\section{What the distance multiset records}\label{sec:geometry} We first express edge distances through landmark positions on the replacement paths. This will isolate the counts that distinguish edges and turn equality of multisets into equations for those counts. Throughout the article, \(n\ge3\) and \(L\ge2\) are integers. Index the original vertices by \(0,\ldots,n-1\). Sums over original vertices and the rows and columns of binary matrices encoding landmark selections use these indices. For distinct originals \(i,j\) and an integer \(0\le t\le L\), let \(x_{ij}(t)\) be the vertex in position \(t\) on the path from \(i\) to \(j\). Thus \(x_{ij}(0)=i\), \(x_{ij}(L)=j\), and \(x_{ij}(t)=x_{ji}(L-t)\). We call \(ij\) the \emph{support} of each edge on this path. Orient an edge so that its nearer original endpoint is \(i\), and write \[ e=x_{ij}(a)x_{ij}(a+1),\qquad 0\le a\le\lfloor(L-1)/2\rfloor,\qquad b=L-1-a. \] Here \(a\) and \(b\) are the distances from \(e\) to \(i\) and \(j\), respectively, and \(a\le b\). We call \(a\) the \emph{edge depth} and put \(\delta=b-a\); either orientation is permitted when \(a=b\). An internal vertex \(x_{ij}(t)\) has \emph{vertex depth} \(\min(t,L-t)\). The distances depend on how the landmark's support meets \(ij\). On \(ij\) itself, the endpoints of \(e\) are in positions \(a\) and \(a+1\), as shown in Fig.~\ref{fig:own-support}. If \(t\le a\), the nearer endpoint is in position \(a\), at distance \(a-t\); if \(t\ge a+1\), it is in position \(a+1\), at distance \(t-(a+1)\). These two cases cover every vertex because \(t\) is an integer. In particular, the distance is zero at either endpoint of \(e\). \begin{figure}[!htbp] \centering\input{figures/own_support_distance.tex} \caption{Distance on the edge's own support. Positions are measured from \(i\). In (a), the nearest endpoint of \(e\) is in position \(a\); in (b), it is in position \(a+1\). Each brace measures the distance to that endpoint. Moving the orange landmark to either endpoint gives distance zero. The paths are schematic.} \label{fig:own-support} \end{figure} \FloatBarrier For the other supports, compare routes through their original endpoints, using distance \(L\) between distinct originals. Fig.~\ref{fig:distance-geometry} shows the two configurations where both endpoint routes can matter. \begin{figure}[!htbp] \centering\input{figures/distance_geometry.tex} \caption{Routes to a landmark on another support. In (a), entering \(ju\) through \(j\) or \(u\) gives lengths \(b+t\) and \(a+L+(L-t)\). In (b), entering the disjoint support through \(u\) or \(v\) gives \(a+L+t\) or \(a+L+(L-t)\). Orange vertices are landmarks. Each complete support has length \(L\); intermediate positions are schematic.} \label{fig:distance-geometry} \end{figure} \FloatBarrier Table~\ref{tab:distances} collects these distances. In its first row, leaving the support of \(e\) cannot shorten the segment drawn in Fig.~\ref{fig:own-support}; in the remaining rows, the shorter endpoint route gives the stated value. \begin{table}[H] \centering\small \caption{Distances from \(e\), with \(a\le b\), \(a+b=L-1\), \(1\le tb\). Their combined contribution to any coefficient is therefore at most two. The exponent \(a+t\) is injective and at most \(L+a-1\). The other negative exponent is \[ L+a+\min(\delta-t,t). \] Its increasing branch starts at \(L+a+1\), above the first negative range, and its decreasing branch is injective. Thus the negative contribution also has magnitude at most two at every coefficient. Subtracting these contributions proves the first bound; the second follows by the triangle inequality. \end{proof} We will recover the edge depth before comparing its endpoint profiles. All distances in Table~\ref{tab:distances} are at most \(L+a+k\). If a depth-\(k\) landmark lies on a disjoint support, it attains this value, so \begin{equation}\label{eq:last-layer} \max\{t:H_e^W(t)>0\}=L+a+k. \end{equation} The constructions in Sections~\ref{sec:symmetric} and~\ref{sec:odd} ensure such a landmark for every support. Equation~\eqref{eq:last-layer} then determines \(a\), and hence \(b\) and \(\delta\). \begin{proposition}\label{prop:ordered} For \(n\ge3\) and \(L\ge2\), \[ \left\lceil\frac{n-1}{2}\right\rceil\le \edim(S_L(K_n))\le n. \] \end{proposition} \begin{proof} The original vertices resolve every edge: the two smallest named distances identify its support endpoints, and either endpoint distance specifies its position. This proves the upper bound. For the lower bound, associate to each landmark its original vertex, or the two endpoints of its support if it is internal. Fewer than \((n-1)/2\) landmarks leave at least two originals unassociated. Interchanging these originals induces a nontrivial automorphism that fixes every landmark and moves an edge of their replacement path. Those two edges have equal named distance records, a contradiction. \end{proof}