Combinatorial Analogs
of Brouwer's Fixed
Point
Theorem on a Bounded Polyhedron
by
Robert M.
Sloan W.P.
No.
1720-85
Freund
October
1985
Abstract
In
this paper, we present a combinatorial
polyhedron for an unrestricted
of the polyhedron,
which
These
the labelling
triangulation
the
function is dual-
to a second theorem on the
is an extension of
results are shown
theorem
When
this theorem specializes
polyhedron,-that
labelling of a
can be interpreted as an extension of
Generalized Sperner lemma.
proper,
integer
theorem on a bounded
Scarf's dual
Sperner lemma.
to be analogs of Brouwer's fixed point
on a polyhedron,
and are
shown to generalize two
combinatorial theorems on the simplotope as well.
The paper contains two other
projective transformation
bounded polyhedron,
equivalent to
polar of
X.
then
lemma
X' =
X if and only
Secondly,
the
that
X =
We present
a
(xERnJAx < e)
is combinatorially
interior of
the
appendix contains a pseudomanifold
in triangulations
Key words:
polyhedron,
I-F
~ I~-·----------~
shows that if
if y is an element of the
researchers
label,
interest.
(xERni(A-eoy)x < e)
construction for a polyhedron and
integer
results of
its dual
based on
triangulation,
simplex.
_aa-~_~-r~_~_~------
that may be of
interest to
primal and dual polyhedra.
pseudomanifold,
fixed-point,
is a
1.
Introduction
In an article published
purely combinatorial lemma
point theorem of
Brouwer
between combinatorial
1928,
Emanuel Sperner
on the n-simplex
that
24],
implied the antipodal
Lusternik and Schnirelman
The connection
theorems was further
who developed a combinatorial lemma
point
[19].
theorems of Borsuk and Ulam,
Kuhn
[15]
and Fan
examined combinatorial results on the n-cube that
fixed
demonstrated a
implied the fixed-
for continuous functions.
theorems and topological
investigated by Tucker
that
in
[5]
and of
later
imply Brouwer's
point theorem.
With the development of fixed-point computation algorithms
stemming from Scarf's seminal work
[21],
of research in combinatorial analogs
analogs of Brouwer's
Sperner lemma
course,
[22],
theorem
there has
of Brouwer's theorem.
on the simplex include
the Generalized Sperner
the original
been a resurgence
Sperner lemma
Scarf's "dual"
[10],
and of
Analogs
of Brouwer's
theorem on the cube include a pair of dual lemmas
presented in
one of which
Talman,
[17].
Recently,
to simplotopes
and Van der Heyden
theorems are special
In this paper,
these combinatorial results have
(see Freund
[18]),
[7]
and van der Laan,
for which the simplex and cubical
cases.
we present a combinatorial theorem on a bounded
polyhedron for an unrestricted
labelling of a triangulation of
polyhedron, which can be interpreted as an extension of
Generalized Sperner
section
[6],
is analogous to the constructive algorithm in van der
Laan and Talman
been extended
[23].
lemma
Such
lemma.
3, theorem 1.
This theorem
is
the main
When the labelling function
the
the
theorem of
is dual-proper.
II1
theorem
1 specializes
polyhedron,
These
that
is an extension of Scarf's
results are
results
contains
In
lemma to
on the
to a second combinatorial
shown in
simplex and
a combinatorial
a bounded
polyhedron.
the section.
on Brouwer's
theorem;
combinatorial
it
proof of
X'
=
(x'ERn(A-eoy)x
lemma
is used
applications
two
< e)
be of
transformation
dual
of theorem 5 is based
of
interest.
lemma,
In section
that shows that
then
is combinatorially equivalent
the
of
Secondly,
construction for
interest to
an extension of Sperner's
the proof
other results
in the proof
elsewhere.
pseudomanifold
theorem 2.
is an open question whether a purely
and only if y is an element of
This
1, and hence of
is a bounded polyhedron,
'
Section 4
We present such an extension as
However,
3, we present a projective
e}
issue of
to
theorem 5 can be demonstrated.
paper contains
if X = (xERnlAx <
the
and their relationship
theorem
section 5, we address the
on
dual Sperner lemma.
simplotope are also shown.
proof of
theorem 5 of
The
section 3,
theorem
researchers
interior of
theorem
the appendix
X.
contains a
and its
in triangulations
2
the polar of
X if
1, but it may also have
a polyhedron
polyhedra.
to
dual
that may
based on primal and
2.
Notation
Let Rn denote real n-dimensional
namely e = (1,
vector of l's,
and outer
product,
let
ISI
S\T
= {xIxES,
be
vectors
respectively.
the cardinality of
denote
denote S\(x}
xT}, and let
in Rn.
If
<v
rank
0
(m+1),
...,
vm>,
... ,
T
j
= <v
,
Let
vk
vj k
If
X be a cell
faces.
i)
u
For two sets S, T,
xSnT}.
If xES,
burden.
Let v,
is
vm ,
...
let
we
...,
vm
denoted
be a real m-dimensional simplex,
a = <vO,
a nonempty
>
and
VM
in Rn,
T
is a finite
o
=
...,
subset
a k-face
i.e.
Let T be a finite collection
their
a set S.
the notational
...
said to
is
...
denote the empty set,
the matrix
simply an m-simplex.
{v Oj ,
Let
then the convex hull of v0
is
and define e to be the
Let x y and xoy denote inner
1).
SAT = {xIxESuT,
by S\x to ease
IVO
has
...,
space,
or
vm>
of
face
or more
is an m-simplex and
(vO,
of
... ,
vm},
then
.
a nonempty bounded polyhedron
of m-simplices a
triangulation of
together with all
in Rn.
of
X if
X,
oET
ii)
iii)
a,
T ET
If a
imply Ca n T ET, and
is an
(m-l)-simplex of T,
is a face of at most two
m-simplices of T.
An abstract complex consists
of
finite nonempty subsets
of
of K O , denoted K,
3
--ol-----
a set of
vertices KO and a set
such that
III
i)
v
KO
E
y
# x c
ii)
An element x of K
1 I denotes
(K O,
x
if x
ii)
E
finite
x
= n,
y
contained in an
n-simplex of
Let
of X
T =
X
or more simply an
E
K with
lyj
= n + 1 and x c y,
of K that contain x.
The boundary of K,
of simplices x E K such that x is
(n-1)-simplex y E K, and y is a subset of exactly one
K.
be an m-cell
and
in Rn,
let T be a finite triangulation
For each nonempty face T of each m-simplex
(vlv is a vertex of T).
Then the
nonempty face of a simplex of T)
called the m-pseudomanifold
of T, define
collection K =
(T{T
is an m-pseudomanifold,
is a
and
is
corresponding to T.
If A and b are a matrix and a vector,
let Ai and bi denote the
ith row and component of A and b respectively, and
denote the
An
then there are at most two
is defined to be the set
K,
implied
if KO is finite.
Let K be an n-pseudomanifold, where n > 1.
denoted
is
n > 1, is a complex K such that
where
n-simplices
the set KO is
complex by K alone.
to denote the
pseudomanifold,
K and
is called an n-simplex,
However, since
K).
K implies there exists
E
then x
or more simply a
Technically, an abstract complex
said to be
An n-dimensional
i)
+ 1,
= n
cardinality.
complex K is
n-pseudomanifold,
and
implies x E K.
Ixi
it is convenient
abstract
K,
E
is called in abstract simplex,
defined by the pair
by K,
(v}
K
E
x E K and
If
simplex.
where
implies
let A
and b8
submatrix and subvector of A and b corresponding to the
rows and components of
A and b
indexed by 8, respectively.
4
and
A vector x is
written x
positive.
y,
lexicographically greater than or equal to y,
if x = y or the first nonzero component of x-y is
A matrix A
matrix B, written A
is lexicographically greater
B,
if
(A-B)i
5
than or equal
0 for every row i of
(A-B).
to a
II'
3.
The Main Theorem
of
Consider a bounded polyhedron X
where A and b are a given
(mxn)-matrix and m-vector,
Let T be a finite triangulation of
vertices of T,
and
Let M =
,
(1,
L(.):K°M
...
be a
of other combinatorial
Our
following assumptions on
A3
(Bounded).
X
X # 0,
row index
under
the simplex,
X ,
cube, and
we will make
some of which will be relaxed
i.e.,
simplotope
the
later on:
there is a vector
interior,
i.e.,
there
exists x
E
X such
< b.
there
<
is no
bi.
(Centered).
interior,
A5
let
in the spirit of and as a generalization
Toward this goal,
X has an
i.e.
There is no redundant constraint governing X,
iM
If
there exists xx
A4
such a labelling function,
is bounded,
(Nonredundant).
and X-b
indices, and
such that XA = 0.
(Solid).
i.e.
denote the set of
interest lies in ascertaining the
theorems on
[5,6,7,9,15,17,18,22,24].
that Ax °
respectively.
labelling function that assigns a constraint
boundary conditions or not,
A2
let K
m) be the set of constraint row
combinatorial implications of
X > 0,
,
X
{xERnjAx < b},
let K be the pseudomanifold corresponding to T.
i to each vertex v of K.
Al
the form X =
and X
X is solid,
such that Aix
X contains
equals 0 or
this means
= bi
= 0 such that XA = Ai
that for every
iM,
and Akx < bk for every k
the origin in
M\i.
its relative
b > 0.
(Centered and Scaled).
interior, and
with Xi
0,
>
the rows
X contains the origin in its
of A have
1.
6
been scaled so
relative
that each bi
Assume
for
the remainder
of this section
nonredundant, and centered and scaled.
XO =
{yERnly=XA,
X°
Furthermore,
all
Then X°
X>O Xb=l}.
can alternately be
xx), whereby X°
is seen to
also a combinatorial dual of
that
Then,
is
X is bounded,
in particular, b=e.
bounded,
be the polar of
reversing mapping from the k-faces of
there
Let
solid, and centered.
described as X°
X, i.e.,
solid,
X
=
< 1 for
{yERnly.x
(see
X°
[20]).
is
is a one-to-one inclusion
(n-k-1)-faces of X° ,
X to the
see
[12].
Because X is nonredundant, each row of A is an extreme point of
X° .
Furthermore,
combination of
point YEX °
(n+l)
extreme
can be expressed as a convex
points of
X,
bounded,
X°
point of
X O,
is solid,
i.e.,
the set of
the above
and y3
is
X° .
rows of
points in X
(n+1) rows of A.
Because X is
X°
in
remarks.
not a regular
In
the
point.
is
that are not
XO has positive measure.
and
A
if y cannot be expressed
and so almost every point
a set of measure zero,
illustrates
n or fewer
i.e.,
X°
is called a regular point of
as a convex combination of
point,
°
every point yX
figure,
yl
a regular
regular are
Figure 1
is a regular
The circled numbers on the
boundary of X in the figure indicate the row constraint index for
the facets indicated.
For a subset acM,
i.e.,
S
have S
define S a
is the convex hull
= X
for a=M, and Sa c
define Gy = {acMlyE
vertices of cells
1 again,
we
(1,2,4},
and
of
see
S}).
S
= {yERny=XAa, X
X
for
all acM.
Then Gy consists of
that contain
plus all
Xba=l},
the rows of A indexed over a.
7
row index sets
of
Referring to figure
four sets
other subsets
We
For every yeXO
the
the point y.
that Gyl consists of the
(1,2,5),
> O,
{1,3,4},
(1,3,5),
of M that contain one
III
X
=
{xeR 2
3
2
1
1
4
1
-1
2
1
Ax < b}, where A =
x2
'
-1
-2
1
3
-2
1
m
(-1/3,1/3)
I3
1/5)
(-1,0)
(1/3,0)
x1
Il
A3
A1
Y1
A4
I
I
I
I
Figure 1
8
A5
of
Likewise, the minimal members of Gy 4 are
these four sets.
{1 ,2,5},
Gy 3
(1,3,5),
and
(1,2,3),
are
Regarding Gy 3 , the minimal members of
1,4,5).
amd (2,3,5).
