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CHAPTER 4. INTRODUCTION TO DUALITY
4.1
A frst example: Shortest paths
Let us start with an example. Recall the shortest path problem from Chapter 1, where we are
given a graph
G
=(
V
,
E
)
, nonnegative lengths
c
e
for all edges
e
2
E
, and two speciFc vertices
s
,
t
2
V
. An
st
path
P
is a sequence of edges
v
1
v
2
,
v
2
v
3
,...,
v
k

2
v
k

1
,
v
k

1
v
k
in
E
such that
v
1
=
s
,
v
k
=
t
, and
v
i
6
=
v
j
for all
i
6
=
j
. We are looking for an
path of
minimum total length
Â
(
c
e
:
e
2
P
)
.
3
2
1
2
2
4
1
4
s
t
a
d
c
b
The Fgure on the right shows an
instance of this problem. Each of the
edges in the graph is labeled by its
length. The thick black edges in the
graph form an
path
P
of total length 7. Is this a shortest path? The answer is “yes”, but how
could you convince your boss?
3
2
1
2
2
4
1
4
s
t
a
d
c
b
({s},3)
({s,a,b,c,d},1)
({s,a,c},2)
({s,a},1)
Here is a nice way of accomplish
ing this. The Fgure on the left shows
the same graph as before together with
a set of four
moats
each of which sep
arates
s
from
t
. Each moat is labeled
by a pair
(
S
i
,
y
i
)
, where
1.
S
i
is the set of vertices of
V
that are on the
s
side of the moat, and
2.
y
i
is the
width
of the moat.
We say that an edge
uv
2
E crosses
a moat
S
i
if
u
is in
S
i
, and
v
is outside, in
V
\
S
i
. ±or
example, edge
ab
in the Fgure on the left crosses moats
{
s
,
a
}
and
{
s
,
a
,
c
}
. We say that a
collection
S
=
{
(
S
1
,
y
1
)
(
S
q
,
y
q
)
}
of moats and their widths is
feasible
if
(a) each moat separates
s
from
t
; i.e.,
s
2
S
i
and
t
2
V
\
S
i
for all
i
, and