ford-fulkerson pathological example · ford-fulkerson pathological example theorem. the...

r =

�5 � 1

2=� r2 = 1 � r

1

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)

2

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

3

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

4

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

5

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

6

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

7

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

8

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

9

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

10

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

11

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)

12

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