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For a prime $p$, fix two bases $U=\{v_1,\dots,v_n\}$ and $W=\{w_1,\dots,w_n\}$ of the vector space $V=\mathbf{F}_p^n$. We may assume $U$ is the standard basis without loss of generality.

For $s_1,\dots, s_n,t_1\dots, t_n\in\mathbf{F}_p\setminus\{0\}$, let $S_i=\{x\in\mathbf{F}_p^n\mid x\cdot v_i=s_i \}$ and $T_i=\{x\in\mathbf{F}_p^n\mid x\cdot w_i=t_i \}$ be the affine planes for each $1\le i\le n$.

I am wondering is there any result studying how many points in $V$ can be covered by $\cup_{i=1}^n (S_i\cup T_i)$, say, $$\min_{\text{ $U$ is the standard basis, $W$ is a base}}\max_{s_1,\dots,s_n,t_1,\dots, t_n\in\mathbf{F}_p\setminus\{0\}}|\cup_{i=1}^n (S_i\cup T_i)|?$$

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If we choose $s$ and $t$ at random then the probability a point $P$ does not lie in the union of the $S_i$ is $(1- (p-1)^{-1})^{k(P)}$ where $k(P)$ is the number of nonzero entries of $P$ and the the probability $P$ does not lie in the union of the $t_i$ is $(1-(p-1)^{-1})^{k'(P)}$ where $k'(P)$ is the number of nonzero coefficients of $P$ expressed in terms of $W$, and these two events are independent, so the probability each point does not lie in the union of the $S_i$ and $T_i$ is $(1-(p-1) ^{-1})^{k(P)+k'(P)}$ and so the union of the $S_i$ and $T_i$ has size, in expectation, at least $$p^n - \sum_{x \in \mathbf F_p^n} (1-(p-1) ^{-1})^{k(x)+k'(x)}.$$ The distribution of values of the functions $k$ and $k's $ is independent of the choice of $W$. The product of two functions, subject to the condition that the distribution of the values of the two functions is fixed (and identical), is maximized when the two functions are equal, so this expectation is at most

$$p^n - \sum_{x \in \mathbf F_p^n} (1-(p-1) ^{-1})^{k(x)+k(x)} = p^n - ( 1 + (p-1) (1-(p-1) ^{-1})^2)^n $$ $$= p^n \left( 1 - \left(\frac{p^2 -3p+3 }{p(p-1)} \right)^n\right)$$

Since this is the expected size it is a lower bound for the maximum size. Since this works for every $W$, it is a lower bound for the min-max.

On the other hand, for $W$ the standard basis, we can't do better than choosing $s_i \neq t_i$ for all $i$ which gives a union of size $$ p^n ( 1- (1- 2p^{-1})^n)$$ so this gives an upper bound for the min-max.

These two bounds are not very far from one another since $$ \frac{p^2 -3p+3 }{p(p-1)} - (1-2 p^{-1} )=\frac{1}{p(p-1)}$$ but both arguments are simple so it's possible either can be improved.

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  • $\begingroup$ Those $s_i,t_i$ are non-zero, which makes the probability not $(1-p^{-1})^n$. $\endgroup$
    – Connor
    Commented Sep 20 at 14:49
  • $\begingroup$ @Connor Sure, this slightly alters the bound. $\endgroup$
    – Will Sawin
    Commented Sep 20 at 21:27
  • $\begingroup$ @Connor Does it make any difference? If there is an uncovered point, you can move it to $0$ by a linear shift, thus making all $s)i$ and $t_i$ to be nonzero. $\endgroup$ Commented Sep 25 at 7:07
  • $\begingroup$ @IlyaBogdanov Clever trick! That shrinks the gap back from $1/p(p-1)$ to $1/p^2$. $\endgroup$
    – Will Sawin
    Commented Sep 25 at 12:43

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