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  pdftitle={Multi-coverings by lattice translates of convex bodies}
}

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\newtheorem{theorem}{Theorem}[section]
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\title{Multi-coverings by lattice translates of convex bodies}

\author{Scott Duke Kominers%
  \thanks{%
    \textsc{Harvard Business School; Department of Economics and Center of
    Mathematical Sciences and Applications, Harvard University; and
    a16z crypto.}
    The analysis in this article was heavily LLM-assisted, building off
    of the construction in~\cite{Lattice2026} iteratively with my guidance,
    mostly using GPT-6 Astra (accessed in part via Poe with the support
    of Quora, where I am an advisor). The problem, final methods, and
    written form are my own; and of course any errors remain my
    responsibility. This work was conducted while I was visiting the
    Technological Innovation, Entrepreneurship, and Strategic Management
    (TIES) Group at the MIT Sloan School of Management; I greatly
    appreciate their hospitality.%
  }%
}

\date{Preliminary Version -- October 8, 2026}

\begin{document}

\maketitle

\begin{abstract}
We show that for every convex body $K\subset\mathbb R^n$ ($n\ge2$),
every $0<\varepsilon<1$, and every prescribed density
$\rho\ge A_\varepsilon n\log n$, there is an arrangement of translates
of $K$ along a single lattice with mean multiplicity exactly $\rho$
and multiplicity at least $(1-\varepsilon)\rho$ everywhere.
Consequently, lattice $k$-fold coverings exist with density at most
$\max\{(1+\varepsilon)k,C_\varepsilon n\log n\}$, where
$C_\varepsilon$ depends only on $\varepsilon$.
In sufficiently large dimension $n$, a version with prescribed common
covolume holds simultaneously for up to $\exp(n^{c_0})$ bodies whose
uniform distributions have a common covariance matrix, with
$c_0>0$ absolute.
The proof builds on OpenAI's horizontal--vertical construction for
single-lattice coverings and uses Gaussian layer averaging and
Laplace bounds for covering counts.
\end{abstract}

\section{Introduction}

Let $K\subset\mathbb R^n$ be a \emph{convex body}, i.e., a compact
convex set with nonempty interior. For a positive integer $k$, a
\emph{$k$-fold covering} (or \emph{multi-covering}) of $\mathbb R^n$
by translates of $K$ is a family in which every point of $\mathbb R^n$
belongs to at least $k$ distinct translates.
We study \emph{lattice $k$-fold coverings}: the family is exactly
$\{K+\lambda:\lambda\in\Lambda\}$ for a full-rank lattice
$\Lambda\subset\mathbb R^n$, with one translate for each lattice point.

The \emph{multiplicity function} and \emph{density} of this lattice
arrangement are
\[
  N_{\Lambda,K}(x)=\sum_{\lambda\in\Lambda}\mathbf1_K(x-\lambda),
  \qquad
  \rho=\frac{\operatorname{vol}_n(K)}{\det\Lambda},
\]
where the \emph{covolume} $\det\Lambda$ is the volume of a fundamental
parallelepiped of $\Lambda$. The density $\rho$ is the mean of
$N_{\Lambda,K}$ on the \emph{lattice torus} $\mathbb R^n/\Lambda$,
equipped with normalized Haar measure. Thus the covering condition
$N_{\Lambda,K}(x)\ge k$ for every $x\in\mathbb R^n$ implies $\rho\ge k$.
We ask how close the density of a lattice $k$-fold covering can come
to this lower bound.

We write $\theta_{\mathrm{lat},k}(K)$ for the infimum of the densities
of such lattice $k$-fold coverings. Thus
$\theta_{\mathrm{lat},k}(K)\ge k$. Moreover, the infimum is attained,
as the compactness argument in Appendix~\ref{sec:attainment} shows.

OpenAI~\cite[Theorem~1.1]{Lattice2026} proved that
$\theta_{\mathrm{lat},1}(K)=O(n\log n)$ uniformly over convex bodies.
If $K+\Lambda_0$ covers space, then an index-$k$ superlattice of
$\Lambda_0$ gives a $k$-fold covering, as each of its $k$ cosets supplies
one covering; this yields density $O(kn\log n)$. We show however that
it is possible to do much better: the dimensional cost need only be
paid once.

\begin{theorem}[Lattice multi-coverings]
\label{thm:multiple}
For every $\varepsilon>0$ there is a constant $C_\varepsilon$ such
that, for every $n\ge2$, every convex body $K\subset\mathbb R^n$, and
every positive integer $k$,
\[
  \theta_{\mathrm{lat},k}(K)
  \le \max\{(1+\varepsilon)k,C_\varepsilon n\log n\}.
\]
\end{theorem}

Thus $\theta_{\mathrm{lat},k}(K)/k\to1$ whenever
$k/(n\log n)\to\infty$, uniformly in the dimension and the body.
Fixed-dimensional asymptotic efficiency follows from elementary
lattice refinement, as we recall below.

The proof gives more: We may prescribe the mean count and make the
minimum count nearly as large, simultaneously for many bodies with
the same covariance matrix.

For a convex body $K$, the \emph{uniform distribution on $K$} is
normalized Lebesgue measure, with density $\mathbf1_K/|K|$, where
$|K|=\operatorname{vol}_n(K)$. Its \emph{barycenter} and
\emph{covariance matrix} are
\[
  b_K=\frac1{|K|}\int_K x\,dx,
  \qquad
  \Sigma_K=\frac1{|K|}
    \int_K(x-b_K)(x-b_K)^{\mathsf T}\,dx,
\]
respectively.

\begin{theorem}[Prescribed covolume]
\label{thm:prescribed}
There is an absolute constant $c_0>0$ such that, for every
$0<\varepsilon<1$, there are constants $A_\varepsilon,n_\varepsilon$
with the following property.
Suppose $n\ge n_\varepsilon$, $1\le N\le\exp(n^{c_0})$, and
$K_1,\ldots,K_N\subset\mathbb R^n$ are convex bodies whose uniform
distributions have a common covariance matrix. If $\Delta>0$
satisfies
\[
  \rho_j:=\frac{\operatorname{vol}_n(K_j)}{\Delta}
  \ge A_\varepsilon n\log n
  \qquad(1\le j\le N),
\]
then there is a full-rank lattice $\Lambda=\Lambda(\Delta)$ with
\[
  \det\Lambda=\Delta,
  \qquad
  N_{\Lambda,K_j}(x)\ge(1-\varepsilon)\rho_j
  \quad(x\in\mathbb R^n,\ 1\le j\le N).
\]
When $N=1$, the conclusion holds for every $n\ge2$.
\end{theorem}

The covariance hypothesis in Theorem~\ref{thm:prescribed} allows a
common linear normalization; it holds, for example, for orthogonal
images of a body with scalar covariance matrix. The bodies may be
nonsymmetric and need not have comparable volumes.

\begin{proof}[Proof of Theorem~\ref{thm:multiple} from
  Theorem~\ref{thm:prescribed}]
Fix $\varepsilon>0$ and $k\ge1$. Put
$\zeta=\varepsilon/(1+\varepsilon)$, and prescribe
\[
  \rho=\max\{(1+\varepsilon)k,A_\zeta n\log n\},
  \qquad
  \Delta=\frac{|K|}{\rho}.
\]
The one-body clause of Theorem~\ref{thm:prescribed} applies in every
dimension $n\ge2$, giving a lattice of density exactly $\rho$ with
\[
  N_{\Lambda,K}(x)\ge(1-\zeta)\rho
  =\frac{\rho}{1+\varepsilon}\ge k
  \qquad(x\in\mathbb R^n).
\]
This proves Theorem~\ref{thm:multiple} with
$C_\varepsilon=A_{\varepsilon/(1+\varepsilon)}$.
\end{proof}

More generally, for prescribed positive integers $k_1,\ldots,k_N$
and $\varepsilon>0$, set $\zeta=\varepsilon/(1+\varepsilon)$ and
choose
\begin{equation}
  \Delta=\min_{1\le j\le N}
  \frac{\operatorname{vol}_n(K_j)}
       {\max\{(1+\varepsilon)k_j,\ A_\zeta n\log n\}}.
  \label{eq:simultaneous-multiplicity}
\end{equation}
For $n\ge n_\zeta$, this gives
$N_{\Lambda,K_j}\ge\rho_j/(1+\varepsilon)\ge k_j$ for every $j$.
The common covolume couples the densities: for some bodies,
$\rho_j$ may greatly exceed
$\max\{(1+\varepsilon)k_j,A_\zeta n\log n\}$.

Note that Theorem~\ref{thm:prescribed} supplies a lattice that may
depend on $\Delta$; it does not assert that one lattice shape works
at all larger densities. We do not optimize the dependence of
$A_\varepsilon,C_\varepsilon$, or $n_\varepsilon$ on the accuracy.

The order of the bound is best-possible in the worst case.

\begin{corollary}[Worst-case order]
\label{cor:worst-case}
Uniformly for $n\ge2$ and positive integers $k$,
\[
  \sup_{K=-K}\theta_{\mathrm{lat},k}(K)
  \asymp\sup_K\theta_{\mathrm{lat},k}(K)
  \asymp k+n\log n,
\]
where the suprema range over origin-symmetric convex bodies and all
convex bodies in $\mathbb R^n$, respectively.
\end{corollary}

\begin{proof}
Li and Liu~\cite[Theorem~1.3]{LiLiu2026} construct, for every
sufficiently large $n$, an origin-symmetric body $K_n$ with
$\theta_{\mathrm{lat},1}(K_n)\ge c_{\mathrm{LL}}n\log n$, for an
absolute $c_{\mathrm{LL}}>0$. Since
$\theta_{\mathrm{lat},k}(K)\ge
\max\{k,\theta_{\mathrm{lat},1}(K)\}$, this gives the lower bound;
the finitely many smaller dimensions follow from
$\theta_{\mathrm{lat},k}(K)\ge k$. Theorem~\ref{thm:multiple} with
$\varepsilon=1$ gives the upper bound.
\end{proof}

\paragraph{Related work.}
Multiple lattice coverings have a classical literature,
see, for example, the work of Cohn~\cite{Cohn1976} and
Groemer~\cite{Groemer1986}.
Asymptotic efficiency in fixed dimension already follows from an
elementary refinement. If an ordinary covering lattice for $K$ has
density $\rho_0$, the argument in
Section~\ref{sec:small-dimensions} gives
\begin{equation}
  \theta_{\mathrm{lat},k}(K)
  \le\bigl(k^{1/n}+\rho_0^{1/n}\bigr)^n
  =k+O_n\!\left(k^{(n-1)/n}\right),
  \label{eq:elementary-multiple}
\end{equation}
where a bound for $\rho_0$ depending only on $n$ makes the estimate
uniform over $K$. The quantitative issue is how large $k$ must be
as the dimension grows. To guarantee density at most
$(1+\varepsilon)k$, the bound~\eqref{eq:elementary-multiple} requires
$k\ge\rho_0[(1+\varepsilon)^{1/n}-1]^{-n}$;
for fixed $\varepsilon$, this sufficient lower bound on $k$ is
comparable to $\rho_0(n/\log(1+\varepsilon))^n$, with comparison
constants depending only on $\varepsilon$.
Theorem~\ref{thm:multiple} guarantees the same relative accuracy as
soon as $k\ge C_\varepsilon n\log n$; the improvement is that a
multiplicity of order $n\log n$ suffices for near-optimal density,
uniformly over convex bodies.

Rogers~\cite{Rogers1957} proved that ordinary coverings by arbitrary
translates have density $O(n\log n)$. Under the single-lattice
restriction, his general bound was
$n^{\log_2\log n+O(1)}$~\cite{Rogers1959}.
Ordentlich, Regev, and Weiss~\cite{ORW2022} obtained the universal
bound $O(n^2)$; Li and Liu~\cite{LiLiu2026} then reduced this to
$n\log n$ times an iterated-logarithmic factor, which
OpenAI~\cite{Lattice2026} was able to remove. Our proof uses local
Gaussian block estimates from the OpenAI construction~\cite[Section~4]{Lattice2026}.
A companion to the OpenAI preprint~\cite{Lattice2026}
states~\cite[Theorem~1.1]{Translative2026} that there are centrally symmetric
bodies requiring density $\Omega(n\log n)$ even for ordinary coverings by
arbitrary translates; we do not use that result here, as the lattice lower
bound of Li and Liu already suffices for Corollary~\ref{cor:worst-case}.

For multiple coverings by arbitrary translates, Frankl, Nagy, and
Nasz\'odi~\cite[Theorem~1.3]{FNN2018} obtained density $O(n\log n)$
for $k\le n(\log n+\log\log n)$.
Nasz\'odi and Polyanskii~\cite[Theorem~1.4]{NP2018} proved the bound
$3.153(1+o(1))\max\{n\log n,k\}$.
Retaining the free parameter in their greedy bound for multiple
transversals~\cite[Theorem~1.1]{NP2018} also gives
$\max\{(1+\varepsilon)k,C^{\mathrm{tr}}_\varepsilon n\log n\}$ for
arbitrary translates. Appendix~\ref{sec:translative-comparison} gives
the deduction, including the geometric reduction for nonsymmetric
bodies.
Thus near-unit efficiency at this scale is already available for
arbitrary translates---Theorem~\ref{thm:multiple} realizes that
efficiency \textit{with a single lattice}.

An \emph{$\varepsilon$-smooth covering} has multiplicity between
$(1-\varepsilon)\rho$ and $(1+\varepsilon)\rho$ everywhere, where
$\rho$ is its mean. Ordentlich, Regev, and
Weiss~\cite[Theorem~1.7]{ORWsmooth2025} construct such coverings at
density $O_\varepsilon(n\log n)$ using a \emph{periodic arrangement},
whose centers form a finite union of translates of one lattice.
For a single lattice, \cite[Theorem~1.2]{ORWsmooth2025} gives a
sufficient density bound cubic in $n$;
Lund~\cite[Theorem~1.7]{Lund2026} obtains
$O(\varepsilon^{-4}n^2)$ in sufficiently large dimension.
Both single-lattice results also bound the failure probability for
a Haar-random lattice; for each fixed failure probability, the
sufficient density thresholds have the stated orders.
In particular, Lund's theorem implies the analogue of
Theorem~\ref{thm:multiple} in sufficiently large dimension with $n^2$
in place of $n\log n$, by taking smoothness error
$\varepsilon/(1+\varepsilon)$ and a fixed failure probability.
Our construction establishes the lower multiplicity inequality
at every prescribed density $\rho\ge A_\varepsilon n\log n$.
Note that we show existence of a specially constructed lattice;
we give neither the upper inequality nor the same random-lattice conclusion.

\paragraph{Proof strategy.}
We describe the proof of Theorem~\ref{thm:prescribed} for one body
in large dimension, at prescribed mean $\rho$. Put
$\delta=\varepsilon/10$. We first work at a slightly smaller mean
$\rho'$, reserving the density cost of a final dilation. The aim is
to make the count at least $(1-4\delta)\rho'$ outside a sufficiently
small set, and then remove that exceptional set.

We use a version of the horizontal--vertical construction of
OpenAI~\cite[Sections~5--6]{Lattice2026}, which builds on the method
of Li and Liu~\cite{LiLiu2026}.
Decompose the ambient space as
$\mathbb R^n=\mathbb R^m\times\mathbb R^D$, where $n=m+D$, with a
large horizontal dimension $m$ and a small vertical dimension $D$;
and write its points as $(x,y)\in\mathbb R^m\times\mathbb R^D$.
We use the term \emph{target} informally for the point currently under
consideration, either in the full space or in the vertical
coordinates being treated.

Put $K[y]=\{x\in\mathbb R^m:(x,y)\in K\}$ for its
\emph{horizontal section} at height $y$.
Let $\Lambda_{\mathrm h}\subset\mathbb R^m$ and $L\subset\mathbb R^D$
be lattices, let $M:\mathbb R^D\to\mathbb R^D$ be invertible, and
let $Z:\mathbb R^D\to\mathbb R^m$ be linear.
The lattice
\[
  \Gamma=\{(h+Z\ell,M\ell):
    h\in\Lambda_{\mathrm h},\ \ell\in L\}
\]
has covering count
\begin{equation}
  N_{\Gamma,K}(x,y)=\sum_{\ell\in L}
  N_{\Lambda_{\mathrm h},K[y-M\ell]}(x-Z\ell).
  \label{eq:count-decomposition}
\end{equation}
Each vertical layer therefore contributes a horizontal section count;
Figure~\ref{fig:layers} illustrates the decomposition in two dimensions.
The map $Z$ determines a \emph{shear}: it translates the horizontal
arrangement in layer $\ell$ by $Z\ell$, keeping all centers in one
lattice.
Each translate retains the shape and orientation of $K$.
The covolume
$\det\Gamma=\det\Lambda_{\mathrm h}\det(ML)$ is independent of $Z$,
so the shear redistributes overlap between layers while preserving
the mean multiplicity. Figure~\ref{fig:translates} shows the
corresponding covering translates. Counts from different layers add,
with distinct centers in each layer.

