
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Face-Edge Handshaking
DEF: The degree of a region F is the number ofedges at its boundary, and is denoted by deg(F ).
THM: Let G be a planar graph with region set R .
Then:
RF
F E )deg(21 ||
Ex: (4+4+4+4+4+4)/2 = 12
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Proving that Q4 is Non-Planar
Now we have enough invariants to prove thatthe 4-cube is non-planar.
1) Count the number of vertices and edges:
|V | = 16 (twice the number for 3-cube)
|E | = 32 (twice the number for 3-cube plusnumber of vertices in 3-cube)
2) Suppose 4-cube were planar so by Euler’sformula number of regions would be:
r = 32-16+2=18
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Proving that Q4 is Non-Planar
3) Calculate the degree of each region
In our case, this give |E | ½·18·4=36
contradicting |E | = 32 !
Thus 4-cube cannot be planar. •
RF
F E )deg(2
1 ||
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Testing if a graph is planar
Circle-Chord Method
Step One: Find a circuit that contains all thevertices of the graph.
Step Two: Draw this circuit as a large circle.
a b c d
e hgfgh
f
bc
e
d
a
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Testing if a graph is planarStep Three: Choose one chord, and decide to draw it either inside or outsidethe circle. If chosen correctly, it will force certain other chords to be drawnopposite to the circle. (Inside if the first chord was drawn outside, and viceversa.)
gh
f
bc
e
d
a
Since the chords could be drawn withoutcrossing, this graph is planar.
a b c d
e hgf
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a
e
f d
c
bh
g
a
f
e
d
c
bh
g
Testing if a graph is planar
• Kuratowski theoremA graph is planar if and only if it does not contain the K5
or the K3,3
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S i Al ith
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Sugiyama Algorithm
Remove cycles
Generate levelling
Sugiyama 1979
Running time O(|V||E|\log|E)
11
3
8
1
5
10 7
4
12
6
9
2
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S i Al ith
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Sugiyama Algorithm
Algorithm
Remove cycles
Generate levelling
11
85
10 7
12
6
9
31 421
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Sugiyama Algorithm
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Sugiyama Algorithm
Algorithm
Remove cycles
Generate levelling
11
8 10
129
31
5 7
4
6
21
2
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Sugiyama Algorithm
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Sugiyama Algorithm
Algorithm
Remove cycles
Generate levelling
1112
3
8
1
5
10
7
42
9
6
1
2
3
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Sugiyama Algorithm
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Sugiyama Algorithm
Algorithm
Remove cycles
Generate levelling
Split long edges
11
3
8 9
1
5
10
7
42
12
6
1
2
3
4
64
Sugiyama Algorithm
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Sugiyama Algorithm
Algorithm
Remove cycles
Generate levelling
Split long edges
Reduce crossings11
6
3
8 9
1
5
10
7
42
12
1
2
3
4
65
Sugiyama Algorithm
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Sugiyama Algorithm
Algorithm
Remove cycles
Generate levelling
Split long edges
Reduce crossings11
6
3
8 9
1
5
10
7
42
12
1
2
3
4
66
Sugiyama Algorithm
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Sugiyama Algorithm
Algorithm
Remove cycles
Generate levelling
Split long edges
Reduce crossings11
6
3
89
1
5
10
7
42
12
1
2
3
4
67
Sugiyama Algorithm
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Sugiyama Algorithm
Algorithm
Remove cycles
Generate levelling
Split long edges
Reduce crossings11
6
3
89
1
5
10
7
42
12
1
2
3
4
68
Sugiyama Algorithm
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Sugiyama Algorithm
Algorithm
Remove cycles Generate levelling
Split long edges
Reduce crossings11
6
3
89
1
5
10
7
42
12
1
2
3
4
69
Sugiyama Algorithm
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Sug ya a go t
Algorithm
Remove cycles Generate levelling
Split long edges
Reduce crossings11
6
3
89
1
5
10
7
42
12
1
2
3
4
70
Sugiyama Algorithm
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g y g
Algorithm
Remove cycles Generate levelling
Split long edges
Reduce crossings11
6
3
89
1
5
10
7
42
12
1
2
3
4
71
Sugiyama Algorithm
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g y g
Algorithm
Remove cycles Generate levelling
Split long edges
Reduce crossings
Join long edges 11
6
3
89
1
5
10
7
42
12
1
2
3
4
72
Sugiyama Algorithm
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g y g
Algorithm
Remove cycles Generate levelling
Split long edges
Reduce crossings
Join long edges
Basic properties
2 level crossing minimisation
Minimum edge set whoseremoval eliminates crossings
11
6
3
89
1 42
10
12
57
1
2
3
4