\section{Fibonacci matrices, six pieces, and ordering proofs}\label{sec:snakes}
\subsection{From a Farey label to an actual matching count}
A snake graph is a chain of square tiles, each attached to the right of or
above the preceding tile. Shared edges are counted once. A perfect matching
covers every vertex exactly once. The Markov snake associated with a reduced
$0<p/q<1$ is built as follows. Take the lower staircase from $(0,0)$ to
$(q,p)$: at each horizontal coordinate $j$, step right and then up to height
$\lfloor pj/q\rfloor$. Write right steps as $a$ and up steps as $b$.
Write this step sequence as $w_{p/q}=azb$.
Remove its first and last letters; start with one square and replace each
$a$ of $z$ by two added squares to the right and each $b$ by two upward.
The resulting snake has $2(p+q)-3$ tiles.

Write $\mu(w)$ for its number of perfect matchings. The established
Markov matching correspondence gives
\begin{equation}\label{eq:wordcount}
\mu(w_{p/q})=m_{p/q}.
\end{equation}
The matching interpretation originated in the REACH work of 2002--2003.
Musiker's memorandum, dated 27 June 2002, proposed the recursive graph model
\cite{Musiker2002}; Itsara, Musiker, Propp, and Viana's manuscript of
10 January 2003 stated and proved the snake-matching correspondence
\cite[Theorem 3.1]{IMPV2003}. Setting all edge weights to one gives the
ordinary perfect-matching counts. Propp's later exposition, first posted
in 2005, records the broader collaborative development \cite{Propp}.
These original contributions precede the later proofs of Aigner's orderings.

For a modern treatment and the Farey-indexed construction, see
\cite[Section 4.1, Theorem 4.2]{RS} and \cite{CS}.
For example, $w_{2/5}=a^3ba^2b$, its trimmed word is $aabaa$, and the
eleven-tile snake has 194 matchings. Auxiliary rearranged words also specify
snakes, counted by $\mu$; they need not be canonical Markov snakes.

\subsection{Six pieces and two connector states}
Figure~\ref{fig:pieces} fixes the five three-tile pieces and the three
one-square variants of type 6. In a cut between pieces, a matching uses
both yellow connector edges or neither: the cut leaves an even number of
vertices on either side. Denote these states by 1 and 0. At a selected
port remove the two vertices already covered by its connector pair.
Direct counting within the pieces gives
\begin{equation}\label{eq:piececounts}
\begin{array}{c|rrrr}
\text{piece}&00&01&10&11\\\hline
1&5&2&2&1\\2&5&3&3&2\\3&5&2&3&1\\4&5&3&2&1\\5&5&2&2&1\\
6\text{i}&2&1&1&1
\end{array}.
\end{equation}
The state order is incoming, outgoing. The one-port end squares 6ii and
6iii have counts $(2,1)$.

For the ordering surgeries, use a fixed one-ended chop: retain tile triples
$(1,2,3),(5,6,7),\ldots$, with tiles $4,8,\ldots$ providing paired
connectors. A one-square remainder occurs exactly when $p+q$ is even and is
counted with the closed end piece. For every connector sequence $\sigma$,
multiply the applicable row entries in \eqref{eq:piececounts}; then sum
over $\sigma$, with outer ports closed. Each matching determines one
sequence and one internal completion per piece, and gluing reverses the
construction. This proves the counting rule, without an independence
assumption. Equivalently it is a sum of terms $5^{k(\sigma)}3^{j(\sigma)}
2^{\ell(\sigma)}$, where the exponents count factors of the indicated
size. For the $2/5$ snake the piece list $2,1,2$ gives
\[
5^3+5\cdot2\cdot3+3\cdot2\cdot5+3^2=194.
\]

\paperfigure{pieces}{.91}{figures/fig3_pieces.pdf}{The six physical pieces.
Gold edges are paired connectors, magenta marks ports, and dots mark tile
vertices. Two-port counts use the order $00,01,10,11$. The central square
6i has counts $(2,1,1,1)$; the one-port end squares have $(2,1)$.
Types 1 and 5 have equal count rows but different physical orientations.}{htbp}

\Needspace{10\baselineskip}
\subsection{Fibonacci-matrix normalization}
The author proposed the Fibonacci-matrix formulation and normalization
approach developed here. We begin with the comparison of three consecutive
blocks and then apply it to complete products of run blocks.