After choosing suitable normalized coordinates, the section volumes
satisfy $|K[w]|\approx |K|\gamma_D(w)$ in the range of residual
heights we will use, where $\gamma_D$ is the standard Gaussian
density. Put $\Delta_{\mathrm h}=\det\Lambda_{\mathrm h}$ and
$\Delta_{\mathrm v}=\det(ML)$, so that
$\rho'=|K|/(\Delta_{\mathrm h}\Delta_{\mathrm v})$.
The mean contribution of layer $\ell$ is therefore approximately
\[
  \frac{|K[y-M\ell]|}{\Delta_{\mathrm h}}
  \approx \rho' w_y(\ell),
  \qquad
  w_y(\ell)=\Delta_{\mathrm v}\gamma_D(y-M\ell).
\]
The full sum $\sum_{\ell\in L}w_y(\ell)$ has mean $1$ over the
vertical torus. We need selected layers whose weights total at
least $1-\delta$ at every height, with each individual weight small.
This is a lower bound relative to the torus mean; the full sum at a
particular height may be much larger.

\begin{figure}[htbp]
  \centering
  \begin{tikzpicture}[
    x=.9cm,
    y=.85cm,
    font=\small,
    layer/.style={line width=1.3pt},
    counted/.style={draw=white,line width=.35pt},
    upperrow/.style={blue!70!black},
    middlerow/.style={teal!75!black},
    lowerrow/.style={orange!85!black}
  ]
    % The reflected translate has three vertices (-7/4,3/2),
    % (9/4,3/2), (3/4,-3/2). Lattice rows are Z + ell/2.
    \node[font=\small\bfseries] at (.1,2.55)
      {Centers in the reflected body};

    \begin{scope}
      \clip (-2.6,-2.2) rectangle (2.8,2.2);
      \fill[black!5]
        (-1.75,1.5)--(2.25,1.5)--(.75,-1.5)--cycle;

      \foreach \ell in {-2,-1,0,1,2}{
        \draw[black!20,dashed] (-2.6,\ell)--(2.8,\ell);
        \foreach \h in {-4,-3,-2,-1,0,1,2,3,4}{
          \fill[black!35]
            ({\h+\ell/2},\ell) circle[radius=1.5pt];
        }
      }

      \draw[black!60]
        (-1.75,1.5)--(2.25,1.5)--(.75,-1.5)--cycle;
      \draw[layer,upperrow] ({-4/3},1)--(2,1);
      \draw[layer,middlerow] (-.5,0)--(1.5,0);
      \draw[layer,lowerrow] ({1/3},-1)--(1,-1);

      \foreach \x/\y/\rowcolor in {%
        -.5/1/upperrow,.5/1/upperrow,1.5/1/upperrow,%
        0/0/middlerow,1/0/middlerow,.5/-1/lowerrow%
      }{
        \filldraw[\rowcolor,counted]
          (\x,\y) circle[radius=2.7pt];
      }
    \end{scope}

    \foreach \rowlevel in {-2,-1,0,1,2}{
      \node[left] at (-2.7,\rowlevel) {$\ell=\rowlevel$};
    }

    \node[fill=white,inner sep=1pt] at (.2,1.77) {$(x,y)-K$};
    \draw[->,>=stealth,black!60] (3.15,.1)--(4.9,.1);
    \node[align=center,font=\footnotesize] at (4.025,.65)
      {shift by\\$-\ell/2$};

    \begin{scope}[xshift=7.3cm]
      \node[font=\small\bfseries] at (.05,2.55)
        {Counts on the horizontal lattice};
      \node[font=\footnotesize] at (.05,1.77)
        {$x-\ell/2-K[y-\ell]$};

      \foreach \ell in {-1,0,1}{
        \draw[black!20,dashed] (-2.6,\ell)--(2.25,\ell);
        \foreach \h in {-2,-1,0,1,2}{
          \fill[black!35] (\h,\ell) circle[radius=1.5pt];
        }
      }

      \draw[layer,upperrow] ({-11/6},1)--(1.5,1);
      \draw[layer,middlerow] (-.5,0)--(1.5,0);
      \draw[layer,lowerrow] ({5/6},-1)--(1.5,-1);

      \foreach \lo/\ell/\rowcolor in {%
        {-11/6}/1/upperrow,-.5/0/middlerow,{5/6}/-1/lowerrow%
      }{
        \draw[layer,\rowcolor]
          ({\lo},{\ell-.1})--({\lo},{\ell+.1});
        \draw[layer,\rowcolor]
          (1.5,{\ell-.1})--(1.5,{\ell+.1});
      }

      \foreach \x/\y/\rowcolor in {%
        -1/1/upperrow,0/1/upperrow,1/1/upperrow,%
        0/0/middlerow,1/0/middlerow,1/-1/lowerrow%
      }{
        \filldraw[\rowcolor,counted]
          (\x,\y) circle[radius=2.7pt];
      }

      \foreach \h in {-2,-1,0,1,2}{
        \node[below,font=\footnotesize] at (\h,-1.3) {$\h$};
      }

      \foreach \ell/\rowcount in {1/3,0/2,-1/1}{
        \node[right] at (2.45,\ell) {$\rowcount$};
      }

      \node[font=\footnotesize] at (2.6,1.77) {count};
      \node at (.05,-2.05) {total multiplicity: $3+2+1=6$};
    \end{scope}
  \end{tikzpicture}
  \caption{\textsl{Counting by horizontal layers.} A translate
    $K+\lambda$ contains $(x,y)$ precisely when
    $\lambda\in(x,y)-K$.
    The downward-pointing triangle is this \emph{reflected translate}:
    reflect $K$ through the origin, then translate by $(x,y)$.
    Here the lattice is $\{(h+\ell/2,\ell):h,\ell\in\mathbb Z\}$.
    Subtracting the shear $\ell/2$ in each row reduces its contribution
    to a count on $\mathbb Z$. Filled colored dots are the counted
    centers. Figure~\ref{fig:translates} shows the actual covering
    translates.}
  \label{fig:layers}
\end{figure}

\begin{figure}[htbp]
  \centering
  \begin{tikzpicture}[
    x=.88cm,
    y=.88cm,
    font=\small,
    upperrow/.style={blue!70!black},
    middlerow/.style={teal!75!black},
    lowerrow/.style={orange!85!black}
  ]
    \node[font=\small\bfseries] at (0,3.3)
      {Shearing the centers};
    \node[font=\footnotesize] at (0,2.83)
      {$(h,\ell)\longmapsto(h+\ell/2,\ell)$};

    \foreach \rowlevel/\rowcolor in {%
      1/upperrow,0/middlerow,-1/lowerrow%
    }{
      \draw[black!18,dashed]
        (-2.65,\rowlevel)--(2.65,\rowlevel);

      \foreach \h in {-2,-1,0,1,2}{
        \draw[black!40,fill=white,line width=.5pt]
          (\h,\rowlevel) circle[radius=2.4pt];
        \fill[black!35]
          ({\h+\rowlevel/2},\rowlevel) circle[radius=1.5pt];
      }

      \node[left,\rowcolor] at (-2.75,\rowlevel)
        {$\ell=\rowlevel$};
    }

    \foreach \h in {-1,0,1}{
      \filldraw[upperrow,draw=white,line width=.35pt]
        ({\h+.5},1) circle[radius=2.8pt];
    }

    \foreach \h in {0,1}{
      \filldraw[middlerow,draw=white,line width=.35pt]
        (\h,0) circle[radius=2.8pt];
    }

    \filldraw[lowerrow,draw=white,line width=.35pt]
      (.5,-1) circle[radius=2.8pt];

    \draw[->,>=stealth,upperrow,line width=.8pt]
      (0,1.43)--(.5,1.43);
    \node[above,upperrow,font=\footnotesize] at (.25,1.46)
      {shift $+1/2$};

    \node[middlerow,font=\footnotesize] at (.25,.43) {fixed};

    \draw[->,>=stealth,lowerrow,line width=.8pt]
      (1,-1.43)--(.5,-1.43);
    \node[below,lowerrow,font=\footnotesize] at (.75,-1.46)
      {shift $-1/2$};

    \draw[black!40,fill=white,line width=.5pt]
      (-1.9,-2.12) circle[radius=2.4pt];
    \node[right,font=\footnotesize] at (-1.7,-2.12)
      {before shear};

    \fill[black!45] (-1.9,-2.55) circle[radius=2pt];
    \node[right,font=\footnotesize] at (-1.7,-2.55)
      {after shear};

    \begin{scope}[xshift=7.6cm]
      \node[font=\small\bfseries] at (.5,3.3)
        {Translates of the original body};

      % All six triangles are translates of the same K,
      % not sheared copies.
      \foreach \cx/\cy/\rowcolor in {%
        -.5/1/upperrow,.5/1/upperrow,1.5/1/upperrow,%
        0/0/middlerow,1/0/middlerow,.5/-1/lowerrow%
      }{
        \path[
          \rowcolor,
          fill,
          fill opacity=.055,
          draw opacity=.8,
          draw,
          line width=.7pt
        ]
          ({\cx-2},{\cy-1.25})--
          ({\cx+2},{\cy-1.25})--
          ({\cx-.5},{\cy+1.75})--cycle;
      }

      \foreach \cx/\cy/\rowcolor in {%
        -.5/1/upperrow,.5/1/upperrow,1.5/1/upperrow,%
        0/0/middlerow,1/0/middlerow,.5/-1/lowerrow%
      }{
        \filldraw[\rowcolor,draw=white,line width=.35pt]
          (\cx,\cy) circle[radius=2.8pt];
      }

      \filldraw[black,draw=white,line width=1pt]
        (.25,.25) circle[radius=3.2pt];
      \draw[black!70,line width=.5pt]
        (.32,.28)--(2.5,.55);
      \node[right,font=\footnotesize,fill=white,inner sep=2pt]
        at (2.5,.55) {$(x,y)$};
      \node[font=\footnotesize] at (.5,-2.67)
        {six translates through the marked point};
    \end{scope}
  \end{tikzpicture}
  \caption{\textsl{From the centers in Figure~\ref{fig:layers} to
    covering translates.} The shear slides rows of centers; each
    translate retains the original upward-pointing triangle.
    The reflection in Figure~\ref{fig:layers} serves only to count
    centers.
    All six displayed translates contain the same marked target, with
    colors recording their rows. A $k$-fold covering requires at least
    $k$ translates from the full arrangement through every point.}
  \label{fig:translates}
\end{figure}

The vertical construction retains nearly the full mean. We group
the vertical coordinates into \emph{blocks}, and use the local
Gaussian-weighted lattice constructions
of~\cite[Section~4]{Lattice2026} in successively smaller blocks.
The choices in one block shift the target in the preceding block,
averaging its available weight.
The last block has no successor, so its weight still depends on the
target. Averaging over all its lattice cosets replaces that weight
by its mean. We keep the last block bounded in dimension, depending
only on the accuracy, so this average has a bounded number of
components. Halving the dimensions makes the accumulated local losses
summable. The original one-fold argument only needs the positive
fraction of the mean supplied by~\cite[Proposition~5.1,
equation~(5.10)]{Lattice2026}; a certificate of near-mean multiplicity
based on selected layers requires their mean to be nearly the full mean.

We also make each selected weight small. A direct use of the
horizontal mean-hole estimate has an additive error of order
$e^{-c m}$, whereas removing the deficient set at the end requires
a proportion of order $e^{-n\log n}$. Many small layers allow us to
combine moderate estimates into a much stronger one. The largest
block controls how small the individual weights are; its dimension
can grow slowly while the last block stays bounded.

To make one horizontal lattice and shear work at every height, we
replace the sections by finitely many slightly smaller translated
representatives. For ordinary coverings, the shear estimate
in~\cite[Lemma~6.1]{Lattice2026} makes an intersection of hole sets
small. For multiple coverings we instead control the sum of the
counts through negative exponentials, which are large where counts
are small. The exponential converts the sum into a product, and
spatial thinning lets the Rogers--Schmidt estimate bound the average
of each factor over horizontal lattices.

Within each terminal coset, the selected indices are \emph{binary by
suffixes}: once the later coordinates are fixed, each coordinate has
at most two values, differing by $1$ when both occur. A uniform shear
column gives the two branches a uniform relative translation.
A fractional moment lets H\"older's inequality handle their shared
earlier columns. One choice of horizontal lattice and shear then
works for all required section patterns. Combining the coset
components and applying exponential Markov's inequality bounds the
deficient set. Adjusting the exponential parameter keeps the
individual tests in the mean-hole estimate's range, while the small
layer weights keep the combined bound effective at arbitrarily large
prescribed densities.

The final \emph{completion} step removes the exceptional set by
enlarging $K$ slightly. If the deficient proportion is small enough,
every target $z\in\mathbb R^n$ has some $w\in K/q$ for which $z-w$
is sufficiently covered.
The same distinct centers then cover $z$ with
$K+K/q=(1+1/q)K$. Put $q=\lceil n/\delta\rceil$.
The dilation factor $s=1+1/q$ costs at most $s^n\le e^\delta$ in
density, so we reserve that cost by starting at mean $\rho'=\rho/s^n$.
After completion, rescaling the lattice restores mean $\rho$ exactly.

The count budget is
\[
  \begin{aligned}
    \text{selected mean}&\ge(1-\delta)^2\rho',\\
    \text{count off the deficient set}&\ge(1-4\delta)\rho',\\
    \text{final count everywhere}&\ge(1-5\delta)\rho
      \ge(1-\varepsilon)\rho.
  \end{aligned}
\]

The two losses in the selected mean come from the section
representatives and the vertical weights. The Laplace estimate
allows for its approximation error and slack in Markov's inequality;
the final dilation accounts for the remaining loss.
Choosing $\rho$ as in the deduction of Theorem~\ref{thm:multiple}
then makes the final lower count at least $k$.

Completion explains the scale $n\log n$. It requires a deficient
proportion smaller than approximately $q^{-n}$, and
$\log(q^n)=n\log n+O_\delta(n)$. The combined Laplace bound makes
that proportion at most $n^{-2n}$ once the density exceeds a
sufficiently large accuracy-dependent multiple of $n\log n$.
The number of section patterns contributes only a lower-order cost.

