\section{Binary set reconstruction on
\texorpdfstring{$C_{2pq}$}{C2pq}}\label{sec:set-reconstruction}

For a subset $A\subset C_n$ write $f_A=\mathbf1_A$ and
$S_A=\supp\widehat f_A$.  Its analytical ordered $k$-deck is exactly the
autocorrelation $M_k(f_A)$ in (3.1); order counts factors, and equivalence in
this subsection means translation only.  Let $\rho_T(A)$ be the least $d$
such that equality of the exact $d$-deck forces a translate of $A$.  This is
the analytical convention used by Keleti--Kolountzakis and, after fixing one
argument at the identity, by Simon \cite{KeletiKolountzakis2006,Simon2017}.

For a nonempty pattern $T\subset C_n$, write $[T]$ for its translation class,
$\mu_A([T])$ for the number of distinct subsets of $A$ in that class, and
$\operatorname{Stab}(T)=\{h\in C_n:T+h=T\}$.  Pebody's 2007 cumulative $d$-deck is the
multiset encoded by the values $\mu_A([T])$ for $1\le |T|\le d$.

\begin{lemma}[Pebody/analytical deck equivalence]
\label{lem:pebody-deck-crosswalk}
Let $1\le d\le n$.  For every $A\subset C_n$, including $A=\varnothing$ and
$|A|<d$, the exact ordered analytical $d$-deck determines Pebody's cumulative
combinatorial $d$-deck and conversely.  More precisely, if
$T=\{0,t_1,\ldots,t_{j-1}\}$ has $j$ distinct elements, then
\[
 M_j(f_A;t_1,\ldots,t_{j-1})
   =|\operatorname{Stab}(T)|\,\mu_A([T]).
\]
Pebody's 2004 count-valued set deck records the left-hand side directly; the
2007 necklace convention records the right-hand multiplicity without the
known stabilizer factor \cite{Pebody2004,Pebody2007}.
\end{lemma}

\begin{proof}
The analytical value counts the translations $x\in C_n$ for which
$x+T\subset A$.  Every distinct translated copy of $T$ contained in $A$ has
exactly $|\operatorname{Stab}(T)|$ such translation witnesses, proving the displayed
identity.  An exact $d$-deck gives every lower order by repeating the zero
shift.  Conversely, in any ordered $d$-tuple, repeated positions collapse to
their set of distinct positions because $\mathbf1_A^m=\mathbf1_A$; the
displayed identity then recovers its analytical value from the cumulative
combinatorial deck.  The same argument gives zero on both sides for every
nonempty pattern when $A=\varnothing$, and it requires no lower bound on
$|A|$.
\end{proof}

The exact deck becomes cumulative for indicators.  If $1\le k\le d$, then
\[
 M_k(f_A;t_1,\ldots,t_{k-1})
 =M_d(f_A;t_1,\ldots,t_{k-1},0,\ldots,0),                 \tag{11.1}
\]
because $f_A^m=f_A$.  This zero-shift identity is false for general weighted
signals and is the first load-bearing binary input.

\begin{proposition}[Supportwise binary crosswalk]
\label{prop:binary-crosswalk}
Every nonempty $A\subset C_n$ satisfies
\[
                         \rho_T(A)\le\depth(S_A).           \tag{11.2}
\]
Equality need not hold.
\end{proposition}

\begin{proof}
Put $d=\depth(S_A)$.  If $d=1$, then $S_A=\{0\}$: a nonzero frequency has a
positive-order kernel relation that degree-one rows cannot generate, and the
zero frequency splits off.  Hence the nonempty binary function $f_A$ is
constant and is reconstructed at order one.

Assume $d\ge2$ and that $B$ has the same exact $d$-deck.  Formula (11.1)
gives equality of the second decks, so Lemma~\ref{lem:support} gives
$S_B=S_A=:S$.  On this common exact support set
$r_s=\widehat f_B(s)/\widehat f_A(s)$.  The Fourier monomial identity and
(11.1) say that $r$ kills $\calL_{\le d}(S)$.  Since
$\calL_{\le d}(S)=K(S)$, Proposition~\ref{prop:ratio} makes $r$ a translation
character.  Thus $B$ is a translate of $A$.
\end{proof}

For strictness, a singleton in $C_{30}$ has full Fourier support, whose
closing depth in the present classifier is three, while all singletons form one translation orbit and
are determined by their first deck.  Hence $\rho_T(A)=1<3=\depth(S_A)$; only
the upper bound in (11.2) is used below.

