Bounding the size of an almost-equidistant set in Euclidean space
BBounding the size of an almost-equidistant set inEuclidean space
Andrey Kupavskii ∗ Nabil H. Mustafa † ‡
Konrad J. Swanepoel § Abstract
A set of points in d -dimensional Euclidean space is almost equidistant if amongany three points of the set, some two are at distance . We show that an almost-equidistant set in R d has cardinality O ( d / ) . A set of lines through the origin of Euclidean d -space R d is almost orthogonal if amongany three of the lines, some two are orthogonal. Erdős asked (see [12]) what is thelargest cardinality of an almost-orthogonal set of lines in R d ? By taking two sets of d pairwise orthogonal lines, we see that d is possible. Rosenfeld [12] showed that d isthe maximum using eigenvalues. This result was subsequently given simpler proofs byPudlák [11] and Deaett [6].In this note we consider the following analogous notion, replacing orthogonal pairs oflines by pairs of point at unit distance. A subset V of Euclidean d -space R d is almostequidistant if among any three points in V , some two are at Euclidean distance . Weinvestigate the largest size, which we denote by f ( d ) , of an almost-equidistant set in R d .Although this is a very natural question, it seems to be much harder than the question ofErdős, which can be refomulated as asking for the largest size of an almost-equidistantset on a sphere of radius / √ in R d . Before stating our main result, we give an overviewof what is known about the size of f ( d ) .Bezdek, Naszódi and Visy [5] showed that f (2) ≤ , and István Talata (personalcommunication, 2007) showed that the only almost-equidistant set in R with points isthe Moser spindle. Györey [7] showed that f (3) ≤ and that there is a unique almost-equidistant set of points in R , a configuration originally considered by Nechushtan [9].The Moser spindle can be generalized to higher dimensions, giving an almost-equidistantset of d + 3 points in R d [4]. (We mention that Bezdek and Langi [4] consideredthe variant of Erdős’s problem where the radius of the sphere is arbitrary instead of ∗ Moscow Institute of Physics and Technology, Ecole Polytechnique Fédérale de Lausanne.Email: [email protected] † Université Paris-Est, Laboratoire d’Informatique Gaspard-Monge, ESIEE Paris, France.Email: [email protected] ‡ The work of Nabil H. Mustafa in this paper has been supported by the grant ANR SAGA (JCJC-14-CE25-0016-01). § Department of Mathematics, London School of Economics and Political Science, London.Email: [email protected] a r X i v : . [ m a t h . C O ] A ug / √ .) A construction of Larman and Rogers [8] shows that f (5) ≥ . Since theredoes not exist a set of d + 2 points in R d that are pairwise at distance , it followsthat f ( d ) ≤ R ( d + 2 , − , where the Ramsey number R ( a, b ) is the smallest n suchthat whenever each edge of the complete graph on n vertices is coloured blue or red,there is either a blue clique of size a or a red clique of size b . Ajtai, Komlós, andSzemerédi [1] proved R ( k,
3) = O ( k / log k ) , which implies the asymptotic upper bound f ( d ) ≤ O ( d / log d ) . Balko et al. [3] generalized the Nechushtan configuration to higherdimensions, giving f ( d ) ≥ d + 4 for all d ≥ . They also obtained the asymptotic upperbound f ( d ) = O ( d / ) by an argument based on Deaett’s paper [6]. Using computersearch and ad hoc geometric arguments, they obtained the following bounds for small d : f (4) ≤ , f (5) ≤ , ≤ f (6) ≤ , ≤ f (7) ≤ , and f (9) ≥ f (8) ≥ . Polyanskii[10] subsequently improved the asymptotic upper bound to f ( d ) = O ( d / ) .In this note we obtain a further improvement to the upper bound. Theorem 1.
An almost-equidistant set in R d has cardinality O ( d / ) . Its proof is based on the approach of Balko et al. [3]. Before giving the proof ofTheorem 1 in Section 3, we collect some lemmas in the next section.
We call a finite subset C of R d (the vertex set of) a unit simplex if the distance betweenany two points in C equals . It is well known that if C is a unit simplex then | C | ≤ d + 1 .Given any finite V ⊂ R d , we define the unit-distance graph G = ( V, E ) on V to be thegraph with vw ∈ E iff (cid:107) v − w (cid:107) = 1 . Thus, C ⊂ V is a unit simplex iff it is a clique in G .We denote the set of neighbours of v ∈ V in G by N ( v ) .The following well-known lemma gives a lower bound for the rank of a square matrixin terms of its entries [2, 6, 11]. Lemma 1.
