30.1 Suspended-Tree Inequalities
489
Then, the inequality (30.1.1) defines a facet
of
CUT
n
.
Proposition 30.1.3
and
Theorem 30.1.4 with the condition (i) are given in Boros
and Hammer
[1993); the alternative condition (ii) of Theorem 30.1.4 is proposed
in Boissin
[1994J.
The
results are presented there in the context of the correlation
polyhedra.
The
proof below is taken from Boros
and
Hammer [1993].
Proof.
We
show
that
the
inequality:
(30.1.5)
2:
PiJ)
2:
0,
ijEE(T)
which corresponds to the inequality (30.1.1) via
the
covariance mapping pointed
at
position
1,
defines a facet of
CORn-I-
b2,.'"
b
n
are integers such
that
b
2
,
•••
,b
p
2:
r,bp+ll
...
,b
n
-1,
T
is
a spanning tree on {2,
...
,p}
and
one
of
the
conditions (i),(ii)
of
Theorem 30.1.4 holds. Denote
the
inequality
(30.1.5) by v
T
P
2:
O.
Let w
T
P
2:
0 be a valid inequality for
CORn-l
such
that
{p
E
CORn-II
v
T
P
=
O}
<;;;
{p E
CORn-l
Iw
T
pO}.
We
show
that
w =
AV
for
some A
>
O.
Let S denote
the
family consisting
of
the sets 8
<;;;
{2,
...
, p} for which
the
subgraph
T[8)
of
T induced by 8 is connected
and
such
that
r + 1
:::;
b(8)
:::;
n - p + r + 1. For
io
E {p +
1,
...
,
n},
let
Sio
denote
the
family consisting of
the
sets 8 U
8',
where 8 E
Sand
8'
{p
+
1,
...
,
n}
\ {
io}
with
18'1
=
b(
8) - r 1.
Set
Sio
:=
{A
U {io} I A E Sio}.
We
prove an intermediary result.
Claim
30.1.6.
For each
io
E
{p
+
1,
...
,n},
the incidence vectors
of
the
mem-
bers
of
Sio
have rank n
2.
Proof
of
Claim 30.1.6.
We
show
that
the space consisting
of
the vectors
that
are
orthogonal to all incidence vectors
of
members
of
Sio
has dimension 1. Consider
the
vector y E
jRn-l
defined by
Yi
:=
bi
(i =
2,
...
,p), Yio
:=
-r
-
1,
and
Yi
:=
-1
(i E
{p
+
1,.
_ .
,n}
\ {io}). Clearly, Y is orthogonal to all incidence
vectors
of
members
of
Si
o
'
Let x E
jRn-1
such
that
Xio
= 0
and
x is orthogonal
to all incidence vectors
of
members
of
Si
o
'
We
show
that
x
O.
Let 8 E
S.
Then,
x(8)
+
x(8')
0 for every subset
8'
of
{p +
1,.
_.,
n}
\ {io} of cardinality
b(S) r
1.
Xi
a for i E
{p
+ 1,
...
,
n}
\ {io}, where
a:=
xeS)
b(S) - r
-1'
Therefore,
(a)
(x + ab)(8) =
a(r
+
1)
for all S E
S.
Suppose, first,
that
the condition (i) from Theorem 30.1.4 holds.
If
S E S
and
if
a node i E {2,
...
,p}
\ 8 is adjacent to a node of 8 in the tree
T,
then
8u
{i} still
belongs
to
S.
Hence,
Xi
+
ab;
= 0 by (a) and, therefore, this relation holds for