
%% if you are submitting an initial manuscript then you should have submission as an option here
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\documentclass[finalversion]{FPSAC2025}
\articlenumber{92}
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\usepackage{amsfonts}
%\usepackage[utf8]{inputenc}
\usepackage{graphics,graphicx}
\usepackage{verbatim}
\usepackage{tikz}
\usepackage{tikz-cd}
%\usepackage[colorinlistoftodos]{todonotes}
\usepackage{pgf}
\usepackage{cancel}
\newtheorem{thm}{Theorem}[section]
\newtheorem{lem}[thm]{Lemma}
\newtheorem{conj}[thm]{Conjecture}
\newtheorem{prop}[thm]{Proposition}
\theoremstyle{remark}
\newtheorem{rem}[thm]{Remark}
\newtheorem{ex}[thm]{Example}
\newtheorem{claim}[thm]{Claim}

\newtheorem{defn}[thm]{Definition}


%case symbols
\newcommand{\Blank}{\mathsf{B}}
\newcommand{\Terminal}{\mathsf{T}}
\newcommand{\Cross}{\mathsf{C}}
%\newcommand{\NCross}{\overline{\mathsf{C}}}
\newcommand{\NCross}{\bcancel{\mathsf{C}}}
%\newcommand{\NCross}{
%\begin{tikzpicture}
%\draw[ultra thick,color=red] (.25,0) arc (0:360:.25);
%\draw[ultra thick,color=red] (0,0)--(135:.25);
%\draw[ultra thick,color=red] (0,0)--(315:.25);
%\node (0,0) {$C$};
%\end{tikzpicture}}

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\newcommand{\Initial}{\mathsf{I}}
%\newcommand{\NPlus}{\overline{\mathsf{P}}}
\newcommand{\NPlus}{\bcancel{\mathsf{P}}}

\newcommand{\Plus}{\mathsf{P}}

%temporary to get rid of syntax errors
\newcommand{\Bnot}{\overline{B}}
\newcommand{\Dnot}{\overline{\text{D}}}
\newcommand{\Jnot}{\overline{\text{J}}}
\newcommand{\Tnot}{\overline{\text{T}}}

\newcommand{\Groth}{\mathfrak{G}}
\newcommand{\Schub}{\mathfrak{S}}
\newcommand{\Gb}{\mathfrak{G}^{(\beta)}}
\newcommand{\id}{\mathrm{id}}
\newcommand{\MBPD}{\mathrm{MBPD}}
\newcommand{\PD}{\mathrm{PD}}

\newcommand{\pop}{\vee}
\newcommand{\RCP}{\mathrm{CP}}%changed 
\newcommand{\red}{\mathrm{red}}
\newcommand{\ua}{\underline{a}}
\newcommand{\ui}{\underline{i}}
\newcommand{\wt}{\mathrm{wt}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\rowpop}{\nabla_r}
\DeclareMathOperator{\rpop}{\mathrm{rpop}}
\newcommand{\es}{e^*}
\newcommand{\fs}{f^*}
\newcommand{\ftarg}{F}
\newcommand{\etarg}{E}
\newcommand{\window}{window}
\newcommand{\Y}{Y}

\newcommand{\XX}{X}
\newcommand{\II}{I}

\newcommand{\term}{\mathrm{T}}
\newcommand{\nonterm}{{\mathrm{\overline{T}}}}

\newcommand{\Daoji}[1]{\todo[size=\tiny,color=purple!30]{#1}}
%\newcommand{\Mark}[1]{\todo[size=\tiny,inline,color=green!30]{#1
%\\ \hfill --- Mark}}
\newcommand{\Mark}[1]{\todo[size=\tiny,color=green!30]{#1}}
\newcommand{\Tianyi}[1]{\todo[size=\tiny,color=pink!30]{#1}}


\newcommand{\btile}{
 \begin{tikzpicture}[x=1em,y=1em,thick,color = blue]
\draw[step=1,gray,thin] (0,0) grid (1,1);
\draw[color=black, thick, sharp corners] (0,0) rectangle (1,1);
\end{tikzpicture}}

\newcommand{\htile}{
\begin{tikzpicture}[x=1em,y=1em,thick,color = blue]
\draw[step=1,gray,thin] (0,0) grid (1,1);
\draw[color=black, thick, sharp corners] (0,0) rectangle (1,1);
\draw(1.0,0.5)--(0.0,0.5);
\end{tikzpicture}
}

\newcommand{\vtile}{
\begin{tikzpicture}[x=1em,y=1em,thick,color = blue]
\draw[step=1,gray,thin] (0,0) grid (1,1);
\draw[color=black, thick, sharp corners] (0,0) rectangle (1,1);
\draw(0.5,1.0)--(0.5,0.0);
\end{tikzpicture}}

\newcommand{\ptile}{
\begin{tikzpicture}[x=1em,y=1em,thick,color = blue]
\draw[step=1,gray,thin] (0,0) grid (1,1);
\draw[color=black, thick, sharp corners] (0,0) rectangle (1,1);
\draw(0.5,1.0)--(0.5,0.0);
\draw(1.0,0.5)--(0.0,0.5);
\end{tikzpicture}}

\newcommand{\rtile}{
\begin{tikzpicture}[x=1em,y=1em,thick,color = blue]
\draw[step=1,gray,thin] (0,0) grid (1,1);
\draw[color=black, thick, sharp corners] (0,0) rectangle (1,1);
\draw(1.0,0.5)--(0.5,0.5)--(0.5,0.0);
\end{tikzpicture}}

\newcommand{\jtile}{
\begin{tikzpicture}[x=1em,y=1em,thick,color = blue]
\draw[step=1,gray,thin] (0,0) grid (1,1);
\draw[color=black, thick, sharp corners] (0,0) rectangle (1,1);
\draw(0.5,1.0)--(0.5,0.5)--(0.0,0.5);
\end{tikzpicture}}

\newcommand{\mtile}{
\begin{tikzpicture}[x=1em,y=1em,thick,color = blue]
\draw[step=1,gray,thin] (0,0) grid (1,1);
\draw[color=black, thick, sharp corners] (0,0) rectangle (1,1);
\draw(0.5,1.0)--(0.5,0.5)--(0.0,0.5);
\node at (0.5,0.5) {$\bullet$};
\end{tikzpicture}}

\newcommand{\bumptile}{
\begin{tikzpicture}[x=1em,y=1em,thick,rounded corners,color = blue]
\draw[step=1,gray,thin] (0,0) grid (1,1);
\draw[color=black, thick, sharp corners] (0,0) rectangle (1,1);
\draw(1.0,0.5)--(0.5,0.5)--(0.5,0.0);
\draw(0.5,1.0)--(0.5,0.5)--(0.0,0.5);
\end{tikzpicture}}

\newcommand{\righttile}{
\begin{tikzpicture}[x=1em,y=1em,thick,rounded corners,color = blue]
\draw[step=1,gray,thin] (0,0) grid (1,1);
\draw[color=black, thick, sharp corners] (0,0) rectangle (1,1);
\draw(0.5,0.5)--(1.0,0.5);
\end{tikzpicture}}

\newcommand{\lefttile}{
\begin{tikzpicture}[x=1em,y=1em,thick,rounded corners,color = blue]
\draw[step=1,gray,thin] (0,0) grid (1,1);
\draw[color=black, thick, sharp corners] (0,0) rectangle (1,1);
\draw(0.0,0.5)--(0.5,0.5);
\end{tikzpicture}}

\newcommand{\uptile}{
\begin{tikzpicture}[x=1em,y=1em,thick,rounded corners,color = blue]
\draw[step=1,gray,thin] (0,0) grid (1,1);
\draw[color=black, thick, sharp corners] (0,0) rectangle (1,1);
\draw(0.5,0.5)--(0.5,1.0);
\end{tikzpicture}}

\newcommand{\downtile}{
\begin{tikzpicture}[x=1em,y=1em,thick,rounded corners,color = blue]
\draw[step=1,gray,thin] (0,0) grid (1,1);
\draw[color=black, thick, sharp corners] (0,0) rectangle (1,1);
\draw(0.0,0.5)--(0.5,0.5);
\end{tikzpicture}}


\newcommand{\Ta}{\btile\rtile}
\newcommand{\Tb}{\htile\ptile}
\newcommand{\Da}{\rtile\ptile}
\newcommand{\Db}{\ptile\jtile}
\newcommand{\Oa}{\otile\rtile}
\newcommand{\Ob}{\htile\jtile}
\newcommand{\Ba}{\btile\vtile}
\newcommand{\Bb}{\rtile\ptile}
\newcommand{\Ca}{\rtile\gtile}
\newcommand{\Cb}{\ptile\htile}
\newcommand{\Na}{\rtile\gtile}
\newcommand{\Nb}{\vtile\rtile}


%% define your title in the usual way
\title[]{A Bijection Between Marked Bumpless Pipedreams and Compatible Pairs}

%% define your authors in the usual way
%% use \addressmark{1}, \addressmark{2} etc for the institutions, and use \thanks{} for contact details
%% for more than 3 names, use the Oxford comma: One Author, Two Author, Red Author, \and Blue Author
\author[Daoji Huang,
Mark Shimozono,
and Tianyi Yu]{\texorpdfstring{Daoji Huang\thanks{\href{mailto:dhuang@ias.edu}{dhuang@ias.edu}. DH was supported by NSF-DMS2202900.}\addressmark{1},
Mark Shimozono\thanks{\href{mshimo@math.vt.edu}{mshimo@math.vt.edu}}\addressmark{2},
\and Tianyi Yu\thanks{\href{mailto:yu.tianyi@uqam.ca}{yu.tianyi@uqam.ca}}\addressmark{3}}{Daoji Huang,
Mark Shimozono,
and Tianyi Yu}}

