ON THE INTERPOINT DISTANCE SUM
INEQUALITY
Interpoint Distance Sum Inequality
YONG XIA AND HONG-YING LIU
LMIB of the Ministry of Education
School of Mathematics and System Sciences
Beihang University, Beijing, 100191,
People’s Republic of China.
EMail: dearyxia@gmail.com
Yong Xia and Hong-Ying Liu
vol. 10, iss. 3, art. 74, 2009
Title Page
liuhongying@buaa.edu.cn
Contents
JJ
II
28 September, 2009
J
I
Communicated by:
P.S. Bullen
Page 1 of 20
2000 AMS Sub. Class.:
51D20, 51K05, 52C26.
Key words:
Combinatorial geometry, Distance geometry, Interpoint distance sum inequality,
Optimization.
Received:
10 April, 2009.
Accepted:
Go Back
Abstract:
Let n points be arbitrarily placed in B(D), a disk in R2 having diameter D.
Denote by lij the Euclidean distance between point i and j. In this paper, we
show
n X
D2
2
min lij
≤
.
j6=i
0.3972
i=1
We then extend the result to R3 .
Full Screen
Close
Contents
1
Introduction
3
2
Main Result
5
3
Extension
13
Interpoint Distance Sum Inequality
Yong Xia and Hong-Ying Liu
vol. 10, iss. 3, art. 74, 2009
Title Page
Contents
JJ
II
J
I
Page 2 of 20
Go Back
Full Screen
Close
1.
Introduction
To estimate upper bounds on the maximum number of simultaneously successful
wireless transmissions and the maximum achievable per-node end-to-end throughput under the general network scenario, Arpacioglu and Haas [1] introduced the
following interesting inequalities. For the sake of clarity in presentation, we use the
notation argminj∈J {Sj } to denote the index of the smallest point in the set {Sj }
(j ∈ J). If there are several smallest elements, we take the first one.
Interpoint Distance Sum Inequality
Yong Xia and Hong-Ying Liu
Theorem 1.1 ([1]). Let B(D) be a disk in R2 having diameter D. Let n points
be arbitrarily placed in B(D). Suppose each point is indexed by a distinct integer
between 1 and n. Let lij be the Euclidean distance between points i and j. Define
the mth closest point to point i, aim , and the Euclidean distance between point i and
the mth closest point to point i, uim , as follows:
ai1 := argmin {lij },
1 ≤ i ≤ n,
j∈{1,2,...,n},
j6=i
aim :=
argmin {lij },
1 ≤ i ≤ n, 2 ≤ m ≤ n − 1,
j∈{1,2,...,n},
m−1
j ∈{i}∪{a
/
ik }k=1
uim := liaim ,
Contents
JJ
II
J
I
1 ≤ i ≤ n, 1 ≤ m ≤ n − 1.
Go Back
Full Screen
n
X
i=1
where
Title Page
Page 3 of 20
Then
(1.1)
vol. 10, iss. 3, art. 74, 2009
u2im ≤
mD2
,
c2
Close
1 ≤ m ≤ n − 1,
√
2
3
c2 := −
≈ 0.3910.
3
2π
We observed [2] that the interpoint distance sum inequality (1.1) can be simply
yet significantly strengthened.
Proposition 1.2. Define B(D), D, n, lij , aim , uim , c2 as in Theorem 1.1. Then
(1.2)
n
X
i=1
(1.3)
n
X
u2im ≤
mD2
,
c2
1 ≤ m < c2 n,
Interpoint Distance Sum Inequality
u2im ≤ nD2 ,
c2 n < m ≤ n − 1.
Yong Xia and Hong-Ying Liu
vol. 10, iss. 3, art. 74, 2009
i=1
The proof follows from (1.1) and the fact that uim ≤ D.
As a direct application, we improved [2] the upper bounds on the maximum number of simultaneously successful wireless transmissions and the maximum achievable per-node end-to-end throughput under the same general network scenario as in
Arpacioglu and Haas [1].
Title Page
Contents
JJ
II
J
I
Page 4 of 20
Go Back
Full Screen
Close
2.
Main Result
In this section, we show that the interpoint distance sum inequality (1.1) when m = 1
can be further improved.
