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\title{Illumination of complex convex bodies}
\author{Alon Schejter, Working research draft}
\date{25 October 2026}
\begin{document}
\maketitle
\begin{abstract}
For every complex $1$-unconditional convex body $K\subset\C^n$, we prove
\[
 \ill(K)\le
 (2^{m+1}-1)\binom nm+\sum_{k=m+1}^{n}\binom nk-(n-m),
 \qquad m=\lceil2n/3\rceil.
\]
This bound is asymptotic to $3^{n+1}/\sqrt{\pi n}$. We further obtain
$\ill^*(K)\le b_*^n$ and $\ill(K)=O(n\log(n+1)b_*^n)$,
where $b_*=2.4805534697\ldots$ is defined by an explicit scalar equation.
For normalized bodies other than $\mathbb D^n$ that contain a full
coordinate $(n-1)$-polydisc, we prove the sharp bound
$\ill(K)\le2^{n+1}-2$. We also prove the sharp bound $14$ for every
normalized complex $1$-unconditional body in $\C^3$ other than
$\mathbb D^3$, and the sharp bound $30$ in $\C^4$ for every
normalized non-polydisc. The four-dimensional proof includes one
finite covering assertion verified by exact integer arithmetic;
its complete verifier is included. Under full coordinate permutation
symmetry, we improve the three-dimensional bound to $12$ by an explicit
twelve-vector construction and an exact finite covering certificate.
For complex Hanner bodies, we compute
both illumination numbers exactly and determine the equality cases in their dimension
bounds. If every extreme point has all coordinates nonzero, we prove
$\ill(K)\le2^{n+1}-1$ and a quantitative strict fractional bound
$\ill^*(K)<2^n$ for normalized non-polydiscs. For real $1$-symmetric
non-cubes we obtain bounds $12$ in dimension four, $22$ in dimension
five, and $5\cdot2^{n-3}$ for every $n\ge6$. We also determine
the leading asymptotic maximum in the real $1$-symmetric non-cube
class: it is $(1+o(1))2^{n-1}$. An explicit binary construction
proves this without an external coding theorem. A connection to
connected domination in cube graphs improves the upper estimate to
$2^{n-1}(1+(1+o(1))/n)$.
\end{abstract}

\noindent\textit{Draft status.} The proofs included here have been checked
internally; external review and literature priority remain outstanding.
Results from the wider research corpus that have not yet been checked
are listed separately in Section~\ref{sec:corpus-ledger}.

\section{Notation and preliminary facts}\label{sec:preliminaries}

Write $[n]=\{1,\ldots,n\}$, $\mathbb D=\{z\in\C:|z|\le1\}$,
and $\R_+=[0,\infty)$. A convex body is a compact convex set with
nonempty interior in the underlying real space. It is complex
$1$-unconditional if it is invariant under independent rotations of
its coordinates. Its Minkowski norm is denoted by $\norm{\cdot}$.
For $z\in\C^n$, write $|z|=(|z_1|,\ldots,|z_n|)$ and
$\supp(z)=\{j\in[n]:z_j\ne0\}$. We also write
\[
 M_x=\{j\in[n]:|x_j|=\|x\|_\infty\},\qquad
 Z_x=\{j\in[n]:x_j=0\}.
\]

A nonzero vector $d$ illuminates $x\in\partial K$ if
$x+td\in\operatorname{int}K$ for some $t>0$. The least cardinality of
a family illuminating every boundary point is $\ill(K)$.
A finite nonnegative Borel measure $\mu$ on $\C^n\setminus\{0\}$
fractionally illuminates $K$ if
\[
 \mu\{d:d\text{ illuminates }x\}\ge1\qquad(x\in\partial K).
\]
The infimum of the total masses of such measures is $\ill^*(K)$.

We call $K$ normalized when $e_j\in\partial K$ for every $j\in[n]$.
Positive diagonal rescaling gives this normalization, and illumination
numbers are invariant under invertible linear maps. Coordinate
contractions preserve membership, as proved below; consequently a
normalized body satisfies $K\subseteq\mathbb D^n$ and contains each
unit coordinate disc.

\begin{lemma}\label{lem:section-contraction}
For a complex $1$-unconditional convex body $K$, the following statements hold.
\begin{enumerate}
\item If $|w_j|\le|z_j|$ for every $j$, then $\norm{w}\le\norm{z}$.
\item For every real linear subspace $E\subseteq\C^n$,
\[
 \operatorname{relint}(K\cap E)=E\cap\operatorname{int}K.
\]
\item Suppose $z\in K$, $J$ is a nonempty subset of
$\supp(z)$, and the point $q$ obtained from $z$ by setting
the coordinates in $J$ to zero belongs to $\operatorname{int}K$.
If $w_j=z_j$ for $j\notin J$ and $|w_j|<|z_j|$ for $j\in J$,
then $w\in\operatorname{int}K$.
\end{enumerate}
\end{lemma}
\begin{proof}
Invariance under all coordinate rotations implies central symmetry.
Because $K$ has nonempty interior, its center $0$ is interior: if
an open Euclidean ball of radius $r$ around $a$ lies in $K$, the same
is true around $-a$, and convexity puts the ball of radius $r$ around
$0$ in $K$. Its Minkowski functional is therefore a norm, with
$\operatorname{int}K=\{z:\norm{z}<1\}$.

Every number $\lambda\in\C$ with $|\lambda|\le1$ is a convex
combination of two unit complex numbers. Explicitly, if
$\lambda=se^{\ii\theta}$, $0\le s\le1$, then
\[
 \lambda=\frac{1+s}{2}e^{\ii\theta}
          +\frac{1-s}{2}(-e^{\ii\theta}).
\]
Thus contracting a single coordinate by $\lambda$ preserves membership
in $K$, by convexity and rotation invariance. Applying these contractions
one coordinate at a time proves coordinatewise monotonicity of membership.
Apply the same fact to every dilation of $K$ to obtain the norm inequality
in part~1. Coordinates with $z_j=0$ also have $w_j=0$, so cause no exception.

The inclusion from right to left in part~2 follows by intersecting an
ambient open ball with $E$. For the reverse inclusion, let
$y\in\operatorname{relint}(K\cap E)$. If $y=0$ there is nothing to prove.
Otherwise relative interior gives an $s>0$ with $(1+s)y\in K\cap E$.
Consequently $\norm{y}\le(1+s)^{-1}<1$, as required.

For part~3 choose $\rho\in(0,1)$ such that
$|w_j|\le\rho|z_j|$ for all $j\in J$; there are finitely many
strict ratios, so one common $\rho$ exists. Put
$y=\rho z+(1-\rho)q$. Then
\[
 \norm{y}\le\rho\norm{z}+(1-\rho)\norm{q}<1.
\]
On $J$ we have $y_j=\rho z_j$, and off $J$ we have $y_j=w_j=z_j$.
Part~1 therefore gives $\norm{w}\le\norm{y}<1$.
\end{proof}

We use the real pairing
\[
 [u,z]=\operatorname{Re}\sum_{j\in[n]}\overline{u_j}z_j,
 \qquad h_K(u)=\max_{z\in K}[u,z].
\]
A direction $d$ illuminates $x\in\partial K$ if and only if
\begin{equation}\label{eq:normal-criterion}
 [u,d]<0\quad\text{for every nonzero }u\text{ with }[u,x]=h_K(u).
\end{equation}
Here is a proof that also covers nonsmooth boundary points. If $x+td\in\operatorname{int}K$
for some $t>0$, then $[u,x+td]<h_K(u)=[u,x]$ for every nonzero supporting
normal at $x$, proving necessity. For sufficiency, suppose no positive step enters
$\operatorname{int}K$. For each integer $j\ge1$, separation at
$x+j^{-1}d\notin\operatorname{int}K$ supplies a Euclidean unit vector $u^j$
with $[u^j,x+j^{-1}d]\ge h_K(u^j)$. Since
$h_K(u^j)\ge[u^j,x]$, we have $[u^j,d]\ge0$.
Pass to a convergent subsequence of these unit vectors, with limit $u$.
The support function is continuous, so $[u,x]\ge h_K(u)$; the reverse
inequality holds because $x\in K$. Thus $u$ is a nonzero supporting normal
at $x$ with $[u,d]\ge0$, contradicting \eqref{eq:normal-criterion}.

For a normalized unconditional body, put $S=\supp(x)$ and $a_j=|u_j|$ for a supporting normal $u$ at $x$. Independent coordinate rotations give $u_j=|u_j|x_j/|x_j|$ whenever $j\in S$:
the maximum of $[u,\cdot]$ on the coordinate-rotation orbit of $x$ is
$\sum_j|u_j||x_j|$, and $x$ already attains that maximum. Thus, writing $h=h_K(u)>0$,
\begin{equation}\label{eq:normalized-normal-estimates}
 h=\sum_{j\in S}a_j|x_j|,
 \qquad a_j\le h\quad(j\in[n]),
 \qquad h\le\sum_{j\in S}a_j.
\end{equation}
The middle inequality follows from the coordinate discs in $K$; the last uses $K\subseteq\mathbb D^n$.

\section{General upper bounds}\label{sec:general-bounds}

\subsection{Directions indexed by coordinate supports}\label{sec:coordinate-supports}

\begin{theorem}\label{thm:support-bound}
For $n\ge1$, put $m=\lceil2n/3\rceil$ and
\begin{equation}\label{eq:support-bound}
 C_n=(2^{m+1}-1)\binom nm+\sum_{k=m+1}^{n}\binom nk-(n-m).
\end{equation}
Every complex $1$-unconditional convex body $K\subset\C^n$ satisfies
$\ill(K)\le C_n$. Moreover,
\[
 C_n\sim\frac{3^{n+1}}{\sqrt{\pi n}}.
\]
\end{theorem}

We use the cyclic construction in the proof of Proposition~2.7
of~\cite{RSS}. For $b\ge1$, put $q_b=2^{b+1}-1$. The vectors
\begin{equation}\label{eq:cyclic-vectors}
 \left\{\left(\exp\frac{2\pi\ii\,2^{j-1}t}{q_b}\right)_{j\in[b]}:
 t\in\mathbb Z/q_b\mathbb Z\right\}
\end{equation}
meet every product of $b$ open semicircles, and hence illuminate
$\mathbb D^b$. We use this published result as an external input.

\begin{lemma}\label{lem:cyclic-projections}
Append to \eqref{eq:cyclic-vectors} the column of frequency $2^b$. Every projection of the resulting $(b+1)$-column matrix onto $b$ columns meets every product of open semicircles.
\end{lemma}
\begin{proof}
Since $2^{b+1}\equiv1\pmod{q_b}$, multiplication of the row parameter $t$ by $2$ cyclically permutes the $b+1$ frequencies. It is a permutation of the rows because $q_b$ is odd. Any omitted column can therefore be moved to the last position, reducing the assertion to \eqref{eq:cyclic-vectors}.
\end{proof}


\begin{lemma}\label{lem:binary-support-tree}
Let a finite rooted tree have nonempty coordinate sets as its nodes. Its root is a set $T$ of cardinality $b$. Each node has at most two children, and each child is obtained by deleting one coordinate from its parent. Then $q_b$ directions suffice, simultaneously for every normalized complex $1$-unconditional body, to illuminate all boundary points whose supports occur in the tree.
\end{lemma}
\begin{proof}
Arrange the vectors of \eqref{eq:cyclic-vectors}, with their additional
coordinate from Lemma~\ref{lem:cyclic-projections}, as rows of a matrix.
Label $b$ columns by $T$, leaving one additional column. At a node with support $S$, let $G$ be the set of rows satisfying fixed
open-semicircle conditions on $b-|S|$ columns outside $S$. Exactly one
column outside $S$ remains unrestricted. At the root, $G$ contains every
row, and the additional column is unrestricted.

The active columns $S$ and the constrained columns together comprise exactly $b$ columns. Lemma~\ref{lem:cyclic-projections} implies that $G$, projected onto $S$, meets every product of open semicircles. Indeed, prescribe the desired semicircles on $S$ and retain all the fixed constraints.

Suppose the children are $S\setminus\{i\}$ and $S\setminus\{j\}$. Split $G$ by imposing, in its unrestricted column, respectively the upper and lower open semicircles. These conditions are disjoint. The previously unrestricted column becomes constrained, and the deleted column $i$ or $j$ becomes the unrestricted column of the corresponding child. Thus the invariant persists. For a single child, use just one of the two semicircles. Rows satisfying neither chosen condition stop at the current node; rows at leaves also stop. Every row consequently visits one nested path
\[
 T=S_0\supset S_1\supset\cdots\supset S_r,
 \qquad |S_{\ell-1}\setminus S_\ell|=1.
\]
The group of a node is exactly the set of rows visiting it. In particular, rows that stop early remain available at that node and all its ancestors.

For a node $S$, define on the torus of unit complex coordinates
\[
 g_S(\zeta)=\max_{p\in G}\min_{j\in S}
                   \bigl[-\operatorname{Re}(\overline{\zeta_j}p_j)\bigr].
\]
The covering property makes $g_S$ positive everywhere. It is continuous,
being a finite maximum of finite minima of continuous functions, so its
minimum on the compact torus is positive. Take $\beta>0$ no larger than
these minima for all the finitely many nodes. Thus, for every node $S$
and every $(\zeta_j)_{j\in S}$ with $|\zeta_j|=1$, some row $p\in G$ satisfies
\begin{equation}\label{eq:tree-inward-inequalities}
 \operatorname{Re}(\overline{\zeta_j}p_j)\le-\beta
 \qquad(j\in S).
\end{equation}
Choose $0<\delta<\min\{1,\beta/n\}$.

For a row with path $S_0,\ldots,S_r$, give amplitude $1$ to the coordinates in $S_r$. The coordinate deleted at step $\ell$ receives amplitude $\delta^{r-\ell+1}$, and coordinates outside $T$ receive zero. Multiply these positive amplitudes by the corresponding entries of the row to obtain $d$. At every visited node $S_\ell$,
\begin{equation}\label{eq:tree-amplitude-gap}
 |d_j|\ge\delta^{r-\ell}\quad(j\in S_\ell),
 \qquad
 |d_j|\le\delta^{r-\ell+1}\quad(j\notin S_\ell).
\end{equation}

Now let $x\in\partial K$ have support $S_\ell$. Select a visiting row satisfying \eqref{eq:tree-inward-inequalities} for $\zeta_j=x_j/|x_j|$, and put $a=\delta^{r-\ell}$. For every supporting normal, \eqref{eq:normalized-normal-estimates} and \eqref{eq:tree-amplitude-gap} give
\begin{align*}
 [u,d]
 &\le -\beta a\sum_{j\in S_\ell}|u_j|
       +\delta a\sum_{j\notin S_\ell}|u_j|\\
 &\le a\bigl[-\beta+(n-|S_\ell|)\delta\bigr]
             \sum_{j\in S_\ell}|u_j|<0.
\end{align*}
The last sum is positive since $h_K(u)>0$. The criterion \eqref{eq:normal-criterion} completes the proof.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:support-bound}]
By diagonal rescaling we may assume that $K$ is normalized.
Write $a_k=\binom nk$ and $a_{n+1}=0$. For $k\in[n-1]$, between the $k$-sets and the $(k+1)$-sets, the Boolean incidence graph has degrees $n-k$ and $k+1$, respectively. We assign a $k$-set to a containing parent, allowing each parent at most two children. Write $N(\mathcal A)$ for the set of neighbors of a vertex family $\mathcal A$.

If $a_k\le2a_{k+1}$, edge counting gives, for every family $\mathcal A$ of $k$-sets,
\[
 |N(\mathcal A)|\ge\frac{n-k}{k+1}|\mathcal A|
 \ge\frac{|\mathcal A|}{2}.
\]
Hall's theorem, after replacing each upper vertex by two slots, assigns every $k$-set to a parent. If $a_k\ge2a_{k+1}$, reverse edge counting gives, for every family $\mathcal B$ of upper vertices,
\[
 |N(\mathcal B)|\ge\frac{k+1}{n-k}|\mathcal B|
 \ge2|\mathcal B|.
\]
To apply Hall's theorem to the duplicated upper vertices, take any set
of slots and let $\mathcal B$ be their underlying vertices. There are
at most $2|\mathcal B|$ slots and at least that many lower neighbors by
the displayed estimate. Thus Hall's theorem fills both slots of every
upper vertex. The unused $k$-sets become roots. The top level is handled separately: $[n]$ is a root and has no parent.
Consequently, including $k=n$ with $a_{n+1}=0$, the numbers of roots are
\[
 r_k=\max\{a_k-2a_{k+1},0\}.
\]
The choices between successive pairs of layers are independent: a vertex's parent and its children are governed by different matchings. Together they form a binary forest containing every nonempty subset exactly once.

The positive root counts occur precisely at $k\ge m=\lceil2n/3\rceil$. Applying Lemma~\ref{lem:binary-support-tree} separately to the trees gives a family of at most
\begin{align*}
 &\sum_{k=m}^{n}(a_k-2a_{k+1})(2^{k+1}-1)\\
 &=(2^{m+1}-1)a_m+\sum_{k=m+1}^{n}a_k.
\end{align*}
directions. Here the last equality follows from
\[
 \sum_{k=m}^n(a_k-2a_{k+1})2^{k+1}=2^{m+1}a_m,
 \qquad
 \sum_{k=m}^n(a_k-2a_{k+1})=a_m-\sum_{k=m+1}^n a_k.
\]


For every $k\ge m$ the number $r_k$ of roots at level $k$ is positive.
Prescribe the nested sets $S_k=[k]$, $m\le k\le n$.
The set $S_n$ is already the top root. For $m\le k<n$, the matching
between levels $k$ and $k+1$ leaves $r_k>0$ lower vertices unmatched.
Apply a coordinate permutation to this matching alone so that one
unmatched vertex becomes $S_k$. Containment and all capacities are
preserved. Whether a $k$-set is a root depends only on this matching;
its children are assigned by the separate matching between levels
$k-1$ and $k$. The permutations therefore impose no compatibility
condition across level pairs. We can make every $S_k$ a root at once.

Use the upper and lower open semicircles at every split in the tree construction.
In each root's matrix, the row indexed by $t=0$ has entry $1$ in every column.
It lies on the boundary of both semicircles, so it stops at the root and
is never needed in a descendant's group.
Take one common positive margin $\beta$ for all groups in the forest,
and choose $0<\delta<\min\{1,\beta/n\}$.

Remove the $n-m+1$ rows indexed by $t=0$ associated with the roots
$S_m,S_{m+1},\ldots,S_n$, and replace them by the single direction
\[
 d_j=
 \begin{cases}
  1,&j\le m,\\
  \delta^{j-m},&j>m.
 \end{cases}
\]
At support $S_k$, its smallest active amplitude is $\delta^{k-m}$,
and every inactive amplitude is at most $\delta^{k-m+1}$.
Whenever the deleted row supplies the margin $\beta$ at that root,
the replacement has positive real coordinates and the required amplitude gap.
Writing $a=\delta^{k-m}$, the same normal estimate gives
\[
 [u,d]\le a[-\beta+(n-k)\delta]\sum_{j\in S_k}|u_j|<0
\]
for every nonzero supporting normal at any point for which that row
was selected. The sum is positive by \eqref{eq:normalized-normal-estimates}.
At all other nodes and for all other selected rows the construction is unchanged.
This saves $n-m=\lfloor n/3\rfloor$ directions and proves the asserted upper bound. To obtain its asymptotic form,
Stirling's formula with $m=2n/3+O(1)$ gives
\[
 2^{m+1}\binom nm\sim\frac{3^{n+1}}{\sqrt{\pi n}}.
\]
All remaining terms in $C_n$ have absolute value at most $2^{n+1}+n$
and are of smaller order, proving the asymptotic statement.
\end{proof}

Taking every nonempty support as a single-vertex tree in
Lemma~\ref{lem:binary-support-tree} gives the comparison bound
\[
 \ill(K)\le\sum_{k=1}^n\binom nk(2^{k+1}-1)
 =2\cdot3^n-2^n-1.
\]

\begin{remark}
The directions just constructed work for every normalized body.
Appendix~\ref{app:simultaneous} gives a lower bound for this stronger
requirement. It does not give a lower bound for the largest illumination
number when the directions may be chosen separately for each body.
\end{remark}

\subsection{An upper bound with a smaller exponential base}
Unconditionality here means invariance under independent rotations of the complex coordinates.
Let \(K\subset\C^n\) be such a convex body, and put
\[
 L=K\cap\R_+^n.
\]
Then \(L\) is compact and convex and is a coordinatewise downset:
if \(y\in L\) and \(0\le x_j\le y_j\) for every \(j\in[n]\), then \(x\in L\).
Its \(n\)-dimensional volume is positive because \(0\in\operatorname{int}K\),
and \(z\in K\) if and only if \((|z_1|,\ldots,|z_n|)\in L\).
In particular, coordinatewise domination of moduli implies domination of the Minkowski norm.
No normalization of the coordinate radii is needed.

Let \(c_*\in(0,1/2)\) be the unique solution of
\begin{equation}\label{eq:cstar}
 \cos(\pi c_*)\cosh(\pi(1-c_*))=1.
\end{equation}
The product on the left is strictly decreasing on \((0,1/2)\), and its endpoint values lie on opposite sides of one.
Numerically,
\[
 c_*=0.4031358373055328\ldots,\qquad
 b_*=c_*^{-1}=2.480553469728144\ldots.
\]

\begin{theorem}\label{thm:main}
For every complex 1-unconditional convex body \(K\subset\C^n\),
\begin{equation}\label{eq:frac}
 \ill^*(K)\le b_*^n,
\end{equation}
and
\begin{equation}\label{eq:explicit}
 \ill(K)\le
 \left\lceil e b_*^n
 \left[2n\log\bigl(1+32b_*^2(n+1)^2\bigr)+1\right]\right\rceil.
\end{equation}
In particular,
\[
 \ill(K)=O\bigl(n\log(n+1)b_*^n\bigr).
\]
Replacing \(b_*\) by \(5/2\) gives the simpler explicit estimate
\begin{equation}\label{eq:simple}
 \ill(K)\le
 \left\lceil e(5/2)^n
 \left[2n\log\bigl(1+200(n+1)^2\bigr)+1\right]\right\rceil.
\end{equation}
\end{theorem}

The minimum of this bound and Theorem~\ref{thm:support-bound} is valid in every dimension.
Numerical comparisons are given in Appendix~\ref{sec:comparison-table}.

