A Note on the No--On-a-Sphere Problem
Abstract
For fixed , we construct subsets of the -dimensional lattice cube of size with no points on a sphere or a hyperplane. This improves the previously best known bound of due to Thiele from 1995.
Keywords and phrases:
General position, no-four-on-a-cirle, -dimensional lattice cubeFunding:
Andrew Suk: Supported by NSF CAREER award DMS-1800746, NSF award DMS-1952786, and NSF grant DMS-2246847.Copyright and License:
![[Uncaptioned image]](x1.png)
2012 ACM Subject Classification:
Mathematics of computing CombinatoricsEditors:
Oswin Aichholzer and Haitao WangSeries and Publisher:

1 Introduction
The famous no-three-in-line problem, raised by Dudeney [5] in 1917 in the special case , asks if it is possible to select points from the lattice square such that no three are collinear. Clearly, one cannot do better than as can be covered by lines. For , many authors have published solutions to this problem obtaining the bound of (e.g. see [6]). However, for large values of , the best known lower bound is due to Hall et al. [9] which contains at least points, for any and 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 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 as any vertical line must contain at most three points. This upper bound was improved by Thiele [19], who showed that at most points can be selected, and moreover, he gave a construction of points from 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 -sphere is a -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 -dimensional sphere in . Again, the maximum number of points that can be selected from the -dimensional lattice cube with no points on a sphere or a hyperplane is at most since we can cover with hyperplanes. In the other direction, Thiele [18] showed that one can select points from with no 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 be a positive integer. Then there is a subset of the -dimensional lattice cube with points with no members on a sphere or a hyperplane.
While it is possible that one can improve this bound to , we make the following more modest conjecture.
Conjecture 2.
Let be a positive integer. Then there is a subset of the -dimensional lattice cube with points with no 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 be a bipartite graph with independent sets and edges . For we define the neighbourhood . Define the set-system with ground set . We say a set is shattered by if for every subset , there is a set such that . The Vapnik-Chervonenkis dimension (VC-dimension) of is the largest integer for which there exists a subset , with , that is shattered by .
The primal shatter function of is defined as
The well-known Sauer-Shelah lemma [14, 16] states that if is the VC-dimension of , then
(1) |
We will need the following result due to Fox, Pach, Sheffer, Suk, and Zahl.
Theorem 3 ([7]).
Let be a bipartite graph with and , such that the set system , with ground set , satisfies for all . Then if is -free, we have
where . In particular,
A subset is a maximal spherical set of if all points in lie on a single sphere in , and no point of can be added to while keeping all points on the sphere . Let denote the set-system of all maximal spherical sets of , with ground set .
Lemma 4.
For , the VC-dimension of the set-system is at most .
Proof.
We proceed by induction on . The base case is trivial since a 0-sphere consists of 2 points. For the inductive step, assume the statement holds for . For sake of contradiction, suppose that a set of points in is shattered by .
Case 1.
Suppose is in general position, that is, no members in lie on a hyperplane. By linear algebra, any points in determines a unique sphere in . However, since is shattered by , there is a sphere that contains . Hence, there is no sphere that contains exactly points from . Contradiction.
Case 2.
Suppose contains a -tuple that lies on a hyperplane in . Since the intersection of a sphere with is a -sphere in , by induction on , cannot be shattered by . Hence, cannot be shattered by , 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 be independent random variables such that and , and let . Then for , we have
and
Next, we estimate the number of points from 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 be a positive integer, and not be all zero with greatest common denominator 1. Let , be a positive integer, and set
If spans a -dimensional subspace, then .
Lemma 7.
The number of -tuples in that lie on a common hyperplane is at most .
Proof.
By the lemma above, the number of -tuples in that lie on a hyperplane which passes through the origin is at most
By symmetry, for any fixed point , the number of -tuples in that lies on a hyperplane which passes through is at most After summing over all points in , the statement follows. Note that Lemma 7 applies to hyperplanes whose intersection with spans a -dimensional subspace. We can assume that any -tuple in lying on a common hyperplane meets this requirement by adding other points from if needed.
Finally, we will use the following result due to Sheffer that bounds the number of points from that lies on a -sphere in , for . See also [10, Sec. 11.2, Equation 11.9], [17, Theorem 2].
Lemma 8 ([15], Lemma 3.2).
Let and . Then there is a positive constant such that every -sphere in contains at most points from .
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 such that very few members of lie on a -sphere or a hyperplane in . We then apply Theorem 3 and ideas from incidence geometry to estimate the number of -tuples in that lie on a common sphere. Finally, we apply the deletion method to find a large subset such that no members of 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 is a power of 2. Indeed, otherwise we can find an integer that is a power of 2 such that , and apply the arguments below to the subcube . This would only change the hidden constant in the term in the exponent. Moreover, we will assume that is sufficiently large.
Fix a positive integer that will be determined later, such that divides . We partition the -dimensional cube into smaller cubes , where each is of the form
where . Hence, .
Consider a random subset obtained by selecting each point in independently with probability . Set , for . Let be the event that the subset satisfies the following properties.
