 From the previous discussion,
we have that:
maximum flow ≤ capacity of min. cut

 To complete the proof, we
show that:
 There
exists
a cut C such that:
capacity of cut C = maximum flow


We will
construct the
cut C
from a maximum flow....
 Assume that
we have found a maximum flow
in the network
Now
construct
the following cut:
 X =
set of
vertices
that can be
reached
from source S
through
a flowaugmenting path
 Y =
all the other vertices

Example:
 Here is a basic network with
a maximum flow:
 Nodes that can
reached
from source S
through
a flowaugmenting path:
 This is the cut that we will
construct:

 We must prove the following:
 The above construction yields a
cut
I.e.:
 The source S
and the sink T
are in different sets !

 The total flow through the
cut is
equal to
the capacity of the
cut

We will prove the above by proving
3 special properties of this
construction
 Property 1:
the construct yields a
cut, meaning:
 S ∈ X
 T ∉ X
(i.e.:
T ∈ Y)

Proof:
 By the way the set
X is formed,
the source S
is always included in X.
I.e.:
 We will show
T ∉ X
(i.e.,
T ∈ Y)
by contradiction
Suppose:
 From the construction of the
cut C:
 X =
all vertices that can be
reached from source S
by a flowaugmenting path

 Because
T ∈ X,
then:
 There exists a
flow augmenting path
from
S → T

which means:
 the
current flow is
not maximum !!!!

 That contradicts the
given fact that the
current flow is
maximum !!!

 Taking stock:
 We have shown that:
 The source S and
the sink T are
in different parts
of the cut C:


Now we will show that:
 Total flow
over the cut is
equal to the
capacity of the cut
Recall:
 Capacity of a cut =
sum of the
capacity of
all
forward edges


 Property 2:
 For each forward edge e
in the
cut C:
Graphically illustrated:

Proof:
 Select an
arbitrary forward edge
A→B of the
cut
 A
∈
X,
and
 B
∈
Y
(i.e., there is
no
flow augmenting path from
S to
B)

 Because
A
∈
X,
there is a flow augmenting path from
S → A:
 If:
f(e) < c(e):
then:
 there will be
a flow augmenting path from
S → B
(by increasing the flow
through A → B) !!!

which mean:
 This contradicts the fact that:
B ∈ Y

 Property 3:
 For each backward edge e
in the cut
C:
Graphically illustrated:

Proof:
 Select an
arbitrary backward edge
B→A of the
cut
 A
∈
X,
and
 B
∈
Y
(i.e., there is
no
flow augmenting path from
S to
B)

 Because
A
∈
X,
there is a flow augmenting path from
S → A:
 If:
f(e) > 0:
then:
 there will be
a flow augmenting path from
S → B
(by increasing the flow
through A → B) !!!
 If we
decrease some
flow on
A→B,
the


which mean:
 This contradicts the fact that:
B ∈ Y

 Therefore, we have the following situation at this particular cut:
And therefore:
total flow = sum of f(e) for every forward edge e (f(e) = c(e) !!!)
= sum of c(e) for every forward edge e
= capacity of the cut

In other words:
 We found a cut that has the
capacity equal to the max. flow !!!

That completes the proof...