%% then use \addressmark to match authors to institutions here
\address{\addressmark{1}School of Mathematics, Institute for Advanced Study, Princeton NJ, USA \\ \addressmark{2}Department of Mathematics, Virginia Tech, Blacksburg VA, USA
\\ \addressmark{3}Laboratoire d'Alg\`ebre, de Combinatoire et d'Informatique Math\'ematique, Universit\'e du\\ Qu\'ebec \`a Montr\'eal, Montreal QC, Canada
}

%% put the date of submission here
\received{\today}

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%\revised{}

%% put your English abstract here, or comment this out if you don't have one yet
%% please don't use custom commands in your abstract / resume, as these will be displayed online
%% likewise for citations -- please don't use \cite, and instead write out your citation as something like (author year)
\abstract{We construct a bijection between marked bumpless pipedreams with compatible pairs, which are in bijection with not-necessarily-reduced pipedreams. This directly unifies various formulas for Grothendieck polynomials in the literature. Our bijection is a generalization of a variant of the bijection of Gao and Huang in the unmarked, reduced case.}


%% put your keywords here, or comment this out if you don't have them yet
\keywords{Grothendieck polynomials,
pipedreams, bumpless pipedreams}

%% you can include your bibliography however you want, but using an external .bib file is STRONGLY RECOMMENDED and will make the editor's life much easier
%% regardless of how you do it, please use numerical citations; i.e., [xx, yy] in the text

%% this sample uses biblatex, which (among other things) takes care of URLs in a more flexible way than bibtex
%% but you can use bibtex if you want
\usepackage[backend=bibtex, giveninits]{biblatex}
\addbibresource{ref.bib}
\appto\bibfont{\setlength{\emergencystretch}{2.1em}}

%% note the \printbibliography command at the end of the file which goes with these biblatex commands

\begin{document}

\maketitle
%% note that you DO NOT have to put your abstract here -- it is generated by \maketitle and the \abstract and \resume commands above

\section{Introduction}
This paper establishes an explicit bijection between compatible pairs
and marked bumpless pipedreams, which are two kinds of combinatorial objects
that yield combinatorial formulas for the \emph{$\beta$-Grothendieck polynomials}
$\{\Gb_w\mid w\in S_\infty := \bigcup_{i=1}^\infty S_n\} \subset \mathbb{Z}[x_1,x_2,\dotsc]$.
%the \emph{$\beta$-Grothendieck polynomials} $\Gb_w$ reside in the polynomial ring $\mathbb{Z}[\beta][x_1, x_2, \ldots]$. 
These polynomials are defined recursively by divided differences in \cite[Section 4]{FK}.

The $\beta$-Grothendieck polynomials indexed by $w \in S_n$ form a Schubert basis for the connective $K$-theory of the flag variety~\cite{H}. 
They have two important specializations:
\textbf{Grothendieck polynomials} $\Groth_w = \Gb_w|_{\beta=-1}$, which represent structure sheaves of Schubert varieties in the flag variety, and 
\textbf{Schubert polynomials} $\Schub_w = \Gb_w|_{\beta=0}$, which represent the cohomology classes of Schubert varieties~\cite{BGG, Dem, LS}.
Notably, $\Schub_w$ is the coefficient of $\beta^0$ in $\Gb_w$, or equivalently the lowest $x$-degree component of $\Gb_w$.

%The $\beta$-Grothendieck polynomials have coefficients in $\mathbb{Z}_{\geq 0}$, which allows for combinatorial formulas. Specifically, 
The polynomials $\Gb_w$ can be expressed as a sum over certain combinatorial objects, where each object contributes one monomial. In such formulas, an object is termed ``reduced'' if it contributes to the lowest $x$-degree component of $\Gb_w$. Restricting to reduced objects yields a combinatorial formula for Schubert polynomials.
This paper involves three such combinatorial formulas for $\Gb_w$:
\begin{itemize}
\itemsep0em
    \item Fomin and Kirillov~\cite[Proposition~3.3]{FK} provided a \textbf{compatible pair} formula for $\Gb_w$, extending the reduced compatible pair formula of Billey, Jockusch, and Stanley~\cite{BJS} for Schubert polynomials.
    \item Knutson and Miller~\cite{KM} interpreted $\Gb_w$ as $K$-polynomials of matrix Schubert varieties (up to change of variables) and showed via Gr\"obner degeneration that they can be computed using \textbf{pipedreams}, generalizing the reduced pipedream formula of Billey and Bergeron~\cite{BB} for Schubert polynomials.
    \item Weigandt~\cite{W} reinterpreted Lascoux's formulas~\cite{Las} for Grothendieck polynomials in terms of alternating sign matrices (ASMs) and obtained the \textbf{marked bumpless pipedream (MBPD)} formula for $\Gb_w$. The reduced version for Schubert polynomials was introduced by Lam, Lee, and Shimozono~\cite{LLS}.
\end{itemize}

It is natural to ask for explicit bijections between these three sets of combinatorial objects. The connection between compatible pairs and pipedreams is classical: Billey and Bergeron~\cite{BB} provided a bijection between reduced compatible pairs and reduced pipedreams, and their bijection generalizes directly to the non-reduced case. On the other hand, Gao and Huang~\cite{GH} constructed a bijection between reduced pipedreams and reduced bumpless pipedreams. The goal of this paper is to extend the bijection of Gao and Huang, furnishing a bijection between pipedreams and marked bumpless pipedreams. This completes the connections between the three objects mentioned above. In a subsequent paper, we will show that our bijection restricts to that of~\cite{GH} in the reduced case.

In fact, long before the discovery of bumpless pipedreams~\cite{LLS}, it was known that the 2-enumeration of alternating sign matrices of size $n$ is $2^{n(n-1)/2}$~\cite{MPR, EKLP, Kuperberg}, which coincides with the number of non-reduced pipedreams of size $n$. Thus, based on Weigandt's observation connecting alternating sign matrices and bumpless pipedreams~\cite{W}, our bijection can be viewed as a bijection between 2-enumerated ASMs and non-reduced pipedreams. This connection was recently explored by Striker and Huang~\cite{HS}, who established a partial bijection between totally symmetric self-complementary plane partitions and ASMs using the Gao--Huang bijection.

The rest of the paper is structured as follows. In \Cref{S: objects}, we define the three combinatorial objects. In \Cref{S: bijection}, we describe our bijection using two pairs of mutually inverse operators $\fs_i, \es_i$ and $f_i, e_i$ on MBPDs. In \Cref{S: ef} we define these operators.

\section{Combinatorial objects}
\label{S: objects}
Fix $n \in \mathbb{Z}_{\geq 0}$. In this section, we define compatible pairs $\RCP(n)$, pipedreams $\PD(n)$, and marked bumpless pipedreams $\MBPD(n)$. Let $\Gamma$ be one of these three sets. Each object $\gamma \in \Gamma$ has an associated permutation $w(\gamma)$ and  composition\footnote{Our compositions are weak, that is, sequences of nonnegative integers that are almost all zero.} $\wt(\gamma)$, called its weight. Let $\Gamma(w)$ be the set of $\gamma \in \Gamma$ such that $w(\gamma) = w$. 
Let $\ell(w)$ be the length of $w$.
Then one may write 
$$
\Groth_w = \sum_{\gamma \in \Gamma(w)}
\beta^{|\wt(\gamma)| - \ell(w)} x^{\wt(\gamma)}.
$$
Here, $|\wt(\gamma)|$ is the sum of entries in $\wt(\gamma)$, and $x^{\wt(\gamma)}$ is the monomial where the power of $x_i$ is the $i\textsuperscript{th}$ entry in $\wt(\gamma)$. In addition, each such $\Gamma(w)$ has a subset $\Gamma^\red(w)$ consisting of reduced $\gamma$, meaning that $|\wt(\gamma)| = \ell(w)$. Thus, 
$$
\Schub_w = \sum_{\gamma \in \Gamma^\red(w)} x^{\wt(\gamma)}.
$$

\subsection{Compatible pairs}
A \textit{biletter} is an ordered pair of integers $(i,a)$ with $1 \le i \le a < n$. A \textit{compatible pair}\footnote{Our definition differs from the definition in the literature
by reversing the order of these biletters.} is a sequence of biletters
\begin{align}\label{E:RCP}
B = ((i_1,a_1),(i_2,a_2),\dotsc,(i_\ell,a_\ell))
\end{align}
which are strictly decreasing in the order given by $(i_1,a_1) > (i_2,a_2)$ if $i_1 > i_2$ or if $i_1 = i_2$ and $a_1 < a_2$. Let $\RCP(n)$ denote the set of compatible pairs. The weight of $B$ is defined by $\wt(B) = (m_1,\dotsc,m_n) \in \mathbb{Z}_{\geq 0}^n$, where $m_i$ is the number of biletters in $B$ of the form $(i,a)$ for some $a$. Let $|B|$ be the length of the sequence $B$, or equivalently $|\wt(B)|$.

The Demazure or $0$-Hecke product $*$ on permutations is the unique monoid structure such that for any simple reflection $s_i$ and permutation $w$, 
\begin{equation*}
    s_i * w = \begin{cases}
        s_i w & \text{if $s_i w > w$}, \\
        w & \text{otherwise}.
    \end{cases}
\end{equation*}
Every $B \in \RCP(n)$ has an associated permutation
$w(B) = s_{a_\ell} * s_{a_{\ell-1}} * \dotsm * s_{a_1}$
with notation as in \eqref{E:RCP}. 
Note that the subscripts of $a$ are \emph{decreasing}.

\subsection{Pipedreams}
Let $[n] = \{1,2,\dotsc,n\}$. A %(not-necessarily-reduced) 
pipedream is a tiling of $\{(i,j) \in [n] \times [n] : i+j \le n+1\}$, such that all entries $(i,j)$ with $i+j = n+1$ have tiles $\jtile$, and all other entries are either $\ptile$ or $\bumptile$. We use the matrix-style notation $D_{i,j}$ for the tile in the $i$-th row and $j$-th column. The weight $\wt(D)$ is the weak composition whose $i\textsuperscript{th}$ entry is the number of $\ptile$ in row $i$.

