Volume 10 (2009), Issue 3, Article 74, 10 pp.
ON THE INTERPOINT DISTANCE SUM INEQUALITY
YONG XIA AND HONG-YING LIU
LMIB OF THE M INISTRY OF E DUCATION ;
S CHOOL OF M ATHEMATICS AND S YSTEM S CIENCES ,
B EIHANG U NIVERSITY, B EIJING , 100191, P.R. C HINA
dearyxia@gmail.com
liuhongying@buaa.edu.cn
Received 10 April, 2009; accepted 28 September, 2009
Communicated by P.S. Bullen
A BSTRACT. 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 .
Key words and phrases: Combinatorial geometry, Distance geometry, Interpoint distance sum inequality, Optimization.
2000 Mathematics Subject Classification. 51D20, 51K05, 52C26.
1. I NTRODUCTION
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.
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 },
j∈{1,2,...,n},
m−1
j ∈{i}∪{a
/
ik }k=1
095-09
1 ≤ i ≤ n, 2 ≤ m ≤ n − 1,
2
YONG X IA AND H ONG -Y ING L IU
1 ≤ i ≤ n, 1 ≤ m ≤ n − 1.
uim := liaim ,
Then
(1.1)
n
X
u2im ≤
i=1
mD2
,
c2
1 ≤ m ≤ n − 1,
where
√
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
n
X
mD2
(1.2)
u2im ≤
, 1 ≤ m < c2 n,
c
2
i=1
(1.3)
n
X
u2im ≤ nD2 ,
c2 n < m ≤ n − 1.
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].
2. M AIN R ESULT
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
D2
u2i1 ≤
.
0.3972
i=1
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},
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
A(B(D) ∩ BS (uim ))
S
fim
:=
,
1 ≤ i ≤ n, 1 ≤ m ≤ n − 1.
A(BS (uim ))
S
Using Figure 2.1, one can obtain the geometrical computation formula: fim
= f (y)|y= uim ,
D
where
r
y 1
2
1
1 1
1
(2.1)
f (y) :=
1 − 2 arccos
+ 2−
− .
2
π
y
2
y
π y
4
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 74, 10 pp.
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I NTERPOINT D ISTANCE S UM I NEQUALITY
3
B (u )
S im
B(D)
uim/2
S
D/2
Figure 2.1: 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. Also
S
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,
A(Bi (uim ) ∩ B(D)) ≥ c2 A(Bi (uim )).
(2.2)
For a given value m, adding the n inequalities in (2.2), we obtain
n
n
X
X
(2.3)
A(Bi (uim ) ∩ B(D)) ≥ c2
A(Bi (uim )), ∀ 1 ≤ m ≤ n − 1.
i=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
Inequalities (2.3) and (2.4) imply
A(B(D)) ≥ c2
n
X
A(Bi (uim )).
i=1
2
Notice that A(B(D)) = πD /4 and A(Bi (ui1 )) = πu2i1 /4. Therefore,
(2.5)
n
X
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,
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 74, 10 pp.
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4
YONG X IA AND H ONG -Y ING L IU
0.5
f(y)
l(y)
0.48
f(y) and l(y)
0.46
0.44
0.42
0.4
0.38
0
0.2
0.4
0.6
0.8
1
y
Figure 2.2: Variations of f (y) and l(y).
−3
1.2
x 10
1
f(y)−l(y)
0.8
0.6
0.4
0.2
0
0
0.2
0.4
0.6
0.8
1
y
Figure 2.3: Variation of f (y) − l(y).
where
f (0) − f (1)
= lim f (y) − f (1) = 0.5 − c2 ≈ 0.1090.
y→0
1−0
Figure 2.2 shows the variation of f (y) and l(y), respectively. Figure 2.3 shows the variation
of f (y) − l(y) with respect to y.
Now we have
u uim
im
S
fim ≥ fim
=f
≥ c2 + k − k
.
D
D
Therefore, for every 1 ≤ i ≤ n and 1 ≤ m ≤ n − 1,
uim
(2.6)
A(Bi (uim ) ∩ B(D)) ≥ (c2 + k)A(Bi (uim )) − k
A(Bi (uim )).
