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Lecture slides by Kevin WayneCopyright © 2005 Pearson-Addison Wesley
Copyright © 2013 Kevin Waynehttp://www.cs.princeton.edu/~wayne/kleinberg-tardos
Last updated on Sep 8, 2013 6:40 AM
7. NETWORK FLOWS I
‣ Ford-Fulkerson pathological example
Intuition. Let r satisfy r2 = 1 – r.
・Initial capacities are { 1, r }.
・After some augmentation, residual capacities are { 1, r, r2 }.
・After some more, residual capacities are { 1, r, r2, r3 }.
・After some more, residual capacities are { 1, r, r2, r3, r4 }.
2
Ford-Fulkerson pathological example
r =
�5 � 1
2=� r2 = 1 � r
1 – r
r – r2
r2 – r3
3
Ford-Fulkerson pathological example
v
u
w
xs 0 / 100
0 / 100 0 / r
0 / 1
network G
0 / 1
t
0 / 100
0 / 100
0 / 100
0 / 100v
u
w
xs
t
sufficiently largethat it won't everbe a bottleneck(we'll suppress)
r =
�5 � 1
2=� r2 = 1 � r
4
Ford-Fulkerson pathological example
v
u
w
xs
0 / r
augmenting path 1: s→v→w→t (bottleneck capacity = 1)
0 / 1
t
v
u
w
xs
t
0 / 1—
1
r =
�5 � 1
2=� r2 = 1 � r
5
Ford-Fulkerson pathological example
v
u
w
xs
augmenting path 2: s→x→w→v→u→t (bottleneck capacity = r)
t
v
u
w
xs
t
0 / r
0 / 1
1 / 1—
1 – r
—r = 1
– r2
—r
r =
�5 � 1
2=� r2 = 1 � r
6
Ford-Fulkerson pathological example
v
u
w
xs
augmenting path 3: s→v→w→x→t (bottleneck capacity = r)
t
v
u
w
xs
t
r / r
1 – r
2 / 1
1 – r / 1
1
—0
r =
�5 � 1
2=� r2 = 1 � r
7
Ford-Fulkerson pathological example
v
u
w
xs
augmenting path 4: s→x→w→v→u→t (bottleneck capacity = r2)
t
v
u
w
xs
t
0 / r
1 / 1—
1 – r2
—r2 =
r – r3
1 – r
2 / 1
1
r =
�5 � 1
2=� r2 = 1 � r
8
Ford-Fulkerson pathological example
v
u
w
xs
augmenting path 5: s→u→v→w→t (bottleneck capacity = r2)
t
v
u
w
xs
t
1 / 1
1 – r2 / 1
—1 – r
2
1
r – r3 /
r
r =
�5 � 1
2=� r2 = 1 � r
9
Ford-Fulkerson pathological example
v
u
w
xs
augmenting path 6: s→x→w→v→u→t (bottleneck capacity = r3)
t
v
u
w
xs
t
1 – r
2 / 1
1 / 1—
1 – r3
1 – r
2 + r3 =
1 –
r4
r – r3 /
rr
r =
�5 � 1
2=� r2 = 1 � r
10
Ford-Fulkerson pathological example
v
u
w
xs
augmenting path 7: s→v→w→x→t (bottleneck capacity = r3)
t
v
u
w
xs
t
r / r
1 – r3 / 1
1
—r – r3 =
1
1 – r
4 / 1
r =
�5 � 1
2=� r2 = 1 � r
11
Ford-Fulkerson pathological example
v
u
w
xs
augmenting path 8: s→x→w→v→u→t (bottleneck capacity = r4)
t
v
u
w
xs
t
r – r3 /
r
1 / 1—
1 – r4
1
1 – r
4 / 1
r – r3 +
r4 =
r –
r5
r =
�5 � 1
2=� r2 = 1 � r
12
Ford-Fulkerson pathological example
v
u
w
xs
augmenting path 9: s→u→v→w→t (bottleneck capacity = r4)
t
v
u
w
xs
t
1 / 1
—1
– r4
1
1 – r4 / 1
r – r5 /
r
r =
�5 � 1
2=� r2 = 1 � r
13
Ford-Fulkerson pathological example
v
u
w
xs
after augmenting path 5: { 1 - r2, 1, r - r3 } (flow = 1 + 2r + 2r2)
t
v
u
w
xs
t
1 / 1
after augmenting path 1: { 1 - r0, 1, r - r1 } (flow = 1)
1 – r
4 / 1
r – r5 /
r
after augmenting path 9: { 1 - r4, 1, r - r5 } (flow = 1 + 2r + 2r2 + 2r3 + 2r4)
r =
�5 � 1
2=� r2 = 1 � r
14
Ford-Fulkerson pathological example
Theorem. The Ford-Fulkerson algorithm may not terminate; moreover, it
may converge a value not equal to the value of the maximum flow.
Pf.
・Using the given sequence of augmenting paths, after (1 + 4k)th such path,
the value of the flow
・Value of maximum flow = 200 + 1. ▪
r =
�5 � 1
2=� r2 = 1 � r
= 1 + 22k�i=1
ri
� 1 + 2��
i=1ri
= 3 + 2r
< 5