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As an
n
-dimensional counterpart of a regular tetrahedron, we define
S
n
and
S
n
as follows:
V
(
S
n
)=
n
{
x
∈{
0
,
1
}
|
x
·
1
≡
0
(mod 2)
}
,
V
(
S
n
)=
n
{
x
∈{
0
,
1
}
|
x
·
1
≡
1
(mod 2)
}
.
n
be two vertices of
C
=conv(
n
). If
Let
x
,
y
∈{
0
,
1
}
{
0
,
1
}
x
and
y
are the two
endpoints of an edge of
C
,weget
1. Therefore, every edge of
C
contains points of both
S
n
and
S
n
. Therefore,
S
n
and
S
n
are imaginary cubes
of
C
that have no e-vertices. Moreover, since both
V
(
S
)and
V
(
S
) contain no
star of
C
,
S
and
S
are MCIs of
C
. Note that
S
n
and
S
n
have the same shape
which is denoted by S
n
.TheshapeS
4
is 16-cell.
Concerning H and T, we define three 0/0.5/1 MCIs of an
n
-cube
C
=
conv(
x
·
1
=
y
·
1
±
n
) as follows:
{−
1
,
1
}
V
(
H
n
)=
n
{
x
∈{−
1
,
1
}
|
x
·
1
≡
0
(mod 3)
}
,
V
(
T
n
)=
n
{
x
∈{−
1
,
1
}
|
x
·
1
≡−
1
(mod 3)
}
,
(1)
V
(
T
n
)=
n
{
x
∈{−
1
,
1
}
|
x
·
1
≡
1
(mod 3)
}
.
By a similar argument, one can see that they define 0/0.5/1 MCIs. Note that
T
n
and
T
n
are similar because we have
T
n
=
−
T
n
. We denote by H
n
and T
n
the shapes of
H
n
and
T
n
, respectively.
These sets of vertices satisfy the following equations. We have
V
(
H
n
+1
)=
V
(
T
n
)
V
(
T
n
)
×{−
1
}∪
×{
1
}
,
V
(
T
n
+1
)=
V
(
H
n
)
V
(
T
n
)
×{−
1
}∪
×{
1
}
,
(2)
V
(
T
n
+1
)=
V
(
T
n
)
V
(
H
n
)
×{−
1
}∪
×{
1
}
.
One can see from (
1
) that each of
H
n
,
T
n
and
T
n
is mapped to itself by a
permutation of the
n
coordinates. Therefore, one can derive from Eq. (
2
)thatfor
n
4, H
n
has 2
n
copies of T
n−
1
1)-simplexes
because each vertex figure of an
n
-cube is a simplex. On the other hand, T
n
has
n
copies of H
n−
1
facets,
n
copies of T
n−
1
facets and some (
n
≥
facets. The other facets are (
n
−
−
1)-simplex facets
4. In the case
n
= 3, the six 2-simplex facets of H
3
coincide with T
2
and
the three H
2
for
n
≥
facets of T
3
degenerate to line segments. Thus, H
3
has twelve T
2
faces and T
3
has eight faces.
One can see that the set of e-vertices of
H
n
,
T
n
and
T
n
are the sets
n
{
x
∈{−
1
,
0
,
1
}
|
x
·
1
≡
0
(mod 3)
,
x
·
x
=
n
−
1
}
,
n
{
x
∈{−
1
,
0
,
1
}
|
x
·
1
≡−
1
(mod 3)
,
x
·
x
=
n
−
1
}
,
and
(3)
n
{
x
∈{−
1
,
0
,
1
}
|
x
·
1
≡
1
(mod 3)
,
x
·
x
=
n
−
1
}
,
respectively.
3.4 Tilings by Imaginary Cubes
As we mentioned above, Hs and Ts form a tiling of three-dimensional Euclidean
space, and 16-cells form a tiling of four-dimensional Euclidean space. We explain
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