\subsubsection{A one-coordinate inequality}
Define a locally finite measure on \(\C\) by
\[
 d\nu_1(z)=\frac{d\rho\,d\theta}{2\pi},\qquad z=\rho e^{\ii\theta},
 \quad \rho\ge0,\quad 0\le\theta<2\pi.
\]
Equivalently its density with respect to plane area is \((2\pi|z|)^{-1}\).
The density is locally integrable. In particular, \(\nu_1(\{|z|\le R\})=R\)
for \(R\ge0\), and \(\nu_1\) has no atom at zero.
Write \(\nu=\nu_1^{\otimes n}\). Then
\begin{equation}\label{eq:volume}
 \nu(K)=\vol_n(L),\qquad \nu(aK)=a^n\nu(K)\quad(a>0).
\end{equation}
The homogeneity degree is \(n\), although the ambient real dimension is \(2n\).

Let
\[
 H(s)=\nu_1\{z:|z|\le1,\ \Re z\le s\}.
\]
For \(-1<s<1\), direct integration gives
\begin{equation}\label{eq:H}
 H(s)=\frac12+\frac{\arcsin s+s\arcosh(1/|s|)}{\pi},
\end{equation}
where the second term is interpreted continuously at \(s=0\).
Also \(H(s)=0\) for \(s\le-1\), and \(H(s)=1\) for \(s\ge1\).
For \(0<s<1\), conditioning on the radius gives
\[
 H(s)=s+\int_s^1\left(\frac12+\frac{\arcsin(s/\rho)}{\pi}\right)d\rho.
\]
Integration by parts yields
\[
 \int_s^1\arcsin(s/\rho)\,d\rho
 =\arcsin s-\frac{\pi s}{2}
   +s\int_s^1\frac{d\rho}{\sqrt{\rho^2-s^2}},
\]
which proves \eqref{eq:H} for positive \(s\). Symmetry and continuity
give the remaining values. Differentiating the resulting expression gives
\[
 H'(s)=\frac1\pi\int_s^1\frac{d\rho}{\sqrt{\rho^2-s^2}}
       =\frac{\arcosh(1/s)}{\pi}.
\]

\begin{lemma}\label{lem:cap}
For every \(0<c\le c_*\),
\begin{equation}\label{eq:cap}
 H(s)\ge c+(1-c)s\qquad(s\le1).
\end{equation}
\end{lemma}
\begin{proof}
On \([0,1]\), the function \(H\) is concave, so
\(H(s)\ge(1+s)/2\ge c+(1-c)s\).
On \([-1,0]\), it is convex, and the minimum of
\(H(s)-(1-c)s\) occurs at
\[
 s=-\sech(\pi(1-c)).
\]
By \eqref{eq:H}, that minimum equals
\[
 \frac12-\frac{\arcsin(\sech(\pi(1-c)))}{\pi}.
\]
It is at least \(c\) precisely when
\(\cos(\pi c)\cosh(\pi(1-c))\ge1\), which holds for \(c\le c_*\).
For \(s\le-1\), the right side of \eqref{eq:cap} is nonpositive.
\end{proof}

The simple choice \(c=2/5\) is certified without numerical approximations.
The needed inequality is
\(\cosh(3\pi/5)\ge1+\sqrt5\).
Using \(\pi>25/8\), the Taylor lower bound gives
\[
 \cosh(3\pi/5)>
 1+\frac{(15/8)^2}{2}+\frac{(15/8)^4}{24}
 =\frac{321729}{98304}>\frac{13}{4}>1+\sqrt5.
\]

\subsubsection{A comparison of product measures}
We first record the measure-comparison fact used below.

\begin{lemma}\label{lem:downset}
Let \(c>0\), and let \(m_1,\ldots,m_n\) be locally finite Borel measures on \([0,\infty)\) such that
\(m_j([0,s])\ge cs\) for every \(j\in[n]\) and \(s\ge0\). For every bounded Borel coordinatewise downset
\(A\subset[0,\infty)^n\),
\[
 (m_1\otimes\cdots\otimes m_n)(A)\ge c^n\vol_n(A).
\]
\end{lemma}
\begin{proof}
In one dimension the cumulative inequality implies
\(\int f\,dm_j\ge c\int f\,ds\) for every nonnegative decreasing function \(f\)
of bounded support. Indeed, apply the layer-cake formula; each superlevel set of \(f\)
is an initial interval. The claim for open initial intervals follows by taking limits from below,
and an initial interval consisting only of zero causes no problem.
Apply this comparison successively to the coordinates of \(1_A\), using Tonelli's theorem.
At each stage the function of every remaining coordinate is decreasing.
The argument allows atoms at zero.
\end{proof}

Fix \(0<c\le c_*\) and a point \(x\in K\). Independent rotations allow us to assume
\(x=r=(r_1,\ldots,r_n)\in L\). Both \(K\) and \(\nu\) are invariant under these rotations.
Define \(F_r:\C^n\to[0,\infty)^n\) by
\begin{equation}\label{eq:F}
 (F_r(d))_j=
 \max\left\{0,\ \frac{|d_j|-(1-c)r_j}{c},\ r_j+\frac{\Re d_j}{c}\right\}.
\end{equation}
When \(r_j=0\), this is simply \(|d_j|/c\).
Putting \(y=F_r(d)\), we have in every coordinate
\begin{equation}\label{eq:constraints}
 |d_j|\le(1-c)r_j+cy_j,\qquad
 \Re d_j\le c(y_j-r_j).
\end{equation}

For \(r_j>0\), let \(m_{r_j}\) be the image measure of \(\nu_1\) under the corresponding
one-coordinate map. For \(s\ge0\), set
\[
 R=(1-c)r_j+cs,\qquad B=c(s-r_j).
\]
Then
\[
 \{d_j:(F_r(d))_j\le s\}=\{|d_j|\le R,\ \Re d_j\le B\}.
\]
The coordinate map is continuous. This preimage is bounded and has finite
\(\nu_1\)-measure, so \(m_{r_j}\) is a locally finite Borel measure.
Since \(R-B=r_j>0\), Lemma~\ref{lem:cap} and the degree-one homogeneity of \(\nu_1\) imply
\begin{equation}\label{eq:cdf}
 m_{r_j}([0,s])=R H(B/R)
 \ge cR+(1-c)B=cs.
\end{equation}
For \(r_j=0\), the pushforward is exactly \(c\) times Lebesgue measure on \([0,\infty)\).
Because \(F_r\) acts separately in each coordinate, the image of
\(\nu_1^{\otimes n}\) under \(F_r\) is the product of these coordinate
image measures. This equality holds first on rectangles and then on Borel
sets by uniqueness of locally finite product measures. Hence
Lemma~\ref{lem:downset}, applied to \((1-\varepsilon)L\), gives
\begin{equation}\label{eq:mass}
 \nu\bigl(F_r^{-1}((1-\varepsilon)L)\bigr)
 \ge c^n(1-\varepsilon)^n\nu(K),\qquad 0<\varepsilon<1.
\end{equation}
Moreover, this preimage lies in \(K\): if \(y\in(1-\varepsilon)L\), then
\(|d|\le(1-c)r+cy\), whose \(K\)-norm is at most \(1-c\varepsilon\).
The closed scaled set \((1-\varepsilon)L\), including its coordinate faces, is intentional;
the measures in \eqref{eq:cdf} may have an atom at zero.

\subsubsection{A uniform estimate for interior points}
\begin{lemma}\label{lem:finite}
Let \(0<c\le c_*\), and let \(D\) have probability distribution \(\nu|_K/\nu(K)\). For \(0<\varepsilon<1\), put
\[
 \tau=\frac{c\varepsilon}{4},\qquad
 \gamma=\frac{c^2\varepsilon^2}{8},\qquad
 p=[c(1-\varepsilon)]^n.
\]
For every \(x\in K\),
\begin{equation}\label{eq:prob}
 \mathbb P\{x+\tau D\in(1-\gamma)K\}\ge p.
\end{equation}
The same distribution, step \(\tau\), contraction \(\gamma\), and lower bound \(p\)
work for every point \(x\).
\end{lemma}
\begin{proof}
Rotate \(x\) to \(r\in L\) and consider the preimage in \eqref{eq:mass}.
We show that every \(d\) in this preimage satisfies the desired finite contraction.
For \(0<\tau\le1\), the two scalar inequalities in \eqref{eq:constraints} imply
\begin{equation}\label{eq:scalar}
 |r_j+\tau d_j|\le (1-\tau c+\tau^2)r_j+(\tau c+\tau^2)y_j.
\end{equation}
Set \(A=r_j+\tau c(y_j-r_j)\), so \(A\ge(1-\tau c)r_j\ge r_j/2\).
Expanding the square and using \eqref{eq:constraints} gives
\begin{align*}
 |r_j+\tau d_j|^2
 &\le A^2+\tau^2\bigl[(1-2c)r_j^2+2cr_jy_j\bigr]\\
 &\le A^2+\tau^2r_j(r_j+y_j).
\end{align*}
On the other hand,
\[
 [A+\tau^2(r_j+y_j)]^2-A^2
 \ge 2A\tau^2(r_j+y_j)
 \ge \tau^2r_j(r_j+y_j).
\]
Taking square roots proves \eqref{eq:scalar}.

Monotonicity of the unconditional norm and its triangle inequality now yield
\begin{align*}
 \norm{x+\tau d}
 &\le(1-\tau c+\tau^2)\norm{r}+(\tau c+\tau^2)\norm{y}\\
 &\le1-\tau c\varepsilon+\tau^2(2-\varepsilon).
\end{align*}
For \(\tau=c\varepsilon/4\), the last expression is
\[
 1-\frac{c^2\varepsilon^2(2+\varepsilon)}{16}
 \le1-\frac{c^2\varepsilon^2}{8}=1-\gamma.
\]
The mass estimate \eqref{eq:mass}, and containment of its preimage in \(K\), prove
\eqref{eq:prob}. Rotation invariance returns the statement to the original \(x\).
\end{proof}

For \(x\in\partial K\), the event in \eqref{eq:prob} consists of illuminating directions.
Consequently \(\mathbb P\{D\text{ illuminates }x\}\ge[c(1-\varepsilon)]^n\)
for every \(\varepsilon>0\). The event on the left is independent of \(\varepsilon\) and is Borel:
it is the union, over positive rational \(q\), of the open sets
\(\{d:x+qd\in\operatorname{int}K\}\). Taking the supremum of the
lower bounds as \(\varepsilon\downarrow0\), the measure
\(c^{-n}\nu|_K/\nu(K)\) fractionally illuminates \(K\).
Its mass at zero is zero, so it defines a measure on the allowed
nonzero directions. Taking \(c=c_*\) proves \eqref{eq:frac}.

\subsubsection{A finite set of illuminating directions}
Let \(\eta=\gamma/2\). Choose a maximal finite \(\eta\)-separated set \(\mathcal N\subset K\)
in the norm of \(K\). Such a set can be formed by successively adding
points at distance at least \(\eta\); the packing estimate below
bounds the number of additions. Maximality makes it an \(\eta\)-net,
meaning that every point of \(K\) has distance at most \(\eta\) from it.
Its cardinality is at most
\begin{equation}\label{eq:net}
 N_0=\left(1+\frac{2}{\eta}\right)^{2n}
     =\left(1+\frac{4}{\gamma}\right)^{2n}.
\end{equation}
Indeed, the translates \(z+(\eta/2)\operatorname{int}K\), for \(z\in\mathcal N\),
are pairwise disjoint and lie in \((1+\eta/2)K\). Comparison of ordinary Lebesgue volumes,
in real dimension \(2n\), gives \eqref{eq:net}.

Choose \(M\) independent vectors distributed as \(D\) in Lemma~\ref{lem:finite}.
For each net point, the probability that all of them fail is at most
\((1-p)^M\le e^{-pM}\). Thus
\[
 M=\left\lceil\frac{\log N_0+1}{p}\right\rceil
\]
makes the probability that any net point is missed at most \(e^{-1}<1\).
There is therefore a choice \(d^1,\ldots,d^M\) such that every \(z\in\mathcal N\)
satisfies \(\norm{z+\tau d^j}\le1-\gamma\) for some \(j\in[M]\).

For arbitrary \(x\in K\), take \(z\in\mathcal N\) with \(\norm{x-z}\le\eta\),
and use the vector serving \(z\). Then
\[
 \norm{x+\tau d^j}\le\norm{x-z}+\norm{z+\tau d^j}
 \le\eta+1-\gamma=1-\gamma/2<1.
\]
These vectors illuminate all of \(\partial K\). The zero vector has probability zero under \(\nu\),
so all sampled vectors are nonzero with probability one. Intersecting this event with the positive-probability covering event gives nonzero illuminating directions.
We have proved the more flexible bound
\begin{equation}\label{eq:eps}
 \ill(K)\le
 \left\lceil
 \frac{2n\log\bigl(1+32/(c^2\varepsilon^2)\bigr)+1}
 {[c(1-\varepsilon)]^n}
 \right\rceil,\qquad 0<\varepsilon<1.
\end{equation}
Take \(\varepsilon=1/(n+1)\) and use
\((1-\varepsilon)^n=(1+1/n)^{-n}\ge e^{-1}\).
This proves \eqref{eq:explicit} with \(c=c_*\), and \eqref{eq:simple} with \(c=2/5\).

\begin{remark}
Choosing \(\varepsilon=1/(n\log n)\), for \(n\ge2\), in \eqref{eq:eps}
gives the sharper asymptotic prefactor
\[
 \ill(K)\le (4+o(1))\,n\log n\,b_*^n.
\]
The finite-dimensional parameter \(\varepsilon\) can also be optimized directly.
Neither refinement changes the exponential base.
\end{remark}


The constant $c_*$ is optimal for the affine inequality \eqref{eq:cap}:
values $c>1/2$ fail at $s=0$, and for $c_*<c\le1/2$ the negative
minimizer used in its proof violates the inequality. This does not
assert optimality of the resulting illumination bound.

\section{A full coordinate hyperplane section}\label{sec:full-coordinate-section}

We now prove a sharp bound for a class defined by an exact coordinate-section
condition. Throughout this section, $n\ge2$ and $K\subset\C^n$ is a normalized
complex $1$-unconditional convex body. Write
$\mathbb D=\{x\in\C:|x|\le1\}$, $[n]=\{1,\ldots,n\}$, and
$\mathbf 1=(1,\ldots,1)$. By a \emph{full coordinate hyperplane section}
we mean that, after permuting coordinates,
\[
 K\cap H=\mathbb D^{n-1}\times\{0\},\qquad H=\{x\in\C^n:x_n=0\}.
\]
Since normalization implies $K\subseteq\mathbb D^n$, this is equivalent to
$\mathbb D^{n-1}\times\{0\}\subseteq K$.

\begin{theorem}\label{thm:full-coordinate-section}
Let $n\ge2$ and let $K\subset\C^n$ be a normalized complex
$1$-unconditional convex body. If $K\ne\mathbb D^n$ and $K$ contains
a full coordinate $(n-1)$-polydisc, then
\begin{equation}\label{eq:full-section-bound}
 \ill(K)\le 2^{n+1}-2.
\end{equation}
This bound is attained, for every $n\ge2$, by
\[
 H_n=\mathbb D^{n-2}\times
 \{(x_{n-1},x_n)\in\C^2:|x_{n-1}|+|x_n|\le1\}.
\]
For $n=2$ the first factor is omitted.
\end{theorem}

We use Lemma~\ref{lem:section-contraction} and the cyclic polydisc construction
\eqref{eq:cyclic-vectors}. The remaining estimates are proved below.

\subsection{Supporting normals}

\begin{lemma}\label{lem:full-section-normal}
Suppose $\mathbb D^{n-1}\times\{0\}\subseteq K$, and let $u\ne0$
be a supporting normal at $x\in\partial K$ for the real pairing
$[u,x]=\operatorname{Re}\sum_j\overline{u_j}x_j$.
Set $b_j=|u_j|$ and $r_j=|x_j|$. Then
\begin{equation}\label{eq:full-section-normal}
 h_K(u)=\sum_{j\in[n]} b_jr_j,
 \qquad
 \sum_{j<n}b_j(1-r_j)\le b_nr_n.
\end{equation}
If $x_j\ne0$, then $u_j=b_jx_j/|x_j|$.
\end{lemma}
\begin{proof}
The orbit of $x$ under independent coordinate rotations lies in $K$.
The maximum of $[u,\cdot]$ on that orbit is $\sum_j b_jr_j$.
Since $x$ itself maximizes $[u,\cdot]$ on all of $K$, equality holds
in each of the scalar inequalities
$\operatorname{Re}(\overline{u_j}x_j)\le b_jr_j$.
This proves both the first equality and the alignment when $x_j\ne0$.
No alignment assertion is needed when $x_j=0$.
Maximizing the same functional on the contained polydisc gives
$h_K(u)\ge\sum_{j<n}b_j$. Subtracting this from the first equality
gives the second inequality in \eqref{eq:full-section-normal}.
\end{proof}

\subsection{Projections onto proper coordinate subsets}

\begin{lemma}\label{lem:proper-projection-margin}
Put $M=2^n-1$ and
\begin{equation}\label{eq:full-section-vectors}
 p^k_j=\exp\left(\frac{2\pi\ii\,2^{j-1}k}{M}\right),
 \qquad 0\le k<M,\quad j\in[n].
\end{equation}
There exists $c_n\in(0,1)$, depending only on $n$, such that for every
nonempty proper subset $S\subset[n]$ and every family of unit complex
numbers $(\zeta_j)_{j\in S}$ there is an index $k$ satisfying
\begin{equation}\label{eq:full-section-margin}
 \operatorname{Re}(\overline{\zeta_j}p^k_j)>c_n
 \qquad(j\in S).
\end{equation}
\end{lemma}
\begin{proof}
First consider a projection onto $n-1$ coordinates. The $n$ column
frequencies in \eqref{eq:full-section-vectors} are $2^a$,
$a=0,\ldots,n-1$, taken modulo $M$. Since $2^n\equiv1\pmod M$,
multiplying all frequencies by a power of $2$ cyclically permutes
these $n$ exponents. This multiplication also just permutes the rows,
because $M$ is odd. Choose the cyclic shift that sends the omitted
exponent to $n-1$. The retained frequencies are then
$1,2,\ldots,2^{n-2}$, up to coordinate order. They are exactly the
cyclic construction \eqref{eq:cyclic-vectors} in dimension $n-1$,
whose denominator is $2^n-1$.

Hence every such projection meets every product of $n-1$ open
semicircles. Every smaller nonempty projection has the same property:
extend its coordinate set to one of size $n-1$ and prescribe arbitrary
open semicircles on the added coordinates. In particular, for any
$(\zeta_j)_{j\in S}$ one row has strictly positive real pairing
with all the $\zeta_j$ on $S$.

To obtain a margin independent of these unit complex numbers, define on the compact torus
$\{(\zeta_j)_{j\in S}:|\zeta_j|=1\}$ the continuous function
\[
 f_S(\zeta)=\max_{0\le k<M}\ \min_{j\in S}
                 \operatorname{Re}(\overline{\zeta_j}p^k_j).
\]
It is continuous because both extrema are over finite sets, and the
preceding paragraph shows $f_S(\zeta)>0$ at every point. Its attained
minimum $\alpha_S$ is therefore positive. There are only finitely
many nonempty proper subsets $S$. Thus
\[
 c_n=\frac12\min_{\varnothing\ne S\subsetneq[n]}\alpha_S
\]
is positive and at most $1/2$. A row attaining the maximum in $f_S$
has every pairing at least $\alpha_S\ge2c_n>c_n$, proving the claim.
\end{proof}

\subsection{The illuminating directions and the boundary cases}

\begin{proof}[Proof of the upper bound in Theorem~\ref{thm:full-coordinate-section}]
Permute coordinates so that the full section is $\{x_n=0\}$, and set
\begin{equation}\label{eq:full-section-parameters}
 \gamma=\norm{\mathbf 1}^{-1},\qquad
 \delta=\frac{c_n(1-\gamma)}2.
\end{equation}
We have $0<\gamma<1$. Indeed, if $\mathbf 1\in K$, coordinatewise
monotonicity would imply $\mathbb D^n\subseteq K$; normalization
would then imply $K=\mathbb D^n$, contrary to the hypothesis.
Thus $\norm{\mathbf 1}>1$, and in particular $\delta>0$.

Every $x\in K$ has at least one coordinate with $|x_j|\le\gamma$.
To see this, suppose $r_j=|x_j|>\gamma$ for all $j$. Choose
$\lambda$ with $\max_j\gamma/r_j<\lambda<1$.
Rotation invariance and Lemma~\ref{lem:section-contraction} give
\[
 1=\norm{\gamma\mathbf 1}
 \le\norm{\lambda(r_1,\ldots,r_n)}
 =\lambda\norm{x}\le\lambda<1,
\]
a contradiction.

Using the vectors from Lemma~\ref{lem:proper-projection-margin}, define
\begin{equation}\label{eq:full-section-directions}
 \begin{split}
 d^{k,0}&=-(p^k_1,\ldots,p^k_{n-1},0),\\
 d^{k,1}&=-(\delta p^k_1,\ldots,\delta p^k_{n-1},p^k_n),
 \qquad 0\le k<M.
 \end{split}
\end{equation}
These are $2M$ nonzero vectors, fixed before choosing the boundary
point. We show that their union illuminates every $x\in\partial K$.

\medskip\noindent\emph{Case 1: $x_n=0$.}
By Lemma~\ref{lem:section-contraction}, $x$ lies on the relative
boundary of $K\cap\{x_n=0\}=\mathbb D^{n-1}\times\{0\}$.
Consequently $\|x\|_\infty=1$, and $M_x=\{j\in[n-1]:|x_j|=1\}$ is nonempty.
Choose $k$ with the margin \eqref{eq:full-section-margin} on $M_x$,
taking $\zeta_j=x_j$. For $j\in M_x$ and $0<t<2c_n$,
\[
 |x_j-tp^k_j|^2
 =1-2t\operatorname{Re}(\overline{x_j}p^k_j)+t^2
 <1-2tc_n+t^2<1.
\]
For every remaining $j<n$, we initially have $|x_j|<1$, so
$|x_j-tp^k_j|<1$ for all sufficiently small positive $t$, by
continuity. This includes coordinates that are zero. There are
finitely many coordinates, so one $t>0$ satisfies all these
inequalities. The last coordinate remains zero. Hence $x+td^{k,0}$
is in the relative interior of the full section, and therefore
in $\operatorname{int}K$ by Lemma~\ref{lem:section-contraction}.