-
1.
.
-
2.
for all .
-
3.
Every -sphere contains less than points of , where .
-
4.
The number -tuples in that lie on a common hyperplane is at most , where .
By Lemma 5 and Markov’s inequality, event holds with probability at least 1/2. Indeed, by Lemma 5, the probability that or is at most Thus, the first property holds with high probability, that is, with probability tending to 1 as approaches infinity. A similar argument follows for the second property. For the third property, let us fix a -sphere in . By Lemma 8, there is a constant such that contains at most points from . By Lemma 5,
Since there are at most -spheres in with at least points from , the union bound implies that the third property holds with high probability. For the fourth property, let denote the number -tuples in that lie on a common hyperplane. By Lemma 7, we have , where . By Markov’s inequality (see [1]), we have
Putting everything together, and setting sufficiently large, event holds with probability at least .
Thus, let us fix with the four properties described above, and set , where . Let be the collection of spheres in , such that each sphere in contains at least points from in general position. Hence, . Let be the set of spheres in that contains at least one point of . Let be the bipartite incidence graph between and . Since the intersection of two distinct spheres in is a -sphere, and every -sphere contains less than points from , each graph is -free.
Let be the set system whose ground set is , and whose sets are , where and . That is,
By Lemma 4, the VC-dimension of is at most . By inequality (1), we have
Hence, we apply Theorem 3 with to conclude that
where .
Given the partition described above, we say that a sphere crosses the subcube if .
Observation 9.
For fixed , every sphere in crosses at most subcubes , where .
Proof.
Let be a large constant that depends only on . We will determine later. For sake of contradiction, suppose there is a sphere in the crosses more than subcubes . Note that we will not make any serious attempts to optimize the constant .111A careful calculation shows that one can take . Another proof follows from a simple inductive argument with .
Let be a subset of points obtained by selecting 1 point from each subcube that crosses. Hence, Let be an auxiliary graph on , where two points in are adjacent if their distance is less than . Since two points in have distance less than only if they lie in adjacent subcubes and , this implies that has maximum degree . By Turan’s theorem, we can find a subset , such that the minimum distance between any two points in is at least . Moreover, .
For each point , consider the spherical cap obtained by intersecting with the ball centered at with radius . Since the minimum distance among the points in is at least , the collection of spherical caps corresponding to the points in will have pairwise disjoint interiors. Moreover, since each spherical cap arises from a ball whose center is on with radius , each spherical cap will have area at least where . Hence, the total area of all of the spherical caps corresponding to the points in is at least
On the other hand, all of the spherical caps on will lie inside of a “large” ball with radius . Hence, the area of all of the spherical caps is at most the surface area of the ball with radius . Since the surface area of is at most , where , we have
By setting sufficiently large, we have a contradiction.
By the observation above, . Putting all of these bounds together and applying Jensen’s inequality, the number of incidences between and is at most
Set to be a power of 2 such that
Note that . From above, the number of incidences between and is at most
where . Let denote the set of -rich spheres in , that is, the set of spheres in with at least points from . Then we have
which implies
Let be the set of spheres in with at least points from , and at most from . Hence
Therefore, the number of -tuples of that lie on a common sphere is at most
We now consider a random subset obtained by selecting each point in independently with probability , where will be determined later. Let be the number of -tuples of that lie on a common sphere or a hyperplane. Then, by linearity of expectation, we have
and
where . Hence, there is a such that for , we have . By Lemma 5, there is a subset of size at least , such that contains at most -tuples on a sphere or a hyperplane. By deleting one point from each -tuple on a sphere or a hyperplane, we obtain a subset of size at least
such that no members in lie on a sphere or a hyperplane.
4 Concluding remarks
Several authors [13, 12, 4] observed that if is prime, then by selecting points from on the “modular moment curve” , for , no points will lie on a hyperplane, and moreover, no points lie on a sphere. Unfortunately, this construction may contain -tuples on a sphere. Nevertheless, one can make the following simple observation.
Theorem 10.
Let be a fixed positive integer. Then there is a subset of the -dimensional lattice cube of size with no points on a hyperplane, and no points on a sphere.
Proof.
Let be prime, and let be points from the -dimensional cube on the moment curve , for . Hence, . For sake of contradiction, suppose there is a sphere in , such that contains points from . Note that this means contains of when viewed as a sphere in . Let the center of be and radius , such that contains points from . Then we have solutions to the equation
On the other hand, by the division algorithm, we have
which implies that as the coefficient of is . However, , which implies , contradiction. Hence, no points in lie on a sphere and no points lie on a hyperplane. If is not prime, we apply Bertrand’s postulate to obtain a prime such that , and apply the argument above to .
Another natural question is to determine the maximum number of points that can be selected from with no points on a sphere, but allowing many points on a hyperplane. Since each point from lies on a sphere centered at the origin with radius , where , we have an upper bound of for this problem. Using Lemma 8 and the probabilistic method, one can show the existence of points from with no on a sphere.
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