Ford-Fulkerson pathological example · Ford-Fulkerson pathological example Theorem. The...
Transcript of Ford-Fulkerson pathological example · Ford-Fulkerson pathological example Theorem. The...
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r =
�5 � 1
2=� r2 = 1 � r
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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
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xs
t
sufficiently largethat it won't everbe a bottleneck(we'll suppress)
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Ford-Fulkerson pathological example
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xs
0 / r
augmenting path 1: s→v→w→t (bottleneck capacity = 1)
0 / 1
t
v
u
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xs
t
0 / 1—
1
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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
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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
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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
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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
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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
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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
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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
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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
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Ford-Fulkerson pathological example
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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)
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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. ▪
= 1 + 2k�
i=1ri
� 1 + 2��
i=1ri
= 3 + 2r
< 5
r =
�5 � 1
2=� r2 = 1 � r