\subsection{The deep-support exclusion}

Fix distinct odd primes $p<q$ and use the CRT coordinates of
Section~\ref{sec:boxes}.  The complete inverse-Fourier dictionary, before the
constant orbit $\Orb1$, is
\[
\begin{array}{c|c|c@{\qquad}c|c|c}
E&\Orb2&(-1)^\varepsilon e &P&\Orb p&U(x)\\
Q&\Orb q&T(y) &P1&\Orb{2p}&(-1)^\varepsilon V(x)\\
Q1&\Orb{2q}&(-1)^\varepsilon W(y)&F&\Orb{pq}&R(x,y)\\
G&\Orb{2pq}&(-1)^\varepsilon Z(x,y)&&&
\end{array}                                                  \tag{11.3}
\]
Here $U,T,V,W$ have zero mean in their displayed prime coordinate, while
$R,Z$ have no zero frequency in either prime coordinate.  Exact presence of
a box means that its displayed component is nonzero.  Every nonempty
indicator also has $\Orb1$, since $\widehat f_A(0)=|A|>0$, and adjoining it
does not change depth by (10.2).  Moreover, $f_A$ is rational-valued, so
(3.3) makes $S_A$ Galois-stable and the profile theorem applies.

Theorem~\ref{thm:profiles} leaves exactly the following bases above depth
four; the mandatory $\Orb1$ is suppressed in the table:
\[
\begin{array}{c|c|c}
&\text{depth five}&\text{depth six}\\ \hline
p,q\ge5&P\cup P1\cup Q1,\ Q\cup P1\cup Q1&P1\cup Q1\\
p=3&P\cup P1\cup Q1,\ Q\cup P1\cup Q1&
\begin{array}{c}P1,\ P1\cup Q,\ P1\cup Q1,\ G,\\
Q\cup G,\ P1\cup G,\ P1\cup Q\cup G
\end{array}
\end{array}                                                   \tag{11.4}
\]
Thus a binary upper bound four is equivalent to excluding every row of
(11.4) as an exact Fourier support.

There is also a useful elementary mask-polynomial obstruction for part of this
list.  General lower bounds for prescribed cyclotomic mask-polynomial divisors
are studied in \cite{KissLabaMarshallSomlai2026}, but no such general bound is
needed here.  For $A\subset C_{2pq}$ put
$A(X)=\sum_{a\in A}X^a$.  Absence of the Fourier orbit of order $s$ is
equivalent to $\Phi_s(X)\mid A(X)$.  For $P1\cup Q1$ the absent-order set is
$\{2,p,q,pq,2pq\}$; when $p=3$, the corresponding sets for $P1$, $G$, and
$P1\cup G$ are respectively
$\{2,p,q,2q,pq,2pq\}$, $\{2,p,q,2p,2q,pq\}$, and
$\{2,p,q,2q,pq\}$.  Every one of these four sets contains $2,p,q$.
The three cyclotomic polynomials are distinct monic irreducibles in
$\mathbb Z[X]$.  By Gauss's lemma their individual divisibilities hold in
$\mathbb Z[X]$, and pairwise coprimality therefore gives
\[
 \Phi_2(X)\Phi_p(X)\Phi_q(X)\mid A(X).
\]
Evaluation at $X=1$ gives
\[
 2pq=\Phi_2(1)\Phi_p(1)\Phi_q(1)\mid A(1)=|A|.
\]
An indicator with any of the four exact supports is nonempty and proper:
$\Orb1$ is mandatory, while the full-group indicator has no nontrivial
Fourier orbit.  Thus $0<|A|<2pq$, contradicting this divisibility.  This direct argument
does not exclude the two bridges or the remaining prime-three rows in
(11.4).  The argument below is retained because it excludes the entire list
uniformly.

\begin{lemma}[Two finite-set rigidity facts]\label{lem:binary-rigidity}
If a nonempty finite $X\subset\C$ satisfies $X+h=X$, then $h=0$.  If
$X,Y\subset\C$ are nonempty finite sets and $|X+Y|\le2$, then one of $X,Y$
is a singleton.
\end{lemma}

\begin{proof}
For the first assertion, iteration of a nonzero $h$ would put infinitely many
distinct points $x+mh$ in $X$.  For the second, if both sets had at least two
points, each would have exactly two.  Two distinct translates of, say, $Y$
would then both equal the same two-point sumset, giving a forbidden nonzero
translation period of $Y$.
\end{proof}

\begin{theorem}[Binary exclusion of every deep support]
\label{thm:binary-deep-exclusion}
No base in (11.4), after restoring $\Orb1$, is the exact Fourier support of a
binary indicator on $C_{2pq}$.
\end{theorem}

\begin{proof}
Assume first $p,q\ge5$.  On the critical base $P1\cup Q1$ the inverse form is
\[
 f(\varepsilon,x,y)=c+(-1)^\varepsilon\bigl(V(x)+W(y)\bigr). \tag{11.5}
\]
Both summands are nonconstant.  Adding the two parity slices shows that
$2c\in\{0,1,2\}$.  The extreme values force $V+W=0$ and contradict exact
support.  Otherwise $c=1/2$ and
$V(C_p)+W(C_q)\subset\{-1/2,1/2\}$.  Lemma
\ref{lem:binary-rigidity} makes one summand constant; its zero mean makes it
zero, again contradicting exact support.

