Given $N$ points $X:=(x_i)_{i \in \{1,..,N\}}$, we now define a score function $S:X \rightarrow \mathbb{N}$ that is $S(X)= \sum_{i=1}^N S(x_i)$ where the score of $S(x_i)$ is

$$S(x_i) = 2* \vert \{x_j; \vert x_i-x_j \vert \in [1,2]\} \vert+ \vert \{x_j; \vert x_i-x_j \vert \in [2,3]\} \vert$$ where $\vert \bullet \vert$ denotes the cardinality of the set. Moreover, we require that for all $i\neq j$ we have $\vert x_i-x_j \vert \ge \frac{1}{2}.$

Question: Is it true that any configuration of $N$ points with maximal possible score is in a domain of diameter $c\sqrt{N}$ for some fixed c?

  • $\begingroup$ Do you assume that your configuration of $N$ points belongs to $\mathbb Z^2$, or you consider all possible configurations of points in $\mathbb R^2$? $\endgroup$ Sep 20, 2020 at 12:45
  • $\begingroup$ @DmitriPanov I allow for any configurations of points. $\endgroup$
    – Sascha
    Sep 20, 2020 at 13:16
  • $\begingroup$ My comment about the integer lattice was only intended to suggest that spacing of order 1 would in general lead to a configuration of size $\sqrt{N}.$ $\endgroup$
    – Sascha
    Sep 20, 2020 at 13:37
  • $\begingroup$ Thanks! And just to double check, $[1,2]$ means the interval $[1,2]$ (not the union of $1$ and $2$) $\endgroup$ Sep 20, 2020 at 13:42
  • $\begingroup$ @DmitriPanov that's right. $\endgroup$
    – Sascha
    Sep 20, 2020 at 16:31

1 Answer 1


What follows below is the new answer to the modified question, where we assume in addition $|x_i-x_j|\ge 1$ (probably one can ask $|x_i-x_j|\ge 1-\varepsilon$ for sufficiently small $\varepsilon$). I want to propose a positive solution of this problem modulo the following guess, which, I hope, is correct.

Guess. Consider the equilateral triangular lattice $E$ with distance $1$ between neighbouring points. Then there are exactly $18$ points on distance at most $2$ from a given one, and $36$ on distance at most $3$. So the score of each point is $54=2*18+(36-18)$. I guess, that for any set $X$ such that any two points are on distance at least $1$, in the punctured $2$-neighbourhood of any point $x\in X$ there are at most $18$ points of $X$. I guess that the same holds for points on distance at most $3$. If this is true then we have the following corollary: for any set $X$ satisfying $|x_i-x_j|\ge 1$ we have $S(x_i)\le 54 $.

So from now on we assume that either the guess is correct, or we are working with a set $X$ such that each point of this set has score at most $54$.

I'll prove that under such condition the constant $c$ exists.

Proof. Note first of all that we can always construct a set $X$ with $N$ points, such that the score of $X$ is $54N-10^{10}\sqrt{N}$. Such a set can be given by intersecting $E$ with a disk of appropriate radius. (one can take a smaller constant than $10^{10}$, but it doesn't matter).

Assume by contradiction, that we constructed a set $X$ maximising the score and such that its diameter is more than $10^{10^{10}}\sqrt{N}$. Take the union of disks of radius $3$ around all points of $X$, and denote this set by $U_3$. It is easy to see that $U_3$ is connected. Indeed, if it is not, we can parallel translate its connected component by pushing one to another and increase this way the score of $X$. So, since the diameter of $X$ is at least $10^{10^{10}}\sqrt{N}$, the perimeter of the exterior boundary of $U_3$ is at least $10^{10^{10}}\sqrt{N}$. We will say that a point of $X$ contributes to the exterior boundary of $U_3$ if it is on distance $3$ from it. It is easy to see that the number of points of $X$ contributing to the exterior boundary is at least $10^{(10^{10}-2)}\sqrt{N}$ (because the length of a radius $3$ circle is $<100$). The final observation is that any point $x$ of $X$ that contributes to the boundary has score less than $54$. This is because the disk of radius $3$ around $x$ has a large sub-region, where points of $X$ can not lie (indeed, take a point $y\in \partial U_3$ on distance $3$ from $x$, then no point on distance less than $3$ from $y$ lies in $X$). Finally, taking into account the guess and the fact that the score of $X$ has to be at least $54N-10^{10}\sqrt{N}$, we get a contradiction.

Old answer

Let's consider two variations of this question. In both cases the answer is yes. In the first case $X$ is any subset of $\mathbb R^2$ in the second it is a subset of $\mathbb Z^2$.

1 We assume first that $X$ is any subset of $\mathbb R^2$. In such case the set with maximal possible score has diameter at most $6$. Let me prove this. Let's first construct a set with score approximatively $\frac{5}{3}N^2$. To do this we put $N/6$ points in each vertice of the regular hexagon with side of length $1$.

Now, suppose we have a set with maximal score and suppose its diameter is more than $6$. We will construct a set with larger score which will give us a contradiction.

So, suppose $X$ has two points $x_i, x_j$ such that $|x_i-x_j|>6$. Let's take two disks of radius $3$ around both points. One of them contains at most $N/2$ points, which means $S(x_i)$ or $S(x_j)$ is at most $N$. Without loss of generality assume $S(x_i)\le N$. On the other hand, we know that $S(X)\ge \frac{5 N^2}{3}$. So, there is a point $x_k$ such that $S(x_k)\ge\frac{5}{3}N$. Move $x_i$ at the place of $x_k$, this will increase the score $S(X)$. Contradiction.

2 What follows is just a sketch of proof. We assume $X\subset \mathbb Z^2$. In such case each point $x_i$ contributes at most $2*8+20=36$ to the sum $S(X)$. Indeed, there are $8$ integer points on distance at most $2$ from a given one, and $20$ on distance in $[2,3]$. From this one can deduce the answer applying the isomperietric inequality to the set that is the union of $2\times 2$ squares with centres at points of $X$. I can give more details, if you want.

  • $\begingroup$ interesting, thank you, indeed I was interested mostly in Case 1. Do you think one can include a penalization saying: All points, are at least a distance $1/2$ away from one another and then still get the $\sqrt{N}$ scaling? Cause now it seems there is nothing in my question that would make the volume grow... $\endgroup$
    – Sascha
    Sep 20, 2020 at 15:29
  • $\begingroup$ I'll need to think about this. This might be harder. I think this depends on whether positioning points in a lattice (for, example Eisensteins one) gives you the best possible score for one vertex. So some variant of the problem (if you fiddle with the intervals [1, 2] and [2, 3] might have a solution using isoperimetric inequality $\endgroup$ Sep 20, 2020 at 16:17

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