This paper investigates the maximum number of points that can be placed within a scaled cross-polytope such that the $l_1$-distance between any two points exceeds a certain threshold, providing exact bounds and constructions in various dimensions.
Contribution
It determines the maximum packing sizes within scaled cross-polytopes for specific ranges of the scale parameter in multiple dimensions, including explicit constructions.
Findings
01
Maximum of 2n points in n-dimensional cross-polytope for certain r
02
Exact maximum points in 3D for specific r ranges
03
Constructive arrangements achieving upper bounds
Abstract
The problem of finding the largest number of points in the unit cross-polytope such that the l1-distance between any two distinct points is at least 2r is investigated for r∈(1−n1,1] in dimensions ≥2 and for r∈(21,1] in dimension 3. For the n-dimensional cross-polytope, 2n points can be placed when r∈(1−n1,1]. For the three-dimensional cross-polytope, 10 and 12 points can be placed if and only if r∈(53,32] and r∈(74,53] respectively, and no more than 14 points can be placed when r∈(21,74]. Also, constructive arrangements of points that attain the upper bounds of 2n, 10, and 12 are provided, as well as 13 points for dimension 3 when r∈(21,116].
Equations198
(x+intK)∩(y+intK)=∅.
(x+intK)∩(y+intK)=∅.
γ(L,K,r):=max{∣D∣∣D⊂L and ∣∣x−y∣∣K≥2r for any x,y∈D,x=y},
γ(L,K,r):=max{∣D∣∣D⊂L and ∣∣x−y∣∣K≥2r for any x,y∈D,x=y},
k(21Cn∗)=max{∣D∣D∪{0} is a packing set for 21Cn∗ and (x+21Cn∗)∩21Cn∗=∅ for any x∈D},
k(21Cn∗)=max{∣D∣D∪{0} is a packing set for 21Cn∗ and (x+21Cn∗)∩21Cn∗=∅ for any x∈D},
k(Cn∗)+1=k(21Cn∗)+1≤γ(Cn∗,21).
k(Cn∗)+1=k(21Cn∗)+1≤γ(Cn∗,21).
k(K)≤3n−1,
k(K)≤3n−1,
k(K)≥n2+n,
k(K)≥n2+n,
k(Cn∗)≥(89)(1−o(1))n
k(Cn∗)≥(89)(1−o(1))n
k(Cn∗)≥1.13488(1−o(1))n
k(Cn∗)≥1.13488(1−o(1))n
γ(Cn∗,r)=1
γ(Cn∗,r)=1
γ(Cn∗,r)=2n
γ(C3∗,r)=10
γ(C3∗,r)=12
12≤γ(C3∗,r)≤14
12≤γ(C3∗,r)≤14
19≤γ(C3∗,r)≤26
M(L,K,m):=sup{r∣∣D∣=m,D⊂L, and∣∣x−y∣∣K≥2r for any x,y∈D,x=y}.
M(L,K,m):=sup{r∣∣D∣=m,D⊂L, and∣∣x−y∣∣K≥2r for any x,y∈D,x=y}.
Vn(r)={v∈Vn∣there exists a p∈Pn(r)∩Sn(r) such that ∣∣v−p∣∣1<2r}
Vn(r)={v∈Vn∣there exists a p∈Pn(r)∩Sn(r) such that ∣∣v−p∣∣1<2r}
∣Pn(r)∩Sn(r)∣=∣Ran(f)∣≤∣Dom(f)∣≤∣Vn(r)∣=2n,
∣Pn(r)∩Sn(r)∣=∣Ran(f)∣≤∣Dom(f)∣≤∣Vn(r)∣=2n,
γ(Cn∗,r)≤2nfor n≥2 and r∈(1−n1,1].
γ(Cn∗,r)≤2nfor n≥2 and r∈(1−n1,1].
γ(Cn∗,r)=2nfor n≥2 and r∈(1−n1,1].
γ(Cn∗,r)=2nfor n≥2 and r∈(1−n1,1].
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TopicsComputational Geometry and Mesh Generation · Optimization and Packing Problems · graph theory and CDMA systems
Full text
The maximum number of points in the cross-polytope that form a packing
set of a scaled cross-polytope
Ji Hoon Chun111The research of the author was supported by the Deutsche Forschungsgemeinschaft (DFG) Graduiertenkolleg ”Facets of Complexity/Facetten der Komplexität” (GRK 2434).
Abstract
The problem of finding the largest number of points in the unit cross-polytope
such that the l1-distance between any two distinct points is
at least 2r is investigated for r∈(1−n1,1]
in dimensions ≥2 and for r∈(21,1] in
dimension 3. For the n-dimensional cross-polytope, 2n points
can be placed when r∈(1−n1,1]. For the three-dimensional
cross-polytope, 10 and 12 points can be placed if and only if
r∈(53,32] and r∈(74,53]
respectively, and no more than 14 points can be placed when r∈(21,74].
Also, constructive arrangements of points that attain the upper bounds
of 2n, 10, and 12 are provided, as well as 13 points for
dimension 3 when r∈(21,116].
1 Introduction
Let K and L be origin-symmetric convex sets in Rn
with nonempty interiors. A set D⊂Rn is a (translative)
packing set for K if, for all distinct x,y∈D,
[TABLE]
For r>0, we consider the problem of finding the maximum number
of points in a packing set D of rK that is contained in L.
