\section{Meridians, addresses, and the alternative tree}\label{sec:indexing}
\subsection{The landscape and its two coordinates}
On the landscape $(\log_{10}M,c)$, the point $(M,c)$ represents the contacts
$f_M(c),f_M(c-1)$. We draw the reflected half $0<c\le1/2$; exchanging the
contacts sends $c$ to $1-c$. A \emph{Vieta arc} keeps one initial contact
height fixed. For example, the branch with height $a$ has
\[
c=\frac{\log\!\left((R_M a+\sqrt{R_M^2a^2-M^2})/M\right)}{\log Q_M}.
\]
This follows by solving
$Q_M^c+Q_M^{-c}=2R_Ma/M$ for $Q_M^c\ge1$ and taking logarithms.
The square root is real exactly when $a\ge M/R_M$, the minimum height
of the profile; reflection supplies the other arm.
A \emph{Vieta trek} follows such arcs, changing the retained contact when
necessary. A \emph{Meridian} is a connected component of terminal host
$H(M,c)=1$. A \emph{phasor} holds the terminal phase $\phi$ fixed. The
alternative tree uses phasors at $\phi=1/2$ to join successive Meridians.

The address phase $x$ on a Meridian is real and unrolled. Its terminal circle
phase is $\phi\equiv x+1/2\pmod1$. Every integer mark therefore closes
arithmetically; the distinguished arrival mark $x=0$ is called the
\emph{helm}. It need not be the minimum of the real profile.

Figure~\ref{fig:landscape} keeps the same configurations in three views.
The alternative tree is in the middle panel. The bottom panel uses smooth
schematic connectors for the Vieta parent relation: node coordinates and
adjacency are exact, but those connectors are not literal level curves.
The two trees each have 55 edges. An unmarked crossing is not a vertex.
\begin{figure}[p]\centering
\includegraphics[height=.82\textheight,width=\linewidth,keepaspectratio]{figures/fig2A_two_trees_landscape.pdf}
\caption{The same 56 configurations with $M\le10^7$. (a) Numerical
geometric descent, with phase encoded by hue and host by color intensity.
The number labels identify the same 20 Markov numbers shown in
Figure~\ref{fig:fareytrees}.
(b) The alternative tree, with Meridian and phasor edges.
(c) The canonical Vieta tree, with simplified connectors retaining exact
endpoints. Red retains the smaller companion, green the larger. Every edge
runs toward larger $M$. Crossings without a node mark are not vertices.}
\label{fig:landscape}\end{figure}

For reproducibility, the field in panel (a) uses the fixed cyclic phase
palette with HSL saturation and lightness
\[
w(H)=\frac1{1+(\ln H/\ln1.05)^2},\qquad S=w(H),\qquad L=1-23w(H)/40.
\]
The background is a numerical illustration, independent of all exact
enumerations and exclusions below.

\subsection{Evaluating an address}
The seed Meridian is
\begin{equation}\label{eq:seed}
\begin{gathered}
a_1(x)=f_1(x+\tfrac12),\qquad b_1(x)=f_1(x-\tfrac12),\\
M_1(x)=3a_1(x)b_1(x)-1.
\end{gathered}
\end{equation}
For integer $x$, its companions are $F_{2x+1},F_{2x-1}$, where
$F_{-j}=(-1)^{j+1}F_j$. Hence
\begin{equation}\label{eq:seedfib}
M_1(x)=2+3F_{2x}^2,
\qquad M_1(0),M_1(1),M_1(2),\ldots=2,5,29,194,\ldots.
\end{equation}
From any ordered state $(a,b,H)$ form the two-sided orbit
\[
u_0=a,\quad u_1=b,\qquad u_{j+1}=3Hu_j-u_{j-1}.
\]
A nonzero index $k$ replaces the state by
\begin{equation}\label{eq:extend}
(u_k,u_{k+1},3u_ku_{k+1}-H).
\end{equation}
The address is written
\[
\alpha=(1,k_m,\ldots,k_1):
\quad k_1\text{ is applied first and written at the right.}
\]
This is the reversed indexing convention: the former $M_{1,a,b,c}(x)$ is
now $M_{1,c,b,a}(x)$. Neither the signs nor $x$ are reversed.
In terms of \eqref{eq:TJ}, with $S=T^{-1}$ and $\mathcal W_k=JS^k$,
\begin{equation}\label{eq:word}
(a_\alpha(x),b_\alpha(x),M_\alpha(x))
=\mathcal W_{k_m}\cdots\mathcal W_{k_1}JT^x(1,1,1).
\end{equation}
Composition acts from right to left. A recursive extension inserts its index
immediately after the leading $1$. For example, $M_{1,1}(0)=13$.
Index zero is forbidden because \eqref{eq:extend} would return the preceding
pole $3ab-H$, rather than create a new component.