Theorem 2.1. Define B(D), D, n, lij , aim , uim , c2 as in Theorem 1.1. Then
n
X
u2i1 ≤
i=1
D2
.
0.3972
Proof. The case n = 2 is trivial to verify since m = 1 and uim ≤ D. So we assume
n ≥ 3. The proof is based on that of Theorem 1.1 [1]. Denote the disk of diameter x
and center i by Bi (x). Define the following sets of disks
Rm := {Bi (uim ) : 1 ≤ i ≤ n},
A(B(D) ∩ BS (uim ))
,
A(BS (uim ))
Yong Xia and Hong-Ying Liu
vol. 10, iss. 3, art. 74, 2009
Title Page
1 ≤ m ≤ n − 1.
First consider the disks in R1 . As shown in [1], all disks in R1 are non-overlapping,
i.e., the distance between the centers of any two disks is smaller than the sum of the
radii of the two disks.
Denote by A(X) the area of a region X. We try to find a lower bound on fim :=
A(B(D) ∩ Bi (uim ))/A(Bi (uim )) for every 1 ≤ i ≤ n and 1 ≤ m ≤ n − 1. Pick any
point S from the boundary of B(D) and consider the overlap ratio
S
fim
:=
Interpoint Distance Sum Inequality
Contents
JJ
II
J
I
Page 5 of 20
Go Back
Full Screen
1 ≤ i ≤ n, 1 ≤ m ≤ n − 1.
S
Using Figure 1, one can obtain the geometrical computation formula: fim
=
f (y)|y= uim , where
D
r
y 1
2
1
1 1
1
(2.1)
f (y) :=
1 − 2 arccos
+ 2−
− .
2
4
π
y
2
y
π y
Close
B (u )
S im
B(D)
uim/2
D/2
S
Interpoint Distance Sum Inequality
Yong Xia and Hong-Ying Liu
vol. 10, iss. 3, art. 74, 2009
Title Page
Contents
Figure 1: Computation of the overlap ratio between B(D) and Bs (uim ).
S
Actually f (y) is a decreasing function of y. We have fim
≥ f (1) due to uim ≤ D.
S
Also fim ≥ fim . Setting c2 := f (1), we obtain the following lower bound on fim for
every 1 ≤ i ≤ n and 1 ≤ m ≤ n − 1,
√
2
3
fim ≥ c2 , where c2 = −
≈ 0.3910.
3
2π
Therefore the area of the parts of the disks in Rm that lie in B(D) is at least c2 A(B(D)).
Hence, for every 1 ≤ i ≤ n and 1 ≤ m ≤ n − 1,
(2.2)
A(Bi (uim ) ∩ B(D)) ≥ c2 A(Bi (uim )).
JJ
II
J
I
Page 6 of 20
Go Back
Full Screen
Close
For a given value m, adding the n inequalities in (2.2), we obtain
(2.3)
n
X
A(Bi (uim ) ∩ B(D)) ≥ c2
i=1
n
X
A(Bi (uim )),
∀ 1 ≤ m ≤ n − 1.
i=1
Since all disks in R1 are non-overlapping, we have
n
X
(2.4)
A(Bi (uim ) ∩ B(D)) ≤ A(B(D)).
i=1
Interpoint Distance Sum Inequality
Yong Xia and Hong-Ying Liu
vol. 10, iss. 3, art. 74, 2009
Inequalities (2.3) and (2.4) imply
A(B(D)) ≥ c2
n
X
A(Bi (uim )).
Title Page
Contents
i=1
Notice that A(B(D)) = πD2 /4 and A(Bi (ui1 )) = πu2i1 /4. Therefore,
n
X
(2.5)
i=1
u2i1 ≤
D2
.
c2
Also, it is easy to see that f (y), defined in (2.1), is a concave function. Then f (y)
has a linear underestimation, denoted by
l(y) := c2 + k − ky,
where
f (0) − f (1)
= lim f (y) − f (1) = 0.5 − c2 ≈ 0.1090.
y→0
1−0
Figure 2 shows the variation of f (y) and l(y), respectively. Figure 3 shows the
variation of f (y) − l(y) with respect to y.
k :=
JJ
II
J
I
Page 7 of 20
Go Back
Full Screen
Close
0.5
f(y)
l(y)
0.48
f(y) and l(y)
0.46
Interpoint Distance Sum Inequality
0.44
Yong Xia and Hong-Ying Liu
vol. 10, iss. 3, art. 74, 2009
0.42
0.4
0.38
Title Page
Contents
0
0.2
0.4
0.6
0.8
1
y
JJ
II
J
I
Figure 2: Variations of f (y) and l(y).