This quantity will be denoted by
[TABLE]
where ∣∣x∣∣K=min{λ∣λ≥0 and x∈λK}
and for a set S, its cardinality is denoted by ∣S∣.
If K=L then we use the notation γ(K,r) as a
shorthand for γ(L,K,r). We will only deal with
the situation where both K and L are the unit cross-polytope
Cn∗={x∈Rn∣∑i=1n∣xi∣≤1}.
A set D in Cn∗ with k points such that the l1-distance
between any two distinct points is greater than or equal to 2r
is equivalent to a packing of rCn∗+D such that each ball
is contained inside the set (1+r)Cn∗. Unless otherwise
specified, we will use “distance” to mean the l1-distance.
The vertices of Cn∗ are the 2n unit vectors {±ei∣i∈{1,…,n}},
so the distance between any two distinct vertices is 2, which implies
that γ(Cn∗,r)=1 for any r>1 and γ(Cn∗,r)≥2n
for r≤1.
The case r=21 is related to the topic of kissing numbers.
The (translative) kissing number k(K) of a convex body
K is the maximum number of translates of K such that no two
translates overlap each other and each translate touches but does
not overlap with K. In the case of 21Cn∗, we
have
[TABLE]
and since k(K) is invariant under the scaling of K,
[TABLE]
For the cross-polytope, it is known that k(C3∗)=18
[11, 16]. This result implies that γ(C3∗,21)≥19,
however, due to the requirement that one point is the origin, it does
not a priori provide an upper bound for any packing set for
21C3∗. An upper bound for the kissing number of
any convex body K is obtained from a result of Hadwiger [9],
[TABLE]
where this inequality is an equality iff K is a parallelepiped.
The cross-polytope is not a parallelepiped, so k(C3∗)≤25,
which results in an upper bound of γ(C3∗,21)≤26.
In the other direction, a well-known result by Swinnerton-Dyer [15]
says that
[TABLE]
and in the case when K is a cross-polytope the lower bound has
been improved to
by [17], which is the best known asymptotic lower bound
for the cross-polytope.
We will work with values of r only in the interval (21,1]
unless otherwise stated. For r∈(1−n1,1],
the upper bound for the number of points in the cross-polytope such
that the distance between any two distinct points is at least 2r
is linear in the dimension of the cross-polytope.
Theorem 1.1**.**
Let n≥2, then γ(Cn∗,r)=2n for any r∈(1−n1,1].
Additionally, (1−n1,1] is the largest possible
interval such that γ(Cn∗,r)=2n for all r
in the interval.
In particular, γ(C3∗,r)=6 for r∈(32,1].
The next theorem is specific to the three-dimensional case.
Theorem 1.2**.**
In dimension 3,
(a)
γ(C3∗,r)=10* for any r∈(53,32],*
(b)
γ(C3∗,r)=12* for any r∈(74,53],
and*
(c)
γ(C3∗,r)≤14* for r∈(21,74].*
For the case n=3 and r∈(21,74],
we could not find exact values of γ(C3∗,r),
but we do have lower bounds. Since γ(C3∗,r′)≥γ(C3∗,r)
for r′<r, it follows immediately from Theorem 1.2 (b) is γ(C3∗,r)≥12
for r∈(21,74]. It is possible to
improve this lower bound for a smaller interval of r.
Proposition 1.3**.**
In dimension 3, γ(C3∗,r)≥13 for r∈(21,116].
These lower bounds are obtained by explicit constructions. It follows
from Theorems 1.1 and 1.2, Proposition 1.3, and the above discussion
that
[TABLE]
Below is a chart of the results for dimension 3 in addition to
the upper and lower bounds for r=21 mentioned above.
A closely related topic is the problem of packing a set with copies
of another set. This problem has been explored mainly in dimension
2. Let K and L be origin-symmetric convex sets with nonempty
interior, then let
[TABLE]
From this definition, M(L,K,m) and γ(L,K,r)
are related by the equation
[TABLE]
The compendium of Goodman, O’Rourke, and Tóth [6]
lists known quantities of M(L,B2,m) for when L
is a square, a circle, and an equilateral triangle and various values
of m, usually small. In three dimensions, M(L,B3,m)
is known for small m and when L is a cube, a cross-polytope,
and a tetrahedron [6]. A related problem
of packings of squares and rectangles in squares is described in [4].
Let Bn be the unit Euclidean ball in n dimensions. Böröczky
Jr. and Wintsche have obtained M(Cn∗,Bn,m)
for n≥3 and m={3,…,2n+1} [1].
Let K be a convex set, B be a bounded convex set, s>0, and
let D(s,K,B) be a packing set of Cn∗ such that
∣{x∈D(s,K,B)∣K+x⊆sB}∣
is maximal among all packing sets of Cn∗. The** **density
of the densest packing of K, or the packing density of K, is
defined to be
[TABLE]
see Definition 4 in Section 20 of [8]
(page 225), and it is independent of B. Then we can set B=Cn∗
and also suppose that K=Cn∗. For s>1, since Cn∗+x⊆sCn∗
iff x∈(s−1)Cn∗, we have ∣{x∈D(s,K,B)∣Cn∗+x⊆sCn∗}∣=∣{x∈D(s,K,B)∣x∈(s−1)Cn∗}∣.