\begin{definition}[Canonical marks]\label{def:canonical}
The root $(1,1,1)$ is separate. For the seed $(1)$ use integer phases $x\ge0$.
For every other address require $k_1>0$, $k_i\ne0$ for $i>1$, and $x\in\Z$.
Evaluation is followed by sorting the two companions.
\end{definition}

\begin{theorem}[Configuration bijection]\label{thm:bijection}
Canonical marks are in bijection with all normalized positive Markov triples
except the root. The displayed value $M_\alpha(x)$ is the largest entry.
\end{theorem}
\begin{proof}
The quadratic argument in Proposition~\ref{prop:closure} shows that replacing
the largest coordinate by its other pole gives a smaller maximum, apart from
the initial repeated entries. This is the forced Vieta reduction of a
configuration and does not assume scalar unicity.

To recover a word, reflect the current maximum $H$ to $P=3ab-H$.
When $P\ge2$, walk its oriented endpoint orbit inward until both endpoints
are below $P$. This edge is unique. Indeed
$u_{j+1}-2u_j+u_{j-1}=(3P-2)u_j>0$, so the entries below $P$ form an interval;
three consecutive entries below $P$ would contradict
$u_{j-1}+u_{j+1}=3Pu_j\ge3P$. Endpoint descent supplies the edge, and its
nonzero signed displacement recovers the last-applied, leftmost index.
Repeat at the smaller pole. At pole one, the orbit reaches $(1,1)$ and fixes
the seed phase.

Conversely, all outward blocks are positive and reduced. If the preceding
pole is $P<H$, the first reflected poles on the two sides are
$(9b^2-2)H-P>6H$ and $(9a^2-2)H-P>6H$. Their farther endpoints exceed
$2H$, and the recurrence makes subsequent outward entries increase.
The reduction therefore undoes every prescribed block. Finally, companion
exchange negates all indices and $x$ simultaneously. The canonical
conditions select one of the two orientations; only $(1,1,2)$ is fixed.
\end{proof}

This bijection is useful computationally as well as geometrically. A
configuration can be retained by its address and reconstructed when a
comparison needs its full integer entries. Section~\ref{sec:verification}
uses the equivalent reduced-fraction labels as compact identities, with
compressed Vieta evaluation. Its coverage therefore counts triples before
testing whether their maxima are distinct.

\subsection{The two parent rules}
At a nonzero phase the alternative parent is the adjacent inward mark,
$(\alpha;x)\mapsto(\alpha;x-\sgn x)$. At a nonseed helm it is
\begin{equation}\label{eq:altparent}
(1,k_m,\ldots,k_2,k_1;0)\longmapsto
(1,k_m,\ldots,k_2;-k_1),
\end{equation}
followed by simultaneous sign reflection if the remaining rightmost index
is negative. An empty tail uses the seed's nonnegative phase. The seed helm
with maximum two has parent $(1,1,1)$.