D
k :=
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http://jipam.vu.edu.au/
I NTERPOINT D ISTANCE S UM I NEQUALITY
5
Adding all the n inequalities in (2.6) for a given m, we obtain
n
X
A(Bi (uim ) ∩ B(D))
i=1
≥ (c2 + k)
n
X
i=1
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
D ≥ (c2 + k)
(2.7)
n
X
n
u2i1
i=1
k X 3
−
u .
D i=1 i1
Now consider the following optimization problem (n ≥ 3):
max
(2.8)
n
X
u3i1
i=1
n
X
u2i1
i=1
(2.9)
s.t.
(2.10)
0 ≤ ui1 ≤ D,
≤
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
(ui1 ) :
, 0 ≤ ui1 ≤ D, i = 1, . . . , n .
u2i1 =
c
2
i=1
Any (ui1 ) with two components lying strictly
0 and D cannot be a vertex. Therefore
j between
k
1
every optimal solution of (2.8) – (2.10) has c2 components with the value D, one component
r
j k
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
3
3
ui1 ≤
D +
−
D3 .
c2
c2
c2
i=1
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
D2 1 + k
u2i1 ≤
i=1
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 74, 10 pp.
j k
1
c2
+
1
c2
c2 1 + k c12
−
j k 32 1
c2
.
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YONG X IA AND H ONG -Y ING L IU
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+k
+ c12 − c12
c2
such that
n
X
D2
.
c+
2
u2i1 ≤
i=1
Iteratively repeating the same approach, we obtain a sequence {c(i) } (i = 1, 2, . . . ), where
c(0) = c2 , c(1) = c+
2 and
0.5
c(i+1) =
(2.13)
1+k
1
c(i)
+
1
c(i)
−
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) > 13 for all i by mathematical induction. We have shown that
c(0) > 13 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)
1
0.5
0.5
32 ≥ 1 + k > 1 + 3k > 3 .
c(i)
To sum up, we obtain 13 < c(i) < 12 , which implies that
formula of c(i+1) (2.13) becomes
0.5
c(i+1) =
1+k 2+
1
c(i)
1
c(i)
= 2. Therefore, the iterative
3 .
−2 2
It is easy to verify that the sequence {c(i) } is monotone increasing with a limit value 0.3972.
3. E XTENSION
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
(3.1)
(3.2)
n
X
i=1
n
X
i=1
(3.3)
n
X
u3i1 ≤
D3
,
0.3168
u3im ≤
mD3
,
c3
u3im ≤ nD3 ,
2 ≤ m < c3 n,
c3 n < m ≤ n − 1,
i=1
where c3 = 0.3125.
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 74, 10 pp.
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I NTERPOINT D ISTANCE S UM I NEQUALITY
7
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
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 non-overlapping, 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. Pick any point S from the
boundary of B(D) and consider the overlap ratio
S
fim
:=
(3.4)
V (B(D) ∩ BS (uim ))
,
V (BS (uim ))
1 ≤ i ≤ n, 1 ≤ m ≤ n − 1.
Using a 3-dimensional version of Figure 2.1, one can obtain the geometrical computation
S
formula: fim
= f (y)|y= uim , where
D
1 3y
− .
2 16
S
Actually f (y) is a decreasing function of y. We have fim
≥ f (1) due to uim ≤ D. Also
S
fim ≥ fim . Setting c3 := f (1), we obtain the following lower bound on fim for every 1 ≤ i ≤ n
and 1 ≤ m ≤ n − 1,
5
fim ≥ c3 , where c3 =
= 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,
f (y) :=
V (Bi (uim ) ∩ B(D)) ≥ c3 V (Bi (uim )).