\medskip\noindent\emph{Case 2: $x_n\ne0$ and $r_j>\gamma$ for all $j<n$.}
The preceding observation forces $r_n\le\gamma<1$. Normalization
gives $\norm{x_ne_n}=r_n<1$, so $q=x_ne_n$ is interior.
All first $n-1$ coordinates of $x$ are nonzero. Choose $k$ with
\eqref{eq:full-section-margin} on $[n-1]$, using $\zeta_j=x_j/r_j$.
For each $j<n$ we have
\[
 |x_j-tp^k_j|^2
 <r_j^2-2tc_nr_j+t^2<r_j^2
 \quad\text{whenever }0<t<2c_nr_j.
\]
Choose $t$ smaller than all these finitely many positive upper bounds.
The point $x+td^{k,0}$ strictly contracts every coordinate in $[n-1]$
and leaves the last one unchanged. Apply part~3 of
Lemma~\ref{lem:section-contraction} with $J=[n-1]$ and $q=x_ne_n$
to conclude that $x+td^{k,0}\in\operatorname{int}K$.

\medskip\noindent\emph{Case 3: $x_n\ne0$ and $r_i\le\gamma$ for some $i<n$.}
Let
\[
 E=Z_x\cup\{i\},\qquad S=[n]\setminus E.
\]
Then $\varnothing\ne E\subseteq[n-1]$, $n\in S$, and every
coordinate of $x$ on $S$ is nonzero. In particular $S$ is a
nonempty proper subset. Also $r_j\le\gamma$ for every $j\in E$.
Choose a single $k$ with \eqref{eq:full-section-margin} on $S$,
using $\zeta_j=x_j/r_j$. This choice uses $x$ and is independent
of its supporting normal.

Now let $u\ne0$ be any supporting normal at $x$, and put $b_j=|u_j|$.
Since $0\le r_j\le1$, all terms on the left below are nonnegative.
Lemma~\ref{lem:full-section-normal} gives
\begin{equation}\label{eq:exceptional-normal-control}
 (1-\gamma)\sum_{j\in E}b_j
 \le\sum_{j\in E}b_j(1-r_j)
 \le\sum_{j<n}b_j(1-r_j)
 \le b_nr_n\le b_n.
\end{equation}
On $S$ the normal aligns with $x$ and the chosen entry has positive
pairing at least $c_n$. On $E$ no alignment or sign estimate is
required: the absolute value of each contribution is at most
$\delta b_j$, including when $x_j=0$. It follows that
\begin{align*}
 [u,d^{k,1}]
 &\le-c_nb_n-c_n\delta\sum_{j\in S\setminus\{n\}}b_j
                    +\delta\sum_{j\in E}b_j\\
 &\le-\left(c_n-\frac{\delta}{1-\gamma}\right)b_n
                    -c_n\delta\sum_{j\in S\setminus\{n\}}b_j\\
 &=-\frac{c_n}{2}b_n
                    -c_n\delta\sum_{j\in S\setminus\{n\}}b_j<0.
\end{align*}
For the final strict inequality, if the last expression were zero,
positivity of $c_n$ and $\delta$ would force $b_n=0$ and
$b_j=0$ for every $j\in S\setminus\{n\}$. Equation
\eqref{eq:exceptional-normal-control} would then force all $b_j$
on $E$ to vanish as well, contradicting $u\ne0$.
Thus the same $d^{k,1}$ pairs negatively with every nonzero supporting
normal at $x$. The illumination criterion
\eqref{eq:normal-criterion} proves that $d^{k,1}$ illuminates $x$.

These three cases exhaust $\partial K$. The parameter $c_n$ was
chosen solely from $n$, and $\delta$ was then fixed from $K$; neither
depends on the boundary point or its supporting normal. We have
constructed at most $2M=2^{n+1}-2$ illuminating directions.
\end{proof}

\subsection{Sharpness and immediate consequences}

\begin{proof}[Proof of sharpness in Theorem~\ref{thm:full-coordinate-section}]
The body $H_n$ in the theorem is normalized, complex $1$-unconditional,
and contains $\mathbb D^{n-1}\times\{0\}$. It is not $\mathbb D^n$,
since $e_{n-1}+e_n\notin H_n$. The upper bound already proved
therefore applies to $H_n$.

For the lower bound consider its two boundary tori
\[
 \begin{split}
 T_-&=\{x:|x_j|=1\ (j<n),\ x_n=0\},\\
 T_+&=\{x:|x_j|=1\ (j\le n-2\text{ or }j=n),\ x_{n-1}=0\}.
 \end{split}
\]
Let $\mathcal V$ be any finite illuminating family for $H_n$, and
let $\mathcal V_-$ and $\mathcal V_+$ consist of the directions
that illuminate at least one point of $T_-$ and $T_+$, respectively.
If $d\in\mathcal V_-$ illuminates $x\in T_-$ at some time $t>0$,
then interior membership gives
\[
 |x_{n-1}+td_{n-1}|+t|d_n|<1.
\]
The reverse triangle inequality gives
$|x_{n-1}+td_{n-1}|\ge1-t|d_{n-1}|$, so
$|d_{n-1}|>|d_n|$. Interchanging the last two coordinates shows
that $d\in\mathcal V_+$ requires $|d_n|>|d_{n-1}|$.
Consequently $\mathcal V_-\cap\mathcal V_+=\varnothing$.

Let $\pi_-$ be projection onto coordinates $1,\ldots,n-1$.
If $d$ illuminates $x\in T_-$ in $H_n$, then
$\pi_-(x+td)\in\operatorname{int}\mathbb D^{n-1}$ for the same
$t$: the first $n-2$ moduli are strictly below one, and the displayed
inequality also makes the $(n-1)$st modulus strictly below one.
Thus $\pi_-(\mathcal V_-)$ illuminates every point of
$(\partial\mathbb D)^{n-1}$.

For completeness, illuminating this torus suffices to illuminate
the entire polydisc, in any dimension $m$. Every point $x\in\mathbb D^m$ is a finite
convex combination $x=\sum_{j\in[N]}\lambda_j y^j$ of
torus points: express each scalar coordinate as a convex combination
of two unit complex numbers and multiply the coordinatewise weights.
Discard terms of zero weight. If a direction $d$ illuminates one
$y^j$, say $y^j+td\in\operatorname{int}\mathbb D^m$,
then
\[
 x+\lambda_j td
 =\lambda_j(y^j+td)+\sum_{a\ne j}\lambda_a y^a
 \in\operatorname{int}\mathbb D^m.
\]
The last inclusion follows coordinatewise from a positive weight on
a vector all of whose coordinate moduli are strictly below one.
The exact polydisc illumination theorem~\cite{RSS} now implies
$|\mathcal V_-|\ge2^n-1$. The projection omitting coordinate $n-1$
gives $|\mathcal V_+|\ge2^n-1$ in exactly the same way.
Since these families are disjoint,
\[
 |\mathcal V|\ge|\mathcal V_-|+|\mathcal V_+|
 \ge2(2^n-1)=2^{n+1}-2.
\]
Together with the upper bound, this proves equality.
\end{proof}

\begin{corollary}\label{cor:full-section-low-dimensions}
Every normalized complex $1$-unconditional body $K\subset\C^2$
with $K\ne\mathbb D^2$ satisfies $\ill(K)\le6$, and this is sharp.
In $\C^3$ a full coordinate bidisc gives the sharp bound $14$;
in $\C^4$ a full coordinate tridisc gives the sharp bound $30$,
in both cases assuming $K$ is not the full polydisc.
\end{corollary}
\begin{proof}
In dimension two every normalized body contains a unit coordinate
disc, so the section hypothesis is automatic. Apply
Theorem~\ref{thm:full-coordinate-section} with $n=2,3,4$.
Its examples $H_2,H_3,H_4$ establish the respective sharpness claims.
\end{proof}

\begin{remark}
The directions \eqref{eq:full-section-directions} depend on $K$
through $\gamma$ and $\delta$. The directions are chosen separately for each body. In dimensions three and four the corollary only
covers the stated section classes. Closeness to $\mathbb D^n$
does not by itself imply the required exact section containment.
\end{remark}


\section{Three complex dimensions}\label{sec:three-dimensions}

\begin{theorem}\label{thm:three-dimensions}
Every normalized complex $1$-unconditional convex body
$K\subset\C^3$ with $K\ne\mathbb D^3$ satisfies
$\ill(K)\le14$. The bound is sharp.
\end{theorem}

Theorem~\ref{thm:full-coordinate-section} already handles bodies
containing a full coordinate bidisc. We construct fourteen directions
whenever at least one coordinate bidisc is missing. The construction
uses two copies of a seven-vector family, each with three directions
tending to each coordinate axis. The partition of the seven rows
comes from an additional coordinate, as in
Lemma~\ref{lem:binary-support-tree}. It is also the two-dimensional
case of the coloring in~\cite[Proposition~4.4]{RSSSym}, with the
omitted row restored.

\subsection{Seven directions in a coordinate plane}

Set $\eta=e^{2\pi\ii/7}$. By Lemma~\ref{lem:cyclic-projections},
every two-coordinate projection of
\[
 \{(\eta^t,\eta^{2t},\eta^{4t}):t\in\{0,\ldots,6\}\}
\]
meets every product of two open semicircles. Replace the last entry
by $h_t=\eta^{3-4t}$. Conjugation and rotation preserve open
semicircles, so the same projection property holds for the columns
$\eta^t,\eta^{2t},h_t$. Partition the rows by the sign of the
imaginary part of the additional coordinate:
\[
 U=\{t:\operatorname{Im}h_t>0\}=\{0,2,4\},\qquad
 V=\{t:\operatorname{Im}h_t<0\}=\{1,3,5\},\qquad h_6=1.
\]
The entries $\eta^t$, $t\in U$, meet every open semicircle: prescribe
that semicircle in the first column and the upper open semicircle
in the additional column. Similarly, the entries $\eta^{2t}$,
$t\in V$, meet every open semicircle, by prescribing the lower
open semicircle in the additional column. All seven rows still
satisfy arbitrary open-semicircle conditions on the two visible
coordinates simultaneously.

For $\varepsilon>0$, define
\begin{equation}\label{eq:seven-directions}
 p^t(\varepsilon)=-(a_t\eta^t,b_t\eta^{2t}),\qquad
 (a_t,b_t)=
 \begin{cases}
 (1,\varepsilon),&t\in U,\\
 (\varepsilon,1),&t\in V,\\
 (1,1),&t=6.
 \end{cases}
\end{equation}
For any $x_1,x_2\ne0$, the two semicircle conditions give a row
with
\[
 \operatorname{Re}(\overline{x_j}p^t_j(\varepsilon))<0
 \qquad(j\in[2]).
\]
Positive amplitudes do not change these signs. Since
\[
 |x_j+sd_j|^2=|x_j|^2+
 2s\operatorname{Re}(\overline{x_j}d_j)+s^2|d_j|^2,
\]
both moduli strictly decrease for a sufficiently small positive
step $s$. As $\varepsilon$ tends to zero, the $U$-rows tend to
three directions on the first axis whose unit entries meet every
open semicircle, and the $V$-rows do the same on the second axis.
Thus the additional coordinate determines both the partition
and the two axis limits.

\subsection{Proof of the bound}

\begin{proof}[Proof of Theorem~\ref{thm:three-dimensions}]
If $K$ contains a full coordinate bidisc, apply
Theorem~\ref{thm:full-coordinate-section}. Otherwise permute the
coordinates so that $e_2+e_3\notin K$. In fact, the following
argument only requires this one missing bidisc. Embed
\eqref{eq:seven-directions} in the two planes by setting
\begin{align*}
 A_\varepsilon&=
 \{(p^t_1(\varepsilon),p^t_2(\varepsilon),0):t\in\{0,\ldots,6\}\},\\
 B_\varepsilon&=
 \{(p^t_1(\varepsilon),0,p^t_2(\varepsilon)):t\in\{0,\ldots,6\}\}.
\end{align*}
We choose one $\varepsilon>0$ for which these fourteen directions
illuminate $K$.

We first record why any three unit entries meeting every open
semicircle on each coordinate axis illuminate a normalized
complex $1$-unconditional body $L\subset\C^2$ different from
$\mathbb D^2$. At a boundary point $x$ with both coordinates
nonzero, their moduli cannot both equal one, since coordinate
contractions and rotations would then give $L=\mathbb D^2$.
Choose a coordinate to decrease so that the unchanged coordinate
has modulus less than one. Setting the chosen coordinate to zero
leaves an interior point of $L$, since every axis point of modulus
less than one is interior. One of the three directions on the
chosen axis strictly decreases that modulus for a small positive
step, and Lemma~\ref{lem:section-contraction}(3) puts the new point
in $\operatorname{int}L$. If only one coordinate is nonzero, decrease
it in the same way; deleting it leaves the interior point zero.
This proves the assertion for all boundary points of $L$.

Now take $L=K\cap\{x_1=0\}$, identified with a body in $\C^2$.
It is normalized and is not a bidisc. The $V$-rows from the two
families tend to three pure directions on axis 2 and three on
axis 3. By the preceding argument, these limits illuminate
$\partial L$ in its plane. Lemma~\ref{lem:section-contraction}(2)
identifies relative interior in the section with ambient interior
intersected with that plane. Thus they also illuminate
$\partial K\cap\{x_1=0\}$ in $K$. The $U$-rows of either family
tend to three pure directions on axis 1 that illuminate its unit
circle. The limiting directions therefore illuminate the compact set
\[
 E=(\partial K\cap\{x_1=0\})
       \cup\{ze_1:|z|=1\}.
\]

There exists $\varepsilon_0>0$ such that
$A_\varepsilon\cup B_\varepsilon$ illuminates all of $E$ whenever
$0<\varepsilon<\varepsilon_0$. Indeed, for each $x\in E$, choose
one limiting direction $d(0)$ and a step $s>0$ with
$x+sd(0)\in\operatorname{int}K$. Openness supplies neighborhoods
of $x$ and $d(0)$ such that the same step enters the interior for
every pair in those neighborhoods. Finitely many of the point
neighborhoods cover $E$. The corresponding finitely many directions
$d(\varepsilon)$ converge to their limits, so the minimum of
their positive admissible upper bounds on $\varepsilon$ gives
one $\varepsilon_0$ valid on all of $E$. Fix
$\varepsilon\in(0,\varepsilon_0)$.

Let $x\in\partial K\setminus E$. Then $x_1\ne0$ and at least one
of $x_2,x_3$ is nonzero. If $x_3=0$, a direction in
$A_\varepsilon$ strictly decreases the first two moduli, and
Lemma~\ref{lem:section-contraction}(3), with the interior point
zero, shows that it illuminates $x$. If $x_2=0$, use
$B_\varepsilon$ instead.

It remains to consider three nonzero coordinates. The moduli
$|x_2|$ and $|x_3|$ cannot both equal one: deleting the first
coordinate and rotating the others would put $e_2+e_3$ in $K$.
If $|x_3|<1$, choose a direction in $A_\varepsilon$ decreasing
the first two moduli. It leaves the third coordinate unchanged,
and $x_3e_3\in\operatorname{int}K$, so
Lemma~\ref{lem:section-contraction}(3) applies. Otherwise
$|x_2|<1$, and a direction in $B_\varepsilon$ works by the same
argument with the interior point $x_2e_2$. These cases exhaust
the boundary and prove $\ill(K)\le14$.

Finally, the body
$\mathbb D\times\{(z,w):|z|+|w|\le1\}$ has illumination number
$14$ by the sharpness part of
Theorem~\ref{thm:full-coordinate-section}. Hence the bound is sharp.
\end{proof}



\section{Four complex dimensions}\label{sec:four-dimensions}

\begin{theorem}\label{thm:four-dimensions}
Every normalized complex $1$-unconditional convex body
$K\subset\C^4$ different from $\mathbb D^4$ satisfies
\[
 \ill(K)\le30.
\]
The bound is sharp. Consequently every complex $1$-unconditional
convex body in $\C^4$ has illumination number at most $31$, and
equality holds precisely when its positive diagonal normalization
is $\mathbb D^4$.
\end{theorem}

The proof separates bodies according to which coordinate bidiscs
they contain. A full coordinate bidisc on $\{i,j\}$ is present
exactly when $e_i+e_j\in K$. Three constructions below cover every
body with a missing bidisc. The remaining case uses a finite
covering assertion, verified by the exact integer program in
Appendix~\ref{app:c4-verifier}. Every geometric argument applies
to all boundary points and all their supporting normals.

\subsection{Two disjoint missing bidiscs}
\begin{proposition}\label{prop:c4-disjoint-missing}
Let $K\subset\C^4$ be normalized and complex $1$-unconditional. If
$e_1+e_2\notin K$ and $e_3+e_4\notin K$, then $\ill(K)\le30$.
\end{proposition}
\begin{proof}
Put
\[
 \gamma=\max_{x\in K}\min\{|x_3|,|x_4|\}<1.
\]
For $j\in\{3,4\}$, use Lemma~\ref{lem:binary-support-tree} with
root $\{1,2,j\}$, children $\{1,j\}$ and $\{2,j\}$, singleton
children $\{1\},\{j\}$ below $\{1,j\}$, and singleton child
$\{2\}$ below $\{2,j\}$. Write $D_j(\delta)$ for its fifteen
directions, embedded with the remaining coordinate zero.
Keep the angular entries and the row partitions fixed and let
$\delta\in(0,1)$ be the amplitude parameter of that construction.
There is $c>0$, independent of $\delta$, such that for every node
$S$ and every choice of coordinate arguments on $S$, a row of its
group is inward with angular margin $c$ at every coordinate of $S$;
if its smallest amplitude on $S$ is $a$, every amplitude outside
$S$ but inside its root is at most $\delta a$.

In each family the rows belonging to the singleton nodes
$\{1\},\{2\}$ converge, as $\delta\to0$, to directions on their
respective coordinate axes, whose unit entries meet every open
semicircle. These limiting directions illuminate
\[
 E=\partial K\cap\{x_3=x_4=0\}.
\]
Indeed, the coordinate section on $\{1,2\}$ is not a bidisc. At a
boundary point with both coordinates nonzero, at least one modulus
is less than one. Decrease the other coordinate, using an inward
axis direction. Deleting that coordinate leaves an axis point of
modulus less than one, which is interior to $K$; strict contraction
toward it therefore enters the interior. If only one coordinate is
nonzero, decrease it instead, with the interior point zero.

The set $E$ is compact. At each of its points choose one limiting
direction and one positive step entering $\operatorname{int}K$.
Openness gives neighborhoods of both the point and that direction
for which the same step enters the interior. Take a finite cover
of $E$ by these point neighborhoods. Convergence of the finitely
many corresponding rows gives $\delta_0>0$ such that
$D_3(\delta)\cup D_4(\delta)$ illuminates $E$ whenever
$0<\delta<\delta_0$. Fix
\[
 0<\delta<\min\left\{\delta_0,1,\frac{c(1-\gamma)}4\right\}.
\]

Let $x\in\partial K\setminus E$. Choose $j\in\{3,4\}$ and the
other index $k$ so that $x_j\ne0$ and $|x_k|\le\gamma$. Such a
choice exists: if one of $x_3,x_4$ vanishes, omit that coordinate;
otherwise omit a coordinate with the smaller modulus. Put
$R=\{1,2,j\}$ and $S=\supp(x)\cap R$. Every subset of $R$
containing $j$ is a node of the chosen tree, so the row property
applies to $S$.

For a nonzero supporting normal $u$ at $x$, put $b_i=|u_i|$,
$r_i=|x_i|$ and $h=[u,x]$. Normal alignment and normalization give
$h=\sum_i b_ir_i>0$ and $b_i\le h$. Consequently
\[
 (1-r_k)h\le\sum_{i\in S}b_ir_i\le\sum_{i\in S}b_i,
 \qquad
 b_\ell\le\frac{1}{1-\gamma}\sum_{i\in S}b_i
 \quad(\ell\in R\setminus S).
\]
The sum over $S$ is positive. Choose a row $d\in D_j(\delta)$
having angular margin $c$ on $S$. This choice depends only on $x$,
not on $u$. Its omitted coordinate is zero, and there are at most
two inactive coordinates inside $R$, whence
\[
 [u,d]\le a\left(-c+\frac{2\delta}{1-\gamma}\right)
                  \sum_{i\in S}b_i<0.
\]
The normal criterion proves that $d$ illuminates $x$. Along with
the compact-set argument for $E$, this proves the bound.
\end{proof}

\subsection{Two present bidiscs and a missing bidisc}

\begin{proposition}\label{prop:c4-two-present-bidiscs}
Let $K\subset\C^4$ be a normalized complex $1$-unconditional convex
body. Suppose that
\[
 e_1+e_3\in K,\qquad e_1+e_4\in K,
 \qquad e_3+e_4\notin K.
\]
Then $\ill(K)\le30$.
\end{proposition}
\begin{proof}
Put
\[
 \gamma=\max_{x\in K}\min\{|x_3|,|x_4|\},\qquad c=1-\gamma.
\]
The maximum exists by compactness. Normalization gives $\gamma\le1$.
Equality would give a point of $K$ with $|x_3|=|x_4|=1$;
coordinate contractions and rotations would then imply
$e_3+e_4\in K$. Thus $c>0$. Every $x\in K$ satisfies
\[
 |x_3|\le\gamma\quad\hbox{or}\quad |x_4|\le\gamma.
\]

Set $\omega=e^{2\pi\ii/15}$ and
\[
 v^t_j=\omega^{2^{j-1}t}
 \qquad(0\le t\le14,\ j\in[4]).
\]
By Lemma~\ref{lem:cyclic-projections}, any three columns meet every
product of three open semicircles. Define the following disjoint sets
of rows:
\[
 \begin{split}
 G&=\{t:\operatorname{Im}v^t_4>0\},\qquad
 H=\{t:\operatorname{Im}v^t_4<0\},\\
 H_1&=\{t\in H:\operatorname{Im}v^t_2>0\},\qquad
 H_3=\{t\in H:\operatorname{Im}v^t_2<0\}.
 \end{split}
\]
Each of $G,H$ has seven elements. The only row with zero imaginary
part in any column is row zero, so $H=H_1\sqcup H_3$ and
$\{0,\ldots,14\}=\{0\}\sqcup G\sqcup H_1\sqcup H_3$.

The rows $G$ meet every product of two open semicircles in columns
$\{1,2\}$, and also in columns $\{2,3\}$. To see this, append to
the two prescribed conditions the upper-open-semicircle condition
on column 4 and use the three-column projection property.
Similarly, $H$ meets every product of two open semicircles in columns
$\{1,3\}$. The entries in column 1 of $H_1$ meet every open
semicircle: prescribe that semicircle in column 1, the upper
semicircle in column 2, and the lower semicircle in column 4.
Prescribing the desired semicircle in column 3 and the lower
semicircles in columns 2 and 4 proves the analogous assertion for
column 3 of $H_3$. The entries in column 2 of $G$ also meet every
open semicircle. Finally, all fifteen rows meet every product of
three open semicircles in columns $\{1,2,3\}$.

