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\title{Illumination of complex 1-unconditional convex bodies}
\author{Working research draft}
\date{1 November 2026}
\begin{document}
\maketitle
\begin{abstract}
We study ordinary and fractional illumination of complex
$1$-unconditional convex bodies. In every complex dimension $n$ we
give an explicit ordinary bound asymptotic to
$3^{n+1}/\sqrt{\pi n}$, and prove
$\ill^*(K)\le b_*^n$ and
$\ill(K)=O(n\log(n+1)b_*^n)$, where
$b_*=2.4805534697\ldots$ is defined by an explicit scalar equation.
For normalized non-polydiscs in dimensions two, three and four,
the sharp ordinary bounds are $6$, $14$ and $30$.
More generally, a full coordinate $(n-1)$-polydisc gives the sharp
bound $2^{n+1}-2$ in every dimension $n\ge2$.
In dimension two we prove the fractional bound $4$ and classify
equality by the two maximal supporting coefficients at the coordinate
axes. If every extreme point has all coordinates nonzero, we obtain
the fractional bound $2^n$, with equality exactly for the polydisc,
and a quantitative strict bound otherwise. Appendices give the exact
sizes of simultaneous illuminating families in dimensions one through
five, the exact finite verification used in dimension four, and a
numerical comparison of the general estimates through dimension $100$.
\end{abstract}

\noindent\textit{Draft status.} The included proofs have been checked
internally. The citations were audited on 1 November 2026; external
review and a definitive priority assessment remain outstanding.
The finite verification used in dimension four is exact. The numerical
comparison is supplementary and is not used in the proofs.

\section{Introduction}\label{sec:introduction}

A complex $1$-unconditional convex body in $\C^n$ is invariant under
independent rotations of its coordinates. Its ordinary illumination
number $\ill(K)$ is the least number of directions that move every
boundary point into the interior. The fractional number $\ill^*(K)$
allows a measure on directions, requiring illuminating mass at least
one at each boundary point. Both notions concern the underlying real
space of dimension $2n$; their precise definitions are given in
Section~\ref{sec:preliminaries}.

Fractional illumination was introduced by Nasz\'odi~\cite{Naszodi}
using finitely supported measures. We use the Borel-measure formulation
of~\cite[Section~1.2]{RSS}. Weighted covering and its relation to
ordinary covering were developed in~\cite{AAS}. In real dimension
$d$, the centrally symmetric fractional bound is $2^d$; in $\C^n$
this gives $4^n$, rather than the complex benchmark $2^n$.

Rotem, Schejter and Slomka proved
$\ill(\mathbb D^n)=2^{n+1}-1$ and
$\ill^*(\mathbb D^n)=2^n$~\cite[Theorems~1.3--1.4]{RSS}.
The ordinary formula was stated earlier by Boltyanski and
Martini~\cite{BM}; the correction of their upper-bound construction
is explained in~\cite[Remark~2.8]{RSS}.
We establish general bounds for the unconditional class and sharp
results in low dimensions and under a coordinate-section condition.
The fractional results distinguish the full-support extreme-point
class from the general two-dimensional class, where non-polydisc
equality cases also occur.

Sun and Vritsiou proved related real results: bounds for
non-parallelepipeds in $\R^3$ and $\R^4$, the full coordinate-hyperplane case,
and the case of extreme points with no zero coordinates
\cite[Theorems~1, 2 and~5]{SV}. Their work on $1$-symmetric bodies
also develops the use of a large inward coordinate with small
remaining coordinates~\cite{SVSym}. These are real predecessors;
the complex assertions below require independent coordinate rotations
and estimates in complex dimension $n$, whose real dimension is $2n$.
The theorem numbers cited from~\cite{SV} refer to its 2024 arXiv version.

The coauthored manuscript~\cite{RSSSym} already gives the bound
$2\cdot3^n-2^n-1$ for complex $1$-unconditional bodies
(Theorem~1.8). Its Theorems~1.5--1.7 treat the $1$-symmetric subclass,
including the ordinary bound $2^{n+1}-2$ for non-polydiscs and the
planar fractional bound $4$. Those results assume
coordinate-permutation invariance. We cite its cyclic projection and
coloring lemmas where they enter the arguments here.