For any non-zero n × n symmetric matrix A = [ a i,j ] , rank( A ) ≥ ( (cid:80) i a i,i ) (cid:80) i,j a i,j . Lemma 2.
Let C be a unit simplex with centroid c = | C | (cid:80) v ∈ C v . Then (cid:107) v − c (cid:107) = 12 (cid:18) − | C | (cid:19) for all v ∈ C ,and (cid:10) v − c, v (cid:48) − c (cid:11) = − | C | for all distinct v, v (cid:48) ∈ C .Proof. We may translate C so that c is the origin o . Write C = { p , . . . , p k } . Bysymmetry, α := (cid:107) p i (cid:107) is independent of i , and β := (cid:104) v i , v j (cid:105) ( i (cid:54) = j ) is independent of i and j . Then (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) k (cid:88) i =1 p i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) = kα + k ( k − β (cid:107) p i − p j (cid:107) = 2 α − β. Solving these two linear equations in α and β , we obtain α = − k and β = − k . Lemma 3.
Let C be a unit simplex with centroid c = | C | (cid:80) v ∈ C v , and let F ⊂ C withcentroid f = | F | (cid:80) v ∈ F v . Then (cid:107) c − f (cid:107) = 12 (cid:18) | F | − | C | (cid:19) . Proof.
Let k = | C | , (cid:96) = | F | . Then (cid:107) f − c (cid:107) = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) (cid:96) (cid:88) v ∈ F ( v − c ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) = 1 (cid:96) (cid:18) (cid:96) ·
12 (1 − k ) − (cid:96) ( (cid:96) − k (cid:19) by Lemma 2 = 12 (cid:18) (cid:96) − k (cid:19) . Let G be the unit-distance graph of the almost-equidistant set V . Thus, the complementof G is K -free, and the non-neighbours of any vertex form a unit simplex. Let C be aclique of maximum cardinality in G . Write k = | C | . Each v ∈ V \ C is a non-neighbourof some point in C , and it follows that | V | ≤ | C | + | C | k = k + k . Thus, without loss ofgenerality, k > d / .We split V up into two parts, each to be bounded separately. Let N = (cid:110) v ∈ V : | N ( v ) ∩ C | ≥ k − k / d − / (cid:111) . Note that k / d − / = O ( d − / ) k . We first bound the complement of N . Consider theset X = { ( u, v ) ∈ C × V \ N : uv / ∈ E ( G ) } . For each v ∈ V \ N , there are more than k / d − / points u ∈ C such that u / ∈ N ( v ) ,hence | X | > k / d − / | V \ N | . On the other hand, for each u ∈ C , the set of non-neighbours of u forms a clique, so has cardinality at most k , and | X | ≤ | C | k = k . Itfollows that | V \ N | < k / d / . (1)Next we estimate | N | . Without loss of generality, k (cid:80) v ∈ C v = o and N = { v , . . . , v n } .We want to apply Lemma 1 to the n × n matrix A = [ (cid:104) v i , v j (cid:105) ] , which has rank at most d . Claim 1.
For each i = 1 . . . , n , (cid:107) v i (cid:107) = + O ( k − / d − / ) , and for each v i v j ∈ E ( G ) , (cid:104) v i , v j (cid:105) = O ( k − / d − / ) . roof of Claim 1. Let C i := N ( v i ) ∩ C and k i := | C i | ≥ k − k / d − / , and c i := k i (cid:80) v ∈ C i v . For any v ∈ C i , (cid:107) v i − c i (cid:107) = (cid:107) v i − v (cid:107) −(cid:107) v − c i (cid:107) . By Lemma 2, (cid:107) v − c i (cid:107) = (cid:16) − k i (cid:17) , hence (cid:107) v i − c i (cid:107) = (cid:115) (cid:18) k i (cid:19) = 1 √ O ( k − ) . By Lemma 3, (cid:107) c i (cid:107) = (cid:115) (cid:18) k i − k (cid:19) ≤ (cid:115) (cid:18) k − k / d − / − k (cid:19) = O ( k − / d − / ) . By the triangle inequality, (cid:107) v i (cid:107) = (cid:107) v i − c i (cid:107) + O ( (cid:107) c i (cid:107) ) = 1 √ O ( k − / d − / ) , and (cid:107) v i (cid:107) = + O ( k − / d − / ) . Also, (cid:104) v i , v j (cid:105) = (cid:107) v i (cid:107) + (cid:107) v j (cid:107) − O ( k − / d − / ) . Claim 2.