These finitely many covering statements have a common positive
margin. More precisely, choose $\beta>0$ such that, for each pair
\[
 \begin{aligned}
 (R,S)\in\bigl\{&
 (\{0,\ldots,14\},\{1,2,3\}),
 (G,\{1,2\}),(G,\{2,3\}),\\
 &(H,\{1,3\}),(H_1,\{1\}),(G,\{2\}),(H_3,\{3\})\bigr\}.
 \end{aligned}
\]
and every family $(\zeta_j)_{j\in S}$ of unit complex numbers, some $t\in R$
satisfies
\[
 \operatorname{Re}(\overline{\zeta_j}v^t_j)\ge\beta
 \qquad(j\in S).
\]
Indeed, for each fixed $(R,S)$ the continuous function
$\max_{t\in R}\min_{j\in S}
\operatorname{Re}(\overline{\zeta_j}v^t_j)$
is positive on the compact torus. The minimum of its positive
minima over the displayed finite list gives such a $\beta$.

Fix
\[
 0<\delta<\min\{1,\beta c/4\}.
\]
For each row define the three amplitudes by
\[
 (a^t_1,a^t_2,a^t_3)=
 \begin{cases}
 (\delta^2,1,\delta),&t\in G,\\
 (1,\delta^2,\delta),&t\in H_1,\\
 (\delta,\delta^2,1),&t\in H_3,\\
 (1,1,1),&t=0,
 \end{cases}
\]
and set
\[
 d^t=-(a^t_1v^t_1,a^t_2v^t_2,a^t_3v^t_3,0),
 \qquad A=\{d^t:0\le t\le14\}.
\]
We prove that $A$ illuminates every $x\in\partial K$ with
$|x_4|\le\gamma$.

Fix such an $x$, put $r_j=|x_j|$ and
$S=\{j\in\{1,2,3\}:r_j>0\}$, and let $u\ne0$ be any supporting
normal at $x$. Write $b_j=|u_j|$ and $h=h_K(u)>0$.
The normalized-normal estimates give
\[
 b_j\le h,\qquad h=\sum_{j\in[4]}b_jr_j
 \le\sum_{j\in S}b_j+\gamma h,
 \qquad
 h\le\frac1c\sum_{j\in S}b_j.
\]
In particular $S\ne\varnothing$ and
$\sum_{j\in S}b_j>0$ for every supporting normal.
At every nonzero coordinate $u_j=b_jx_j/r_j$.
We select the row below using only $x$, and show its strict inward
inequality for every supporting normal $u$.

If $S=\{1,2,3\}$, take a row having the common margin on all three
coordinates. All three amplitudes are positive, so
\[
 [u,d^t]\le-\beta\sum_{j\in[3]}a^t_jb_j<0.
\]

If $S=\{2,3\}$, select $t\in G$ with the margin on both active
coordinates. If $S=\{1,3\}$, select $t\in H$ with the margin on
those coordinates. In either case the sole inactive coordinate
among the first three has amplitude $\delta^2$, and both active
amplitudes are at least $\delta$. Consequently
\[
 [u,d^t]\le-\beta\delta\sum_{j\in S}b_j+\delta^2h
 \le\left(-\beta\delta+\frac{\delta^2}{c}\right)
        \sum_{j\in S}b_j<0.
\]

Suppose $S=\{1,2\}$. The two full bidiscs in the hypothesis give
$b_1+b_4\le h$ and $b_1+b_3\le h$. Since $r_3=0$,
\[
 (1-r_4)b_4\le b_2r_2-b_1(1-r_1)\le b_2,
 \qquad b_4\le\frac{b_2}{c},
\]
and hence
\[
 b_3\le b_2r_2+b_4r_4-b_1(1-r_1)
 \le b_2+\gamma b_4\le\frac{b_2}{c}.
\]
Choose $t\in G$ with the margin on coordinates 1 and 2. Then
\[
 [u,d^t]\le-\beta\delta^2b_1-\beta b_2+\delta b_3
 \le-\beta\delta^2b_1+
          \left(-\beta+\frac\delta c\right)b_2<0,
\]
because $b_1+b_2>0$.

Finally, suppose $S=\{i\}$ is a singleton. Choose a row from
$H_1$, $G$, or $H_3$ according as $i=1$, $i=2$, or $i=3$,
with the margin on coordinate $i$. Its amplitude there is one,
and its other two amplitudes are at most $\delta$. Thus
\[
 [u,d^t]\le-\beta b_i+2\delta h
 \le\left(-\beta+\frac{2\delta}{c}\right)b_i<0.
\]
All nonempty subsets $S$ have now been covered, with a single row
chosen before the supporting normal. The normal criterion proves
the asserted illumination by $A$.

Let $P$ interchange coordinates 3 and 4, and put $B=P A$.
The body $PK$ satisfies the same hypotheses and has the same value
of $\gamma$. Applying the preceding argument to $PK$ and then
applying $P$ back shows that $B$ illuminates every boundary point
of $K$ with $|x_3|\le\gamma$. Every boundary point satisfies one of the two
coordinate inequalities. Therefore $A\cup B$ illuminates $K$,
and $|A\cup B|\le30$.
\end{proof}

\subsection{A coordinate triple containing all its bidiscs}

\begin{proposition}\label{prop:c4-missing-star}
Let $K\subset\C^4$ be normalized and complex $1$-unconditional. If
\[
 e_3+e_4\notin K,\qquad e_1+e_2,e_1+e_4,e_2+e_4\in K,
\]
then $\ill(K)\le30$.
\end{proposition}
\begin{proof}
Set
\[
 \gamma=\max_{x\in K}\min\{|x_3|,|x_4|\}<1.
\]
The strict inequality follows from compactness, coordinate
contractions and $e_3+e_4\notin K$. For a nonzero supporting normal
$u$ at $x$, write $r_j=|x_j|$, $b_j=|u_j|$ and $h=[u,x]$.
The normal alignment and normalization give
\[
 h=\sum_{j\in[4]}b_jr_j>0,\qquad b_j\le h\quad(j\in[4]).
\]
Whenever $e_i+e_j\in K$, we also have $b_i+b_j\le h$.

Put $\zeta=e^{2\pi\ii/15}$ and $v^t_j=\zeta^{2^{j-1}t}$ for
$0\le t\le14$ and $j\in[4]$. Every three-coordinate projection of
these fifteen rows meets every product of three open semicircles.
Define
\[
 G=\{t:\operatorname{Im}v^t_4>0\},\qquad H=\{t:\operatorname{Im}v^t_4<0\}.
\]
Both sets have seven elements and their complement is $\{0\}$.
Prescribing the upper semicircle in column $4$ shows that the
$G$-rows meet every product of two open semicircles in columns
$\{1,3\}$ and in columns $\{2,3\}$. Prescribing the lower
semicircle gives the same property for the $H$-rows in columns
$\{1,2\}$. These statements also imply their restrictions to
single columns.

For a positive parameter $\delta$, define
\[
 d^{t,A}=\begin{cases}
 - (\delta v^t_1,\delta v^t_2,v^t_3,0),&t\in G,\\
 - (v^t_1,v^t_2,\delta v^t_3,0),&t\in H,\\
 - (1,1,1,0),&t=0.
 \end{cases}
\]
Positive amplitudes preserve the three-coordinate covering property
of all fifteen rows.

Use Lemma~\ref{lem:binary-support-tree} for the tree with root
$\{1,2,4\}$, children $\{1,4\}$ and $\{2,4\}$, singleton children
$\{1\},\{4\}$ below $\{1,4\}$, and singleton child $\{2\}$ below
$\{2,4\}$. Denote its fifteen directions by $d^{t,B}$, embedded
in coordinates $1,2,4$, with coordinate $3$ zero.
The construction supplies the following property for every nonempty
$S\subseteq\{1,2,4\}$ containing $4$: for any coordinate arguments
on $S$, one row has a positive angular margin on $S$, and its
smallest active amplitude $a$ is at least $1/\delta$ times every
inactive amplitude within $\{1,2,4\}$. Take a common positive
margin $c\in(0,1)$ for these finitely many covering properties and the
$G,H$ properties above, and fix
\[
 0<\delta<\frac{c(1-\gamma)}4.
\]
Here the rows are chosen from their coordinate arguments, independently
of the supporting normal.

First suppose $r_3\le\gamma$ and $x_4\ne0$, and put
$S=\supp(x)\cap\{1,2,4\}$. Since $b_3\le h$,
\[
 (1-r_3)h\le\sum_{j\in S}b_jr_j\le\sum_{j\in S}b_j.
\]
The latter sum is positive. Every inactive normal coefficient is
therefore at most $(1-\gamma)^{-1}\sum_{j\in S}b_j$. Choose the
row $d^{t,B}$ just described. There are at most two inactive
coordinates within $\{1,2,4\}$, so
\[
 [u,d^{t,B}]\le a\left(-c+\frac{2\delta}{1-\gamma}\right)
                  \sum_{j\in S}b_j<0.
\]
This row illuminates $x$.

Next suppose $x_3\ne0$ and $r_4\le\gamma$. If $x_1,x_2\ne0$,
choose a direction $d^{t,A}$ inward in all three of its nonzero coordinates.
Their normal coefficients have positive sum, since $b_4\le h$
and $r_4<1$. Hence $[u,d^{t,A}]<0$.

If $x_1=0$ and $x_2\ne0$, the full bidiscs on $\{2,4\}$ and
$\{1,2\}$ yield
\begin{align*}
 (1-r_4)b_4&\le b_3r_3-b_2(1-r_2)\le b_3,\\
 b_1&\le b_3r_3+b_4r_4-b_2(1-r_2)
       \le\frac{b_3}{1-r_4}.
\end{align*}
Choose a $G$-row inward on $\{2,3\}$. Then
\[
 [u,d^{t,A}]\le
 \left(-c+\frac{\delta}{1-\gamma}\right)b_3-c\delta b_2<0.
\]
For strictness, if $b_3=0$ the displayed bounds imply $b_1=b_4=0$,
and consequently $b_2>0$. If $x_2=0$ and $x_1\ne0$, interchange
$1$ and $2$, using the full bidiscs on $\{1,4\}$ and $\{1,2\}$.

If $x_1=x_2=0$, the full bidiscs on $\{1,4\}$ and $\{2,4\}$ give
\[
 b_1,b_2\le b_3r_3-(1-r_4)b_4\le b_3.
\]
Here $b_3>0$, since $r_4<1$. A $G$-row inward on coordinate $3$
satisfies $[u,d^{t,A}]\le(-c+2\delta)b_3<0$.

Finally suppose $x_3=x_4=0$. Choose an $H$-row inward on all
nonzero coordinates among $1,2$. A zero coordinate among these two
has zero normal coefficient. Indeed, if $x_2=0$, then
$h=b_1r_1\ge b_1+b_2$ and $r_1\le1$ force $b_2=0$; the other
case is identical. Moreover $b_3\le h\le b_1+b_2$, so
\[
 [u,d^{t,A}]\le(-c+\delta)(b_1+b_2)<0.
\]

These cases exhaust $\partial K$. If $x_4\ne0$ and $r_3\le\gamma$,
the first case applies. If $x_4\ne0$ and $r_3>\gamma$, the
definition of $\gamma$ gives $r_4\le\gamma$ and $x_3\ne0$, so the
second case applies. When $x_4=0$, either $x_3\ne0$ and the second
case applies, or the last case applies. Thus the fifteen vectors $d^{t,A}$
and fifteen vectors $d^{t,B}$ illuminate $K$.
\end{proof}

\subsection{All coordinate bidiscs}

We first record a finite construction on four coordinates. Put
\[
 v^k_j=\exp\left(\frac{2\pi\ii\,2^{j-1}k}{31}\right),
 \qquad k\in[30],\quad j\in[4],\qquad c=\sin(\pi/124).
\]
Define two distinct indices $m(k),q(k)\in[4]$ by the following table.
\begin{center}
\small
\begin{tabular}{c|rrrrrrrrrrrrrrr}
$k$&1&2&3&4&5&6&7&8&9&10&11&12&13&14&15\\\hline
$m(k)$&1&2&1&4&3&2&3&4&1&4&1&4&3&2&4\\
$q(k)$&4&4&2&3&4&3&2&2&4&2&3&2&1&4&1
\end{tabular}

\medskip
\begin{tabular}{c|rrrrrrrrrrrrrrr}
$k$&16&17&18&19&20&21&22&23&24&25&26&27&28&29&30\\\hline
$m(k)$&2&1&2&4&2&3&3&1&3&2&3&4&1&2&1\\
$q(k)$&1&4&1&3&1&2&4&2&1&3&4&3&3&4&4
\end{tabular}
\end{center}

\begin{lemma}\label{lem:c4-finite-construction}
For every nonempty proper $S\subset[4]$ and every family of unit
complex numbers $(a_j)_{j\in S}$, there is a row $k\in[30]$ such that
\begin{equation}\label{eq:c4-first-cover}
 m(k)\in S,\qquad
 \operatorname{Re}(\overline{a_j}v^k_j)>c\quad(j\in S).
\end{equation}
For every family $(a_j)_{j\in[4]}$ of unit complex numbers and every
two-element set $L\subset[4]$, there is a row $k\in[30]$ satisfying
either
\begin{equation}\label{eq:c4-second-cover-full}
 \operatorname{Re}(\overline{a_j}v^k_j)>c\quad(j\in[4]),
\end{equation}
or
\begin{equation}\label{eq:c4-second-cover-pair}
 q(k)\in L,\qquad
 \operatorname{Re}(\overline{a_j}v^k_j)>c\quad(j\ne q(k)).
\end{equation}
\end{lemma}
\begin{proof}
For a unit complex number $a_j$, the condition
$\operatorname{Re}(\overline{a_j}z)>c$ defines an open arc on the
unit circle of length $61/124$ turns. Since $61/124>15/31$, this
arc contains fifteen consecutive points of the uniform grid with
denominator $31$. Indeed, the first grid point strictly after the
left endpoint lies at most $1/31$ turns after it, so the fifteenth
lies at most $15/31$ turns after it, strictly before the right
endpoint. If the first residue is $b_j\in\{0,\ldots,30\}$, every
retained row in
\[
 W(j,b_j)=\{k\in[30]:(2^{j-1}k-b_j)\bmod31<15\}
\]
satisfies the required strict inequality in coordinate $j$.

The integer verifier in Appendix~\ref{app:c4-verifier} checks
\begin{equation}\label{eq:c4-integer-first}
 \left(\bigcap_{j\in S}W(j,b_j)\right)
 \cap\{k:m(k)\in S\}\ne\varnothing
\end{equation}
for every nonempty proper $S$ and every choice of its block starts.
These are exactly
$4\cdot31+6\cdot31^2+4\cdot31^3=125054$ checks, and they prove
\eqref{eq:c4-first-cover}.

For all $31^4=923521$ choices of four block starts, the verifier
checks that either the full intersection is nonempty, or
\begin{equation}\label{eq:c4-integer-second}
 \#\left\{j\in[4]:
 \left(\bigcap_{i\ne j}W(i,b_i)\right)
 \cap\{k:q(k)=j\}\ne\varnothing\right\}\ge3.
\end{equation}
The first alternative proves \eqref{eq:c4-second-cover-full}.
In the second alternative, the three or more eligible indices meet
every pair $L$, which proves \eqref{eq:c4-second-cover-pair}.
There are $9341$ empty full intersections; the cardinality in
\eqref{eq:c4-integer-second} is three in $1320$ cases and four in
$8021$ cases. The displayed verifier exhausts these finite sets,
uses integer arithmetic only, and returns the stated counts.
The open-arc reduction proves the assertions for all continuous
arguments, including endpoint configurations.
\end{proof}

\begin{lemma}\label{lem:c4-all-bidiscs}
Suppose $K\subset\C^4$ is a normalized complex $1$-unconditional
convex body containing every coordinate bidisc. If $K\ne\mathbb D^4$,
then $\ill(K)\le30$.
\end{lemma}
\begin{proof}
If $K$ contains a full coordinate tridisc, apply
Theorem~\ref{thm:full-coordinate-section}. Otherwise define
\begin{equation}\label{eq:c4-bidisc-parameters}
 \gamma=\max_{i\in[4]}\norm{\mathbf1-e_i}^{-1}<1,
 \qquad \delta=\frac{c(1-\gamma)}4,
 \qquad \tau=\frac{c\delta(1-\gamma)}2.
\end{equation}
Every coordinate triple of every $x\in K$ contains an index $j$
with $|x_j|\le\gamma$. To see this, if $T$ is a coordinate triple
and $r=\min_{j\in T}|x_j|$, monotonicity gives
$r\norm{\mathbf1_T}\le\norm{x}\le1$, hence $r\le\gamma$.
In particular, a point of full support has at least two coordinates
of modulus at most $\gamma$.

Fix the thirty directions
\begin{equation}\label{eq:c4-bidisc-directions}
 d^k_j=-a^k_jv^k_j,\qquad
 a^k_j=\begin{cases}
 1,&j=m(k),\\
 \tau,&j=q(k),\\
 \delta,&j\notin\{m(k),q(k)\}.
 \end{cases}
\end{equation}
Here $0<\tau<\delta<1$. We show that they illuminate $K$.

Fix $x\in\partial K$, put $r_j=|x_j|$, and let $u\ne0$ be a
supporting normal at $x$. With $b_j=|u_j|$ and $h=h_K(u)$, normal
alignment and containment of the coordinate bidiscs give
\begin{equation}\label{eq:c4-bidisc-normals}
 h=\sum_{j\in[4]}b_jr_j>0,\qquad
 h\ge b_i+b_j\quad(i\ne j),\qquad h\ge b_j.
\end{equation}
At a nonzero coordinate, $u_j=b_jx_j/r_j$. Thus a row satisfying
$\operatorname{Re}(\overline{x_j/r_j}v^k_j)>c$ contributes at most
$-ca^k_jb_j$ to $[u,d^k]$. Any unchecked coordinate contributes
at most $a^k_jb_j$. In each case below the row is chosen from $x$
before the supporting normal is considered.

Suppose first that $x$ has full support. Choose a pair $L$ on
which $r_j\le\gamma$ and apply the second assertion of
Lemma~\ref{lem:c4-finite-construction}. A row satisfying
\eqref{eq:c4-second-cover-full} has negative pairing with every
nonzero supporting normal. Otherwise put $q=q(k)\in L$. From
$h\ge b_q$ we obtain
\[
 (1-\gamma)b_q\le(1-r_q)b_q
 \le\sum_{j\ne q}b_jr_j\le\sum_{j\ne q}b_j.
\]
All amplitudes outside $q$ are at least $\delta$, so
\[
 [u,d^k]\le-c\delta\sum_{j\ne q}b_j+\tau b_q
 \le-\frac{c\delta}{2}\sum_{j\ne q}b_j<0.
\]
The final sum cannot vanish: otherwise $u\ne0$ would imply
$h=b_qr_q<b_q$, contradicting \eqref{eq:c4-bidisc-normals}.

Suppose $x$ has exactly three nonzero coordinates. Let $i$ be the
missing coordinate, choose $t\in\supp(x)$ with $r_t\le\gamma$,
and put $R=\supp(x)\setminus\{t\}$. Apply
\eqref{eq:c4-first-cover} to $R$. Write $m=m(k)\in R$ and let
$l$ be the other index of $R$. The bidisc inequalities for
$\{t,l\}$ and $\{i,l\}$ give
\[
 b_t(1-r_t)+b_l(1-r_l)\le b_mr_m,\qquad
 b_i\le b_mr_m+b_tr_t-b_l(1-r_l).
\]
Consequently
\[
 b_t\le\frac{b_m}{1-\gamma},\qquad
 b_i\le b_m+\frac{\gamma b_m}{1-\gamma}
       =\frac{b_m}{1-\gamma}.
\]
The unchecked coordinates $i,t$ have amplitude at most $\delta$.
It follows that
\[
 [u,d^k]\le-cb_m-ca^k_lb_l+\delta(b_i+b_t)
 \le-\frac c2b_m-ca^k_lb_l<0.
\]
For strictness, $b_m=b_l=0$ would force $b_t=b_i=0$ by the
preceding estimates. The amplitude $a^k_l$ may equal $\tau$;
its positivity suffices.

Suppose $x$ has exactly two nonzero coordinates. Apply
\eqref{eq:c4-first-cover} to their pair, and denote them by
$m=m(k)$ and $l$. For either missing coordinate $i$, the bidisc
inequality for $\{i,l\}$ implies
\[
 b_i\le b_mr_m-b_l(1-r_l)\le b_m.
\]
Therefore
\[
 [u,d^k]\le-(c-2\delta)b_m-ca^k_lb_l<0.
\]
Indeed, $c-2\delta\ge c/2>0$, and $h>0$ ensures that at least
one of the active coefficients $b_m,b_l$ is positive.

Finally, if $x$ has only one nonzero coordinate $j$, then $r_j=1$:
the normalized axis section is the unit disc and its relative
interior lies in $\operatorname{int}K$. Now $h=b_j$, and the
bidisc inequality for $\{i,j\}$ forces $b_i=0$ for every $i\ne j$.
The row supplied by \eqref{eq:c4-first-cover} for the singleton
$\{j\}$ has negative pairing with every nonzero supporting normal.
There is no empty-support boundary point because $0\in\operatorname{int}K$.

In every case one of the fixed directions has negative pairing
with all nonzero supporting normals at $x$. The strict normal
criterion proves the claim.
\end{proof}


\subsection{Completion of the proof}

\begin{proof}[Proof of Theorem~\ref{thm:four-dimensions}]
Suppose first that $K\ne\mathbb D^4$. If every coordinate bidisc
is present, Lemma~\ref{lem:c4-all-bidiscs} applies. Otherwise let
\[
 \mathcal E=\{\{i,j\}\subset[4]:i\ne j,\ e_i+e_j\notin K\}.
\]
If two members of $\mathcal E$ are disjoint, a coordinate
permutation puts us in Proposition~\ref{prop:c4-disjoint-missing}.

It remains that $\mathcal E$ is nonempty and its members pairwise
intersect. Either all of them contain a common index, or they are
exactly the three edges of a triangle. Indeed, if there are at
least two edges, write two of them as $\{a,b\}$ and $\{a,c\}$.
An edge avoiding $a$ must be $\{b,c\}$. Once all three edges
are present, every further edge would be disjoint from one of
them. A family with only one edge has a common index as well.

