Abstract 1 Introduction 2 VC-dimension theory 3 Proof of Theorem 1 4 Concluding remarks References

A Note on the No-(d+2)-On-a-Sphere Problem

Andrew Suk Department of Mathematics, University of California San Diego, La Jolla, CA, USA Ethan Patrick White ORCID Department of Mathematics, University of Illinois Urbana-Champaign, Urbana, IL, USA
Abstract

For fixed d3, we construct subsets of the d-dimensional lattice cube [n]d of size n3d+1o(1) with no d+2 points on a sphere or a hyperplane. This improves the previously best known bound of Ω(n1d1) due to Thiele from 1995.

Keywords and phrases:
General position, no-four-on-a-cirle, d-dimensional lattice cube
Funding:
Andrew Suk: Supported by NSF CAREER award DMS-1800746, NSF award DMS-1952786, and NSF grant DMS-2246847.
Ethan Patrick White: supported by an NSERC PDF Award.
Copyright and License:
[Uncaptioned image] © Andrew Suk and Ethan Patrick White; licensed under Creative Commons License CC-BY 4.0
2012 ACM Subject Classification:
Mathematics of computing Combinatorics
Related Version:
Full Version: https://arxiv.org/abs/2412.02866
Editors:
Oswin Aichholzer and Haitao Wang

1 Introduction

The famous no-three-in-line problem, raised by Dudeney [5] in 1917 in the special case n=8, asks if it is possible to select 2n points from the lattice square [n]2 such that no three are collinear. Clearly, one cannot do better than 2n as [n]2 can be covered by n lines. For n52, many authors have published solutions to this problem obtaining the bound of 2n (e.g. see [6]). However, for large values of n, the best known lower bound is due to Hall et al. [9] which contains at least (32ε)n points, for any ε>0 and n sufficiently large.

The similar no-four-on-a-circle problem, raised by Erdős and Purdy in 1981 (see [8]), asks to determine the maximum number of points that can be selected from [n]2 such that no four are on a circle. Here, a line is also considered to be a degenerate circle. Again, we have the trivial upper bound of 3n as any vertical line must contain at most three points. This upper bound was improved by Thiele [19], who showed that at most 52n32 points can be selected, and moreover, he gave a construction of n4 points from [n]2 with no four on a circle (or a line).

In this paper, we study the no-four-on-a-circle problem in higher dimensions (Problem 4 in Chapter 10 of [3]). A k-sphere is a k-dimensional sphere. Thus, a 0-sphere is a pair of points, a 1-sphere is a circle, and etc. For simplicity, we will simply use the term sphere when referring to a (d1)-dimensional sphere in d. Again, the maximum number of points that can be selected from the d-dimensional lattice cube [n]d with no d+2 points on a sphere or a hyperplane is at most (d+1)n since we can cover [n]d with n hyperplanes. In the other direction, Thiele [18] showed that one can select Ω(n1d1) points from [n]d with no d+2 points on a sphere or a hyperplane, providing the first non-trivial construction for this problem. In this note, we prove the following.

Theorem 1.

Let d3 be a positive integer. Then there is a subset of the d-dimensional lattice cube [n]d with n3d+1o(1) points with no d+2 members on a sphere or a hyperplane.

While it is possible that one can improve this bound to Ω(n), we make the following more modest conjecture.

Conjecture 2.

Let d3 be a positive integer. Then there is a subset of the d-dimensional lattice cube [n]d with Ω(ndd+1) points with no d+2 members on a sphere or a hyperplane.

Our paper is organized as follows. In Section 2, we recall several results from VC-dimension theory that will be used in the proof of Theorem 1. In Section 3, we prove Theorem 1. We conclude our paper with several remarks and related problems. For sake of clarity, we systemically remove floor and ceilings whenever they are not crucial. All logarithms are in based 2 unless stated otherwise.

2 VC-dimension theory

In this section, we recall several results from VC-dimension theory that will be used in the proof of Theorem 1. First we introduce some terminology.