Page 8 of 20
Now we have
Go Back
S
fim ≥ fim
=f
u
im
D
≥ c2 + k − k
uim
.
D
Close
Therefore, for every 1 ≤ i ≤ n and 1 ≤ m ≤ n − 1,
(2.6)
A(Bi (uim ) ∩ B(D)) ≥ (c2 + k)A(Bi (uim )) − k
Full Screen
uim
A(Bi (uim )).
D
−3
1.2
x 10
1
f(y)−l(y)
0.8
0.6
Interpoint Distance Sum Inequality
Yong Xia and Hong-Ying Liu
0.4
vol. 10, iss. 3, art. 74, 2009
0.2
0
0
0.2
0.4
0.6
0.8
Title Page
1
y
Contents
Figure 3: Variation of f (y) − l(y).
Adding all the n inequalities in (2.6) for a given m, we obtain
JJ
II
J
I
Page 9 of 20
n
X
Go Back
A(Bi (uim ) ∩ B(D))
i=1
≥ (c2 + k)
n
X
i=1
Full Screen
n
k X
A(Bi (uim )) −
uim A(Bi (uim )),
D i=1
∀1 ≤ m ≤ n − 1.
Using (2.4) and the facts A(B(D)) = πD2 /4 and A(Bi (ui1 )) = πu2i1 /4, we obtain
(2.7)
2
D ≥ (c2 + k)
n
X
i=1
n
u2i1
k X 3
−
u .
D i=1 i1
Close
Now consider the following optimization problem (n ≥ 3):
n
X
(2.8)
max
u3i1
i=1
(2.9)
s.t.
n
X
i=1
(2.10)
0 ≤ ui1 ≤ D,
u2i1 ≤
D2
c2
i = 1, . . . , n.
The objective function (2.8) is strictly convex and the feasible region defined by (2.9)
– (2.10) is also convex. Since n ≥ 3 and 2 < c12 < 3, the inequality (2.9) holds at
any of the optimal solutions. Therefore the optimal solutions of (2.8) – (2.10) must
occur at the vertices of the set
(
)
n
2
X
D
, 0 ≤ ui1 ≤ D, i = 1, . . . , n .
(ui1 ) :
u2i1 =
c
2
i=1
Any (ui1 ) with two components lying strictly between 0jandk D cannot be a vertex.
Therefore every optimal solution of (2.8) – (2.10) has c12 components with the
r
j k
value D, one component with the value c12 − c12 D and the others are zeros,
where bxc is the largest integer less than or equal to x. Then the optimal objective
value is
32
1
1
1
3
D +
−
D3 .
c2
c2
c2
In other words, we have proved for valid ui1 that
32
n
X
1
1
1
u3i1 ≤
D3 +
−
D3 .
c
c
c
2
2
2
i=1
Interpoint Distance Sum Inequality
Yong Xia and Hong-Ying Liu
vol. 10, iss. 3, art. 74, 2009
Title Page
Contents
JJ
II
J
I
Page 10 of 20
Go Back
Full Screen
Close
Now (2.7) becomes
(2.11)
D 2 ≥ c2
n
X
u2i1 + k
i=1
n
X
u2i1 −
i=1
32 ! !
1
1
1
D2 .
+
−
c2
c2
c2
Then we have
n
X
D
u2i1
2
1+k
j k
1
c2
≤
+
c2 1 +
i=1
1
c2
k c12
−
j k 32 1
c2
Interpoint Distance Sum Inequality
.