Our first general estimate is Theorem~\ref{thm:support-bound}:
with $m=\lceil2n/3\rceil$,
\[
 \ill(K)\le C_n
 =(2^{m+1}-1)\binom nm+\sum_{k=m+1}^{n}\binom nk-(n-m),
 \qquad C_n\sim\frac{3^{n+1}}{\sqrt{\pi n}}.
\]
Its proof shares directions along trees of coordinate supports,
improving the separate-support construction of~\cite[Theorem~1.8]{RSSSym}.
A different construction gives Theorem~\ref{thm:main}:
\[
 \ill^*(K)\le b_*^n,\qquad
 \ill(K)\le\left\lceil e b_*^n
 \left[2n\log\bigl(1+32b_*^2(n+1)^2\bigr)+1\right]\right\rceil,
\]
where $b_*=c_*^{-1}$ and $c_*\in(0,1/2)$ is determined by
\[
 \cos(\pi c_*)\cosh(\pi(1-c_*))=1.
\]
The two arguments and their auxiliary estimates are kept together in
Section~\ref{sec:general-bounds}.

After positive diagonal normalization, a body contains each unit
coordinate disc and is contained in $\mathbb D^n$. If it contains
a full coordinate $(n-1)$-polydisc and is not $\mathbb D^n$,
Theorem~\ref{thm:full-coordinate-section} gives
$\ill(K)\le2^{n+1}-2$. Its sharpness examples are
\[
 H_n=\mathbb D^{n-2}\times
 \{(z,w)\in\C^2:|z|+|w|\le1\}.
\]
The coordinate-section hypothesis is automatic in dimension two.
Theorems~\ref{thm:three-dimensions} and~\ref{thm:four-dimensions}
remove it in dimensions three and four. Consequently the largest
ordinary illumination numbers of normalized non-polydiscs in
dimensions two, three and four are $6$, $14$ and $30$; when the
polydisc is included, they are $7$, $15$ and $31$.

For fractional illumination, Theorem~\ref{thm:planar-fractional-equality}
proves $\ill^*(K)\le4$ for every normalized unconditional body in
$\C^2$, with equality exactly when $K=\mathbb D^2$ or
$a(K)b(K)\ge1/2$. The coefficients $a(K)$ and $b(K)$ are the
maximal coefficients in the supporting inequalities
$|z_1|+a(K)|z_2|\le1$ and $b(K)|z_1|+|z_2|\le1$ on $K$.
In particular, the already defined body $H_2$ has fractional number
$4$ without being a linear image of the bidisc; see the remark after
Theorem~\ref{thm:planar-fractional-equality}. This contradicts the
equality characterization in~\cite[Conjecture~1.2]{RSS}, while
remaining consistent with its proposed upper bound.
In arbitrary dimension, Theorem~\ref{thm:full-support} proves that
if every extreme point has all coordinates nonzero, then
$\ill^*(K)\le2^n$, with equality exactly for $\mathbb D^n$.
It also gives an explicit improvement below $2^n$ for each
normalized non-polydisc in that class.

The main estimates allow the directions to depend on the body.
Appendix~\ref{app:simultaneous} considers the distinct quantity $U_n$,
the smallest size of one family illuminating every normalized body,
and proves $U_1=3$, $U_2=7$, $U_3=21$, $U_4=60$ and $U_5=155$.
Appendix~\ref{app:c4-verifier} supplies the finite verification used
in dimension four, and Appendix~\ref{sec:comparison-table} compares
the general bounds numerically.

\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 $K\subseteq\mathbb D^n$.

For completeness we include the polydisc calculation
from~\cite[Theorem~1.3]{RSS}.
\begin{lemma}\label{lem:fractional-polydisc}
For every $n\ge1$, $\ill^*(\mathbb D^n)=2^n$.
\end{lemma}
\begin{proof}
Let $\mathbb T=\{z\in\C:|z|=1\}$ and let $\sigma$ be product
normalized arc-length measure on $\mathbb T^n$. At a boundary point
$x$ of the polydisc, a direction $d$ illuminates $x$ precisely when
\[
 \operatorname{Re}(\overline{x_j}d_j)<0
 \qquad\text{for every }j\text{ with }|x_j|=1.
\]
Indeed, expansion of $|x_j+td_j|^2$ proves necessity at each active
coordinate and sufficiency for all sufficiently small positive $t$;
coordinates with modulus less than one remain interior for such $t$.
If there are $k$ active coordinates, the displayed conditions have
$\sigma$-probability $2^{-k}\ge2^{-n}$. Thus $2^n\sigma$
fractionally illuminates the entire polydisc.