In the common-index case, permute coordinates so that this index
is $3$ and one missing edge is $\{3,4\}$. The bidiscs on
$\{1,2\}$, $\{1,4\}$ and $\{2,4\}$ are present, so
Proposition~\ref{prop:c4-missing-star} applies. In the triangle
case, label its vertices $2,3,4$. The bidiscs on $\{1,3\}$ and
$\{1,4\}$ are present and the one on $\{3,4\}$ is missing;
Proposition~\ref{prop:c4-two-present-bidiscs} applies. These cases
exhaust all possibilities and prove $\ill(K)\le30$.

The normalized non-polydisc
\[
 H_4=\mathbb D^2\times
       \{(z_3,z_4)\in\C^2:|z_3|+|z_4|\le1\}
\]
has illumination number $30$ by the sharpness argument in
Theorem~\ref{thm:full-coordinate-section}. Thus the bound is sharp.
The exact polydisc theorem gives $\ill(\mathbb D^4)=31$.
Finally, positive diagonal rescaling preserves illumination
numbers and normalizes every complex $1$-unconditional body,
which proves the last assertion.
\end{proof}

\section{$1$-symmetric bodies in dimension three}\label{sec:symmetric-three}

A complex $1$-unconditional body is \emph{$1$-symmetric} if it is
also invariant under every coordinate permutation. The bound
$\ill(K)\le14$ for non-polydiscs in this class follows from
\cite[Theorem~1.5]{RSSSym}. The following construction improves it.
It uses the reduction underlying \cite[Proposition~2.8]{RSSSym};
we include a complete geometric proof using supporting normals.

\begin{theorem}\label{thm:symmetric-three}
Every complex $1$-symmetric convex body $K\subset\C^3$ which is
not a complex affine image of $\mathbb D^3$ satisfies
\[
 \ill(K)\le12.
\]
\end{theorem}
\begin{proof}
Put $\zeta=\exp(2\pi\ii/13)$ and define
\begin{equation}\label{eq:symmetric-three-rows}
 v^r=(\zeta^r,\zeta^{3r},\zeta^{4r})\quad(r\in[12]),\qquad
 (f(1),\ldots,f(12))=(2,1,1,3,1,2,2,1,3,1,3,2).
\end{equation}
We first verify that, with $\beta=\sin(\pi/26)>0$, for every
nonempty proper $S\subset[3]$ and every $(z_j)_{j\in S}$ with
$|z_j|=1$, there is $r\in[12]$ such that
\begin{equation}\label{eq:symmetric-three-cover}
 f(r)\in S,\qquad
 \operatorname{Re}(\overline{z_j}v_j^r)\le-\beta
 \quad(j\in S).
\end{equation}

All residues in the following verification are taken modulo $13$.
Set $(a_1,a_2,a_3)=(1,3,4)$ and
$B_s=\{s,s+1,\ldots,s+5\}$. A nonempty residue set meets every
$B_s$ if and only if its largest cyclic gap is at most six: a
gap of length $g$ leaves exactly $g-1$ consecutive missing residues.
For singleton supports, the sets
$\{a_jr:f(r)=j\}$ and their largest gaps are
\[
\begin{array}{c|c|c}
 j&\{a_jr:f(r)=j\}&\text{largest gap}\\\hline
 1&\{2,3,5,8,10\}&5\\
 2&\{3,5,8,10\}&6\\
 3&\{3,5,10\}&6
\end{array}
\]
For $i<j$, define
\[
 R_{ij}(s)=\{a_jr:a_ir\in B_s,\ f(r)\in\{i,j\},\ r\in[12]\}.
\]
Their largest cyclic gaps, calculated directly from
\eqref{eq:symmetric-three-rows}, are
\[
\begin{array}{c|rrrrrrrrrrrrr}
 (i,j)\backslash s&0&1&2&3&4&5&6&7&8&9&10&11&12\\\hline
 (1,2)&6&6&6&4&4&4&6&6&6&6&6&6&6\\
 (1,3)&4&4&4&5&6&5&4&4&4&5&6&6&5\\
 (2,3)&6&6&6&6&6&4&6&6&6&6&6&6&6
\end{array}
\]
Every entry is at most six. Thus for any prescribed blocks
$B_{s_j}$, $j\in S$, a row satisfies
$f(r)\in S$ and $a_jr\in B_{s_j}$ for every $j\in S$.
This finite assertion comprises $3\cdot13+3\cdot13^2=546$ cases;
the accompanying verifier checks all of them with integer arithmetic.

For each unit $z$, the closed arc
$\{w:|w|=1,\ \operatorname{Re}(\overline z w)\le-\beta\}$
has angular length $12\pi/13$, exactly six spacings between
successive thirteenth roots of unity. It therefore contains six
consecutive roots. Applying the finite assertion to one such
block in each coordinate proves \eqref{eq:symmetric-three-cover},
including all endpoint cases.

Scale $K$ so that $e_j\in\partial K$ for $j\in[3]$. This uses
one common factor, since $K$ is $1$-symmetric. Then
$K\subset\mathbb D^3$ and there is a unique $\rho>0$ with
$\rho(1,1,1)\in\partial K$. Coordinate contractions give
$\rho\mathbb D^3\subset K$, so $0<\rho<1$: equality would give
$K=\mathbb D^3$. Fix
\begin{equation}\label{eq:symmetric-three-amplitudes}
 \Delta=\frac{\beta}{6},\qquad
 \delta=\min\left\{\frac{\Delta}{2},
             \frac{\beta\Delta(1-\rho)}{2\rho}\right\}.
\end{equation}
For every $r\in[12]$ define a direction by
\begin{equation}\label{eq:symmetric-three-directions}
 d_j^r=
 \begin{cases}
 v_j^r,&j=f(r),\\
 \delta v_j^r,&j=3\ne f(r),\\
 \Delta v_j^r,&j\ne f(r),\ j\ne3.
 \end{cases}
\end{equation}
These twelve nonzero directions are fixed before choosing a
boundary point.

Let $x\in\partial K$. For any nonzero supporting normal $u$ at
$x$, write $b_j=|u_j|$ and $t_j=|x_j|$. Maximizing independently
over coordinate rotations shows that
\[
 u_j=b_j\frac{x_j}{|x_j|}\quad(x_j\ne0),\qquad
 h_K(u)=\operatorname{Re}\sum_{j=1}^3\overline{u_j}x_j
       =\sum_{j=1}^3b_jt_j>0.
\]
There is no restriction on the argument of $u_j$ when $x_j=0$.
Swapping any two coordinate moduli of $x$ and choosing their
arguments to maximize the same functional gives
\begin{equation}\label{eq:symmetric-three-order}
 (b_i-b_j)(t_i-t_j)\ge0.
\end{equation}
Indeed, the swapped point belongs to $K$, and subtracting its
functional value from $h_K(u)$ gives the displayed inequality.

Suppose first that $M_x\ne[3]$. Apply
\eqref{eq:symmetric-three-cover} with $S=M_x$ and
$z_j=x_j/|x_j|$, and put $k=f(r)\in M_x$ for the chosen row.
Write $\alpha_j=|d_j^r|$. Then $\alpha_k=1$ and
$0<\alpha_j\le\Delta$ for $j\ne k$. For $j\notin M_x$,
\eqref{eq:symmetric-three-order} gives $b_j\le b_k$, including
when $x_j=0$. Consequently
\begin{align*}
 \operatorname{Re}\sum_{j=1}^3\overline{u_j}d_j^r
 &\le-\beta b_k-\beta\sum_{j\in M_x\setminus\{k\}}\alpha_jb_j
       +\Delta\sum_{j\notin M_x}b_j\\
 &\le-(\beta-3\Delta)b_k
       -\beta\sum_{j\in M_x\setminus\{k\}}\alpha_jb_j<0.
\end{align*}
For the strict inequality, $\beta-3\Delta=\beta/2>0$.
If $b_k=0$, all coefficients outside $M_x$ vanish, and some
coefficient on $M_x\setminus\{k\}$ is positive because $u\ne0$.
The row was chosen using $x$ alone, so this holds for every
supporting normal at $x$.

If $M_x=[3]$, all coordinate moduli equal $\rho$. Apply
\eqref{eq:symmetric-three-cover} with $S=\{1,2\}$ and
$z_j=x_j/|x_j|$ for $j\in\{1,2\}$. Then
$k=f(r)\in\{1,2\}$, $\alpha_1,\alpha_2\ge\Delta$, and
$\alpha_3=\delta$. The third unit coordinate disc is contained
in $K$, hence
\[
 b_3\le h_K(u)=\rho(b_1+b_2+b_3),\qquad
 b_3\le\frac{\rho}{1-\rho}(b_1+b_2).
\]
In particular $b_1+b_2>0$. By
\eqref{eq:symmetric-three-amplitudes},
\[
 \operatorname{Re}\sum_{j=1}^3\overline{u_j}d_j^r
 \le-\beta\Delta(b_1+b_2)+\delta b_3
 \le-\frac{\beta\Delta}{2}(b_1+b_2)<0.
\]
Again the chosen row works for every supporting normal. The
normal criterion \eqref{eq:normal-criterion} now shows that the
directions \eqref{eq:symmetric-three-directions} illuminate $K$.
\end{proof}

Thus the bound $2^{n+1}-2$ is not sharp in the $1$-symmetric
class when $n=3$. It is sharp when $n=2$, since the complex
$\ell_1$ unit ball has illumination number six. In dimension
three that ball has illumination number nine by
Theorem~\ref{thm:hanner-formulas}, so the largest illumination
number among non-polydisc $1$-symmetric bodies lies between
nine and twelve. The construction does not establish that
twelve is sharp, or settle sharpness in higher dimensions.


\section{Complex Hanner bodies}\label{sec:hanner}

A \emph{complex Hanner body} is obtained recursively from the unit
disc $\mathbb D\subset\C$ by taking Cartesian products and
$\ell_1$ sums on disjoint coordinate blocks. For bodies $A$ and $B$
with Minkowski norms $\|\cdot\|_A$ and $\|\cdot\|_B$, these operations
are
\begin{align*}
 A\times B&=\{(x,y):\max\{\|x\|_A,\|y\|_B\}\le1\},\\
 A\oplus_1B&=\{(x,y):\|x\|_A+\|y\|_B\le1\}.
\end{align*}
Every resulting body $H\subset\C^n$ is normalized and complex
$1$-unconditional. Write $\operatorname{ext}H$ for its extreme points
and define
\[
 \mathcal S(H)=\{\supp(x):x\in\operatorname{ext}H\},\qquad
 F_H(t)=\sum_{S\in\mathcal S(H)}t^{|S|}.
\]
These definitions depend on $H$, not on a choice of recursive
description. For a nonempty $S\subseteq[n]$, put
\begin{align*}
 E_S&=\{z\in\C^n:z_j=0\text{ for }j\notin S\},\\
 \mathbb D^S&=\{z\in E_S:|z_j|\le1\text{ for }j\in S\},
\end{align*}
and let
\[
 T_S=\{z\in E_S:|z_j|=1\text{ for }j\in S\}
\]
be its coordinate torus.

\begin{theorem}\label{thm:hanner-formulas}
For every complex Hanner body $H$,
\begin{equation}\label{eq:hanner-formulas}
 \ill(H)=2F_H(2)-F_H(1),\qquad \ill^*(H)=F_H(2).
\end{equation}
The polynomial is computed recursively by
\begin{equation}\label{eq:hanner-recursion}
 F_{\mathbb D}(t)=t,\qquad
 F_{A\times B}(t)=F_A(t)F_B(t),\qquad
 F_{A\oplus_1B}(t)=F_A(t)+F_B(t).
\end{equation}
\end{theorem}

We first describe the extreme points and explain why directions
illuminating different tori must be counted separately.

\begin{lemma}\label{lem:hanner-extreme}
For a complex Hanner body $H$,
\[
 \operatorname{ext}H=\bigsqcup_{S\in\mathcal S(H)}T_S,
 \qquad H=\operatorname{conv}(\operatorname{ext}H).
\]
Every point of $H$ is a finite convex combination of these extreme
points. Moreover, $H\cap E_S=\mathbb D^S$ for every
$S\in\mathcal S(H)$, and \eqref{eq:hanner-recursion} holds.
\end{lemma}
\begin{proof}
The extreme points of a disc are its unit circle, and each point
of the disc is a convex combination of two circle points. We prove
the remaining assertions inductively, using
\begin{align}
 \operatorname{ext}(A\times B)
   &=\operatorname{ext}A\times\operatorname{ext}B,
       \label{eq:hanner-product-extreme}\\
 \operatorname{ext}(A\oplus_1B)
   &=(\operatorname{ext}A\times\{0\})
        \cup(\{0\}\times\operatorname{ext}B).
       \label{eq:hanner-sum-extreme}
\end{align}
For \eqref{eq:hanner-product-extreme}, if one coordinate block is
not extreme, a nontrivial convex decomposition in that block gives
one in the product. Conversely, a convex decomposition of a point
whose two blocks are extreme must be constant in each block.

For \eqref{eq:hanner-sum-extreme}, an extreme point lies on the
boundary, where $\|x\|_A+\|y\|_B=1$. If both blocks are nonzero,
the identity
\[
 (x,y)=\|x\|_A\left(\frac{x}{\|x\|_A},0\right)
       +\|y\|_B\left(0,\frac{y}{\|y\|_B}\right)
\]
is a nontrivial convex decomposition. Thus an extreme point has
one block zero, and its other block must be extreme in the
corresponding factor. Conversely, suppose $x\in\operatorname{ext}A$
and $(x,0)$ is a convex combination, with both weights positive,
of $(x^1,y^1),(x^2,y^2)\in A\oplus_1B$. Since $x^1,x^2\in A$,
extremality gives $x^1=x^2=x$. Now $\|x\|_A=1$, so the sum
constraint forces $y^1=y^2=0$. This proves extremality of $(x,0)$;
the other branch is identical.

Finite convex representations also persist under both operations.
For a product, multiply the coefficients in finite convex
representations of its two blocks. For a sum, use
\[
 A\oplus_1B=\operatorname{conv}\bigl((A\times\{0\})
                                \cup(\{0\}\times B)\bigr)
\]
and then the representations in the factors. The displayed
identity follows directly from the norm inequality: for a point
with $\alpha=\|x\|_A$, $\beta=\|y\|_B$ and $\alpha+\beta\le1$,
use weights $\alpha,\beta,1-\alpha-\beta$ on the two normalized
blocks and zero, omitting zero weights. The reverse inclusion is
convexity. The induction therefore proves the finite convex-hull
assertion as well as the extreme-point identities.

At a product, the supports are precisely $S\cup R$ with
$S\in\mathcal S(A)$ and $R\in\mathcal S(B)$; their intersections
with the two coordinate blocks recover $S$ and $R$ uniquely.
At a sum, the support families are united without repetitions,
since their nonempty sets lie in disjoint blocks. The two identities
above now give the asserted disjoint union of tori and the
polynomial recursion. The supports themselves may intersect;
the tori for distinct supports are disjoint.

Finally, $T_S\subset H$ implies
$\mathbb D^S=\operatorname{conv}T_S\subset H\cap E_S$.
The convex-hull identity follows by taking products of finite
circle representations in the coordinates of $S$. The reverse
inclusion follows from $H\subseteq\mathbb D^n$.
\end{proof}

\begin{lemma}\label{lem:hanner-distinct-tori}
If $S,R\in\mathcal S(H)$ are distinct, no direction illuminates
both a point of $T_S$ and a point of $T_R$.
\end{lemma}
\begin{proof}
We induct on the recursive construction. The assertion is vacuous
for a disc. For $H=A\times B$, a direction illuminating $(x,y)$
has projections illuminating $x$ in $A$ and $y$ in $B$, since
interior membership in the product requires interior membership
in both factors. Distinct supports differ in at least one factor,
where the induction hypothesis rules out a common direction.

For $H=A\oplus_1B$, a direction $d=(v,w)$ illuminating $(x,0)$,
with $\|x\|_A=1$, satisfies for some $s>0$
\[
 1>\|x+sv\|_A+s\|w\|_B
   \ge1-s\|v\|_A+s\|w\|_B.
\]
Thus $\|v\|_A>\|w\|_B$. Illuminating a point $(0,y)$ in the
other branch requires $\|w\|_B>\|v\|_A$, which is impossible
for the same direction. If both tori lie in the same branch,
projection of an illuminating direction illuminates the
corresponding points of that factor. The induction hypothesis
again applies.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:hanner-formulas}]
The recursion follows from Lemma~\ref{lem:hanner-extreme}. We
first justify the reduction to extreme points for both illumination
numbers. For $x\in\partial H$, let
\[
 \mathcal I_H(x)=\{d\in\C^n\setminus\{0\}:
                         d\text{ illuminates }x\}.
\]
By Lemma~\ref{lem:hanner-extreme}, write
$x=\sum_{k\in[M]}\lambda_k y^k$, where
$y^k\in\operatorname{ext}H$, $\lambda_k>0$ and $\sum_k\lambda_k=1$.
Fix one index $k$ before choosing a direction. If
$y^k+sd\in\operatorname{int}H$, then
\[
 x+\lambda_k sd
   =\lambda_k(y^k+sd)+\sum_{h\ne k}\lambda_hy^h
       \in\operatorname{int}H.
\]
Indeed, its $H$-norm is at most
$\lambda_k\|y^k+sd\|_H+1-\lambda_k<1$. Consequently
\begin{equation}\label{eq:hanner-extreme-transfer}
 \mathcal I_H(y^k)\subseteq\mathcal I_H(x).
\end{equation}
A set illuminating every extreme point therefore illuminates
the boundary. Likewise, a measure assigning mass at least one
to $\mathcal I_H(y)$ for each extreme point $y$ assigns mass
at least one to $\mathcal I_H(x)$ for every boundary point $x$.
No choice of the extreme point depending on the direction is needed.

For the ordinary lower bound, take any finite illuminating set
$D$ and, for each $S\in\mathcal S(H)$, let $D_S$ consist of its
directions illuminating at least one point of $T_S$. These groups
are pairwise disjoint by Lemma~\ref{lem:hanner-distinct-tori}.
If $d$ illuminates $x\in T_S$, then $H\subseteq\mathbb D^n$
implies $|x_j+sd_j|<1$ for $j\in S$ at some positive step.
Expanding the square gives
\begin{equation}\label{eq:hanner-inward}
 \operatorname{Re}(\overline{x_j}d_j)<0\qquad(j\in S).
\end{equation}
Thus the projections of $D_S$ illuminate every point of $T_S$
in the polydisc $\mathbb D^S$. They also illuminate its entire
boundary by the finite convex-combination argument above, applied
to that polydisc. Every such projection is nonzero; even if some
projections coincide, their number is at most $|D_S|$. The exact
polydisc theorem~\cite[Theorem~1.4]{RSS} therefore gives
\[
 |D|\ge\sum_{S\in\mathcal S(H)}|D_S|
       \ge\sum_{S\in\mathcal S(H)}(2^{|S|+1}-1).
\]

For the matching upper bound, embed an optimal illuminating set
of $\mathbb D^S$ in $E_S$ for each $S\in\mathcal S(H)$, with
zero coordinates outside $S$. It has $2^{|S|+1}-1$ directions.
Lemma~\ref{lem:hanner-extreme} gives $H\cap E_S=\mathbb D^S$,
and Lemma~\ref{lem:section-contraction}(2) identifies the relative
interior of this section with $E_S\cap\operatorname{int}H$.
The embedded directions therefore illuminate $T_S$ in $H$.
Their union illuminates all extreme points, and hence all of
$\partial H$ by \eqref{eq:hanner-extreme-transfer}. This proves
\[
 \ill(H)=\sum_{S\in\mathcal S(H)}(2^{|S|+1}-1)
        =2F_H(2)-F_H(1).
\]

For the fractional lower bound, define the direction sets
\[
 R_S=\bigcup_{x\in T_S}\mathcal I_H(x)
       \qquad(S\in\mathcal S(H)).
\]
They are open and pairwise disjoint. Openness follows because,
for each fixed positive step, the condition
$x+sd\in\operatorname{int}H$ is open jointly in $(x,d)$;
taking the union over steps gives an open illumination relation.
Its sections $\mathcal I_H(x)$ and their unions $R_S$ are open.
Disjointness is Lemma~\ref{lem:hanner-distinct-tori}.

Let $\sigma_S$ be Haar probability measure on $T_S$, that is,
the product of the uniform probability measures on its coordinate
circles. For any fixed direction $d$, the necessary conditions
\eqref{eq:hanner-inward} hold on a set of $\sigma_S$-measure
$2^{-|S|}$ if all $d_j$, $j\in S$, are nonzero, and on an empty
set otherwise. Each nonzero coordinate imposes one open
semicircle, of measure $1/2$. It follows that
\[
 \sigma_S\{x\in T_S:d\in\mathcal I_H(x)\}
       \le2^{-|S|}\mathbf1_{R_S}(d).
\]
Here $\mathbf1_{R_S}$ is the indicator function of $R_S$.
For a fractional illuminating measure $\mu$, the open relation
is Borel measurable, so Tonelli's theorem yields
\begin{align*}
 1
 &\le\int_{T_S}\mu(\mathcal I_H(x))\,d\sigma_S(x)\\
 &=\int_{\C^n\setminus\{0\}}
       \sigma_S\{x\in T_S:d\in\mathcal I_H(x)\}\,d\mu(d)\\
 &\le2^{-|S|}\mu(R_S).
\end{align*}
Summing over the disjoint $R_S$ gives
$\mu(\C^n\setminus\{0\})\ge\sum_S2^{|S|}=F_H(2)$.