Every such diagram $D$ can be viewed as $n$ pipes entering from the top and exiting to the left. We label the pipe entering from column $c$ as pipe $c$. We trace the pipes from top to left with the proviso: if two pipes have already crossed, then their subsequent crossings are ignored. Then we read the labels on the left edge of $D$ from top to bottom, obtaining $w(D) \in S_n$.

\begin{rem} \label{R:CP and PD} 
There is a weight-preserving bijection 
$\RCP(n) \to \PD(n)$ that sends the compatible pair $B = ((i_1,a_1),\dotsc,(i_\ell,a_\ell))$ to the pipedream with crossings at positions $(i_j, a_j - i_j + 1)$ for $1 \le j \le \ell$. This was proved in \cite{BB} for the reduced case, but the proof works in the non-reduced setting. 
\end{rem}

\subsection{Marked Bumpless Pipedreams}
We work with a set of tiles
\[
\btile,\quad \htile, \quad \vtile,\quad \ptile,\quad \rtile,\quad \jtile,\quad \mtile
\]
named blank, horizontal, vertical, plus, R, J, and marked J. We call $\btile$ and $\mtile$ \textit{heavy}, and the other tiles \textit{light}. 

A \textit{marked bumpless pipedream} is an $[n] \times [n]$ matrix with entries in the above set of tiles such that the pipes are ``connected''. 
Moreover, every column has a pipe
entering from the bottom but not top,
and every row has a pipe exiting from the right
but not left. Let $\MBPD(n)$ be the set of $[n] \times [n]$ marked bumpless pipedreams. The weight $\wt(D)$ is the weak composition whose $i\textsuperscript{th}$ entry is the number of heavy tiles in row $i$.

We label the pipe entering from column $i$ as pipe $i$. We trace the pipes from bottom to right with the proviso: if two pipes have already crossed, their subsequent crossings are ignored. Then we read the labels on the right edge of $D$ from top to bottom, obtaining $w(D) \in S_n$. For every $w \in S_n$, the \textit{Rothe bumpless pipedream} $D_w$ is the unique element of $\MBPD_w$ whose $i$-th pipe turns only at position $(i, w(i))$ for all $i$. Clearly, $D_\id$ is the only element of $\MBPD(n)$ with no heavy tiles. 
See \Cref{S: Example} for 
an example of a MBPD. 

\section{Describing the bijection}
\label{S: bijection}
In this section we present our main result:
a bijection between $\RCP(n)$ and $\MBPD(n)$
that preserves $w(\cdot)$ and $\wt(\cdot)$.
Our bijection is recursive, 
requiring several operators on $\RCP(n)$ 
and $\MBPD(n)$. We also describe the effect of the operators
on $w(\cdot)$ and $\wt(\cdot)$.
Let $\textbf{e}_i$ be the composition
whose $i\textsuperscript{th}$
entry is $1$ and other entries are $0$. 

\begin{defn}
\label{D: reduce to B_id}
Let $B_\id \in \RCP(n)$ be the empty biword.
For any other $B \in \RCP(n)$,
the following two  steps may be iterated to reduce $B$ to $B_\id$. 
Let $(i,a)$ the first biletter in $B$.
\begin{enumerate}
\item[] (CP1) If $i=a$ remove the first biletter $(i,i)$ from $B$.
Write $\XX_i(B)$ for the resulting biword. Then $w(B) = w(\XX_i(B))* s_i$
and $\wt(B) = \wt(\XX_i(B)) + \textbf{e}_i$.
\item[] (CP2) If $i<a$, replace the first biletter of $B$ by $(i+1,a)$.
Call $\uparrow B$ the result of this operation.
We have $w(B) = w(\uparrow B))$
and $\wt(B) = \wt(\uparrow B) + \textbf{e}_i -  \textbf{e}_{i+1}$.
\end{enumerate}
\end{defn}

%\begin{rem}
%\label{R: Effect on RCP}
%We summarize the effect of the two operators
%on $w(\cdot)$ and $\wt(\cdot)$.
%\begin{enumerate}
%\item[] (RCP1) We have 
%\item[] (RCP2) 
%\end{enumerate}
%\end{rem}

The operators $X_i$ and $\uparrow$ on $\RCP(n)$ 
have corresponding left inverses $\II_i$ 
and $\downarrow$.
\begin{enumerate}
\item[] (RCP1) Let $\II_i(B)$ be the result of prepending $(i,i)$ to $B$.
\item[] (RCP2) Let $\downarrow{}B$
be the result of changing the first biletter of $B$
from $(i,a)$ to $(i-1, a)$.
\end{enumerate}
Apparently, 
these two operators are not defined on arbitrary
elements of $\RCP(n)$.

The bijection $\Phi:\MBPD(n)\to\RCP(n)$ is defined 
recursively by specifying the operations on $\MBPD(n)$ which correspond to $\XX_i$ and $\uparrow$ on $\RCP(n)$.
They are the operators $\fs_i$
and $f_i$
whose definitions
are postponed to the next
section. 
For the base case, 
let $\Phi(D_\id) = B_\id$.
For $D \in \MBPD(n)$ with 
at least one heavy tile,
let $(i,j)$ be such that $D_{i,j}$ is the heavy tile 
with $i$ minimum and then $j$ maximum.
%Correspondingly, 
%we consider two cases:
\begin{enumerate}
\item[] (Case 1) There is no
$\jtile$ or $\mtile$ 
in row $i+1$ to the right of
column $j$.
In this case
the operator $\fs_i$ is defined
on $D$.
We define $$\Phi(D) := \II_i \circ \Phi \circ \fs_i(D).$$
\item[] (Case 2) Otherwise
the operator $f_i$ is defined
on $D$.
We define 
$$\Phi(D) := \downarrow \circ \Phi \circ f_i(D).$$
\end{enumerate}

Diagrammatically
we may describe $\Phi$ as 
\[
\begin{tikzcd}
  D \arrow[r,"\Phi"] \arrow[d,swap,"f_i^*"] & \Phi(D) \\
  D' \arrow[r,swap,"\Phi"] & \Phi(D') \arrow[u,swap,"\text{$\II_i$}"] \\
\end{tikzcd}
\quad\quad
\begin{tikzcd}
  D \arrow[r,"\Phi"] \arrow[d,swap,"f_i"] & \Phi(D) \\
  D' \arrow[r,swap,"\Phi"] & \Phi(D') \arrow[u,swap,"\text{$\downarrow$}"] \\
\end{tikzcd}
\]
\vskip-2em
\noindent where the diagram on the left (resp. right) represents
case $1$ (resp. $2$). 
We present an example of $\Phi(\cdot)$
in \Cref{S: Example}.
Since $\II_i$ and $\downarrow$ are not
defined on all elements of $\RCP(n)$,
the well-definedness of $\Phi$ is highly nontrivial,
involving a delicate analysis of the operators $\fs_i$
and $f_i$.
\begin{prop}
The map $\Phi: \MBPD(n) \to \RCP(n)$
is well-defined.
\end{prop}
To see $\Phi$
preserves $w(\cdot)$ and $\wt(\cdot)$,
we just need to describe how
$\fs_i$ and $f_i$ 
change $w(\cdot)$ and $\wt(\cdot)$.
\begin{prop}
\label{P: Effect on MBPD}
If $\fs_i(D)$ is defined
then $w(D) = w(\fs_i(D))*s_i$
and $\wt(D) = \wt(\fs(D)) + \textbf{e}_i$.
If $f_i(D)$ is defined
then $w(D) = w(f_i(D))$
and $\wt(D) = \wt(f_i(D)) + \textbf{e}_i -  \textbf{e}_{i+1}$.
\end{prop}

Compared with the effect
of $\XX_i$ and $\downarrow$,
\Cref{P: Effect on MBPD}
shows that $\Phi$ preserves $w(\cdot)$
and $\wt(\cdot)$.

The operators $\fs_i$ and $f_i$
have left inverses 
$\es_i$ and $e_i$ respectively,
which are also defined in the next section.
The bijection $\Psi: \MBPD(n)\to\RCP(n)$
is defined analogously.
As the base case, 
let $\Psi(B_\id) = D_\id$.
For non-empty $B \in \MBPD(n)$,
we let $(i,a)$ be its first biletter.
We consider the two cases as in \Cref{D: reduce to B_id}:
\begin{itemize}
\itemsep0em
\item[] (Case 1) 
We define $\Psi(B) := \es_i \circ \Psi \circ \XX_i(B)$.
\item[] (Case 2) 
We define $\Psi(B) := e_i \circ \Psi\circ \uparrow(B)$.
\end{itemize}

Similarly, 
the well-definedness
of $\Psi$ is non-trivial. 
Our main result is: 
\begin{thm}
The maps $\Phi: \RCP(n)\to \MBPD(n)$
and $\Psi:\MBPD(n)\to\RCP(n)$ are well-defined
and mutually inverse maps that 
preserve both $w(\cdot)$ and $\wt(\cdot)$.
\end{thm}


\section{Describing the operators on \texorpdfstring{$\MBPD(n)$}{MBPD(n)}}
\label{S: ef}

In this section,
we describe the operators
$\fs_r$, $f_r$, $\es_r$ and $e_r$
on $\MBPD(n)$.
There will be nine cases for $\fs_r, f_r$
and nine cases for $\es_r, e_r$.
Readers might refer to \Cref{F: 9 cases}
for a visual illustration of the cases. 

\subsection{Some tile notation}
Given $D\in \MBPD(n)$ and intervals $I,J\subset [n]$ we denote by $D_{I,J}$ the submatrix of entries $D_{i,j}$ for $i\in I$ and $j\in J$. We write $D_{r,J}$ when $I=\{r\}$ is a singleton row index. 