For the reverse bound, regard $\sigma$ as a measure on the boundary
torus. A fixed direction illuminates $\sigma$-mass $2^{-n}$ if all
its coordinates are nonzero, and zero otherwise, by the same strict
coordinate inequalities. If $\mu$ is a fractional illuminating
measure, Tonelli's theorem for the open illumination relation gives
\[
 1\le\int_{\mathbb T^n}\mu\{d:d\text{ illuminates }x\}\,d\sigma(x)
 =\int\sigma\{x:d\text{ illuminates }x\}\,d\mu(d)
 \le2^{-n}\mu(\C^n\setminus\{0\}).
\]
Taking the infimum over $\mu$ proves the claim.
\end{proof}

\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.

The following projection property is
\cite[Lemma~4.3]{RSSSym}; we recall its short proof.
\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~\cite{Hall}, 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}

\begin{remark}
The directions just constructed work for every normalized body.
Appendix~\ref{app:simultaneous} determines the minimum size under this
stronger requirement in dimensions one through five. 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}
The passage from a covering measure to finitely many directions uses
the standard sampling and net argument; compare the weighted-covering
rounding theorem and its proof in~\cite[Theorem~1.6 and Section~2.6]{AAS}.
We give the argument with the interior margin from Lemma~\ref{lem:finite},
which determines the constants here.
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
$\mathbb D^{n-1}\times\{0\}\subseteq K$.

The real counterpart of the following upper bound is
\cite[Theorem~2]{SV}. Here the section is a complex polydisc, and
the sharp comparison value is $2^{n+1}-2$.
\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}

The low-dimensional real counterpart is~\cite[Theorem~1]{SV}.
Within the complex $1$-symmetric subclass, the following upper bound
also follows from~\cite[Theorem~1.5]{RSSSym}; the assertion here
requires no coordinate-permutation invariance.
\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}

As in dimension three, related bounds hold for real unconditional
bodies~\cite[Theorem~1]{SV} and for the complex $1$-symmetric
subclass~\cite[Theorem~1.5]{RSSSym}. We treat all complex
$1$-unconditional bodies, including those without permutation symmetry.
\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{Fractional illumination in two complex dimensions}\label{sec:fractional-equality}

We give the fractional bound and its complete equality criterion in
complex dimension two. The upper bound for the $1$-symmetric
subclass appears in~\cite[Theorem~1.7]{RSSSym}; the criterion here
allows different supporting coefficients at the two axes.
All bodies in the following theorem are normalized.

\begin{theorem}\label{thm:planar-fractional-equality}
Let $K\subset\C^2$ be a normalized complex $1$-unconditional convex body, and define
\begin{align}
 a(K)&=\max\{a\ge0:|z_1|+a|z_2|\le1\text{ for every }z\in K\},\notag\\
 b(K)&=\max\{b\ge0:b|z_1|+|z_2|\le1\text{ for every }z\in K\}.
 \label{eq:planar-axis-slopes}
\end{align}
Then $\ill^*(K)\le4$, and
\begin{equation}\label{eq:planar-fractional-equality}
 \ill^*(K)=4
 \quad\Longleftrightarrow\quad
 K=\mathbb D^2\ \text{ or }\ a(K)b(K)\ge\frac12.
\end{equation}
\end{theorem}
\begin{proof}
Write $a=a(K)$ and $b=b(K)$. The feasible sets in
\eqref{eq:planar-axis-slopes} are closed, contain zero and are
contained in $[0,1]$, by the coordinate discs in $K$. Thus the
maxima exist. Let $\sigma_j$ denote uniform probability measure
on the circle $\{\xi e_j:|\xi|=1\}$.

A nonzero supporting normal at $e_1$ has the form $(s,w)$,
where $s>0$ and $|w|\le as$. Indeed, its support value is $s$;
coordinate rotations then give
$s|z_1|+|w||z_2|\le s$ throughout $K$. Conversely, every vector
$(1,w)$ with $|w|\le a$ supports $K$ at $e_1$. The normal
criterion, followed by a coordinate rotation, gives
\begin{align}
 d\in\mathcal I_K(\xi e_1)
 &\quad\Longleftrightarrow\quad
 \operatorname{Re}(\overline\xi d_1)+a|d_2|<0,\notag\\
 d\in\mathcal I_K(\xi e_2)
 &\quad\Longleftrightarrow\quad
 \operatorname{Re}(\overline\xi d_2)+b|d_1|<0,
 \label{eq:planar-axis-illumination}
\end{align}
where $\mathcal I_K(x)$ is the set of directions illuminating $x$.