Next, scale this set by a factor of s−11 to get ∣{x∈D(s,K,B)∣x∈(s−1)Cn∗}∣={x∈s−11D(s,K,B)x∈Cn∗}=γ(Cn∗,s−11).
Now let r=s−11. It follows from the definition of the
packing density that
[TABLE]
Hence the packing density of Cn∗ is related to γ(Cn∗,r)
in the sense that γ(Cn∗,r)(1−1+r1)n∼δ(Cn∗)
as r→0.
We now mention some related results involving circle packings in a
circle and sphere packings in a cylinder. For the problem of sphere
packing inside a cylinder of fixed width in three dimensions, Fu et
al. [5] predict that as
the radius of the spheres approach zero, densest packings resemble
the face-centered cubic lattice—a densest sphere packing in three
dimensions [2]—except for the spheres that
are near the walls of the cylinder. In the case of dimension two the
densest circle packing is generated by the hexagonal lattice [2].
Hopkins, Stillinger, and Torquato [10]
provide examples of this phenomenon for dense packings of circles
inside a large circle under the condition that the large circle has
the same center as one of the small circles. Schürmann [13, 14]
has shown that under certain conditions the best finite packings of
strictly convex bodies can only be obtained using nonlattice packings.
Other dense arrangements of k circles within a large circle include
modified wedge hexagonal packings and curved hexagonal packings [10],
which are the best known packings for some values of k [12].
Basic facts about convexity and the cross-polytope can be found in
books such as the ones from Gruber [7], Ziegler [18],
and Coxeter [3], and about packings are in Conway
and Sloane [2], Gruber [7], and
Zong [19]. Additional details on the kissing number are
also in Zong [19].
Section 2 of this paper provides the notation and preliminaries that
will be used for the rest of the text. Section 3 contains the proof
of the n-dimensional case, Theorem 1.1. Section 4 proves the equalities
and upper bound present in the three parts of the 3-dimensional
case, Theorem 1.2, introducing additional notation as needed. Theorem
1.3 is proved in Section 5, and finally Section 6 presents a gallery
of diagrams related to these lower bounds.
2 General notation and preliminaries
Here we introduce notation that will be used over the course of this
paper. For a given r>0, let Pn(r)⊂Cn∗
be a packing set of rCn∗. For any polytope K, let vert(K)
be the set of its vertices. For a fixed n∈N, define
sets Vn and Sn(r) as follows:
[TABLE]
and
[TABLE]
Therefore Sn(r) is the set of all points in Cn∗
that are of distance <2r from some vertex of Cn∗.
For p∈Rn and r>0, we use the notation
[TABLE]
to denote the interior of the cross-polytope centered at p
and scaled by the factor r.
The following lemma is necessary for the general n-dimensional
case.
Lemma 2.1**.**
Let r∈(0,1]. For each j∈{1,…,n},
[TABLE]
where X is the closure of X, and similarly for −ej
instead of ej.
Proof.
Without loss of generality we take the ej case. Let
y=∑i=1nyiei∈Cn∗∩C(ej,2r),
then ∣∣y∣∣1≤1 and ∣∣y−ej∣∣1≤2r.
Then the distance from y to (1−r)ej
is
[TABLE]
If yj+r−1≥0 then since ∑i=1n∣yi∣≤1,
[TABLE]
Similarly, if yj+r−1<0 then since ∣∣y−ej∣∣1≤2r,
[TABLE]
So ∣∣y−(1−r)ej∣∣1≤r,
or in other words, y∈{x∈Rn∣∣∣x−(1−r)ej∣∣1≤r}=C((1−r)ej,r).
∎
3 Proof of Theorem 1.1 (the n-dimensional case)
In this section we assume that n≥2. We will show that for any
r∈(1−n1,1] and any packing set Pn(r),
the number of points in Pn(r)∩Sn(r)
is bounded above by the number of vertices of Cn∗. Then ∣Pn(r)∣≤2n
and this inequality is true for all Pn(r), so γ(Cn∗,r)≤2n.
As mentioned in the introduction, the set of vertices Vn⊂Cn∗
is a packing set of rCn∗, which means that 2n is also
a lower bound, and so γ(Cn∗,r)=2n.
Lemma 3.1**.**
Let r∈(1−n1,1], then Cn∗=Sn(r).
Proof.
By definition, Sn(r)⊆Cn∗, and so it
remains to show the reverse inclusion. Let x∈Cn∗
and without loss of generality it can be assumed that x
is in the convex hull of 0,e1,…,en.
Then x=∑i=1nxiei with 0≤xi≤1
and ∑i=1nxi≤1. Then there exists some j∈{1,…,n}
such that xj≥n1∑i=1nxi, so
[TABLE]
So every point in Cn∗ is within distance 2r from some
vertex of Cn∗.
∎
The next lemma will be crucial for showing that the number of points
in Pn(r)∩Sn(r) is bounded above
by the number of vertices of Cn∗. It is a uniqueness condition
which shows that if a point p∈Pn(r)∩Sn(r)
is close to a vertex v of Cn∗, specifically ∣∣p−v∣∣1<2r,
then no other point in Pn(r) can be close to v.
Lemma 3.2**.**
Let r∈(0,1]. If a vertex v of Cn∗
has the property that v∈C(p,2r)∩C(q,2r)
for some p,q∈Pn(r)∩Sn(r),
then p=q.
Proof.