Let G=(P,Q,E) be a bipartite graph with independent sets P,Q and edges EP×Q. For qQ we define the neighbourhood NQ(q)={pP:(p,q)E}. Define the set-system ={NG(q):qQ} with ground set P. We say a set SP is shattered by if for every subset BS, there is a set A such that AS=B. The Vapnik-Chervonenkis dimension (VC-dimension) of (P,) is the largest integer d for which there exists a subset SP, with |S|=d, that is shattered by .

The primal shatter function of (P,) is defined as

π(z)=maxPP,|P|=z|{AP:A}|.

The well-known Sauer-Shelah lemma [14, 16] states that if d0 is the VC-dimension of , then

π(z)i=0d0(zi). (1)

We will need the following result due to Fox, Pach, Sheffer, Suk, and Zahl.

Theorem 3 ([7]).

Let G=(P,Q,E) be a bipartite graph with |P|=m and |Q|=n, such that the set system 1={N(q):qQ}, with ground set P, satisfies π1(z)czd0 for all z. Then if G is Kt,t-free, we have

|E(G)|c1(mn11/d0+n),

where c1=c1(c,d0,t). In particular, c110ct2d0+1(d0logd0)d0.

A subset A[n]d is a maximal spherical set of [n]d if all points in A lie on a single sphere S in d, and no point of [n]d can be added to A while keeping all points on the sphere S. Let 𝒮n,d denote the set-system of all maximal spherical sets of [n]d, with ground set [n]d.

Lemma 4.

For d2, the VC-dimension of the set-system ([n]d,𝒮n,d) is at most d+1.

Proof.

We proceed by induction on d. The base case d=1 is trivial since a 0-sphere consists of 2 points. For the inductive step, assume the statement holds for d<d. For sake of contradiction, suppose that a set Q of d+2 points in [n]d is shattered by 𝒮n,d.

Case 1.

Suppose Q is in general position, that is, no d+1 members in Q lie on a hyperplane. By linear algebra, any d+1 points in Q determines a unique sphere in d. However, since Q is shattered by 𝒮n,d, there is a sphere that contains Q. Hence, there is no sphere that contains exactly d+1 points from Q. Contradiction.

Case 2.

Suppose Q contains a (d+1)-tuple Q that lies on a hyperplane h in d. Since the intersection of a sphere with h is a (d2)-sphere in h, by induction on d, Q cannot be shattered by 𝒮n,d. Hence, Q cannot be shattered by 𝒮n,d, which is a contradiction.

3 Proof of Theorem 1

In this section, we prove Theorem 1. First, let us recall several results. The first is the well-known Chernoff inequality (see [11, Theorem 2.8]).

Lemma 5 (Chernoff’s inequality).

Let X1,,Xn be independent random variables such that (Xi=1)=q and (Xi=0)=1q, and let X=iXi. Then for 0<δ<1, we have

(X(1+δ)qn)eδ2qn3,

and

(X(1δ)qn)eδ2qn3.

Next, we estimate the number of points from [n]d that can lie on a sphere or a hyperplane. For the latter, we will use the following result due to Balogh and White.

Lemma 6 ([2]).

Let d2 be a positive integer, and a1,,ad not be all zero with greatest common denominator 1. Let s=maxi{|ai|}, ns be a positive integer, and set

={(x1,,xd)d:i=1daixi=0}.

If [n]d spans a (d1)-dimensional subspace, then |[n]d|3dnd1/s.

Lemma 7.

The number of (d+2)-tuples in [n]d that lie on a common hyperplane is at most O(nd2+d1).

Proof.

By the lemma above, the number of (d+1)-tuples in [n]d that lie on a hyperplane which passes through the origin is at most

s(nd1s)d+1sd1=O(nd21).