Suppose first that $K\ne\mathbb D^2$. We claim that
$\mu_0=2\sigma_1+2\sigma_2$ fractionally illuminates $K$.
A boundary point with one zero coordinate belongs to one of the
two unit axis circles, where the inward half of that same direction
circle has $\mu_0$-mass one. Now let
$x=(r_1\xi_1,r_2\xi_2)\in\partial K$, with $r_1,r_2>0$ and
$|\xi_1|=|\xi_2|=1$. At least one coordinate, say $r_j$, is
strictly less than one; otherwise unconditionality would give
$K=\mathbb D^2$. Write $k$ for the other index. Every nonzero
supporting normal has the form $u_l=c_l\xi_l$, $c_l\ge0$.
If $c_k=0$, then $c_j>0$ and
$h_K(u)=r_jc_j<c_j$, contradicting the coordinate disc in $K$.
Thus $c_k>0$ for every such normal. Every direction on the
$k$th coordinate circle with
$\operatorname{Re}(\overline{\xi_k}d_k)<0$ illuminates $x$.
This again gives mass one and proves the claim. For
$K=\mathbb D^2$, the value four follows from
Lemma~\ref{lem:fractional-polydisc}.

Assume $ab\ge1/2$. Regard
$\nu=2\sigma_1+2\sigma_2$ as a measure on $\partial K$.
For a direction with both coordinates nonzero, put
$t=|d_2|/|d_1|$. By \eqref{eq:planar-axis-illumination}, the
proportions of the two axis circles illuminated by $d$ are
\[
 q_1=\begin{cases}\pi^{-1}\arccos(at),&at<1,\\0,&at\ge1,\end{cases}
 \qquad
 q_2=\begin{cases}\pi^{-1}\arccos(b/t),&b/t<1,\\0,&b/t\ge1.\end{cases}
\]
If either proportion is zero, their sum is at most $1/2$.
Otherwise set $v=at$ and $w=b/t$. Then
$v^2+w^2\ge2vw=2ab\ge1$, so
$\arccos v+\arccos w\le\pi/2$. Hence $q_1+q_2\le1/2$
in every case. If one coordinate of $d$ is zero, it illuminates
half of the corresponding axis circle and none of the other one,
since $a,b>0$. Consequently every direction illuminates
$\nu$-mass at most one. For any fractional illuminating measure
$\mu$, Tonelli's theorem for the open illumination relation gives
\begin{equation}\label{eq:planar-dual-measure}
 4\le\int_{\partial K}\mu(\mathcal I_K(x))\,d\nu(x)
 =\int\nu\{x:d\in\mathcal I_K(x)\}\,d\mu(d)
 \le\mu(\C^2\setminus\{0\}).
\end{equation}
Together with the upper bound this proves equality.