We say \textit{$D_{r,[b,c]}$ is a pipe segment} if the interior part 
$D_{r,[b+1,c-1]}$ consists solely of tiles $\htile$ and $\ptile$.
A \textit{kink} is a pipe segment $D_{r,[b,c]}$ such that $D_{r,b}=\rtile$ and $D_{r,c}=\jtile$.
\[
\begin{tikzpicture}[x=1.5em,y=1.5em,thick,color = blue]
\draw[step=1,gray,thin] (0,0) grid (7,1);
\draw[color=black, thick, sharp corners] (0,0) rectangle (7,1);
\draw(0.5,0)--(0.5,0.5)--(6.5,0.5)--(6.5,1.0);
\draw(1.5,0)--(1.5,1);
\draw(3.5,0)--(3.5,1);
\node [color=black] at (-.5,.5) {$r$};
\node [color=black] at (.5,1.5) {$b$};
\node [color=black] at (6.5,1.5) {$c$};
\end{tikzpicture}
\]

By definition, a non-blank tile
$D_{r, [b, b]}$ is always a pipe segment.
A \textit{light sequence} in $D\in \MBPD(n)$ is a set of tiles of the form $D_{r,[b,c]}$ 
which consists solely of light tiles. 
A light sequence $D_{r,[b,c]}$ is \textit{paired} if $D_{r,b}$ is not 
connected to the left
and $D_{r, c}$ is not connected to the right.

For a row index $r$ and column indices $1\le b<d\le n$,
the $(r,[b,d])$-droop and $(r,[b,d])$-undroop are operations that change an MBPD into another,
only changing tiles in the ``(un)droop rectangle'' $[r,r+1] \times [b,d]$.
We say that $D\in\MBPD(n)$ admits the \textit{$(r,[b,d])$-droop} if 
\begin{itemize}
\itemsep0em
    \item The droop rectangle contains only light tiles except possibly a $\btile$
    at $(r+1,d)$. 
    \item $D_{r,[b,d]}$ is a pipe segment.
    \item $D_{r+1,[b+1,d-1]}$ is a paired light sequence.
    \item $D_{r,b}\ne \htile$ and $D_{r,d}\ne \ptile$.
\end{itemize}
If $D$ admits the $(r,[b,d])$-droop then we may produce $D'\in\MBPD(n)$
as follows; we say that $D\to D'$ is the $(r,[b,d])$-droop.

%In the following pictures we only draw the parts of the upper right and lower left corner tiles $D_{r,d}$ and $D_{r+1,b}$ that change during the (un)droop; the other parts remain the same.

The pipe segment $D_{r,[b,d]}$, which connects down to $D_{r+1,b}$,  ``droops'' to a pipe segment $D'_{r+1,[b,d]}$ which is connected upwards to $D'_{r,d}$. In columns $x$ of $D$ for $b<x<d$, the vertical pipes ($x$ such that $b<x<d$ and $D_{r,x}=D_{r+1,x}=\ptile$) are unchanged. Each kink in row $r+1$ in $D$ between columns $b$ and $d$, is shifted up into row $r$ in $D'$. A shifted kink is shaded gray in the following picture of a droop.
\[
\begin{tikzpicture}[x=1.5em,y=1.5em,thick,color = blue]
\draw[step=1,gray,thin] (0,0) grid (7,2);
\draw[color=black, thick, sharp corners] (0,0) rectangle (7,2);
\draw [color=gray, fill=gray] (1,0) rectangle (5,1);
%\draw[color=gray, fill=gray] (4,1) rectangle (5,2);
\draw(6.5,1.5)--(0.5,1.5)--(.5,.5);
\draw(1.5,0.0)--(1.5,0.5)--(4.5,0.5)--(4.5,2.0);
\draw(2.5,0.0)--(2.5,2.0);
\draw(5.5,0.0)--(5.5,2.0);
\node [color=black] at (-.5,1.5) {$r$};       
\node [color=black] at (-1,.5) {$r+1$};
\node[color=black] at (.5,2.5) {$b$};
\node[color=black] at (6.5,2.5) {$d$};
\node [color=black] at (3.5,-1) {$D$};
\node[color=black] at (7.6,1) {$\rightarrow$};
\end{tikzpicture} 
\begin{tikzpicture}[x=1.5em,y=1.5em,thick,color = blue]
\draw[step=1,gray,thin] (0,0) grid (7,2);
\draw[color=black, thick, sharp corners] (0,0) rectangle (7,2);
%\draw[color=gray,fill=gray] (1,0) rectangle (2,2);
\draw[color=gray,fill=gray] (1,1) rectangle (5,2);
\draw(6.5,1.5)--(6.5,0.5)--(.5,.5);
\draw(1.5,0.0)--(1.5,1.5)--(4.5,1.5)--(4.5,2.0);
\draw(2.5,0.0)--(2.5,2.0);
\draw(5.5,0.0)--(5.5,2.0);
\node [color=black] at (3.5,-1) {$D'$};
\end{tikzpicture}
\]
Notice that the tile $D_{r,b}$ might change from 
$\rtile$ to $\btile$ or $\ptile$ to $\jtile$.
The tile $D_{r+1, d}$ might change from 
$\rtile$ to $\ptile$ or $\btile$ to $\jtile$.

By definition, $D\to D'$ is a \textit{$(r,[b,d])$-undroop} if and only if 
$D'\to D$ is a $(r,[b,d])$-droop. 

Explicitly, a \textit{$(r,[b,d])$-undroop} $D\to D'$ is defined when 
\begin{itemize}
\itemsep0em
    \item The undroop rectangle contains only light tiles except possibly a $\btile$ 
    at $(r,b)$.
    \item $D_{r+1,[b,d]}$ is a pipe segment.
    \item $D_{r,[b+1,d-1]}$ is a paired light sequence.
    \item $D_{r+1,d}\ne \htile$, $D_{r+1,b}\ne\ptile$.
\end{itemize}


Finally, for $D\in \MBPD(n)$ the subdiagram $D_{[r,r+1],[b,d]}$ is a \textit{doublecross} if 
$D_{r,[b,d]}$ and $D_{r+1,[b,d]}$ are pipe segments, $D_{r,b}=\rtile$, 
and $D_{r+1,d}=\jtile$.
\[
\begin{tikzpicture}[x=1.5em,y=1.5em,thick,color = blue]
\draw[step=1,gray,thin] (0,0) grid (5,2);
\draw[color=black, thick, sharp corners] (0,0) rectangle (5,2);
\draw (.5,0)--(.5,1.5)--(5,1.5);
\draw (0,.5)--(4.5,.5)--(4.5,2);
\draw (1.5,0)--(1.5,2);
\draw (2.5,0)--(2.5,2);
\node[color=black] at (-.5,1.5) {$r$};
\node[color=black] at (-1.0,0.5) {$r+1$};
\node[color=black] at (0.5,2.5) {$b$};
\node[color=black] at (4.5,2.5) {$d$};
\end{tikzpicture}
\]
It is not hard to deduce that 
if $D_{[r, r+1], [b,d]}$
is a double cross, 
then $D_{r+1, b} = D_{r, d} = \ptile$.
This is so named because the pipe through
$(r,b)$ and the pipe through $(r+1, d)$
cross twice at $(r,d)$ and $(r+1,b)$.

\begin{rem}\label{R:doublecrosses unique}
There is at most one doublecross $D_{[r,r+1],[b,d]}$ for each fixed pair $(r,b)$ 
and also at most one such doublecross for a fixed pair $(r,d)$.
\end{rem}

\subsection{Describing \texorpdfstring{$\fs_r$}{f*r} and \texorpdfstring{$f_r$}{fr}}
\label{S: F}
Let $D\in \MBPD(n)$. Say that $(r,c)$ is an \textit{$f$-target} of $D$ if the conditions (f1), (f2), and (f3) hold.
\begin{itemize}
\itemsep0em
    \item[(f1)] $D_{r,c}$ is the rightmost heavy tile in row $r$.
    \item[(f2)] There is an index $c'>c$ such that $D_{r+1,c'}=\jtile$.
    Let $c'$ be minimum with this property.
    \item[(f3)] All tiles $D_{r+1,j}$ are light for $j<c'$.
\end{itemize}
We say that $(r,c)$ is an \textit{$f^*$-target} of $D$ if the conditions (f1), (f*2), and (f3) hold, where
\begin{itemize}
\itemsep0em
    \item[(f*2)] There is no $\jtile$ nor $\mtile$ in row $r+1$ to the right of column $c$. In this case let $c'$ be the maximum index such that $D(r,c')=\rtile$.
\end{itemize}
By abuse of language we will also say that $(D,(r,c))$ is an $f$-target (resp. $f^*$-target)
to mean that $D\in\MBPD(n)$ and $(r,c)$ is an $f$-target (resp. $f^*$-target) of $D$.
We will write $F$ to mean either $f$ or $f^*$.
So $(D,(r,c))$ is an $F$-target means it is either an $f$-target or $f^*$-target.

We define the \textit{\window} of an $F$-target $(D,(r,c))$
to be the two-row rectangle $[r,r+1] \times [b,c']$ where 
\begin{align}
    \label{E:f b} 
    b &= \begin{cases}
        c & \text{if $D_{r,c}=\btile$} \\
        \text{maximum $b<c$ with $D_{r,b}=\rtile$} & \text{if $D_{r,c}=\mtile$}
    \end{cases}
\end{align}
and $c'$ is defined by (f2) in the case of an $f$-target and (f*2) in the case of an $f^*$-target.

There are two trichotomies for an $F$-target $(D,(r,c))$, 
giving nine cases in all. They are called left and right because they describe the left and right sides of the window respectively.