By symmetry, for any fixed point p[n]d, the number of (d+1)-tuples in [n]d that lies on a hyperplane which passes through p is at most O(nd21). After summing over all points in [n]d, the statement follows. Note that Lemma 7 applies to hyperplanes whose intersection with [n]d spans a (d1)-dimensional subspace. We can assume that any (d+2)-tuple in [n]d lying on a common hyperplane meets this requirement by adding other points from [n]d if needed.

Finally, we will use the following result due to Sheffer that bounds the number of points from [n]d that lies on a k-sphere in d, for kd1. See also [10, Sec. 11.2, Equation 11.9], [17, Theorem 2].

Lemma 8 ([15], Lemma 3.2).

Let d2 and kd1. Then there is a positive constant c=c(d) such that every k-sphere in d contains at most nk1+c/loglogn points from [n]d.

Before we prove Theorem 1, let us give a brief outline of the argument. First, we use the probabilistic method and Lemmas 7 and 8 to obtain a large subset A[n]d such that very few members of A lie on a (d2)-sphere or a hyperplane in d. We then apply Theorem 3 and ideas from incidence geometry to estimate the number of r-tuples in A that lie on a common sphere. Finally, we apply the deletion method to find a large subset A′′A such that no d+2 members of A′′ lie on a sphere or a hyperplane. We now flesh out all of the details of the proof.

Proof of Theorem 1.

Without loss of generality, we can assume that n is a power of 2. Indeed, otherwise we can find an integer n<n that is a power of 2 such that n>n/2, and apply the arguments below to the subcube [n]d. This would only change the hidden constant in the o(1) term in the exponent. Moreover, we will assume that n is sufficiently large.

Fix a positive integer D<n that will be determined later, such that D divides n. We partition the d-dimensional cube [n]d into Dd smaller cubes Q1,,QDd, where each Qi is of the form

{i1nD+1,,(i1+1)nD}××{idnD+1,,(id+1)nD}.

where i1,,id{0,,D1}. Hence, |Qi|=(n/D)d.

Consider a random subset A[n]d obtained by selecting each point in [n]d independently with probability n3d. Set Pi=QiA, for i=1,,Dd. Let 𝒲 be the event that the subset A[n]d satisfies the following properties.

  1. 1.

    n3/2|A|2n3.

  2. 2.

    n3Dd/2|Pj|2n3Dd for all 1jDd.

  3. 3.

    Every (d2)-sphere contains less than nc1/loglogn points of A, where c1=c1(d).

  4. 4.

    The number (d+2)-tuples in A that lie on a common hyperplane is at most c2n2d+5, where c2=c2(d).

By Lemma 5 and Markov’s inequality, event 𝒲 holds with probability at least 1/2. Indeed, by Lemma 5, the probability that |A|>2n3 or |A|<n3/2 is at most en33+en312. Thus, the first property holds with high probability, that is, with probability tending to 1 as n approaches infinity. A similar argument follows for the second property. For the third property, let us fix a (d2)-sphere S in d. By Lemma 8, there is a constant c=c(d) such that S contains at most nd3+c/loglogn points from [n]d. By Lemma 5,

[|SA|2nc/loglogn]enc/loglogn3.

Since there are at most nd2 (d2)-spheres in d with at least d points from [n]d, the union bound implies that the third property holds with high probability. For the fourth property, let X denote the number (d+2)-tuples in A that lie on a common hyperplane. By Lemma 7, we have 𝔼[X]cn2d+5, where c=c(d). By Markov’s inequality (see [1]), we have

[X>10cn2d+5]<𝔼[X]10cn2d+5<110.

Putting everything together, and setting c1,c2 sufficiently large, event 𝒲 holds with probability at least 1/2.

Thus, let us fix A[n]d with the four properties described above, and set t=2nc3/loglogn, where c3=max{c1,c2}. Let 𝒮 be the collection of spheres in d, such that each sphere in 𝒮 contains at least d+1 points from A in general position. Hence, |𝒮|=O(|A|d+1). Let 𝒮j𝒮 be the set of spheres in 𝒮 that contains at least one point of Pj=QjA. Let Gj=(Pj,𝒮j,Ej) be the bipartite incidence graph between Pj and 𝒮j. Since the intersection of two distinct spheres in d is a (d2)-sphere, and every (d2)-sphere contains less than t points from A, each graph Gj is Kt,2-free.