With $F_0=0,F_1=1$ and $F_{-j}=(-1)^{j+1}F_j$, the basic odd-index identity is
\begin{equation}\label{eq:odd-fibonacci}
F_{2k-1}F_{2k+3}-F_{2k+1}^{\,2}=1.
\end{equation}
For example, $5\cdot34-13^2=1$.
Consider the author's three consecutive Fibonacci blocks:
\begin{equation}\label{eq:fibAB}
\begin{aligned}
A&=\mat{F_{2k+3}}{F_{2k+1}}{F_{2k+1}}{F_{2k-1}},\\
B&=\mat{F_{2k+1}}{F_{2k-1}}{F_{2k-1}}{F_{2k-3}},\\
C&=\mat{F_{2k-1}}{F_{2k-3}}{F_{2k-3}}{F_{2k-5}},\qquad k\ge2.
\end{aligned}
\end{equation}
Applying \eqref{eq:odd-fibonacci} twice gives precisely the upper-left
comparison used in normalization:
\begin{equation}\label{eq:AC-BB}
\begin{aligned}
(AC)_{11}-(B^2)_{11}
&=\bigl(F_{2k+3}F_{2k-1}-F_{2k+1}^2\bigr)\\
&\quad+\bigl(F_{2k+1}F_{2k-3}-F_{2k-1}^2\bigr)=2.
\end{aligned}
\end{equation}
The same identities in the other entries give
\[
AC-B^2=\mat2602,\qquad CA-B^2=\mat2062.
\]
Thus replacing $AC$ or $CA$ by $B^2$ decreases the upper-left entry
even inside a longer product: the surrounding Fibonacci blocks have
nonnegative entries and positive upper-left entries. For $k=3$ this reads
$(AC)_{11}=34\cdot5+13\cdot2=196>194=13^2+5^2=(B^2)_{11}$.

\subsubsection{Run counts and general index shifts}
To apply the same comparison at arbitrary run lengths, index the blocks by
\begin{equation}\label{eq:fibrun}
\mathcal A_t=\mat{F_{2t+3}}{F_{2t+1}}{F_{2t+1}}{F_{2t-1}},
\qquad t\ge0.
\end{equation}
Thus $A=\mathcal A_k$, $B=\mathcal A_{k-1}$ and $C=\mathcal A_{k-2}$
in \eqref{eq:fibAB}. The four boundary counts of a straight block of
$2t+1$ tiles are $F_{2t+3},F_{2t+1},F_{2t+1},F_{2t-1}$:
these follow by conditioning on the first edge of a ladder, giving the
Fibonacci recurrence; at $t=0$ they are $(2,1,1,1)$.
Gluing successive runs with their paired connector states therefore gives,
for $w=a^{h_1}b\cdots a^{h_p}b$ with $h_1\ge1$ and $h_j\ge0$,
\begin{equation}\label{eq:fibword}
\mu(w)=[\mathcal A_{h_1-1}\mathcal A_{h_2}\cdots\mathcal A_{h_p}]_{11}.
\end{equation}
The first run is shortened by the removed initial $a$; the final $b$
closes the other end. Equivalently, a block-index list $(t_1,\ldots,t_p)$
represents $w=a^{t_1+1}ba^{t_2}b\cdots a^{t_p}b$, with
$t_1+\cdots+t_p=q-1$.

Fibonacci addition gives the scalar index-shifting identity
\[
F_{2r+1}F_{2s+1}-F_{2r-1}F_{2s+3}=F_{2(r-s)-2}.
\]
Apply it to the two summands of each product entry to obtain
\begin{equation}\label{eq:runbalance}
\mathcal A_r\mathcal A_s-\mathcal A_{r-1}\mathcal A_{s+1}
=2\mat{F_{2(r-s)-2}}{F_{2(r-s)}}{F_{2(r-s)-4}}{F_{2(r-s)-2}},
\quad r\ge1,\ s\ge0.
\end{equation}
Consequently, whenever two adjacent indices differ by at least two,
moving one unit from the larger to the smaller strictly decreases the
whole product's upper-left entry. The case $r=s+2$ is exactly
\eqref{eq:AC-BB}. The only remaining local comparison has adjacent
indices differing by one; it is an interchange of the two blocks.