With common covariance, the same coordinates, horizontal lattice,
and shear work for all bodies. Each body gives its own covering by
$\Lambda$-translates, with multiplicity compared with its mean
$|K_j|/\det\Lambda$; these means need not agree.
Figure~\ref{fig:common-covariance} illustrates the simultaneous
covering condition using the same lattice as
Figure~\ref{fig:translates}: it gives a $3$-fold covering by the
triangle and a $5$-fold covering by an ellipse with the same
covariance.\footnote{%
  We verify that the illustrated coverings have the claimed
  multiplicities everywhere in the plane. For the triangle in
  Figures~\ref{fig:layers}--\ref{fig:translates}, the inverse shear
  $(u,v)\mapsto(u-v/2,v)$ sends $\Gamma$ to $\mathbb Z^2$
  and the triangle to a translate of
  $\operatorname{conv}\{(0,0),(4,0),(0,3)\}$. This contains the three
  disjoint half-open unit squares with lower-left corners $(0,0)$,
  $(1,0)$, and $(0,1)$, each a fundamental domain, giving three
  distinct covering centers everywhere. For the ellipse in
  Figure~\ref{fig:common-covariance}, a horizontal section
  at vertical displacement $t$ from its center has length
  $2\sqrt{(2/3)(4-2t^2)}$. The two nearest distinct lattice rows to
  the center of any reflected translate have distances at most $1/2$
  and $1$. Their sections therefore have lengths greater than $3$ and
  $2$, and hence contain at least five lattice points altogether.%
}

\begin{figure}[htbp]
  \centering
  \begin{tikzpicture}[
    x=.88cm,
    y=.88cm,
    font=\small,
    layer/.style={line width=1.3pt},
    counted/.style={draw=white,line width=.35pt},
    upperrow/.style={blue!70!black},
    middlerow/.style={teal!75!black}
  ]
    \node[font=\small\bfseries] at (3.9,4.05)
      {The ellipse uses the same lattice as the triangle};
    \node[font=\small] at (3.9,3.3)
      {$\displaystyle
        \Sigma_{K_1}=\Sigma_{K_2}
        =\begin{pmatrix}
          49/72&-1/12\\
          -1/12&1/2
        \end{pmatrix}$};

    \node[font=\small\bfseries] at (0,2.6)
      {Centers for the ellipse};
    \node[font=\footnotesize] at (0,2.15)
      {$(x,y)-K_2$};

    \path[fill=black!5,draw=black!65]
      plot[domain=0:360,samples=100,smooth cycle]
      ({5/12+sqrt(8/3)*cos(\x)-sqrt(2)/6*sin(\x)},
       {.5+sqrt(2)*sin(\x)});

    \begin{scope}
      \clip (-2.6,-1.7) rectangle (2.75,2.05);

      \foreach \rowlevel in {-1,0,1}{
        \draw[black!20,dashed]
          (-2.6,\rowlevel)--(2.75,\rowlevel);
        \foreach \h in {-4,-3,-2,-1,0,1,2,3,4}{
          \fill[black!35]
            ({\h+\rowlevel/2},\rowlevel) circle[radius=1.5pt];
        }
      }
    \end{scope}

    \foreach \rowlevel in {-1,0,1}{
      \node[left,font=\footnotesize] at (-2.65,\rowlevel)
        {$\ell=\rowlevel$};
    }

    \draw[layer,upperrow]
      ({1/3-sqrt(7/3)},1)--({1/3+sqrt(7/3)},1);
    \draw[layer,middlerow]
      ({.5-sqrt(7/3)},0)--({.5+sqrt(7/3)},0);

    \foreach \x in {-.5,.5,1.5}{
      \filldraw[upperrow,counted]
        (\x,1) circle[radius=2.7pt];
    }

    \foreach \x in {-1,0,1,2}{
      \filldraw[middlerow,counted]
        (\x,0) circle[radius=2.7pt];
    }

    \foreach \rowlevel/\rowcount in {1/3,0/4,-1/0}{
      \node[right,font=\footnotesize] at (2.85,\rowlevel)
        {$\rowcount$};
    }

    \node[font=\footnotesize] at (3,1.6) {count};
    \node[font=\footnotesize] at (0,-2.15)
      {local count: $3+4=7$};

    \begin{scope}[xshift=7.9cm]
      \node[font=\small\bfseries] at (.35,2.6)
        {The ellipse covering};

      % These seven translates correspond to the seven colored centers
      % at left. Their barycenters are b+lambda; the colored dots mark
      % lambda itself.
      \foreach \cx/\cy/\rowcolor in {%
        -.5/1/upperrow,.5/1/upperrow,1.5/1/upperrow,%
        -1/0/middlerow,0/0/middlerow,1/0/middlerow,2/0/middlerow%
      }{
        \path[
          \rowcolor,
          fill,
          fill opacity=.035,
          draw opacity=.75,
          draw,
          line width=.65pt
        ]
          plot[domain=0:360,samples=100,smooth cycle]
          ({\cx-1/6+sqrt(8/3)*cos(\x)-sqrt(2)/6*sin(\x)},
           {\cy-.25+sqrt(2)*sin(\x)});
      }

      \foreach \cx/\cy/\rowcolor in {%
        -.5/1/upperrow,.5/1/upperrow,1.5/1/upperrow,%
        -1/0/middlerow,0/0/middlerow,1/0/middlerow,2/0/middlerow%
      }{
        \filldraw[\rowcolor,counted]
          (\cx,\cy) circle[radius=2.7pt];
      }

      \filldraw[black,draw=white,line width=1pt]
        (.25,.25) circle[radius=3.2pt];
      \draw[black!70,line width=.5pt]
        (.32,.28)--(2.5,1.48);
      \node[right,font=\footnotesize,fill=white,inner sep=2pt]
        at (2.5,1.48) {$(x,y)$};
      \node[font=\footnotesize] at (.35,-2.15)
        {seven translates through the marked point};
    \end{scope}

    \node[font=\small,align=center] at (4,-3.02)
      {The same $\Gamma$ gives both coverings everywhere:\\[3pt]
       $N_{\Gamma,K_1}\ge3$\quad and\quad $N_{\Gamma,K_2}\ge5$.};
  \end{tikzpicture}
  \caption{\textsl{Simultaneous coverings using the same lattice
    $\Gamma$ and shear as in Figure~\ref{fig:translates}.}
    The seven colored centers give the ellipse translates through
    $(x,y)=(1/4,1/4)$ shown on the right. Here
    $K_1=\operatorname{conv}\{(-2,-5/4),(2,-5/4),(-1/2,7/4)\}$,
    $b=(-1/6,-1/4)$, and
    $K_2=\{u:(u-b)^{\mathsf T}\Sigma^{-1}(u-b)\le4\}$,
    so their uniform distributions have the displayed covariance
    $\Sigma$.
    The full families $\{K_j+\lambda:\lambda\in\Gamma\}$ have
    respective global multiplicities at least $3$ and $5$; the
    displayed local counts are $6$ and $7$.}
  \label{fig:common-covariance}
\end{figure}

\paragraph{Roadmap and conventions.}
Sections~\ref{sec:counts}--\ref{sec:vertical} establish the thinning
estimate, section comparisons, and vertical patterns;
Section~\ref{sec:covering-transfer} assembles the covering.
Section~\ref{sec:vertical-spike} shows that the count is close to
its mean in $L^1$ but has large pointwise peaks.
The appendixes establish attainment and the translative comparison.

All logarithms are natural unless a base is displayed. The notation
$|A|$ denotes volume, cardinality, or normalized torus measure,
according to context. Named constants retain their stated values.
Implied constants are absolute unless otherwise indicated; asymptotics
in the covering construction keep the accuracy fixed.

\section{Thinning and Laplace bounds}
\label{sec:counts}

To bound small lattice counts, we convert the Rogers--Schmidt
mean-hole estimate into a Laplace bound by \emph{independent thinning}.
Retaining each of $r$ points independently with probability $1-e^{-t}$,
where $t>0$, leaves none with probability $e^{-tr}$. We realize this
operation through a random region independent of the lattice; small
spatial cells make its tests at distinct lattice points independent.

Write $\mu_d$ for invariant probability measure on the space of
\emph{unimodular lattices} (i.e., those of covolume~$1$) in
$\mathbb R^d$, and
$\pi_\Lambda:\mathbb R^d\to\mathbb R^d/\Lambda$ for the quotient map.
For a Borel set $A\subset\mathbb R^d$, its \emph{uncovered proportion}
is $h_\Lambda(A)=1-|\pi_\Lambda(A)|$.
We use the mean-hole estimate of Rogers~\cite{Rogers1958}
(see also Schmidt~\cite{Schmidt1958}), in the form recorded
in~\cite[Theorem~2.3]{ORW2022}. With
\[
  \eta_d=\frac d4\log\frac{27}{16}-3\log d,
\]
there is an absolute constant $C_{\mathrm{RS}}$ such that
\begin{equation}
  \left|\int h_\Lambda(A)\,d\mu_d(\Lambda)-e^{-|A|}\right|
  \le C_{\mathrm{RS}}e^{-\eta_d}
  \qquad(|A|\le\eta_d).
  \label{eq:hole}
\end{equation}
The estimate applies to arbitrary Borel sets, as the thinning
argument requires. We assume $\eta_d>0$ throughout this section;
in the applications $d\to\infty$ and $\eta_d\asymp d$.

For a bounded nonnegative integrable Borel function $f$, set
\[
  S_{\Lambda,f}(x)=\sum_{\lambda\in\Lambda}f(x-\lambda).
\]
For $f=\mathbf1_K$, each summand records whether $x\in K+\lambda$,
so $S_{\Lambda,\mathbf1_K}=N_{\Lambda,K}$, the case used for
horizontal sections in the covering argument. Integration over a
fundamental domain gives
\begin{equation}
  \int_{\mathbb R^d/\Lambda}S_{\Lambda,f}(x)\,dx
  =\frac1{\det\Lambda}\int_{\mathbb R^d}f.
  \label{eq:mean-sum}
\end{equation}
Thus the sum is finite almost everywhere; it is finite everywhere
when $f$ has compact support.

For $t>0$, we call $x\mapsto e^{-tS_{\Lambda,f}(x)}$ an
\emph{exponential test} for small values of the sum. Its integral
over the normalized torus is the \emph{Laplace transform} of the sum
at a uniform point. For any threshold $a>0$,
\[
  \bigl|\{x:S_{\Lambda,f}(x)<a\}\bigr|
  \le e^{ta}\int_{\mathbb R^d/\Lambda}e^{-tS_{\Lambda,f}(x)}\,dx.
\]
A small Laplace transform thus bounds the proportion of points at
which the sum falls below $a$.

For a weighted sum, retain a point $y$ with probability
$1-e^{-tf(y)}$. The integral $v_t$ of this probability is the expected
volume of the thinned region. The following lemma transfers the
mean-hole estimate whenever $v_t<\eta_d$. Since $v_t\le t\int f$,
this condition can be met by choosing $t$ sufficiently small, even
when $\int f$ is large.

\begin{lemma}[Functional thinning]
\label{lem:thinning}
Let $f$ be as above and $t>0$. If
$v_t=\int_{\mathbb R^d}(1-e^{-tf})<\eta_d$, then
\begin{equation}
  \left|
    \int\!\int_{\mathbb R^d/\Lambda}
      e^{-tS_{\Lambda,f}(x)}\,dx\,d\mu_d(\Lambda)-e^{-v_t}
  \right|
  \le C_{\mathrm{RS}}e^{-\eta_d}.
  \label{eq:laplace}
\end{equation}
\end{lemma}

\begin{proof}
First suppose $f$ is compactly supported. Put $q=1-e^{-tf}$ and
partition space into half-open cubes of side $h$. In each cube meeting
the support, choose an independent uniform random variable
$U_Q\in[0,1]$. Define
\[
  A_h=\bigcup_Q\{y\in Q:q(y)>U_Q\}.
\]
This is a bounded Borel set sampled independently of the lattice.
All averages below are measurable: writing $\Lambda=g\mathbb Z^d$
and $x=gz$ for $z\in[0,1)^d$, avoidance is a countable intersection
of Borel conditions in $(U_Q)_Q$, $g$, and $z$. Thus Fubini applies.

In this proof, $\mathbb E$, $\mathbb P$, and $\operatorname{Var}$
refer to the independent random variables $(U_Q)_Q$ used to construct
$A_h$; averages over lattices are written explicitly as integrals
against $\mu_d$.
The volumes $V_Q=|A_h\cap Q|$ are independent, with
$\mathbb EV_Q=\int_Qq$ and $V_Q^2\le h^dV_Q$. Hence
\[
  \mathbb E|A_h|=v_t,
  \qquad
  \operatorname{Var}|A_h|\le h^dv_t.
\]
Thus $|A_h|\to v_t$ in probability. Since $v_t<\eta_d$,
$\mathbb P\{|A_h|>\eta_d\}\to0$, and
$\mathbb E e^{-|A_h|}\to e^{-v_t}$.

Fix a unimodular lattice $\Lambda$. Once $\sqrt d\,h$ is smaller
than the length of its shortest nonzero vector, distinct points of
$x-\Lambda$ lie in distinct cubes for every $x$. Their retention
tests are then independent, giving
\[
  \mathbb E h_\Lambda(A_h)
  =\int_{\mathbb R^d/\Lambda}
    \prod_{\lambda\in\Lambda}(1-q(x-\lambda))\,dx
  =\int_{\mathbb R^d/\Lambda}e^{-tS_{\Lambda,f}(x)}\,dx.
\]
Only finitely many factors differ from $1$. The hole measures lie
in $[0,1]$, so dominated convergence handles the lattice-dependent
cutoff in $h$ when averaging over $\mu_d$.
On $\{|A_h|\le\eta_d\}$ we can apply \eqref{eq:hole}; on its
complement the absolute difference in that estimate is at most $1$.
Averaging over $A_h$ therefore gives
\[
  \left|
    \mathbb E\!\int h_\Lambda(A_h)\,d\mu_d(\Lambda)
    -\mathbb E e^{-|A_h|}
  \right|
  \le C_{\mathrm{RS}}e^{-\eta_d}
    +\mathbb P\{|A_h|>\eta_d\}.
\]
Letting $h\to0$ proves \eqref{eq:laplace}.

For general $f$, apply the compactly supported result to
$f_R=f\mathbf1_{\{\|y\|\le R\}}$. The sums increase to
$S_{\Lambda,f}$ and the corresponding $v_t(R)$ increase to
$v_t<\eta_d$.
Bounded convergence then proves the assertion.
\end{proof}

For an indicator $f=\mathbf1_A$, the comparison term in
\eqref{eq:laplace} is $e^{-(1-e^{-t})|A|}$, the Laplace transform of
a Poisson random variable of mean $|A|$. More generally, $e^{-v_t}$
is the Laplace transform of $\sum_{z\in\mathcal X}f(z)$ for a
unit-intensity Poisson point process $\mathcal X$. Thus the lemma
compares exponential tests averaged over the lattice and a uniform
point of its torus, in the range $v_t<\eta_d$.
The covering proof combines these transforms before estimating an
exceptional set; it requires no independence between section counts.

\section{Common Gaussian sections}
\label{sec:sections}

The preceding Laplace bound concerns a fixed region. To treat all
required heights and bodies in one probabilistic choice, we refine
\cite[Lemma~2.1]{Lattice2026} to find common coordinates in which the
section volumes, divided by the body volumes, have an approximately
Gaussian profile. Finitely many representative bodies then fit
inside the sections after translation, with arbitrarily small
relative volume loss. Their counts bound the actual section counts
from below.

Write $\gamma_D(y)=(2\pi)^{-D/2}e^{-\|y\|^2/2}$ for the
\emph{standard Gaussian density}. A probability distribution is
\emph{isotropic} if its mean is zero and its covariance matrix is
the identity.
Uniform distributions on convex bodies are log-concave. The
pointwise marginal theorem of Eldan and
Klartag~\cite[Theorem~1]{EldanKlartag2008} supplies absolute positive
constants $c_1,c_2,c_3,c_4,C_E$ such that, for an isotropic log-concave
probability distribution on $\mathbb R^n$ and $1\le D\le n^{c_1}$,
the orthogonal projection onto a Haar-random $D$-dimensional subspace
$E$ has a density $f_E$ satisfying
\begin{equation}
  \left|\frac{f_E(y)}{\gamma_D(y)}-1\right|
  \le C_En^{-c_3}
  \qquad(\|y\|\le n^{c_4})
  \label{eq:marginal}
\end{equation}
with probability at least $1-C_Ee^{-n^{c_2}}$.
Fix an absolute constant
\begin{equation}
  0<c_0<\min(c_2,1/2).
  \label{eq:family-exponent}
\end{equation}
For $K\subset\mathbb R^m\times\mathbb R^D$, write
$K[y]=\{x\in\mathbb R^m:(x,y)\in K\}$. For a family of bodies,
$K_j[y]$ denotes the section of the $j$-th body.

\begin{lemma}[Common Gaussian sections]
\label{lem:sections}
Fix $0<\delta<1/10$ and $A_{\mathrm{sec}}>0$.
For all sufficiently large $n$, depending only on these two
parameters, the following holds. Let
$K_1,\ldots,K_N\subset\mathbb R^n$ be convex bodies whose uniform
distributions have a common covariance matrix, with
$1\le N\le e^{n^{c_0}}$. Choose a vertical dimension $D$ and a
radius $R$ satisfying
\[
  1\le D\le A_{\mathrm{sec}}(\log\log n)^2,
  \qquad
  1\le R,
  \qquad
  R^2\le A_{\mathrm{sec}}D\log(D+1).
\]
After centering the bodies separately and applying one common linear
map, there are coordinates
$\mathbb R^n=\mathbb R^m\times\mathbb R^D$, with $m=n-D$, common to
all the bodies. Write $V_j=|K_j|$ in these coordinates. For each
body $K_j$ there are:
\begin{itemize}
  \item an affine map $T_j:\mathbb R^D\to\mathbb R^m$, written
    $T_j(y)=A_jy+t_j$;
  \item a family of at most $(3m^2)^D$ representative convex bodies
    $J_{j,a}\subset\mathbb R^m$.
\end{itemize}
For each $j$ and every height $y$ with $\|y\|\le R$, we can choose a
label $a$ such that the translated representative lies in the section
and has the required volume:
\begin{equation}
  \begin{gathered}
    J_{j,a}+T_j(y)\subset K_j[y],\\
    (1-\delta)V_j\gamma_D(y)
    \le |J_{j,a}|\le V_j\gamma_D(y).
  \end{gathered}
  \label{eq:section-labels}
\end{equation}
The label may depend on the body and the height; the finite families
and affine maps are chosen once.
\end{lemma}

Whenever label $a$ serves height $y$, the containment transfers
covering counts for any horizontal lattice:
\[
  N_{\Lambda_{\mathrm h},K_j[y]}(x)
  \ge N_{\Lambda_{\mathrm h},J_{j,a}}\bigl(x-T_j(y)\bigr).
\]
Every center counted on the right is also counted on the left.

\begin{proof}[Proof of Lemma~\ref{lem:sections}]
Center each body and apply the inverse square root of their common
covariance matrix, making every uniform distribution isotropic.
Put $R_{\mathrm{out}}=2DR+2$. For sufficiently large $n$, we have
$D\le n^{c_1}$, $R_{\mathrm{out}}<n^{c_4}$, and
$C_Ee^{n^{c_0}-n^{c_2}}<1$. A union bound in \eqref{eq:marginal}
therefore gives one subspace for which the marginal estimate holds
for every body.