Let j be the set system whose ground set is Pj, and whose sets are SPj, where S𝒮j and |SPj|t. That is,

j={SPj:S𝒮j,|SPj|t}.

By Lemma 4, the VC-dimension of j is at most d+1. By inequality (1), we have

πj(z)=O(zd+1).

Hence, we apply Theorem 3 with t=nc3/loglogn to conclude that

|Ej|nc4/loglogn(|Pj||𝒮j|dd+1+|𝒮j|),

where c4=c4(d).

Given the partition [n]d=Q1QDd described above, we say that a sphere S crosses the subcube Qi if SQi.

Observation 9.

For fixed d2, every sphere S in d crosses at most c5Dd1 subcubes Qi, where c5=c5(d).

Proof.

Let c5=c5(d) be a large constant that depends only on d. We will determine c5 later. For sake of contradiction, suppose there is a sphere S in d the crosses more than c5Dd1 subcubes Qi. Note that we will not make any serious attempts to optimize the constant c5.111A careful calculation shows that one can take c5=2O(d). Another proof follows from a simple inductive argument with c5=dO(d).

Let T[n]dS be a subset of points [n]dS obtained by selecting 1 point from each subcube Qi that S crosses. Hence, |T|c5Dd1. Let 𝒢=(T,E) be an auxiliary graph on T, where two points in T are adjacent if their distance is less than n/D. Since two points in T have distance less than n/D only if they lie in adjacent subcubes Qi and Qj, this implies that 𝒢 has maximum degree 3d1. By Turan’s theorem, we can find a subset TT, such that the minimum distance between any two points in T is at least n/D. Moreover, |T|(c5/3d)Dd1.

For each point pT, consider the spherical cap Cp obtained by intersecting S with the ball Bp centered at p with radius n/(2D). Since the minimum distance among the points in T is at least n/D, the collection of spherical caps corresponding to the points in T will have pairwise disjoint interiors. Moreover, since each spherical cap arises from a ball whose center is on S with radius n/(2D), each spherical cap will have area at least c6(n/D)d1 where c6=c6(d). Hence, the total area of all of the spherical caps corresponding to the points in T is at least

|T|c6(n/D)d1c5c63dnd1.

On the other hand, all of the spherical caps on S will lie inside of a “large” ball B with radius dn. Hence, the area of all of the spherical caps is at most the surface area of the ball B with radius dn. Since the surface area of B is at most c7nd1, where c7=c7(d), we have

c7nd1c5c63dnd1.

By setting c5 sufficiently large, we have a contradiction.

By the observation above, j=1Dd|𝒮j|=O(Dd1|𝒮|). Putting all of these bounds together and applying Jensen’s inequality, the number of incidences between A and 𝒮 is at most

j=1Dd|Ej| nc4/loglognj=1Dd(|Pj||𝒮j|dd+1+|𝒮j|)
2n3+c4/loglognDdj=1Dd|𝒮j|dd+1+nc4/loglognj=1Dd|𝒮j|
2n3+c4/loglognDdDdd+1(j=1Dd|𝒮j|)dd+1+c5Dd1nc4/loglogn|𝒮|
2c5dd+1n3+c4/loglognDdd+1|𝒮|dd+1+c5Dd1nc4/loglogn|𝒮|.

Set D to be a power of 2 such that

n3(d+1)d2+d1|𝒮|1d2+d1D2n3(d+1)d2+d1|𝒮|1d2+d1.

Note that |𝒮||A|d+1. From above, the number of incidences between 𝒮 and A is at most

n3(d21)d2+d1+c8/loglogn|𝒮|d2d2+d1,

where c8=c8(d). Let 𝒮r𝒮 denote the set of r-rich spheres in 𝒮, that is, the set of spheres in 𝒮 with at least r points from A. Then we have

r|𝒮r|n3(d21)d2+d1+c8/loglogn|𝒮r|d2d2+d1,

which implies

|𝒮r|n3(d+1)+c8/loglognrd2+d1d1.