2,4).
let T be a finite triangulation of X, let K be the
Now
pseudoma nifold corresponding to T, and
the set of vertices of K, to M, the set of
from K,
function
For a simplex aoK,
constrai nt row indices of X.
L(a)
A,I - I
Gus)
I
lrlmla"
_
_-
_,1
~__
ItlMIAiX=Di
The mapping C(-)
C(({x}).
w
__c%
or all x.
let
For a given subset S of X, define
for some vEK).
= { iEMli=L(v)
=
let L(-):K°-M be a labelling
.
_
. . ._
_
- _s..
- ve_
l
_
For a point xtX, aeiine
the
identifies
"carrier"
1__%v
OLX)
=
hyperplanes of
the set S or point x.
notation in hand, we can state
With the above
Theorem
1.
Let X be a polyhedron that
nonredundant,
and centered and
triangulation of
and
X, let
is bounded,
for
solid,
the pseudomanifold cor responding to T,
K be
any regular
theorem:
Let T be a finite
scaled.
let L():K°MM be a labelling function.
(i)
our main
point yEx 0 ,
simplices acK such that
Then
there are an
(L(o)
u C(a))EGy ,
odd number of
and hence at
least one.
aEK such that
To illustrate
figure
K° .
of
1.
there
is at lea st one simplex
u C(o))EGy.
let us continue with the
Figure 2 shows a triangulation T of
Regarding yl,
K for which
(1,2,5)),
L((w,v))
(L(a)
the theorem,
X° ,
int
for any point yE
(ii)
(L(v)
namely
=
a regular
(1,3),
(t},
C(a))
point of
E
Gyl =
X° ,
five simplices
((1,3,4), (1,3 ,5},
(a),
(f,g,k},
=
(5),
and hence
9
X an d a labelling of
there are
(w,v},
C({w,v))
example of
and
(p,q ,u}.
1,2,4},
Note
that
li
O
0
3
0
2
1
L
5
1
3
4
®
1
5
Figure 2
10
(L({w,v})
u C({u,v})) = {1,3,5} E Gyl.
three simplices aEK for which (L(o)
{1,3,5},
{1,4,5}},
namely {p,q,u},
Regarding y 4 , there are
u C(r))
E
Gy4 =
{{1,2,4},
{w,v), and {x,t,z}.
In the case
of the pentagon X in figure 1, theorem 1 actually makes eleven
assertions about the oddness of certain instances of labels, one
assertion for each of the eleven regions composing X° .
The assertions of theorem 1 do not depend on any special
restrictions of the labelling L(-)
restrict the labelling L(-)
stronger form of theorem 1.
on the boundary of X.
If we
on the boundary of X, we can obtain a
A labelling L(-):K°.M is called
dual proper if L(v)EC(v) for all vax, vEK °.
If L(-)
is dual-
proper, L(v) must index a binding constraint at v if v lies on the
boundary of
X.
This restriction was first introduced by Scarf
for the simplex.
The denotation here
is consistent with the notion
of a dual proper labelling as used in [7].
is said to be bridgeless if for each aET,
faces of X that meet a is nonempty.
Figure 3, for n=2.
A triangulation T of X
the intersection of all
This concept is illustrated in
In the figure, each of the simplices a1,
a3 fails the intersection property.
bridgeless, then no simplex
2,
Essentially, if T is
of T meets too many faces of X that
are disparate.
If L(.)
[22]
is dual-proper and T is bridgeless, we have the
following stronger version of theorem 1:
11
and
III
12
Cases where the intersection of the faces that meet a are empty
Figure 3
12
Theorem 2.
Let
nonredundant,
X be a polyhedron that is bounded,
and centered and
triangulation of
T.
X and
labelling function on K°
proper and T is bridgeless,
(i)
Let T be a finite
let K be the pseudomanifold
L(-):K°-M be a
Let
scaled.
solid,
for any regular
.
corresponding to
If L(-)
is dual-
then:
point yEXO,
there are an odd number of
simplices aEK such that L(c)
E
Gy, and hence at
least
one.
X° ,
for any point y E int
(ii)
there
is at least one simplex
aEK such that L(a) E Gy.
Theorem 2 can be deduced from theorem 1 as follows:
Proof of
Theorem 2:
Assuming theorem
X° ,
that for each y E int
Suppose not.
C(a)#0.
Then
a EX,
must satisfy L(v)EC(v).
every facet
Fi
Fi
for
for iC(a).
for
every
x E n
iEa
ba,'a =
But since
Aax = b.
and y =
y
Thus
(l+e)y
xEX,
y x
E
({xE X
iL(o).
Thus n Fi
i a
i.e.
1
If L(v)=i,
=
kaAa.
IAix=bi}.
every
facet
we have a meets Fi
0, because T is bridgeless.
Since aEGy,
there exists
However, y x = XaAx
X° ,
there exists e
> 0
X
and
(1+e)y-x
>
=
+e
< 1, contradicting (1+e)y-x
_---..
----__.
-_-.
meets the
Therefore a meets
u C(a)),
[X]
13
---------
then v, and hence a,
int
theorem is proved.
= 0.
(L(e) u C(a))EGy and
Furthermore, a meets
#
to show
whereby each vertex v of a
Denoting a=(L(a)
ia.
Fi,
by Fi
it suffices
(L(a) u C(o))EGy, then C(o)
there exists aEK such that
Because C(c)#A,
facet Fi defined
that if
1 is true,
However,
1.
ka
0 for which
Xba =
such that
1.
>
=
Let
Thus C)
1.
(y + ey)
E X° .
for any yEX ° ,
=
,
and the
II
Theorems 1 and 2 (without the oddness assertion) are equivalent
to
the fixed point theorem
Brouwer's theorem
of L.E.J.
on a bounded polyhedron.
bounded polyhedron, and let
Then
there
Brouwer
f():
XX
[2],
Let
stated below:
X be a nonempty
be a continuous function.
exists a fixed point of f(-),
i.e.
a point x*EX such that
= x*.
f(x*)
In
order to demonstrate the equivalence of
Brouwer's
theorems
theorem, we will use the following lemma,
equilivance of
polyhedral representations
and 2 to
which relates
the
under projective
transformation.
Projective Transformation Lemma.
Let X =
(xERnAx < b)
be a
polyhedron that is bounded,
solid, and centered and scaled,
and let
X°
=
{yERnly=XA,
=
X° ,
the set
X'
=
{x'ERn{(A-eoy)x
X' °
= X
-
faces of
y.
X'
'
< b)
and is
1}.
For any given y
X if
Since yE int
and only if T'
simplices a'
X° ,
yx
maps faces
= g(a)
< 1 for all
xx.
X onto
sets
are, of
and g-l(.)
, where
Consider the
It is
easy to verify
is given by g-l(x')
X' are combinatorially equivalent.
X' follows
to affine sets and
and g-l(.)
= bi,
Thus X and
is a triangulation of
X'
X' continuously, that xX satisfies Aix=bi
(A-eoy)ig(x)
x'/(l+y-x').
T is a
for every aT.
that g(-)
if
X onto the
is a triangulation of
XX', given by g(x) = x/(1-y.x).
and only
of
Furthermore,
mapping g(.):
maps
int
is combinatorially equivalent to X, and
inclusion preserving.
is the collection of
PROOF:
b
The mapping g(x) = x/(1-y-x)
triangulation of
T'
X > O,
from the fact that g(-)
convex sets
to convex sets.
=
That T'
maps affine
The mappings g(')
course, projective transformations.
14
if
[X]
(without the oddness assertion)
Proof of theorem
from Brouwer's
theorem:
Let X, T,
given,
K'
define X' and T' as
be the pseudomanifold
L(g-l(v'))
for
arg min
z 'EX'
where
is continuous,
in T'
that contains
Then
+ h(x')
=
Xy(A-eoy)y
x',
-X 8(A
-
eoy) B,
and nonzero solution.
= 0,
Upon
sum to unity, we have XaA =
has
u C(a'))
all
(L(a) u C(a))
= a
E
for
Let a'
.
let Y = L(a'),
some X
y, Xa >
B = C(a'),
O,
_____
_____
and
-
> O.
Furthermore, h(x')
=
Xyey
= 1.
Therefore,
= 0 has a nonnegative
multipliers X
, Xae
= 1.
so that they
Thus aE Gy and
ET defined by a
Gy, proving the result.
problem of h(-)
15
_I _
= ALI(v,)-y, and
that x'
rescaling the
problem on f().
_IY_1
=
be the smallest simplex
whereby Xa(A-eoy)
the stationary point
let
the Euclidean norm.
The construction of the function f(.) was
to convert
be
Define f(x') =
X'.
whereby the simplex a
= a.
X
is continuous
x'
and
of
define h(v')
11-112 denotes
particular Xy >
for some
int
and define L'(v')
the Karush-Kuhn-Tucker conditions state
XB(A-eoy)B + Xy(A-eoy)y
(L(a')
v'EK'° ,
f(-)
so contains a fixed point
a = y u B.
x'
corresponding to T',
For each
llz'-x'+h(x')112 ,
Because h(-)
a'
v'EK °' .
Let y
in the projective transformation lemma,
in a PL manner over
extend h(-)
and
and K be given as in theorem 1.
L(-)
= g-l(a')
[XI
introduced by Eaves
to a fixed point
[3]
III
Proof of
solid, nonredundant, and centered and scaled,
is bounded,
that
Let
L(.) be a labelling function on K
which Ai(v-f(v))
index i for
EK such that
a simplex
that
Ag(x-f(x))
there
> 0,
> 0, Aa(x-f(x))
and
> O.
the property that
f(x))
> O.
to y,
point of
regular
(L(a) u C(a))
E
X, a
C(x))
a.
exists
(8 u
x
However,
E
Gy,
E
Gy.
such an index
bounded,
X
.
Then there exists
Taking a limit of
However,
because
since a
y is
E
Gy, there exists Xa
regular Gy =
Gz for all
theorem and
oddness assertion) can be accomplished
above arguments.
16
> 0 with
= XaAa(xz
> 0 for all z sufficiently
proving Brouwer's
The equivalence of Brouwer's
Because Ay(x-f(x))
Thus y(x-f(x))
Thus z(x-f(x))
whereby x-f(x) = O,
a such that
and 8
Let Y = C(x).
eXa = 1 and XaAa = y.
sufficiently close to y.
close
X is
any
such a as the mesh of T goes to zero, we
sequences of
conclude
so that L(v) equals
defined
Because
> O.
Let y be a given
must exist.
corresponding to T.
X and K be the pseudomanifold
triangulation of
and
Let T be a finite
function.
XX be a continuous
let f(.):
X be a polyhedron
Let
Brouwer's theorem from Theorem 1:
theorem.
theorem 2
[X]
(without the
in a manner that parallels the
Relation of
Theorems
1 and
2 to combinatorial
results on the simplex
and simplotope.
We show how theorems
1 and
simplex and the simplotope.
the simplex,
[22],
of
and
The
namely Sperner's
an odd number of
simplotope,
specially
to known results
three major combinatorial
lemma
[23],
the Generalized Sperner lemma
However, when
The
2 specialize
Scarf's dual Sperner
[10],
three
results
are extended to
the cube and
the oddness assertion disappears, and the dimension of the
labelled simplices
simplices of interest
is reduced
(see
Herein,
simplex and
1 and 2 for particular values of y E X,
assertion holds on
regular
point
in X ° ,
regular point
Let S n
does
= {xERnI
(1,
...
as instances
of
theorems
we will see that the
oddness assertion on the simplotope
not hold, precisely because y is not a
x
< e,
-e-x
l 1}.
