# Favourite distances in 3-space

Let S be a set of n points in Euclidean 3-space. Assign to each x∈ S a distance r(x)>0, and let e_r(x,S) denote the number of points in S at distance r(x) from x. Avis, Erdős and Pach (1988) introduced the extremal quantity f_3(n)=max∑_x∈ Se_r(x,S), where the maximum is taken over all n-point subsets S of 3-space and all assignments r S→(0,∞) of distances. We show that if the pair (S,r) maximises f_3(n) and n is sufficiently large, then, except for at most 2 points, S is contained in a circle C and the axis of symmetry L of C, and r(x) equals the distance from x to C for each x∈ S∩L. This, together with a new construction, implies that f_3(n)=n^2/4 + 5n/2 + O(1).

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## 1 Introduction

Let be a set of points in the -dimensional Euclidean space . We write for the Euclidean distance between and , and for the distance from to the finite set . Let be a choice of a positive number for each point in . Define the favourite distance digraph on determined by to be the directed graph on the set with arcs

 →Er(S):={(x,y):x,y∈S and d(x,y)=r(x)}.

Let denote the cardinality of the set . Define

 fd(n):=max{|→Er(S)|:S⊂Rd,|S|=n and r:S→(0,∞)}.

The problem of determining was originally introduced by Avis, Erdős and Pach [2], who showed that

 n24+3n2≤f3(n)≤n24+an2−b

for some constants . Our main result is the following asymptotic improvement:

###### Theorem A.

For all sufficiently large ,

The upper bound follows from the following structural result and the lower bound from a construction in Section 2. A finite set of points in and a function are called a suspension if is contained in the union of some circle and its axis of symmetry , and satisfies for all . If then and are uniquely determined by . The next result states that if and are extremal, then is a suspension except for at most points.

###### Theorem B.

Let be finite and be such that . If is sufficiently large then for some with , is a suspension with circle and symmetry axis .

We conjecture that the exceptional set is empty if is large. We prove the above theorem using the following stability result, which states that if is almost extremal, then it is a suspension up to points.

###### Theorem C.

For each there exists and such that if and is a set of points with , then for some with , is a suspension with circle and symmetry axis , and .

In the paper [8] we determined for all and sufficiently large depending on . See also Erdős and Pach [6]. Csizmadia [4] determined the maximum number of furthest distance pairs in , where we fix . Favourite and furthest distances in the plane have been considered by Avis [1] and Edelsbrunner and Skiena [5].

In the next section we make a careful construction which proves the lower bound of Theorem A. In Section 3 we give a relatively straightforward induction proof of the upper bound (Theorem 3). Using the ideas of this proof we then prove the main theorems in Section 4.

## 2 Suspensions

In this section we consider some properties of favourite distance digraphs on suspensions and estimate the maximum of

taken over all suspensions of points. Define and for any . Without loss of generality, we let the radius of the circle of the suspension be and we identify with the real line so that the centre of is . Write and . Then for each . Consider the subdigraph . For any arc , we have . If we solve for we obtain , hence each vertex has at most two out-neighbours in . If we solve for , we obtain , hence each vertex has at most one in-neighbour in (and if then cannot have any in-neighbour). Therefore, . Next, note that if , then each vertex has exactly one in-neighbour, and it follows that each connected component of consists of a directed cycle of length at least together with binary trees where each binary tree is attached to the cycle at its root and its arcs are directed away from the root.

We can use angles to give a simple description of the dynamics of the predecessor and successors of a vertex in . For each , let . Then is the angle between the ray from to a point on the circle and the ray from in the positive direction on . Then simple properties of angles in circles give that , and (Figure 1).

Thus, if we consider the binary expansion of , then is the left-shift of , is the right-shift of with a added to the left, and is the right-shift of with a added to the left. It follows that the angles corresponding to the vertices of a directed cycle in are all rational multiples of such that for any two vertices and of the cycle there exist such that . In the extremal case where has arcs, this property holds for any two vertices in the same connected component of .

###### Proposition 1.

For any suspension with points, .

###### Proof.

Let . Thus . We estimate by decomposing it as follows:

 er(S)=er(L,C)+er(L)+er(C,L)+er(C).

Since each point in is joined to each point in , . As shown above, each point in has at most one in-neighbour in , so . A sphere intersects a line in at most two points, hence . Furthermore, a sphere with centre on intersects in at most two points, which gives . Therefore,

 er(S) ≤ℓc+ℓ+4c=(ℓ+4)(c+1)−4 =⌊n2+10n+94⌋=⌈n24+5n2⌉+2.\qed

For sufficiently large , we can almost attain the bound in Proposition 1.

###### Theorem 2.

For each there is a suspension with points such that .

###### Proof.

Consider the full binary tree consisting of and its successors obtained by repeatedly applying and (Figure 2).