\subsubsection{Comparing the surrounding Fibonacci blocks}
For a finite list of block indices, define the ratio of its two open counts by
\[
\rho(u_1,\ldots,u_m)=
\frac{[\mathcal A_{u_1}\cdots\mathcal A_{u_m}]_{21}}
     {[\mathcal A_{u_1}\cdots\mathcal A_{u_m}]_{11}},
\qquad \rho(\varnothing)=0.
\]
All these ratios lie in $[0,1)$. Direct multiplication gives
\begin{equation}\label{eq:fib-ratio}
\rho(t,U)=
\frac{F_{2t+1}+F_{2t-1}\rho(U)}
     {F_{2t+3}+F_{2t+1}\rho(U)}.
\end{equation}
Its right-hand side increases strictly with $\rho(U)$, since
\eqref{eq:odd-fibonacci} gives $\det\mathcal A_t=1$.
Moreover, another application of that identity gives
\[
\frac{F_{2t+1}}{F_{2t+3}}-
\frac{F_{2t+3}+F_{2t+1}}{F_{2t+5}+F_{2t+3}}
=\frac{1}{F_{2t+3}(F_{2t+5}+F_{2t+3})}>0.
\]
Hence $\rho(t+1,U)<\rho(t,V)$ for any two finite tails $U,V$.
Two lists can therefore be compared by cancelling equal initial indices:
at the first unequal pair, the larger index gives the smaller ratio;
an exhausted list has the smaller ratio than any nonempty remainder.

For adjacent blocks in positions $j,j+1$, the outward lists are
$(t_{j-1},\ldots,t_1)$ on the left and $(t_{j+2},\ldots,t_p)$ on the right.
At $r=s+1$, \eqref{eq:runbalance} becomes
\[
\mathcal A_{s+1}\mathcal A_s-\mathcal A_s\mathcal A_{s+1}
=2\mat01{-1}0.
\]
After multiplication by the surrounding blocks, its upper-left entry has
the sign of
\begin{equation}\label{eq:flip}
\rho(t_{j+2},\ldots,t_p)-\rho(t_{j-1},\ldots,t_1).
\end{equation}
Indeed, it equals this difference times twice the positive upper-left
entries of the two surrounding products. This proves the needed
surrounding-block test entirely in the same Fibonacci matrices.

\begin{lemma}[Fibonacci-matrix normalization]\label{lem:normalization}
For coprime positive $p,q$, every word $w=azb$ with $q$ letters $a$ and
$p$ letters $b$ satisfies $\mu(w_{p/q})\le\mu(w)$.
\end{lemma}
\begin{proof}
For $p=1$ there is only the word $a^qb$. Assume $p>1$ and use the
block-index list in \eqref{eq:fibword}. Its prefix discrepancies are
\[
\delta_j=qj-p(1+t_1+\cdots+t_j)\quad(1\le j\le p),
\qquad \delta_0=\delta_p=0.
\]
Changing $(t_j,t_{j+1})$ to $(t_j+1,t_{j+1}-1)$ decreases only
$\delta_j$, by $p$; the reverse change increases only $\delta_j$, by $p$.
We show that an appropriate change always decreases the matching count
until $-p\le\delta_j\le0$ for every $j$.

The sign checks use just a prefix-sum identity. Put $h_1=t_1+1$ and
$h_i=t_i$ for $i>1$, the actual run lengths. Whenever $j-m\ge0$ and
$j+m\le p$,
\begin{equation}\label{eq:prefix-pair}
\delta_{j-m}+\delta_{j+m}
=2\delta_j+p\left(
\sum_{i=j-m+1}^{j}h_i-\sum_{i=j+1}^{j+m}h_i\right).
\end{equation}
In the block indices, the left sum has the extra $1$ precisely when it
includes $t_1$.

\emph{First remove positive discrepancies.}
Choose the rightmost largest $\delta_j>0$ and transfer one unit from
$t_{j+1}$ to $t_j$. Comparison with its two neighboring discrepancies
gives $h_j<q/p<h_{j+1}$; coprimality makes these inequalities strict.
Thus the transfer is allowed and $t_{j+1}\ge t_j+1$; for $j=1$ the
gap is at least two. Gaps of at least two are handled by
\eqref{eq:runbalance}.

If $t_{j+1}=t_j+1$, compare the outward lists. A first unequal pair with
the left index larger would make the parenthesis in
\eqref{eq:prefix-pair} nonnegative: the central difference is $-1$,
the equal outward indices cancel, and the first unequal difference is
at least $1$. This contradicts maximality and the choice of the rightmost
maximum. If the left list were exhausted first, its extra initial $1$
would instead give $\delta_{2j}=2\delta_j>\delta_j$, again impossible
(including simultaneous exhaustion, when $2j=p$).
The left outward ratio is therefore strictly larger than the right one.
Reversing the interchange in \eqref{eq:flip} shows that the selected
transfer strictly decreases the count. It also decreases
$\sum_{j=1}^{p-1}\delta_j$. There are only finitely many nonnegative
block-index lists of sum $q-1$, so this phase ends with all $\delta_j\le0$.