Comparing with (2.5), we actually obtain a new c+
2:
c2 1 + k c12
+
j k (2.12)
c2 =
j k 32 ≈ 0.3957
1
1
1+k
+ c2 − c12
c2
such that
n
X
D2
.
c+
2
u2i1 ≤
i=1
(2.13)
c
1+k
1
c(i)
+
1
c(i)
Contents
JJ
II
J
I
Go Back
0.5
=
Title Page
Page 11 of 20
Iteratively repeating the same approach, we obtain a sequence {c(i) } (i = 1, 2, . . . ),
where c(0) = c2 , c(1) = c+
2 and
(i+1)
Yong Xia and Hong-Ying Liu
vol. 10, iss. 3, art. 74, 2009
−
1
32 .
c(i)
Clearly, we can conclude that c(i) < 21 for all i since the denominator above is greater
than 1. Secondly, we prove that c(i) > 31 for all i by mathematical induction. We
Full Screen
Close
have shown that c(0) >
1
3
and c(1) > 13 . Now assume c(i) > 13 . Since
32 1
1
1
1
1
1
1
+ (i) − (i)
≤ (i) + (i) − (i)
= (i) ,
(i)
c
c
c
c
c
c
c
we have
0.5
c(i+1) =
1+k
1
c(i)
+
1
c(i)
−
1
c(i)
0.5
0.5
1
32 ≥ 1 + k > 1 + 3k > 3 .
c(i)
0.5
Yong Xia and Hong-Ying Liu
vol. 10, iss. 3, art. 74, 2009
To sum up, we obtain 13 < c(i) < 12 , which implies that
iterative formula of c(i+1) (2.13) becomes
c(i+1) =
Interpoint Distance Sum Inequality
1+k 2+
1
c(i)
1
c(i)
= 2. Therefore, the
3 .
−2 2
It is easy to verify that the sequence {c(i) } is monotone increasing with a limit value
0.3972.
Title Page
Contents
JJ
II
J
I
Page 12 of 20
Go Back
Full Screen
Close
3.
Extension
Theorem 3.1. Let B(D) be a sphere in R3 having diameter D. Let n points be
arbitrarily placed in B(D). lij , aim , uim are similarly defined as in Theorem 1.1.
Then
n
X
(3.1)
i=1
n
X
(3.2)
i=1
n
X
(3.3)
u3i1 ≤
D3
,
0.3168
Interpoint Distance Sum Inequality
Yong Xia and Hong-Ying Liu
u3im
mD3
≤
,
c3
u3im ≤ nD3 ,
2 ≤ m < c3 n,
c3 n < m ≤ n − 1,
vol. 10, iss. 3, art. 74, 2009
Title Page
Contents
i=1
where c3 = 0.3125.
JJ
II
Proof. To begin with, we prove the first inequality (3.1). The case n = 2 is trivial
since m = 1 and uim ≤ D. So we assume that n ≥ 3. The proof is based on that of
Theorem 1.1 [1]. Denote the sphere of diameter x and center i by Bi (x). Define the
following sets of spheres
J
I
Rm := {Bi (uim ) : 1 ≤ i ≤ n},
1 ≤ m ≤ n − 1.
First consider the spheres in R1 . As shown in [1], all spheres in R1 are nonoverlapping, i.e., the distance between the centers of any two spheres is smaller
than the sum of the radii of the two spheres.
Denote by A(X) the volume of a region X. We try to find a lower bound on
fim := V (B(D) ∩ Bi (uim ))/V (Bi (uim )) for every 1 ≤ i ≤ n and 1 ≤ m ≤ n − 1.
Page 13 of 20
Go Back
Full Screen
Close
Pick any point S from the boundary of B(D) and consider the overlap ratio
(3.4)
S
:=
fim
V (B(D) ∩ BS (uim ))
,
V (BS (uim ))
1 ≤ i ≤ n, 1 ≤ m ≤ n − 1.
Using a 3-dimensional version of Figure 1, one can obtain the geometrical comS
putation formula: fim
= f (y)|y= uim , where
D
Interpoint Distance Sum Inequality
1 3y
f (y) := − .
2 16
Yong Xia and Hong-Ying Liu
vol. 10, iss. 3, art. 74, 2009
S
Actually f (y) is a decreasing function of y. We have fim
≥ f (1) due to uim ≤ D.
S
Also fim ≥ fim . Setting c3 := f (1), we obtain the following lower bound on fim for
every 1 ≤ i ≤ n and 1 ≤ m ≤ n − 1,
fim ≥ c3 ,
where
c3 =
5
= 0.3125.