Let be the vertex set of any subtree with vertices that contains at least the points , and . We then have . We next show that we can choose points on such that .

For the vertices of , choose the vertices of squares inscribed in . For each of these vertices , define . Since , each has out-degree . If is not divisible by , choose two of the squares to be such that one vertex of one square and one vertex of the other square are at distance . This distance is chosen so that there exists a point on such that , where is the point . See Figure 3.

The same holds for any of the other three vertices of the square with vertex inscribed in . We then add vertices of this square to and define for each of these vertices , to obtain a set of exactly points such that . Then

 er(S) =er(L,C)+er(L)+er(C,S) =ℓc+ℓ−1+4c=(ℓ+4)(c+1)−5 =⌊(n+5)24⌋−5=⌈n24+5n2⌉+1.\qed

If the upper bound of Proposition 1 is attained, it would have to be because of a very special algebraic coincidence, and we believe that this is not possible. The following observation may help to prove this. For a point let be the angle , where satisfies . Then . If also where , then it follows from by elementary trigonometrical relations that . It follows from a result of M. Newman [7] that there are only two solutions to this equation with both and rational multiples of in the interval , namely and . The first solution corresponds to our construction in Theorem 2, while the second solution correspond to an analogous construction with inscribed regular hexagons, which turns out to be worse than our construction.

## 3 A simple upper bound

We will use the Kővari-Sós-Turán Theorem in the standard form and in a form for directed graphs. See for instance [3, Theorem 2.2].

###### Kővari-Sós-Turán Theorem I.

Let be a bipartite graph with parts of sizes and . Suppose that does not contain a complete bipartite graph with parts of size and , with the part of size contained in and the part of size contained in . Then .

Denote by the digraph with and

 →E={(ai,bj):i=1,…,r;j=1,…,s}.
###### Kővari-Sós-Turán Theorem II.

Let be a directed graph with vertices. Suppose that does not contain a copy of . Then .

###### Proof.

Apply the Kővari-Sós-Turán Theorem I to the bipartite double cover of . ∎

The following upper bound improves the error term from [2]. The proof uses an induction argument which, although conceptually simple, needs some computation.

###### Theorem 3.

There exists such that for all and for any set of points in and any function , .

###### Proof.

Fix a constant , to be determined later. The Kővari–Sós–Turán Theorem II guarantees the existence of a such that any directed graph on vertices and with at least edges contains a . We can ensure that the theorem holds for all by taking large enough. Assume next that and that the theorem holds for sets of up to points.

Without loss of generality, . Let and be the two classes of a contained in chosen such that is maximal among all such , where , and . Then lies on a circle and on its axis of symmetry , and for each . That is, forms a suspension of maximum cardinality among all those contained in with . Let and write . We will bound by writing it as the following sum and then bounding each term separately.

 er(S) =er(L,C)+er(C,L)+er(T,L)+er(L)+er(C) +er(L,T)+er(T,C)+er(C,T)+er(T).

See Figure 4.

Since each vertex on is joined to each vertex on , we have . A sphere and a line intersects in at most points, hence and . Each vertex on has at most one in-neighbour on , hence . A circle and a sphere with centre on the circle intersect in at most points, hence .

Suppose that some has at least two in-neighbours , say. It then follows that lies on the intersection of the spheres with centre and radius , , which is the circle . This contradicts the maximality of . Therefore, .

Suppose that some has at least three out-neighbours , say. Then necessarily , which contradicts the maximality of . Therefore, .

There is no from to , otherwise there would be a from to to , which is not realisable in . By the Kővari-Sós-Turán Theorem I (applied to the reverse bipartite graph from to ), .

Finally, we estimate by the induction hypothesis:

 er(S) ≤ℓc+2c+2t+ℓ+2c+t+2t+(2c+21/3tc2/3)+t24+At5/3 =(ℓ+6)(c+1)−6+t24+At5/3+5t+21/3tc2/3 =(ℓ+c+72)2−(ℓ−c+52)2+t24+At5/3+5t−6+21/3tc2/3 ≤(n−t+72)2+t24+At5/3+5t−6+21/3t(n−t)2/3 =14(n2−2nt+2t2+14n+6t+25+4At5/3)+21/3t(n−t)2/3.

Since ,

 14n+6t+25≤14n+6(n−2p)+25=20n−12p+25<20n,

hence,

 er(S)≤14(n2−2nt+2t2+20n+4At5/3)+21/3t(n−t)2/3,

which will be , thus finishing the induction step, if

 −2nt+2t2+20n+4At5/3+4⋅21/3t(n−t)2/3≤4An5/3. (1)

We next show that (1) holds if is sufficiently large. Since , we can ensure that is large by choosing large.

First suppose that . The inequality (1) will follow if

 −2nt+2t2+20n+4⋅21/3t(n−t)2/3≤0,

which is equivalent to

 (n−t)2/3((n−t)1/3−24/3)≥10n/t.