Without loss of generality, let v=ej for some
j∈{1,…,n}, then by hypothesis ej∈C(p,2r)∩C(q,2r).
It suffices to show that ∣∣p−q∣∣1<2r
since the distance between two distinct points in Pn,r must
be 2r or greater. . Then in turn, p,q∈C(ej,2r).
Since C(ej,2r) is open there exists a r′<r
(r′ depends on p and q) such that p,q∈C(ej,2r′).
Then it follows from Lemma 2.1 applied to Cn∗∩C(ej,2r′)
that
[TABLE]
so ∣∣p−q∣∣1<2r.
∎
The following lemma will be used both here and in the 3-dimensional
cases in the next section.
Lemma 3.3**.**
Let r∈(0,1], then
[TABLE]
Proof.
Define Vn(r) by
[TABLE]
(this set may be empty) and a map f:Vn(r)→Pn(r)∩Sn(r)
where f(v) is the point p∈Pn(r)∩Sn(r)
such that ∣∣v−p∣∣1<2r.
First we need to show that f is well-defined. Let p,q∈Pn(r)∩Sn(r)
be points such that v∈C(p,2r)∩C(q,2r)
for some v∈Vn, then p=q by
Lemma 3.2, which justifies the use of the words “the point” in
the definition of f. From the definition of Sn(r),
every point p∈Pn(r)∩Sn(r)
has the property that there is some v∈Vn such that
∣∣p−v∣∣1<2r, so f
is surjective. Both the domain and range of f are finite sets,
so the cardinality of the range can be bounded above by
[TABLE]
completing the proof.
∎
Now we prove Theorem 1.1. With the preparation above, the proof is
mostly a matter of putting together earlier lemmas.
Proof of Theorem 1.1*.*
Assume that Pn(r) is nonempty, otherwise ∣Pn(r)∣=0
and there is nothing to prove. Since r∈(1−n1,1],
it follows from Lemma 3.1 that Cn∗=Sn, so ∣Pn(r)∩Sn(r)∣
is nonempty. Then Lemma 3.3 shows that ∣Pn(r)∣=∣Pn(r)∩Sn(r)∣≤2n.
This inequality holds for any Pn(r), so
[TABLE]
The upper bound of 2n is achieved by Vn={±ei∣i∈{1,…,n}}
as a packing set of rCn∗, so
[TABLE]
The interval r∈(1−n1,1] cannot be extended
in either direction, because γ(Cn∗,r)=1 for
r>1 and in Proposition 5.1 we construct a packing set of rCn∗,
for r≤1−n1, with 2n+2 points in Cn∗. For
such r, Sn(r)⊊Cn∗ and specifically
the centroid of each facet is not in Sn(r) (cf. Subsection
4.1), so the set consisting of the 2n vertices of Cn∗
and the two centroids on opposing facets of Cn∗ is a packing
set of rCn∗. Therefore, r∈(1−n1,1]
is the largest possible interval such that γ(Cn∗,r)=2n
is true.
∎
4 Proof of Theorem 1.2 (the 3-dimensional case)
When r≤32, the set S3(r) no longer
covers all of C3∗, so unlike the n-dimensional case above,
the proofs for the three-dimensional cases require consideration of
the remainder C3∗\S3(r).
4.1 Notation and preliminaries for dimension 3
Here we collect some lemmas and notation for the three-dimensional
cases. Let r∈[21,32]. Recall that
[TABLE]
and
[TABLE]
For any σ1,σ2,σ3∈{−1,1},
define the following subsets of R3:
[TABLE]
Figure 4.1. The grey cross-polytope is the set C3∗,
the blue spheres are the points of V(2011,(1,1,1)),
and the purple spheres are the points of V(2011,(σ1,σ2,σ3))
for σ1,σ2,σ3∈{−1,1} and
not all equal to 1.
The endpoints of the range r∈[21,32]
are 32 and 21. For r=32 the set
reduces to
[TABLE]
the centroid of the facet conv{e1,e2,e3},
and when r=21 the set is
[TABLE]
which contains the midpoints of the edges of the facet conv{e1,e2,e3}.
Subsets defined using midpoints of edges are used to solve the related
problems of finding upper bounds for k(C3∗) and
M(Cn∗,Bn,m) for some values of n and m.
To find the kissing number of the cross-polytope, Larman and Zong
[11] divided the boundary of the cross-polytope
into the union of 18 subsets including sets of the form
[TABLE]
where m is a midpoint of an edge in V3, and showed
that each subset could contain the center of at most one cross-polytope,
resulting in k(C3∗)≤18. Another method to prove
that k(C3∗)≤18 was used by Talata [16],
who showed that any packing set achieving a kissing number of 18
must consist of six points on the vertices, six points on the midpoints
of the edges of two opposing facets, and the remaining points on the
hexagon passing through the midpoints of the other edges. Böröczky
Jr. and Wintsche [1] use sets defined by vertices
and midpoints of edges to determine an upper bound for M(Cn∗,Bn,m)
where n≥3 and m∈{4,…,2n}.
For a packing set P3(r) and a set conv(V(r,(σ1,σ2,σ3))),
σ1,σ2,σ3∈{−1,1}, call
conv(V) a blocked set of P3(r) if
conv(V(r,(σ1,σ2,σ3)))
does not contain any points of P3(r).
First we show that C3∗\S3(r) can
be written in terms of V(r,(σ1,σ2,σ3)).
