\section{Geometric descent and arithmetic closure}\label{sec:geometry}
\subsection{Moving along a profile and changing its host}
The catenary profile \eqref{eq:profile} turns two elementary operations
on Markov states into movements of the contacts. The first operation is
a unit step along a fixed profile. Directly from its powers,
\begin{equation}\label{eq:unitrec}
\begin{aligned}
f_h(x+1)+f_h(x-1)
 &=\frac{h}{2R_h}(Q_h+Q_h^{-1})(Q_h^x+Q_h^{-x})\\
 &=3h f_h(x).
\end{aligned}
\end{equation}
Thus the height beyond a pair is obtained by multiplying the nearer
height by $3h$ and subtracting the farther height. This is the usual
Vieta replacement of a companion entry, expressed on the curve.

The second operation keeps the contact heights fixed and changes the
host. For any real $x$, put $A=f_h(x)$ and $B=f_h(x+1)$.
Their possible hosts are the roots of
$X^2-3ABX+A^2+B^2=0$. One root is $h$, so the other is
\begin{equation}\label{eq:otherhost}
 h,\qquad K=3AB-h=\frac{A^2+B^2}{h}.
\end{equation}

For descent, select an edge straddling the minimum of the profile.
We record its right endpoint as $c$, with $0\le c<1$, and its ordered
contact heights as $A=f_h(c)$ and $B=f_h(c-1)$.
This is the same pair used in the introduction, since $f_h(c-1)=f_h(1-c)$.
We retain the full interval of $c$ here to keep track of orientation.
Seek a wider profile through these same two heights, still one unit
apart, allowing a horizontal translation. Smaller $h$ gives a wider profile. If the other host satisfies
$K<h$, fit that profile and walk the unit lattice inward until the selected
edge again straddles its vertex. Otherwise stop. Figure~\ref{fig:1A} shows
one complete step. The fit is explicitly
\begin{equation}\label{eq:fit}
v=\frac{\log\!\bigl((Q_K A-B)/K\bigr)}{\log Q_K},\qquad
f_K(v)=A,\quad f_K(v-1)=B.
\end{equation}
Indeed, $Q_K f_K(v)-f_K(v-1)=KQ_K^v$, so taking logarithms
recovers $v$ directly. The Markov relation gives
\[
(Q_K A-B)(B-Q_K^{-1}A)=K^2.
\]
Both factors are positive: they have the same sign, and their sum is
$2R_KA>0$. This also fixes the real branch of the logarithm and verifies
the fitted contact heights by direct substitution.
The walk replaces $v$ by $v-\lfloor v\rfloor$. Its orientation is retained:
reflection exchanges the contacts and sends phase to its negative modulo one.

\paperfigure{1A}{1}{figures/fig1_descent.pdf}{Geometric descent from
$(1,2,5)$. The wider profile passes through the same unit-spaced contacts.
On $f_1$ their coordinates are $1/2,3/2$; one inward step selects
$-1/2,1/2$, with heights $1,1$. The curve is fixed during the walk.}{htbp}

\begin{proposition}[Uniqueness of the fit]\label{prop:fit}
A marked edge has at most one distinct compatible profile, hence at most one
wider fit. For fixed $0<a<b$, a same-arm fit
$f_h(u-1)=a$, $f_h(u)=b$ with $u>1$ is unique if it exists.
\end{proposition}
\begin{proof}
The hosts are the two roots of $X^2-3ABX+A^2+B^2$.
The other root belongs to the geometric domain because
$(3K-2)AB=K^2+(A-B)^2>0$.
For a same-arm fit the horizontal distance between heights $a,b$ is
\[
\frac1{\log Q_h}\int_a^b\frac{dy}{\sqrt{y^2-h^2/R_h^2}}.
\]
When $a>2/3$ and the lower height is in range, $Q_h$ strictly increases
and $h/R_h=2h/\sqrt{9h^2-4}$ strictly decreases with $h$.
Thus the prefactor and the integrand strictly decrease with $h$.
The distance can therefore equal one at most once. The profile is
strictly increasing on that arm, which also determines $u$. If $a\le2/3$
there is no fit.
\end{proof}

\subsection{Termination}
Write $\mathcal D(M,c)=(H,\phi)$ for the terminal host and oriented phase.
\begin{theorem}[Finite descent]\label{thm:termination}
Every straddling input with $M\ge1$ terminates. On $1\le M\le R$ the number
of host changes is uniformly bounded. The terminal map is continuous into
$(2/3,\infty)\times\R/\Z$ and analytic on cells with a fixed strict itinerary.
\end{theorem}
\begin{proof}
Temporarily sort the endpoints as $a\le b$. Inward walking does not increase
either endpoint. Every current host is at most $R$, and the discriminant in
\eqref{eq:edgeconic} gives
\[
(9h^2-4)a^2-4h^2=(2b-3ha)^2\ge0.
\]
Consequently the endpoints are at least
$a_R=2R/\sqrt{9R^2-4}>2/3$. At an accepted change,
$a^2+b^2<h^2$, so the endpoints are also below $h\le R$.
Using either endpoint as the host in the same discriminant identity bounds
the new host below by $a_R$. The initial host is at least $1>a_R$.