The left trichotomy for the $F$-target $(D,(r,c))$ asserts
that exactly one of $\Blank$, $\Cross$, or $\NCross$ holds.
\begin{itemize}
\itemsep0em
    \item[($\Blank$)] (Blank): This holds if $D_{r,c}=\btile$.
\end{itemize}
Otherwise, we know $D_{r,c}=\mtile$. 
There are two more cases:
\begin{itemize}
\itemsep0em
    \item[($\Cross$)] (Crossing): %$D_{r+1,b}=\ptile$ and 
    $D_{r+1,[b,c]}$ is a pipe segment.
    This is so named because in this case the pipe through $(r,c)$ goes to the left and then crosses vertically with the above pipe segment at $D_{r+1,b}=\ptile$.
    \item[($\NCross$)] (Noncrossing): $D_{r+1,[b,c]}$ is not a pipe segment.
\end{itemize}

The right  trichotomy says that for an $F$-target $(D,(r,c))$, 
exactly one $\Terminal$, $\DCross$, or $\Ordinary$ holds.
\begin{itemize}
\itemsep0em
    \item [($\Terminal$)] (Terminal) The tile $(r,c)$ is an $f^*$-target of $D$.
    \item[($\DCross$)] (Doublecross): The tile $(r,c)$ is an $f$-target of $D$. Moreover, there is a doublecross $D_{[r,r+1], [d,c']}$ for some $d$ with $c<d<c'$.
    \item[($\Ordinary$)] (Ordinary): Cases $\Terminal$ and $\DCross$ do not hold.
\end{itemize}

Given an $\ftarg$ move, define $\rho$ as
$d$ in Case $\DCross$ or $c'$ otherwise. 
We define $F_r(D)\in \MBPD(n)$ to be the result
of the following steps:
\begin{itemize}
\itemsep0em
\item If $D_{r,c}=\mtile$ remove the marking.
\item In Cases $\Blank$ and $\Cross$ perform the $(r,[c,\rho])$-undroop.
\item If the tile at $(r+1,c')$ is $\jtile$, mark it.
\end{itemize}

\subsection{Describing \texorpdfstring{$\es_r$}{e*r} and \texorpdfstring{$e_r$}{er}}
We say the tile $(r+1, c)$ is an \textit{$e$-target} of $D\in \MBPD(n)$ if the following are satisfied. 
\begin{itemize}
\itemsep0em
\item[(e1)] It is the leftmost heavy tile on row $r+1$.
\item[(e2)] There is an index $c' < c$ such that $D_{r,c'}=\rtile$
and $D_{r+1,[c',c]}$ is not a pipe segment.
Let $c'$ be maximum with this property.
\item[(e3)] On the right of $(r, c')$ in row $r$ there are no heavy tiles. 
\end{itemize}
We say the tile $(r+1,c)$ is an \textit{$e^*$-target} of $D\in \MBPD(n)$ if the 
conditions (e*1), (e2), and (e3) are satisfied, where
the condition (e1) for an $e$-target has been replaced by the condition:
\begin{itemize}
\itemsep0em
\item[(e*1)] There are no heavy tiles in row $r+1$,
and $c$ is the largest so that $D_{r,c}$ or $D_{r+1, c}$ is $\rtile$.
\end{itemize}

Again we abuse language by saying that an 
\emph{$e$-target} (resp. $e^*$-target) is a pair $(D,(r+1,c))$ where $D\in\MBPD(n)$
and $(r+1,c)$ is an $e$-target (resp. $e^*$-target) of $D$. Similar to the $F$-case, an $E$-target is either an $e$-target or $e^*$-target.
We define the \emph{\window } of an $E$-target $(D,(r+1,c))$ to be the two-row rectangle
$[r, r+1] \times [c', c]$
with $c'$ as in (e2). 

Let $(D,(r+1,c))$ be an $E$-target.
We have 9 cases with three choices for the first symbol and three for 
the second symbol.

Here is the right  trichotomy for $E$-moves.
Exactly one of them holds for an $E$-target.
\begin{itemize}
\itemsep0em
\item $\Initial$ (Initial): $(r+1, c)$ is an $e^*$-target.
\item $\Plus$ (Plus): $(r+1, c)$ is an $e$-target and $D_{r,c}=\ptile$. 

\item $\NPlus$ (No plus): $(r+1, c)$ is an $e$-target and $D_{r, c}\ne\ptile$. 
\end{itemize}

Define the \emph{right droop column} $\rho$ of the $E$-target $(D,(r+1,c))$ by
\begin{align}\label{E:right droop column}
  \rho = \begin{cases} 
  c & \text{in Cases $\Initial$ and $\NPlus$} \\
  \max\bigg\{d' \mid \text{$d'<c$ and $D_{r+1,d'}=\rtile$}\bigg\} & \text{in Case $\Plus$.} 
  \end{cases}
\end{align}

The left  trichotomy for $\etarg$-moves
are the following three cases, exactly one of which holds for any $E$-target $(D,(r+1,c))$
with window $[r,r+1]\times [c',c]$.

\begin{itemize}
\itemsep0em
\item $\Leftturn$ (Left turn): $D_{r,[c',c]}$ is not a pipe segment. That is, the pipe leaving the $\rtile$ at $(r,c')$ to the right, must turn upwards (make a left turn) before arriving at $(r,c)$.
\item $\DCross$ (Doublecross): $D_{r,[c',c]}$ is a pipe segment and 
the tile $(r,c')$ is the top-left corner of a doublecross.
\item $\Straight$ (Straight): $D_{r,[c',c]}$ is a pipe segment
and the tile $(r,c')$ is not the top-left corner of a doublecross.
\end{itemize}

Define the \emph{left droop column} $\lambda$ of the $E$-target $(D,(r+1,c))$ by
\begin{align}\label{E:left droop column}
  \lambda = \begin{cases} 
  c' & \text{in Case $\Straight$} \\
  \min \bigg\{\text{$d\mid d>c'$ such that $D_{r+1,d}=\jtile$}\bigg\} & \text{in Case $\DCross$}\\
  \min \bigg\{\text{$d\mid d>c'$ such that $D_{r,d}=\jtile$}\bigg\} & \text{in Case $\Leftturn$.}
  \end{cases}
\end{align}

We shall now define $E$-moves, which are the inverses of $F$-moves.
Given an $E$-target $(D,(r+1,c))$, define $E_r(D)\in\MBPD(n)$ to be the result of the
following steps:
\begin{itemize}
\itemsep0em
\item If $D_{r+1,c}=\mtile$, remove the marking.
\item In Cases $\Straight$ and $\DCross$, perform the
$(r,[\lambda,\rho])$-droop.
\item If $(r, \lambda)$ is $\jtile$, mark it.
\end{itemize}


\subsection{\texorpdfstring{$\ftarg$}{F}-moves and \texorpdfstring{$\etarg$}{E}-moves are mutually inverses}
\newcommand{\boxit}[1]{\parbox{.85in}{#1}}

We give a diagrammatic summary of the $F$- and $E$-moves. Small examples of \window{s} of the nine cases for $f$ and $f^*$ are pictured 
in \Cref{F: 9 cases}. 
% For both the $F$- and $E$-moves the target column is marked $c$
% and the cotarget column is marked $c'$. 
The green square has coordinate $(r,\lambda)$
and the red square has coordinate $(r+1,\rho)$. In $F$-move cases other than $\NCross$
and $E$-move cases other than  $\Leftturn$, the green and red squares are the corners of the 
(un)droop window. The parts of tiles not pictured
remain the same. Small examples of the nine cases for $e$ and $e^*$ are pictured 
in \Cref{F: 9 cases}.
\newcommand{\BT}{
\begin{tikzpicture}[x=1.2em,y=1.2em,thick,color = blue]
\draw[step=1,gray,thin] (0,0) grid (4,2);
\draw[color=black, thick, sharp corners] (0,0) rectangle (4,2);
\draw(1.5,2.0)--(1.5,1.0);
\draw(2.5,2.0)--(2.5,1.0);
\draw(4.0,1.5)--(3.5,1.5)--(3.5,1.0);
\draw(1.0,0.5)--(0.5,0.5);
\draw(1.5,1.0)--(1.5,0.0);
\draw(2.0,0.5)--(1.0,0.5);
\draw(2.5,1.0)--(2.5,0.0);
\draw(3.0,0.5)--(2.0,0.5);
\draw(3.5,1.0)--(3.5,0.0);
\draw(4.0,0.5)--(3.0,0.5);
\draw[step=1,red,thick] (3,0) grid (4,1);
\draw[step=1,green,thick] (0,1) grid (1,2);
\end{tikzpicture}
}
\newcommand{\BD}{
\begin{tikzpicture}[x=1.2em,y=1.2em,thick,color = blue]
\draw[step=1,gray,thin] (0,0) grid (4,2);
\draw[color=black, thick, sharp corners] (0,0) rectangle (4,2);
\draw(1.5,2.0)--(1.5,1.0);
\draw(3.0,1.5)--(2.5,1.5)--(2.5,1.0);
\draw(3.5,2.0)--(3.5,1.0);
\draw(4.0,1.5)--(3.0,1.5);
\draw(1.0,0.5)--(0.5,0.5);
\draw(1.5,1.0)--(1.5,0.0);
\draw(2.0,0.5)--(1.0,0.5);
\draw(2.5,1.0)--(2.5,0.0);
\draw(3.0,0.5)--(2.0,0.5);
\draw(3.5,1.0)--(3.5,0.5)--(3.0,0.5);
\draw[step=1,red,thick] (2,0) grid (3,1);
\draw[step=1,green,thick] (0,1) grid (1,2);
\end{tikzpicture}
}
\newcommand{\BO}{
\begin{tikzpicture}[x=1.2em,y=1.2em,thick,color = blue]
\draw[step=1,gray,thin] (0,0) grid (4,2);
\draw[color=black, thick, sharp corners] (0,0) rectangle (4,2);
\draw(1.5,2.0)--(1.5,1.0);
\draw(2.5,2.0)--(2.5,1.0);
\draw(3.5,1.5)--(3.5,1.0);
\draw(1.0,0.5)--(0.5,0.5);
\draw(1.5,1.0)--(1.5,0.0);
\draw(2.0,0.5)--(1.0,0.5);
\draw(2.5,1.0)--(2.5,0.0);
\draw(3.0,0.5)--(2.0,0.5);
\draw(3.5,1.0)--(3.5,0.5)--(3.0,0.5);
\draw[step=1,red,thick] (3,0) grid (4,1);
\draw[step=1,green,thick] (0,1) grid (1,2);
\node[color=black] at (0.5,-.508) {$c$};
\node[color=black] at (3.5,-.4) {$c'$};
\end{tikzpicture}
}
\newcommand{\CT}{
\begin{tikzpicture}[x=1.2em,y=1.2em,thick,color = blue]
\draw[step=1,gray,thin] (0,0) grid (4,2);
\draw[color=black, thick, sharp corners] (0,0) rectangle (4,2);
\draw(1.0,1.5)--(0.5,1.5)--(0.5,1.0);
\draw(1.5,2.0)--(1.5,1.5)--(1.0,1.5);
\node at (1.5,1.5) {$\bullet$};
\draw(2.5,2.0)--(2.5,1.0);
\draw(4.0,1.5)--(3.5,1.5)--(3.5,1.0);
\draw(0.5,1.0)--(0.5,0.0);
\draw(1.0,0.5)--(0.0,0.5);
\draw(2.0,0.5)--(1.0,0.5);
\draw(2.5,1.0)--(2.5,0.0);
\draw(3.0,0.5)--(2.0,0.5);
\draw(3.5,1.0)--(3.5,0.0);
\draw(4.0,0.5)--(3.0,0.5);
\draw[step=1,red,thick] (3,0) grid (4,1);
\draw[step=1,green,thick] (1,1) grid (2,2);
\end{tikzpicture}
}