For each i = 1 , . . . , n , n (cid:88) j =1 v i v j / ∈ E ( G ) (cid:104) v i , v j (cid:105) = O ( k / d − / ) . Proof of Claim 2.
The non-neighbours N \ N ( v i ) of v i form a unit simplex with cardinality t := | N \ N ( v i ) | ≤ k and with centroid c , say. If t = d + 1 , remove one point v j fromthe unit simplex, which decreases the sum by (cid:104) v i , v j (cid:105) = O (1) . Thus, without loss ofgenerality, t ≤ d , and there exists a point p ∈ R d such that p − c is orthogonal to theaffine hull of N ( v i ) , (cid:107) p − c (cid:107) = 1 / √ t , the set { v j − p : v j ∈ N \ N ( v i ) } is orthogonal, and (cid:107) v j − p (cid:107) = 1 / √ for each non-neighbour v j of v i . Then, by the finite Bessel inequality, (cid:88) v j ∈ N \ N ( v i ) (cid:104) v i , v j − p (cid:105) ≤ (cid:107) v i (cid:107) , hence by applying Cauchy-Schwarz a few times, (cid:88) v j ∈ N \ N ( v i ) (cid:104) v i , v j (cid:105) = (cid:88) v j ∈ N \ N ( v i ) (cid:0) (cid:104) v i , v j − p (cid:105) + (cid:104) v i , p − c (cid:105) + (cid:104) v i , c (cid:105) (cid:1) ≤ (cid:88) v j ∈ N \ N ( v i ) (cid:0) (cid:104) v i , v j − p (cid:105) + (cid:104) v i , p − c (cid:105) + (cid:104) v i , c (cid:105) (cid:1) ≤ (cid:18) (cid:107) v i (cid:107) + t (cid:107) v i (cid:107) (cid:107) p − c (cid:107) + t (cid:107) v i (cid:107) (cid:107) c (cid:107) (cid:19) ≤ (cid:107) v i (cid:107) + t (cid:107) v i (cid:107) (cid:107) c (cid:107) ) . (2)4y Claim 1, (cid:107) v i (cid:107) = + O ( k − / d − / ) and (cid:107) c (cid:107) = (cid:13)(cid:13)(cid:13) t (cid:88) v j ∈ N \ N ( v i ) v j (cid:13)(cid:13)(cid:13) = 1 t (cid:18) (cid:88) v j ∈ N \ N ( v i ) (cid:107) v j (cid:107) + (cid:88) v j ,v j (cid:48) ∈ N \ N ( v i ) v j (cid:54) = v j (cid:48) (cid:10) v j , v j (cid:48) (cid:11)(cid:19) ≤ t (cid:18) t (cid:18)
12 + O ( k − / d − / ) (cid:19) + t ( t − O ( k − / d − / ) (cid:19) = 12 t + O ( k − / d − / ) . Therefore, t (cid:107) c (cid:107) = 12 + O ( tk − / d − / ) = O ( k / d − / ) . Substitute this back into (2) to finish the proof of Claim 2.We now finish the proof of the theorem. By Claim 2, n (cid:88) j =1 (cid:104) v i , v j (cid:105) = (cid:107) v i (cid:107) + (cid:88) v j ∈ N ( v i ) (cid:104) v i , v j (cid:105) + O ( k / d − / )= nO ( k − / d − / ) + O ( k / d − / ) by Claim 1.Also by Claim 1, (cid:80) ni =1 (cid:107) v i (cid:107) = Ω ( n ) . Therefore, by Lemma 1, d ≥ rank( A ) ≥ (cid:16)(cid:80) ni =1 (cid:107) v i (cid:107) (cid:17) (cid:80) ni,j =1 (cid:104) v i , v j (cid:105) = Ω ( n ) n (cid:0) nO ( k − / d − / ) + O ( k / d − / ) (cid:1) , hence n = O ( nk − / d / ) + O ( k / d / ) . Since O ( k − / d / ) = o (1) , it follows that | N | = n = O ( k / d / ) . Recalling (1), we obtain that | V | = | N | + | V \ N | = O ( k / d / ) = O ( d / ) . References [1] Miklós Ajtai, János Komlós, and Endre Szemerédi,
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