For the upper bound, view each $\sigma_S$ instead as a probability
measure on directions in $E_S$, still with unit moduli on $S$
and zeros outside $S$, and put
\[
 \mu=\sum_{S\in\mathcal S(H)}2^{|S|}\sigma_S.
\]
All these directions are nonzero. At any $x\in T_S$, independent
uniform coordinates satisfy \eqref{eq:hanner-inward} with
probability exactly $2^{-|S|}$. For directions supported on $S$,
these inequalities are sufficient: a sufficiently small positive
step enters the relative interior of $\mathbb D^S$, hence the
interior of $H$. Therefore
$2^{|S|}\sigma_S(\mathcal I_H(x))=1$. The measure $\mu$
illuminates every extreme point and, by
\eqref{eq:hanner-extreme-transfer}, every boundary point.
Its total mass is $F_H(2)$, proving the fractional formula.
\end{proof}

\begin{corollary}\label{cor:hanner-equality}
Let $H\subset\C^n$ be a complex Hanner body, and put
\[
 B=\mathbb D\oplus_1\mathbb D
   =\{(z,w)\in\C^2:|z|+|w|\le1\}.
\]
Then $\ill^*(H)\le2^n$, with equality precisely when, after a
coordinate permutation,
\[
 H=\mathbb D^p\times B^r,\qquad
 p,r\in\mathbb Z_{\ge0},\quad p+2r=n.
\]
Factors with exponent zero are omitted. For these bodies,
\begin{equation}\label{eq:hanner-equality-family}
 F_H(t)=2^rt^{p+r},\qquad \ill(H)=2^{n+1}-2^r.
\end{equation}
Moreover, $\ill(H)\le2^{n+1}-1$, with equality precisely for
$H=\mathbb D^n$. For $n\ge2$ and $H\ne\mathbb D^n$,
\[
 \ill(H)\le2^{n+1}-2,
\]
with equality precisely for coordinate permutations of
$\mathbb D^{n-2}\times B$.
\end{corollary}
\begin{proof}
We prove $F_H(2)\le2^n$ by induction, retaining the equality
conditions. It holds for a disc. A product of factors $A_1,A_2$
of dimensions $a,b\ge1$ satisfies
$F_{A_1\times A_2}(2)=F_{A_1}(2)F_{A_2}(2)\le2^{a+b}$, with equality if
and only if both factors attain equality. At a sum,
\[
 F_{A_1\oplus_1A_2}(2)=F_{A_1}(2)+F_{A_2}(2)
        \le2^a+2^b\le2^{a+b}.
\]
The last inequality is equivalent to $2^{-a}+2^{-b}\le1$.
Both terms are at most $1/2$, so equality occurs exactly when
$a=b=1$. In that case the factors are discs and their sum is $B$.
Inductively, equality is possible exactly for products of discs
and copies of $B$, up to coordinate permutation. Conversely,
every such product attains equality. The recursion gives
$F_B(t)=2t$ and then \eqref{eq:hanner-equality-family}.

Let $c_H=F_H(1)=|\mathcal S(H)|$. Its recursion multiplies at a
product and adds at a sum, starting from one for a disc. Thus
$c_H=1$ if and only if $H=\mathbb D^n$: a sum has at least two
supports, while a product has one only when both factors do.
The exact formula therefore gives
\[
 \ill(H)=2F_H(2)-c_H\le2^{n+1}-c_H.
\]
This proves the ordinary bound and its polydisc equality case.
If $H\ne\mathbb D^n$, then $c_H\ge2$, proving the stricter bound.
Equality in that bound implies
\[
 2\bigl(2^n-F_H(2)\bigr)+c_H=2.
\]
The first term is nonnegative and $c_H\ge2$, so
$F_H(2)=2^n$ and $c_H=2$. The classification already proved
gives $H=\mathbb D^p\times B^r$, after a coordinate permutation,
with $c_H=2^r$. Consequently $r=1$ and $p=n-2$.
Conversely, this body attains the stated value by
\eqref{eq:hanner-equality-family}.
\end{proof}

For example, the complex $\ell_1$ unit ball, obtained as the
$\ell_1$ sum of $n$ discs, has $F_H(t)=nt$ and hence
$\ill(H)=3n$ and $\ill^*(H)=2n$. In dimension four,
$B\times B$ has illumination numbers $28$ and $16$, whereas
$\mathbb D^2\times B$ has $30$ and $16$.
All numerical conclusions are preserved by invertible complex
linear maps: directions are mapped bijectively and illuminating
measures are pushed forward with unchanged total mass. The
coordinate descriptions in Corollary~\ref{cor:hanner-equality}
refer to the recursively defined class above; for its linear
images they become statements of complex linear equivalence.


\section{Extreme points with no zero coordinates}\label{sec:full-support}

The condition considered here is the complex counterpart of the
condition in Sun and Vritsiou's Theorem~5~\cite{SV}. Extreme points
are understood in the underlying real vector space.

\begin{theorem}\label{thm:full-support}
Let $K\subset\C^n$ be a normalized complex $1$-unconditional convex
body such that
\begin{equation}\label{eq:full-support-extreme}
 x\in\operatorname{ext}K\quad\Longrightarrow\quad
 \supp(x)=[n].
\end{equation}
Then
\[
 \ill(K)\le2^{n+1}-1,\qquad \ill^*(K)\le2^n.
\]
If $K\ne\mathbb D^n$, put
\[
 \gamma=\max\{s\ge0:s(1,\ldots,1)\in K\}
        =\|(1,\ldots,1)\|_K^{-1}.
\]
Then $0<\gamma<1$ and
\begin{equation}\label{eq:full-support-fractional}
 \ill^*(K)\le
 \left(2^{-n}+
 \frac{\arcsin((1-\gamma)/2)}{\pi\,3^{n-1}}\right)^{-1}<2^n.
\end{equation}
Consequently, within the normalized class defined by
\eqref{eq:full-support-extreme}, equality $\ill^*(K)=2^n$ holds
exactly for $K=\mathbb D^n$.
\end{theorem}

\begin{proof}
We first record the reduction to extreme points for both assertions.
Every boundary point $y$ is a finite convex combination of extreme
points. Choose one summand $x$ with positive coefficient $\lambda$.
If $\lambda<1$, write $y=\lambda x+(1-\lambda)z$ with $z\in K$.
Whenever $x+td\in\operatorname{int}K$, convexity gives
\[
 y+\lambda td=\lambda(x+td)+(1-\lambda)z
       \in\operatorname{int}K.
\]
The case $\lambda=1$ is immediate. Thus every direction illuminating
$x$ illuminates $y$. In particular, a family or measure illuminating
every extreme point illuminates the whole boundary.

Set $N=2^{n+1}-1$ and $\zeta=\exp(2\pi\ii/N)$. The cyclic
construction for the polydisc~\cite[Proposition~2.7]{RSS} gives
the directions
\[
 d^k=-(\zeta^k,\zeta^{2k},\ldots,\zeta^{2^{n-1}k}),
 \qquad 0\le k<N,
\]
with the following property: for every
$\xi\in\mathbb T^n$, where $\mathbb T=\{w\in\C:|w|=1\}$,
some $k$ satisfies
\[
 \operatorname{Re}(\overline{\xi_j}d^k_j)<0
 \qquad(j\in[n]).
\]
Fix an extreme point $x$ and write $x_j=r_j\xi_j$, where
$r_j>0$ and $\xi_j\in\mathbb T$. Every nonzero supporting normal
at $x$ has the form $u_j=b_j\xi_j$, with $b_j\ge0$.
The chosen direction therefore satisfies
\[
 [u,d^k]=\sum_{j\in[n]}b_j
           \operatorname{Re}(\overline{\xi_j}d^k_j)<0
\]
for every such normal. The normal criterion proves the ordinary
bound, followed by the reduction above.

Let $\sigma$ be the product of normalized arc-length measures on
$\mathbb T^n$. For the same extreme point $x$, the set
\[
 A_x=\{d\in\mathbb T^n:
       \operatorname{Re}(\overline{\xi_j}d_j)<0
       \text{ for every }j\in[n]\}
\]
consists of illuminating directions and has measure $2^{-n}$.
Thus $2^n\sigma$ fractionally illuminates $K$.

Assume now that $K\ne\mathbb D^n$. Since $0\in\operatorname{int}K$
and $K\subseteq\mathbb D^n$, we have $0<\gamma\le1$.
If $\gamma=1$, coordinate rotations and contractions of
$(1,\ldots,1)\in K$ give $\mathbb D^n\subseteq K$,
a contradiction. For every $x\in K$, coordinatewise monotonicity
also gives $\min_j|x_j|\le\gamma$.

For each extreme point $x$, choose $j_0$ with $r_{j_0}\le\gamma$.
This choice depends only on $x$, not on a supporting normal.
Since the unit coordinate discs lie in $K$, every nonzero
supporting normal at $x$ satisfies
\[
 b_{j_0}\le\sum_{j\in[n]}r_jb_j,
 \qquad
 (1-\gamma)b_{j_0}
 \le\sum_{j\ne j_0}b_j=:S.
\]
Here $S>0$: if $S=0$, then $\gamma<1$ forces $b_{j_0}=0$,
contrary to $u\ne0$. Define
\[
 B_x=\left\{d\in\mathbb T^n:
 \begin{array}{l}
 0<\operatorname{Re}(\overline{\xi_{j_0}}d_{j_0})
          <(1-\gamma)/2,\\
 \operatorname{Re}(\overline{\xi_j}d_j)<-1/2
          \quad(j\ne j_0)
 \end{array}\right\}.
\]
Every direction in $B_x$ satisfies, for every nonzero supporting
normal at $x$,
\[
 [u,d]<\frac{(1-\gamma)b_{j_0}-S}{2}\le0.
\]
Strictness follows from $S>0$, including when $b_{j_0}=0$.
Thus $B_x$ also consists of illuminating directions.
It is disjoint from $A_x$, and independence of the coordinates gives
\[
 \sigma(B_x)=
 \frac{\arcsin((1-\gamma)/2)}{\pi}\left(\frac13\right)^{n-1}.
\]
Indeed, $\cos\theta<-1/2$ occupies one third of the circle,
and $0<\cos\theta<a$ occupies the proportion $\arcsin(a)/\pi$.
This lower bound for the measure of illuminating directions is
uniform over all extreme points and therefore over the entire
boundary, by the reduction at the start of the proof. The
illumination relation is open in $(x,d)$, so all its direction
sections are Borel. Scaling $\sigma$ by the reciprocal of
$\sigma(A_x)+\sigma(B_x)$ proves
\eqref{eq:full-support-fractional}. Finally,
$\ill^*(\mathbb D^n)=2^n$ follows from
Theorem~\ref{thm:hanner-formulas}, completing the equality assertion.
\end{proof}

The ordinary bound is also the full-support specialization of the
earlier construction by coordinate supports. The fractional estimate
does not give the strict ordinary inequality
$\ill(K)\le2^{n+1}-2$; that assertion remains unresolved here for
the general class \eqref{eq:full-support-extreme}.


\section{An improvement for real $1$-symmetric bodies}\label{sec:real-symmetric-improvement}

In this section, a real $1$-symmetric convex body is invariant under
all coordinate sign changes and permutations. The proper-support
condition in \cite[Proposition~2.9]{RSSSym} admits smaller
families than the $3\cdot2^{n-2}+1$ pairs constructed there.

\begin{theorem}\label{thm:real-symmetric-improvement}
Let $n\ge3$, and let $K\subset\R^n$ be a real $1$-symmetric convex body that is not
an affine image of $[-1,1]^n$. Then
\[
 \ill(K)\le
 \begin{cases}
  6,&n=3,\\
  12,&n=4,\\
  22,&n=5,\\
  5\cdot2^{n-3},&n\ge6.
 \end{cases}
\]
In particular, $\ill(K)\le3\cdot2^{n-2}$ for every $n\ge3$,
so the bound $3\cdot2^{n-2}+1$ is not sharp in any dimension
$n\ge4$.
\end{theorem}

We first give the geometric implication, including a common choice
of the magnitudes of all directions.

\begin{lemma}\label{lem:real-proper-support}
Let $n\ge2$ and let $\mathcal B\subset\{-1,1\}^n\times[n]$
be finite. Suppose that for every nonempty proper $S\subset[n]$
and every $a\in\{-1,1\}^S$ there is $(b,k)\in\mathcal B$ with
\begin{equation}\label{eq:real-proper-support}
 k\in S,\qquad b_j=-a_j\quad(j\in S).
\end{equation}
If $K\subset\R^n$ is $1$-symmetric and is not an affine cube,
then $\ill(K)\le|\mathcal B|$.
\end{lemma}
\begin{proof}
Scale $K$ by one common factor so that $e_j\in\partial K$ for
every $j\in[n]$. Coordinate contractions follow from convexity
and sign invariance. Thus $K\subset[-1,1]^n$, and there is a
unique $\rho>0$ with $\rho(1,\ldots,1)\in\partial K$.
Moreover, $\rho[-1,1]^n\subset K$, whence $0<\rho<1$;
equality would give $K=[-1,1]^n$. Set
\begin{equation}\label{eq:real-symmetric-amplitudes}
 \Delta=\frac1{2n},\qquad
 \delta=\min\left\{\frac\Delta2,
                  \frac{\Delta(1-\rho)}{2\rho}\right\}>0.
\end{equation}
For every $(b,k)\in\mathcal B$ define $d^{b,k}\in\R^n$ by
\begin{equation}\label{eq:real-symmetric-directions}
 d^{b,k}_j=
 \begin{cases}
  b_j,&j=k,\\
  \delta b_j,&j=n\ne k,\\
  \Delta b_j,&j\ne k,\ j\ne n.
 \end{cases}
\end{equation}
These directions are fixed before choosing a boundary point.

Fix $x\in\partial K$. For a nonzero supporting normal $u$ at
$x$, put $c_j=|u_j|$ and $t_j=|x_j|$. Maximizing over the
coordinate sign changes gives
\[
 u_j=c_j\operatorname{sign}(x_j)\quad(x_j\ne0),\qquad
 h_K(u)=\sum_{j=1}^n c_jt_j>0.
\]
At a zero coordinate of $x$, the sign of $u_j$ is unrestricted.
Exchanging the moduli of coordinates $i,j$ of $x$ and then
choosing signs that maximize the same functional gives
\begin{equation}\label{eq:real-normal-order}
 (c_i-c_j)(t_i-t_j)\ge0.
\end{equation}
Indeed, the exchanged point belongs to $K$, and subtracting
its functional value from $h_K(u)$ gives this inequality.

Suppose $M_x\ne[n]$. Choose $(b,k)$ using
\eqref{eq:real-proper-support} with $S=M_x$ and
$a_j=\operatorname{sign}(x_j)$. Write
$\alpha_j=|d^{b,k}_j|$. Then $\alpha_k=1$,
$0<\alpha_j\le\Delta$ for $j\ne k$, and
\eqref{eq:real-normal-order} gives $c_j\le c_k$ for
$j\notin M_x$, including the zero coordinates of $x$.
Consequently
\begin{align*}
 \langle u,d^{b,k}\rangle
 &\le-c_k-\sum_{j\in M_x\setminus\{k\}}\alpha_jc_j
       +\Delta\sum_{j\notin M_x}c_j\\
 &\le-(1-n\Delta)c_k
       -\sum_{j\in M_x\setminus\{k\}}\alpha_jc_j<0.
\end{align*}
For strictness, $1-n\Delta=1/2$. If $c_k=0$, every
coefficient outside $M_x$ vanishes, and some coefficient on
$M_x\setminus\{k\}$ is positive because $u\ne0$.

If $M_x=[n]$, then every $t_j=\rho$. Choose $(b,k)$ using
\eqref{eq:real-proper-support} with $S=[n-1]$ and
$a_j=\operatorname{sign}(x_j)$. Thus $k\ne n$,
$\alpha_j\ge\Delta$ for $j<n$, and $\alpha_n=\delta$.
Since $[-e_n,e_n]\subset K$,
\[
 c_n\le h_K(u)=\rho\sum_{j=1}^n c_j,
 \qquad
 c_n\le\frac\rho{1-\rho}\sum_{j=1}^{n-1}c_j.
\]
The sum on the right is positive. Hence
\[
 \langle u,d^{b,k}\rangle
 \le-\Delta\sum_{j=1}^{n-1}c_j+\delta c_n
 \le-\frac\Delta2\sum_{j=1}^{n-1}c_j<0.
\]
In both cases the pair $(b,k)$ depends on $x$ alone, and the
inequality holds for every nonzero supporting normal at $x$.
The supporting-normal criterion for illumination therefore
shows that the directions \eqref{eq:real-symmetric-directions}
illuminate $K$.
\end{proof}

Let $Q_n$ denote the graph with vertex set $\{0,1\}^n$, in
which two vertices are adjacent when exactly one coordinate
differs. We write $v\oplus e_j$ for the vertex obtained from
$v$ by changing coordinate $j$. A vertex cover is a set meeting
every edge; equivalently, its complement is independent.

\begin{lemma}\label{lem:real-cube-assignment}
Let $A\subset\{0,1\}^n$ be a vertex cover, and let
$T:A\to A$ satisfy
\[
 T(v)=v\oplus e_{k(v)}\qquad(v\in A)
\]
for some $k(v)\in[n]$. Suppose every directed cycle of $T$
uses every coordinate direction. Then the family
\[
 \mathcal B(A,T)=
 \left\{\big((2v_j-1)_{j\in[n]},k(v)\big):v\in A\right\}
\]
satisfies \eqref{eq:real-proper-support} and has cardinality
$|A|$.
\end{lemma}
\begin{proof}
Fix $S$ and $a$ as in \eqref{eq:real-proper-support}. The
vertices satisfying $2v_j-1=-a_j$ for every $j\in S$ form a
proper face $F$ of positive dimension. It contains an edge,
so $A\cap F\ne\varnothing$. If $k(v)\notin S$ for every
$v\in A\cap F$, then $T(A\cap F)\subset A\cap F$.
Iterating $T$ from any vertex of this finite nonempty set
would give a directed cycle contained in $F$. Such a cycle
cannot use any coordinate in $S$, a contradiction. Thus some
$v\in A\cap F$ has $k(v)\in S$, which proves the assertion.
\end{proof}

\begin{lemma}\label{lem:real-cube-extension}
Suppose $A\subset\{0,1\}^n$ is a vertex cover and
$T:A\to A$ sends every vertex to an adjacent vertex. Suppose
$T$ has exactly one directed cycle, and that cycle uses all
$n$ coordinates. Then there is a vertex cover
$A'\subset\{0,1\}^{n+1}$ of cardinality $2|A|$ and an
adjacent-vertex map $T':A'\to A'$ with exactly one directed
cycle, using all $n+1$ coordinates.
\end{lemma}
\begin{proof}
Choose an edge $a\to b$ of the directed cycle with
$b=a\oplus e_1$. Define
\begin{equation}\label{eq:real-cube-extension-set}
 A'=(A\times\{0\})\cup((A\oplus e_1)\times\{1\}),
 \qquad A\oplus e_1=\{v\oplus e_1:v\in A\}.
\end{equation}
Writing $I=\{0,1\}^n\setminus A$, its complement is
$(I\times\{0\})\cup((I\oplus e_1)\times\{1\})$.
Both layers are independent. A vertical edge in the
complement would require $v\in I\cap(I\oplus e_1)$,
contradicting the independence of $I$. Thus $A'$ is a
vertex cover, and $|A'|=2|A|$.

The complete successor assignments are
\begin{equation}\label{eq:real-cube-extension-map}
 \begin{aligned}
 T'(v,0)&=(T(v),0) &&(v\in A\setminus\{a\}),\\
 T'(v\oplus e_1,1)&=(T(v)\oplus e_1,1)
                         &&(v\in A\setminus\{a\}),\\
 T'(a,0)&=(a,1),&\qquad T'(b,1)&=(b,0).
 \end{aligned}
\end{equation}
The two exceptional targets belong to $A'$ because $a,b\in A$
and $a=b\oplus e_1$. All assigned successors are adjacent.
Before the two exceptions, the graph consists of two copies
of the original directed cycle with their incoming trees.
The exceptions remove the edges $(a,0)\to(b,0)$ and
$(b,1)\to(a,1)$ and join the resulting two directed paths
into one cycle. All other vertices still reach this cycle.
The new cycle contains both copies of every original cycle
vertex and uses a vertical edge. It takes both values in
each coordinate, and therefore uses every coordinate
direction. This also supplies an edge in coordinate $1$
for the next application.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:real-symmetric-improvement}]
For the initial sets below, identify $v\in\{0,1\}^m$ with
the integer $\sum_{j=1}^m2^{j-1}v_j$. Thus integers differing
by exclusive-or with $2^{j-1}$ are adjacent in coordinate $j$.
The weight of a vertex is the sum of its binary coordinates.

In dimension three, take $A_3=\{1,2,3,4,5,6\}$ and let
$T_3$ follow the cyclic list
\[
 (1,3,2,6,4,5).
\]
The two missing vertices are $0$ and $7$, so $A_3$ is a
vertex cover. Consecutive vertices, including the last and
first, are adjacent, and the cycle uses all three coordinates.
Repeatedly applying Lemma~\ref{lem:real-cube-extension} gives
$3\cdot2^{n-2}$ vertices for every $n\ge3$.

For dimension five, let $O_5$ be the set of all odd-weight
vertices, and define
\begin{equation}\label{eq:real-five-seed}
 \begin{gathered}
 E_5=\{6,9,12,23,24,27\},\qquad A_5=O_5\cup E_5,\\
 C_5=(4,6,7,23,19,27,26,24,25,9,8,12).
 \end{gathered}
\end{equation}
The complement of $A_5$ consists of even-weight vertices and
is independent. The list $C_5$ is a simple cycle using all
five coordinates and containing every vertex of $E_5$.
Every odd-weight vertex has a neighbor in $E_5$, as seen
from the following complete neighbor lists:
\[
\begin{array}{c|l}
v\in E_5&\{v\oplus e_j:j\in[5]\}\\\hline
6&7,4,2,14,22\\
9&8,11,13,1,25\\
12&13,14,8,4,28\\
23&22,21,19,31,7\\
24&25,26,28,16,8\\
27&26,25,31,19,11
\end{array}
\]
Here and in the next table the entries use the integer
encoding just defined. Their union is exactly $O_5$.
Define $T_5(v)$ to be the next vertex of $C_5$ when $v$
occurs on that cycle, and otherwise to be the smallest
integer in $E_5$ adjacent to $v$. This defines all assignments.
Every vertex outside $C_5$ reaches it in one step. Hence
Lemmas~\ref{lem:real-cube-assignment} and
\ref{lem:real-proper-support} give $\ill(K)\le16+6=22$
when $n=5$.

For dimension six, let $O_6$ be the set of all odd-weight
vertices and define
\begin{equation}\label{eq:real-six-seed}
 \begin{gathered}
 E_6=\{0,5,23,24,43,46,51,60\},\qquad A_6=O_6\cup E_6,\\
 C_6=(0,16,24,56,60,44,46,42,43,59,51,55,23,7,5,1).
 \end{gathered}
\end{equation}
Again $A_6$ is a vertex cover because its complement has
even weight. Consecutive vertices in $C_6$, including the
last and first, are adjacent. This simple cycle uses all six
coordinates and contains all of $E_6$. The neighbor lists are
\[
\begin{array}{c|l}
v\in E_6&\{v\oplus e_j:j\in[6]\}\\\hline
0&1,2,4,8,16,32\\
5&4,7,1,13,21,37\\
23&22,21,19,31,7,55\\
24&25,26,28,16,8,56\\
43&42,41,47,35,59,11\\
46&47,44,42,38,62,14\\
51&50,49,55,59,35,19\\
60&61,62,56,52,44,28
\end{array}
\]
Their union is exactly $O_6$. Define $T_6$ by following $C_6$
on the cycle and, at every other vertex, choosing the smallest
adjacent integer in $E_6$. Thus $|A_6|=32+8=40$, and every
vertex reaches the unique cycle $C_6$.