\emph{Then remove discrepancies below $-p$.}
Choose the leftmost smallest $\delta_j<-p$ and transfer one unit from
$t_j$ to $t_{j+1}$. For $j>1$, the neighboring comparisons give
$h_j>q/p>h_{j+1}$; for $j=1$, the inequality $\delta_1<-p$ gives
$t_1>q/p$, while $t_2<q/p$. Thus $t_j\ge t_{j+1}+1$ and the transfer
is allowed. Again a gap of at least two is settled by
\eqref{eq:runbalance}.

In the gap-one case, a first unequal outward pair with the left index
smaller would make the parenthesis in \eqref{eq:prefix-pair} nonpositive,
contradicting minimality and the leftmost choice. If that pair reaches
$t_1$, its extra $1$ instead gives
$\delta_{2j}\le2\delta_j+p<\delta_j$, the same contradiction.
If the right list were exhausted first, the identity would give
$\delta_{2j-p}=2\delta_j+p<\delta_j$.
Simultaneous exhaustion would give $\delta_j=-p$.
Thus the left outward ratio is strictly smaller than the right one,
and \eqref{eq:flip} proves a strict decrease in count.
The changed discrepancy is $\delta_j+p<0$, so positivity is not
reintroduced. This phase increases the bounded sum of discrepancies
by $p$ each time and terminates with $-p\le\delta_j\le0$.

Finally, $\delta_j\equiv qj\pmod p$. For $1\le j<p$, coprimality
leaves exactly one possible value in $[-p,0]$, forcing
\[
1+t_1+\cdots+t_j=\left\lceil\frac{jq}{p}\right\rceil.
\]
These are precisely the canonical run lengths. Every transfer has
decreased the upper-left product entry, proving the assertion.
\end{proof}

For the canonical word, $h_j=\lceil jq/p\rceil-\lceil(j-1)q/p\rceil$.
With $k=1+\lfloor(q-1)/p\rfloor$, the factors are the author's $A,B$
from \eqref{eq:fibAB}, and the end factors are $B$.
For example, $m_{3/8}=[BAB]_{11}$ with $k=3$.
A similar normalization result appears in \cite[Proposition 21]{LPTV};
the proof above is self-contained in the author's Fibonacci matrices.


\subsection{The three Aigner orderings}
Fixed-numerator monotonicity was proved by Rabideau and Schiffler
\cite{RS}; all three orderings were proved by LPTV, Lee--Li--Rabideau--Schiffler,
and McShane \cite{LPTV,LLRS,McShane}. The author's Fibonacci matrices prove normalization and the fixed-numerator
ordering. His six pieces prove the remaining fixed-sum comparison and
the fixed-denominator ordering, using normalization to return to a
canonical Markov snake.
\begin{theorem}[Established monotonicities]\label{thm:aigner}
For $i>0$, whenever both fractions are reduced and lie in $(0,1)$,
\begin{align*}
m_{p/q}&>m_{(p+i)/(q-i)}&&\text{(fixed sum)},\\
m_{p/q}&<m_{p/(q+i)}&&\text{(fixed numerator)},\\
m_{p/q}&<m_{(p+i)/q}&&\text{(fixed denominator)}.
\end{align*}
\end{theorem}
\Needspace{6\baselineskip}
\subsubsection{Fixed numerator: Fibonacci matrices}
\begin{proof}
Fibonacci recurrence gives, for $t\ge1$,
\[
\mathcal A_t-\mathcal A_{t-1}=\mat{F_{2t+2}}{F_{2t}}{F_{2t}}{F_{2t-2}}.
\]
All entries are nonnegative and the upper-left is positive. Start with
$w_{p/(q+i)}$ and delete $i$ internal letters $a$, retaining the first.
Each deletion changes one factor $\mathcal A_t$ to $\mathcal A_{t-1}$ with $t\ge1$.
Surrounding factors are positive, so the matching count strictly falls.
The resulting word ends at $(q,p)$; normalization gives
$m_{p/q}\le\mu(w_*)<m_{p/(q+i)}$.
\end{proof}

\Needspace{6\baselineskip}
\subsubsection{Fixed sum: six pieces}
\begin{proof}
At the same physical cuts, the changes $2\to3$ or $4$, then $3$ or
$4\to5$, increase no entry of \eqref{eq:piececounts}; the changed factors
are $3\to2$ or $2\to1$. Rotation $5\leftrightarrow1$ preserves the row.
Selecting the changed connector and no others exhibits a strictly smaller
positive summand whenever an equal-direction join becomes unequal.