Then S n
is an n-dimensional
By defining
we can write
K the
these
in X ° .
and b =
An =
Sn,
[18]).
the simplex precisely because y is a
and the
the cube)
and
by
simplotope theorems
oddness
[7]
specially
stems from the constructive proofs of
theorems.
casting the
simplex.
lemma
all assert the existence
inability to assert that there are an odd number of
(and hence
results on
simplices with certain label configurations.
these
labelled
simplotope
on the
n as
= (xERniAnx < b.
n
Let T be a triangulation of
pseudomanifold corresponding to T,
m}
=
(1,
= (yly=XAn, X > 0,
...
,
n+1},
eX =
1)
because m = n+l.
and L(-):Ko-M,
For X =
Sn ,
where
the set
is an n-simplex that contains the
17
M
X0
=
i
origin,
y = 0
int
and any y
X
is a regular point
Because Sn is bounded,
we can apply theorem
simplices
u
C(¢) =
lemma
is a regular
X °,
in
solid,
...
[10],
=
and G
1, and assert that
, n+1}.
in X° .
(M} =
In particular,
((1,
...
,
n+))}.
nonredundant, and centered and scaled,
EK with the property that
(1,
point
there are an odd number of
u
(L(o)
C(a))
E G,
i.e.,
L(a)
This is precisely the Generalized Sperner
and is seen
to follow as a specific instance of
theorem
1.
Now suppose
vE2Sn,
L(v) =
i must
that no simplex
bi},
i=1,
...
that the labelling L(-)
,
of T meets
n+1.
the intersection
bridgeless,
be chosen so
every
Then it
of all
faces
Sperner
(1,
= bi.
that Aiv
facet Fi
can be shown
of S n
i.e. for each
Furthermore,
of S n ,
where
Fi
suppose
= (xeSnlAix=
that for any simplex
of T,
that meet ca is nonempty, i.e.
whereby the conditions of theorem 2 are satisfied.
there exists an odd number of
L(a) =
is dual-proper,
...
lemma
,
This
n+1}.
[22], and
simplices
rEK such that L(a) E
latter result
it is seen to
Thus
G,
i.e.
Scarf's dual
is precisely
follow as a
T is
specific instance of
theorem 2.
We now turn our attention to
simplotope
S is defined to be the
Sm 1
x
x Smp,
mj
1, j=1,
...
...
theorems
on the
simplotope.
cross-product of
A
n simplices,
S
=
where, for simplicity, we will assume that each
,
p.
Any point xES
is a vector
in RN,
where
p
£
N =
mj,
and x can be written as
x =
(x 1 ;
...
;
xP),
where each
j=1
each
xJ ERmj,
vectors
x,
j=1,
j=1,
...
...
,
,
p.
p, and x is
the concatenization
Defining A n as above,
(N+p)x(N) matrix:
18
of the n
let us define A as
the
Aml
0
0
Amp
A =
where Amj
I
is as described previously.
Then S can be described as S = (xERnlAx
Define M = {(j,k)lj=l,
...,
..., p, k=l,
indexed by the ordered pairs
where bRN+P and b=e.
b
mj+l}.
The
rows of A can be
M where row (j,k) of A is in
(j,k)
j-1
Likewise, a vector X E RN+P
(mi+1) + k) of A.
fact row number (
i=1
will be indexed by the ordered pairs
(j,k) E M.
Let T be a
triangulation of S, let K be the pseudomanifold corresponding to T, and
let L(.):K°M be a labelling function.
For X = S, X is bounded, solid,
nonredundant, and centered and scaled,
and so the conditions of theorem
1 are met.
We have X°
X > 0} and y=O E X
To see this,
= {yERnly=XA,
eX=l,
However, y=O is not a regular point of
pick any one index j from among j
Y
(1, ..., p}.
Then set
0
j
i
if
1/(mj+l)
for each (i,k) E M, and note that X > O,
define aj=((j,l),
lajl
= mj+1
point of X°O.
... ,
< N+1,
Thus,
so
(j,mj+l)},
e-X=l, and XA=O=y.
we see that 0 E Sj,
long as n > 1.
by theorem 1.
if i = j,
Thus y=O
If we
but
is not a regular
we can only assert that there
19
III
However G
Thus
=
that
(L(a)
{(t c Ml a
=
for some
of K
that
one simplex o
exists at least
({ cM
I y ES)}
=
there exists a simplex o
(j,mj+l)}
or lemma
for some j
3.2 of
(1,
K such
of
.,
such
j
Th is
P)
(2,2),
with
X°
=
G
(L()
(1,2),
{(2,1),
is
u C(1))
Gy,
E
the diamond shown
{( 1,1),
(2,1)),
(2,2 ), (1,2))
is the convex hull of
regular
(1,2),
na mely 01,
...
th e points
(1,O),
in the fi gure.
2 =l,
L(v)
must
{xESIA(j,k)x=b(j,k)),
can be shown
for
that the
being the case,
As
06
p}.
the
This
be
the
(-1,0)
1 of
[7]
(2
The set
figure.
(0,1)
figure shows,
chosen so that Aix=bi.
logic
all
(j,k)
aj,
requirements
and
(0 , -1)
y=O i s not a
of
facet' F(j,k)
for any
theorem
The Sperner
instance
lemma, and
[6],
lemma
its extension
to
theorem on a bounded
this
...,
p.
2 are met.
Then
This
to show that
for some j E (1,
[7],
the simplotope
and
paper,
polyhedron,
the Generalized Sperner lemma,
we
that
simplex.
20
[7,17],
theorems 1 or 2.
but this derivation fails to carry over
of
j=1,
of theorem 2 of this paper.
can be derived from
last section
i.e.
=
is precisely theorem 2 of
does not appear to be a specific instance of
Sperner's
aj
proper,
Furthermore,
employed herein can be used
latter statement
is a specific
on the
....
six simpli ces of S
in the
there exists a simplex a E K such that L(¢)
In the
),
With
and n=2.
(2,2)),
There are
suppose that no simplex a E K meets each
see
({(j,
p}}.
point.
for each vS,
thus
{1,.. . ,
precisely theorem
Suppose now that the labelling L():K°-M is dual
it
E
[18].
(((1,1),
(1,1)},
j
EG-.
(L(a) u C(c))
Figure 4 illustrates t he theore m for m 1=m
y=O,
u C(a))
to the simplotope.
present another combinatorial
specifies to Sperner's
lemma
2
X,
1
K1
r)
1
Yl
/
-1
Figure 4
21
111
Theorem 1
Relaxing the Assumptions of
Our final
the
remarks
to Brouwer's
proof of
theorem
1 in the
X°
x°
int X,
E
-x 0 ,
X'
scaled, and so the
is just
the
set X
Ax°)=1},
(yERnly=XA, e-X=1,
and for a c M,
Thus theorem
X
centered and scaled,
= (yERnly=XA,
centered
In this case,
> O,
X
X-(b-
> 0 Xa(b-AxO)a=l}.
can be modified to
dimension of
E rel
[8].
Furthermore, by
include the
int
X.
Then
in order
X can be found using,
Once x°
E rel
int
Bx < b},
scaling the rows of
been
X has
where Dx°=d
(B,b), we can
Let C be any matrix whose rows form an
basis for the
{yERnJy=UD+B,
X' now is
scaled.
k be the
the methodology in
> 0.
X by
the case when X is neither solid nor
and let
ensure that s=b-Bx°=e.
orthonormal
For any given
1 apply.
theorem
X can be rewritten as X = {xERnlDx=d,
s=b-Bx °
the case when X =
as
> O)
theorem 2)
and scale X, a point x
for example,
but the definition
a translation of
= (yeRnly=XaAa,X=
centered and
let us consider
Next,
given,
S
1 (and hence
case when X is not
to center
assertions of
in the
The assumptions that X is
where Ai = Ai/(bi-AixO).
and
and
b-Ax ° }
= {xERnlAx
({xRnIAx < e},
=
will be seen
can be eliminated,
and can alternatively be written
X' =
as
is solid but not centered and scaled.
b)
<
simplices,
Let us first consider
must then be changed.
{xERnlAx
the counting arguments
next section.
solid and centered and scaled
of
The assumption that X is bounded
and to
theorem,
of paths of
regarding endpoints
relaxing
this section are concerned with
presented earlier.
assumptions
is central
of
subspace N={xERnlDx=O},
s-X=l, X > 0).
Then the
22
and
let X°
=
transformation f(.)=XRk
given
by f(x)=Cx-Cx °
maps
X onto X'={ZERkIBCTz
transformation g(y)=Cy maps the set X °
X > O,
{vERklv=XBCT,
between
each given
X s=l}.
X °.
of
theorem 1.
For a given point
The sets
regular
point of
X
This
the subset
YEXO,
{yERnly=CTv + DTU
conform to the conditions of
there
is a unique point vEX'
The point y in
X°is
°
called a
of X' ° .
define Gy=Gv where v is uniquely
relation y=CTv+DTu for some u.
transformation,
and scaling,
=
if y=CTv+DTu and v is a regular point
Furthermore, for any yEX ° ,
determined by the
into the set X '°
X' and X' °
that y=CTv+DTu for some u.
and the
is a one-to-one correspondence
X' ° and
point v in
for some u}
such
There
s},
together
can be used
with the above
to prove
the
remarks
on centering
following extension of
theorem
1:
Theorem 3.
in Rn
x°
Let X be a nonempty bounded polyhedron of dimension k
that is nonredundant, of the form X={xERnlDx=d,
be a given point
T be a finite
of B,
rel
int
triangulation of
corresponding to T.
rows
in
and
X, and
X and
Let
let s=b-Bx°> 0 be given.
Let
let K be the pseudomanifold
Let L():K°M, where M={1,
let C(a)={iEMIBix=bi
for all
X ° be defined as in the remarks above.
(i)
Bx < b}.
..., m}
xEa)
indexes the
for each cET.
Then:
If y is a regular point of
X° ,
there exists an odd
number of simplices oET with the property that
C(0))
(ii)
E Gy,
and hence
If y E rel
simplex
int
ET with the
Let
(L(a) u
at least one.
X° ,
then there exists at
property that
23
least one
(L(a) u C(a))
E Gy.
[X]
III
Theorem 3 obviously
remarks
implies theorem 1 as a special case.
outline how to
prove theorem
1, using the transformation f(-).
combinatorial
convenience.
either
the
of
3 as a consequence of
Theorem 3 is
theorem we will consider.
the nonredundancy assumption.
Because
This assumption
still retains
is retained for
do not contribute to
the geometric or combinatorial properties of
a polyhedron,
away does not detract from the
the results.
24
theorem
the most general
The theorem
redundant constraints
fact they are assumed
The above
generality
4.
A Combinatorial Proof of
This
ideas
section contains a combinatorial proof of
that are easy
dimensions,
two dimensions.
X°
X and
in order
to motivate the 'proof along
we start by showing an example of the proof
in
case.
two dimensions
be as shown in figure 1,
in Figure 2, and
let T and L(.-)
be as
let K be the pseudomanifold corresponding
to
Define K to be the pseudomanifold consisting of simplices aEK
"joined"
with the indices
K = (ala
Ko = K
,
u
{1,
a
AK =
...,
For each
iEM.
= C(x)
{(BI
iEM,
of C(o),
(a
= K
shown
let
with the
property that L(a)E
suffices
to
Now let 8
jOB}.
c M =
#Gy
{1,...,5}
For example,
and
in Fgure 5.
Note that
for some xK}.
X °,
show that
aEK),
u M.
extend L():K°-M to
For each y
i.e.
C(a)),
m)
The construction of K is
for
The
In higher
We then proceed to the more general
Example of proof in
Let
to follow in two dimensions.
Hence,
more intuitive lines,
T.
1.
they become more encumbered due to the possible presence
degeneracy in X.
shown
theorem
behind the proof derive from relatively straightforward
concepts
of
Theorem 1
L(-):K°-M by the association L(i)=i
#Gy
denote the number of
Gy.
In order
to prove
is odd for all regular
with
IBl=n=2.
for B = (1,3),
25
theorem 1,
points y
Let RB =
RB = ((1,2,3),
simplices aK
{Bu{j},
(1,3,4),
it
X.
jEM,
(1,3,5)).