The alternative parent deletes the \emph{rightmost} index at a helm.
Recursive evaluation removes the \emph{leftmost} index after $1$.
These operations serve different purposes. The depth
\begin{equation}\label{eq:depth}
d(\alpha;x)=1+\sum_{i=1}^m(|k_i|+1)+|x|
\end{equation}
decreases by one under either parent edge, with root depth zero.
Thus the parent rule defines a rooted tree on the same configurations.
The ordinary outward Vieta moves instead are
\begin{equation}\label{eq:vietachildren}
(a,b,M)\longmapsto(a,M,3aM-b),\qquad
(a,b,M)\longmapsto(b,M,3bM-a).
\end{equation}
For example, $29\to194$ is a Meridian edge, whereas $29\to169$ is a Vieta
edge. Figure~\ref{fig:fareytrees} compares the two sets of links.

\paperfigure{fareytrees}{.96}{figures/fig2B_farey_trees.pdf}{Two trees on
the same 20 configurations: the 16 with $M\le3000$ and the four additional
values $6466,7561,14701,37666$. Each node displays its reduced standard
Farey label and maximum. Red and green denote the Vieta moves; blue and
gold denote Meridian and phasor steps.}{p}

\paperfigure{fan}{1}{figures/fig2C_reflection_fan.pdf}{The orbit underlying
one recursive extension. Reflecting $H$ across $(u_j,u_{j+1})$ gives
$N_j=3u_ju_{j+1}-H$; $N_0$ is the preceding pole. For source
$(a,b,H)=(1,5,13)$ at $M_{1,1}(0)$, the preceding pole is two,
$(u_{-2},u_{-1},u_2,u_3,u_4)=(1325,34,194,7561,294685)$, and
$(N_{-2},N_{-1},N_1,N_2,N_3)=(135137,89,2897,4400489,6684339842)$.
The right panel locates these reflections in the Vieta tree.}{htbp}

\subsection{Real components and phasor edges}
\begin{proposition}\label{prop:realcomponents}
The canonical real families in \eqref{eq:word}, modulo companion reflection,
are the nonterminal connected components of $H(M,c)=1$. Their terminal phase is
$x+1/2$ modulo one. The separate terminal component is $M=1$.
\end{proposition}
\begin{proof}
The seed companions are at least $2/\sqrt5$, and $M_1\ge2$ is strictly
larger than either. More generally, let $h\ge1$ and
$2/\sqrt5\le a,b\le h$. The first outward endpoint satisfies
$3hb-a\ge(6/\sqrt5-1)h>h$; its reflected pole is larger still.
The reversed orbit gives the same conclusion on the other side.
Every outward block thus has a strict inverse descent to its source and
preserves the terminal phase.

Conversely, reverse any descent ending at host one. Its last host change
produces the seed with $x=v-1/2$, and every preceding walk is nonzero:
a zero walk would simply have stopped at the smaller host. These strict
blocks uniquely recover a locally constant word. The unrolled $x$ joins
the final phase charts continuously. The seed is injective as an ordered
curve, as its endpoint difference is strictly monotone; each block is
invertible. The positive Laurent expansion proved in
Proposition~\ref{prop:laurent} makes every family proper. The inverse charts
are relatively open in the level set, and properness makes the families
closed. Each connected family therefore fills a component. Reflection
gives exactly the canonical identifications.
\end{proof}

The phasor edges are obtained by applying the same outward blocks to
\begin{equation}\label{eq:basearc}
\left(\frac h{\sqrt{3h-2}},\frac h{\sqrt{3h-2}},h\right),\qquad 1\le h\le2.
\end{equation}
This symmetric arc joins $(1,1,1)$ to $(1,1,2)$. Its other host is
$2h/(3h-2)$, and both possible terminal fits have phase $1/2$.
Its endpoints satisfy the bounds used in the proposition, so the outward
blocks preserve that phase. The displayed phasors consequently arise from
the descent itself, rather than interpolation between integer nodes.