It remains to prove strictness when $K\ne\mathbb D^2$ and
$ab<1/2$. Choose $t>0$ such that
\[
 \sqrt2\,b<t<\frac{1}{\sqrt2\,a},
\]
where the right endpoint means infinity if $a=0$.
Let $\sigma$ be the probability measure on
$d=(\eta_1,t\eta_2)$, with independent uniform unit complex
numbers $\eta_1,\eta_2$. The two axis-circle illumination
probabilities are
\[
 p_1=\pi^{-1}\arccos(at)>\frac14,\qquad
 p_2=\pi^{-1}\arccos(b/t)>\frac14.
\]
We obtain a uniform strict improvement at all remaining boundary
points. Fix, in this order,
\[
 \gamma=\max\{s\ge0:(s,s)\in K\}<1,\qquad
 m=\min\{1,t\},\qquad M=\max\{1,t\},\qquad
 c=\frac{(1-\gamma)m}{2}.
\]
The strict inequality for $\gamma$ follows from
$K\ne\mathbb D^2$. At a full-support boundary point
$x=(r_1\xi_1,r_2\xi_2)$ choose $j$ with $r_j\le\gamma$,
which exists by coordinatewise monotonicity, and let $k\ne j$.
Every supporting normal as above satisfies
\[
 c_j\le h_K(u)=r_jc_j+r_kc_k\le\gamma c_j+c_k,
 \qquad (1-\gamma)c_j\le c_k.
\]
In particular $c_k>0$ for every nonzero normal. Directions
with both aligned real parts negative illuminate $x$, and have
$\sigma$-probability $1/4$. A disjoint additional set is
\[
 0<\operatorname{Re}(\overline{\xi_j}d_j)<c,
 \qquad
 \operatorname{Re}(\overline{\xi_k}d_k)<-m/2.
\]
On this set, for every nonzero supporting normal,
\[
 [u,d]<c c_j-\frac m2c_k\le0.
\]
The first inequality is strict because $c_k>0$. Its probability
is at least
\[
 \varepsilon=\frac{1}{3\pi}
      \arcsin\!\left(\frac{(1-\gamma)m}{2M}\right)>0:
\]
the first coordinate condition has probability at least
$\pi^{-1}\arcsin(c/M)$, and the second at least $1/3$.
Both bounds are independent of $x$ and of the chosen index.
Thus every boundary point is illuminated with probability at least
\[
 p=\min\{p_1,p_2,1/4+\varepsilon\}>1/4.
\]
The measure $p^{-1}\sigma$ fractionally illuminates $K$ and has
mass $p^{-1}<4$, completing the proof.
\end{proof}

\begin{remark}\label{rem:planar-conjecture}
For $H_2=\{(z,w):|z|+|w|\le1\}$, the supporting coefficients are
$a(H_2)=b(H_2)=1$. Theorem~\ref{thm:planar-fractional-equality}
therefore gives $\ill^*(H_2)=4$. This is a counterexample to the
equality clause of~\cite[Conjecture~1.2]{RSS}.
Indeed,
\[
 \operatorname{ext}H_2=(\mathbb T\times\{0\})
                       \cup(\{0\}\times\mathbb T),
 \qquad \operatorname{ext}(\mathbb D^2)=\mathbb T^2,
 \qquad \mathbb T=\{z\in\C:|z|=1\}.
\]
A boundary point of $H_2$ with both coordinates nonzero is the
nontrivial convex combination, with weights $|z|$ and $|w|$, of
$(z/|z|,0)$ and $(0,w/|w|)$. Every point on either displayed axis
circle is extreme: its unit coordinate forces the same coordinate
in both terms of any convex decomposition, and then the other
coordinate must vanish. In the bidisc, extremality is equivalent
to both coordinates being extreme in the disc, which gives the torus.
The former extreme-point set is disconnected and the latter is
connected. Since an invertible affine map carries extreme points
homeomorphically onto extreme points, these bodies are not affinely
equivalent. This conclusion concerns the equality clause only.
\end{remark}

\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. Having no zero
complex coordinate does not mean having no zero real coordinate
in $\R^{2n}$, so that real theorem does not directly imply the
fractional bound below.

\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
\[
 \ill^*(K)\le2^n.
\]
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 reduce the fractional assertion to extreme points, using
the convex-combination argument of~\cite[Lemma~2.1]{RSS}.
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.

Fix an extreme point $x$ and write $x_j=r_j\xi_j$, where
$r_j>0$ and $\xi_j\in\mathbb T$, with
$\mathbb T=\{w\in\C:|w|=1\}$. Every nonzero supporting normal
at $x$ has the form $u_j=b_j\xi_j$, with $b_j\ge0$.

Let $\sigma$ be the product of normalized arc-length measures on
$\mathbb T^n$. For the same extreme point $x$, the set
\[
 A_x=\{d\in\mathbb T^n:
       \operatorname{Re}(\overline{\xi_j}d_j)<0
       \text{ for every }j\in[n]\}
\]
has measure $2^{-n}$. For $d\in A_x$, every nonzero supporting normal
satisfies
\[
 [u,d]=\sum_{j\in[n]}b_j
       \operatorname{Re}(\overline{\xi_j}d_j)<0,
\]
because its coefficients are nonnegative and not all zero.
Thus every direction in $A_x$ illuminates $x$.
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
Lemma~\ref{lem:fractional-polydisc}, completing the equality assertion.
\end{proof}

\appendix
\section{Simultaneous families in dimensions one through five}\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 differs from bounding $\ill(K)$ separately for each $K$.

\begin{proposition}\label{prop:simultaneous-small}
\[
 U_1=3,\quad U_2=7,\quad U_3=21,\quad U_4=60,\quad U_5=155.
\]
\end{proposition}
\begin{proof}
Fix $n\in\{1,2,3,4,5\}$, put $m=\lceil2n/3\rceil$, and write
$L_n=(2^{m+1}-1)\binom nm$. We show that
$L_n\le U_n\le C_n$, where $C_n$ is defined in
\eqref{eq:support-bound}.