Use orthonormal coordinates $E^\perp\times E$ and put
$\varepsilon_n=C_En^{-c_3}$. For the uniform distribution on $K_j$,
the projection density is $f_{j,E}(y)=|K_j[y]|/V_j$ almost everywhere.
The section-volume function is upper semicontinuous: compactness
gives
$\limsup_{\nu}\mathbf1_{K_j}(x,y_\nu)\le\mathbf1_{K_j}(x,y)$
when $y_\nu\to y$, and all sections have a common bounded horizontal
support, so reverse Fatou applies. Its almost-everywhere Gaussian
lower bound therefore extends to every height:
\begin{equation}
  |K_j[y]|\ge(1-\varepsilon_n)V_j\gamma_D(y)
  \qquad(\|y\|\le R_{\mathrm{out}}).
  \label{eq:section-lower}
\end{equation}

To compare nearby sections, choose affine centers as in
\cite[Lemma~2.1]{Lattice2026}. Let $v_0,\ldots,v_D$ be the vertices
of a regular simplex centered at $0$ with inradius $2R$. Their
norms are $2DR<R_{\mathrm{out}}$, so \eqref{eq:section-lower}
lets us choose $z_{j,i}\in K_j[v_i]$. Define the affine map $T_j$ by
$T_j(v_i)=z_{j,i}$. If $y=\sum_i\alpha_i v_i$ is in the simplex,
where $\alpha_i\ge0$ and $\sum_i\alpha_i=1$, convexity gives
\[
  (T_j(y),y)=\sum_i\alpha_i(z_{j,i},v_i)\in K_j.
\]
Since the simplex contains the ball of radius $2R$, we obtain
\[
  0\in\widehat K_j[y]:=K_j[y]-T_j(y)
  \qquad(\|y\|\le2R).
\]
Segments toward those origins let us move between nearby heights
while shrinking the horizontal section by a controlled factor.

To control both the contraction loss and the Gaussian variation,
choose a maximal $R/m^2$-separated set $\{y_a\}$ in the ball of
radius $R$, using the same net for every body. It has at most
$(3m^2)^D$ elements, so its logarithmic cardinality is $O(D\log n)$.
Every height in the ball lies within $R/m^2$ of a net point.
Convexity gives
\[
  (1-m^{-2})\widehat K_j[y_a]\subset\widehat K_j[y]
  \qquad(\|y-y_a\|\le R/m^2).
\]
Indeed, $y=(1-m^{-2})y_a+m^{-2}z$ for
$z=y_a+m^2(y-y_a)$, where $\|z\|\le2R$.
The contraction retains a fraction
$(1-m^{-2})^m\ge1-m^{-1}$ of the volume. By
\eqref{eq:section-lower}, its volume is therefore at least
$(1-\varepsilon_n)(1-m^{-1})V_j\gamma_D(y_a)$.
For sufficiently large $n$, we may shrink it further about zero to
obtain a contained convex body $J_{j,a}$ of volume exactly
$(1-\delta/2)V_j\gamma_D(y_a)$.
For $y$ and $y_a$ in the radius-$R$ ball with
$\|y-y_a\|\le R/m^2$,
\[
  \left|\log\frac{\gamma_D(y_a)}{\gamma_D(y)}\right|
  \le R^2/m^2\le\delta/8.
\]
Consequently $|J_{j,a}|/(V_j\gamma_D(y))$ lies between
$(1-\delta/2)e^{-\delta/8}\ge1-\delta$ and
$(1-\delta/2)e^{\delta/8}\le1$, as required.
\end{proof}

Separate body translations preserve the covering requirements, and
the common linear map multiplies all volumes and covolumes by the
same factor. The body-dependent maps $T_j$ are used only to compare
sections; they do not change the ambient coordinates of the covering
problem.

\section{Averaging the last Gaussian block}
\label{sec:vertical}

We now choose the vertical lattice and, at each height, finitely many
layers whose normalized Gaussian weights total at least $1-\delta$,
with each weight small. Multiplying the Gaussian weights by the
vertical covolume makes their full sum have torus mean $1$.

A \emph{block} is a group of coordinates, viewed as one factor
$\mathbb R^b$ in a product decomposition of the vertical space.
The standard Gaussian density factors over these groups. We call
the local lattice construction in one factor, weighted by
$\gamma_b$, a \emph{Gaussian block}. We assemble the block estimates
of~\cite[Section~4]{Lattice2026} in successively smaller dimensions
and finish with an exact average over the last block's cosets.

\paragraph{One block.}
For a block of dimension $b$, put $h(b)=\sqrt{(9/4)\log b}$ and
measure heights in units of $h(b)$.
The block lattice will have the form
$L_b=\mathbb Z^b+\mathcal C_b/p(b)$, where $p(b)$ is a prime and
$\mathcal C_b\subset\mathbb F_{p(b)}^b$ is a one-dimensional subspace.
Since $[L_b:\mathbb Z^b]=p(b)$, the lattice $h(b)L_b$ in the original
height units has covolume $h(b)^b/p(b)$. For a target
$t\in\mathbb R^b$, choose a cube base point $\lambda\in L_b$ with
$\tau=t-\lambda\in[0,1)^b$. The $2^b$ corners $\lambda+\sigma$,
$\sigma\in\{0,1\}^b$, of the cube $\lambda+[0,1]^b$ are candidate
layers. Their residuals lie in $[-1,1]^b$, and their Gaussian weights
split into a total weight and a product probability law.
Specifically, define
\begin{equation}
  \begin{gathered}
    f_b(v)=h(b)\gamma_1(h(b)v)
      +h(b)\gamma_1\bigl(h(b)(v-1)\bigr)
      \quad(0\le v\le1),\\
    F_b(\tau)=\prod_{s=1}^b f_b(\tau_s),
    \qquad
    P_{b,\tau}(\sigma)=\prod_{s=1}^b
      \frac{h(b)\gamma_1(h(b)(\tau_s-\sigma_s))}{f_b(\tau_s)}.
  \end{gathered}
  \label{eq:block-folded}
\end{equation}
Then
\[
  \frac{h(b)^b}{p(b)}
    \gamma_b\bigl(h(b)(\tau-\sigma)\bigr)
  =\frac{F_b(\tau)}{p(b)}P_{b,\tau}(\sigma).
\]
The factor $F_b(\tau)/p(b)$ is the normalized \emph{folded weight}
of the cube; $P_{b,\tau}$ is its \emph{Gaussian-weighted bit law}.
A coordinate is \emph{balanced} if both bit probabilities are at most
$e/(1+e)$, equivalently $h(b)^2|\tau_s-1/2|\le1$. Many balanced
coordinates spread the folded weight over many corners.

We write
\[
  \mathfrak m(b)=
  \frac{\int_0^1 f_b(v)\log f_b(v)\,dv}
       {\int_0^1 f_b(v)\,dv}.
\]
The denominator is $\mathbb P\{|X|\le h(b)\}$ for standard normal
$X$; its $b$-th power is $1-O(b^{-1/8})$, so truncation to
$[-h(b),h(b)]^b$ retains nearly all the Gaussian mass.
The quantity $\mathfrak m(b)$ is the mean logarithmic folded density;
it determines the prime window below.

We now record the precise consequences of the Gaussian block
estimates in~\cite[Lemmas~4.1, 4.2, and~4.4]{Lattice2026} that we use.
The following lemma constructs a block lattice and a function $g_b$
recording the normalized Gaussian weight available in the current
block. Its mean is close to $1$, although it may vanish at some
targets. Wherever $g_b$ is positive, the lemma selects a cube with
small individual corner weights and fixes a bit law at a nearby
grid point, with a small comparison loss at the actual residual.
The same bits sample the preceding block's function $g_0$, preserving
almost all of its mean uniformly over translates.

\begin{lemma}[Gaussian block input]
\label{lem:gaussian-block-input}
There are absolute constants $B_{\mathrm{blk}},C_{\mathrm{blk}}>0$
and $c_{\mathrm{at}}>0$ with the following property. For every
integer $b\ge B_{\mathrm{blk}}$, put $h(b)=\sqrt{(9/4)\log b}$.
There is a prime $p(b)$ such that
\begin{equation}
  b\mathfrak m(b)-4b^{14/25}
  \le\log p(b)
  \le b\mathfrak m(b)-3b^{14/25}.
  \label{eq:block-prime-window}
\end{equation}
Moreover, $\mathfrak m(b)=\log h(b)+O(1)$, and hence
\begin{equation}
  \log p(b)=b\log h(b)+O(b)=O(b\log\log b),
  \label{eq:v-prime}
\end{equation}
with absolute implied constants. Put
$\mathcal G_b=\mathbb F_{p(b)}^{b}$ and
$\xi(b)=b h(b)^2/p(b)$.
Suppose $H$ is a finite abelian group with $\log|H|\le b^4$ and
$g_0:H\to[0,\infty)$ satisfies
\[
  \tfrac12\le\overline g_0\le2,
  \qquad
  g_0\le\exp(5b^{28/45}),
\]
where bars denote uniform means. Then there are shifts
$w_1,\ldots,w_b\in H$, a one-dimensional subspace
$\mathcal C_b\subset\mathcal G_b$, and a function
$g_b:\mathcal G_b\to[0,\infty)$, constant on $\mathcal C_b$-cosets,
such that
\begin{equation}
  |\overline g_b-1|\le C_{\mathrm{blk}}b^{-1/100},
  \qquad
  g_b\le\exp(5b^{14/25}).
  \label{eq:v-local-means}
\end{equation}
Write $L_b=\mathbb Z^b+\mathcal C_b/p(b)$, using integer
representatives for $\mathcal C_b$.
For every $t\in\mathbb R^b$ with $g_b(x)>0$, where
$x=\lfloor p(b)t\rfloor\bmod p(b)$, there are $\lambda\in L_b$
and a probability law $\pi_t$ on $\{0,1\}^b$ such that, for every
$\sigma\in\{0,1\}^b$,
\begin{align}
  \|t-\lambda-\sigma\|_\infty
  &\le1,
  \label{eq:v-local-residual}\\
  e^{-\xi(b)}g_b(x)\pi_t(\sigma)
  &\le\frac{h(b)^b}{p(b)}
    \gamma_b\bigl(h(b)(t-\lambda-\sigma)\bigr)
    \le e^{-c_{\mathrm{at}}b^{7/10}},
  \label{eq:v-local-weight}
\end{align}
and, simultaneously for every $u\in H$,
\begin{equation}
  \sum_{\sigma\in\{0,1\}^b}\pi_t(\sigma)
    g_0\left(u-\sum_{s=1}^b\sigma_sw_s\right)
  \ge(1-C_{\mathrm{blk}}b^{-1/100})\overline g_0.
  \label{eq:v-local-sampling}
\end{equation}
There is also a base version with no $H,g_0$, or shifts, for which
\eqref{eq:v-local-means}--\eqref{eq:v-local-weight} hold.
\end{lemma}

\begin{proof}[Derivation from the block estimates
  in~{\cite[Section~4]{Lattice2026}}]
\emph{Grid selection.}
The prime window and the estimate for $\mathfrak m(b)$ are given by
\cite[Lemma~4.1, equation~(4.3)]{Lattice2026}.
For a grid point $z\in\mathcal G_b$, use its representative in
$\{0,\ldots,p(b)-1\}^b$ and put $t_z=z/p(b)$. It is \emph{eligible}
if
\[
  |\log F_b(t_z)-b\mathfrak m(b)|\le b^{14/25},
  \qquad
  \#\{s:h(b)^2|t_{z,s}-1/2|\le1\}\ge b^{7/10}.
\]
Thus eligibility controls the total folded weight and guarantees
many balanced bits.

The construction in~\cite[Lemma~4.2]{Lattice2026} selects common
shifts $w_s$ and a subset $U_b$ of eligible points at which these
bits sample $g_0$ uniformly over its translate:
\[
  \sum_{\sigma\in\{0,1\}^b}P_{b,t_z}(\sigma)
    g_0\left(u-\sum_{s=1}^b\sigma_sw_s\right)
  \ge(1-C_{\mathrm{blk}}b^{-1/100})\overline g_0
  \qquad(z\in U_b,\ u\in H);
\]
these are the \emph{usable} grid points
of~\cite[Lemma~4.2]{Lattice2026}. The hypothesis on $g_0$
in~\cite[Lemma~4.2]{Lattice2026} allows the stated cap, as
$(10/9)(14/25)=28/45$.

The construction in~\cite[Lemma~4.2]{Lattice2026} also supplies a line
$\mathcal C_b$ such that
\begin{equation}
  g_b(x)=\max_{c\in\mathcal C_b}
    \frac{F_b(t_{x-c})}{p(b)}\mathbf1_{U_b}(x-c)
  \label{eq:block-coset-maximum}
\end{equation}
satisfies \eqref{eq:v-local-means};
see~\cite[equations~(4.11)--(4.12)]{Lattice2026}.
For the base version, take $U_b$ to be the full eligible set and use
the final clause of~\cite[Lemma~4.2]{Lattice2026}.

The negative offset in \eqref{eq:block-prime-window} is deliberate.
For a suitable line, cosets containing competing usable points
contribute little to the total folded weight. The coset maximum
therefore retains nearly the mean of the coset sum; this is the
argument in \cite[equations~(4.15)--(4.17)]{Lattice2026}.
The resulting near-$1$ estimate is an average over cosets, not a
pointwise estimate for $g_b$.

\emph{From grid points to targets.}
For a target $t$ with $g_b(x)>0$, put $\nu_t=\lfloor p(b)t\rfloor$
and $x=\nu_t\bmod p(b)$. Choose $z\in x-\mathcal C_b$ attaining the
maximum in \eqref{eq:block-coset-maximum}; then $z\in U_b$. Define
\[
  \begin{gathered}
    \lambda=\frac{\nu_t-z}{p(b)}\in L_b,
    \qquad
    \pi_t=P_{b,t_z},\\
    \tau=t-\lambda=\frac{z+\{p(b)t\}}{p(b)}
    \in t_z+[0,p(b)^{-1})^b.
  \end{gathered}
\]
Here fractional parts are taken coordinatewise. These choices agree
with \cite[Lemma~4.4, equations~(4.18)--(4.19)]{Lattice2026}, and
$g_b(x)=F_b(t_z)/p(b)$. The law $\pi_t$ is evaluated at the grid
point $t_z$, not at the actual residual $\tau$; it depends on $t$
but not on $u$. Its sampling bound is therefore simultaneous for
every $u$.
Since $\tau\in[0,1)^b$, all corners have residuals in $[-1,1]^b$.

Moving from $t_z$ to $\tau$ changes the logarithm of each Gaussian
corner weight by at most $\xi(b)$; see
\cite[equation~(4.6)]{Lattice2026}. Consequently
\[
  e^{-\xi(b)}g_b(x)\pi_t(\sigma)
  \le\frac{h(b)^b}{p(b)}\gamma_b\bigl(h(b)(\tau-\sigma)\bigr)
  \le e^{\xi(b)}g_b(x)\pi_t(\sigma)
  \qquad(\sigma\in\{0,1\}^b).
\]

Eligibility and the prime window give
$g_b(x)\le\exp(5b^{14/25})$.
The balanced coordinates give
$\pi_t(\sigma)\le(e/(1+e))^{b^{7/10}}$ for every $\sigma$. Hence
\[
  e^{\xi(b)}g_b(x)\pi_t(\sigma)
  \le\exp\bigl(\xi(b)+5b^{14/25}\bigr)
    \left(\frac{e}{1+e}\right)^{b^{7/10}}
  \le e^{-c_{\mathrm{at}}b^{7/10}},
\]
with fixed absolute $c_{\mathrm{at}}>0$ after choosing
$B_{\mathrm{blk}}$ sufficiently large. Here $14/25<7/10$, and
\eqref{eq:v-prime} makes $\xi(b)$ smaller than every fixed negative
power of $b$. This proves the residual, weight, and sampling
conclusions, including the base version.
\end{proof}

We will assemble these blocks into a vertical lattice of the form
$ML$, where points of $L$ encode the successive choices and the
invertible map $M$ places them in Euclidean space. The shear argument
requires the following branching structure on these indices.