\subsection{Degree and the exact Farey dictionary}\label{subsec:dictionary}
Attach endpoint count vectors $u=(1,0)$ and $v=(0,1)$ to the seed. Under
an index $k$ update them by
\begin{equation}\label{eq:countupdate}
(u,v)\longmapsto
\begin{cases}
((k-1)(u+v)+v,\ k(u+v)+v),&k>0,\\
(|k|(u+v)+u,\ (|k|-1)(u+v)+u),&k<0.
\end{cases}
\end{equation}
Write their sum as $(r,s)$. Their determinant remains $\pm1$ and their
unordered residues modulo two remain $\{(1,0),(0,1)\}$. Thus $r,s$ are
positive coprime odd integers. Canonical orientation gives $r\le s$.
The \emph{Meridian degree} is $D=r+s$. It is the extreme Laurent exponent
in $\rho^x$, where $\rho=(3+\sqrt5)/2$; the degree in $\rho^{2x}$ is $D/2$.

\begin{proposition}[Helm classification]\label{prop:helm}
Canonical Meridians correspond bijectively to coprime positive odd pairs
$r\le s$. Their helm labels are the standard Farey fractions $r/s$.
\end{proposition}
\begin{proof}
The endpoint vectors are the two Stern--Brocot parents: their sum is $(r,s)$
and their determinant has absolute value one. Conversely, the parents of
an odd/odd pair have opposite coordinate parities. Their componentwise
absolute difference is a smaller positive primitive odd/odd pair, except
at $(1,1)$. The difference has one componentwise sign; an opposite sign
would give the initial pair. Iteration reaches $(1,1)$.

To recover the signed indices, put $n_\alpha=(r,s)$ and let $\alpha^-$
remove the leftmost index after $1$. Consecutive difference vectors have
determinant of absolute value two, so three consecutive vectors satisfy
$n=2c\,n^-+\varepsilon n^{--}$ with $c\ge1$ and $\varepsilon=\pm1$.
Determinant division gives an integer coefficient, parity makes it even,
and decreasing coordinate sums fix its magnitude and sign. At the bottom
use $(1,1),(-1,1)$; canonical orientation fixes the first index positive.
The magnitude is fixed by the ratio of successive degrees, which lies in
$[2c,2c+1)$ or $(2c-1,2c]$ according to the sign; equality occurs only
at the first applied index. Reversing the process gives precisely
$n_{1,k,k_m,\ldots,k_1}=2|k|n_\alpha+\sgn(kk_m)n_{\alpha^-}$.
For the seed use $k_0=1$ and $n_{\varnothing}=(-1,1)$, so the first
extension has $(r,s)=(2k_1-1,2k_1+1)$. This recovers the unique address.
Mediant addition identifies the label.
\end{proof}

\begin{lemma}[Phase dictionary]\label{lem:dictionary}
For helm $(r,s)$ and degree $D=r+s$, the standard Farey label at phase $x$ is
\[
\begin{array}{c|c|c}
x&(p,q)&p+q\\\hline
0&(r,s)&D\\
1&(r,D)&D+r\\
-1&(s,D)&D+s\\
n\ge2&(D,(n-1)D+r)&nD+r\\
-n\le-2&(D,(n-1)D+s)&nD+s.
\end{array}
\]
The negative seed phases are omitted.
\end{lemma}
\begin{proof}
At seed phases $0,1,n\ge2$, the endpoint bases are respectively
$((1,0),(0,1))$, $((1,1),(0,1))$, and $((1,n),(1,n-1))$.
The count updates are linear, so substitution gives the table; negative
phase exchanges the bases.
For the inverse, odd/odd $p,q$ give $(r,s,x)=(p,q,0)$.
If $q$ is even, sort $p,q-p$ and choose $+1$ when $p$ is the smaller entry,
otherwise $-1$. If $p$ is even, write $q=tp+u$ with $0<u<p$; sort $u,p-u$
and choose $+(t+1)$ or $-(t+1)$ in the same way. A tie is the seed and uses
the nonnegative representative. These formulas recover every mark.
\end{proof}
In particular, degree can be read directly from a reduced label:
\begin{equation}\label{eq:degree-from-label}
D=\begin{cases}p+q,&p,q\text{ odd},\\q,&q\text{ even},\\p,&p\text{ even}.
\end{cases}
\end{equation}
This dictionary connects the Meridian and corridor collision windows below.