Since , it is sufficient to have

 (n−t)2/3((n−t)1/3−24/3)≥20,

which holds if . We can ensure this by requiring that .

Next consider the remaining case where . Since is increasing on , we have . Also, and . Therefore, to derive (1), it is sufficient to show that

 20n5/3+4A(n/2)5/3+4⋅21/3(n/2)5/3≤4An5/3.

This inequality is equivalent to

 A≥5+2−4/31−2−5/3,

which can be ensured. This finishes the induction step. ∎

###### Corollary 4.

There exists , and such that any set of points with can be partitioned into a suspension and a remainder set such that , , and

 ≤er(S)+(ℓ−c+52)2 ≤14(n2−2nt+2t2+14n+6t+25+4At5/3)+21/3t(n−t)2/3 (2) ≤14n2+An5/3.
###### Proof.

Proceed as in the proof of Theorem 3, but instead of applying an induction hypothesis to , apply Theorem 3 itself to . ∎

## 4 Proof of the main theorems

###### Proof of Theorem C.

Without loss of generality, by requiring , we may assume that . By Corollary 4 there exist , , such that any set of at least points can be partitioned into a suspension with cardinalities and and a remainder set of cardinality such that (2) holds. Combine (2) with the lower bound to obtain

 2nt−2t2 ≤4δn2+14n+6t+25+4At5/3+4⋅21/3t(n−t)2/3 <5δn2

for sufficiently large depending on and . The resulting quadratic inequality in implies that

 t<(1−√1−10δ2)nort>(1+√1−10δ2)n.

Thus, either or is small. We next show that is not small. Recall from the proof of Theorem 3 that the suspension was chosen such that is maximised subject to . This implies that the favourite distance digraph does not contain a , hence by the Kővari-Sós-Turán Theorem II,

 er(S)≤(n−t−1)1/pn2−1/p+(p−1)n.

Combine this with to obtain

 t+1≤(1−(14−δ−p−1n)p)n.

If , then

 1+√1−10δ2<1−(14−δ−p−1n)p,

which gives a contradiction if is sufficiently small and sufficiently large, both depending on . Therefore,

 t<(1−√1−10δ2)n<εn

if is sufficiently small depending on .

The bound

 er(S)+(ℓ−c+52)2≤n24+An5/3

from Corollary 4, together with the lower bound gives

 (ℓ−c+52)2≤δn2+An5/3<2δn2

for sufficiently large depending on and . Then for sufficiently large depending on , hence for sufficiently small depending on . It follows that and . ∎

###### Proof of Theorem B.

Let with and satisfy . Since by Theorem 2, Corollary 4 gives that for each there exists and such that for all , consists of a spindle except for an exceptional set such that

 er(S)≤14(n2−2nt+2t2+14n+6t+25+4At5/3)+21/3t(n−t)2/3.

By Theorem C we can also assume that by making sufficiently large. It follows that

 n24+5n2≤14(n2−2nt+2t2+14n+6t+25+4At5/3)+21/3t(n−t)2/3,

hence

 0 ≤2n(2−t)+2t2+6t+25+4At5/3+4⋅21/3t(n−t)2/3 ≤2n(2−t)+2t2+Btn2/3:=g(t)

for sufficiently large depending on . It is easy to check that for each sufficiently large fixed , is decreasing on and negative for . It follows that . ∎

###### Proof of Theorem A.

The lower bound is given by Theorem 2. Consider an extremal example with sufficiently many points so that Theorem B can be applied.

If is empty, then is a suspension, and Proposition 1 provides the upper bound.

If , then and

 er(S) =er(L,C)+er(C,L)+er(C)+er(L)+er(L,T)+er(C,T)+er(T,L∪C) ≤ℓc+2c+2c+ℓ+1+c+4=(ℓ+5)(c+1) ≤⌊(ℓ+c+62)2⌋=⌈n24+5n2⌉+6,

where the separate estimates are made as in the proof of Theorem 3.

Similarly, if , then and

 er(S) =er(L,C)+er(C,L)+er(C)+er(L) =+er(L,T)+er(C,T)+er(T,L∪C)+er(T) ≤ℓc+2c+2c+ℓ+2+2c+2⋅4+2=(ℓ+6)(c+1)+6 ≤⌊(ℓ+c+72)2⌋+6=⌈n24+5n2⌉+12.\qed

## References

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• [2] D. Avis, P. Erdős, and J. Pach, Repeated distances in space, Graphs Combin. 4 (1988), 207–217.
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• [7] M. Newman, Some results on roots of unity, with an application to a diophantine problem, Aequationes Math. 2 (1969), 163–166.
• [8] K. J. Swanepoel, Favorite distances in high dimensions, In: Thirty Essays in Geometric Graph Theory, ed. J. Pach, Algorithms and Combinatorics 29, Springer 2013. pp. 499–519.