Lemma 4.1**.**
Let r∈(21,32]. For any σ1,σ2,σ3∈{−1,1},
define the following subsets of C3∗:
[TABLE]
Then R(r,(σ1,σ2,σ3))=conv(V(r,(σ1,σ2,σ3)))
and
[TABLE]
Proof.
Without loss of generality, assume that σ1=σ2=σ3=1,
and we will show that R(r,(1,1,1))=conv(V(r,(1,1,1))).
The set R(r,(1,1,1)) is the subset of the
unit cross-polytope with all nonnegative coordinates and excluding
the sets C(ei,2r) for i∈{1,2,3},
and the set conv(V(r,(1,1,1)))
is the intersection of the inequalities −x1−x2+x3≥−(2r−1),
x1−x2−x3≤−(2r−1), −x1+x2−x3≤−(2r−1),
and x1+x2+x3≤1, since the four points in V(r,(1,1,1))
satisfy each inequality. We will show that R(r,(1,1,1))
is also the intersection of these inequalities.
Figure 4.2. The grey cross-polytope is the set C3∗,
the green cross-polytopes are the sets C(x,2011)
for x∈V3, and the blue spheres are the points of
V(2011,(1,1,1)).
Let x∈R(r,(1,1,1)). Then x∈C3∗
so x1+x2+x3≤1, and in addition, x∈/C(e1,2r)
so
[TABLE]
Similarly, x∈/C(e2,2r) and
x∈/C(e3,2r) so −x1+x2−x3≤−(2r−1)
and x1+x2−x3≥2r−1. That proves R(r,(1,1,1))⊆conv(V(r,(1,1,1))).
For the converse, let x∈conv(V(r,(1,1,1))).
Since 21≤r≤32, both 2r−1≥0 and 1−r≥0,
so the four points in V(r,(1,1,1)) all have
nonnegative coordinates. Also, ∣∣v∣∣1≤1
for all v∈V(r,(1,1,1)), and since
x is in the convex hull of V(r,(1,1,1)),
it is also true that x1+x2+x3≤1. Then x1,x2,x3≥0
and x1+x2+x3≤1 imply that x∈Cn∗.
Also, x satisfies x1−x2−x3≤−(2r−1),
then
[TABLE]
where the last inequality holds because x2,x3≥0 and 0≤x1≤21,
so x∈/C(e1,2r). Similarly,
−x1+x2−x3≤−(2r−1) and −x1+x2−x3≤−(2r−1)
so x∈/C(e2,2r) and x∈/C(e3,2r).
Hence
[TABLE]
To complete the proof of the lemma, note that for any x∈R3,
let σi=∣xi∣xi if xi=0
and σi=1 if xi=0, then σ1x1,σ2x2,σ3x3≥0,
so C3∗\S3(r) is indeed covered by
all the R(r,(σ1,σ2,σ3)),
σ1,σ2,σ3∈{−1,1}, resulting
in
[TABLE]
∎
From this lemma, the cross-polytope C3∗ is the union of S3(r)
and the eight regions conv(V(r,(σ1,σ2,σ3))).
By Lemma 3.3,
[TABLE]
but for r∈(0,32], some points of P3(r)
may be contained in one or more of the sets conv(V(r,(σ1,σ2,σ3))).
For r∈(21,32], each V(r,(σ1,σ2,σ3))
cannot contain more than one point of P3(r), which
gives an upper bound of ∣P3(r)∣≤14
proved in Subsection 4.4. When r∈(74,32],
the required minimum distance between points of P3(r)
is large enough so that the presence of a point of P3(r)
in one set conv(V(r(σ1,σ2,σ3)))
may imply that one other set conv(V(r(σ1′,σ2′,σ3′))),
(σ1,σ2,σ3)=(σ1′,σ2′,σ3′),
cannot contain any points in P3(r). Then it is possible
to obtain an upper bound of 12, and the proof in Subsection 4.3
uses a more complicated argument involving the position of p
in V(r,(σ1,σ2,σ3)).
In Subsection 4.2 we prove that when r∈(53,32],
a point p∈P3(r)∩V(r,(σ1,σ2,σ3))
implies that three other sets of the form conv(V(r(σ1,σ2,σ3)))
cannot contain any points in P3(r).
4.2 Proof of Theorem 1.2 (a) (the r∈(53,32]
case)
Lemma 4.2**.**
Let r∈(53,32] and p∈P3(r).
If p∈conv(V(r,(σ1,σ2,σ3)))
for any σ1,σ2,σ3∈{1,1},
then conv(V(r,(−σ1,σ2,σ3))),
conv(V(r,(σ1,−σ2,σ3))),
and conv(V(r,(σ1,σ2,−σ3)))
are blocked sets of P3(r).
Proof.
Without loss of generality, assume that σ1=σ2=σ3=1.
To show that conv(V(r,(−σ1,σ2,σ3)))
is a blocked set of P3(r), it suffices to show that
∣∣p−y′∣∣1<2r for all
[TABLE]
then by the convexity of V(r,(−1,1,1)),
the statement ∣∣p−y∣∣1<2r
holds true for any y∈conv(V(r,(−1,1,1))).
