The fixed-length theorem leaves two quantitative questions: which
finite orders admit a resolving set, and what is lost by restricting
the available landmark positions? We collect the existence ranges
first, then combine the constructions and counting bounds developed
below to compare constrained selections. We again allow all integers
\(n\ge3\) and \(L\ge2\), with the conventions of
Section~\ref{sec:geometry}.

\subsection{Existence at finite orders}
To locate finite orders where a resolving set exists, we combine the
known cycle case with explicit thresholds from the constructions.
For \(n=3\), we have \(S_L(K_3)=C_{3L}\), where \(C_m\) is the cycle
on \(m\) vertices. Since \(\edimm(C_m)=3\) for \(m\ge7\)
\cite[Theorem 12]{ikhlaq}, every \(L\ge3\) works at \(n=3\);
the remaining bounds concern larger orders.

\begin{corollary}\label{cor:length-four}
For every \(n\ge210\),
\[
 \edimm(S_4(K_n))<18n\lceil\sqrt n\rceil.
\]
\end{corollary}
\begin{proof}
Apply Corollary~\ref{prop:even} with \(k=2\).
For \(210\le n\le225\), its condition is \(n\ge14\cdot15\).
For \(n\ge226\), put \(Q=\lceil\sqrt n\rceil\ge16\). Then
\(n\ge(Q-1)^2+1\ge14Q\), because \(Q^2-16Q+2\ge0\).
\end{proof}

\begin{theorem}\label{thm:uniform}
For every \(n\ge222\) and every \(L\ge5\), there is a resolving set
using only internal landmarks of depths one and two, of size less than
\[
 24n^{3/2}+18n.
\]
\end{theorem}
The proof is given in Appendix~\ref{app:uniform}, starting with
four-coordinate Sidon labels (Lemma~\ref{uniform:lem:sidon}) and
showing that balancing preserves their distinct pair sums
(Lemma~\ref{uniform:lem:robust}). Proposition~\ref{uniform:prop:decode}
then recovers each edge from its histogram; the recovery cases are
collected in Table~\ref{uniform:tab:decoding}.
Both the constant and the order threshold are independent of \(L\).
The finite prime intervals needed for the threshold \(222\) appear
in Table~\ref{uniform:tab:primes}.

For smaller orders at long lengths,
Proposition~\ref{prop:long-path-probability} gives a resolving set
using at most half of all vertices when \(n\ge9\) and \(L\ge12\).
Its proof bounds the probability of equal histograms through
a forest of distance levels (Lemma~\ref{lem:distance-forest}).
Fig.~\ref{fig:distance-forest} in Appendix~\ref{app:refinements}
shows how this forest separates the constraints on a random selection.
Fig.~\ref{fig:existence-ranges} collects these bounds with thresholds
uniform over the indicated lengths. The length-dependent conditions in
Corollary~\ref{prop:even} and Proposition~\ref{prop:odd} cover additional
pairs \((n,L)\).

\begin{figure}[htbp]
\centering\input{figures/existence_ranges.tex}
\caption{Uniform existence ranges. Each line starts at an included
order and continues to the right. The points at \(n=3\) use the known
cycle case. In the bottom row, the two constructions overlap for
\(n\ge222\). The scale groups orders by the four displayed thresholds.
Unmarked intervals receive no assertion from this summary;
length-dependent conditions can cover further pairs.
At length two no resolving set exists for \(n\ge3\).}
\label{fig:existence-ranges}
\end{figure}

\subsection{The cost of restricting the selection}
The uniform construction uses only the first two depths, whatever the
path length. For fixed \(L\ge5\), comparing it with the restricted-depth
count in Section~\ref{supp:exact} shows the cost of this choice:
the minimum keeps exponent \(3/2\), even when further depths permit
a smaller exponent.

\begin{corollary}\label{cor:depth-two}
For each fixed \(L\ge5\), the minimum size among resolving sets whose
internal landmarks have depth at most two is \(\Theta(n^{3/2})\).
For \(n>33\), at least \(F_{32,2}(n)\) internal landmarks are needed,
where \(F_{Q,d}\) is the minimum total cost of integer vectors defined
in Section~\ref{supp:exact}.
\end{corollary}
\begin{proof}
In Theorem~\ref{thm:active-depths}, set \(s=2\), so \(Q_s=32\).
It gives the finite lower bound \(F_{32,2}(n)\) for \(n>33\), whose
order is \(n^{3/2}\) by~\eqref{eq:lattice-asymptotic} with \(d=2\).
Theorem~\ref{thm:uniform} gives the matching upper order.
\end{proof}

For comparison, Corollary~\ref{cor:symmetric-odd} gives
\(\Theta_k(n^{1+2/k})\) for symmetric profiles at length \(2k+1\),
when \(k\ge3\). Its lower bound already follows by distinguishing
only the central edges, as Proposition~\ref{prop:central-lower} shows.
The unrestricted order is
\(\Theta_k(n^{1+1/k})\). Thus using further depths helps, and allowing
opposite profiles to differ is also essential to attain the sharp odd
exponent.

The comparison with named distances follows from Theorem~\ref{thm:main}:
Proposition~\ref{prop:ordered} bounds \(\edim(S_L(K_n))\) between two
linear functions of \(n\), so, for every fixed \(L\ge3\),
\begin{equation}\label{eq:ratio}
 \frac{\edimm(S_L(K_n))}{\edim(S_L(K_n))}
 =\Theta_L\!\left(n^{1/\lfloor L/2\rfloor}\right).
\end{equation}
This is the factor required to replace named reports by a multiset.

The growth exponents leave the optimal leading constants open.
Proposition~\ref{prop:refined-three} improves the lower constant at length
three to \(c_*=0.080208390067\ldots\), giving
\[
 \edimm(S_3(K_n))\ge c_*n^2-O(n).
\]
The upper constant supplied by Theorem~\ref{short:thm:three} is \(1/2\).
Proposition~\ref{prop:biased-selection} also gives bounds using
sparse independent selections, for \(L\ge14\) and the explicit order
threshold stated there. These bounds complement the fixed-length
theorem when the length grows with \(n\).