(3.5)
For a given value m, adding the n inequalities in (3.5), we obtain
(3.6)
n
X
V (Bi (uim ) ∩ B(D)) ≥ c3
n
X
i=1
V (Bi (uim )),
∀1 ≤ m ≤ n − 1.
i=1
Since all spheres in R1 are non-overlapping, we have
n
X
(3.7)
V (Bi (uim ) ∩ B(D)) ≤ V (B(D)).
i=1
Inequalities (3.6) and (3.7) imply
V (B(D)) ≥ c3
n
X
V (Bi (uim )).
i=1
3
Notice that V (B(D)) = πD /6 and V (Bi (ui1 )) = πu3i1 /6. Therefore,
n
X
(3.8)
u3i1 ≤
i=1
Defining k =
3
16
D3
.
c3
= 0.1875, we have
fim ≥
S
fim
=f
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 74, 10 pp.
u
im
D
≥ c3 + k − k
uim
.
D
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YONG X IA AND H ONG -Y ING L IU
Therefore, for every 1 ≤ i ≤ n and 1 ≤ m ≤ n − 1,
V (Bi (uim ) ∩ B(D)) ≥ (c3 + k)V (Bi (uim )) − k
(3.9)
uim
V (Bi (uim )).
D
Adding the n inequalities in (3.9) for a given m, we obtain
(3.10)
n
X
V (Bi (uim ) ∩ B(D))
i=1
≥ (c3 + k)
n
X
i=1
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
D3 ≥ (c3 + k)
(3.11)
n
X
n
u3i1 −
i=1
k X 4
u .
D i=1 i1
Now consider the following optimization problems (n ≥ 3):
max
(3.12)
n
X
u4i1
i=1
n
X
(3.13)
s.t.
(3.14)
0 ≤ ui1 ≤ D,
u3i1 ≤
i=1
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
0 and D cannot be a vertex. Therefore
j between
k
1
every optimal solution of (3.12) – (3.14) has c3 components with the value D, one component
r
j k
with the value c13 − 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
In other words, we have proved for valid ui1 that
43
n
X
1
1
1
u4i1 ≤
D4 +
−
D4 .
c
c
c
3
3
3
i=1
Now (3.11) becomes
(3.15)
D 3 ≥ c3
n
X
i=1
u3i1 + k
n
X
u3i1 −
i=1
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 74, 10 pp.
43 ! !
1
1
1
+
−
D3 .
c3
c3
c3
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I NTERPOINT D ISTANCE S UM I NEQUALITY
Then we have
n
X
D
3
1+k
j k
1
c3
u3i1 ≤
+
1
c3
−
9
j k 43 1
c3
c3 (1 + k c13 )
i=1
.
Comparing with (3.8), we actually obtain a new c+
3:
c3 1 + k c13
j k (3.16)
c+
3 =
j k 43 ≈ 0.3156
1
+ c13 − c13
1+k
c3
such that
n
X
u3i1 ≤
i=1
D3
.
c+
3
Iteratively repeating the same approach, we obtain a sequence {c(i) } (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) < 31 for all i. We prove this by mathematical
1 induction. We have
1
1
(0)
(i)
c = 0.3125 < 3 . Now assume that c < 3 , which also implies c(i) ≥ 3. Then based on
(3.17), we have
0.5
c(i+1) =
1 43 1
1
1 + k c(i) + c(i) − c(i)
≤
1
0.5
0.5
1 ≤
< .
1 + 3k
3
1 + k c(i)
Secondly, we prove
1
4
for all i by mathematical induction. We have shown c(0) > 14 . 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) >
we have
0.5
c(i+1) =
1+k
1
c(i)
+
1
c(i)
−
1
c(i)
0.5
0.5
1
43 ≥ 1 + k > 1 + 4k > 4 .
c(i)
1 To sum up, we obtain 14 < c(i) < 31 , which implies that c(i)
= 3. Therefore, the iterative
(i+1)
formula (2.13) of c
becomes
0.5
c(i+1) =
4 .
1
1 + k 2 + c(i)
−3 3
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
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 74, 10 pp.
http://jipam.vu.edu.au/
10
YONG X IA AND H ONG -Y ING L IU
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
V (Bi (uim ) ∩ B(D)) ≤ mV (B(D)).
i=1
It follows that
3
mD ≥ c3
n
X
u3i1 .
i=1
The last inequality (3.3) directly follows from the fact uim ≤ D.
R EFERENCES
[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.
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 74, 10 pp.
http://jipam.vu.edu.au/