The calculations are as follows:
[TABLE]
[TABLE]
and similarly
[TABLE]
and
[TABLE]
By the symmetry of V(r,(−1,1,1)), V(r,(1,−1,1)),
and V(r,(1,1,−1)), it follows that ∣∣p−y∣∣1<2r
for any y∈V(r,(−1,1,1))∪V(r,(1,−1,1))∪V(r,(1,1,−1)),
and so these three sets are blocked sets of P3(r).
∎
If P3(r)∩(C3∗\S3(r))=∅
then trivially every set of the form conv(V(r,(σ1,σ2,σ3))),
σ1,σ2,σ3∈{−1,1}, is a
blocked set of P3(r). Otherwise, the above lemma
implies that for any given P3(r), three of the eight
sets of the form conv(V(r,(σ1,σ2,σ3)))
are blocked sets of P3(r). Therefore,
[TABLE]
However, it is possible to lower the 5 to a 4 with the following
argument.
Lemma 4.3**.**
Let r∈(53,32]. Then for any P3(r),
there exist at least four blocked sets of P3(r).
Proof.
Let p∈P3(r)∩(C3∗\S3(r)).
Without loss of generality assume that there is a p∈V(r,(1,1,1)).
Then by Lemma 4.3, V(r,(−1,1,1)), V(r,(1,−1,1)),
and V(r,(1,1,−1)) are blocked sets of P3(r).
Consider the set V(r,(−1,−1,−1)). If it is
a blocked set, then there is nothing more to prove. If it is not,
then again by Lemma 4.3, V(r,(1,−1,−1)),
V(r,(−1,1,−1)), and V(r,(−1,−1,1))
are blocked sets of P3(r), resulting in a total of
six blocked sets.
∎
With this lemma we can prove Theorem 1.2 (a).
Proof of Theorem 1.2 (a)**.
Let r∈(53,32]. As in the proof
of Theorem 1.2 (c), we split up P3(r) into P3(r)∩S3(r)
and P3(r)∩(C3∗\S3(r)),
then
[TABLE]
By Lemma 4.4,
[TABLE]
which, when combined with the previous inequality, gives
[TABLE]
This inequality holds for any P3(r), so
[TABLE]
From Proposition 5.2 below, there is a 10-point packing set for
rC3∗ contained in C3∗. So this upper bound is the
best possible, giving
[TABLE]
∎
4.3 Proof of Theorem 1.2 (b) (the r∈(74,53]
case)
The following additional notation will be used in this section. For
each r>0 and σ1,σ2,σ3∈{−1,1},
define the following sets V(r,(σ1,σ2,σ3)):
[TABLE]
They have the property that
[TABLE]
and the numbering of these subsets is so that the set V(r,(σ1,σ2,σ3),i)
contains the point in the set
[TABLE]
that is furthest away from the vertex σiei.
Lemma 4.4**.**
Let r∈(74,53] and p∈P3(r).
If x∈conv(V(r,(σ1,σ2,σ3)))
then there is a blocked set conv(V(r,(σ1′,σ2′,σ3′)))
of P3(r), with (σ1,σ2,σ3)
and (σ1′,σ2′,σ3′) differing
by exactly one coordinate.
Proof.
Without loss of generality, assume that σ1=σ2=σ3=1,
then p is in one of the subsets conv(V(r,(1,1,1),i))
for i∈{1,2,3}. Assume that p∈conv(V(r,(1,1,1),1))
and write p=∑i=13piei. We will
show that ∣∣x−y∣∣1<2r
for any x∈V(r,(1,1,1),1) and
y∈V(r,(−1,1,1)), and then by
the convexity of conv(V(r,(1,1,1),1))
and conv(V(r,(−1,1,1))), it follows
that ∣∣p−y∣∣1<2r for
any y∈conv(V(r,(−1,1,1))),
which shows that conv(V(r,(−1,1,1)))
is a blocked set of P3(r). This approach is similar
to the proof of the previous lemma, but the same approach cannot be
used here as the second calculation in the proof of Lemma 4.3 ends
with 6−8r<2r, which is not true for r≤53. There
are 20 different combinations of points but not all of them need
to be explicitly checked. To keep track of the cases, we use the following
grid:
[TABLE]
For each k∈{1,…,20}, case k
corresponds to the calculation of ∣∣x−y∣∣1,
where x is the element of V(r,(1,1,1),1)
in the same row as k and y is the element of V(r,(−1,1,1))
in the same column as k. For example, (2r−1,2r−1,2r−1)T−(−(2r−1),2r−1,2r−1)T1
will be calculated in case 1 below. Cases that are similar to previous
cases will be pointed out as they arise.
∎
Since 2r−1,1−r<21<r, it immediately follows that
[TABLE]
2. 2.
[TABLE]
3. 3.
By symmetry, this case is similar to case 2.
4. 4.
[TABLE]
5. 5.
[TABLE]
6. 6.
[TABLE]
7. 7.
This case follows from case 6 due to symmetry.
8. 8.
This case follows from the same argument used in case 1, that 2r−1,1−r<21<r.
9. 9.
[TABLE]
10. 10.
[TABLE]
11. 11.
[TABLE]
12. 12.
[TABLE]
13. 13.
By symmetry, this case is similar to case 9.
14. 14.
By symmetry, this case is similar to case 11.
15. 15.
By symmetry, this case is similar to case 10.
16. 16.
By symmetry, this case is similar to case 12.
17. 17.
[TABLE]
18. 18.
[TABLE]
19. 19.
By symmetry, this case is similar to case 18.
20. 20.