©
P
The pseudomanifold K
Figure 5
26
Let *#R
be the
ER B ,
and let q
that
L(o)
number of simplices
later
the
be the number of
= 8.
parity of q
implies,
and Tolle
is the
states that the
in particular,
RB
is odd, and
(ii)
if B
K, 161=2,
R6
is even.
= 6,
whereby #R8
odd.
{w,x,s},
simplices a E
even number.
{x,t,z},
and
= 8
Thus q 8 =O,
let
simplices qEK with L()
eleven regions
regular point of
i.e.
is odd,
namely
8 {
if so,
K,
=
1,4),
E R,
The
second
there can be no
then o = L(o)
= 8,
R8
and hence
namely
EaK.
(4,p,u},
an odd number.
a
is an
aK.
{a,3,4),
Tk,
X °.
now shown in Figure 6,
k=l,...,11.
The re
{f,g,k},
Also,
Gy = R,
whereby #Gy
for any y E
where
_
11
subdivided in to
For any y E int T 1 ,
={4,5}.
T 1,
is odd, because R
1111_
y is a
Gy = ((1,4,5),
Because
27
10_1
t hen
{t,1,2}.
Now consider the set X ° ,
(3,4,5)),
R,
E
that if
(for
so,
Note that
an even number, and hence
As an example,
and
let B={4,5}.
{e,j,i),
from the fact
(if
= 1, an odd number,
EK with L(a)
K with L(a)
contradiction).
This
EK, then
K with L(a)=
E
Thus q
As an example,
{e,j,2),
statement follows
four
from the fact that if
is no other simplex
There are five simplices
(x,t,z),
then
a contradiction).
is
and
B
that
K, 11=2, then
and there
[16], and
parity of #R
IBl=n.
, with
if 8 E
a=L(a)
#R 8
[11],
(i)
L(B)=B,
the
first introduced by Kuhn
same for any given
The first statement follows
are
simplices oE2K with the property
A parity argument,
used by Gould
EK with the property that L(a)
= Gy.
B E
K,
This
by
(2,4,5)
(i) abo ve,
proves
A2
A3
A1
A4
A5
0
The subdivided cell X
Figure 6
28
y E int T 1.
theore m 1 for all
E Gy are
For y E int T1,
those simplices
for wh ich
L(a)
(e,j,2 }.
The main fact that has been used is that all y E int
"s ufficiently close"
are
n
j 4,5
i.e.
(x,t,z),
K
{e,i,j}, and
1
to the face <A 4 ,A 5 > so that y E int
{{(4,5,j})
j4, j5, JEM),
G y = R4,5 }
next will
show that
if
and Tj
of
adjacent regions T i
parity of
this
#Gz.
Since the
will mean
k=2,...,11,
follows
X°,
that the parity of
#Gy
parity of
proving assert ion (i)
is odd
for y E
of theorem 1.
consider ainy two adjacent regions T i
and
4
5.
int
and
{1,3,4 ), (3,4,5),
= {{1,2,4),
(2,4,5),
(2 ,3,4},
(1,3,4 )}, {1,3,5),
({{1,3,5), (2,3,5),
Furthermore,
,Aj>,
c R
j
the
Note that
2,3,5))}}.
from T
is
It is no coincidence
that
E.very
(3 ,5) contains either t4 or T5 but nlot
co ntain
Thus GyAGz
But because the line
segment sepa rating T4 from
=
R
3
(3,5),
t5,
segment
thlen any a
3,5),
i.e. a
,5).
For any collection D
29
_P1_·ll______l__l___111_____
s5 ,
la st two statements.
that R{3 ,5) c GyAGz.
in GyAGz must
3 ,5})
15 that s eparates T 4
in e ach of
in X0 ,
= R (3,5)
segm ent <A 3 ,A 5 >.
is the unique line
that lies
GyAGz
5
This shows
<A 3 ,A 5 >
line
(3,5) appears
simplex <A 3 ,A
(3,4,5))
the face of T 4 n
generated by the
both.
tk,
and T ,i
Fo r any y E int T 4 and z
(2 ,3,4},
set
i, then
Assertion (ii)
{2,4,5),
=
int
is odd for y E int
Gy
((1,2,4),
Gy&Gz
#Gy equals the
from a closure argiument.
for example
Gy =
y and z are in the interior of
that the pa:rity of
Therefore,
the
(s,x,w),
conv <A 4 ,A,Aj>, whereby Gy =
We
Gz
(p,u,4},
E
whereby
of
subsets
of M,
that L(o) E D.
Gy =
#Gy
D denote the number of
let
EK such
Note that
(Gy\Gz)
(Gy n
u
= #(Gy\Gz)
Gz),
whereby
+ #(Gy n Gz),
Similarly,
two sets are disjoint.
because these
simplicies
we have:
(Gz\Gy) + #(Gy n Gz).
*Gz =
We obtain:
#Gy
-
*Gz = *(Gy\Gz) - #(Gz\Gy)
= *(Gy\Gz)
+ #(Gz\Gy)
= *(GyAGz)
-
= *R{ 3 , 5 }
However,
*#R
is even,
i.e.
3
2*(Gz\Gy).
tj
X°
XB
> O,
X °,
#Gy
-
Proof
separates
whereby B
= #RS
#Gz
-
of Theorem
Let
X, T,
of A.
Note
regular point
loss
of
figures
Ti
(n-1)-simplex S
j.
which
#Gz
1 and 2.
Gy and #Gz have
regions T i
there
and Tj are adjacent,
i from
#Gy -
This completes the
to adjacent
=
is
and
a
{yERnJy=XA8,
Furthermore,
Finally, GyAGz = RB.
2#(Gy\Gz),
SB
cannot
lie
Therefore
is an even number.
1
L(.),
centrally regular
rows
3K.
V
Therefore
same parity.
interior
B such that the
e.XB=1},
on
If
are as follows:
index set
unique
K.
leading to the proof that
same parity if y and z are
of
the
1 for the example of
The important facts
the
{3,5} 0
because
Gy and #Gz have
of Theorem
proof
2#(Gz\Gy)
2#(Gz\Gy)
is even,
, 5}
-
and K be as given in theorem 1.
if y=O does not lie
in
the affine hull of
that if X is centrally regular,
X° .
of generality,
X is said to be
then y=O must
The remark below allows us to assume,
that
X is centrally regular.
30
n or fewer
be a
without
Remark 1.
regular,
If theorem
1 is true when the polyhedron
then theorem 1 is true
X is centrally
independent of X being centrally
regular.
Proof:
Suppose X is not
hypotheses of theorem
y E X
does not
of A.
Let y be any
Now let
X' =
Xe=l}.
lie
centrally regular,
Then X°
1.
(x'ERn](A=eoy)x '
X° .
of any set of n or
T'and K to
the pseudomanifold
L(g-l(v')).
Therefore,
maps
fewer rows
Then y must be a regular point in
X' °
< b),
= (y'ERnly'=X(A-eoy),
From the projective transformation lemma,
given by g(x) = x/(1-y x)
X satisfies the
is solid, and almost every point
in the affine hull
such point
but that
X onto X',
and maps
T to the triangulation
define L'(v')
Notice that X'°=X°-y, and that 0 is a regular point
because
the system
equivalent to our original
this new system,
theorem
We therefore will
X',
T',
L'(
O,
the mapping g('):X-X'
For each v'EK' ° ,
K'.
X >
X.
), and K'
=
in
X' ° .
is combinatorially
system and by hypothesis theorem
1 is true for the original system.
1 is true for
[X]
assume for the remainder of this section
that
X is
centrally regular.
X is
1al > n+1
that
The
said to be nondegenerate if Ax=b~ has no solution xEX when
and
cM.
In order to prove theorem
X is nondegenerate.
remark below
lists
1, we will
This assumption will be
relaxed
first assume
subsequently.
some of the properties that are consequences of
the property of nondegeneracy in conjunction with the assumption that
is centrally regular.
Remark
2.
If X satisfies the assumptions of
nondegenerate
and centrally regular,
31
then:
theorem 1 and X is
X
III
(a)
every extreme point
(b)
every facet of
X°
a facet of
X° , where
= n+l,
every a c M with
a
(d)
for
every B c M with
1I1
= n,
Sa is an n-simplex.
S
is
an
affine hull contains no other rows Ai,
To prove
(b),
must
be
whereby
cell,
IBI
181
n,
= n,
must
S,
IBI
>
for a particular
and hence
n,
Aix
IBI
= 1 for each
Z
= n,
let a
i
c M,
AT
ai
is an
]
32
Because 0
e
int X ° ,
= b,
and Ax
(n-l)-
and F is an (n-1)X and
(yERnlyTx =
Thus S
1)
let
= C(x).
supports
is a face of
rows, the rows of A
(n-1)-simplex.
= n+1.
F =
Hence
(x,e).
But since F is an
iEB.
have linearly independent
whereby So
This hyperplane
Therefore xEx,
=1.
and the hyperplane
(a),
since
(c),
= F.
let x be an extreme point of
affinely independent,
To prove
X
X is nondegenerate.
since
from
Then there exists a
e} and Aix < 8 for iEM\B.
=
Conversely,
and contains
Since A
e)
positive and we can assume
we must have
simplex.
Then
(ilAix
X°.
such that H n
H = (ylyTx =
can be written as
=
X°
least n facets
X is n.
F be a facet of
let
supporting hyperplane H of
8
i E M\8.
X must meet at
point x of
an extreme
of X, because the dimension of
where
(n-l)-simplex, whose
is a direct consequence of the definition of nondegeneracy
and the fact that
SB ,
S
= C(x).
for
(a)
point xEX,
X, and for every extreme
of
(c)
Proof:
where B = C(x) for
(n-l)-simplex S,
is an
extreme point
some
X.
X meets exactly n facets of
of
If the matrix
are
X°
X° .
is
does not have rank n+l,
Aax
then there exists
= ee, a contradiction.
Therefore
(x,e)
(0,0) such
that
Z has full rank, whereby the set
Sa is an n-simplex.
For
any
(d),
i E M\S,
affine hull
let B c M,
whereby S
of S
Our next
18
= n,
= S u {i}
then a
(n-l)-simplex and Ai
is an
satisfies
does not
(c) for
lie
in the
[XI
.
task is
to
construct the extended pseudomanifold K,
defined by
= K
Ko
K
u M
= (
EKOo
This construction
is
#
a C (
,
u
C(a) )
for some
figure 5.
illustrated in
rEK).
We have the following
lemma:
Lemma
(
1.
If
X is nondegenerate,
c MIB c C(x) for some x
if and only
PROOF:
K and
X,
if So is a facet
from part
are as stated.
(b)
of
(n-l)-simplex) of
resembles
second statement of
in nature and
X.
the lemma follows
[X]
polyhedra as
well.
the construction of an antiprism
combinatorial
=
of the appendix,
The construction of K will be generalized later
include degenerate bounded
K
(n-1)-simplex in aK
B is an
and so by Corrollary Al
The
remark 2.
B # *}.
(and an
X is nondegenerate,
K
K is an n-pseudomanifold, and
This
in this section to
construction
in Broadie
1],
but
is
so does not depend on the geometric
projection property used in his work.
We now extend L(-):K°-M to L(-):K°-M, by defining L(i)=i for iEM.
For each
c M,
{B0=n,
collection D of subsets
define R
of M,
let
=
(
u
{j}jjEM\B}.
For
any
D denote the number of simplices
33
i
a
E
K with the property that L(a)
D.
E
We have the followihng
result:
Lemma
2:
Let
IBl=n.
c M with
(a)
If
B E
a K,
then
(b)
If
0
B
a K,
then *#R
PROOF:
Let q
#R s
the parity of
q
is the
RB
Consider a regular
Furthermore, because y
thus y
two
E
int
n
S a,
aEGy
regular
is
that the parity of
c M with
I=n.
If
is odd.
8 0
If
and
RB and
E
XO.