The trimmed canonical word begins and ends with $a$, contains no $bb$,
and has $q-p-1$ equal joins, all $aa$. At its rightmost equal join,
interchange $a,b$ strictly to the right. That suffix is alternating, so
only the chosen join changes. Exactly one prescribed piece changes by
the stated rules, while any rotated pieces keep their rows. Each surgery
strictly decreases the count. Odd-numbered operations replace one
horizontal step by a vertical step in net; even-numbered operations keep
the totals. After $2i$ operations the endpoint is $(q-i,p+i)$.
There are enough joins because $p+i<q-i$.
For the resulting auxiliary word $w_*$, normalization gives
$m_{(p+i)/(q-i)}\le\mu(w_*)<m_{p/q}$.
A similar tail comparison appears in \cite[Proposition 22]{LPTV};
here the comparison is proved using the author's six physical pieces.
\end{proof}

\paperfigure{surgery}{.98}{figures/fig4_surgery.pdf}{Fixed-sum surgery on
the same five pieces. The canonical $2/9$ snake has 9077 matchings; after
two changes the auxiliary snake has 7661; normalization gives the $3/8$
snake with 7561. The intermediate first change, not shown, has 8263.
Gold pairs mark cuts and teal dots the starting corners.}{htbp}



\Needspace{6\baselineskip}
\subsubsection{Fixed denominator: six pieces}
\begin{proof}
For the fixed-denominator comparison, refine the same six-piece chop into
single-square pieces of type 6, retaining odd-numbered tiles and using the
even tiles as paired connectors. A retained square has counts
$(2,1,1,1)$ for opposite ports and $(2,1,1,0)$ for adjacent ports.
The zero forbids covering their common corner twice. Summing an intervening
state regroups these squares into the original three-tile pieces.

Insert one internal $b$, adding two vertical squares. The affected local
counts are the following complete table, with outside ports fixed:
\[
\begin{array}{c|rrrr|rrrr}
&\multicolumn{4}{c|}{\text{before}}&\multicolumn{4}{c}{\text{after}}\\
\text{directions}&00&01&10&11&00&01&10&11\\\hline
a,a&2&1&1&1&5&2&2&1\\
b,b&2&1&1&1&5&3&3&2\\
a,b&2&1&1&0&5&3&2&1\\
b,a&2&1&1&0&5&2&3&1
\end{array}.
\]
The endpoint caps have the corresponding rows with an outer port closed;
the empty trimmed word has the increase $2\to5$. No outside count falls,
and the all-zero choice witnesses a strict increase. Thus the two added
squares compensate for every changed direction.

Starting from $w_{(p+i)/q}$, delete $i$ internal $b$'s and reverse this
comparison. At least $i$ are available. Normalizing the resulting word
to $(q,p)$ proves $m_{p/q}\le\mu(w_*)<m_{(p+i)/q}$.
\end{proof}

\subsection{Folding an actual snake at its center}
The same local counts also compare families sharing a long initial sequence
of pieces. Represent the four counts by
\[
R_1=R_5=\mat5221,\quad R_2=\mat5332,\quad
R_3=\mat5231,\quad R_4=\mat5321.
\]
Attaching piece $i$ to an open-count row $(x,y)$ replaces it by $(x,y)R_i$.
This is the sum over the common connector state, not an additional
assumption about matchings.

For a helm, use the symmetric central chop: retain the middle square as
6i and group retained squares in pairs outward on each side. For odd/odd
$p,q$, each half has $(p+q-2)/2$ retained squares. If this number is odd,
one square remains at each outer end, giving pieces 6ii and 6iii with
initial counts $(2,1)$. Otherwise the half starts at $(1,0)$ and consists
of complete three-tile pieces. This repartition requires rebuilding the
piece list; one cannot move a cut and keep its old boundary counts.

