Selected Solutions for Chapter 26:
Maximum Flow
Solution to Exercise 26.211
For any two vertices
u
and
±
in
G
, we can define a flow network
G
u±
consisting
of the directed version of
G
with
s
D
u
,
t
D
±
, and all edge capacities set to
1
.
(The flow network
G
u±
has
V
vertices and
2
j
E
j
edges, so that it has
O.V /
vertices
and
O.E/
edges, as required. We want all capacities to be 1 so that the number of
edges of
G
crossing a cut equals the capacity of the cut in
G
u±
.) Let
f
u±
denote a
maximum flow in
G
u±
.
We claim that for any
u
2
V
, the edge connectivity
k
equals min
±
2
V
Nf
u
g
fj
f
u±
jg
. We’ll
show below that this claim holds. Assuming that it holds, we can find
k
as follows:
EDGECONNECTIVITY
.G/
k
D 1
select any vertex
u
2
G:
V
for
each vertex
±
2
G:
V
N f
u
g
set up the flow network
G
u±
as described above
find the maximum flow
f
u±
on
G
u±
k
D
min
.k;
j
f
u±
j
/
return
k
The claim follows from the maxflow mincut theorem and how we chose capac
ities so that the capacity of a cut is the number of edges crossing it. We prove
that
k
D
min
±
2
V
Nf
u
g
fj
f
u±
jg
, for any
u
2
V
by showing separately that
k
is at least this
minimum and that
k
is at most this minimum.
±
Proof that
k
±
min
±
2
V
Nf
u
g
fj
f
u±
jg
:
Let
m
D
min
±
2
V
Nf
u
g
fj
f
u±
jg
. Suppose we remove only
m
N
1
edges from
G
. For
any vertex
±
, by the maxflow mincut theorem,
u
and
±
are still connected.
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 Spring '10
 .
 Flow network, Maximum flow problem, Maxflow mincut theorem, Menger's theorem, Ford–Fulkerson algorithm, min fjfu17 jg

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