A finite nonempty subset of $\mathbb R^d$ is \emph{binary by suffixes}
if, on reading coordinates from last to first, each coordinate has
at most two possible values once the later coordinates are fixed,
and these values differ by $1$ when both occur. Thus the two choices
may depend on the later coordinates. Such a set has at most $2^d$
points.\footnote{%
  For example, $\{(0,0),(1,0),(3,1),(4,1)\}$ is binary by suffixes:
  the last coordinate is $0$ or $1$, and the first-coordinate choices
  are $\{0,1\}$ or $\{3,4\}$ accordingly.%
}
Nonempty subsets and slices obtained by fixing and deleting final
coordinates retain the property.
In dimension zero the pattern consists of the empty tuple.

\begin{proposition}[Vertical patterns]
\label{prop:vertical}
For every $\delta\in(0,1/10)$ there are integers
$B_\delta,I_\delta\geq1$ with the following property. For every
integer $b_1\geq B_\delta$ there are a dimension $D$, a radius
$R\geq1$, a full-rank lattice $L\subset\mathbb R^D$ containing
$\mathbb Z^D$, and an invertible linear map $M$ such that, writing
$\Delta_{\mathrm v}=\det(ML)$,
\[
  b_1\leq D\leq2b_1,
  \qquad
  R^2\leq5b_1\log(b_1+1),
\]
and every $y\in\mathbb R^D$ has a finite nonempty pattern
$\mathcal P_y\subset L$ satisfying
\begin{align}
  \|y-M\ell\|
  &\leq R
  &&(\ell\in\mathcal P_y),
  \label{eq:v-radius}\\
  \sum_{\ell\in\mathcal P_y}
    \Delta_{\mathrm v}\gamma_D(y-M\ell)
  &\geq1-\delta,
  \label{eq:v-total}\\
  \max_{\ell\in\mathcal P_y}
    \Delta_{\mathrm v}\gamma_D(y-M\ell)
  &\leq\exp(-c_{\mathrm{at}}b_1^{7/10}).
  \label{eq:v-atom}
\end{align}
The pattern is a disjoint union of at most $I_\delta$ patterns that
are binary by suffixes. There are fixed finite sets
$\mathcal A_1,\ldots,\mathcal A_D$ with
$\log|\mathcal A_s|=O(D\log(D+2))$ such that
$\mathcal P_y\subset\prod_{s=1}^D\mathcal A_s$ whenever
$y\in M[0,1)^D$.
The implied constant in the alphabet bound is absolute.
\end{proposition}

\begin{proof}
Iterating Lemma~\ref{lem:gaussian-block-input} gives a lower bound
for the total selected weight that still depends on the last block's
target. We first establish this recursive bound, then remove the
dependence by averaging over the last block's cosets.

\emph{Block construction.}
For a sufficiently large cutoff $B$, set
$b_{i+1}=\lfloor b_i/2\rfloor$ while this is at least $B$, and stop
at $b_r$. Then
\begin{equation}
  B\leq b_r<2B,
  \qquad
  b_i\geq2b_{i+1},
  \qquad
  D:=\sum_{i=1}^r b_i\leq2b_1.
  \label{eq:v-block-sizes}
\end{equation}
A large initial block gives small individual weights, while a bounded
final block limits the number of terminal cosets. Halving connects
these scales with summable losses and, as we check next, permits
successive applications of the block input.

Fix one prime $p(b)$ as in Lemma~\ref{lem:gaussian-block-input}
for each dimension $b\ge B$. In block $i$, of dimension $b_i$,
abbreviate $h_i=h(b_i)$, $p_i=p(b_i)$, and
$\mathcal G_i=\mathbb F_{p_i}^{b_i}$.
Apply Lemma~\ref{lem:gaussian-block-input} successively, starting
with its base version and then using $H=\mathcal G_{i-1}$ and
$g_0=g_{i-1}$.
For sufficiently large $B$, its hypotheses hold at every step.
Equation~\eqref{eq:v-local-means} places the preceding mean in
$[1/2,2]$. Together with \eqref{eq:v-prime} and the fact that
$b_{i-1}\le2b_i+1\le3b_i\le b_i^{10/9}$, we thus obtain
\[
  \begin{gathered}
    g_{i-1}\le\exp(5b_{i-1}^{14/25})
      \le\exp(5b_i^{28/45}),\\
    \log|\mathcal G_{i-1}|
    =O(b_{i-1}^2\log\log b_{i-1})\le b_i^4.
  \end{gathered}
\]

We obtain lines $\mathcal C_i\subset\mathcal G_i$, functions $g_i$,
and, for $i<r$, coupling matrices $W_i$ of size $b_i\times b_{i+1}$:
represent the shifts in $\mathcal G_i$ by columns in
$\{0,\ldots,p_i-1\}^{b_i}$ and divide by $p_i$.
Thus $p_iW_i$ is integral, while every entry of $W_i$ lies in $[0,1)$.

Use block coordinates
$\mathbb R^D=\prod_{i=1}^r\mathbb R^{b_i}$; subscripts on vectors
denote blocks, and floors and reductions modulo a prime are taken
coordinatewise.
The block lattices
$L_i^0=\mathbb Z^{b_i}+\mathcal C_i/p_i$ are the usual
\emph{Construction~A} preimages of the lines $\mathcal C_i$ over
$\mathbb F_{p_i}$, scaled by $p_i^{-1}$; they have covolume
$p_i^{-1}$.
Put $L^0=\prod_iL_i^0$ and define
\[
  (M\ell)_i=h_i(\ell_i+W_i\ell_{i+1})\quad(i<r),
  \qquad
  (M\ell)_r=h_r\ell_r,
  \qquad
  \Delta_{\mathrm v}^0=\det(ML^0)=\prod_i\frac{h_i^{b_i}}{p_i}.
\]

The map is triangular: changing $\ell_i$ moves only its own block
and, when present, the preceding block. We can therefore choose
points from the last block backwards, leaving the later choices
undisturbed and absorbing each new block's effect into a translation
of the earlier target. The bits of block $i$ sample the preceding
weight $g_{i-1}$ at shifted grid points, preserving nearly its mean.

\emph{The recursive weight estimate.}
For $i\le r$, retain only the first $i$ blocks of $M$, omitting their
interaction with block $i+1$. Denote the resulting map by $M^{[i]}$
and put $d_i=\sum_{j\le i}b_j$. The lattice
$M^{[i]}(\prod_{j\le i}L_j^0)$ has covolume
$\Delta_{\mathrm v}^{[i]}=\prod_{j\le i}h_j^{b_j}/p_j$.

We claim that, for every target $y\in\mathbb R^{d_i}$, there is a
possibly empty pattern $P_y^{[i]}\subset\prod_{j\le i}L_j^0$ with
\[
  \sum_{\ell\in P_y^{[i]}}
    \Delta_{\mathrm v}^{[i]}\gamma_{d_i}(y-M^{[i]}\ell)
  \ge\alpha_i g_i(x_i),
  \qquad
  x_i=\lfloor p_iy_i/h_i\rfloor\bmod p_i.
\]
The pattern is binary by suffixes when nonempty, and each of its
points satisfies the block residual bounds
$\|(y-M^{[i]}\ell)_j\|_\infty\le h_j$.
The coefficients are defined by
\begin{equation}
  \alpha_1=e^{-\xi_1},
  \qquad
  \alpha_i=\alpha_{i-1}e^{-\xi_i}
    (1-C_{\mathrm{blk}}b_i^{-1/100})\overline g_{i-1},
  \qquad
  \xi_i=\xi(b_i)=b_ih_i^2/p_i.
  \label{eq:v-coefficient-recurrence}
\end{equation}

If $g_i(x_i)=0$, we take the empty pattern. Otherwise, we apply the
block input at $t=y_i/h_i$, obtaining $\lambda_i$ and $\pi_t$.
For $i=1$, take all $\lambda_1+\sigma$ and sum
\eqref{eq:v-local-weight}.

For $i>1$, attach to each last-block choice
$\ell_i=\lambda_i+\sigma$ the preceding pattern with modified target
\[
  \widetilde y_{i-1}
  =y_{i-1}-h_{i-1}W_{i-1}(\lambda_i+\sigma),
\]
leaving the earlier target blocks unchanged. Its last grid index is
\begin{equation}
  x-p_{i-1}W_{i-1}\sigma\pmod{p_{i-1}},
  \qquad
  x=\left\lfloor
    p_{i-1}y_{i-1}/h_{i-1}
    -p_{i-1}W_{i-1}\lambda_i
  \right\rfloor\pmod{p_{i-1}}.
  \label{eq:v-grid-update}
\end{equation}
Here $p_{i-1}W_{i-1}\sigma$ is integral, so it commutes with the
coordinatewise floor. The vector involving $\lambda_i$ need not be
integral; no compatibility between consecutive primes is required.

By induction and \eqref{eq:v-local-weight}, the total weight is at
least
\[
  \alpha_{i-1}e^{-\xi_i}g_i(x_i)
  \sum_{\sigma\in\{0,1\}^{b_i}}\pi_t(\sigma)
    g_{i-1}(x-p_{i-1}W_{i-1}\sigma)
  \ge\alpha_i g_i(x_i),
\]
where the last step is \eqref{eq:v-local-sampling}. Thus the
recurrence has no further normalization loss.

Different bit choices give different last blocks. Once a suffix of
later blocks is fixed, each current block lies in one translate
$\lambda_j+\{0,1\}^{b_j}$. This proves the scalar suffix property.
The residual bounds follow from \eqref{eq:v-local-residual}; the
normalized Gaussian factors multiply over the blocks, so
\eqref{eq:v-local-weight} gives an individual weight at most
\[
  \prod_{j\le i}e^{-c_{\mathrm{at}}b_j^{7/10}}
  \le e^{-c_{\mathrm{at}}b_1^{7/10}}.
\]

Taking $i=r$ and writing $P_y^0=P_y^{[r]}$ gives
\begin{equation}
  \sum_{\ell\in P_y^0}
    \Delta_{\mathrm v}^0\gamma_D(y-M\ell)
  \ge\alpha_rg_r\bigl(\lfloor p_ry_r/h_r\rfloor\bmod p_r\bigr).
  \label{eq:v-inherited-pattern}
\end{equation}

\emph{Accumulated loss.}
Equation~\eqref{eq:v-prime} makes $\xi_i$ smaller than every fixed
negative power of $b_i$. Together with \eqref{eq:v-local-means} and
$b_{i-1}\ge b_i$, this gives a fixed absolute
$C_{\mathrm{step}}>0$ such that
$\alpha_1\ge1-C_{\mathrm{step}}b_1^{-1/100}$ and
$\alpha_i/\alpha_{i-1}\ge1-C_{\mathrm{step}}b_i^{-1/100}$.

Set
$C_{\mathrm{sum}}=C_{\mathrm{step}}/(1-2^{-1/100})$.
The dyadic sizes give
$\sum_i b_i^{-1/100}\le B^{-1/100}/(1-2^{-1/100})$.
With $B$ large enough all factors below are nonnegative, so
\[
  \alpha_r
  \geq\prod_i(1-C_{\mathrm{step}}b_i^{-1/100})
  \geq1-C_{\mathrm{step}}\sum_i b_i^{-1/100}
  \geq1-C_{\mathrm{sum}}B^{-1/100}
  =:\kappa_B.
\]

\emph{Averaging over cosets.}
Choose $B=B_\delta$ sufficiently large that
\[
  \kappa_B(1-C_{\mathrm{blk}}B^{-1/100})\geq1-\delta.
\]
Although $\overline g_r$ is close to $1$, the value of $g_r$ at a
particular target need not be. The last block has bounded dimension,
so retaining all its cosets requires only a bounded number of binary
components. Enlarge only the last factor of $L^0$:
\[
  L=\left(\prod_{i<r}L_i^0\right)
    \times p_r^{-1}\mathbb Z^{b_r},
  \qquad
  \iota=[L:L^0]=p_r^{b_r-1},
  \qquad
  \Delta_{\mathrm v}=\iota^{-1}\Delta_{\mathrm v}^0.
\]

Let $\mathcal V$ be integer representatives for
$\mathcal G_r/\mathcal C_r$ and write
$q_v=(0,\ldots,0,v/p_r)$. For each target put
\[
  P_v=q_v+P^0_{y-Mq_v},
  \qquad
  \mathcal P_y=\bigcup_{v\in\mathcal V}P_v.
\]
Distinct $q_v$ represent distinct cosets of $L^0$ in $L$, so the
components are disjoint. Translation preserves their residuals and
suffix property.

Although $q_v$ is supported in the last block, $Mq_v$ also changes
block $r-1$ when $r>1$. This causes no loss:
\eqref{eq:v-inherited-pattern} holds for every earlier target, and
its right side depends only on the last block. In that block the
integer vector $v$ changes the grid index from
$x_r=\lfloor p_ry_r/h_r\rfloor\bmod p_r$ to $x_r-v$. Hence
\begin{equation}
  \sum_{\ell\in\mathcal P_y}
    \Delta_{\mathrm v}\gamma_D(y-M\ell)
  \geq\frac{\kappa_B}{\iota}
    \sum_{v\in\mathcal V}g_r(x_r-v)
  =\kappa_B\overline g_r
  \geq1-\delta.
  \label{eq:v-coset-average}
\end{equation}
The equality holds because $g_r$ is constant on
$\mathcal C_r$-cosets.

The union is therefore nonempty. Each normalized individual weight
is an inherited weight divided by $\iota$, which proves
\eqref{eq:v-atom}. Omitting empty components leaves at most $\iota$
components. Since $B_\delta\le b_r<2B_\delta$, we may fix
\[
  I_\delta=\max_{B_\delta\le b<2B_\delta}p(b)^{\,b-1},
\]
so that $\iota\le I_\delta$ uniformly in the initial block size.
Also $L$ contains $\mathbb Z^D$, and the inherited residual bounds
give
\[
  R^2=\sum_i b_ih_i^2
  \le\tfrac94D\log b_1
  \le\tfrac92b_1\log b_1
  \le5b_1\log(b_1+1).
\]

\emph{The finite coordinate sets.}
To choose one shear for all heights, we need finitely many possible
patterns when $y\in M[0,1)^D$.
For $y=Mz$ with $z\in[0,1)^D$, put $e_i=z_i-\ell_i$.
The residual bounds imply
\[
  \|e_r\|_\infty\leq1,
  \qquad
  \|e_i\|_\infty\leq1+D\|e_{i+1}\|_\infty.
\]
Hence each coordinate of $\ell$ is bounded by
$Q_{\mathrm{coord}}=1+(D+2)^D$.
Since $\ell_i\in p_i^{-1}\mathbb Z^{b_i}$, including in the enlarged
last block, each scalar coordinate has at most
$2Q_{\mathrm{coord}}p_i+1$ possible values.
Since $b_i\leq D$, equation~\eqref{eq:v-prime} gives the claimed bound.
\end{proof}

\section{Coverings with prescribed covolume}
\label{sec:covering-transfer}

We now assemble the covering from the vertical patterns of
Proposition~\ref{prop:vertical} and the representative bodies
$J_{j,a}$ for the horizontal sections of $K_j$ supplied by
Lemma~\ref{lem:sections}.
We choose one horizontal lattice and shear for these representatives,
then use completion to remove the deficient set.
The shear estimate extends~\cite[Lemma~6.1]{Lattice2026}
to nonnegative functions. For a pattern $P$, a fractional moment of
order $1/|P|$ lets H\"older's inequality handle dependence between
branches; we later apply H\"older again to average over the common
horizontal lattice.