\begin{proposition}[Folded matching counts]\label{prop:fold}
For a reduced $0<p<q$, let $(x,y)$ be the open counts of one reflected
half under the central chop. If $p,q$ are odd, then
\begin{equation}\label{eq:fold}
m_{p/q}=2x^2+2xy+y^2=(x+y)^2+x^2.
\end{equation}
If $p+q$ is odd, then $m_{p/q}=x^2+y^2$.
In either case $\gcd(x,y)=1$; a nonempty half has positive counts.
\end{proposition}
\begin{proof}
The trimmed staircase sequence is a palindrome: for $0<j<q$,
$\lfloor p(q-j)/q\rfloor=p-1-\lfloor pj/q\rfloor$.
For odd/odd labels the two halves meet at an opposite-port central square,
whose four counts are $2,1,1,1$. The reflected halves contribute
$x^2,xy,yx,y^2$, giving \eqref{eq:fold}. For opposite parities they meet
across a single connector and must use its same state, giving $x^2+y^2$.
Every full piece table and every exposed-square table has determinant
$\pm1$, so its integral inverse preserves the gcd of an open row.
Starting at $(1,0)$ proves primitivity. The first square gives $(2,1)$,
and every subsequent permitted table preserves positivity.
\end{proof}
The seed $1/1$ consists only of its central square. The two examples in
Figure~\ref{fig:centre} give
$m_{1/3}=2\cdot2^2+2\cdot2\cdot1+1^2=13$ and
$m_{1/5}=2\cdot5^2+2\cdot5\cdot3+3^2=89$.
\paperfigure{centre}{.96}{figures/fig5_centred_helm.pdf}{Central and end
pieces in actual helms. At $1/3$, central square 6i has the end squares
6ii and 6iii as its halves, with counts $(2,1)$. At $1/5$, each half is a
type-2 triple with counts $(5,3)$. Folding produces 13 and 89.}{htbp}

\subsection{A common prefix leaves one quadratic comparison}
Two folded graphs with the same outer boundary convention may share a
prefix with open counts $(x,y)$. Sum over all remaining states, leaving
that cut open. The two centers give positive definite symmetric tables
$G,\widetilde G$ with determinant one and nonnegative off-diagonal entries.
For a helm they have the form $SQS^{\mathsf T}$ with $Q=\mat2111$;
for opposite-parity labels they have form $SS^{\mathsf T}$.
Thus
\begin{equation}\label{eq:quadratic}
m-\widetilde m=\alpha x^2+2\beta xy+\gamma y^2,
\qquad G-\widetilde G=\mat\alpha\beta\beta\gamma.
\end{equation}
\begin{proposition}[At most one primitive common prefix]\label{prop:quadratic}
For a nonempty common part and $G\ne\widetilde G$, equality requires
$\alpha\gamma<0$ and $\beta^2-\alpha\gamma$ to be an integer square.
There is then at most one primitive positive row $(x,y)$ giving equality.
That row must still arise from both actual prescribed snakes.
\end{proposition}
\begin{proof}
A center table $\mat a b b c$ has $ac-b^2=1$ and $b\ge0$.
If its two diagonal differences from another table have the same weak
sign, then the difference of $b=\sqrt{ac-1}$ has that sign too. A nonzero
difference cannot vanish at a positive row. Opposite diagonal signs are
therefore necessary. The resulting quadratic in $y/x$ has one positive
root. An integer zero obeys
$(\gamma y+\beta x)^2=(\beta^2-\alpha\gamma)x^2$;
a rational square root of an integer is integral. A positive rational
root has only one primitive positive representative.
\end{proof}

For an empty prefix, test its actual boundary row. If $G=\widetilde G$,
the criterion supplies no exclusion. Equal count tables do not certify
equal physical graphs: types 1 and 5 already show why orientations and the
prescribed sequence of horizontal and vertical steps must be retained.

A useful uniform exclusion follows directly modulo three. The physical
piece tables satisfy $R_iQR_i^{\mathsf T}\equiv Q\pmod3$, whereas a
single retained square gives $-Q$. A helm tail with $d$ retained squares
therefore has center table $(-1)^dQ\pmod3$. Opposite parities give a
nonzero difference at every primitive row because the discriminant of $Q$
is $-1$, a nonsquare modulo three. For equal parity and nonzero difference,
let $h=\min_{i,j}v_3((G-\widetilde G)_{ij})$ and write
$3^{-h}(G-\widetilde G)=\mat\alpha\beta\beta\gamma$.
Then $h\ge1$. Expanding the equal determinants and reducing modulo three
gives $\gamma\equiv\alpha+\beta$; the normalized discriminant is one or
two modulo three. If it is two, every primitive shared row has count
difference of valuation exactly $h$. These are exclusions for specified
tails with arbitrary actual common prefixes, not a classification of all
possible snake-graph collisions.