\section{Realizing endpoint profiles}\label{sec:realization}
The constructions in Sections~\ref{sec:symmetric} and~\ref{sec:odd}
use two standard sufficient conditions for realizing chosen landmark
counts. The first produces binary matrices for different endpoint
profiles; the second handles symmetric selections, including middle
vertices. We recall their short proofs and describe how each realization
yields an ordinary landmark set.

\begin{lemma}\label{lem:margins}
Let \(r_i,c_i\in\{0,\ldots,n-1\}\), for \(0\le i<n\), have common sum \(E\), and suppose \(r_i\le M_r\), \(c_i\le M_c\), with nonnegative integer bounds. If
\[
(M_r+1)M_c\le E,
\]
there is a binary \(n\times n\) matrix with zero diagonal, row sums \(r_i\), and column sums \(c_i\).
\end{lemma}
\begin{proof}
This sufficient condition is due to Burstein and Rubin~\cite{burstein}; we give the flow argument. If \(E=0\), use the zero matrix. Otherwise \(M_c>0\). The bipartite flow criterion with diagonal entries forbidden~\cite{fulkerson} requires, for each set \(I\) of \(s\) rows,
\[
\sum_{i\in I}r_i\le\sum_{j=0}^{n-1}\min\{c_j,s-\ind_{j\in I}\}.
\]
For \(1\le s\le M_c\), the right side is at least
\[
\sum_j\min(c_j,s)-s\ge\frac{sE}{M_c}-s\ge sM_r.
\]
For \(s\ge M_c+1\), it equals \(E\). For the empty row set, both sides are zero. All cuts therefore have sufficient capacity, and integral flow supplies the binary matrix.
\end{proof}

Given such a matrix for positions \(t,L-t\), where \(t\) is an integer with \(1\le t<L/2\), select \(x_{ij}(t)\) when its \((i,j)\)-entry is one. Equivalently, on support \(ij\) with \(i<j\), its two relevant entries select positions \(t\) and \(L-t\). The row sums become \(D_i(t)\), and the column sums become \(D_i(L-t)\).

\begin{lemma}\label{lem:graphical}
Let \(m,M\) be positive integers. Suppose positive integers
\(d_0,\ldots,d_{n-1}\) have even sum and satisfy
\[
m\le d_i\le M<n,\qquad 4nm\ge(M+m+1)^2.
\]
Then they are the degrees of a simple graph.
\end{lemma}
\begin{proof}
This is the sufficient condition of Zverovich and Zverovich~\cite[Theorem 6]{zverovich1992}, also recalled in~\cite[Theorem 2]{cloteaux}. Order the degrees nonincreasingly and use the Erd\H{o}s--Gallai inequalities. For \(s\le m\), their right side is \(s(n-1)\), which dominates the sum of the first \(s\) degrees. For \(s>m\), the right side minus that sum is at least
\[
s(s-1)+(n-s)m-sM
=s^2-(M+m+1)s+nm\ge0.
\]
The last inequality follows from the assumed discriminant bound. Thus all the inequalities hold.
\end{proof}

For a simple graph with degrees \(d_i\), selecting both positions \(t,L-t\) on each of its edges gives \(D_i(t)=D_i(L-t)=d_i\). If \(L=2k\), selecting just the middle vertex on each graph edge gives \(D_i(k)=d_i\). In that case the total number of middle landmarks is \(\frac12\sum_i d_i\).