458
Chapter
28.
Hypermetric Inequalities
Observe
that
,(b)
and
a(b) have
the
same parity. Trivially,
,(b)
:S
a(b). One
can
check
that
,(b):S
ml1xlb;l.
•
In
particular,
,(b)
= 1 if a(b) =
1,
and
,(b)
= 0 if a(b) =
O.
Therefore,
the
gap inequalities (28.4.1) with a(b) = 1 are precisely
the
hypermetric inequalities,
while those with
a(b) = 0 are
the
negative type inequalities. Using
Lemma
28.4.3
we
can
identify which gap inequalities arise as switchings
of
hypermetric
or
neg-
ative type inequalities.
Lemma
28.4.4.
Let
b E
zn.
Then, the inequality (28.4.1) can
be
obtained by
switching
from
a hypermetric inequality (resp. from a negative type inequality)
if
and
only
if
,(b)
= 1 (resp.
,(b)
=
0)
or, equivalently,
if
there exists a subset
S
~
Vn
such that
b(S)
=
~(a(b)
-
1)
(resp.
b(S)
=
~a(b)).
I
Hence,
the
orbits
of
faces
of
CUT~
defined by hypermetric inequalities consist
of
those faces
of
CUT~
defined by inequalities
of
type (28.4.1) for which
,(b)
= 1
holds.
In
other
words,
Chapter
28
is
devoted to
the
class
of
the
gap inequalities
(28.4.1) with
,(b)
= 1.
We
remind
that
no inequality (28.4.1) with gap
,(b)
= 0
is
facet defining
(by Corollary 6.1.4).
In
the
case
of
a gap
,(b)
= 1, large classes
of
inequalities
(28.4.1) defining facets have been presented in Section 28.2.
The
gap inequalities
(28.4.1) seem difficult to study,
in
the
case
of
a gap
,(b)
2
2.
In
particular,
we
do
not
know any example
of
an
integer sequence b E
zn
with
,(b)
2 2
and
for
which
the
gap inequality (28.4.1) defines a facet
of
CUT~.
Thus,
the
following
problem
is open.
Problem
28.4.5.
Does there exist an integer sequence b with gap
,(b)
2 2
for
which the gap inequality: Q(b)T x
:S
!(a
2
(b) -
,2(b))
defines a facet
of
the
cut
polytope?
Only
the
following
is
known. Laurent
and
Poljak
[1996b]
show
2
that,
if
the
components
of
b take only two distinct values (in absolute value)
and
if
,(b)
2
2,
then
the
gap
inequality (28.4.1)
is
not facet inducing.
In
addition,
the
gap inequalities present the following difficulty:
It
is
an
NP-
hard
problem
to
compute
the gap
of
a sequence b E
zn
and, thus, to
compute
the
exact
right-hand
side
of
the
inequality (28.4.1). Indeed, given b E
zn,
deciding
whether
,(b)
= 0
amounts
to deciding whether b can
be
partitioned
into two
subsequences
of
equal sums; this
is
the
partition
problem, which
is
known to be
NP-complete (Garey
and
Johnson
[1979]).
In
order to avoid this difficulty, one may consider
the
following weaker in-
equality:
2Laurent
and
Poljak
[1996bJ also
present
a
characterization
of
the
sequences b for which
(28.4.1) is
facet
defining
that
involves
only
conditions expressed in
the
n-dimensional
space
instead
of
the
dimension
(~)
in
which
the
cut
polytope
CUT~
lives.