On $P\cup P1\cup Q1$ one has
\[
 f(\varepsilon,x,y)=c+U(x)+(-1)^\varepsilon(V(x)+W(y)).    \tag{11.6}
\]
For fixed $x$, the sum of the parity slices is $2(c+U(x))$.  If it were zero
or two, both binary slices would be fixed for every $y$ and their difference
would make $W$ constant, contrary to exact $Q1$.  The sum is therefore one
for every $x$, so $U$ is constant; its zero mean gives $U=0$, contrary to
exact $P$.

The other bridge has
$f=c+T(y)+(-1)^\varepsilon(V(x)+W(y))$.  Now the parity-slice sum is
$2(c+T(y))$, and the same argument, using exact $P1$ to keep $V$
nonconstant, forces the zero-mean term $T$ to vanish.  This also proves both
bridge exclusions when $p=3$.

It remains to treat the seven $p=3$ critical bases.  Identify
$C_2\times C_3$ with $C_6$.  If a binary $h:C_6\to\{0,1\}$ has Fourier
support in $\Orb1\cup\Orb6$, vanishing of the order-three frequencies gives
\[
 h_0+h_3=h_1+h_4=h_2+h_5=:s.                              \tag{11.7}
\]
For $s=0,2$ the function is constant.  For $s=1$ it has three ones, whereas
vanishing at the order-two frequency would require
$h_0+h_2+h_4=h_1+h_3+h_5$, impossible because the two integer sides have odd
total.  Thus every such slice is constant.

Write $\mathsf A=P1$, $\mathsf B=Q1$, $\mathsf C=Q$, and $\mathsf D=G$.
For each fixed $y\in C_q$, all six nonexceptional branches have slice support
contained in $\Orb1\cup\Orb6$.  To make the completeness check visible, their
inverse forms and contradictions are listed separately:
\begin{center}
\small
\begin{tabularx}{0.96\textwidth}{@{}lYY@{}}
\toprule
Base & Slice form, with $y$ fixed & Component killed by slice constancy\\
\midrule
$\mathsf A$ & $c+u(t)$ & $u$, hence exact $\mathsf A$\\
$\mathsf A\cup\mathsf C$ & $c+s(y)+u(t)$ & $u$, hence exact $\mathsf A$\\
$\mathsf D$ & $c+z(t,y)$ & $z$, hence exact $\mathsf D$\\
$\mathsf C\cup\mathsf D$ & $c+s(y)+z(t,y)$ & $z$, hence exact $\mathsf D$\\
$\mathsf A\cup\mathsf D$ & $c+u(t)+z(t,y)$ & both $\mathsf A$ and $\mathsf D$\\
$\mathsf A\cup\mathsf C\cup\mathsf D$ &
$c+s(y)+u(t)+z(t,y)$ & both $\mathsf A$ and $\mathsf D$\\
\bottomrule
\end{tabularx}
\end{center}
In the last two rows, Fourier transformation in $y$ separates the zero
$y$-frequency $\mathsf A$ component from the nonzero $\mathsf D$ components.
The only branch not in the table is $\mathsf A\cup\mathsf B$.

On that exceptional base,
\[
 f(t,y)=c+u(t)+(-1)^t w(y),                                \tag{11.8}
\]
where $u$ has exact support $\Orb6=\{1,5\}$ and $w$ is nonconstant.  For
each even $t$, the nonconstant set $c+u(t)+w(C_q)$ is contained in
$\{0,1\}$ and hence equals it.  Comparing two even values and applying the
first part of Lemma~\ref{lem:binary-rigidity} shows that $u$ is constant on
even residues; the odd residues give the same conclusion.  Thus
$u(t+2)=u(t)$.  A period-two function on $C_6$ has Fourier support in
$\{0,3\}=\Orb1\cup\Orb2$, disjoint from $\Orb6$, so $u=0$, the final
contradiction.
\end{proof}

\subsection{Exact reconstruction and the reflection quotient}

\begin{proof}[Proof of Theorem~\ref{thm:set-reconstruction}]
Every nonempty binary exact support has depth at most four by Theorem
\ref{thm:binary-deep-exclusion} and the complete profile theorem.  Proposition
\ref{prop:binary-crosswalk} therefore gives
$r_{\rm set}(C_{2pq})\le4$; the empty set is determined by its first deck.