[TABLE]
Proof.
Hence conv(V(r,(−1,1,1))) is
a blocked set of P3(r). By symmetry, if p∈conv(V(r,(1,1,1),2))
or p∈conv(V(r,(1,1,1),3))
then calculations similar to the above can be performed with y∈conv(V(r,(1,−1,1)))
or y∈conv(V(r,(1,1,−1)))
respectively.
∎
Using the above lemma and the same argument as after Lemma 4.3 in
the last subsection, we have
[TABLE]
However, just like in the previous subsection it is possible to lower
the 7 to a 6 with the following argument.
Lemma 4.5**.**
Let r∈(74,53]. Then for any P3(r),
there exist at least two blocked sets of P3(r).
Proof.
Let conv(V(r,(σ1,σ2,σ3))),
σ1,σ2,σ3∈{−1,1} be a blocked
set of P3(r) and consider the set conv(V(r,(−σ1,−σ2,−σ3))).
If conv(V(r,(−σ1,−σ2,−σ3)))
is a blocked set of P3(r) then we are done. Otherwise,
by Lemma 4.3 there must be a blocked set conv(V(r,(σ1′,σ2′,σ3′)))
of P3(r) such that (−σ1,−σ2,−σ3)
and (σ1′,σ2′,σ3′) differ by
exactly one coordinate. Then (σ1′,σ2′,σ3′)=(σ1,σ2,σ3),
which means that conv(V(r,(σ1,σ2,σ3)))
and conv(V(r,(σ1′,σ2′,σ3′)))
are two distinct blocked sets of P3(r).
∎
The proof of Theorem 1.2 (b) is virtually identical to the proof of
Theorem 1.2 (a).
Proof of Theorem 1.2 (b)**.
Let r∈(74,53]. As in the proof
of Theorem 1.2 (c), we split up P3(r) into P3(r)∩S3(r)
and P3(r)∩(S3∗\S3(r)),
then
[TABLE]
By Lemma 4.6,
[TABLE]
which, when combined with the previous inequality, gives
[TABLE]
This inequality holds for any P3(r), so
[TABLE]
From Proposition 5.3 below, there is a 12-point packing set for
rC3∗ contained in C3∗. So this upper bound is the
best possible, giving
[TABLE]
∎
4.4 Proof of Theorem 1.2 (c) (the r∈(21,74]
case)
For r∈(21,74], we will use an approach
that has similarities to Larman and Zong [11] and
Böröczky Jr. and Wintsche [1] in that the maximum
distance between any two points in conv(V(r,(σ1,σ2,σ3)))
is less than 2r. Then each conv(V(r,(σ1,σ2,σ3)))
can contain at most one point of P3(r), and since
S3∗\S3(r)=⋃σ1,σ2,σ3∈{−1,1}conv(V(r,(σ1,σ2,σ3))),
the number of points of P3(r) in S3∗\S3(r)
is bounded above by 8.
Lemma 4.6**.**
Let r∈(21,74] and σ1,σ2,σ3∈{−1,1}.
For any two points x,y∈conv(V(r,(σ1,σ2,σ3))),
∣∣x−y∣∣1<2r.
Proof.
Without loss of generality, let σ1=σ2=σ3=1,
then x,y∈conv(V(r,(1,1,1))).
It suffices to show that the distance between any two points in V(r,(1,1,1))
is less than 2r, then the conclusion for all points in conv(V(r,(1,1,1)))
follows by the convexity of conv(V(r,(1,1,1))).
We also assume that the two points are distinct. Suppose that neither
point is (2r−1,2r−1,2r−1)T, where vT
is the transpose of v, then both points are permutations
of (1−r,1−r,2r−1)T, so the distance between
the two points is
[TABLE]
If one of the points is (2r−1,2r−1,2r−1)T,
then the other point must be a permutation of (1−r,1−r,2r−1)T,
so the distance between the two points is
[TABLE]
∎
Below is the proof for Theorem 1.2 (c).
Proof of Theorem 1.2 (c)**.
Let r∈(21,74]. By Lemma 3.3,
[TABLE]
Write P3(r) as the union of two sets P3(r)∩S3(r)
and P3(r)∩(S3∗\S3(r)),
whose cardinalities can be individually bounded above. In particular,
by Lemma 4.1 the latter set can be expressed as
[TABLE]
An immediate consequence of Lemma 4.5 is that
[TABLE]
for all σ1,σ2,σ3∈{−1,1},
which, when combined with the previous inequality, gives
[TABLE]
This inequality holds for any P3(r), so
[TABLE]
∎
We are not able to find the exact value of γ(C3∗,r)
for such r, but some lower bounds are in Section 5.
5 Constructive lower bounds including the proof of Proposition 1.3
In contrast to the upper bounds, the lower bounds are all obtained
by explicit constructions of points in the cross-polytope. For n=3
and r∈(21,32], all of the constructions
shown here contain the six points of V3 and the remaining points
are in the union of the eight sets conv(V(r,(σ1,σ2,σ3))).
There are no claims of uniqueness made here; more than one set of
points may achieve the lower bounds of Theorem 1.3.
The calculations in the proofs below can be performed by hand or using
a computer.
Proposition 5.1**.**
Let qn=(n1,…,n1)T∈Rn.
Then Vn∪{±qn}⊂C3∗
and for r∈(0,1−n1],
[TABLE]
is a packing set of rCn∗.