\begin{lemma}[Functional shear]
\label{lem:shear}
Let $\Lambda_{\mathrm h}\subset\mathbb R^m$ be a full-rank lattice,
and equip $\mathbb T=\mathbb R^m/\Lambda_{\mathrm h}$ with normalized
Haar measure.
Let $P\subset\mathbb R^D$ be binary by suffixes, and let
$\phi_\ell:\mathbb T\to[0,\infty)$, $\ell\in P$, be bounded
measurable functions.
Choose the columns of a linear map $Z:\mathbb R^D\to\mathbb R^m$
independently and uniformly in a fixed fundamental parallelepiped
of $\Lambda_{\mathrm h}$, and put
\[
  G_P(Z)=\int_{\mathbb T}\prod_{\ell\in P}
    \phi_\ell\bigl(x-\pi_{\Lambda_{\mathrm h}}(Z\ell)\bigr)\,dx.
\]
Then
\begin{equation}
  \mathbb E G_P^{1/|P|}
  \le\prod_{\ell\in P}
    \left(\int_{\mathbb T}\phi_\ell\right)^{1/|P|}.
  \label{eq:shear-root}
\end{equation}
\end{lemma}

\begin{proof}
We induct on the number of coordinates. In dimension zero $P$ is a
singleton and equality holds. If the last coordinate has only one
value, its column only translates the product, and induction applies.
Otherwise let $c$ and $c+1$ be its two values. Deleting the last
coordinate from the points whose last coordinate is $c+i$, and
retaining their functions, gives slices
$P_0,P_1\subset\mathbb R^{D-1}$.
They are binary by suffixes, and $|P_0|+|P_1|=|P|$. Write $z$ for
the last column of $Z$ and $Z'$ for the others, and put
\[
  \psi_i(x)=\prod_{\ell'\in P_i}\phi_{(\ell',c+i)}
    \bigl(x-\pi_{\Lambda_{\mathrm h}}(Z'\ell')\bigr)
  \qquad(i=0,1),
\]
so that $G_{P_i}(Z')=\int_{\mathbb T}\psi_i$.
Translation invariance of the torus integral gives, for real $c$,
\begin{align*}
  G_P(Z)
  &=\int_{\mathbb T}
    \psi_0(x-cz)\,\psi_1\bigl(x-(c+1)z\bigr)\,dx\\
  &=\int_{\mathbb T}\psi_0(x)\,\psi_1(x-z)\,dx,
\end{align*}
where translations are taken modulo $\Lambda_{\mathrm h}$. Since
$\pi_{\Lambda_{\mathrm h}}(z)$ is uniform on $\mathbb T$ and
independent of $Z'$, Fubini's theorem gives
$\mathbb E[G_P\mid Z']=G_{P_0}(Z')\,G_{P_1}(Z')$.
Conditional Jensen then gives
\[
  \mathbb E\bigl[G_P^{1/|P|}\bigm| Z'\bigr]
  \le\bigl(G_{P_0}G_{P_1}\bigr)^{1/|P|}
  =\bigl(G_{P_0}^{1/|P_0|}\bigr)^{|P_0|/|P|}
   \bigl(G_{P_1}^{1/|P_1|}\bigr)^{|P_1|/|P|}.
\]
The weights $|P_0|/|P|$ and $|P_1|/|P|$ sum to $1$. Taking
expectations over $Z'$ and applying H\"older's inequality with
exponents $|P|/|P_0|$ and $|P|/|P_1|$ gives
\[
  \mathbb E G_P^{1/|P|}
  \le\left(\mathbb E G_{P_0}^{1/|P_0|}\right)^{|P_0|/|P|}
     \left(\mathbb E G_{P_1}^{1/|P_1|}\right)^{|P_1|/|P|};
\]
the inductive hypothesis then proves \eqref{eq:shear-root}.
\end{proof}

Below, the shear estimate and Lemma~\ref{lem:thinning} will give a
lattice whose covering count meets the required lower bound outside
a small exceptional set. To complete the argument, we slightly
enlarge $K$ to remove this exceptional set. If $x-w$ has enough
centers for some $w\in K/q$, those same centers cover $x$ after the
dilation $K+K/q=(1+1/q)K$. The following volume condition guarantees
such a displacement for every $x$, extending~\cite[Lemma~6.2]{Lattice2026}
to multiplicity with the same quantitative hypothesis.

\begin{lemma}[Completion with multiplicity]
\label{lem:completion}
Let $K\subset\mathbb R^n$ be a convex body and $\Gamma$ a full-rank
lattice. Let $\kappa>0$ be real and $q\ge1$ an integer. If
\[
  \omega=|\{x\in\mathbb R^n/\Gamma:N_{\Gamma,K}(x)<\kappa\}|
  <(1-\omega)q^{-n},
\]
then $N_{\Gamma,(1+1/q)K}(x)\ge\kappa$ for every $x$.
\end{lemma}

\begin{proof}
Every $x$ with $N_{\Gamma,K}(x)\ge\kappa>0$ lies in $K+\Gamma$, so
the image $\pi_\Gamma(K)$ has measure at least $1-\omega$.
Multiplication by $q$ on the torus maps $\pi_\Gamma(K/q)$ onto this
image and increases measure by at most $q^n$: partition a fundamental
parallelepiped into $q^n$ pieces on which this map is injective.
Thus $|\pi_\Gamma(K/q)|\ge q^{-n}(1-\omega)>\omega$.
For every $x$, some $w\in K/q$ therefore has
$N_{\Gamma,K}(x-w)\ge\kappa$. Choose distinct
$\lambda_1,\ldots,\lambda_{\lceil\kappa\rceil}\in\Gamma$ with
$x-w-\lambda_i\in K$.
Then $x-\lambda_i\in K+K/q=(1+1/q)K$ for every $i$, by convexity.
\end{proof}

\subsection{Parameters and mean section counts}
\label{sec:parameters}

This subsection fixes the parameters for
Lemma~\ref{lem:uniform-laplace}; the large-dimensional proof of
Theorem~\ref{thm:prescribed} is completed in
Section~\ref{sec:large-dimensional-completion}.

Fix $0<\varepsilon<1$ and convex bodies
$K_1,\ldots,K_N\subset\mathbb R^n$ whose uniform distributions have
a common covariance matrix, where $1\le N\le\exp(n^{c_0})$, and
$\Delta>0$.
Assume $n\ge n_\varepsilon$ and
$\rho_j=|K_j|/\Delta\ge A^{\mathrm{large}}_\varepsilon n\log n$
for every $j$. The constant $A^{\mathrm{large}}_\varepsilon$ is
specified below; all dimension requirements depend only on
$\varepsilon$ and the fixed input constants.
Section~\ref{sec:small-dimensions} then enlarges the density constant
to cover the remaining one-body cases.

\emph{Scales.}
Put $\delta=\varepsilon/10$ and
\[
  q=\lceil n/\delta\rceil,
  \qquad
  s=1+1/q,
  \qquad
  s^n\le e^\delta.
\]
We first construct a lattice of covolume $s^n\Delta$ with a small
deficient set, then complete the covering and divide the lattice
by $s$.
This reserves the final dilation's density cost.

Use Proposition~\ref{prop:vertical} at accuracy $\delta$ with
$b_1=\lceil(\log\log n)^2\rceil$. Write its lattice and map as $L,M$,
and put $\Delta_{\mathrm v}=\det(ML)$, $m=n-D$, and $I=I_\delta$.
Then
\begin{equation}
  D\le2b_1,
  \qquad
  \tau_n=e^{-c_{\mathrm{at}}b_1^{7/10}},
  \qquad
  2^D=n^{o(1)},
  \qquad
  \frac{\eta_m}{\tau_n n\log n}\longrightarrow\infty.
  \label{eq:transfer-scales}
\end{equation}
Indeed $b_1^{7/10}\asymp(\log\log n)^{7/5}$ grows faster than
$\log\log n$, so $\tau_n\log n\to0$.
Fix
\[
  A^{\mathrm{large}}_\varepsilon
  =\frac{3e^\delta I_\delta}{\delta^2}.
\]

For all sufficiently large $n$, Lemma~\ref{lem:sections} applies
with $A_{\mathrm{sec}}=5$ and gives common horizontal--vertical
coordinates and affine section maps $T_j(y)=A_jy+t_j$.
Multiply $\Delta$ by the absolute determinant of the common linear
map; the ratios $|K_j|/\Delta$ are unchanged.

Keep the notation $K_j,\Delta$ for the normalized data, and put
\[
  V_j=|K_j|,
  \qquad
  \Delta_{\mathrm h}=\frac{s^n\Delta}{\Delta_{\mathrm v}},
  \qquad
  \rho'_j=\frac{V_j}{s^n\Delta},
  \qquad
  u_{j,a}=\frac{|J_{j,a}|}{\Delta_{\mathrm h}}.
\]
Here $\rho'_j$ is the mean count of the temporary arrangement of
covolume $s^n\Delta$, whereas $\rho_j=s^n\rho'_j$ is the prescribed
final mean. The quantity $u_{j,a}$ is the mean lattice count for the
representative body $J_{j,a}$.

\emph{Mean section counts.}
Fix a body $j$ and a height $y$. For each $\ell\in\mathcal P_y$,
choose the label $a_y(\ell)$ from \eqref{eq:section-labels} at height
$y-M\ell$. This residual has norm at most $R$ by \eqref{eq:v-radius}.
Writing $w_y(\ell)=\Delta_{\mathrm v}\gamma_D(y-M\ell)$ for the
normalized vertical weight, the bounds on the representative volumes
give
\[
  (1-\delta)\rho'_j w_y(\ell)
  \le u_{j,a_y(\ell)}
  \le\rho'_j w_y(\ell).
\]
Thus $w_y(\ell)$ gives the representative's fraction of the full
mean up to the displayed $1-\delta$ loss. Combining these inequalities
with \eqref{eq:v-total} and \eqref{eq:v-atom} gives
\begin{equation}
  \sum_{\ell\in\mathcal P_y}u_{j,a_y(\ell)}
  \ge(1-\delta)^2\rho'_j,
  \qquad
  u_{j,a_y(\ell)}\le\tau_n\rho'_j.
  \label{eq:total-load}
\end{equation}

For body $j$, retain only labels with
$u_{j,a}\le\tau_n\rho'_j$.
Every selected residual still has a retained label by
\eqref{eq:total-load}, and the uniform upper bound on these mean
counts allows us to apply thinning to every labeled pattern
considered below.

\emph{Labeled patterns.}
A \emph{labeled pattern} for body $j$ consists of a nonempty set
$P\subset\mathcal A_1\times\cdots\times\mathcal A_D$, binary by
suffixes, and a retained label $a(\ell)$ at each point. Here the
finite coordinate sets come from Proposition~\ref{prop:vertical}.
Every nonempty component of $\mathcal P_y$ for $y\in M[0,1)^D$,
labeled by $a_y$, belongs to this family.

For a fixed body, a labeled point has at most
$\exp(O(D\log n+D^2\log(D+2)))$ possible descriptions. Each pattern
has at most $2^D$ points, so the total number $Q$ of labeled
patterns, counting all $N$ bodies, satisfies
\begin{equation}
  \log Q
  \le\log N
    +O\!\left(2^D[D\log n+D^2\log(D+2)]\right)
  =o(\sqrt m).
  \label{eq:transfer-entropy}
\end{equation}

\emph{Laplace parameters.}
For each body put
\begin{equation}
  \beta_j=\delta\min\left\{
    1,\frac{\eta_m}{\tau_n\rho'_j}
  \right\}.
  \label{eq:laplace-scales}
\end{equation}
Then $\beta_j\le\delta$ and
$\beta_ju_{j,a}\le\delta\eta_m$ for every retained label, keeping
each exponential test within the range of thinning. The total
exponent satisfies
$\beta_j\rho'_j=\delta\min\{\rho'_j,\eta_m/\tau_n\}$.

\subsection{A Laplace bound at every height}

Lemma~\ref{lem:thinning} controls a single function's Laplace
transform on average over lattices. Here we apply it to the
representative indicators and use the shear estimate to select one
horizontal lattice and shear that work for all labeled patterns.
The section containments and vertical weight bounds then give the
following estimate for the full covering count, at every height and
for every body.

\begin{lemma}[Uniform Laplace bound]
\label{lem:uniform-laplace}
With the setup and parameters of Section~\ref{sec:parameters}, for
all sufficiently large $n$ there are a lattice
$\Lambda_{\mathrm h}\subset\mathbb R^m$ of covolume
$\Delta_{\mathrm h}$ and a linear map
$Z:\mathbb R^D\to\mathbb R^m$ such that the lattice
\[
  \Gamma=\{(\lambda+Z\ell,M\ell):
    \lambda\in\Lambda_{\mathrm h},\ \ell\in L\},
\]
of covolume
$\Delta_{\mathrm h}\Delta_{\mathrm v}=s^n\Delta$, satisfies
\begin{equation}
  \int_{\mathbb R^m/\Lambda_{\mathrm h}}
    e^{-(\beta_j/I)N_{\Gamma,K_j}(x,y)}\,dx
  \le\exp\!\left[
    2^D\sqrt m-(1-\delta)^3\frac{\beta_j}{I}\rho'_j
  \right]
  \label{eq:total-laplace}
\end{equation}
for every $j$ and every $y\in\mathbb R^D$. The integral uses
normalized Haar measure on the horizontal torus.
\end{lemma}

The three factors $1-\delta$ come from the representative volumes,
the selected vertical weight, and the linearization of
$1-e^{-\beta_j}$. The coset components need not form one binary
pattern, so H\"older combines their bounds at parameter $\beta_j/I$.
The parameter $\beta_j$ contains a factor $\delta$, and the Markov
step uses the gap of at least $\delta\rho'_j$ between
$(1-\delta)^3\rho'_j$ and the deficient-count cutoff
$(1-4\delta)\rho'_j$. Together these give the factor
$I_\delta/\delta^2$ in the density constant.

\begin{proof}[Proof of Lemma~\ref{lem:uniform-laplace}]
\emph{One representative.}
Choose $\Lambda_{\mathrm h}$ with the distribution obtained by
scaling $\mu_m$ by $\Delta_{\mathrm h}^{1/m}$. Conditional on
$\Lambda_{\mathrm h}$, choose the columns of $Z$ independently and
uniformly in a fundamental parallelepiped, using a measurable choice
of lattice bases. For each retained label $a$ of body $j$, put
\[
  F_{j,a}(\Lambda_{\mathrm h})
  =\int_{\mathbb R^m/\Lambda_{\mathrm h}}
    e^{-\beta_jN_{\Lambda_{\mathrm h},J_{j,a}}(x)}\,dx.
\]

Apply Lemma~\ref{lem:thinning} to
$\mathbf1_{\Delta_{\mathrm h}^{-1/m}J_{j,a}}$.
Since
\[
  (1-e^{-\beta_j})u_{j,a}
  \ge(1-\beta_j/2)\beta_ju_{j,a}
  \ge(1-\delta)\beta_ju_{j,a},
  \qquad
  (1-e^{-\beta_j})u_{j,a}\le\delta\eta_m<\eta_m,
\]
we obtain
\begin{align}
  \mathbb E_{\Lambda_{\mathrm h}}F_{j,a}(\Lambda_{\mathrm h})
  &\le e^{-(1-\delta)\beta_ju_{j,a}}
    +C_{\mathrm{RS}}e^{-\eta_m}\notag\\
  &\le(1+C_{\mathrm{RS}})
    e^{-(1-\delta)\beta_ju_{j,a}},
  \label{eq:horizontal-laplace}
\end{align}
where the last step uses
$(1-\delta)\beta_ju_{j,a}\le\delta\eta_m\le\eta_m$.

\emph{One pattern.}
The affine section centers satisfy
\[
  T_j(y-M\ell)+Z\ell=T_j(y)+(Z-A_jM)\ell.
\]
For a labeled pattern $P$ of body $j$, define
\[
  G_P(\Lambda_{\mathrm h},Z)
  =\int_{\mathbb R^m/\Lambda_{\mathrm h}}
    \exp\!\left[
      -\beta_j\sum_{\ell\in P}
      N_{\Lambda_{\mathrm h},J_{j,a(\ell)}}
        \bigl(x-(Z-A_jM)\ell\bigr)
    \right]dx.
\]

Conditional on $\Lambda_{\mathrm h}$, apply
Lemma~\ref{lem:shear} to
$\phi_\ell(x)=
e^{-\beta_jN_{\Lambda_{\mathrm h},J_{j,a(\ell)}}(x+A_jM\ell)}$.
Translation preserves its integral
$F_{j,a(\ell)}(\Lambda_{\mathrm h})$.
All factors $F_{j,a(\ell)}$ depend on the same lattice.
H\"older's inequality with $|P|$ equal exponents nevertheless gives
\begin{align}
  \mathbb E_{\Lambda_{\mathrm h},Z}G_P^{1/|P|}
  &\le\mathbb E_{\Lambda_{\mathrm h}}\prod_{\ell\in P}
    F_{j,a(\ell)}(\Lambda_{\mathrm h})^{1/|P|}\notag\\
  &\le\prod_{\ell\in P}
    \bigl(
      \mathbb E_{\Lambda_{\mathrm h}}F_{j,a(\ell)}
    \bigr)^{1/|P|}\notag\\
  &\le(1+C_{\mathrm{RS}})
    \exp\!\left[
      -\frac{(1-\delta)\beta_j}{|P|}
      \sum_{\ell\in P}u_{j,a(\ell)}
    \right].
  \label{eq:joint-moment}
\end{align}

Markov's inequality, applied to $G_P^{1/|P|}$, shows that each
labeled pattern violates
\begin{equation}
  G_P(\Lambda_{\mathrm h},Z)
  \le e^{|P|\sqrt m}
    \exp\!\left[
      -(1-\delta)\beta_j\sum_{\ell\in P}u_{j,a(\ell)}
    \right]
  \label{eq:joint-choice}
\end{equation}
with probability at most $(1+C_{\mathrm{RS}})e^{-\sqrt m}$.
By \eqref{eq:transfer-entropy},
$(1+C_{\mathrm{RS}})Qe^{-\sqrt m}<1$, so one pair
$(\Lambda_{\mathrm h},Z)$ satisfies every bound. Fix such a pair,
and define $\Gamma$ as in the statement.