Let 𝒯i𝒮 be the set of spheres in 𝒮 with at least 2i points from A, and at most 2i+1 from A. Hence

|𝒯i||𝒮2i|=n3(d+1)+c8/loglogn2id2+d1d1.

Therefore, the number of (d+2)-tuples of A that lie on a common sphere S𝒮 is at most

i=1log|A||𝒯i|(2i+1d+2)n3(d+1)+c8/loglogni=1log|A|2i1d1n3(d+1)+c8/loglogn.

We now consider a random subset AA obtained by selecting each point in A independently with probability p, where p will be determined later. Let X be the number of (d+2)-tuples of A that lie on a common sphere or a hyperplane. Then, by linearity of expectation, we have

𝔼[X]n3(d+1)+c8/loglognpd+2+O(n2d+5)pd+2n3(d+1)+c9/loglognpd+2,

and

𝔼[|A|]=p|A|n3p/2,

where c9=c9(d). Hence, there is a c10=c10(d) such that for p=n3d/(d+1)c10/loglogn, we have 𝔼[X]<𝔼[|A|]/4. By Lemma 5, there is a subset AA of size at least p|A|/2, such that A contains at most p|A|/4 (d+2)-tuples on a sphere or a hyperplane. By deleting one point from each (d+2)-tuple on a sphere or a hyperplane, we obtain a subset A′′A of size at least

|A|p/4n3d+12c10/loglogn=n3d+1o(1),

such that no d+2 members in A′′ lie on a sphere or a hyperplane.

4 Concluding remarks

Several authors [13, 12, 4] observed that if n is prime, then by selecting n points from [n]d on the “modular moment curve” (x,x2,,xd)modn, for x=1,,n, no d+1 points will lie on a hyperplane, and moreover, no 2d+1 points lie on a sphere. Unfortunately, this construction may contain (d+2)-tuples on a sphere. Nevertheless, one can make the following simple observation.

Theorem 10.

Let d2 be a fixed positive integer. Then there is a subset of the d-dimensional lattice cube [n]d of size Ω(n) with no d+1 points on a hyperplane, and no 2d points on a sphere.

Proof.

Let n be prime, and let A[n]d be points from the d-dimensional cube on the moment curve (x,x2,,xd)modn, for 1xn/(4d). Hence, |A|=n/(4d). For sake of contradiction, suppose there is a sphere S in d, such that S contains 2d points from A. Note that this means S contains 2d of [n]d when viewed as a sphere in 𝔽nd. Let the center of S𝔽nd be (c1,,cd) and radius r, such that S contains 2d points from A. Then we have 2d solutions s1,s2,,s2d to the equation

(xc1)2+(x2c2)2+(xdcd)2r2=0modn.

On the other hand, by the division algorithm, we have

(xc1)2+(x2c2)2+(xdcd)2r2=(xs1)(xs2)(xs2d)modn,

which implies that s1++s2d=0modn as the coefficient of x2d1 is 0modn. However, sin/(4d), which implies s1++s2d0modn, contradiction. Hence, no 2d points in A lie on a sphere and no d+1 points lie on a hyperplane. If n is not prime, we apply Bertrand’s postulate to obtain a prime n<n such that n>n/2, and apply the argument above to [n]d.

Another natural question is to determine the maximum number of points that can be selected from [n]d with no d+2 points on a sphere, but allowing many points on a hyperplane. Since each point from [n]d lies on a sphere centered at the origin with radius t, where t=1,2,,dn2, we have an upper bound of d(d+1)n2 for this problem. Using Lemma 8 and the probabilistic method, one can show the existence of n24/(d+1)o(1) points from [n]2[n]d with no d+2 on a sphere.

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