\section{Restrictions and constants in the label constructions}

The label constructions leave two quantitative questions at odd
lengths: the cost of symmetric profiles and the scale needed for
asymmetric ones. We answer the first by matching a Sidon construction
with a central-edge count, and improve the second by bounding the
correction terms more closely.

\subsection{Symmetric profiles at odd lengths}
To determine the cost of symmetric profiles at odd lengths, we match
a construction with the central-edge lower bound in
Proposition~\ref{prop:central-lower}. Integer Sidon labels keep the
endpoint sums distinct, and writing them in a suitable base turns
them into short coordinate vectors.

\begin{corollary}\label{cor:symmetric-odd}
Let \(L=2k+1\ge5\), suppose \(n>2\), choose a prime \(P\) with \(n\le P<2n\),
and put \(Q=\lceil(2P^2)^{1/k}\rceil\).
If \(n\ge14Q\), there is a resolving set with symmetric profiles and
size less than \(12knQ\).
For every fixed \(k\ge3\), the minimum size of a resolving set with
symmetric profiles in \(S_{2k+1}(K_n)\) is
\(\Theta_k(n^{1+2/k})\).
\end{corollary}
\begin{proof}
For \(i=0,\ldots,n-1\), take the integer labels
\[
 z_i=2Pi+(i^2\bmod P)\in[0,2P^2).
\]
These are the classical integer Sidon labels of Erd\H{o}s and Tur\'an~\cite{erdos}.
For completeness, an equality \(z_i+z_j=z_u+z_v\) first gives
\(i+j=u+v\) over the integers: the difference between two residue sums
has absolute value below \(2P\).
The residue sums then agree, so reduction modulo \(P\) gives
\(i^2+j^2=u^2+v^2\).
Since \(P\) is odd, the products agree as well.
The two unordered pairs are the roots of the same monic quadratic
over \(\mathbb F_P\), and the indices \(0,\ldots,n-1\) are distinct
modulo \(P\).

Write each \(z_i\) in base \(Q\), using \(k\) digits.
The resulting vectors belong to \(\{0,\ldots,Q-1\}^k\) because
\(Q^k\ge2P^2\). Equality of their vector pair sums would imply
equality of the corresponding integer pair sums. Hence they form a
Sidon set, and Proposition~\ref{prop:symmetric-codes} applies.

For fixed \(k\ge3\), Bertrand's postulate provides the stated prime,
and \(Q\le8^{1/k}n^{2/k}+1\).
Thus \(n\ge14Q\) eventually holds.
The symmetric central-edge bound with \(s=k\) in
Proposition~\ref{prop:central-lower} supplies the matching
lower order.
\end{proof}


\subsection{A smaller scale for odd-length labels}
The construction in Proposition~\ref{prop:odd} remains valid with the fixed scale
\(K=7\) if its correction terms are bounded independently of \(k\).
We obtain this improvement by checking the nonzero columns among the
last \(k\) positions of each incidence matrix; within each family,
their distance coordinates are all distinct.

\begin{lemma}\label{lem:odd-correction}
Let \(k\ge1\) and \(0\le r\le k\) be integers, and put \(L=2k+1\).
Use the integer coefficient matrices \(\mathcal A,\mathcal B_r\)
from~\eqref{eq:incidence-columns}, after removing \(z^a\),
with \(\delta=2r\).
For \(r=0\), put \(\mathcal B_0=\mathcal A\).
If vectors \(u,v\in\mathbb Z^{2k}\) vanish in coordinates \(1,\ldots,k\) and
satisfy \(\|u\|_\infty,\|v\|_\infty\le1\), where this norm is the
largest absolute coordinate, then
\[
 \|\mathcal A u+\mathcal B_r v\|_\infty\le2
 \qquad(0\le r\le k).
\]
\end{lemma}
\begin{proof}
In the last \(k\) coordinates the columns of \(\mathcal A\) are
\[
 \mathcal A_t=e_t-e_{2L-t},\qquad k+1\le t\le2k.
\]
All \(2k\) displayed endpoints are distinct: positive exponents lie in
\([k+1,2k]\), and negative ones lie in \([2k+2,3k+1]\).
Consequently \(\|\mathcal A u\|_\infty\le1\).

For \(r>0\), the only nonzero columns of \(\mathcal B_r\) among these
coordinates are
\[
 (\mathcal B_r)_t=e_{t+2r}-e_{2L-t},
 \qquad k+1\le t\le2k-r.
\]
Their positive exponents are distinct and at most \(2k+r\).
Their negative exponents are distinct and at least \(2k+r+2\).
Thus all endpoints within this restricted family are again distinct,
and \(\|\mathcal B_r v\|_\infty\le1\).
The family is empty if \(r=k\).
For \(r=0\), the same estimate follows from \(\mathcal B_0=\mathcal A\).
The triangle inequality proves the assertion.
\end{proof}



\begin{proposition}\label{supp:prop:odd-seven}
Let \(L=2k+1\), \(k\ge2\). Choose a prime \(q>k\) with \(q^k\ge2n\),
and put \(m=7(q-1)+1\). If \(n>12\) and \(2n\ge9m+3\), then
\[
 \edimm(S_{2k+1}(K_n))<k(2m+1)n.
\]
\end{proposition}
\begin{proof}
Use the labels and balanced degrees of Proposition~\ref{prop:odd},
with \(K=7\). The margin, size, and depth arguments are unchanged.
In~\eqref{eq:rounding}, each correction difference is supported on
the last \(k\) coordinates and has norm at most one.
Lemma~\ref{lem:odd-correction} bounds its image by two.
The support error adds at most four, so the total is at most \(6<7\).
The integer divisibility argument therefore gives
\eqref{eq:exact-label-map}, and the same recovery proves the claim.
\end{proof}