16
Therefore the area of the parts of the disks in Rm that lie in B(D) is at least c3 A(B(D)).
Hence, for every 1 ≤ i ≤ n and 1 ≤ m ≤ n − 1,
V (Bi (uim ) ∩ B(D)) ≥ c3 V (Bi (uim )).
(3.5)
Title Page
Contents
JJ
II
J
I
Page 14 of 20
Go Back
For a given value m, adding the n inequalities in (3.5), we obtain
(3.6)
n
X
V (Bi (uim ) ∩ B(D)) ≥ c3
i=1
n
X
Full Screen
V (Bi (uim )),
∀1 ≤ m ≤ n − 1.
i=1
Since all spheres in R1 are non-overlapping, we have
(3.7)
n
X
i=1
V (Bi (uim ) ∩ B(D)) ≤ V (B(D)).
Close
Inequalities (3.6) and (3.7) imply
V (B(D)) ≥ c3
n
X
V (Bi (uim )).
i=1
Notice that V (B(D)) = πD3 /6 and V (Bi (ui1 )) = πu3i1 /6. Therefore,
n
X
(3.8)
D3
≤
.
c3
u3i1
i=1
Interpoint Distance Sum Inequality
Yong Xia and Hong-Ying Liu
vol. 10, iss. 3, art. 74, 2009
Defining k =
3
16
= 0.1875, we have
fim ≥
S
fim
=f
u
im
≥ c3 + k − k
D
Therefore, for every 1 ≤ i ≤ n and 1 ≤ m ≤ n − 1,
(3.9)
uim
.
D
Title Page
Contents
V (Bi (uim ) ∩ B(D)) ≥ (c3 + k)V (Bi (uim )) − k
uim
V (Bi (uim )).
D
Adding the n inequalities in (3.9) for a given m, we obtain
JJ
II
J
I
Page 15 of 20
(3.10)
n
X
Go Back
V (Bi (uim ) ∩ B(D))
i=1
≥ (c3 + k)
n
X
i=1
Full Screen
n
k X
V (Bi (uim )) −
uim V (Bi (uim )),
D i=1
∀1 ≤ m ≤ n − 1.
Using (3.7) and the facts V (B(D)) = πD3 /6 and V (Bi (ui1 )) = πu3i1 /6, we have
(3.11)
3
D ≥ (c3 + k)
n
X
i=1
n
u3i1
k X 4
−
u .
D i=1 i1
Close
Now consider the following optimization problems (n ≥ 3):
max
(3.12)
n
X
i=1
n
X
u3i1
i=1
(3.13)
s.t.
(3.14)
0 ≤ ui1 ≤ D,
u4i1
D3
≤
c3
i = 1, . . . , n.
The objective function (3.12) is strictly convex and the feasible region defined by
(3.13) – (3.14) is also convex. Since n ≥ 3 and 2 < c13 < 3, the inequality (3.13)
holds at any of the optimal solutions. Therefore the optimal solutions of (3.12) –
(3.14) must occur at vertices of the set
(
)
n
X
D3
3
(ui1 ) :
ui1 =
, 0 ≤ ui1 ≤ D, i = 1, . . . , n .
c3
i=1
Any (ui1 ) with two components lying strictly between 0 jandk D cannot be a vertex.
Therefore every optimal solution of (3.12) – (3.14) has c13 components with the
r
j k
1
value D, one component with the value c3 − c13 D and the others are zeros,
where bxc is the largest integer less than or equal to x. Then the optimal objective
value is
43
1
1
1
4
D +
−
D4 .
c3
c3
c3
Interpoint Distance Sum Inequality
Yong Xia and Hong-Ying Liu
vol. 10, iss. 3, art. 74, 2009
Title Page
Contents
JJ
II
J
I
Page 16 of 20
Go Back
Full Screen
Close
In other words, we have proved for valid ui1 that
43
n
X
1
1
1
4
4
D +
−
D4 .
ui1 ≤
c
c
c
3
3
3
i=1
Now (3.11) becomes
(3.15)
D 3 ≥ c3
n
X
u3i1 + k
i=1
n
X
u3i1 −
i=1
43 ! !