Proof.
Any points x,y∈Vn∪{±qn},
x=y, have the property that ∣∣x∣∣1≤1
and ∣∣x−y∣∣1≤2(1−n1),
so Vn∪{±qn}⊂C3∗
is a packing set of rCn∗ for r≤1−n1.
∎
Proposition 5.2**.**
Let
[TABLE]
Then V3∪Q10⊂C3∗ and for r∈(0,32],
[TABLE]
is a packing set of rC3∗.
Proof.
Any points x,y∈V3∪Q10, x=y,
have the property that ∣∣x∣∣1≤1
and ∣∣x−y∣∣1≤34,
so V3∪Q10⊂C3∗ is a packing set of rCn∗
for r≤32.
∎
Proposition 5.3**.**
Let
[TABLE]
Then V3∪Q12+∪(−Q12+)⊂C3∗
and for r∈(0,53],
[TABLE]
is a packing set of rC3∗.
Proof.
Any points x,y∈V3∪Q12+∪(−Q12+),
x=y, have the property that ∣∣x∣∣1≤1
and ∣∣x−y∣∣1≤56,
so V3∪Q12+∪(−Q12+)⊂C3∗
is a packing set of rCn∗ for r≤53.
∎
Finally we consider the case r∈(0,116]. The
construction below differs from the previous constructions as there
are no obvious large-scale symmetries.
Proposition 5.4**.**
Let
[TABLE]
Then V3∪Q13⊂C3∗ and for r≤116,
[TABLE]
is a packing set of rC3∗.
Proof.
Any points x,y∈V3∪Q13, x=y,
have the property that ∣∣x∣∣1≤1
and ∣∣x−y∣∣1≤1112,
so V3∪Q13⊂C3∗ is a packing set of rCn∗
for r≤116.
∎
We do not know if this result can be improved, either in the sense
of a 13-point configuration for some r>116 or a 14-point
configuration for r=116. Regarding the first avenue for
improvement, the upper end of the range r∈(0,116]
cannot be raised without moving the points of V3∪Q13.
As for the second, according to the proof of Theorem 1.2 (c), at
most eight points in any P3(116) can be
in the sets conv(V(116,(σ1,σ2,σ3)))
for all σ1,σ2,σ3∈{−1,1}.
The packing set V3∪Q13 contains points in each set of
the form conv(V(116,(σ1,σ2,σ3))),
σ1,σ2,σ3∈{−1,1}, except
for conv(V(116,(1,1,1))),
see Figure 6.5. Since
[TABLE]
and the distances from each point in this set to 111(−1,5,5)T∈V3∪Q13
are
[TABLE]
[TABLE]
[TABLE]
and
[TABLE]
it follows that the distance from any point in conv(V(116,(1,1,1)))
to conv(V(116,(1,1,1)))
is less than 1. Therefore, a 14-point packing set of 116C3∗
is not possible without moving one or more of the points in the subset
[TABLE]
Proof of Proposition 1.3*.*
** By Proposition 5.4, the set V3∪Q13 is a subset
of Cn∗ with 13 points and is a packing set for rC3∗
where r∈(21,116]. Therefore
[TABLE]
∎
6 Diagrams of cross-polytope packings
Below are graphs showing the unit cross-polytope with cross-polytopes
of radius r around each point of V3, V3∪Q10,
V3∪Q12+∪(−Q12+), and V3∪Q13.
For each diagram except the first one, the value of r in the diagram
is the largest possible for that configuration of points. In each
diagram the grey cross-polytope in the middle is C3∗.
V3: 6 points in C3∗
V3 is a packing set of rC3∗ for all 0<r≤1.
Figure 6.1. The green cross-polytopes represent the sets C(x,109)
for x∈V3.
V3∪Q10: 10 points in C3∗
V3∪Q10 is a packing set of rC3∗ for all 0<r≤32.
Figures 6.2 (left) and 6.3 (right). The green cross-polytopes
represent the sets C(x,32)
for x∈V3 and the blue cross-polytopes represent
the sets C(y,32) for y∈Q10.
V3∪Q12+∪(−Q12+):
12 points in C3∗
V3∪Q12+∪(−Q12+) is a packing set
of rC3∗ for all 0<r≤53.
Figures 6.4 (left) and 6.5 (right). The green cross-polytopes
represent the sets C(x,53)
for x∈V3 and the blue cross-polytopes represent
the sets C(y,53) for y∈V3∪Q12+∪(−Q12+).
V3∪Q13: 13 points in C3∗
V3∪Q13 is a packing set of rC3∗ for all 0<r≤116.
Figures 6.6 (top left), 6.7 (top right), 6.8 (bottom left),
and 6.9 (bottom right). The green cross-polytopes represent the sets
C(x,116) for x∈V3,
the blue cross-polytopes represent C(111(−1,5,5),116),
C(111(5,−1,5),116),
and C(111(5,5,−1),116),
the purple cross-polytopes represent C(111(−5,−2,4),116),
C(111(−5,4,−2),116),
and C(111(4,−2,−5),116),
and the magenta cross-polytope represents C(111(−3,−5,−3),116).
7 Acknowledgements
Thanks to Martin Henk for his advice, suggestions, and feedback, and
Fei Xue for feedback and discussions, especially with organizing and
simplifying the proof of the n=3 and r∈(53,32]
case.
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