\newcommand{\CD}{
\begin{tikzpicture}[x=1.2em,y=1.2em,thick,color = blue]
\draw[step=1,gray,thin] (0,0) grid (4,2);
\draw[color=black, thick, sharp corners] (0,0) rectangle (4,2);
\draw(1.0,1.5)--(0.5,1.5)--(0.5,1.0);
\draw(1.5,2.0)--(1.5,1.5)--(1.0,1.5);
\node at (1.5,1.5) {$\bullet$};
\draw(3.0,1.5)--(2.5,1.5)--(2.5,1.0);
\draw(3.5,2.0)--(3.5,1.0);
\draw(4.0,1.5)--(3.0,1.5);
\draw(0.5,1.0)--(0.5,0.0);
\draw(1.0,0.5)--(0.0,0.5);
\draw(2.0,0.5)--(1.0,0.5);
\draw(2.5,1.0)--(2.5,0.0);
\draw(3.0,0.5)--(2.0,0.5);
\draw(3.5,1.0)--(3.5,0.5)--(3.0,0.5);
\draw[step=1,red,thick] (2,0) grid (3,1);
\draw[step=1,green,thick] (1,1) grid (2,2);
\end{tikzpicture}}

\newcommand{\CO}{
\begin{tikzpicture}[x=1.2em,y=1.2em,thick,color = blue]
\draw[step=1,gray,thin] (0,0) grid (4,2);
\draw[color=black, thick, sharp corners] (0,0) rectangle (4,2);
\draw(1.0,1.5)--(0.5,1.5)--(0.5,1.0);
\draw(1.5,2.0)--(1.5,1.5)--(1.0,1.5);
\node at (1.5,1.5) {$\bullet$};
\draw(2.5,2.0)--(2.5,1.0);
\draw(3.5,1.5)--(3.5,1.0);
\draw(0.5,1.0)--(0.5,0.0);
\draw(1.0,0.5)--(0.0,0.5);
\draw(2.0,0.5)--(1.0,0.5);
\draw(2.5,1.0)--(2.5,0.0);
\draw(3.0,0.5)--(2.0,0.5);
\draw(3.5,1.0)--(3.5,0.5)--(3.0,0.5);
\draw[step=1,red,thick] (3,0) grid (4,1);
\draw[step=1,green,thick] (1,1) grid (2,2);
\node[color=black] at (1.5,-.508) {$c$};
\node[color=black] at (3.5,-.4) {$c'$};
\end{tikzpicture}
}

\newcommand{\NT}{
\begin{tikzpicture}[x=1.2em,y=1.2em,thick,color = blue]
\draw[step=1,gray,thin] (0,0) grid (4,2);
\draw[color=black, thick, sharp corners] (0,0) rectangle (4,2);
\draw(1.0,1.5)--(0.5,1.5)--(0.5,1.0);
\draw(1.5,2.0)--(1.5,1.5)--(1.0,1.5);
\node at (1.5,1.5) {$\bullet$};
\draw(2.5,2.0)--(2.5,1.0);
\draw(4.0,1.5)--(3.5,1.5)--(3.5,1.0);
\draw(0.5,1.0)--(0.5,0.5);
\draw(2.0,0.5)--(1.5,0.5)--(1.5,0.0);
\draw(2.5,1.0)--(2.5,0.0);
\draw(3.0,0.5)--(2.0,0.5);
\draw(3.5,1.0)--(3.5,0.0);
\draw(4.0,0.5)--(3.0,0.5);
\draw[step=1,green,thick] (1,1) grid (2,2);
\draw[step=1,red,thick] (3,0) grid (4,1);
\end{tikzpicture}
}

\newcommand{\ND}{
\begin{tikzpicture}[x=1.2em,y=1.2em,thick,color = blue]
\draw[step=1,gray,thin] (0,0) grid (4,2);
\draw[color=black, thick, sharp corners] (0,0) rectangle (4,2);
\draw(1.0,1.5)--(0.5,1.5)--(0.5,1.0);
\draw(1.5,2.0)--(1.5,1.5)--(1.0,1.5);
\node at (1.5,1.5) {$\bullet$};
\draw(3.0,1.5)--(2.5,1.5)--(2.5,1.0);
\draw(3.5,2.0)--(3.5,1.0);
\draw(4.0,1.5)--(3.0,1.5);
\draw(0.5,1.0)--(0.5,0.5);
\draw(2.0,0.5)--(1.5,0.5)--(1.5,0.0);
\draw(2.5,1.0)--(2.5,0.0);
\draw(3.0,0.5)--(2.0,0.5);
\draw(3.5,1.0)--(3.5,0.5)--(3.0,0.5);
\draw[step=1,green,thick] (1,1) grid (2,2);
\draw[step=1,red,thick] (2,0) grid (3,1);
\end{tikzpicture}
}

\newcommand{\NO}{
\begin{tikzpicture}[x=1.2em,y=1.2em,thick,color = blue]
\draw[step=1,gray,thin] (0,0) grid (4,2);
\draw[color=black, thick, sharp corners] (0,0) rectangle (4,2);
\draw(1.0,1.5)--(0.5,1.5)--(0.5,1.0);
\draw(1.5,2.0)--(1.5,1.5)--(1.0,1.5);
\node at (1.5,1.5) {$\bullet$};
\draw(2.5,2.0)--(2.5,1.0);
\draw(3.5,1.5)--(3.5,1.0);
\draw(0.5,1.0)--(0.5,0.5);
\draw(2.0,0.5)--(1.5,0.5)--(1.5,0.0);
\draw(2.5,1.0)--(2.5,0.0);
\draw(3.0,0.5)--(2.0,0.5);
\draw(3.5,1.0)--(3.5,0.5)--(3.0,0.5);
\draw[step=1,green,thick] (1,1) grid (2,2);
\draw[step=1,red,thick] (3,0) grid (4,1);
\node[color=black] at (1.5,-.508) {$c$};
\node[color=black] at (3.5,-.4) {$c'$};
\end{tikzpicture}

}

% the reverse case entries
\newcommand{\daa}{
\begin{tikzpicture}[x=1.2em,y=1.2em,thick,color = blue]
\draw[step=1,gray,thin] (0,0) grid (4,2);
\draw[color=black, thick, sharp corners] (0,0) rectangle (4,2);
\draw(1.0,1.5)--(0.5,1.5)--(0.5,1.0);
\draw(1.5,2.0)--(1.5,1.0);
\draw(2.0,1.5)--(1.0,1.5);
\draw(2.5,2.0)--(2.5,1.0);
\draw(3.0,1.5)--(2.0,1.5);
\draw(4.0,1.5)--(3.5,1.5)--(3.0,1.5);
\draw(0.5,1.0)--(0.5,0.5);
\draw(1.5,1.0)--(1.5,0.0);
\draw(2.5,1.0)--(2.5,0.0);
\draw(4.0,0.5)--(3.5,0.5)--(3.5,0.0);
\draw[step=1,red,thick] (3,0) grid (4,1);
\draw[step=1,green,thick] (0,1) grid (1,2);
\end{tikzpicture}
}

\newcommand{\dab}{
\begin{tikzpicture}[x=1.2em,y=1.2em,thick,color = blue]
\draw[step=1,gray,thin] (0,0) grid (4,2);
\draw[color=black, thick, sharp corners] (0,0) rectangle (4,2);
\draw(1.0,1.5)--(0.5,1.5)--(0.5,1.0);
\draw(1.5,2.0)--(1.5,1.0);
\draw(2.0,1.5)--(1.0,1.5);
\draw(2.5,2.0)--(2.5,1.0);
\draw(3.0,1.5)--(2.0,1.5);
\draw(4.0,1.5)--(3.5,1.5)--(3.0,1.5);
\draw(0.5,1.0)--(0.5,0.0);
\draw(1.0,0.5)--(0.0,0.5);
\draw(1.5,1.0)--(1.5,0.5)--(1.0,0.5);
\draw(2.5,1.0)--(2.5,0.0);
\draw(4.0,0.5)--(3.5,0.5)--(3.5,0.0);
\draw[step=1,red,thick] (3,0) grid (4,1);
\draw[step=1,green,thick] (1,1) grid (2,2);
\end{tikzpicture}
}

