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This is a follow-up question, to a question I asked earlier.

See Algebraic curve intersecting square-grid.

Consider $n^d$ unit hypercubes in $d$-dimensional Euclidean space tightly packed in the canonical way.

Let $f \in \mathbb{Z}^d$ be a vector, then we define the hypercube $c_f$ as $$ c_f = \{ f + x \in \mathbb{R}^d : 0\leq x_i \leq 1, \forall i =1,\ldots,d\} $$ We consider all hypercubes with $1\leq f_i\leq n$.

Let us now consider a polynomial $p$ in $d$ variables with maximum degree $\Delta$.

How many hypercubes can $p$ intersect, in terms of $\Delta$, $n$ and $d$?

(We say $p$ intersects a set $S$ if $\exists \ x \in S : p(x) = 0$.)

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Let us start with the case $d = 1$. In this case, we have a univariate polynomial $p(x)$ and we ask how many unit intervals it can hit at most. The answer $\Delta$ is well known.

Now we tackle the case $d = 2$, for pedagogical purposes. Let us assume that $c = \{ x\in \mathbb{R}^2: p(x) = 0\}$ has only one connected component. (Here is a gap in the argument, that we don't want to fill for the moment.) Note that $c$ is a one-dimensional object and we can think of $c$ as going in and out of a component. So whenever $c$ does this, we visited one more component. In two dimensions, our cubes are bounded by $n+1$ horizontal and vertical lines. It is sufficient to count the number of times such a line is hit. Now the set $c \cap \ell$, for a line $\ell$ can be described by a univariate polynomial of max degree $\Delta$. Thus by the case $d=1$, each line is intersected at most $\Delta$ times. This gives an upper bound of $(2n+2)(\Delta) \leq 3n\Delta$. (We assume $n\geq 2$.) See also the answer of Dmitri Panov (Algebraic curve intersecting square-grid)

Now let us go to the general induction step $(d-1) \rightarrow d$. Again, let us assume that $c = \{ x\in \mathbb{R}^2: p(x) = 0\}$ has only one connected component. All the hypercubes are bounded by $dn+d$ hyperplanes. Every intersection of $c$ to a hypercube is witnessed by an intersection of $c$ to at least one of the hyperplanes. Now consider one of the hyperplanes $H$ and consider the induced grid arrangement $A$ on $H$. By induction, at most $(d-1)!(n+1)^{d-2}\Delta$ of the cells of $A$ are visited by $c$. Thus in total among all the hyperplanes at most $(dn+d) \cdot (d-1)!(n+1)^{d-2}\Delta \leq d!(n+1)^{d-1}\Delta$ of the induced cells are touched by $c$. (We assume $n\geq 2$.) Thus also at most $3n^{d-1}\Delta$ full dimensional cubes are visited by $c$.

Additionally, if $c$ has more components, some of the components are completely contained in some of the hypercubes. But the number of components is bounded by a function of $\Delta$ and $d$ and independent of $n$.

So the answer will be $d!(n+1)^{d-1}\Delta + f(d,\Delta)$.

Remark: We have to take care of the $f(d,\Delta)$ also in the induction step. :(. This proof is not complete yet. It would be nice to absorb it somehow in the first term.

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