Then
y
n
aeGy
int
al
S,
= n+1
K
aK,
for every
Sa is an n-cell,
is a regular
for every a
point
E
E
Gy.
Gy, and
and every element
in X.°
If y,z are
then define the relationship yzz
if Gy=Gz.
is an equivalence relation on the regular points of
because Gy can take on only a finite number of
this equivalence relation divides the regular points y of
into p mutually disjoint
distinct values of y,
and
(int
k)
n
sets of
the form int
( n
aEGy
Sa
), for p
k
= Y,...,Yp.
T
For k=l,...,p, define
in X ° ,
[16]
[X]
regular,
whereby
K with the property
introduced by Kuhn
states
for any
point yE
in X ° ,
points
Furthermore,
values,
X°
[11],
is even.
the interior of this cell
The relation z
X0° .
same
first
E
IBI=n, then L(B)=S, and q=1, whereby #R 8
then q=O, and hence
of
simplices
A parity argument,
later used by Gould and Tolle
and
is even.
be the number of
that L(a) = B.
and
is odd,
Ti
=
k =
n
Sa ,
aEGy k
for all
argument easily demonstrates that
and each T k
ik.
p
u
Tk =
k=l
34
is thus an n-cell
Furthermore, a limiting
X°.
Figure 6
H
Ti
F
*y'
Y
Figure 7
35
II1
collection M = {T1 .....
Eaves
Let X satisfy the assumptions of
Let M =
X is nondegenerate.
X° ,
then Gy&Gz
n-cells
of
PROOF:
In order
T i,
and
and
let y' be a given
y E
F,
there exists a
int
Because y E rel
> O,
some XB
hyperplane of
yx
>
y + tx
l(d),
Ti,
E
Aj
int
of
subsets of M.
and 8 f
1B0=n,
j#i.
Let T i
aX °.
i.
Then
i =
rel
E
and y E rel
int S.
B c a,
Because
1B0=n,
Thus y = X\gA,
for
Let H be the
H can be represented
(see Figure 7);
(0,0) and unique up to
scalar
Because H is asupporting
of generality, we can assume that
y E rel
int
S,
and y E
t > 0 and sufficiently small.
Ajx
}) and N = {jEMIAjx
let
S.
is uniquely determined.
S
it
int F,
Sa.
n
aEGy'
property that y E
c H, Ax = ee.
Because
K.
then either F
be given,
Let y
there exists a unique subset
i.
is a
c M,
int
with the
H for any jM\B, i.e.
P = {jEMIAjx >
regular,
E Gy'
for all
Ti
X° )
in adjacent
that F
then without loss
yET
for all
element
F,
SB
Because
(M,
if y and z lie
}) for some (x,e) #
{yERnly.x =
multiple.
Then
(M,X ° ) is a subdivided n-manifold,
in Rn containing
unique hyperplane
as H =
,...Tp)}.
for some
suppose
= 1, and X
eXB
1
an n-cell rj,
that y E S,
the property
with
(see
suppose
theorem 1 and
if F is a facet of an n-cell Ti,
or F is a facet of
a facet of
= R
show that
show that
suffices to
Xo ,
to
(T
Furthermore,
subdivided n-manifold.
F be
X°
k}) constitutes a PL subdivision of
[4]).
Lemma 3.
c
that the
It is our aim to prove
remarks.
the above
illustrates
#
rel
From remark
8 for any JEM\B.
< 8).
Then,
P, N,
int F,
and
Let
are disjoint
Furthermore, because we can assume that X is centrally
the set
{ilAix =
}) can contain at most n elements, whereby
36
8
e}
= (ijAix =
and M = P
B.
N
y + tx
int S
for t > 0
and sufficiently small
for each
and sufficiently small.
small,
(8 u
{j})
y,*
sufficiently small.
Thus for y = y + tx,
jP.
for jP.
But
since Gy=Gy, for all
Also,
it is easy
for jEN.
E S u{jfor all
sufficiently small.
y E int S a
that
B ~ a,
and t > 0 and
sufficiently
small,
Bu{j})
S
(Gyl\(
S{j
(y-tx)
Suppose
sufficiently small.
If
8
Thus y
sufficiently small.
E
G
c
that
a,
then y E Sy for some Y c a,
F.
4 a,
Also,
y,
Ti
=
F for all
tx for t > 0
jEN, whereby
(u{j})
Furthermore,
since
for t > 0
and
Therefore,
aGy for y=y-tx and t > 0 and
Sa.
whereby y + tx E
aGy,.
8
If y EaSa, and
Y8, and hence y is not
and hence
and sufficiently
for y=y-tx and t > 0 and
then y E
int Sa,
t > O
E Sa for all aGy,
(y-tx)
for any jEP.
}
Bu{j}))
sufficiently small.
interior of
8
sufficiently small.
( U
jjEN
j EP
such
(8 uj))
for all
to verify SBufj}n
E Gy for y=y-tx and t > 0 and
for all aEGy,
Ti
t > 0
Now consider y=y -
Then y-tx
Gy 2
y + tx E
Thus,
(Bu(j}) 0 6G y
JEN, whereby
and
G
E
For all t > O and sufficiently small,
a,
in the relative
int Sa
for all
Thus we have for all t>
t > 0 and
0
and sufficiently small:
Gy = Gy'
for y = y -
tx,
\ ( U B{j})
jEP
X° )
(M,
sufficiently small.
There
is
a facet
(
U
jEN
u{j}),
and y is therefore a regular point of
the unique n-cell of
regular point
u
of
of Tk.
X° ,
thus exists
and Tk =
and from
containing y=y
n
aOEG
(*),
Sa .
we have
37
t > 0
(*)
XO.
Let Tk be
+ tx for t > 0
and
such that y=y -
Thus Tk n H = Ti
n
tx is a
H, and so F
li
GyAGy
=
u
(B
u(j))
=
jEPuN
Finally,
note that
would be a facet
The last
(u{(j})
B cannot be an element
of
X° ,
hyperplane for
u
X
= RB.
jEM\B
(by lemma
which cannot
of
K,
for otherwise SB
1) and H would be a supporting
be true.
[X]
intermediate result we will need to prove
theorem 1
under nondegeneracy is:
Lemma 4.
If X satisfies the assumptions
nondegenerate,
some kE
tk for
PROOF:
X°
of
Let
, by lemma
for all
t > 0
all
t > 0 and
{j}
EM\B,
for y =
Thus
and y
k E
Thus S
and
(b),
Thus ABx = e
let T =
and
n
JEM\8
Ssu(j
X is
)
for all
y-tx,
is a regular
such that
t > 0 and
} .
small.
al
ia,
(y -
Then
whereby T is
t > 0 and sufficiently
iB
Suppose
= n+1.
X° ,
jM\B.
tx) E SBu{j}
an n-cell and
If a
whereby, y
(y-tx) E
#
Bu{j}
Sa .
(y-tx)
Sa
for
for some
But
sufficiently small, a contradiction.
and t > 0 and sufficiently small,
point of
(n-1)-simplex)
and Ajx < 1 for all
S{j
and
(and is an
there is an extreme point x of
n
sufficiently small,
(1,...,p}.
IBl=n,
is a facet
By remark 2
there exists
S
E2K with
and sufficiently small,
int T for all
E
1.
int S,
rel
(y-tx)
IBl=n.
= C(x).
Let y E
each
theorem
{l,...,p}.
BE3K,
X such that
E
then for
of
whereby T
[X]
38
Gy =
= Tk for some
u
Bu{j},
jEM\8
We now have:
Proof of
Let
theorem
EOK with
1 when
1Bl=n.
X is nondegenerate:
Then
by lemma 4,
n
Sou{j)
=
k
for some ke
jEM\8
{
By
T
Therefore for every y E int
,...,p}.
lemma
2, #RS
n-cells Ti,
Tj,
z E int
by
Tj,
is odd,
GyAGz
and hence #Gy
= R
lemma 3.
Gy = (Gy\Gz)
Similarly, we have
+
Gz),
(Gy
+ #(Gy n Gz).
= #(Gy\Gz)
-
even,
+ #(Gz\Gy)
-
-
int
i,
two sets are disjoint.
2 #(Gz\Gy)
2 #(Gz\Gy)
2 #(Gz\Gy).
by lemma
2.
Gy and #Gz
Thus
#Gz have
the same parity for all y E int Ti,
for any
#Gy
have the same
two adjacent n-cells T i
Therefore,
one T k ,
for any y E
- #(Gz\Gy)
parity.
for at least
two adjacent n-
whereby
Gz = #(Gz\Gy)
= #(Gy&Gz)
is
Bj=n,
because these
=#R
Ouj) = R.
We have:
= #Gy\Gz)
RB
K,
n
jEM\3
For any
n Gz),
Therefore, #Gy - #Gz
However,
, Gy =
is odd.
for some B
u (Gy
*Gy = #(Gy\Gz)
k
is odd,
j,
and
Z E int Tj.
by choosing Tk =
n
#Gy
and
Furthermore,
SBU{j}
where
jEM\B
BEaK.
8
Thus
Gy must be odd for
k = X°
is a connected set.
all
y
int Tk for all k,
E
But y
E
int
T
k for some k
because
if and only
k=l
if y is a regular
X° .
point of
Thus
Gy
is odd for all regular points
X° .
of
Therefore,
E
aEK,
K such
if y is regular,
that L(;)
and L()
u C(a)
there exists an odd number of
E Gy.
Each simplex
= L(o)
for
of simplices aEK with the
all aEK.
property L(a)
39
is of
Thus
the form
simplices
u C(a) where
there exists an odd number
u C(a) E Gy.
This proves
III
(i) of
assertion
closure
1.
theorem
argument.
combinatorial equivalence
point of
X,
X is nondegenerate.
combinatorial
result,
the more general
{zERnlz=aEa,
point
of
Z if
define G z
every facet of
rows of
Z is
E is
(i)
every row of
(ii)
z=O E int Z and
lal
an extreme
z=O does not
<
n,
zZ
case.
is
X > O, X-e=l}.
define T
=
said to be a regular
For every zZ, we
a
< n.
be
special
point of
lie
1 for
if
Z,
in the affine hull
ot Ta for
and
no hyperplane H in Rn meets more
than n extreme
pseudomanifold with vertex
(i)
BEaK
implies B c M and T
(ii)
if
c M and TB
the
and that
in proving theorem
aid
set KO
say that K agrees with Z if
We have
row Ai of A is
suggest a more general
A point
said to
This
(n-1)-simplex,
For each a c M,
E.
Ta for any a c M with
Let K be a finite
We
is an
form Z = {zERnlz=XE,
the
Xaea=l}.
XO.
fact that every
whose development will
= (a c MlzETa).
(iii)
X°
These observations
index the
any a with
K whose boundary
the boundary of
(degenerate or nondegenerate)
a > O,
z
to
is driven by the
Let Z be a polyhedron of
Let M ={l,...,m}
has depended
1 when X is nondegenerate
equivalence
a combinatorial
an extreme
from an elementary
follows
to create a pseudomanifold
critically on being able
bears
(ii)
[X]
of theorem
The proof
Assertion
is a face of
is a face of
following result:
40
Z,
Z,
then BEK.
and
points
M.
of Z.
L(o)EGz.
define X =
If Z is special,
PROOF:
bounded,
nondegenerate, and
centrally regular.
X°
X > O,
=
{yERnjy=XE,
have X°=Z,
where
pertains.
Furthermore,
only depends
on the
(The statement
Therefore Remark 2
lemma 2
and not on how K was
X.
3 and 4 hold true.
z is a regular point
i.e.
of
[X]
lemma 5 can be regarded as a combinatorial
the No Retraction
of
see this,
suppose
vertices of
given in the
Z is as
of T consist
set of vertices
is a row of
E.
If we
collection K of all
let a
According
to
Any
Z.
p
>,
{v,°.
simplex
let f(Ei)=Ei.
k
v ..