\newcommand{\dac}{
\begin{tikzpicture}[x=1.2em,y=1.2em,thick,color = blue]
\draw[step=1,gray,thin] (0,0) grid (4,2);
\draw[color=black, thick, sharp corners] (0,0) rectangle (4,2);
\draw(1.0,1.5)--(0.5,1.5)--(0.5,1.0);
\draw(1.5,2.0)--(1.5,1.5)--(1.0,1.5);
\draw(2.5,2.0)--(2.5,1.0);
\draw(4.0,1.5)--(3.5,1.5)--(3.5,1.0);
\draw(0.5,1.0)--(0.5,0.5);
\draw(2.0,0.5)--(1.5,0.5)--(1.5,0.0);
\draw(2.5,1.0)--(2.5,0.0);
\draw(3.0,0.5)--(2.0,0.5);
\draw(3.5,1.0)--(3.5,0.0);
\draw(4.0,0.5)--(3.0,0.5);
\draw[step=1,red,thick] (3,0) grid (4,1);
\draw[step=1,green,thick] (1,1) grid (2,2);
\end{tikzpicture}
}

\newcommand{\dba}{
\begin{tikzpicture}[x=1.2em,y=1.2em,thick,color = blue]
\draw[step=1,gray,thin] (0,0) grid (4,2);
\draw[color=black, thick, sharp corners] (0,0) rectangle (4,2);
\draw(1.0,1.5)--(0.5,1.5)--(0.5,1.0);
\draw(1.5,2.0)--(1.5,1.0);
\draw(2.0,1.5)--(1.0,1.5);
\draw(3.0,1.5)--(2.0,1.5);
\draw(3.5,2.0)--(3.5,1.0);
\draw(4.0,1.5)--(3.0,1.5);
\draw(0.5,1.0)--(0.5,0.5);
\draw(1.5,1.0)--(1.5,0.0);
\draw(3.0,0.5)--(2.5,0.5)--(2.5,0.0);
\draw(3.5,1.0)--(3.5,0.5)--(3.0,0.5);
\node at (3.5,0.5) {$\bullet$};
\draw[step=1,red,thick] (2,0) grid (3,1);
\draw[step=1,green,thick] (0,1) grid (1,2);
\end{tikzpicture}
}

\newcommand{\dbb}{
\begin{tikzpicture}[x=1.2em,y=1.2em,thick,color = blue]
\draw[step=1,gray,thin] (0,0) grid (4,2);
\draw[color=black, thick, sharp corners] (0,0) rectangle (4,2);
\draw(1.0,1.5)--(0.5,1.5)--(0.5,1.0);
\draw(1.5,2.0)--(1.5,1.0);
\draw(2.0,1.5)--(1.0,1.5);
\draw(3.0,1.5)--(2.0,1.5);
\draw(3.5,2.0)--(3.5,1.0);
\draw(4.0,1.5)--(3.0,1.5);
\draw(0.5,1.0)--(0.5,0.0);
\draw(1.0,0.5)--(0.0,0.5);
\draw(1.5,1.0)--(1.5,0.5)--(1.0,0.5);
\draw(3.0,0.5)--(2.5,0.5)--(2.5,0.0);
\draw(3.5,1.0)--(3.5,0.5)--(3.0,0.5);
\node at (3.5,0.5) {$\bullet$};
\draw[step=1,red,thick] (2,0) grid (3,1);
\draw[step=1,green,thick] (1,1) grid (2,2);
\end{tikzpicture}
}

\newcommand{\dbc}{
\begin{tikzpicture}[x=1.2em,y=1.2em,thick,color = blue]
\draw[step=1,gray,thin] (0,0) grid (4,2);
\draw[color=black, thick, sharp corners] (0,0) rectangle (4,2);
\draw(1.0,1.5)--(0.5,1.5)--(0.5,1.0);
\draw(1.5,2.0)--(1.5,1.5)--(1.0,1.5);
\draw(3.0,1.5)--(2.5,1.5)--(2.5,1.0);
\draw(3.5,2.0)--(3.5,1.0);
\draw(4.0,1.5)--(3.0,1.5);
\draw(0.5,1.0)--(0.5,0.5);
\draw(2.0,0.5)--(1.5,0.5)--(1.5,0.0);
\draw(2.5,1.0)--(2.5,0.0);
\draw(3.0,0.5)--(2.0,0.5);
\draw(3.5,1.0)--(3.5,0.5)--(3.0,0.5);
\node at (3.5,0.5) {$\bullet$};
\draw[step=1,red,thick] (2,0) grid (3,1);
\draw[step=1,green,thick] (1,1) grid (2,2);
\end{tikzpicture}
}

\newcommand{\dca}{
\begin{tikzpicture}[x=1.2em,y=1.2em,thick,color = blue]
\draw[step=1,gray,thin] (0,0) grid (4,2);
\draw[color=black, thick, sharp corners] (0,0) rectangle (4,2);
\draw(1.0,1.5)--(0.5,1.5)--(0.5,1.0);
\draw(1.5,2.0)--(1.5,1.0);
\draw(2.0,1.5)--(1.0,1.5);
\draw(2.5,2.0)--(2.5,1.0);
\draw(3.0,1.5)--(2.0,1.5);
\draw(3.5,1.5)--(3.0,1.5);
\draw(0.5,1.0)--(0.5,0.5);
\draw(1.5,1.0)--(1.5,0.0);
\draw(2.5,1.0)--(2.5,0.0);
\draw[step=1,red,thick] (3,0) grid (4,1);
\draw[step=1,green,thick] (0,1) grid (1,2);
\node[color=black] at (3.5,-.508) {$c$};
\node[color=black] at (0.5,-.4) {$\lambda$};
\end{tikzpicture}
}

\newcommand{\dcb}{
\begin{tikzpicture}[x=1.2em,y=1.2em,thick,color = blue]
\draw[step=1,gray,thin] (0,0) grid (4,2);
\draw[color=black, thick, sharp corners] (0,0) rectangle (4,2);
\draw(1.0,1.5)--(0.5,1.5)--(0.5,1.0);
\draw(1.5,2.0)--(1.5,1.0);
\draw(2.0,1.5)--(1.0,1.5);
\draw(2.5,2.0)--(2.5,1.0);
\draw(3.0,1.5)--(2.0,1.5);
\draw(3.5,1.5)--(3.0,1.5);
\draw(0.5,1.0)--(0.5,0.0);
\draw(1.0,0.5)--(0.0,0.5);
\draw(1.5,1.0)--(1.5,0.5)--(1.0,0.5);
\draw(2.5,1.0)--(2.5,0.0);
\draw[step=1,red,thick] (3,0) grid (4,1);
\draw[step=1,green,thick] (1,1) grid (2,2);
\node[color=black] at (3.5,-.508) {$c$};
\node[color=black] at (1.5,-.4) {$\lambda$};
\end{tikzpicture}
}

\newcommand{\dcc}{
\begin{tikzpicture}[x=1.2em,y=1.2em,thick,color = blue]
\draw[step=1,gray,thin] (0,0) grid (4,2);
\draw[color=black, thick, sharp corners] (0,0) rectangle (4,2);
\draw(1.0,1.5)--(0.5,1.5)--(0.5,1.0);
\draw(1.5,2.0)--(1.5,1.5)--(1.0,1.5);
\draw(2.5,2.0)--(2.5,1.0);
\draw(3.5,1.5)--(3.5,1.0);
\draw(0.5,1.0)--(0.5,0.5);
\draw(2.0,0.5)--(1.5,0.5)--(1.5,0.0);
\draw(2.5,1.0)--(2.5,0.0);
\draw(3.0,0.5)--(2.0,0.5);
\draw(3.5,1.0)--(3.5,0.5)--(3.0,0.5);
\node at (3.5,0.5) {$\bullet$};
\node[color=black] at (3.5,-.508) {$c$};
\node[color=black] at (1.5,-.4) {$\lambda$};
\draw[step=1,red,thick] (3,0) grid (4,1);
\draw[step=1,green,thick] (1,1) grid (2,2);
\end{tikzpicture}
}



\begin{figure}
\[
\begin{array}{ccccc}
& \Blank & \Cross & \NCross & \\
\Terminal & \boxit{\BT} & \boxit{\CT} & \boxit{\NT}\\[5mm]
\DCross & \boxit{\BD} &\boxit{\CD} &\boxit{\ND}\\[5mm]
\Ordinary & \boxit{\BO} &\boxit{\CO} &\boxit{\NO}\\[5mm]
\end{array}
\begin{array}{ccccc}
&\Straight & \DCross& \Leftturn& \\
\Initial& \boxit{\daa} & \boxit{\dab} & \boxit{\dac}\\[5mm]
 \Plus &\boxit{\dba} &\boxit{\dbb} &\boxit{\dbc}\\[5mm]
\NPlus &\boxit{\dca} &\boxit{\dcb} &\boxit{\dcc}\\[5mm]
\end{array}
\]
\caption{9 Cases for $\ftarg$-moves and $\etarg$-moves}
\label{F: 9 cases}
\end{figure}


From the figures, we deduce the following results,
which are a refined way of saying that 
$\ftarg$-moves and $\etarg$-moves are inverses. 


\begin{prop}
\label{P: EF}
Let $D$ have $F$-target $(r,c)$
with $\window$ $[r, r+1] \times [b, c']$.
\begin{itemize}
\itemsep0em
\item Then $F_r(D)$ has $\etarg$-target $(r+1, c')$
with the same window.
\item The case of $(D,(r,c))$ 
corresponds
to the case of $(F_r(D),(r+1,c'))$ 
according to \Cref{F: 9 cases}.
For instance, 
if $(D,(r,c))$ is in case $\Blank\Terminal$,
then $(F_r(D),(r+1,c'))$ is in case $\Straight\Initial$.
\item We have $E_r (F_r(D)) = D$.
\end{itemize}
\end{prop}

\begin{prop}
\Cref{P: EF} holds after switching
the roles of $\ftarg$ and $\etarg$.
\end{prop}