For the reverse inequality put $k=pq\ge15$.  Keleti--Kolountzakis prove that,
in $C_{2k}$,
\[
 \{0\}\cup\{3,\ldots,k-1\}\cup\{k+1,k+2\},\qquad
 \{0,1\}\cup\{3,\ldots,k-1\}\cup\{k+2\}                \tag{11.9}
\]
have the same analytical three-deck but are not translates
\cite[Theorem~2.22]{KeletiKolountzakis2006}.  Hence order three is
insufficient and the value is four.
\end{proof}

Keep the deck oriented, but now call two outputs equivalent when one is a
translation or a reflected translation of the other; denote the resulting
number by $r_{\rm set}^{T/I,{\rm or}}$.  This changes the output quotient, not
the data being measured.

\begin{corollary}[Oriented deck modulo translations and reflections]
\label{cor:set-reconstruction-ti}
For distinct odd primes $p<q$,
\[
              r_{\rm set}^{T/I,{\rm or}}(C_{2pq})=4.       \tag{11.10}
\]
\end{corollary}

\begin{proof}
Theorem~\ref{thm:set-reconstruction} gives the upper bound.  For the lower
bound set $m=q$, $\bar n=2p$, and represent $C_{2pq}$ by $qx+i$ with
$x\in C_{2p}$ and $i\in C_q$.  In $C_{2p}$ put
\[
 H_A=\{0,2,\ldots,2p-2\},\qquad H_B=\{0,1,\ldots,p-1\}.
\]
In $C_q$ let $J_B=\{0,1,-1\}$, $J_A=C_q\setminus J_B$, and set
$\ell_{-1}=1$, $\ell_i=0$ otherwise.  Define
$D_i=H_A+\ell_i$ for $i\in J_A$ and $D_i=H_B+\ell_i$ for
$i\in J_B$, and
\[
 \mathcal A=\{qx+i:x\in D_i,\ i\in C_q\},\qquad
\mathcal A'=\{qx+i:x\in D_i+2\mathbf 1_{\{i=0\}},\ i\in C_q\}. \tag{11.11}
\]

This is Simon's Proposition~3.11 with $i_0=0$ and $\delta=2$.  Its only
nonvacuous equation is at $j=1$:
$\ell_{-1}+\ell_1=1=2\ell_0+2-1$ in $C_{2p}$, so the two sets have equal
combinatorial three-decks.  They have size $pq$, and Simon's Corollary~1.41
therefore transfers equality to the oriented analytical three-deck.  Since
$q$ is prime and $|J_B|=3$, $J_B$ is not a subgroup coset; the consequence
of Proposition~3.13 excludes translations.  At $j=2$ its reflection test has
$\ell_{-2}+\ell_2=0$ but right side $1$ modulo two, excluding reflected
translations.  Thus order three fails even in the coarser output quotient,
proving (11.10) \cite[Corollary~1.41 and Propositions~3.11, 3.13]{Simon2017}.
\end{proof}

\subsection{Load-bearing controls and theorem boundary}

The binary hypothesis cannot be replaced even by a three-letter alphabet.
The function $(0,0,1,2,2,1)$ on $C_6$ has exact Fourier support
$\Orb1\cup\Orb6$: its coefficients are $6$ at zero and
$-3\pm i\sqrt3$ at the two order-six frequencies, with every other
coefficient zero.  Thus the binary slice argument at (11.7) is genuinely an
idempotent two-value phenomenon.

Nor does Theorem~\ref{thm:binary-deep-exclusion} collapse the weighted signal
problem.  The integer functions
\[
 (14,7,-7,-14,-7,7),\qquad (2,-11,-13,-2,11,13)            \tag{11.12}
\]
on $C_6$ have exact support $\Orb6$, agree at every autocorrelation through
order five, and are not translates.  Their all-zero-shift sixth moments are
$15529668$ and $13196868$, respectively.  Direct Fourier multiplication
therefore retains the depth-six obstruction outside the binary class.

Finally, the two Keleti--Kolountzakis sets in (11.9) satisfy
$F=(k+2)-E$: they are reflected translates.  Their published pair proves the
translation-only lower bound but cannot prove Corollary
\ref{cor:set-reconstruction-ti}.  The separate Simon specialization
(11.11), including the failed reflection test, is therefore logically
necessary rather than a second presentation of the same witness.

\begin{proofboundary}
The result is restricted to binary indicators on $C_{2pq}$ with $p<q$
distinct odd primes.  It asserts neither a ternary or weighted analogue nor
all even squarefree orders, orders with more than two odd prime factors,
noise stability, or noncyclic reconstruction.  The Simon corollary keeps the
analytical deck oriented.  Pebody's general even-order conjecture remains a
separate problem outside this theorem.
\end{proofboundary}