Minimal number of points and lines so that the number of incidence pairs is $n$

Question

Let's define points and and lines in a plane in the following way. We have a set $\mathcal{P}$ whose elements we call points and a set $\mathcal{L}$ which contains such subsets of $\mathcal{P}$ so that for any two distinct points in $\mathcal{P}$ there is exactly one set in $\mathcal{L}$ which contains both of those points. We call elements of $\mathcal{L}$ lines. Incidence pair is an ordered pair $(p, L)$ where $p$ is a point, $L$ is a line and $p \in L$. A plane is defined by the pair $(\mathcal{P}, \mathcal{L})$.

I am interested in the following problem. Let's denote with $\mathcal{N}_{(\mathcal{P}, \mathcal{L})}$ all such collections of points and lines in the plane $(\mathcal{P}, \mathcal{L})$ for which the number of incidence pairs between these points and lines is $n$ which is a given number, that is $\mathcal{N}_{(\mathcal{P}, \mathcal{L})} = \{(A, B), A \subseteq \mathcal{P}, B \subseteq \mathcal{L}, \sum_{L\in B}|L\cap A| = n\}$. Now let $m_{(\mathcal{P}, \mathcal{L})} = \min(|A|+|B|, (A, B) \in \mathcal{N}_{(\mathcal{P}, \mathcal{L})})$, that is the smallest number of points and lines in the plane so that there are $n$ incidence pairs between them. Which plane has the minimal $m_{(\mathcal{P}, \mathcal{L})}$ and for which collection of points and lines is that minimum achieved in that plane?

An asymptotically optimal solution would suffice for me. Also, I am assuming that it is quite possible that for different families of $n$ the constructions of optimal solutions vary significantly. I am looking for a construction for such an unbounded family of $n$ for which the optimal solution is asymptotically minimal amongst all subsets of $\mathbb{N}$.

Let's clarify what I mean with an example. The plane with the least points and lines that I could come up with is for $n = (k + 1)(k^2 + k + 1)$ where you can take the entire finite projective plane of order $k$ which contains $2\cdot(k^2 + k + 1)$ points and lines. So the number of lines and points is of order $\Theta(n^{\frac{2}{3}})$. However, I don't know whether my solution is optimal for the given family of $n$. Furthermore, even if this is optimal for the given family of $n$, I don't know whether for some other family of $n$ there is an asymptotically better solution.

Answer 1

Given a natural number $n$, let $C(n)$ be the minimal $\mathcal{N}_{(\mathcal{P}, \mathcal{L})}$. Observe first that the function $C(n)$ is nondecreasing. Indeed, suppose there are a plane $(\mathcal{P}, \mathcal{L})$ and sets $A \subseteq \mathcal{P}$ and $B \subseteq \mathcal{L}$ such that $\sum_{L\in B}|L\cap A|\ge n$. It is easy to check that we can for each $L\in B$ pick a subset $L'$ of $L\cap A$ such that $\sum_{L\in B}|(L\setminus L')\cap A|=n$. Put $B'=\{L\setminus L':L\in B\}$ and $$\mathcal{L}'=\mathcal{L}\setminus B\cup\{L\setminus L':L\in B\}\cup\bigcup_{L\in B,\,L'\ne\varnothing} \{\{x,y\}:x\in L, y\in L'\}.$$ Then $(\mathcal{P}, \mathcal{L'})$ is a plane and $\sum_{L\in B'}|L\cap A|=n$.

Next, we show that $$C(n)\ge \frac{3\sqrt[3]{2}}2n^{2/3}+\left(\frac{\sqrt[3]{4}}{4}-1\right)n^{1/3}-0.27\dots.$$

Let $n$ be a fixed natural number and $(A,B)\in \mathcal{N}_{(\mathcal{P}, \mathcal{L})}$. Consider $|A|\times |B|$ axes-aligned rectangle whose rows are indexed by elements of $A$ and columns are indexed by elements of $B$. If $p\in A$, $L\in B$ and $p\in L$ then color the $(p,L)$th cell of the rectangle. Since for any two distinct points in $A$ there is at most one $L\in B$ containing these points, no four centers of distinct colored cells form vertices of an axes-aligned rectangle. According to this my answer, $2n\le m+\sqrt{D}$, where $m$ is the smallest of the numbers $|A|$ and $|B|$, $k$ is the other of these numbers, and $D=4mk^2-4mk+m^2$. Therefore $m\ge 2n$ or $D\ge (2n-m)^2$, that is $mk^2-mk\ge n^2-mn$. Then $m\ge \frac{n^2}{k^2-k+n}$ and thus $m+k\ge \frac{n^2}{k^2-k+n}+k$. When the minimum of the right-hand side is attained then its derivative with respect to $k$ equals $0$, that is $k=1$ or $k^3-k^2+2kn-2n^2=0$. The latter equation suggests that $k\approx \sqrt[3]{2}n^{2/3}$ and so the minimum of the right-hand side is about $\frac{3\sqrt[3]{2}}2n^{2/3}$, which provides the respective lower bound for $m_{(\mathcal{P}, \mathcal{L})}$ and thus for $C(n)\$