\subsection{Example of \texorpdfstring{$\ftarg$}{F}-moves
and the bijection \texorpdfstring{$\Phi$}{Phi}}
\label{S: Example}
Suppose we want to apply $\Phi$ on $D$, 
which is the leftmost MBPD below. 
We apply $f_3$ followed by $\fs_4$,
obtaining $D_1$ and $D_2$.
In the following diagrams the red square is the target and the violet square is the opposite corner of the undroop.
\[
\parbox{1.25in}{\begin{tikzpicture}[x=1em,y=1em,thick,color = blue]
\draw[step=1,gray,thin] (0,0) grid (6,6);
\draw[color=black, thick, sharp corners] (0,0) rectangle (6,6);
\draw(4.0,5.5)--(3.5,5.5)--(3.5,5.0);
\draw(5.0,5.5)--(4.0,5.5);
\draw(6.0,5.5)--(5.0,5.5);
\draw(3.0,4.5)--(2.5,4.5)--(2.5,4.0);
\draw(3.5,5.0)--(3.5,4.5)--(3.0,4.5);
\draw(5.0,4.5)--(4.5,4.5)--(4.5,4.0);
\draw(6.0,4.5)--(5.0,4.5);
\draw(2.0,3.5)--(1.5,3.5)--(1.5,3.0);
\draw(2.5,4.0)--(2.5,3.5)--(2.0,3.5);
\node at (2.5,3.5) {$\bullet$};
\draw(4.0,3.5)--(3.5,3.5)--(3.5,3.0);
\draw(4.5,4.0)--(4.5,3.0);
\draw(5.0,3.5)--(4.0,3.5);
\draw(6.0,3.5)--(5.0,3.5);
\draw(1.0,2.5)--(0.5,2.5)--(0.5,2.0);
\draw(1.5,3.0)--(1.5,2.0);
\draw(2.0,2.5)--(1.0,2.5);
\draw(3.0,2.5)--(2.0,2.5);
\draw(3.5,3.0)--(3.5,2.0);
\draw(4.0,2.5)--(3.0,2.5);
\draw(4.5,3.0)--(4.5,2.5)--(4.0,2.5);
\draw(6.0,2.5)--(5.5,2.5)--(5.5,2.0);
\draw(0.5,2.0)--(0.5,1.0);
\draw(1.5,2.0)--(1.5,1.0);
\draw(3.0,1.5)--(2.5,1.5)--(2.5,1.0);
\draw(3.5,2.0)--(3.5,1.0);
\draw(4.0,1.5)--(3.0,1.5);
\draw(5.0,1.5)--(4.0,1.5);
\draw(5.5,2.0)--(5.5,1.0);
\draw(6.0,1.5)--(5.0,1.5);
\draw(0.5,1.0)--(0.5,0.0);
\draw(1.5,1.0)--(1.5,0.0);
\draw(2.5,1.0)--(2.5,0.0);
\draw(3.5,1.0)--(3.5,0.0);
\draw(5.0,0.5)--(4.5,0.5)--(4.5,0.0);
\draw(5.5,1.0)--(5.5,0.0);
\draw(6.0,0.5)--(5.0,0.5);
\draw[color=red, thick, sharp corners] (2,3) rectangle (3,4);
\draw[color=violet, thick, sharp corners] (3,2) rectangle (4,3);
\node[color=black] at (3,-.5) {$D$};
\end{tikzpicture}}
\overset{\DCross\Cross}\to 
\parbox{1.25in}{\begin{tikzpicture}[x=1em,y=1em,thick,color = blue]
\draw[step=1,gray,thin] (0,0) grid (6,6);
\draw[color=black, thick, sharp corners] (0,0) rectangle (6,6);
\draw(4.0,5.5)--(3.5,5.5)--(3.5,5.0);
\draw(5.0,5.5)--(4.0,5.5);
\draw(6.0,5.5)--(5.0,5.5);
\draw(3.0,4.5)--(2.5,4.5)--(2.5,4.0);
\draw(3.5,5.0)--(3.5,4.5)--(3.0,4.5);
\draw(5.0,4.5)--(4.5,4.5)--(4.5,4.0);
\draw(6.0,4.5)--(5.0,4.5);
\draw(2.0,3.5)--(1.5,3.5)--(1.5,3.0);
\draw(2.5,4.0)--(2.5,3.0);
\draw(3.0,3.5)--(2.0,3.5);
\draw(4.0,3.5)--(3.0,3.5);
\draw(4.5,4.0)--(4.5,3.0);
\draw(5.0,3.5)--(4.0,3.5);
\draw(6.0,3.5)--(5.0,3.5);
\draw(1.0,2.5)--(0.5,2.5)--(0.5,2.0);
\draw(1.5,3.0)--(1.5,2.0);
\draw(2.0,2.5)--(1.0,2.5);
\draw(2.5,3.0)--(2.5,2.5)--(2.0,2.5);
\draw(4.0,2.5)--(3.5,2.5)--(3.5,2.0);
\draw(4.5,3.0)--(4.5,2.5)--(4.0,2.5);
\node at (4.5,2.5) {$\bullet$};
\draw(6.0,2.5)--(5.5,2.5)--(5.5,2.0);
\draw(0.5,2.0)--(0.5,1.0);
\draw(1.5,2.0)--(1.5,1.0);
\draw(3.0,1.5)--(2.5,1.5)--(2.5,1.0);
\draw(3.5,2.0)--(3.5,1.0);
\draw(4.0,1.5)--(3.0,1.5);
\draw(5.0,1.5)--(4.0,1.5);
\draw(5.5,2.0)--(5.5,1.0);
\draw(6.0,1.5)--(5.0,1.5);
\draw(0.5,1.0)--(0.5,0.0);
\draw(1.5,1.0)--(1.5,0.0);
\draw(2.5,1.0)--(2.5,0.0);
\draw(3.5,1.0)--(3.5,0.0);
\draw(5.0,0.5)--(4.5,0.5)--(4.5,0.0);
\draw(5.5,1.0)--(5.5,0.0);
\draw(6.0,0.5)--(5.0,0.5);
\draw[color=red, thick, sharp corners] (4,2) rectangle (5,3);
\draw[color=violet, thick, sharp corners] (5,1) rectangle (6,2);
\node[color=black] at (3,-.5) {$D_1$};
\end{tikzpicture}}\overset{\Terminal\Cross}\to
\parbox{1.25in}{\begin{tikzpicture}[x=1em,y=1em,thick,color = blue]
\draw[step=1,gray,thin] (0,0) grid (6,6);
\draw[color=black, thick, sharp corners] (0,0) rectangle (6,6);
\draw(4.0,5.5)--(3.5,5.5)--(3.5,5.0);
\draw(5.0,5.5)--(4.0,5.5);
\draw(6.0,5.5)--(5.0,5.5);
\draw(3.0,4.5)--(2.5,4.5)--(2.5,4.0);
\draw(3.5,5.0)--(3.5,4.5)--(3.0,4.5);
\draw(5.0,4.5)--(4.5,4.5)--(4.5,4.0);
\draw(6.0,4.5)--(5.0,4.5);
\draw(2.0,3.5)--(1.5,3.5)--(1.5,3.0);
\draw(2.5,4.0)--(2.5,3.0);
\draw(3.0,3.5)--(2.0,3.5);
\draw(4.0,3.5)--(3.0,3.5);
\draw(4.5,4.0)--(4.5,3.0);
\draw(5.0,3.5)--(4.0,3.5);
\draw(6.0,3.5)--(5.0,3.5);
\draw(1.0,2.5)--(0.5,2.5)--(0.5,2.0);
\draw(1.5,3.0)--(1.5,2.0);
\draw(2.0,2.5)--(1.0,2.5);
\draw(2.5,3.0)--(2.5,2.5)--(2.0,2.5);
\draw(4.0,2.5)--(3.5,2.5)--(3.5,2.0);
\draw(4.5,3.0)--(4.5,2.0);
\draw(5.0,2.5)--(4.0,2.5);
\draw(6.0,2.5)--(5.0,2.5);
\draw(0.5,2.0)--(0.5,1.0);
\draw(1.5,2.0)--(1.5,1.0);
\draw(3.0,1.5)--(2.5,1.5)--(2.5,1.0);
\draw(3.5,2.0)--(3.5,1.0);
\draw(4.0,1.5)--(3.0,1.5);
\draw(4.5,2.0)--(4.5,1.5)--(4.0,1.5);
\draw(6.0,1.5)--(5.5,1.5)--(5.5,1.0);
\draw(0.5,1.0)--(0.5,0.0);
\draw(1.5,1.0)--(1.5,0.0);
\draw(2.5,1.0)--(2.5,0.0);
\draw(3.5,1.0)--(3.5,0.0);
\draw(5.0,0.5)--(4.5,0.5)--(4.5,0.0);
\draw(5.5,1.0)--(5.5,0.0);
\draw(6.0,0.5)--(5.0,0.5);
\node[color=black] at (3,-.5) {$D_2$};
\end{tikzpicture}}
\]

Suppose we know $\Phi(D_2)$
is the following compatible pair:
$$((3,5),(2,2),(2,5),(1,1),(1,3),(1,5)).$$
Then the recurrence definition of 
$\Phi(\cdot)$,
we have
$$
\Phi(D_1) = \II_4(\Phi(D_2))= 
((4,4),(3,5),(2,2),(2,5),(1,1),(1,3),(1,5)),
$$
$$
\Phi(D) = \downarrow(\Phi(D_1))= 
((3,4),(3,5),(2,2),(2,5),(1,1),(1,3),(1,5)).
$$
\printbibliography


\end{document}