\emph{All components at one height.}
Fix a body $j$ and a height $y\in M[0,1)^D$. Write
$\mathcal P_y=P_1\sqcup\cdots\sqcup P_I$, allowing empty components,
and define the contribution of a component by
\[
  S_i(x)=\sum_{\ell\in P_i}
    N_{\Lambda_{\mathrm h},J_{j,a_y(\ell)}}
      \bigl(x-T_j(y)-(Z-A_jM)\ell\bigr).
\]
By \eqref{eq:count-decomposition} and the section inclusions,
$N_{\Gamma,K_j}(x,y)\ge\sum_iS_i(x)$.
All counted centers are distinct: different $\ell$ have different
vertical coordinates because $M$ is invertible.

Translation invariance and \eqref{eq:joint-choice} give
\[
  \int e^{-\beta_j S_i(x)}\,dx
  \le\exp\!\left[
    |P_i|\sqrt m
    -(1-\delta)\beta_j\sum_{\ell\in P_i}u_{j,a_y(\ell)}
  \right].
\]
For an empty component both sides are $1$.

H\"older's inequality over the $I$ components, the bound
$|P_i|\le2^D$, and \eqref{eq:total-load} now give
\begin{align*}
  \int e^{-(\beta_j/I)N_{\Gamma,K_j}(x,y)}\,dx
  &\le\prod_{i=1}^I
    \left(\int e^{-\beta_j S_i(x)}\,dx\right)^{1/I}\\
  &\le\exp\!\left[
    2^D\sqrt m-(1-\delta)^3\frac{\beta_j}{I}\rho'_j
  \right].
\end{align*}
Here all integrals are over the normalized horizontal torus.

\emph{All heights.}
Let $y\in\mathbb R^D$ be arbitrary, and write $y=y'+Mz$ with
$y'\in M[0,1)^D$ and $z\in\mathbb Z^D\subset L$. Since
$(Zz,Mz)\in\Gamma$, we have
$N_{\Gamma,K_j}(x,y)=N_{\Gamma,K_j}(x-Zz,y')$, and translation
invariance of the horizontal integral transfers the bound from $y'$
to $y$.
\end{proof}

\subsection{Completion}
\label{sec:large-dimensional-completion}

\begin{proof}[Proof of Theorem~\ref{thm:prescribed} in sufficiently
  large dimension]
Take $\Lambda_{\mathrm h}$, $Z$, and $\Gamma$ from
Lemma~\ref{lem:uniform-laplace}.
Call a point $(x,y)$ \emph{deficient} for body $j$ if
$N_{\Gamma,K_j}(x,y)<(1-4\delta)\rho'_j$.
Since $(1-\delta)^3-(1-4\delta)\ge\delta$, exponential Markov and
\eqref{eq:total-laplace} give, for every $j$ and every $y$,
\begin{equation}
  \left|
    \{x:N_{\Gamma,K_j}(x,y)<(1-4\delta)\rho'_j\}
  \right|
  \le\exp\!\left[
    2^D\sqrt m-\frac{\delta^2}{I}
    \min\left\{\rho'_j,\frac{\eta_m}{\tau_n}\right\}
  \right].
  \label{eq:transfer-small-hole}
\end{equation}
By \eqref{eq:transfer-scales}, $2^D\sqrt m=o(n\log n)$ and
$\delta^2\eta_m/(I\tau_n)\ge3n\log n$ for sufficiently large $n$.
The hypothesis
$\rho_j\ge A^{\mathrm{large}}_\varepsilon n\log n$ and our choice
of $A^{\mathrm{large}}_\varepsilon$ also give
$\delta^2\rho'_j/I\ge3n\log n$.
Thus \eqref{eq:transfer-small-hole} is at most $n^{-2n}$.

Let $F_{\mathrm h},F_{\mathrm v}$ be fundamental parallelepipeds for
$\Lambda_{\mathrm h},ML$. Their product is a fundamental domain for
$\Gamma$: reduce vertically, including the shear, and then
horizontally. The count is a locally finite sum of indicators of
compact translates, hence measurable. Fubini therefore converts the
heightwise estimate into a full deficient proportion at most
$n^{-2n}$, without requiring the auxiliary patterns to vary measurably
with height.

Since $q=O_\delta(n)$, for sufficiently large $n$ we have
$n^{-2n}<(1-n^{-2n})q^{-n}$. Lemma~\ref{lem:completion} applies
with $\kappa=(1-4\delta)\rho'_j$ and this same $q$. The identity
$N_{s^{-1}\Gamma,K_j}(x)=N_{\Gamma,sK_j}(sx)$ shows that
$\Lambda=s^{-1}\Gamma$ has exactly the prescribed covolume $\Delta$
and
\[
  N_{\Lambda,K_j}
  \ge(1-4\delta)\rho'_j
  \ge(1-4\delta)e^{-\delta}\frac{V_j}{\Delta}
  \ge(1-5\delta)\frac{V_j}{\Delta}
  \ge(1-\varepsilon)\frac{V_j}{\Delta}.
\]
Undo the common linear normalization. Choose $n_\varepsilon\ge3$
large enough for all the preceding dimension requirements. They
depend only on $\varepsilon$ and the fixed constants from the input
lemmas, so this completes the large-dimensional case.
\end{proof}

\subsection{The one-body case in all dimensions}
\label{sec:small-dimensions}

It remains to prove the one-body assertion of
Theorem~\ref{thm:prescribed} in the finitely many dimensions
$2\le n<n_\varepsilon$.
We start from a \emph{1-fold covering lattice} $\Lambda_0$, meaning
that $K+\Lambda_0=\mathbb R^n$. Such a lattice always exists: the
elementary affine normalization below gives one of density at most
$(2n)^n$. The subsequent contraction argument works for any initial
covering lattice.

\begin{proof}[Proof of Theorem~\ref{thm:prescribed} in the remaining
  dimensions]
Choose a maximum-volume simplex contained in $K$, which exists by
compactness, and map it to $\operatorname{conv}(0,e_1,\ldots,e_n)$.
Write $\widetilde K$ for the image of $K$.
Replacing the simplex's $i$-th nonzero vertex by any
$x\in\widetilde K$ shows that $|x_i|\le1$, so
$\widetilde K\subset[-1,1]^n$. Also
$[0,1/n]^n\subset\widetilde K$. This cube is a fundamental cell of
$(1/n)\mathbb Z^n$, so the translates of $\widetilde K$ along that
grid cover space, with density at most
$2^n/(1/n)^n=(2n)^n$; pulling the grid back by the linear part of the
normalization gives a covering lattice $\Lambda_0$ for the original
body: the translation part does not affect whether the translates
cover all of space.
Volume and covolume scale by the same factor, so
$\rho_0=|K|/\det\Lambda_0\le(2n)^n$.

To contract $\Lambda_0$, we use the identity
\[
  N_{\Lambda_0/s,K}(x)=N_{\Lambda_0,sK}(sx),
  \qquad s>1.
\]
Thus we may equivalently enlarge the body and count along the
original lattice. Choose a measurable fundamental domain
$F\subset K$ for $\Lambda_0$:
enumerate the lattice and, for each point in a half-open fundamental
parallelepiped, use the first translate placing it in $K$.
The translates $F+\lambda$ partition space. For $s>1$, every such
tile meeting $x-(s-1)K$ has $\lambda\in x-sK$: a point in the
intersection can be written as $f+\lambda=x-(s-1)u$, with
$f,u\in K$, and convexity gives $f+(s-1)u\in sK$.
These tiles cover $x-(s-1)K$, each has volume $\det\Lambda_0$, and
all their indices are counted by $N_{\Lambda_0,sK}(x)$.
Comparing volumes gives
\[
  N_{\Lambda_0,sK}(x)\ge\rho_0(s-1)^n.
\]
Given the prescribed density $\rho=|K|/\Delta>\rho_0$, set
$s=(\rho/\rho_0)^{1/n}$ and $\Lambda=\Lambda_0/s$.
Then $\det\Lambda=\Delta$ and
\[
  N_{\Lambda,K}(x)
  \ge\rho\bigl(1-(\rho_0/\rho)^{1/n}\bigr)^n
  \ge(1-\varepsilon)\rho
\]
provided
\[
  \rho\ge\rho_0
    \bigl[1-(1-\varepsilon)^{1/n}\bigr]^{-n}.
\]
The sufficient lower bound on $\rho$ depends only on $n$ and
$\varepsilon$ after replacing $\rho_0$ by $(2n)^n$. To express
it in the form $\rho\ge A_\varepsilon n\log n$ simultaneously
in all remaining dimensions, we set
\[
  A^{\mathrm{small}}_\varepsilon
  =\max_{2\le d<n_\varepsilon}
    \frac{(2d)^d}{d\log d}
    \bigl[1-(1-\varepsilon)^{1/d}\bigr]^{-d},
  \qquad
  A_\varepsilon
  =\max\{A^{\mathrm{large}}_\varepsilon,
         A^{\mathrm{small}}_\varepsilon\}.
\]
This finite maximum depends only on $\varepsilon$, and its $d=n$
term ensures that $\rho\ge A_\varepsilon n\log n$ satisfies the
density condition for the contraction argument in every remaining
small dimension.
Since $A_\varepsilon\ge A^{\mathrm{large}}_\varepsilon$, the
large-dimensional proof, including its simultaneous conclusion,
also applies.
\end{proof}

\section{Upper multiplicity and smooth coverings}
\label{sec:vertical-spike}

Theorem~\ref{thm:prescribed} keeps the multiplicity everywhere above
nearly its mean. A \emph{smooth covering} also requires an upper
bound: for $0<\varepsilon<1$, an arrangement of density $\rho$ is
$\varepsilon$-smooth if
\[
  (1-\varepsilon)\rho\le N_{\Lambda,K}(x)
  \le(1+\varepsilon)\rho
  \qquad\text{for every }x\in\mathbb R^n.
\]
For the lattices constructed here,
Proposition~\ref{prop:vertical-spike} below shows that the maximum
normalized count grows with dimension. This is a limitation of their
vertical structure; it does not exclude smooth coverings by other
lattices.

The lower bound does, however, control the average deviation. Write
$\mathcal N=N_{\Lambda,K}$ and $\rho=|K|/\det\Lambda$.
Since $\mathcal N/\rho$ has mean $1$, the bound
$\mathcal N\ge(1-\varepsilon)\rho$ gives
\begin{equation}
  \int_{\mathbb R^n/\Lambda}
    \left|\frac{\mathcal N(x)}\rho-1\right|\,dx
  =2\int_{\mathbb R^n/\Lambda}
    \left(1-\frac{\mathcal N(x)}\rho\right)_+\,dx
  \le2\varepsilon,
  \label{eq:lower-l1}
\end{equation}
where the integrals use normalized Haar measure and
$a_+=\max\{a,0\}$.
This $L^1$ bound allows large counts on small sets. In our
construction, one horizontal layer can already have mean count far
above the mean of the full arrangement. A shear translates that
layer's covering pattern without changing its horizontal mean, so
some point must still have high multiplicity. The small-atom bound
in Proposition~\ref{prop:vertical} controls only the selected
pattern and is compatible with such a peak in the full sum.

\begin{proposition}[Upper multiplicity from a single layer]
\label{prop:vertical-spike}
Let $K\subset\mathbb R^m\times\mathbb R^D$ be a convex body,
$\Lambda_{\mathrm h}\subset\mathbb R^m$ and $L\subset\mathbb R^D$
full-rank lattices, $M:\mathbb R^D\to\mathbb R^D$ invertible linear,
and $Z:\mathbb R^D\to\mathbb R^m$ linear. Put
\begin{gather*}
  \Gamma=\{(\lambda+Z\ell,M\ell):
    \lambda\in\Lambda_{\mathrm h},\ \ell\in L\},\\
  \Delta_{\mathrm h}=\det\Lambda_{\mathrm h},
  \qquad
  \Delta_{\mathrm v}=\det(ML),
  \qquad
  \rho=\frac{|K|}{\Delta_{\mathrm h}\Delta_{\mathrm v}}.
\end{gather*}
Then
\begin{equation}
  \frac{\sup_{x,y}N_{\Gamma,K}(x,y)}{\rho}
  \ge\Delta_{\mathrm v}\frac{\sup_y|K[y]|}{|K|}.
  \label{eq:vertical-spike-average}
\end{equation}
For the vertical lattices in Proposition~\ref{prop:vertical}, with
cutoff $B$ fixed and $b_1\to\infty$,
\begin{equation}
  \log\bigl(\Delta_{\mathrm v}\gamma_D(0)\bigr)
  =\bigl(\tfrac12+o(1)\bigr)D.
  \label{eq:vertical-spike-growth}
\end{equation}
\end{proposition}

\begin{proof}
For every height $y$, normalized horizontal integration gives
\[
  \int_{\mathbb R^m/\Lambda_{\mathrm h}}
    N_{\Gamma,K}(x,y)\,dx
  =\frac1{\Delta_{\mathrm h}}
    \sum_{\ell\in L}|K[y-M\ell]|
  \ge\frac{|K[y]|}{\Delta_{\mathrm h}}.
\]
The sum is finite because $K$ is compact, and translation invariance
removes $Z$ from each integral. Divide by $\rho$ and take the
supremum over $y$ to obtain \eqref{eq:vertical-spike-average}.

It remains to compute the vertical covolume of our construction.
Recall the folded density $f_b$ and its logarithmic mean
$\mathfrak m(b)$ from Section~\ref{sec:vertical}.
If $X$ is standard normal conditioned on $|X|\le h(b)$, then
$\psi_b(X)=X/h(b)\pmod1$, represented in $[0,1)$, has density
$f_b/\int_0^1f_b$. Thus
\begin{align*}
  \mathfrak m(b)
  &=\mathbb E\log f_b(\psi_b(X))\\
  &=\log h(b)-\tfrac12\log(2\pi)-\tfrac12\mathbb E X^2
    +\mathbb E\log\bigl(1+e^{-h(b)^2/2+h(b)|X|}\bigr)\\
  &=\log h(b)-\tfrac12\log(2\pi e)+o(1).
\end{align*}
Indeed $\mathbb E X^2\to1$, and the last logarithmic summand tends
to zero pointwise. The inequality
$-h(b)^2/2+h(b)|X|\le X^2/2$ bounds it by $\log2+X^2/2$.
Dominated convergence under the unconditioned normal law, divided
by $\mathbb P(|X|\le h(b))\to1$, proves the last equality.

The prime window \eqref{eq:block-prime-window} gives
$\log p(b)=b\mathfrak m(b)+O(b^{14/25})$.
Since
$\Delta_{\mathrm v}=\iota^{-1}\prod_i h_i^{b_i}/p_i$ with
$\iota=p_r^{b_r-1}$, it follows that
\begin{align*}
  \log\bigl(\Delta_{\mathrm v}\gamma_D(0)\bigr)
  &=\sum_i b_i\left(
    \log h_i-\mathfrak m(b_i)-\tfrac12\log(2\pi)
  \right)
    +O\left(\sum_i b_i^{14/25}\right)-\log\iota\\
  &=\tfrac D2+o(D).
\end{align*}
Here $\log\iota=O_B(1)$ and the dyadic sizes give
$\sum_i b_i^{14/25}=O(b_1^{14/25})=o(D)$.
For the remaining errors, put
$e(b)=\log h(b)-\mathfrak m(b)-\tfrac12\log(2\pi e)$, so
$e(b)\to0$.
Given $\xi>0$, choose $b_*>B$ such that $|e(b)|\le\xi$ for
$b\ge b_*$.
The blocks with $b_i<b_*$ have total dimension at most $2b_*$,
hence
\[
  \left|\sum_i b_i e(b_i)\right|
  \le\xi D+2b_*\max_{B\le b<b_*}|e(b)|.
\]
First let $b_1\to\infty$ with $B,b_*$ fixed, and then let $\xi\to0$.
This proves that their sum is $o(D)$.
\end{proof}

The covolume is therefore $(2\pi e)^{D/2+o(D)}$. A Gaussian residual
with squared norm $D+o(D)$ has normalized weight $e^{o(D)}$, whereas
the zero residual has weight $e^{D/2+o(D)}$. The exponential peak
reflects the difference between the Gaussian density at its mode
and its density at typical radius.