Starting from $(A_6,T_6)$, the assignments
\eqref{eq:real-cube-extension-set}--\eqref{eq:real-cube-extension-map}
give $40\cdot2^{n-6}=5\cdot2^{n-3}$ vertices in each
dimension $n\ge6$. Apply Lemma~\ref{lem:real-cube-assignment}
and then Lemma~\ref{lem:real-proper-support}. Together with
the dimension-three construction and its first extension,
this proves all the asserted bounds.
\end{proof}

These constructions establish upper bounds for each individual
body. They do not prove that $12$, $22$, or $5\cdot2^{n-3}$ is
attained in the real $1$-symmetric class.

\begin{proposition}\label{prop:real-symmetric-examples}
For $n\ge3$, the real $1$-symmetric bodies
\[
 B_1^n=\left\{x\in\R^n:\sum_{j=1}^n|x_j|\le1\right\},\qquad
 P_n=\left\{x\in[-1,1]^n:\sum_{j=1}^n|x_j|\le n-1\right\}
\]
are not affine cubes, and
\[
 \ill(B_1^n)=2n,\qquad \ill(P_n)=2^{n-1}.
\]
\end{proposition}
\begin{proof}
At a vertex $\sigma e_j$ of $B_1^n$, where $\sigma\in\{-1,1\}$,
a direction $d$ illuminates the vertex if and only if
\[
 \sigma d_j+\sum_{i\ne j}|d_i|<0.
\]
This follows by evaluating the norm along a sufficiently short
positive step. Necessity for any illuminating step follows from
convexity, which permits replacing it by a smaller step.
The inequality forces $|d_j|>\sum_{i\ne j}|d_i|$ and fixes
the sign of $d_j$. One direction therefore illuminates at most
one vertex. The $2n$ coordinate directions illuminate all $2n$
vertices, proving the first equality. We use here the elementary
fact that illuminating every vertex of a polytope illuminates
its boundary: in a convex representation of a boundary point,
move any vertex of positive weight into the interior and keep
the other vertices fixed.

The vertices of $P_n$ are exactly the signed permutations of
$(1,\ldots,1,0)$. To see this, a point with
$\sum_j|x_j|<n-1$ has a coordinate of modulus less than one
that can be perturbed in both directions while preserving both
constraints. If equality holds and two coordinate moduli lie
strictly between zero and one, transfer a small amount from one
modulus to the other, retaining the signs; again the point is
not extreme. There cannot be exactly one nonintegral modulus,
since the sum of the moduli is the integer $n-1$. The remaining
possibility is precisely $n-1$ coordinates of modulus one and
one zero. Conversely, in a convex decomposition of such a point,
the cube constraints fix all its unit coordinates in every term,
and the sum constraint then forces the remaining coordinate to
be zero. Thus each displayed point is a vertex.

Fix a direction $d$ and the index $j$ of the zero coordinate of
one of these vertices. Illumination requires $d_i$ to have sign
opposite to the vertex coordinate for every $i\ne j$, by the
active cube constraints. These requirements determine at most
one vertex with zero coordinate $j$; if a required $d_i$ is
zero there is no such vertex. Consequently $d$ illuminates at
most $n$ vertices. There are $n2^{n-1}$ vertices, so
$\ill(P_n)\ge2^{n-1}$.

For the reverse inequality take
\[
 D=\left\{d\in\{-1,1\}^n:\prod_{j=1}^n d_j=1\right\}.
\]
At a vertex $x$ with $x_j=0$, prescribe $d_i=-x_i$ for $i\ne j$
and choose the unique remaining sign making $d\in D$.
For $0<t<1$,
\[
 \|x+td\|_\infty<1,\qquad
 \sum_{i=1}^n|x_i+td_i|=n-1-(n-2)t<n-1.
\]
Thus $d$ illuminates $x$, and the $2^{n-1}$ members of $D$
illuminate $P_n$. Both bodies are $1$-symmetric directly from
their definitions. Their vertex counts, respectively $2n$
and $n2^{n-1}$, differ from $2^n$ for $n\ge3$, excluding
affine cubes.
\end{proof}

The bound six in dimension three is therefore sharp. In
dimensions four and five the largest illumination number in
the non-cube $1$-symmetric class lies respectively in $[8,12]$
and $[16,22]$. For $n\ge6$ the corresponding interval is
$[2^{n-1},5\cdot2^{n-3}]$. No equality classification at the
new upper bounds is asserted.


\subsection{Asymptotic sharpness}

For $n\ge3$, let $M_n$ be the largest illumination number among
real $1$-symmetric convex bodies in $\R^n$ that are not affine cubes.
The maximum exists: this class is nonempty, and its illumination
numbers are positive integers bounded by
Theorem~\ref{thm:real-symmetric-improvement}.

\begin{theorem}\label{thm:real-symmetric-asymptotic-explicit}
For every $n\ge3$ and every power of two $m=2^r\le n$, $r\ge1$,
\begin{equation}\label{eq:real-asymptotic-explicit}
 M_n\le 2^{n-1}+\frac{2^{n-1}+2^{m-1}}m+n-2.
\end{equation}
Consequently,
\begin{equation}\label{eq:real-asymptotic-sharp}
 M_n\sim2^{n-1}\qquad(n\longrightarrow\infty).
\end{equation}
\end{theorem}
\begin{proof}
Index the first $m$ coordinates by $a\in\mathbb F_2^r$ in
increasing binary order; the first coordinate has label $0$.
For $v\in\mathbb F_2^m$, define
\[
 p(v)=\sum_a v_a\in\mathbb F_2,\qquad
 s(v)=\sum_a v_a a\in\mathbb F_2^r,\qquad
 C=\{v:p(v)=0,\ s(v)=0\}.
\]
The columns $(1,a)$ span $\mathbb F_2^{r+1}$: they include
$(1,0)$, and adding this column to $(1,e_i)$ gives $(0,e_i)$.
Thus
\[
 c:=|C|=2^{m-r-1}=\frac{2^{m-1}}m.
\]
Every odd-weight vertex $u$ of $Q_m$ has precisely one neighbor
in $C$, namely
\[
 q(u)=u\oplus e_{s(u)}.
\]
Flipping that coordinate changes its parity and makes its syndrome
zero. Conversely, the syndrome specifies the only coordinate whose
flip can do so. Distinct elements of $C$ have Hamming distance at
least four: the distance is even, and a difference in exactly two
distinct coordinates would have nonzero syndrome.

The graph on the odd vertices of $Q_m$ joining vertices at Hamming
distance two is connected. Indeed, partition the coordinates in
which two odd vertices differ into pairs and flip one pair at a time.
For adjacent vertices $u,u'$ in this graph,
\[
 d_H(q(u),q(u'))\le1+2+1=4.
\]
Every element of $C$ is the image of an odd neighbor. It follows
that the graph on $C$ joining pairs at distance four is connected:
images of the preceding paths have consecutive vertices either
equal or at distance four. Choose its breadth-first spanning tree
rooted at $0$, resolving choices by increasing binary order.
For each tree edge from a parent $v$ to a child $w$, let $a,b$ be
the first two coordinate labels in which they differ, and define
\[
 w_{vw}=v\oplus e_a\oplus e_b.
\]
This vertex is even and has distance two from both endpoints.
Hence at most $c-1$ additional even vertices connect the centers
in $C$ by paths with odd intervening vertices. When $c=1$ the
tree has no edges and no such vertices are needed.

Put $\ell=n-m$. For $z\in\mathbb F_2^\ell$, write $p(z)$ for its
parity, taking $p(z)=0$ if $\ell=0$. Define
\[
 E=\{(v\oplus p(z)e_0,z):v\in C,\ z\in\mathbb F_2^\ell\},
 \qquad
 W=\{(w_{vw},0):vw\text{ is a tree edge}\}.
\]
Both sets consist of even vertices, and
\[
 |E|=2^\ell c=\frac{2^{n-1}}m,\qquad |W|\le c-1.
\]
Let $O_n$ denote all odd vertices of $Q_n$. Every $(u,z)\in O_n$
is adjacent to $E$: flip the first-block coordinate labeled $s(u)$.
The resulting first block has syndrome zero and parity $p(z)$,
so adding $p(z)e_0$ gives an element of $C$.

For each fixed $v\in C$, changing one coordinate of $z$ changes
exactly two coordinates of $(v\oplus p(z)e_0,z)$. The corresponding
even vertices are therefore joined by a path of length two with
an odd intermediate vertex. The vertices in $W$ connect these
sets through their copies with $z=0$. Every odd vertex has a
neighbor in $E$, so the induced graph on
\[
 A_0=O_n\cup E\cup W
\]
is connected.

In the ordinary coordinate order $1,\ldots,n$, define
\begin{equation}\label{eq:real-full-coordinate-cycle}
 v^i=\sum_{h=1}^{i}e_h\quad(0\le i\le n),\qquad
 v^{n+i}=\sum_{h=i+1}^{n}e_h\quad(1\le i\le n-1).
\end{equation}
The list $(v^0,\ldots,v^{2n-1})$, closed by $v^{2n}=v^0$,
is a simple cycle: successive edges flip coordinates
$1,\ldots,n,1,\ldots,n$. It has $n$ even vertices, including
$v^0=0\in E$. Put
\[
 A=A_0\cup\{v^0,\ldots,v^{2n-1}\}.
\]
All odd cycle vertices already belong to $O_n$, and hence
\[
 |A|\le2^{n-1}+2^\ell c+(c-1)+(n-1)
      =2^{n-1}+\frac{2^{n-1}+2^{m-1}}m+n-2.
\]
The complement of $A$ consists of even vertices and is independent,
so $A$ is a vertex cover. Its induced graph is connected, since
every added even cycle vertex has an odd neighbor in $O_n$.

Let $h(u)$ be graph distance in $Q_n[A]$ from $u$ to the cycle
vertex set $V=\{v^0,\ldots,v^{2n-1}\}$.
Define the successor map by
\begin{equation}\label{eq:real-distance-successor}
 \begin{aligned}
 T(v^i)&=v^{i+1} &&(i\bmod 2n),\\
 T(u)&=\min\{w\in A:w\sim u,\ h(w)=h(u)-1\}
   &&(u\in A\setminus V),
 \end{aligned}
\end{equation}
where the minimum uses binary order. Connectedness ensures that
the set is nonempty. Distances decrease until the displayed cycle
is reached, so it is the only directed cycle, and it uses all
coordinates. Lemmas~\ref{lem:real-cube-assignment} and
\ref{lem:real-proper-support} prove
\eqref{eq:real-asymptotic-explicit}.

Choose the largest power of two $m\le n$. Since $m>n/2$,
\[
 1\le\frac{M_n}{2^{n-1}}
 \le1+\frac2m+\frac{n-2}{2^{n-1}}
 <1+\frac4n+\frac{n-2}{2^{n-1}}.
\]
The lower bound is supplied by $P_n$ in
Proposition~\ref{prop:real-symmetric-examples}. The two bounds
give \eqref{eq:real-asymptotic-sharp}.
\end{proof}

The next argument improves the error term using a published
graph theorem. A set $D$ dominates a graph if every vertex outside
$D$ has a neighbor in $D$. Write $\gamma_c(Q_d)$ for the minimum
size of such a set whose induced graph is connected.

\begin{proposition}\label{prop:real-connected-domination}
For every $n\ge3$,
\begin{equation}\label{eq:real-connected-domination}
 M_n\le2^{n-1}+\gamma_c(Q_{n-1})+n-1.
\end{equation}
Consequently,
\begin{equation}\label{eq:real-asymptotic-error}
 M_n\le2^{n-1}\left(1+\frac{1+o(1)}n\right)
 \qquad(n\longrightarrow\infty).
\end{equation}
\end{proposition}
\begin{proof}
Choose a minimum connected dominating set $D\subset Q_{n-1}$.
Translate it so that $0\in D$, and put
\[
 E_D=\{(y,p(y)):y\in D\}\subset Q_n.
\]
Every odd vertex has the form $(z,1-p(z))$. If $z\in D$,
its last-coordinate neighbor belongs to $E_D$. Otherwise take
a neighbor $y\in D$ of $z$. Then $p(y)=1-p(z)$, so the
corresponding $(y,p(y))\in E_D$ is again adjacent to the odd
vertex. Adjacent elements of $D$ lift to even vertices at distance
two and are joined by a path with an odd intermediate vertex.
Thus $O_n\cup E_D$ is connected.

Add the cycle \eqref{eq:real-full-coordinate-cycle}. At most
$n-1$ further even vertices are needed, since $0\in E_D$.
Its union with $O_n\cup E_D$ is a connected vertex cover, so
the successor map \eqref{eq:real-distance-successor} and the
same two lemmas prove \eqref{eq:real-connected-domination}.
Griggs proves
\[
 \gamma_c(Q_d)=(1+o(1))\frac{2^d}{d}
 \qquad(d\longrightarrow\infty)
\]
in \cite[Theorems~2 and~4]{Griggs}.
Substitute $d=n-1$ and use $n-1=o(2^{n-1}/n)$ to obtain
\eqref{eq:real-asymptotic-error}.
\end{proof}

The sufficient proper-support condition has a stricter lower
bound than the known lower examples for individual bodies.

\begin{proposition}\label{prop:real-sign-family-size}
Let $t_n$ be the least size of a family
$\mathcal B\subset\{-1,1\}^n\times[n]$ satisfying
\eqref{eq:real-proper-support}. Then, for $n\ge3$,
\[
 \left\lceil\frac{n2^{n-1}}{n-1}\right\rceil
 \le t_n\le2^{n-1}+\gamma_c(Q_{n-1})+n-1.
\]
In particular,
\[
 t_n=2^{n-1}\left(1+\frac1n+o\!\left(\frac1n\right)\right).
\]
\end{proposition}
\begin{proof}
Consider only supports $S=[n]\setminus\{j\}$. There are
$n2^{n-1}$ requirements $(S,a)$. A fixed pair $(b,k)$ meets
exactly $n-1$ of them: the omitted coordinate can be any
$j\ne k$, and then $a=-b|_S$ is prescribed. Counting incidences
gives $(n-1)|\mathcal B|\ge n2^{n-1}$. This remains valid
when several labels are used at the same sign vector, since
they are distinct pairs.

The preceding construction supplies a family satisfying
\eqref{eq:real-proper-support} of size at most the stated upper
bound. Griggs's asymptotic formula then gives
\[
 \frac{2^{n-1}}{n-1}
 \le t_n-2^{n-1}
 \le (1+o(1))\frac{2^{n-1}}{n-1}+n-1,
\]
which proves the assertion.
\end{proof}

The asymptotic equality for $t_n$ does not imply the same
second-order term for $M_n$: the sign-family condition is
sufficient for illumination, and need not describe an optimal
illuminating family for a particular body. In particular,
\eqref{eq:real-asymptotic-sharp} settles the leading asymptotic
value, while the exact maxima $M_n$ for fixed $n\ge4$ remain
undetermined here.


\section{Research status}\label{sec:corpus-ledger}

The following table records the results and remaining claims in the research corpus.
\begin{center}
\begin{tabular}{p{0.48\textwidth}|p{0.22\textwidth}|p{0.20\textwidth}}
Claim & Present status & Treatment here\\
\hline
Sharp bounds $6$ in $\C^2$ and $14$ in $\C^3$ for normalized non-polydiscs
& internally verified, including sharpness
& Corollary~\ref{cor:full-section-low-dimensions} and
Theorem~\ref{thm:three-dimensions}\\
A non-polydisc containing a full coordinate $(n-1)$-polydisc has
$\ill(K)\le 2^{n+1}-2$
& internally verified, including sharpness
& full proof in Section~\ref{sec:full-coordinate-section}\\
Every normalized non-polydisc in $\C^4$ has $\ill(K)\le30$
& internally verified; one covering assertion computationally certified
& full proof in Section~\ref{sec:four-dimensions}; exact verifier in
Appendix~\ref{app:c4-verifier}\\
Every non-polydisc complex $1$-symmetric body in $\C^3$ has $\ill(K)\le12$
& internally verified; finite covering computationally certified
& full proof and certificate in Section~\ref{sec:symmetric-three}\\
Real $1$-symmetric non-cubes: bounds $12$, $22$, and $5\cdot2^{n-3}$
& internally verified; explicit finite starting families checked exactly
& full proof in Section~\ref{sec:real-symmetric-improvement}\\
Real $1$-symmetric non-cubes: $M_n\sim2^{n-1}$
& internally verified; self-contained proof of the leading term
& Theorem~\ref{thm:real-symmetric-asymptotic-explicit}; sharper error in
Proposition~\ref{prop:real-connected-domination}\\
Exact ordinary and fractional formulas for complex Hanner bodies
& internally verified, including equality cases
& full proof in Section~\ref{sec:hanner}\\
Failure of one cyclic construction in dimension five
& reported failure of this construction & verification pending
\end{tabular}
\end{center}

The ordinary complex $1$-unconditional problem is settled here
through dimension four. Its sharp dimension-only bound remains open
in this corpus in dimensions at least five. The general estimates proved above are weaker than the conjectured sharp bound.
For bodies whose extreme points have no zero coordinates,
Theorem~\ref{thm:full-support} gives the conjectured ordinary upper
bound and a strict fractional bound for non-polydiscs. The strict
ordinary inequality $\ill(K)\le2^{n+1}-2$ remains unresolved here
for that class beyond the cases already covered above.
The source inventory, theorem dependency ledger, and dated research log accompanying this
file record the provenance and audit state of every item.

\appendix
\section{One family for all normalized bodies}\label{app:simultaneous}

Let $U_n$ be the least cardinality of a finite family of directions
that illuminates every normalized complex $1$-unconditional convex
body in $\C^n$. The family must be chosen independently of the body.
This is a stronger requirement than bounding $\ill(K)$ separately
for each $K$.

\begin{proposition}\label{prop:simultaneous}
Put $m=\lceil2n/3\rceil$ and
\[
 L_n=(2^{m+1}-1)\binom nm,\qquad
 B_n=L_n+\sum_{k=m+1}^n\binom nk.
\]
With $C_n$ defined in \eqref{eq:support-bound},
\begin{equation}\label{eq:universal-sandwich}
 L_n\le U_n\le C_n\le B_n,
\end{equation}
and
\begin{equation}\label{eq:universal-asymptotic}
 U_n\sim\frac{3^{n+1}}{\sqrt{\pi n}}.
\end{equation}
\end{proposition}
\begin{proof}
The directions in the proof of Theorem~\ref{thm:support-bound}
depend only on $n$, the finite matchings and cyclic vectors, and their
common positive margin. The normal estimates use only normalization
of $K$. Hence the same family works for every normalized $K$, proving
$U_n\le C_n$.

For every $m$-set $S\subset[n]$, when $m<n$, consider the normalized body
\[
 K_S=\{z\in\C^n:
       \|z_S\|_\infty+\|z_{S^c}\|_\infty\le1\}.
\]
At a point $x$ of its coordinate torus
\[
 |x_j|=1\ (j\in S),\qquad x_j=0\ (j\notin S),
\]
the directional derivative of its norm is
\[
 \max_{j\in S}\operatorname{Re}(\overline{x_j}d_j)
       +\|d_{S^c}\|_\infty.
\]
Illumination forces this derivative to be negative: for a convex function, an interior point along the ray forces a negative one-sided derivative at its initial point. Consequently
\[
 \operatorname{Re}(\overline{x_j}d_j)
       <-\|d_{S^c}\|_\infty\quad(j\in S),
 \qquad
 \min_{j\in S}|d_j|>\max_{j\notin S}|d_j|.
\]
Thus the $m$ largest coordinate magnitudes of $d$ determine $S$ uniquely. The groups of directions serving these tori for different $m$-sets are disjoint. Within the group for $S$, projection must illuminate the $m$-polydisc torus, and hence the polydisc, requiring at least $2^{m+1}-1$ directions by the exact polydisc illumination theorem~\cite{RSS}. Summation gives $U_n\ge L_n$. If $m=n$, the polydisc itself gives this conclusion.

Finally, $B_n-L_n\le2^n$. Stirling's formula, with $m=2n/3+O(1)$, yields
\[
 2^{m+1}\binom nm\sim\frac{3}{\sqrt{\pi n}}\,3^n.
\]
The same asymptotic holds for $C_n=B_n-(n-m)$; squeezing between the lower and upper bounds proves the claim.


\end{proof}

\begin{corollary}
\[
 U_1=3,\quad U_2=7,\quad U_3=21,\quad U_4=60,\quad U_5=155.
\]
\end{corollary}
\begin{proof}
For $n=1,2$, $B_n-L_n=0$ and $n-m=0$. For $3\le n\le5$,
$B_n-L_n=1=n-m$. Thus $C_n=L_n$ in all five dimensions, and
\eqref{eq:universal-sandwich} gives the stated values.
\end{proof}

\begin{center}
\begin{tabular}{r|r|r|r}
$n$ & $2\cdot3^n-2^n-1$ & $C_n$ & $L_n$\\
\hline
2 & 13 & 7 & 7\\
3 & 45 & 21 & 21\\
4 & 145 & 60 & 60\\
5 & 453 & 155 & 155\\
6 & 1393 & 470 & 465\\
8 & 12865 & 3563 & 3556\\
10 & 117073 & 30653 & 30600\\
15 & 28665045 & 6149077 & 6147141\\
20 & 6972520225 & 1270070614 & 1270048920
\end{tabular}
\end{center}

These are general estimates, not assertions of the best possible bound for
each individual body or each dimension. In particular, exact values of $U_n$
concern the simultaneous normalized problem. The same asymptotic
$C_n\sim 3^{n+1}/\sqrt{\pi n}$ holds, and the final additive refinement does
not affect it.



