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As generalizations of a Sierpinski tetrahedron, fractal imaginary cubes such
that an IFS that induces the fractal is composed of
k
2
homothetic transforma-
tions of scale 1
/k
are studied [
2
]. Sierpinski tetrahedron is the only such shape
for
k
= 2. In the case
k
= 3, there are two such fractal shapes H
∞
and T
∞
whose convex hulls are H and T, respectively (Fig.
5
(b,c)). In particular, H
∞
is
a double imaginary cube. In the following, we explain these fractal imaginary
cubes and their higher-dimensional counterparts.
For
k
n
n
i
=1
,
2
,...,k
n−
1
be an IFS such that
f
i
(
i
=1
,
2
,...,k
n−
1
) are homothetic transformations with the scale 1
/k
.Let
X
I
be the fractal object obtained as the fixedpoint of the contraction map
F
I
on
≥
2, let
I
=
{
f
i
:
R
ₒ
R
|
}
n
,the
sequence
B, F
I
(
B
)
,F
I
(
B
)
,...
converges to
X
I
with respect to the Hausdorff
metric. Here,
f
m
is the
m
-times repetition of
f
.
n
defined by (
4
). Since
X
I
is the fixedpoint of
F
I
, for any
B
H
∈H
Lemma 10.
Let
C
be an
n
-cube and let
(
A
i
;
i
=0
,
1
,...
)
be a sequence of
imaginary
n
-cubes of
C
. If the sequence
(
A
i
;
i
=0
,
1
,...
)
converges to
A
with
respect to the Hausdorff metric, then
A
is also an imaginary cube of
C
.
Proof.
For each projection
p
from
C
to a hyperplane containing a facet of
C
,
p
(
A
i
)for
i
=0
,
1
,...
are the same (
n
−
1)-cube
p
(
C
). Since
p
induces a continuous
H
n
to
H
n−
1
,
p
(
A
) is also equal to
p
(
C
).
map from
Proposition 11.
Let
I
be an IFS as above. The limit
X
I
is an imaginary cube
of an
n
-cube
C
if and only if
F
I
(
C
)
is an imaginary cube of
C
.
Proof.
Suppose that
X
I
is an imaginary cube of
C
.Wehave
C
Ↄ
X
I
and the
converges to
X
I
. Therefore, all of
F
I
(
C
)are
imaginary cubes of
C
. In particular,
F
I
(
C
) is an imaginary cube of
C
. Conversely,
if
F
I
(
C
) is an imaginary cube of
C
, then all of
F
I
(
C
) are imaginary cubes of
C
by
induction, and the limit
X
I
is also an imaginary cube of
C
from Lemma
10
.
F
I
(
C
)
sequence
C
Ↄ
F
I
(
C
)
Ↄ
···
The fractal object
X
I
has the similarity dimension
n
1. Note that
F
I
(
C
)isan
imaginary cube of
C
if and only if
f
i
(
C
)(
i
=1
,
2
,...,k
n−
1
)are
n
-cubes obtained
by cutting
C
into
k
n
n
-cubes of the same size and selecting
k
n−
1
of them so that
they form an imaginary
n
-cube. Such a selection of
k
n−
1
−
cubes corresponds to
an (
n
1)-dimensional Latin hypercube of order
k
. See, for example, [
9
] for the
notion of a Latin hypercube.
−
4.2 Higher-Dimensional Extensions of the Sierpinski Tetrahedron
1
2
(
C
+
n
). We set
P
n
a
n
.
Let
C
be the
n
-cube conv(
{
0
,
1
}
=
{
a
}
)for
a
∈{
0
,
1
}
There are the following two ways of selecting 2
n−
1
P
n
n
-cubes from
{
a
|
a
∈
n
{
0
,
1
}
}
to form an imaginary cube.
S
n
=
P
n
n
,
∪{
a
|
a
∈{
0
,
1
}
a
·
≡
}
,
1
0
(mod 2)
S
n
=
P
n
n
,
∪{
a
|
a
∈{
0
,
1
}
a
·
≡
}
.
1
1
(mod 2)
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