Fix the accuracy, so the block cutoff $B$ is fixed as $n\to\infty$.
By \eqref{eq:section-lower}, the normalized bodies in the
large-dimensional construction of Section~\ref{sec:covering-transfer}
satisfy $|K[0]|/|K|\ge(1-o(1))\gamma_D(0)$.
Thus their supremum multiplicity divided by density is at least
$\exp((1/2+o(1))D)$, independently of the horizontal lattice and
shear.
Replacing $\Gamma$ by $\Gamma/s$ multiplies the right side of
\eqref{eq:vertical-spike-average} by $s^{-D}$.
In the completion step $s=1+O_\varepsilon(1/n)$ and $D=o(n)$, so
this factor tends to $1$. Consequently the final covering lattice
$\Lambda$ satisfies
\[
  \frac{\sup_x N_{\Lambda,K}(x)}{|K|/\det\Lambda}
  \ge\exp\bigl((1/2+o(1))D\bigr)
  \longrightarrow\infty.
\]
Undoing the linear normalization preserves this ratio. Thus these
lattices eventually violate every fixed upper smoothness bound,
even though their lower counts remain close to the mean everywhere.

It remains open whether a different construction gives two-sided
$\varepsilon$-smooth lattice coverings at density
$O_\varepsilon(n\log n)$, matching the periodic benchmark discussed
in the introduction. Another question is whether the density constant
$A_\varepsilon$ in Theorem~\ref{thm:prescribed} can be chosen to grow
at most polynomially in $1/\varepsilon$. The present proof does not
provide such a bound.

\appendix

\section{Attainment of the optimal lattice density}
\label{sec:attainment}

\begin{proposition}[Attainment of the infimum]
For every convex body $K\subset\mathbb R^n$ and positive integer $k$,
there is a lattice $k$-fold covering by translates of $K$ with
density $\theta_{\mathrm{lat},k}(K)$.
\end{proposition}

\begin{proof}
A cube contained in $K$ gives a one-fold covering on a sufficiently
fine lattice; passing to an index-$k$ superlattice gives a $k$-fold
covering.
Choose a sequence $\Lambda_j$ of $k$-fold covering lattices whose
densities tend to $\theta_{\mathrm{lat},k}(K)$. The densities are
bounded above by a finite constant, so their covolumes, equal to
$|K|$ divided by those densities, are bounded below by some
$\Delta_*>0$.
Fix $R>0$ with $\|z\|\le R$ for all $z\in K$.
Since $K+\Lambda_j=\mathbb R^n$, every point of $\mathbb R^n$ is
within distance $R$ of $\Lambda_j$.

Each $\Lambda_j$ contains $n$ linearly independent vectors of norm
at most $3R$. Indeed, given their current span, choose a unit vector
$u$ perpendicular to it. A lattice point within distance $R$ of
$2Ru$ lies outside that span and has norm at most $3R$.
Let $U_j$ have the selected vectors as columns. Then
\[
  \Delta_*\le|\det U_j|\le(3R)^n,
  \qquad
  [\Lambda_j:U_j\mathbb Z^n]\le(3R)^n/\Delta_*.
\]
There are only finitely many superlattices of $\mathbb Z^n$ of
bounded index, as a superlattice of index $r$ lies in
$r^{-1}\mathbb Z^n$.
The matrices $U_j$ are bounded, with determinants bounded away from
zero. After passing to a subsequence, therefore,
\[
  U_j\longrightarrow U,
  \qquad
  U_j^{-1}\Lambda_j=\Lambda_*,
\]
where $U$ is invertible and $\Lambda_*$ is fixed.

Set $\Lambda=U\Lambda_*$. For each fixed $x$, boundedness of $K$
and of $U_j^{-1}$ means that only finitely many $v\in\Lambda_*$
can satisfy $x-U_jv\in K$, uniformly in $j$. Along a subsequence
attaining the limsup of the counts, the same finite set of covering
vectors occurs infinitely often, by the pigeonhole principle.
Closedness of $K$ preserves each of these vectors in the limit,
giving
\[
  N_{\Lambda,K}(x)
  \ge\limsup_{j\to\infty}N_{\Lambda_j,K}(x)
  \ge k.
\]
Thus $K+\Lambda$ is a $k$-fold covering. Continuity of the covolume
shows that its density is $\theta_{\mathrm{lat},k}(K)$.
\end{proof}

\section{The comparison with arbitrary translates}
\label{sec:translative-comparison}

In its set-cover formulation, the greedy bound of Nasz\'odi and
Polyanskii~\cite[Theorem~1.1]{NP2018} controls the size of an integral
multicover in terms of the fractional one-fold covering number,
with a free parameter $0<\lambda<1$. Their geometric
bound~\cite[Theorem~1.4]{NP2018} uses a fixed parameter; choosing it
according to $\varepsilon$ gives the near-unit bound below. We record
the deduction for nonsymmetric bodies and distinct translation
vectors.
The density of a periodic arrangement is its number of centers in
one period, multiplied by $|K|$ and divided by the period volume.

\begin{proposition}[Periodic multi-coverings]
\label{prop:translative-comparison}
For every $\varepsilon>0$ there is $C^{\mathrm{tr}}_\varepsilon>0$
such that every convex body $K\subset\mathbb R^n$, $n\ge2$, admits
a periodic $k$-fold covering by distinct translates of density at most
\[
  \max\{(1+\varepsilon)k,C^{\mathrm{tr}}_\varepsilon n\log n\}
  \qquad(k\ge1).
\]
\end{proposition}

\begin{proof}
We first show that, for $0<t,\lambda<1$, the infimum of the densities
of such coverings is at most
\begin{equation}
  (1-t)^{-n}\frac{1-\lambda^k}{1-\lambda}
  \left[1+n\log\frac4t-(k-1)\log\lambda\right].
  \label{eq:translative-density}
\end{equation}
A suitable translate of $K$ has $S=K\cap(-K)$ with
$|S|\ge2^{-n}|K|$. Indeed,
\[
  \int_{2K}|K\cap(z-K)|\,dz=|K|^2,
  \qquad
  |2K|=2^n|K|,
\]
so some $z$ gives an intersection of at least this volume; replace
$K$ by $K-z/2$. Thus $S$ is a symmetric convex body contained in $K$.

\emph{A finite covering problem.}
Let $C_R=[-R,R]^n$, and choose a maximal packing of translates of
$(t/2)S$ with centers in a finite set $\mathcal A_R\subset C_R$.
Maximality gives $C_R\subset\mathcal A_R+tS$. If centers of this
packing lie in $x+(1-t)K$, their disjoint interiors lie in
$x+(1-t/2)K$. Consequently,
\begin{equation}
  \sup_x\#\bigl(\mathcal A_R\cap[x+(1-t)K]\bigr)
  \le\frac{(1-t/2)^n|K|}{(t/2)^n|S|}
  \le(4/t-2)^n
  <(4/t)^n.
  \label{eq:translative-degree}
\end{equation}

We first cover the finite set $\mathcal A_R$ fractionally: assign
nonnegative weights to translates of $(1-t)K$ so that the total
weight covering each point is at least $1$. Giving translation
vectors in $C_R-(1-t)K$ Lebesgue weight $1/|(1-t)K|$ does this:
the vectors covering a fixed $a\in\mathcal A_R$ form
$a-(1-t)K$, so their total weight is exactly $1$.
The total weight of all translation vectors is
\[
  \frac{|C_R-(1-t)K|}{(1-t)^n|K|}
  =(1+o_{R\to\infty}(1))
    \frac{|C_R|}{(1-t)^n|K|}.
\]

There are only finitely many \emph{incidence patterns}, namely the
subsets $\mathcal A_R\cap(z+(1-t)K)$. Thus this is a finite covering
problem: combine the weights of all translation vectors with the
same pattern, discard empty patterns, and cap weights at $1$ if
necessary. Its largest set has cardinality bounded by
\eqref{eq:translative-degree}. Applying
\cite[Theorem~1.1]{NP2018} in the dual covering formulation therefore
bounds the number of selected translates by this fractional weight
times
\[
  \frac{1-\lambda^k}{1-\lambda}
  \left[1+n\log\frac4t-(k-1)\log\lambda\right].
\]

The greedy construction may select a translate more than once.
Since $(1-t)K+tS\subset K$, the corresponding translates of $K$
cover $C_R$ at least $k$ times, counting these repetitions. Repeat
the finite list with period lattice $2R\mathbb Z^n$. This gives a
periodic covering of the whole space whose density, counting
repetitions, is at most $(1+o_{R\to\infty}(1))$ times
\eqref{eq:translative-density}.

\emph{Distinct translates.}
Repetitions can be removed with arbitrarily small density cost.
Apply the construction to $(1-\alpha)K$, where $0<\alpha<1$, and
choose $r>0$ with $rB_2^n\subset K$. Perturb each translation vector
in the finite period list by a vector $e$ with $\|e\|<\alpha r$,
choosing distinct representatives modulo the period lattice. This
is possible by avoiding finitely many congruence classes at each
choice. The inclusion
\[
  (1-\alpha)K-e
  \subset(1-\alpha)K+\alpha K
  =K
\]
shows that the new, distinct translates of $K$ retain every covering
provided by the old list.

Their density, measured using $K$, is $(1-\alpha)^{-n}$ times the
old density measured using $(1-\alpha)K$. Letting $R\to\infty$ and
then $\alpha\downarrow0$ proves the claimed infimum bound for
distinct translates.

\emph{Choice of parameters.}
For $0<\varepsilon\le1$, set $a=\varepsilon/16$, $t=a/n$, and
$\lambda=1-a$. Bernoulli's inequality and
$-\log(1-a)\le a/(1-a)$ bound \eqref{eq:translative-density} by
\[
  \frac{k}{(1-a)^2}
  +\frac{1+n\log(64n/\varepsilon)}{a(1-a)}
  \le(1+\varepsilon/2)k
  +\frac{32}{\varepsilon}
    \bigl[1+n\log(64n/\varepsilon)\bigr].
\]
The last inequality uses $(1-a)^{-2}\le1+8a$ for
$0\le a\le1/16$.

Since $n\ge2$, the second term is bounded by a constant depending
only on $\varepsilon$ times $n\log n$.
Increasing that constant also absorbs the arbitrarily small
approximation to the infimum. Thus there is an actual periodic
covering by distinct translates of density at most
\[
  (1+\varepsilon/2)k+C^{\mathrm{add}}_\varepsilon n\log n
\]
for some $C^{\mathrm{add}}_\varepsilon>0$.

Split into the cases
$k\ge2C^{\mathrm{add}}_\varepsilon n\log n/\varepsilon$ and
$k<2C^{\mathrm{add}}_\varepsilon n\log n/\varepsilon$.
The desired maximum follows with
$C^{\mathrm{tr}}_\varepsilon
=2(1+\varepsilon)C^{\mathrm{add}}_\varepsilon/\varepsilon$.
For $\varepsilon>1$, use the case $\varepsilon=1$.
\end{proof}

\providecommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\thinspace}
\providecommand{\MR}{\relax\ifhmode\unskip\space\fi MR }
% \MRhref is called by the amsart/book/proc definition of \MR.
\providecommand{\MRhref}[2]{%
  \href{http://www.ams.org/mathscinet-getitem?mr=#1}{#2}
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\providecommand{\href}[2]{#2}
\begin{thebibliography}{10}

\bibitem{Cohn1976}
M.~J. Cohn, \emph{Multiple lattice covering of space}, Proc. London Math. Soc.
  (3) \textbf{32} (1976), no.~1, 117--132,
  \href{https://doi.org/10.1112/plms/s3-32.1.117}{doi:10.1112/plms/s3-32.1.117}.

\bibitem{EldanKlartag2008}
R.~Eldan and B.~Klartag, \emph{Pointwise estimates for marginals of convex
  bodies}, J. Funct. Anal. \textbf{254} (2008), no.~8, 2275--2293,
  \href{https://doi.org/10.1016/j.jfa.2007.08.014}{doi:10.1016/j.jfa.2007.08.014}.

\bibitem{FNN2018}
N.~Frankl, J.~Nagy, and M.~Nasz\'odi, \emph{Coverings: variations on a result
  of {R}ogers and on the {E}psilon-net theorem of {H}aussler and {W}elzl},
  Discrete Math. \textbf{341} (2018), no.~3, 863--874,
  \href{https://doi.org/10.1016/j.disc.2017.11.017}{doi:10.1016/j.disc.2017.11.017}.

\bibitem{Groemer1986}
H.~Groemer, \emph{Multiple packings and coverings}, Studia Sci. Math. Hungar.
  \textbf{21} (1986), no.~1--2, 189--200,
  \href{https://real-j.mtak.hu/5463/1/StudScientMath_21.pdf}{Journal archive}.

\bibitem{LiLiu2026}
H.~Li and X.~Liu, \emph{Nearly sharp bounds for lattice coverings by convex
  bodies}, preprint, version~2, 2026,
  \href{https://arxiv.org/abs/2607.28429v2}{arXiv:2607.28429v2}.

\bibitem{Lund2026}
B.~Lund, \emph{Two-sided linear hashing and quadratic density bounds for smooth
  lattice coverings}, preprint, 2026,
  \href{https://arxiv.org/abs/2609.20351}{arXiv:2609.20351}.

\bibitem{NP2018}
M.~Nasz\'odi and A.~Polyanskii, \emph{Approximating set multi-covers}, European
  J. Combin. \textbf{67} (2018), 174--180,
  \href{https://doi.org/10.1016/j.ejc.2017.08.001}{doi:10.1016/j.ejc.2017.08.001}.

\bibitem{Lattice2026}
{OpenAI}, \emph{A single-lattice covering bound of order $n\log n$}, OpenAI Math
  Release preprint,
  \href{https://github.com/openai/math/blob/main/CONTENTS.md}{catalog
  family~092}, 2026,
  \href{https://github.com/openai/math/blob/main/preprints/A-single-lattice-covering-bound-of-order-n-log-n-September-23-2026/paper.pdf}{version
  dated September 23}, released on October 6; all source references use this
  version, identified by PDF SHA-256
  \nolinkurl{c7dc74641f89bce3f456410280429a47b3e214365a9467b494e14ee6b2348cbb}.

\bibitem{Translative2026}
\bysame, \emph{Translative covering densities of order $n\log n$}, OpenAI Math
  Release preprint, \href{https://github.com/openai/math/blob/main/CONTENTS.md}
	{catalog family~092}, 2026,
  \href{https://github.com/openai/math/blob/main/preprints/Translative-covering-densities-of-order-n-log-n-September-23-2026/paper.pdf}{version
  dated September 23}, released on October 6.

\bibitem{ORW2022}
O.~Ordentlich, O.~Regev, and B.~Weiss, \emph{New bounds on the density of
  lattice coverings}, J. Amer. Math. Soc. \textbf{35} (2022), no.~1, 295--308,
  \href{https://doi.org/10.1090/jams/984}{doi:10.1090/jams/984}.

\bibitem{ORWsmooth2025}
\bysame, \emph{Bounds on the density of smooth lattice coverings}, J. Anal.
  Math. \textbf{156} (2025), 301--326,
  \href{https://doi.org/10.1007/s11854-025-0367-2}{doi:10.1007/s11854-025-0367-2}.

\bibitem{Rogers1957}
C.~A. Rogers, \emph{A note on coverings}, Mathematika \textbf{4} (1957), no.~1,
  1--6,
  \href{https://doi.org/10.1112/S0025579300001030}{doi:10.1112/S0025579300001030}.

\bibitem{Rogers1958}
\bysame, \emph{Lattice coverings of space: The {M}inkowski--{H}lawka theorem},
  Proc. London Math. Soc. (3) \textbf{8} (1958), no.~3, 447--465,
  \href{https://doi.org/10.1112/plms/s3-8.3.447}{doi:10.1112/plms/s3-8.3.447}.

\bibitem{Rogers1959}
\bysame, \emph{Lattice coverings of space}, Mathematika \textbf{6} (1959),
  no.~1, 33--39,
  \href{https://doi.org/10.1112/S002557930000190X}{doi:10.1112/S002557930000190X}.

\bibitem{Schmidt1958}
W.~M. Schmidt, \emph{The measure of the set of admissible lattices}, Proc.
  Amer. Math. Soc. \textbf{9} (1958), no.~3, 390--403,
  \href{https://doi.org/10.1090/S0002-9939-1958-0096638-X}{doi:10.1090/S0002-9939-1958-0096638-X}.

\end{thebibliography}

\end{document}