\section{Exact lattice counts and restricted depths}\label{supp:exact}

To turn the profile count into a finite lower bound, we fill
integer-vector layers in increasing order of their coordinate sum.
For positive integers \(Q,d\) and a nonnegative integer \(N\), let
\(F_{Q,d}(N)\) be the minimum total coordinate sum of \(N\) vectors
in \(\mathbb Z_{\ge0}^d\), with each vector allowed at most \(Q\)
occurrences. This allowance accounts for endpoints with the same
vector that can still be distinguished by their local choices.
The minimization ignores the cost of those local selections;
in particular, \(F_{Q,d}(0)=0\).

\Needspace{12\baselineskip}
\begin{lemma}\label{lem:cost}
Let \(\ell\) be the smallest nonnegative integer satisfying
\(Q\binom{\ell+d}{d}\ge N\). Then
\begin{equation}\label{eq:lattice}
 F_{Q,d}(N)
 =Qd\binom{\ell+d-1}{d+1}
  +\ell\left(N-Q\binom{\ell+d-1}{d}\right).
\end{equation}
For fixed \(Q,d\),
\begin{equation}\label{eq:lattice-asymptotic}
 F_{Q,d}(N)=
 \frac d{d+1}\left(\frac{d!}{Q}\right)^{1/d}
 N^{1+1/d}+O_{Q,d}(N).
\end{equation}
Here \(\binom uv=0\) for nonnegative integers \(u<v\).
\end{lemma}
\begin{proof}
There are \(\binom{r+d}{d}\) nonnegative integer vectors of coordinate
sum at most \(r\). This is the Ehrhart polynomial of the standard simplex~\cite{beck}. A minimum fills each layer below \(\ell\) with
\(Q\) copies of every vector, and then uses the required number of
vectors in layer \(\ell\). The completed layers cost
\[
 Q\sum_{r=0}^{\ell-1}r\binom{r+d-1}{d-1}
 =Qd\binom{\ell+d-1}{d+1}.
\]
This gives the exact formula. As \(N\) tends to infinity,
\[
 N=\frac{Q}{d!}\ell^d+O_{Q,d}(\ell^{d-1}),\qquad
 F_{Q,d}(N)=\frac{Qd}{(d+1)!}\ell^{d+1}
                 +O_{Q,d}(\ell^d).
\]
Eliminating \(\ell\) proves the asymptotic formula.
\end{proof}


\Needspace{12\baselineskip}
\begin{theorem}[The number of active depths]
\label{thm:active-depths}
Let \(s\) be an integer with \(1\le s\le\lfloor L/2\rfloor\), put \(Q_s=2^{2s+1}\), and suppose
\(n>Q_s+1\). If an edge multiset resolving set in \(S_L(K_n)\) has
all its internal landmarks at depth at most \(s\), then its number \(B\)
of internal landmarks satisfies
\[
 B\ge F_{Q_s,s}(n).
\]
Original vertices may also be selected. In particular, for fixed \(s\),
\[
 B\ge a_s n^{1+1/s}-O_s(n),\qquad
 a_s=\frac{s}{s+1}\left(\frac{s!}{2^{2s+1}}\right)^{1/s}.
\]
The lower bound is uniform over all integers \(L\ge2s\).
\end{theorem}
\begin{proof}
Fix an original vertex \(i\), and let \(e_{ij}\) be the edge adjacent
to \(i\) on support \(ij\). The far-endpoint columns of its histogram are
\[
 B_t(z)=
 \begin{cases}
 z^{L-1+t}-z^{L+t},&1\le t\le\lfloor L/2\rfloor,\\
 0,&\lfloor L/2\rfloor<t<L.
 \end{cases}
\]
Consequently its histogram has the form
\[
 H_{e_{ij}}(z)=C_i(z)+\alpha_j(z^{L-1}-z^L)
 +\sum_{t=1}^{\lfloor L/2\rfloor}
 D_j(t)(z^{L-1+t}-z^{L+t})+E_{e_{ij}}(z),
\]
where \(C_i\) is independent of \(j\), and the support correction
\(E_{e_{ij}}\) from~\eqref{eq:local-error} depends only on the
landmark bits on support \(ij\).

First suppose \(L>2s\). The only allowed internal positions are
\(1,\ldots,s,L-s,\ldots,L-1\). The second group lies above
\(\lfloor L/2\rfloor\), so assign to \(j\) the profile
\[
 p_j=(D_j(1),\ldots,D_j(s)).
\]
For fixed \(i,p_j\), the displayed histogram is determined by
\(\alpha_j\) and the \(2s\) bits on support \(ij\). There are at most
\(2^{2s+1}=Q_s\) possibilities.

If \(L=2s\), use instead
\[
 p_j=(D_j(1),\ldots,D_j(s-1),\lfloor D_j(s)/2\rfloor).
\]
The additional parity bit of \(D_j(s)\) recovers its unrounded value.
Together with \(\alpha_j\) and the \(2s-1\) support bits, this again
gives at most \(2^{2s+1}\) histograms.

No profile class can have more than \(Q_s\) members. For a proper class
with more than \(Q_s\) members, take \(i\) outside it and apply the preceding bound to
its targets. If the class contains all \(n\) original vertices, any
origin sees \(n-1>Q_s\) targets. Either case would give a collision.

For \(L>2s\), every internal landmark contributes once to
\(\sum_j\|p_j\|_1\), so this sum is \(B\). For \(L=2s\), every
nonmiddle landmark still contributes once, while a middle landmark
is counted at both endpoints before the division by two. Hence
\[
 \sum_j\|p_j\|_1
 =B-\frac12\bigl|\{j:D_j(s)\text{ is odd}\}\bigr|\le B.
\]
The multiplicity bound gives \(B\ge F_{Q_s,s}(n)\), and
Lemma~\ref{lem:cost} gives the asymptotic statement.
\end{proof}

\begin{corollary}\label{cor:exact-lower}
For \(L\ge2\), put \(k=\lfloor L/2\rfloor\).
If \(n>2^{2k+1}+1\), every resolving set in \(S_L(K_n)\) has at least
\(F_{2^{2k+1},k}(n)\) internal landmarks. For each fixed \(L\),
\[
 \operatorname{edim}_{\rm m}(S_L(K_n))
 \ge a_k n^{1+1/k}-O_k(n)
\]
whenever a resolving set exists. In particular \(a_2=1/6\).
\end{corollary}
\begin{proof}
Every internal vertex has depth at most \(k\), so
Theorem~\ref{thm:active-depths} applies with \(s=k\).
Its finite bound and asymptotic constant give the assertion; at
\(k=2\), the constant is \((2/3)(2/32)^{1/2}=1/6\).
\end{proof}