1
1
1
+
−
D3 .
c3
c3
c3
Interpoint Distance Sum Inequality
Yong Xia and Hong-Ying Liu
vol. 10, iss. 3, art. 74, 2009
Then we have
n
X
D
3
1+k
u3i1 ≤
j k
1
c3
+
c3 (1 +
i=1
1
c3
−
j k 43 1
c3
.
k c13 )
Title Page
Contents
c+
3:
Comparing with (3.8), we actually obtain a new
c3 1 + k c13
j k (3.16)
c+
3 =
j k 43 ≈ 0.3156
1
1
1+k
+ c3 − c13
c3
JJ
II
J
I
Page 17 of 20
Go Back
such that
n
X
i=1
u3i1 ≤
D3
.
c+
3
Full Screen
Close
(i)
Iteratively repeating the same approach, we obtain a sequence {c } (i = 1, 2, . . . ),
where c(0) = c3 , c(1) = c+
3 and
0.5
(3.17)
c(i+1) =
1 43 .
1
1
1 + k c(i) + c(i) − c(i)
First we conclude that c(i) < 13 for all i. We prove this by mathematical induction.
1 We have c(0) = 0.3125 < 13 . Now assume that c(i) < 13 , which also implies c(i)
≥
3. Then based on (3.17), we have
0.5
c(i+1) =
1+k
≤
1
c(i)
+
1
c(i)
−
1
43 c(i)
0.5
1
0.5
1 ≤
< .
1 + 3k
3
1 + k c(i)
Interpoint Distance Sum Inequality
Yong Xia and Hong-Ying Liu
vol. 10, iss. 3, art. 74, 2009
Secondly, we prove
1
4
for all i by mathematical induction. We have shown c(0) > 41 . Now assume c(i) > 14 .
Since
43 1
1
1
1
1
1
1
≤ (i) + (i) − (i)
+ (i) − (i)
= (i) ,
(i)
c
c
c
c
c
c
c
c(i) >
Title Page
Contents
JJ
II
J
I
Page 18 of 20
we have
c
(i+1)
0.5
=
1+k
1
c(i)
+
1
c(i)
−
1
c(i)
1
0.5
0.5
43 ≥ 1 + k > 1 + 4k > 4 .
c(i)
Go Back
Full Screen
Close
To sum up, we obtain 14 < c(i) < 13 , which implies that
iterative formula (2.13) of c(i+1) becomes
c(i+1) =
0.5
1+k 2+
1
c(i)
43 .
−3
1
c(i)
= 3. Therefore, the
It is easy to verify that the sequence {c(i) } is monotone increasing with a limit value
0.3168.
Next, consider the spheres in Rm for every 2 ≤ m ≤ n − 1. In this case, there can
be overlaps between some pairs of spheres in Rm . However, as shown in [1], any
arbitrarily chosen point within B(D) can belong to at most m overlapping spheres
from Rm . Then for every 2 ≤ m ≤ n − 1, we have
n
X
Interpoint Distance Sum Inequality
V (Bi (uim ) ∩ B(D)) ≤ mV (B(D)).
i=1
Yong Xia and Hong-Ying Liu
vol. 10, iss. 3, art. 74, 2009
It follows that
3
mD ≥ c3
n
X
u3i1 .
Title Page
i=1
The last inequality (3.3) directly follows from the fact uim ≤ D.
Contents
JJ
II
J
I
Page 19 of 20
Go Back
Full Screen
Close
References
[1] O. ARPACIOGLU AND Z.J. HAAS, On the scalability and capacity of planar
wireless networks with omnidirectional antennas, Wirel. Commun. Mob. Comput., 4 (2004), 263–279.
[2] Y. XIA AND H.Y. LIU, Improving upper bound on the capacity of planar wireless
networks with omnidirectional antennas, in Baozong Yuan and Xiaofang Tang
(Eds.) Proceedings of the IET 2nd International Conference on Wireless, Mobile
& Multimedia Networks, (2008), 191–194.
Interpoint Distance Sum Inequality
Yong Xia and Hong-Ying Liu
vol. 10, iss. 3, art. 74, 2009
Title Page
Contents
JJ
II
J
I
Page 20 of 20
Go Back
Full Screen
Close