\clearpage
\section{The four-coordinate finite verification}\label{app:c4-verifier}

The following Python program verifies the finite assertions
\eqref{eq:c4-integer-first} and \eqref{eq:c4-integer-second}.
The finite reduction to these assertions is proved in
Lemma~\ref{lem:c4-finite-construction}. All sets of rows are
represented by integer bit masks; every loop ranges over the
entire finite set indicated. The program requires only Python's
standard library. Running it with Python 3.12.14 gives
\texttt{125054 923521 9341 [0, 0, 0, 1320, 8021]}.

\begin{verbatim}
from itertools import combinations, product

m = [1,2,1,4,3,2,3,4,1,4,1,4,3,2,4,
     2,1,2,4,2,3,3,1,3,2,3,4,1,2,1]
q = [4,4,2,3,4,3,2,2,4,2,3,2,1,4,1,
     1,4,1,3,1,2,4,2,1,3,4,3,3,4,4]
assert len(m) == len(q) == 30
assert all(1 <= a <= 4 and 1 <= b <= 4 and a != b
           for a, b in zip(m, q))
w = [[sum(1 << (k-1) for k in range(1,31)
          if ((1 << j)*k-a) % 31 < 15)
      for a in range(31)] for j in range(4)]
main = [sum(1 << k for k in range(30) if m[k] == j+1)
        for j in range(4)]
tiny = [sum(1 << k for k in range(30) if q[k] == j+1)
        for j in range(4)]

tested = 0
for size in (1,2,3):
    for S in combinations(range(4), size):
        allowed = 0
        for j in S:
            allowed |= main[j]
        for starts in product(range(31), repeat=size):
            candidates = (1 << 30)-1
            for j, a in zip(S, starts):
                candidates &= w[j][a]
            assert candidates & allowed
            tested += 1
assert tested == 125054

holes = 0
histogram = [0]*5
for a,b,c,d in product(*w):
    if a & b & c & d:
        continue
    holes += 1
    candidates = [b&c&d, a&c&d, a&b&d, a&b&c]
    count = sum(bool(candidates[j] & tiny[j])
                for j in range(4))
    assert count >= 3
    histogram[count] += 1
assert holes == 9341
assert histogram == [0,0,0,1320,8021]
print(tested, 31**4, holes, histogram)
\end{verbatim}


\clearpage
\section{Numerical comparison through dimension 100}
\label{sec:comparison-table}
The table compares the following estimates. Here $c<c_*$ is the positive
rational approximation recorded in the accompanying data file, and $b=1/c$:
\begin{align*}
 A_n&=2\cdot3^n-2^n-1,\\
 C_n&=(2^{m+1}-1)\binom nm+\sum_{k=m+1}^{n}\binom nk-(n-m),
       \qquad m=\lceil2n/3\rceil,\\
 E_n&=\left\lceil e b^n
       \left[2n\log\bigl(1+32b^2(n+1)^2\bigr)+1\right]\right\rceil,\\
 T_n&=\left\lceil
       \frac{2n\log\bigl(1+32/(c^2\varepsilon_n^2)\bigr)+1}
       {[c(1-\varepsilon_n)]^n}\right\rceil.
\end{align*}
Here $A_n$ is the sum of the bounds for individual supports,
$C_n$ is the bound in Theorem~\ref{thm:support-bound},
$E_n$ evaluates the explicit formula with $b=1/c$, and $T_n$ is \eqref{eq:eps} with a
numerically chosen value of $\varepsilon_n$.
The chosen parameters approximately minimize the real-valued expression
before taking its ceiling; validity needs only $0<\varepsilon_n<1$.

The first dimension in this table with $E_n<C_n$ is $n=51$;
the first with $T_n<C_n$ is $n=47$.
Each remains smaller through $n=100$, the end of this table.
The smaller of $C_n$ and $T_n$ is bold.
The last column is $T_n/C_n$: values below one mean $T_n<C_n$.
These compare the displayed estimates, not necessarily the best known bounds
in each individual dimension.

Numbers below $10^7$ are printed exactly. Larger counts are rounded
\emph{upward} to six significant digits for display, so the displayed
counts remain upper bounds; ratios are ordinary rounded approximations.
Admissibility of the rational $c$ was checked by exact rational lower bounds
on $\cos(\pi c)\cosh(\pi(1-c))$. The formulas remain valid for this
smaller value of $c$.
Integer ceilings agree when evaluated at 120 and 170 decimal digits.
The integer evaluations are numerical and are not used in the proofs.
Full computed counts and the chosen $\varepsilon_n$ are recorded in
the accompanying data file.

\begingroup
\footnotesize
\setlength{\tabcolsep}{3pt}
\renewcommand{\arraystretch}{1.12}
\begin{longtable}{r r r r r r}
\hline
$n$ & $A_n$ & $C_n$ & $E_n$ & $T_n$ & $T_n/C_n$\\
\hline
\endfirsthead
\hline
$n$ & $A_n$ & $C_n$ & $E_n$ & $T_n$ & $T_n/C_n$\\
\hline
\endhead
\hline
\multicolumn{6}{r}{Continued on the next page}\\
\endfoot
\hline
\endlastfoot
1 & $3$ & $\boldsymbol{3}$ & $97$ & $56$ & $18.67$\\
2 & $13$ & $\boldsymbol{7}$ & $518$ & $308$ & $44$\\
3 & $45$ & $\boldsymbol{21}$ & $2047$ & $1221$ & $58.14$\\
4 & $145$ & $\boldsymbol{60}$ & $7103$ & $4225$ & $70.42$\\
5 & $453$ & $\boldsymbol{155}$ & $22891$ & $13555$ & $87.45$\\
6 & $1393$ & $\boldsymbol{470}$ & $70353$ & $41470$ & $88.23$\\
7 & $4245$ & $\boldsymbol{1329}$ & $209211$ & $122774$ & $92.38$\\
8 & $12865$ & $\boldsymbol{3563}$ & $607225$ & $354861$ & $99.6$\\
9 & $38853$ & $\boldsymbol{10711}$ & $1729986$ & $1007086$ & $94.02$\\
10 & $117073$ & $\boldsymbol{30653}$ & $4856875$ & $2817201$ & $91.91$\\
11 & $352245$ & $\boldsymbol{84379}$ & $1.34743\times10^{7}$ & $7789603$ & $92.32$\\
12 & $1058785$ & $\boldsymbol{253240}$ & $3.70155\times10^{7}$ & $2.13328\times10^{7}$ & $84.24$\\
13 & $3180453$ & $\boldsymbol{731819}$ & $1.00851\times10^{8}$ & $5.79540\times10^{7}$ & $79.19$\\
14 & $9549553$ & $\boldsymbol{2049513}$ & $2.72846\times10^{8}$ & $1.56369\times10^{8}$ & $76.3$\\
15 & $2.86651\times10^{7}$ & $\boldsymbol{6149077}$ & $7.33709\times10^{8}$ & $4.19428\times10^{8}$ & $68.21$\\
16 & $8.60280\times10^{7}$ & $\boldsymbol{1.78895\times10^{7}}$ & $1.96264\times10^{9}$ & $1.11929\times10^{9}$ & $62.57$\\
17 & $2.58150\times10^{8}$ & $\boldsymbol{5.06892\times10^{7}}$ & $5.22569\times10^{9}$ & $2.97352\times10^{9}$ & $58.66$\\
18 & $7.74579\times10^{8}$ & $\boldsymbol{1.52071\times10^{8}}$ & $1.38569\times10^{10}$ & $7.86814\times10^{9}$ & $51.74$\\
19 & $2.32400\times10^{9}$ & $\boldsymbol{4.44521\times10^{8}}$ & $3.66100\times10^{10}$ & $2.07461\times10^{10}$ & $46.67$\\
20 & $6.97253\times10^{9}$ & $\boldsymbol{1.27008\times10^{9}}$ & $9.64068\times10^{10}$ & $5.45279\times10^{10}$ & $42.93$\\
21 & $2.09187\times10^{10}$ & $\boldsymbol{3.81023\times10^{9}}$ & $2.53123\times10^{11}$ & $1.42909\times10^{11}$ & $37.51$\\
22 & $6.27580\times10^{10}$ & $\boldsymbol{1.11768\times10^{10}}$ & $6.62813\times10^{11}$ & $3.73568\times10^{11}$ & $33.42$\\
23 & $1.88278\times10^{11}$ & $\boldsymbol{3.21332\times10^{10}}$ & $1.73137\times10^{12}$ & $9.74214\times10^{11}$ & $30.32$\\
24 & $5.64843\times10^{11}$ & $\boldsymbol{9.63995\times10^{10}}$ & $4.51251\times10^{12}$ & $2.53514\times10^{12}$ & $26.3$\\
25 & $1.69455\times10^{12}$ & $\boldsymbol{2.83529\times10^{11}}$ & $1.17370\times10^{13}$ & $6.58396\times10^{12}$ & $23.22$\\
26 & $5.08367\times10^{12}$ & $\boldsymbol{8.19082\times10^{11}}$ & $3.04702\times10^{13}$ & $1.70679\times10^{13}$ & $20.84$\\
27 & $1.52511\times10^{13}$ & $\boldsymbol{2.45725\times10^{12}}$ & $7.89653\times10^{13}$ & $4.41713\times10^{13}$ & $17.98$\\
28 & $4.57534\times10^{13}$ & $\boldsymbol{7.24241\times10^{12}}$ & $2.04313\times10^{14}$ & $1.14136\times10^{14}$ & $15.76$\\
29 & $1.37261\times10^{14}$ & $\boldsymbol{2.10030\times10^{13}}$ & $5.27839\times10^{14}$ & $2.94491\times10^{14}$ & $14.02$\\
30 & $4.11782\times10^{14}$ & $\boldsymbol{6.30090\times10^{13}}$ & $1.36176\times10^{15}$ & $7.58813\times10^{14}$ & $12.04$\\
31 & $1.23535\times10^{15}$ & $\boldsymbol{1.86027\times10^{14}}$ & $3.50857\times10^{15}$ & $1.95276\times10^{15}$ & $10.5$\\
32 & $3.70604\times10^{15}$ & $\boldsymbol{5.41168\times10^{14}}$ & $9.02878\times10^{15}$ & $5.01937\times10^{15}$ & $9.275$\\
33 & $1.11182\times10^{16}$ & $\boldsymbol{1.62351\times10^{15}}$ & $2.32076\times10^{16}$ & $1.28875\times10^{16}$ & $7.938$\\
34 & $3.33544\times10^{16}$ & $\boldsymbol{4.79993\times10^{15}}$ & $5.95884\times10^{16}$ & $3.30547\times10^{16}$ & $6.886$\\
35 & $1.00064\times10^{17}$ & $\boldsymbol{1.39998\times10^{16}}$ & $1.52846\times10^{17}$ & $8.46976\times10^{16}$ & $6.05$\\
36 & $3.00190\times10^{17}$ & $\boldsymbol{4.19994\times10^{16}}$ & $3.91676\times10^{17}$ & $2.16825\times10^{17}$ & $5.163$\\
37 & $9.00568\times10^{17}$ & $\boldsymbol{1.24319\times10^{17}}$ & $1.00279\times10^{18}$ & $5.54581\times10^{17}$ & $4.461$\\
38 & $2.70171\times10^{18}$ & $\boldsymbol{3.63392\times10^{17}}$ & $2.56518\times10^{18}$ & $1.41730\times10^{18}$ & $3.9$\\
39 & $8.10511\times10^{18}$ & $\boldsymbol{1.09018\times10^{18}}$ & $6.55652\times10^{18}$ & $3.61925\times10^{18}$ & $3.32$\\
40 & $2.43154\times10^{19}$ & $\boldsymbol{3.23015\times10^{18}}$ & $1.67454\times10^{19}$ & $9.23531\times10^{18}$ & $2.859$\\
41 & $7.29460\times10^{19}$ & $\boldsymbol{9.45971\times10^{18}}$ & $4.27366\times10^{19}$ & $2.35493\times10^{19}$ & $2.489$\\
42 & $2.18838\times10^{20}$ & $\boldsymbol{2.83792\times10^{19}}$ & $1.08994\times10^{20}$ & $6.00083\times10^{19}$ & $2.115$\\
43 & $6.56514\times10^{20}$ & $\boldsymbol{8.41588\times10^{19}}$ & $2.77789\times10^{20}$ & $1.52815\times10^{20}$ & $1.816$\\
44 & $1.96955\times10^{21}$ & $\boldsymbol{2.46866\times10^{20}}$ & $7.07545\times10^{20}$ & $3.88916\times10^{20}$ & $1.575$\\
45 & $5.90863\times10^{21}$ & $\boldsymbol{7.40598\times10^{20}}$ & $1.80107\times10^{21}$ & $9.89217\times10^{20}$ & $1.336$\\
46 & $1.77259\times10^{22}$ & $\boldsymbol{2.19791\times10^{21}}$ & $4.58201\times10^{21}$ & $2.51470\times10^{21}$ & $1.144$\\
47 & $5.31777\times10^{22}$ & $6.45634\times10^{21}$ & $1.16505\times10^{22}$ & $\boldsymbol{6.38921\times10^{21}}$ & $0.9896$\\
48 & $1.59533\times10^{23}$ & $1.93691\times10^{22}$ & $2.96073\times10^{22}$ & $\boldsymbol{1.62251\times10^{22}}$ & $0.8377$\\
49 & $4.78599\times10^{23}$ & $5.75201\times10^{22}$ & $7.52029\times10^{22}$ & $\boldsymbol{4.11826\times10^{22}}$ & $0.716$\\
50 & $1.43580\times10^{24}$ & $1.69177\times10^{23}$ & $1.90924\times10^{23}$ & $\boldsymbol{1.04481\times10^{23}}$ & $0.6176$\\
51 & $4.30739\times10^{24}$ & $5.07531\times10^{23}$ & $4.84487\times10^{23}$ & $\boldsymbol{2.64950\times10^{23}}$ & $0.522$\\
52 & $1.29222\times10^{25}$ & $1.50809\times10^{24}$ & $1.22889\times10^{24}$ & $\boldsymbol{6.71587\times10^{23}}$ & $0.4453$\\
53 & $3.87665\times10^{25}$ & $4.44049\times10^{24}$ & $3.11567\times10^{24}$ & $\boldsymbol{1.70161\times10^{24}}$ & $0.3832$\\
54 & $1.16300\times10^{26}$ & $1.33215\times10^{25}$ & $7.89606\times10^{24}$ & $\boldsymbol{4.30966\times10^{24}}$ & $0.3235$\\
55 & $3.48899\times10^{26}$ & $3.96044\times10^{25}$ & $2.00031\times10^{25}$ & $\boldsymbol{1.09109\times10^{25}}$ & $0.2755$\\
56 & $1.04670\times10^{27}$ & $1.16729\times10^{26}$ & $5.06543\times10^{25}$ & $\boldsymbol{2.76130\times10^{25}}$ & $0.2366$\\
57 & $3.14009\times10^{27}$ & $3.50186\times10^{26}$ & $1.28226\times10^{26}$ & $\boldsymbol{6.98571\times10^{25}}$ & $0.1995$\\
58 & $9.42026\times10^{27}$ & $1.04158\times10^{27}$ & $3.24472\times10^{26}$ & $\boldsymbol{1.76668\times10^{26}}$ & $0.1696$\\
59 & $2.82608\times10^{28}$ & $3.07266\times10^{27}$ & $8.20785\times10^{26}$ & $\boldsymbol{4.46643\times10^{26}}$ & $0.1454$\\
60 & $8.47824\times10^{28}$ & $9.21797\times10^{27}$ & $2.07557\times10^{27}$ & $\boldsymbol{1.12882\times10^{27}}$ & $0.1225$\\
61 & $2.54347\times10^{29}$ & $2.74291\times10^{28}$ & $5.24691\times10^{27}$ & $\boldsymbol{2.85199\times10^{27}}$ & $0.104$\\
62 & $7.63041\times10^{29}$ & $8.09811\times10^{28}$ & $1.32598\times10^{28}$ & $\boldsymbol{7.20349\times10^{27}}$ & $0.08895$\\
63 & $2.28913\times10^{30}$ & $2.42944\times10^{29}$ & $3.34992\times10^{28}$ & $\boldsymbol{1.81891\times10^{28}}$ & $0.07487$\\
64 & $6.86737\times10^{30}$ & $7.23180\times10^{29}$ & $8.46070\times10^{28}$ & $\boldsymbol{4.59151\times10^{28}}$ & $0.06349$\\
65 & $2.06022\times10^{31}$ & $2.13667\times10^{30}$ & $2.13627\times10^{29}$ & $\boldsymbol{1.15873\times10^{29}}$ & $0.05423$\\
66 & $6.18064\times10^{31}$ & $6.41001\times10^{30}$ & $5.39245\times10^{29}$ & $\boldsymbol{2.92342\times10^{29}}$ & $0.04561$\\
67 & $1.85419\times10^{32}$ & $1.90876\times10^{31}$ & $1.36082\times10^{30}$ & $\boldsymbol{7.37376\times10^{29}}$ & $0.03863$\\
68 & $5.56257\times10^{32}$ & $5.64328\times10^{31}$ & $3.43322\times10^{30}$ & $\boldsymbol{1.85943\times10^{30}}$ & $0.03295$\\
69 & $1.66878\times10^{33}$ & $1.69299\times10^{32}$ & $8.65953\times10^{30}$ & $\boldsymbol{4.68773\times10^{30}}$ & $0.02769$\\
70 & $5.00632\times10^{33}$ & $5.04293\times10^{32}$ & $2.18365\times10^{31}$ & $\boldsymbol{1.18153\times10^{31}}$ & $0.02343$\\
71 & $1.50190\times10^{34}$ & $1.49187\times10^{33}$ & $5.50511\times10^{31}$ & $\boldsymbol{2.97733\times10^{31}}$ & $0.01996$\\
72 & $4.50568\times10^{34}$ & $4.47560\times10^{33}$ & $1.38756\times10^{32}$ & $\boldsymbol{7.50087\times10^{31}}$ & $0.01676$\\
73 & $1.35171\times10^{35}$ & $1.33355\times10^{34}$ & $3.49653\times10^{32}$ & $\boldsymbol{1.88931\times10^{32}}$ & $0.01417$\\
74 & $4.05512\times10^{35}$ & $3.94730\times10^{34}$ & $8.80907\times10^{32}$ & $\boldsymbol{4.75777\times10^{32}}$ & $0.01205$\\
75 & $1.21654\times10^{36}$ & $1.18419\times10^{35}$ & $2.21887\times10^{33}$ & $\boldsymbol{1.19789\times10^{33}}$ & $0.01012$\\
76 & $3.64961\times10^{36}$ & $3.52935\times10^{35}$ & $5.58782\times10^{33}$ & $\boldsymbol{3.01536\times10^{33}}$ & $0.008544$\\
77 & $1.09489\times10^{37}$ & $1.04523\times10^{36}$ & $1.40691\times10^{34}$ & $\boldsymbol{7.58889\times10^{33}}$ & $0.00726$\\
78 & $3.28465\times10^{37}$ & $3.13569\times10^{36}$ & $3.54165\times10^{34}$ & $\boldsymbol{1.90958\times10^{34}}$ & $0.00609$\\
79 & $9.85393\times10^{37}$ & $9.34791\times10^{36}$ & $8.91379\times10^{34}$ & $\boldsymbol{4.80413\times10^{34}}$ & $0.005139$\\
80 & $2.95618\times10^{38}$ & $2.76975\times10^{37}$ & $2.24305\times10^{35}$ & $\boldsymbol{1.20842\times10^{35}}$ & $0.004363$\\
81 & $8.86853\times10^{38}$ & $8.30925\times10^{37}$ & $5.64334\times10^{35}$ & $\boldsymbol{3.03906\times10^{35}}$ & $0.003657$\\
82 & $2.66056\times10^{39}$ & $2.47767\times10^{38}$ & $1.41958\times10^{36}$ & $\boldsymbol{7.64168\times10^{35}}$ & $0.003084$\\
83 & $7.98168\times10^{39}$ & $7.34452\times10^{38}$ & $3.57029\times10^{36}$ & $\boldsymbol{1.92118\times10^{36}}$ & $0.002616$\\
84 & $2.39451\times10^{40}$ & $2.20336\times10^{39}$ & $8.97794\times10^{36}$ & $\boldsymbol{4.82918\times10^{36}}$ & $0.002192$\\
85 & $7.18351\times10^{40}$ & $6.57141\times10^{39}$ & $2.25725\times10^{37}$ & $\boldsymbol{1.21370\times10^{37}}$ & $0.001847$\\
86 & $2.15506\times10^{41}$ & $1.94877\times10^{40}$ & $5.67430\times10^{37}$ & $\boldsymbol{3.04988\times10^{37}}$ & $0.001565$\\
87 & $6.46516\times10^{41}$ & $5.84629\times10^{40}$ & $1.42619\times10^{38}$ & $\boldsymbol{7.66279\times10^{37}}$ & $0.001311$\\
88 & $1.93955\times10^{42}$ & $1.74398\times10^{41}$ & $3.58406\times10^{38}$ & $\boldsymbol{1.92499\times10^{38}}$ & $0.001104$\\
89 & $5.81865\times10^{42}$ & $5.17380\times10^{41}$ & $9.00552\times10^{38}$ & $\boldsymbol{4.83511\times10^{38}}$ & $0.0009345$\\
90 & $1.74560\times10^{43}$ & $1.55214\times10^{42}$ & $2.26246\times10^{39}$ & $\boldsymbol{1.21430\times10^{39}}$ & $0.0007823$\\
91 & $5.23678\times10^{43}$ & $4.63098\times10^{42}$ & $5.68314\times10^{39}$ & $\boldsymbol{3.04916\times10^{39}}$ & $0.0006584$\\
92 & $1.57104\times10^{44}$ & $1.37436\times10^{43}$ & $1.42737\times10^{40}$ & $\boldsymbol{7.65558\times10^{39}}$ & $0.000557$\\
93 & $4.71311\times10^{44}$ & $4.12306\times10^{43}$ & $3.58447\times10^{40}$ & $\boldsymbol{1.92185\times10^{40}}$ & $0.0004661$\\
94 & $1.41394\times10^{45}$ & $1.23038\times10^{44}$ & $9.00026\times10^{40}$ & $\boldsymbol{4.82396\times10^{40}}$ & $0.0003921$\\
95 & $4.24180\times10^{45}$ & $3.65268\times10^{44}$ & $2.25959\times10^{41}$ & $\boldsymbol{1.21069\times10^{41}}$ & $0.0003315$\\
96 & $1.27254\times10^{46}$ & $1.09581\times10^{45}$ & $5.67215\times10^{41}$ & $\boldsymbol{3.03815\times10^{41}}$ & $0.0002773$\\
97 & $3.81762\times10^{46}$ & $3.27055\times10^{45}$ & $1.42368\times10^{42}$ & $\boldsymbol{7.62311\times10^{41}}$ & $0.0002331$\\
98 & $1.14529\times10^{47}$ & $9.71253\times10^{45}$ & $3.57292\times10^{42}$ & $\boldsymbol{1.91251\times10^{42}}$ & $0.0001969$\\
99 & $3.43586\times10^{47}$ & $2.91376\times10^{46}$ & $8.96564\times10^{42}$ & $\boldsymbol{4.79761\times10^{42}}$ & $0.0001647$\\
100 & $1.03076\times10^{48}$ & $8.69779\times10^{46}$ & $2.24952\times10^{43}$ & $\boldsymbol{1.20337\times10^{43}}$ & $0.0001384$\\
\end{longtable}
\endgroup

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\end{thebibliography}

\end{document}