, i
if
Z continuously and
lemma 5,
f(-)
Then
the
are
ET can be written
E K,
and each
then the
that agrees with Z.
°
,
let
f(v)=EL(v),
is extended to
and
a PL function
leaves the boundary fixed.
for each regular point zEZ,
41
Z.
ip},
i
To
T be a
where each vi
then for each vK
Then
e.g.).
where K
u K,
such o is a pseudomanifold
is a labelling K°O-M,
f(.) maps Z into
=
facet of
{EiliEM}
= <v,...,vk,Ei,......Ei
form
for each Ei,
=
[13],
and let
lemma,
refine any
of K
interior of
T in the
(see Hirsch
Theorem
Z that does not
triangulation of
If L(-)
of Z,
X is nonredundant
there exists an odd number of simplices aEK with the
property that L(a)EGz.
in the
lemmas
Finally,
X is solid
because the proof of
that K agrees with
T.
Then
In this case, we
eX=l}.
valid,
if z is a regular point
Therefore,
version
2 is
lemma
fact
from K and
constructed
X° ,
(xeRnjEx < e}.
and because Z is special,
and centered and scaled,
Ei
EK with the property
then there are an odd number of simplices
that
of
z is a regular point in
iEM, and
that L(i)=i for each
labelling such
Z,
K agrees with Z, and L(-):KROM is a given
If Z is special,
Lemma 5.
there exists an
II1
for at
z=f(x)
corresponding to
Theorem.)
Retraction
then the preceding proof of
When X is degenerate,
In particular,
valid.
not
(To see this,
> n.
Then,
at least n+2
X is
but this set
EK,
contains
is not an n-simplex in K.)
so
elements, and
C(x)
(x}
x}EK,
since
suppose
X for which
of
let x be an extreme point
degenerate, and
theorem 1 is not
the construction of K does
X is degenerate,
if
in an n-pseudomanifold.
result
IC(x)l
maps Z onto Z, proving the No
Thus f(-)
.
that
simplex
is the real
where
one xa,
least
But this means
that L(-)EGz.
simplices aEK such
odd number of
The typical method for side-stepping degeneracy is to perturb the
constant coefficients of
our case,
In
infinitesimals.
adverse consequences.
properties
such as
this.
amend T so
X.
In any
again
of
X
,
the
is undesirable
the combinatorial
alter
in a combinatorial analysis
X,
if T is a triangulation of
it
is unclear how to
is a triangulation of
the perturbed
which
T may change,
combinatorial structure of
is undesirable.
perturbation of X is performed
hand side coefficient b i
to
bi
+
i
.
We will
by changing
instead
by using this same construction in a dual form.
however,
is said
C(x)=a.
of
X has
such a perturbation of
however,
the amended version
case,
X by a vector of
of
The perturbation will
which
Also,
that
The usual
X° ,
constraints
the
is to repair K.
to be consistent
We proceed as follows.
if there
exists xE
For any matrix D or vector d,
leading zero columns or components
B=[b,I].
If a c M is consistent,
let
42
right-
perturb
Our first task,
A subset a
c M
X with the property that
(D)
or
of D or d,
a subset
each
(d) denote the number
respectively.
Let
B c a is said to be a basis for a if AX+Y=B has a solution X, Y with
Y
= O,
0, (YM) > 1, Y
i.e.,
(Ye) = m+l,
and
I8[
= rank Aa.
Instead of constructing the pseudomanifold K by joining each
simplex a of K with its carrier set C(a), we now construct K by
of K with every subset B of its carrier set C(a) that
joining each
forms a basis for this carrier set.
We obtain the following
theorem:
Theorem 4.
Let X be a solid, bounded, and nonredundant polyhedron.
Let T be a triangulation of X and let K be the pseudomanifold
corresponding to T.
Let K°=K°
M, and define K = (a c Kola #
= a i 8, where aEK and B is a basis for C(a)).
n-pseudomanifold, and
some XE X}.
K = (B c MB#o,
A,a
Then K is an
and B is a basis for a = C(x) for
[X].
The proof of this theorem is rather laborious, and so is relegated
to the appendix.
The boundary elements of K correspond in a natural
way to subsets of faces of X °,
in a manner that we will soon see.
Theorem 4 thus gives a constructive procedure for triangulating the
boundary of X°
The procedure of joining simplices a of X with bases
B c M is similar to the construction of an antiprism, see Broadie
l1].
This construction of K is also closely related to the construction of a
primal-dual pair of subdivided manifolds, as in Kojima and Yamamoto
[14], although K is combinatorial while the primal-dual pair of
manifolds is not.
43
II1
Our next
for iEM,
x
°
task is
and define A
{(yRn y=XA,
=
to perturb the
to be the
n-cell
X.
matrix whose
X > 0, e.X=l}
and
ith
row is Ai.
= {yERnly=Xa A,
S
) i
Ai/(I+c
Define A
Define
O,
Xa
ea*ma=l }.
Lemma
6.
If a c M and
solution for all
PROOF:
and
values of
.
.n+l
trivial
.
and
The proof
is by contradiction.
be n distinct values of
If
e
in all
c M
Therefore let
for which Ax=eea has a non-
of these
solutions,
I
1
has a solution
Let Q be the
[c]=(1,c,c2,....,m).
1 ...
1
then by
. .
m
.
'
2
2
rank of Ba.
However,
the rank of
cannot have a solution,
It only remains
that
to
because
it has a solution with
solution
(x,e)=(x,0),
matrix
l
cm
n+l
since m > n,
But,
the rank of Q is n+l, as
A
an
is the
is at most n, whence AaX=BaQ
the rank of BaQ
show that
=cl,...,en+l where
n+l
there exists a solution X to AaX=BaQ.
induction argument establishes
for
1
CI
m
1
nontrivial
if a
then Ax=eea can only have a solution for at most n
Therefore Aax=B[c]
solution,
that
we can assume that Aax=ea has a solution for all
c=c1,...n+1.
Then
ea has no nontrivial
sufficiently small positive
solution.
rescaling,
B=[e,I]
then Ax =
n+l,
We will actually prove a stronger statement,
al=n+l,
I.....
a
is also n+1.
if Ax=eea has a nontrivial
e#O.
Suppose that Ax=eea has a
and suppose that
44
there is no
.
solution to Ax=ea
XaAa=O,
Xa-ea=l,
hull
S.
of
i.e.
the zero vector
1BI
C
_
affine combinations
£
hull of SB.
n and
<
of
the
with the property
is an element of
has dimension at most n-1.
a such that
fewer
there exists X a
Denoting this affine hull
whereby H
of
Then
by HE,
the affine
we also have Aa x=0,
there exists a subset 8
Thus
S
the affine hull of
is spanned by
_ Therefore HE
rows A
that
is the affine
But y=O £ He, and hence y=O lies in the affine hull of n or
rows of So,
and hence
X is centrally regular,
of
S
This
.
and the proof
contradicts
the assumption that
is now complete.
[X]
We also have:
Lemma
(a)
7.
XO°
For all
sufficiently small positive c,
is special,
(b)
°
XO
agrees with K,
(c)
any
regular point
y
is a regular
PROOF:
Let
X={xERnlAcx
[C]=(,c,E
2
small
,...,cm).
positive
points
,
of
points of X°O.
X,
extreme points
follows
from
of
the
{xERnIAx
X°O
of
£
X°O.
and
if
where B=[e,I],
and
sufficiently
Furthermore,
of A are all
because
positive
.
of A
are all
y=O E int X° ,
Finally,
lai
extreme
y=O E int
X°O
,
is
> n+l,
A
fact that BE2K if and
45
for
x=eea has no
no hyperplane meets
special,
X°
because the conditions
> 0 and sufficiently small.
positive
Thus
B[c]),
are all simplices.
whenever a c M and
small
X,
if and only if ySa.
<
whereby the rows Ai
solution for all
all sufficiently
yES a
of
XE is nondegenerate for all
rows Ai
Furthermore,
lemma 6 are met,
nontrivial
=
b}
is a regular point
X° ,
point of
the faces
all sufficiently small
of
y of X°
Because
X is nonredundant, the
extreme
and
proving
Thus for
more than n
(a).
Part
(b)
only if AX+W=B has a solution
since
iI
with W=O
and and WM\ B
with w=O
and wM\B
only if
if and
i.e.
.
positive
Our last
Let a
K have
E
and a1
B1B82.
2,
#
then we
exists X 1,
X1AL
2,
Ul,
+ U1AB
1
U2
if
exists
this
1)
-i=(oi U
and 82
> O.
E
means L(a)
<
n and
if
EGy.
C(c)
E Gy and a2 = (2
u 82)
=1=o2 and
Let L=L(a).
C C(a).
181
and only
EGy if
Suppose
is of
Then
Gy.
and because
Now L(a)
B c C(a),
Then there
such that:
(1)
= y
2 is a basis
u C(O)) E Gy.
c C(o),
EK and
(2)
X2AL + U2AB 2 = y
Because
(L(a)
that satisfy L(a) E Gy, and
cannot have a1=a2.
c C(o)
1
K
E
there is a one to one
Then
the property that L(a)
Because
show that
Then
of X °O.
> 1, whereby a#.
ja
u B) EGy.
thus must
satisfy the assumptions of theorem
simplices
form a=o u B where
IL(a)I=n+1,
We
point
that satisfy
E K
simplices
(L()
note that y
IBl=n, for all sufficiently
B c M,
and L(.)
y be a regular
correspondence between
the
(c),
[X]
X, T, K,
Let
1, and let
PROOF:
X > 0, X-e=l},
intermediary result is:
8.
Lemma
{yERnly=XA,
; if and only
positive
For part
if and
if y meets no So for B c M,
if and only
no S,
and so y meets
IBl=n,
small
X°
point of
is a regular
X °O.
is a face of
S
positive c;
{yERnjy=XA C,
is a face of
O, XB-e=l}
gB >
has a solution
XBBB[c]=I} is a face of
> O,
sufficiently small
X > 0, XB[e]=1} for all
{yly=XSAC,
sufficiently small
> 0 for all
{yERn-y=jAS, XB
only if
if Ax+w=B[c]
if and only
O;
for a
= C(a)
i=1,2,
a sqaure matrix T such that A
46
2
=
111=1021
AB 1
and there
, and Wt must
be
E Gy
if
nonsingular, whereby A
1Bil=n-t,
i=1,2.
ILuBil=n+l.
-lA2.
Because y
Combining
(X 1 -X 2 )
(1)
+ (u 1 -u 2 n)
AL
#
, or U1-U2T
#
If
> O,
and
i
(2)
there must exist X 1 ,
A81X1
= Be
1
A82X1 + Y1
=
A82X2 = B
2
AS1X2 + Y2
= Be
Therefore B
1
-l1Yl
X2
= A 81
Because
u 1,
= A
X 2,
1Y
U2 > 0,
i=1,2,
Y 1,
< t+l and
and
for otherwise y
is
above, we obtain
(1)
or
(2),
(3)
to reduce
and UlT-l=u 2 -
is a basis for a = C(8) and
Y2 such
the
violating the fact
Thus X 1 =X 2 and U=u2,
Si
each 8i
a,
that
0, Y 1 # 0.
, Y1
Y2
= T-
X1
1
-
1
> 0,
ILI
Ll=t+l,
then we could use
2' Y1
B8
then
(3)
number of positive components in
that y is a regular point.
,
point,
A81=
O,
Furthermore, because
t=dim
is a regular
Furthermore, X i
not regular.
If X 1 -X 2
=
1
1
A8
1
= BS 1 -
0
U 1Y 2
L
2
=
X1
0.
-1B -8
- 1Y 1 ,
-
Y2
#
= U 2TY
2
-lY
2
1
= T- 1 A 8 2 X2
and hence Y 2 =
= -2Y1
--
-
IY1.
O0, a contradiction.
Thus 81=82.
Next,
let 6
E
K and
(L(o)
C())
= L(a).
We need
to find a basis 8
E Gy.