In fact, we can provide the announced concrete lower bound as follows. Let $r=\sqrt[3]{n}$ be the cubic root of $n$. The cubic polynomial $f(k)=k^3-k^2+2kn-2n^2$ has the positive derivative $f'(k)=3k^2-2k+n$, so $f(k)$ increases and so has a unique real root $k_0$. Since $$f\left(\sqrt[3]{2}r^2\right)=\sqrt[3]{2}r^2\left(2n-\sqrt[3]{2}r^2\right)>0,$$ we have $k_0<\sqrt[3]{2}r^2$. On the other hand, it can be checked (I did it via Mathcad) that

$$f\left(\sqrt[3]{2}r^2-\frac 23r\right)=$$ $$\left(2\sqrt[3]{2}-2\sqrt[3]{4}\right)r^5+ \left(\frac 43\cdot\sqrt[3]{2}-\sqrt[3]{4}-\frac 43\right)r^4+ \left(\frac 43\cdot\sqrt[3]{2}-\frac 8{27}\right)r^3- \frac 49r^2<0,$$ because $2\sqrt[3]{2}-2\sqrt[3]{4}=-0.65\dots$, $\frac 43\cdot\sqrt[3]{2}-\sqrt[3]{4}-\frac 43=-1.24\dots$, but $\frac 43\cdot\sqrt[3]{2}-\frac 8{27}=1.38\dots$. So $k_0>\sqrt[3]{2}r^2-\frac 23r$.

Next $$\frac{n^2}{k_0^2-k_0+n}+k_0=\frac{n^2+k_0^3-k_0^2+nk_0}{k_0^2-k_0+n}=\frac{n^2+2n^2-nk_0}{k_0^2-k_0+n}=$$ $$\frac{3n^2-nk_0}{\frac{2n^2}{k_0}-2n+n}=\frac{3nk_0-k_0^2}{2n-k_0}=k_0\left(1+\frac{n}{2n-k_0}\right).$$

Since when $1\le k\le 2n$ then the function $k\left(1+\frac{n}{2n-k}\right)$ increases and $k_0>\sqrt[3]{2}r^2-\frac 23r$, using Mathcad we obtain

$$\frac{n^2}{k_0^2-k_0+n}+k_0>\left(\sqrt[3]{2}r^2-\frac 23r\right) \left(1+\frac{n}{2n-\left(\sqrt[3]{2}r^2-\frac 23r\right)}\right)=$$ $$\frac{3\sqrt[3]{2}}2r^2+\left(\frac{\sqrt[3]{4}}{4}-1\right)r+\frac 14-\frac{\sqrt[3]2}3+ \frac{\left(2-\frac {9\sqrt[3]{4}}2-\frac {9\sqrt[3]{2}}4\right)r+2\sqrt[3]{2}-\frac 32}{18r^2-9\sqrt[3]{2}r+6}.$$

I checked via Mathcad that the last summand has positive derivative when $r\ge 1$, so it increases and thus

$$\frac{n^2}{k_0^2-k_0+n}+k_0>\frac{3\sqrt[3]{2}}2r^2+\left(\frac{\sqrt[3]{4}}{4}-1\right)r+\frac 14-\frac{\sqrt[3]2}3+ \frac{\left(2-\frac {9\sqrt[3]{4}}2-\frac {9\sqrt[3]{2}}4\right)1+2\sqrt[3]{2}-\frac 32}{18\cdot 1-9\sqrt[3]{2}\cdot 1+6}=$$ $$\frac{3\sqrt[3]{2}}2n^{2/3}+\left(\frac{\sqrt[3]{4}}{4}-1\right)n^{1/3}-0.27\dots.$$

And so the latter lower bound holds also for $m_{(\mathcal{P}, \mathcal{L})}$ and so for $C(n)$.

To provide an upper bound for $C(n)$ observe that a Steiner system $S(2,k,m)$ straightforwardly provides a bound $$C\left(\frac{m(m-1)}{k-1}\right)\le m+\frac{m(m-1)}{k(k-1)}.$$ Such a Steiner system exists when $(k,m)$ equals $(q, q^2)$ or $(q+1,q^2+q+1)$, where $q$ is a power of a prime number (unfortunately, we still cannot construct respective Steiner systems for other $q$). This yields the inequalities $C(q^3+q^2)\le 2q^2+q$ and $C((q^2+q+1)(q+1))\le 2q^2+2q+2$, respectively.

I expect to obtain from them the bound $C(n)\le 2n^{2/3}+O(n^{2/3})$ as follows. According to [BHP], there exists a natural number $x_0$, which, with enough effort, can be determined effectively, such that for all $x > x_0$ the interval $[x-x^{0.525},x]$ contains prime numbers. It follows that for all sufficiently big natural $n$ there exists a prime $q$ such that $n\le q^3+q^2\le ? $. Then $$C(n)\le C(q^3+q^2)\le 2q^2+q\le ? $$

References

[BHP] R. Baker, G. Harman, J. Pintz, The difference between consecutive primes II, Proc. Lond. Math. Soc., (3) Ser. 83 (2001) 532-562.