If the new host $K$ is at least $b$, its edge already straddles the vertex,
because both endpoints are below $f_K(1)$; the next candidate is the old,
larger host, so descent stops. Otherwise
\[
\frac hK>\frac hb=3a-\frac Kb>3a-1\ge3a_R-1>1.
\]
All changes except possibly the last therefore contract the host by a fixed
factor. An upper bound is
$1+\lceil\log(R/a_R)/\log(3a_R-1)\rceil$.

At a boundary $K=h$, the fitted coordinate in \eqref{eq:fit} is the original
$c$, so the accepted and stopped maps agree. At a phase cut,
$K(h,0)=h(9h^2+4)/(9h^2-4)>h$, and a neighborhood is stopped.
The uniform bound on compositions then gives continuity, with phase taken
modulo one. The strict formulas are analytic.
\end{proof}

\subsection{Which inputs close arithmetically?}
The two reversible state operations are
\begin{equation}\label{eq:TJ}
T(a,b,h)=(3ha-b,a,h),\qquad J(a,b,h)=(a,b,3ab-h).
\end{equation}
They and their inverses are integer polynomial maps. A terminal state has
the nonnegative defect
\begin{equation}\label{eq:defect}
\begin{split}
Hf_H(\phi)f_H(\phi-1)-1
={}&\frac{(H-1)^2(H+2)}{3H-2}\\
&+\frac{H^3}{4R_H^2}
\bigl(Q_H^{\phi-1/2}-Q_H^{1/2-\phi}\bigr)^2.
\end{split}
\end{equation}
For $0\le\phi<1$ it vanishes exactly at $(H,\phi)=(1,1/2)$.
\begin{proposition}[Arithmetic closure]\label{prop:closure}
A straddling input with $M\ge1$ reaches $(1,1/2)$ if and only if its three
entries are positive integers. Modulo companion exchange, these inputs are
exactly the normalized configurations \eqref{eq:markov}.
\end{proposition}
\begin{proof}
Expanding the two profiles gives
\[
Hf_H(\phi)f_H(\phi-1)-1
=\frac{H^3}{4R_H^2}
\bigl(Q_H^{2\phi-1}+Q_H^{1-2\phi}+3H\bigr)-1.
\]
Replace the first two powers by
$\bigl(Q_H^{\phi-1/2}-Q_H^{1/2-\phi}\bigr)^2+2$.
The constant term is $H^3/(3H-2)-1=(H-1)^2(H+2)/(3H-2)$,
which proves \eqref{eq:defect}. Reversing a finite
closure itinerary from $(1,1,1)$ by \eqref{eq:TJ} preserves integrality.
Conversely, integral descent stays integral. At an integral terminal state
order the entries as $u\le v\le w$. There is no coordinate-decreasing
mutation. But for $Q(X)=X^2-3uvX+u^2+v^2$,
\[
Q(v)=u^2+(2-3u)v^2\le3u^2(1-u)\le0.
\]
If neither root is below $w\ge v$, equality is forced, giving $u=v=1$;
terminality then gives $w=1$. Finally, a straddling edge has no decreasing
endpoint mutation, so its host is a largest entry. Its ordered split is
uniquely recovered by \eqref{eq:fit}.
\end{proof}

\begin{figure}[p]
\begin{minipage}{\linewidth}
\noindent Both terminal coordinates matter. Figure~\ref{fig:1B} shows inputs with
terminal host one but different terminal phases. A crossing of the height
level $H=1$ alone does not certify integrality. The marked configurations
follow the host itineraries $37666\to5\to1$, $7561\to5\to1$, and
$14701\to2\to1$, all ending at phase $1/2$.
\end{minipage}
\par\medskip\centering
\includegraphics[width=.90\linewidth]{figures/fig1B_phase_readouts.pdf}
\caption{Terminal host and
phase for $M=37666,7561,14701$. Height is $H$ on a logarithmic scale and
hue is oriented phase. Left: $0\le c\le1/2$, with gray where an initial
endpoint is below one. Right: the admissible portion enlarged. Stars mark
the exact integral starts; dots are other numerically located $H=1$
crossings. These numerical plots illustrate the closure criterion.}
\label{fig:1B}\end{figure}