Because
(6 u a)
(6 u a)
the
suppose
property that X 6 A
E
Gy,
+ XaAa = y,
E
Gy.
Let a = C(o),
for a with the property that
there exists X6
and e 6 X 6
+ eXa =
> 0,
1.
Xk
Let B =
linear
[b,I] and
let
> 0 with
Furthermore,
there exists x E X with the property that C(x) = C(a) = a,
ba.
and
i.e. Ax
z=XaAa and consider the following lexico-
programs:
47
=
III
lex
max
z.X
lex
min
XaBa
P
subject
to AaX+Y
Y
Note that X = (x;O),
feasible for D,
(X*,
=
Y*),
a
for D,
Y = (0;I)
K.
Let v*
Y =
+ Xaea = 1, and so
L(a)
with
(L(a)
Gy.
odd.
1.
Then
7.
X°
Thus
Gy,
*
= X 6 A6
=
(L(a) u 8)
Let X, T,
for
all
c
K,
c M with
L(a)
simplices aEK-with
By lemma 8,
the property
that
48
Then
is feasible
Thus
8
+ Xe
is
B
=
for every a
E
K
= (a u B) E K and
satisfy the assumptions
z
of
X°O
X° ,
then y
point of
X,
is a regular
and hence #Gy
simplices of aEK with the
there are an odd number of
(L(a) u C(a))
of
agrees with K,
is a regular
sufficiently small positive c,
EGy.
Therefore,
[X]
if
there are an odd number of
property that
Thus
sufficiently positive c,
If y is a regular point
for all
= Xea.
E Gy.
and L(-)
by lemma 5,
is
_
+ XaAa = y, and X6e
there exists
1.
Therefore,
is odd.
point of
E
c a.
is feasible for P and Xa=Xa
This completes the proof.
of Theorem
by lemma
#Gz
C(a))
= X
of P and D.
> zx = XaAax = Xaba > v 1 , whereby v
X6e
t > 0
solutions
with basis
*
+ XSA
theorem
basic optimal
be the optimal value
(O;I)
=
feasible for P and Xk
respectively,
a basis for a, and X6A6
Proof
is
_
v
L(a) E
X
whereby there exists
because X = (x;O),
*
subject to
to XA+Y
0
k
to P and D,
(o u B) E
D
Ba
EGy.
[X]
is
A Combinatorial
5.
Sperner's
Theorem on a Bounded Polyhedron that Generalizes
lemma
Theorems
results on
shown to generalize combinatorial
and 2 have been
the simplex and
and dual-proper
labels,
simplotope that have unrestricted labels
respectively.
In this section, we
theorem that generalizes the results on the simplex and
proper
labels,
including Sperner's
Let X, T,
and L(-)
(yCRnly=XA, X > 0, X e=l}.
Dy =
{(a,8) E MxM
Theorem 5.
pseudomanifold
with the
PROOF:
scaled.
E
property that
Let
l}.
=
X, T,
L(.),
E
int
X °,
let
solid, nonredundant,
Let T be a triangulation of X,
int
XO, there exists at
(L(a)),
C(a))
least one simplex aEK
Dy.
E
and K be given as in theorem 5.
X' and T'
lemma,
pseudomanifold corresponding to T',
L(v')
be the
= L(g-l(v'))
for v'EK ° ',
projective transformation
AL'(v,)+y, and
f(x')
= arg min
'EX'
Because
h'(-)
x', and
and
llz'-x'+h(x')112 ,
Let a'
let a = L(o'),
B = C(a).
X°
where g(.)
and define
is as defined in the
For each v'EK °' ,
define h'(v')
in a PL manner over all of X'.
is continuous,
fixed point x'.
Let y E int
as in the projective transformation
lemma.
extend h(-)
let K be the
let L(-):K°-M be a labelling
be given, and define
let K'
let
We have:
b} be bounded,
corresponding to T, and
Then if y
function.
For any y
eX
X = ({xRnAx
and centered and
[23].
- XaAa = y has a solution X 8 , Xa such that
I XBA
> 0, and ea-Xa
Let
simplotope for
satisfy the assumptions of theorem 1, and
X° =
> 0, X a
lemma
present a
where
n.112 denotes
=
Define
the Euclidean norm.
f'(.) is continuous and so contains a
be the smallest
= C(a').
Let
simplex
= gl(a).
The Karush-Kuhn-Tucker conditions
49
' in T'
that contains
Then a = L(a)
state that
li
x'-x'+h(x')
= XB(A-eoy)
-XxAa -
X 8 AB
so
-
XaAa = (e-X
=
(L(6),
Section
the
the
existence of
inward
normal
normal
from the
of an inward normal
point of
a fixed
in
the proof
of h(-)
vEK ° ,
index of
L(v)
section 2,
let
and so
there
E Dy
Let a
T be
of h(-)
see Eaves
is just
-h(-)
of
in
the
is equivalent
[3].
When y=O,
and the existence
an outward
n
n
XA ,
n
(XaA)T > 0
is proper a n
Thus A (An)T <
Sn ,
a triangulation of
L(-)
K be
This
All
the
is said to be proper if for
an element of
M\C(v),
for vEK ° .
of v,
The conditions
i.e.
L(v)
of
S = C(a);
X0 > 0, Xa
.
Note
0 and so 0O
thus a
= M
> 0,
(L(o),
then there exists
eBX
+ e
X A (An)T Xa =
(1,...,n+1),
is precisely Sperner's
50
the
theorem 5 are met,
that for any i,
=
is
the
For X = S n , the set
cEK with the property that
=
n
whereby XaAa = 0,
(1,...,n+}l.
let S n ,
and L(-):K°-M be a labelling
= L(o),
such that XA=
< 0.
=
X° .
exists a simplex
for y=O.
Because L(-)
Ai-Aj
presented
eX=l) is an n-simplex that contains
so y=O E int
and
L(a)
is the index of
(yeRny=XAn, X > O,
XB
we see that
lemma derives from theorem 5,
a nonbinding contraint
origin,
Xa,
is one,
the same as the existence of
is
where M = (l,...,n+1).
function,
C(a))
f('),
above
pseudomanifold corresponding to T,
=
both vectors
existence of an outward normal
show that Sperner's
be defined as in
X°
Therefore,
of h'(-).
To
each
-= 1.
[X]
The existence
function h'(-)
of an
ea-X
=
After normalizing the vectors X8 and Xa
component of
EDy.
Furthermore, h(x')
theorem 1 using Brouwer's theorem,
2, derives
function h.
+ eX~)y.
B
C(a))
The proof of
to the
> 0.
y for some particular Xa > O,
that the sum of
(a,B)
for some XB
B
X
=
1.
j E M,
ij,
(XaAn)
and so
lemma, without
the
oddness assertion.
The logic used above can also be used to prove
(see also van der Laan and Talman
lemma to the
[17]),
theorem 3 of
[7]
which generalizes Sperner's
simplotope.
Theorem 5 does not contain an assertion of the oddness of
number of simplices under
consideration.
The basic constructs
the
used to
prove theorem 1 combinatorially do not appear to carry over directly
to
the case
It is an open question whether there exists
of theorem 5.
a combinatorial proof of
theorem 5 which asserts the existence of
odd number of simplices aEK
for which
regular.
51
?·___I
_____
(L(o),C(o))
EDy,
when y
is
an
li
A Pseudomanifold Extension Theorem
Appendix A.
X = {xERIAx < b} be bounded, solid, and nonredundant, let T be
Let
a triangulation of X, and let K be the pseudomanifold corresponding to T.
We wish to
Let M = {l,...,m} be the set of constraint row indices.
construct an n-pseudomanifold K, where each n-simplex a of K consists of
of C(a), the carrier indices of a.
a simplex a of K together with a subset
In order to construct K for arbitrary polyhedra, we need to work with
the lexicographic system AX + Y = B, Y
For a given
O, where B = [b,I].
vector v, define (v) to be the number of leading zeroes of v.
For a
matrix V, let (V) denote the number of leading zero columns of V.
An
index set a c M is said to be consistent if there exists xx with C(x)=a.
A subset 8 c a is said to be a basis for a if there exists X, Y such that
AX + Y = B, Y
0, (Y)> 1, and (Ye)= m+l, and
[8
=
rank (A). Let us
construct K as follows:
-o
let K = K
u M, where M = {l,...,m}, and
let K =
c a u
f6 ~
c, aa E K, a = C(a), and
is a basis for a}.
Our aim is to prove:
Theorem 4.
Let T be a triangulation
Let X be solid, bounded, and nonredundant.
of X and let K be the pseudomanifold corresponding to T.
define K = {a c K
a
, a c a u 5, where a E K and
Then K is an n-pseudomanifold, and
K = {8 c M
some a, where a is consistent}.
52
Let K = K u M, and
is a basis for C(a)}.
, and
is a basis for
When X is nondegenerate, we have:
Corollary Al.
Let X = {x
E
RnAx
and nondegenerate polyhedron.
<
b} be a solid, bounded, nonredundant,
Let T be a triangulation of X, and let K
-o
be the pseudomanifold corresponding to T.
K = {a c K
a ;r,
a c a u
an n-pseudomanifold, and
, where a
K = {
E
c M I a4$,
Proof of Corollary Al from Theorem 4.
Let K
K and
=
=
u M, and define
C(a)}.
Then K is
= C(x) for some xx}.
If X is nondegenerate, then if
a is consistent, the rows of A. are linearly independent, the only
basis
for a is
=a, and the result then follows.
[X]
In order to prove theorem 4, we proceed as follows.
it is assumed that X is solid, bounded, and nonredundant.
53
Throughout,
i
Proposition Al. If a is consistent, then there exists X*,Y*such that
At+ Y*= B
PROOF:
Y
0, and (Y*)
a
1, i.e., the first column of Y
a
is zero.
Consider the lexico-linear program:
lex min Z = e Y
a
s.t. AX + Y = B
Y
where e
P ,
O
is the vector with (e )j equal to one if
otherwise.
j
a
E
and equal to zero
Because a is consistent, there exists x, y such that
Ax + y = b, y
0,
note that X, Y
is
and y
= 0.
Then define
feasible for P.
X = [x,O], Y = [y,I],
Furthermore, since
e Y
0
a
and
for any
feasible X, Y, the above has an optimal solution, X*, Y*.
If (eaY*)1 > 0,
then there exists
by duality.
Thus
0 < (r*B)
(eaY*)l = 0,
rr*< ea
= r*b = r*Ax +
whereby (Ya*)
Lemma Al. Let X, Y satisfy
= {i
PROOF:
Yi = 01.
= 0, (*B)
7T*A
e y = 0,
> O,
1
a contradiction.
Thus
1.
AX + Y = B , Y
and let a = {i
0,
aB
B'
such that
l B I = rank(A ).
First note that
we must have rank(A)
IYil
' is a basis for a
To see this, observe that
= I B I , rank(Aa) 2 IB I , but since rank(A )
=
IB I
.
= 0},
A X = B
I
,
I B I = rank(A 8) < rank(A ) . And if
Therefore
B, then 1B'I = rank (AS,) < rank (Aa).
Let
proved.
c = rank(A ) -
B I.
If
c = 0, then let
Suppose the lemma is true for rank(A ) -
consider the case
T
,T*y
Then there exists
whereby since rank(B)
8'
such that
= {ilYil > 0.
rank(Aa) Because
IB
I = c.
Let
c > 0, there exists
54
' =
, and the lemma is
B8 I = 0,..., c-l1, and
6 = a\
j E 6
,
and
T =
M\a, i.e.,
with A independent
J
,
of the rows ofA
AMd = 1 .
J
d E Rn
There thus exists
.
such that Ad = 0, and
Now, let
Y.;Y.)
v = lex min
for some
Y
i
E
M .
A.d>0
1
We note that
(v) = (Yi)
Y'
whereby
Y
1
B-AX',
= 0,
and
(Y )
we have that
one where
a' with
Y'
i
>
a
1,
i
0,
PROOF:
Because
j = 1
O,. {a=
j = 2.
