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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