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.

For $n=1,2$ we have $m=n$, so $C_n=L_n$.
For $3\le n\le5$ we have $m=n-1$ and
\[
 C_n-L_n=\sum_{k=m+1}^n\binom nk-(n-m)=1-1=0.
\]
Thus $U_n=L_n=C_n$ in these five dimensions. Substitution gives
the stated values.
\end{proof}

\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 separate-support bound
from~\cite[Theorem~1.8]{RSSSym},
$C_n$ 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. This rounding
does not decrease the computed counts; 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.
Agreement at two precisions is not interval certification of each
integer ceiling. 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

\begin{thebibliography}{9}
\bibitem{AAS}
S.~Artstein-Avidan and B.~A.~Slomka,
\emph{On weighted covering numbers and the Levi-Hadwiger conjecture},
Israel J. Math. \textbf{209} (2015), no.~1, 125--155.
\href{https://doi.org/10.1007/s11856-015-1213-5}{doi:10.1007/s11856-015-1213-5}.
For the locators used here, see
\href{https://arxiv.org/abs/1310.7892v2}{arXiv:1310.7892v2}.

\bibitem{BM}
V.~Boltyanski and H.~Martini,
\emph{Illumination of direct vector sums of convex bodies},
Studia Sci. Math. Hungar. \textbf{44} (2007), no.~3, 367--376.
\href{https://doi.org/10.1556/SScMath.44.2007.3.6}{doi:10.1556/SScMath.44.2007.3.6}.

\bibitem{Hall}
P.~Hall,
\emph{On representatives of subsets},
J. London Math. Soc. \textbf{s1-10} (1935), no.~1, 26--30.
\href{https://doi.org/10.1112/jlms/s1-10.37.26}{doi:10.1112/jlms/s1-10.37.26}.

\bibitem{Naszodi}
M.~Nasz\'odi,
\emph{Fractional illumination of convex bodies},
Contrib. Discrete Math. \textbf{4} (2009), no.~2, 83--88.
\href{https://doi.org/10.55016/ojs/cdm.v4i2.62022}{doi:10.55016/ojs/cdm.v4i2.62022}.

\bibitem{RSS}
L.~Rotem, A.~Schejter, and B.~A.~Slomka,
\emph{The complex Illumination problem},
Combinatorica \textbf{46}, article~3 (2026).
\href{https://doi.org/10.1007/s00493-025-00195-7}{doi:10.1007/s00493-025-00195-7}.
The theorem, proposition and remark numbers cited here refer to
the published article.

\bibitem{RSSSym}
L.~Rotem, A.~Schejter and B.~Slomka,
\emph{Illumination of Real and Complex 1-Symmetric Convex Bodies},
unpublished manuscript, October 2026, 22 pages.
The locators used here refer to the version with Theorems~1.5--1.8,
Lemma~4.3 and Proposition~4.4.

\bibitem{SV}
W.~R.~Sun and B.-H.~Vritsiou,
\emph{Illuminating $1$-unconditional convex bodies in $\mathbb R^3$
and $\mathbb R^4$, and certain cases in higher dimensions},
Canad. J. Math., accepted manuscript posted online 4 August 2025.
\href{https://doi.org/10.4153/S0008414X25101260}{doi:10.4153/S0008414X25101260}.
The theorem numbers used here refer to
\href{https://arxiv.org/abs/2407.11331v1}{arXiv:2407.11331v1} (2024).

\bibitem{SVSym}
W.~R.~Sun and B.-H.~Vritsiou,
\emph{On the illumination of $1$-symmetric convex bodies},
\href{https://arxiv.org/abs/2407.10314v1}{arXiv:2407.10314v1} (2024).
\end{thebibliography}

\end{document}