AX +Y
Define
D
1
and
=B,y
a
D
=
,
= -Y
D
k
==
k.
Y
= 0,
=Y
Upon setting
k
O
Y>-O
k 0
=
Y
Let
=
k E
Then the choice of j
there exists X, Y
i
S u {i},
=
.
Let
is unique.
such that
, and
Now suppose there are two such
j
k
such that
Then, by reordering if necessary assume that
Then there exist
= A(X-X )
D
j
Yil=O},,
iI
is a basis for a.
and
O.
be a basis for a.
is a basis for a,
S I = rank(A B) = rank(A).
\ k uj
Y'
X' = X + do v,
and 5' is a basis for a.
be a basis for a , with
AX +Y = B, Y
Let
= c- 1 . We thus have reduced the problem to
Lemma A2. Let a be consistent and
\kuj
0.
< c , which by induction, means that there exists
a
,
v
It then follows that
and
T
I
whereby
i
A.d
(Y) =
rank(A )v
Y
=
rank(A )- 1
a'
and i C 8,
A.d
1,
O,
al
-
Y ,
i E S\k
X,Y
, X,
0,
a=
,
,
1,2.
=
= 1,2
55
\ku
=
{iY
Then we have:
1,2 ,
£ = 1,2,
y2, such that:
= 01,
= 1,2
III
Also, because
rank(A) = rank(Aa) , and
unique matrix
DiQ.=
A.(X-X
i
1
i
a
such that
)
=
IA(X-X
DQ <O
1 D
T D9
9,
Dk
9,R
)
= Aa .
=
iA
D
91
g
z, we can write
1 =
+ D1
=2 + 2
Now, Y2
D1
'r2
r
Y1
=I
1
, whereby
2
1
Thus
A
= 'iR k
Y2
=
ri =1ik
Tik
for all
iiDka .
<0
= 1,2.
Also, because
~~~~~~~~~~~~~~~~~~~~~9
7-Dk
PIrr
Thus,
2n 1 Y
'2
71
2
, we obtain
B
\ ku l
u2
0\u~
{2}'
matrix
A \k u{l1 u{2} has rank equal to
rank(B
\ ku {1} u{2} ) =
LemmaA3
we have
Y
Y1
=2
r1
7T
2
, whereby
2
1
2
=y =Y =0
1
1
2
y
\k u{l}
u{2}) X1
(5~~~\k{1}u
determined.
a,
7r
= Y2
Y
E
fi D I
, ,=,2.
=1,2
9
1
=Y2 + -r D
Using a parallel argument with
Y2
forall
DkQ
?O, we must have
k
and
i
If we then define
9,
DD
I = rank(A) , there exists a
For any
ikk
k
, then we have
Because
EA
I
B I + 1,
But since
I1 I = rank(Aa) ,
a contradiction.
1,2 Ea , the
but
Thus j is uniquely
[
If a is consistent and
is a basis for a,
and
i E
, then
exactly one of the following statements are true:
i)
ii)
there exists
j
a,
j
(5 \ i) is a basis for some
i
such that
a' c a ,
56
B \ iu j
is a basis for a,
where a' is consistent.
or
PROOF:
Let d be chosen so that
vector.
There exists
a = {kYkl =
A d
a
A d
a
or
O
Case 1
A d
a
Upon examining
= O}.
(v) = (Y.)
and
for all
= 0
Yk
Let
1 .
X' = X + do v, Y' = B - AX' .
k E 6 \ iuj .
v = A
A.d
We must have
j E a .
A d $ 0,
Furthermore, since
>
unit
we can either have
Aad,
and we have
k
d
k
Let v = lex min
Ad> 0
for some j .
th
is the i-
We have two cases:
0 .
0 .
i
, where e
AX + Y = B, Y a 0,
such that
X, Y
= {kY
k
,
i
Ad = -e
Also, (Y )
Then
> 1,
O,
Y'
and
=
{kJYk
O} =
\ iuj .
Thus the conditions of (i) are satisfied.
Aad < O.
Case 2
v = lex min
Let
(This set is n ot empty, since
Akd
Akd> O
_
Y.
otherwise X is unbounded.)
(v) = (Yj)
=
Let
0.
kA dYk
Yi
X' =X +
aJ
Y
Yj 2A.d >0
J
,
=
for some
A.d
j
,Y'
= B-AX' .
.
For
k E
\ i,
an I hence
a,
d ov
Y'nd
Then fo:r
Yk
=
.
k E M,
Also,
and
Y
(Yi) =
and
v
Then
0, i.e., Yil >
O.
Likewise, for
21
k
a,
(Yk)
,
i.e.,
Ykl
>
0
.
Define
6' = B\ i,
a'
= {iJYil
=
O} .
there is a unique HI such that
is a basis for a,
Since
a'c a.
Then
=
Akd
EA
= A
implies
Akd =
.
Now for each
Akd = O.
AB d = -ki
k E a',
Likewise
= 0
Yk
k E a'
if and only if
combination of the rows of Aa\ i.
6 \i·
=k
A.d Yj
3
if and only if
k E a' .
, whereby
Ad = O.
> 1
Now
Thus each Ak is a linear
Thus B' is a basis for a'.
57
(Yk)
III
It only remains to show that (i) and (ii) cannot take place
simultaneously.
If (i) holds, then there exists X, Y , such that
AX + Y = B, Y
O,
AX' + Y' = B , Y'
Defining
to
D8
1A
{kYk = O} , a
= Aa
Y
wheereby si.nce
D
=
O=Y
Now,
i = {k| Yk=
we have
IDk =
Ak would
Dk =
we must have
NkiDi
k1.kiD for all
for all
ki = 0
a' = {kYkl =O} ,
and
Di0
Ki< 0} =
k
k E a , and
ki >
a',
and
a' za
If
Yk
X, Y
A , \a
AaNa
AX +
= B,
A.d>0
1
Dil> 0,
=
Yk +
a'
O0 ,
(otherwise
whereby since
kiDi
a'
Y'
D = A(X-X') =Y'-Y,
i
we have
Thus
Eji< 0
from above.
Thus
that is a basis for a'.
' 8
VY
0,
rank(A ,) = rank(A ) + 1,
=
a
{iil = 01 ,
is independent of the rows of A
is independent of the rows of A
given, and let d be any vector such that
v = lex min
O,<
j
0
then there exists
satisfy
=-Yj
AX' + Y' = B,
Now, since
k E a\
for all
PROOF:
Then some element of
with
Again letting
= Yk + Dk
' is uniquely determined.
element of
X', Y'
a' c a.
Furthermore,
Let
D
j
and a' and a are consistent, and
is a basis for a,
and
ji< 0
, contradicting the fact that
exactly one of (i) and (ii) holds.
Lemma A4
0,
k E a.
for
,
kE a, because
for
kIkiDi,
1
have independent rows) we must have
Lot
I
{k E a
} ,
I be the unique solution
D
If (i.i) also holds, then there exists
8\
Let
1
jiDi,
..
and X', Y' with
= {klYkl = 0} .
kA(X- X') =
) =
=0-0=0.
~\i
B\i
= O} ,
{kYkl
D = Y'- Y.
Dk = Ak(x-
Then
=
\ iuj = {kIY'k = 0} ,
O,
D - A(X- X') , we have
= Y'
\i
=
Let
Asd = O,
= {ilYi= 0}
, whereby some
j E a' \a
Then let
Ajd = 1.
Yk
{Y
fosoe
kEa\a.
Tn
Akd
for some
k E a' \
Then upeon setting
k
58
.
p
be
Aid
_
= X + do v
X
Y' = B - AX',
we have
Y
= Yor
,
Akd
Y
uk =
.
B' = 8 uk
Now
ukca',
rank(A
all j
°
uk) = rank(A ,) , whereby
is a basis for a'.
Now suppose that
B1
B
Then consider 81.
u {l} , 8
1\ {1} u {2}
i E s
there does not exist
B1\{l}
are both bases for a' that contain 8.
B1 =
ease of notation, suppose
But
and
Y
k
.~
ac a',
1\ {i}
{1,2} c a' \ a.
, where
is a basis for a',
such that
is a basis for
= B u{2
For
whereby by lemma 3,
is a basis for some
a contradiction.
Thus
ac a'
1 is uniquely
determined.
With lemmas Al, A2, A3, and A4 as preparation, we can now prove Theorem 4.
PROOF of Theorem 4:
Let
and B is a basis for a.
T E K
-Da
with
TUBEK, and
=
6c a U
be a simple:x of K,
Then if
k = rank(A-') , there exists an (n-k)-simplex
C(T) = C(a) = a.
such that
=
I[uB1 = n- k+1+ k
n+l .
a E K,
where
6 c a u
Thus,
cT u
a
C(a) ,
, and
Thus every simplex of K is
contained in an n-simplex of K.
Now let
v E K
.
Then
{v}
K,
and let
there exists X, Y such that AX + Y = B ,
Y
with
8 = {iYi = O} ,
B
Then
1B'1
v
E
a,
Thus,
= rank(At,)
C(a) = a,
ou B'
0,
a = C(v)
and (Y )2 1 .
there exists
B'
= rank(Aa) .
Now there exists
By lemma Al,
such that B' is a basis for a.
and a is an (n-k)-simplex, where
is an n-simplex of K, and
By proposition Al,
.
a
K
such that
k = rank(As,) =
{v}c ac a u'
B'1.
Thus every vertex
of K is a zero simplex of K.
Likewise let
exists
x
X
i
with
thus exists X, Y with
M.
Then, since X has no redundant equations, there
Ax + y = b,
AX + Y = B,
y
and by lemmaAl, we may assume
O,
Y
Yi = 0,
smallest simplex of T that contains x.
59
Yi = 0,
yj > 0
for
j
i .
There
O, {i} = {Yjly= 0} , by proposition Al,
Yj
O,
for
j
i.
Let a be the
Then C(a) = {i}, and a is an
III
a I = n.
thus
(n-l)-simplex;
a zero-simplex of K . Thus
au
basis for a,
and
v E K
for some
Then a E K
Case 1 s = v E a.
a'
exists
and
that
\v
,
E
for some i
/ a
Ka, there exists a
o
v' E K
a \v uv'
such that
v' tv,
,
a
- \ VEKa
such that
, and
is a k-simplex in K
a\ vE a K
If
and t is uniquely determined.
t = v'
Thus
If
s
= {o E KIC(a) - a},
, a k-pseudomanifold where K
and where n-k = rank(A ), and a is a k-simplex.
unique
s = i E M
, i.e.,
s E
, or
and
It is our aim to show
s .
t
is a
a u
s
Regarding s, we have either
that the choice of t is unique.
s = v
Now let
a u B \ s u t , where
consider an (n-l)-simplex
and
C(a) = a
Thus
8
= rank(A ) =
rank(A)
, whereby {i} is
is precisely the set of zero-simplices of K.
K
be an n-simplex of K .
Now let
i.e.,
a u {i} E K
Therefore
, then there
is a (k-l)-pseudomanifold,
Ka ,
rank(A ,) = n-k+l. By lemma A4, there exists a unique
B' = B u{j}
Thus
is a basis for a'.
t = j
j EM
such
and is uniquely
determined.
Case 2
s = i
. From lemma A3, either there exists
E
to lemma A2, or
\ i
i ,
and j is uniquely determined according
is a basis for a
such that B \ i u j
j Ea , j
is a basis for some a' c a, where a' is consistent.
the former case,
t = j.
Ka , that contains
oEK
In the latter case, let
Then
T
T
In
be the unique simplex in
= a U {v} for a unique v E K .
Thus t = v
and t is uniquely determined.
Because t can never attain more than one value, K is an n-pseudomanifold.
Furthermore, the only instance where t cannot exist occurs when a is an
X , and
extreme point of
aK = {6cM t
,
B ,
B
s = v
where
a
=
{v} .
= n,
and
= n,
and B is a basis for some a,
a is consistent },
60
Thus
where
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