<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">


\junk{
%{\bf AK got this to compile but it's surely not close to being right}

\tilecool\bhwire\hwire{1}{1}{\es}{1}{1}{\es} $\mapsto$
        \tilecool\nowire\nowire{\es}{\es}{\es}{\es}{\es}{\es},
\tilecool\en\hwire{1}{1}{1}{\es}{1}{\es} $\mapsto$ 
        \tilecool\are\bnowire{\es}{\es}{1}{1}{\es}{\es},
\tilecool\bnowire\jay{1}{\es}{\es}{\es}{\es}{1} $\mapsto$
        \tilecool\hwire\sea{\es}{1}{\es}{1}{1}{1},
\tilecool\bnowire\nowire{\es}{\es}{\es}{\es}{\es}{\es} $\mapsto$
        \tilecool\hwire\hwire{1}{1}{\es}{1}{1}{\es}.
}

%This tells us that there does not exist a weight preserving completely local bijection
%between decorated $O\atop B$ sections and decorated $B\atop O$ sections which separates every $O\atop B$ section $D$ at every position. The
%property of separating every $O\atop B$ section at every position gives only a little information
%beyond implying that $\partial \phi_y (D)= \partial D$ and in fact
%Unfortunately there does not exist any completely local weight preserving bijection
%between $O\atop B$ sections and $B\atop O$ sections which preserves the boundary data.

 % Unfortunately, there does not exist a completely local bijection between decorated $O\atop B$ sections and decorated $B\atop O$ sections which preserves weights and connectivity. 

%\begin{tikzpicture}[scale=0.4]
%	\node at (0,0) {\tilecol\nowire\jay};  \node at (1,0) {\tilecol\nowire\nowire}; \node at (2,0) {\nowire\nowire};\node at (3,0) {\tilecol\en\nowire}
%\end{tikzpicture}

%This example \textcolor{red}{TODO: finish example and explain it properly} demonstrates the impossibility of a completely local bijection. In fact, the obstructions to having a completely local bijection all look something like one of the following examples:

%\begin{tikzpicture}[scale=0.4]
%	\node at (0,0) {\tilecol\bump\cross}; \node at (1,0) {\tilecol\cross\bump};
%\end{tikzpicture}

\junk{Dealing with this class of obstructions, as we do in the following lemma, allows us to obtain the weight preserving row swapping bijection we desire by splitting up a pair of rows into these obstructive pieces and everything else, dealing with the obstructions, and applying the tile by tile map that worked for all $(-y)$-decoration case to everything else. }



\junk{The bijection in Theorem \ref{thm:justy} is not weight-preserving for
arbitrary decorated $C\atop B$ sections because of the four instances
(\fbox{boxed} in the table above)
where a weighty tile switches from the classic row to the bumpless
row or vice versa.}
  Although the bijection in Theorem \ref{thm:justy} is not weight-preserving for
 	arbitrary decorated $C\atop B$ section, most of the replacement rules in the bijection described in Theorem \ref{thm:justy} will still be a part of the more general bijection we present in Theorem \ref{bijection}. The leftover columns will be the ones appearing in the seven \fbox{boxed} and \dashbox{dash-boxed} rules in Theorem \ref{thm:justy}. To define the row swapping bijection for arbitrary decorated $C\atop B$ sections we will (1) organize these leftover tiles into chunks which are described in Definition \ref{def:obstructive}, (2)  define the bijection on these chunks in Theorem \ref{obthm}, and finally (3) show that everything outside those chunks can be dealt with using the local replacement rules from Theorem \ref{thm:justy}.


	
\begin{Definition}\label{def:obstructive}
\junk{We say a $C\atop B$ section is \defn{obstructive} if it is of the following form:
\begin{enumerate}
	\item its leftmost column is \tilecol\bnowire\nowire, \tilecol\bnowire\jay, or \tilecol\sea\bump
	\item the first tile is followed by some (possibly 0) number of \tilecol\bnowire\nowire tiles
	\item  if the first column was not \tilecol\sea\bump, then the (possibly empty) string of blank columns is followed by a single \tilecol\sea\are column.
	 
	\item the \tilecol\sea\are column, or the \tilecol\sea\bump column if that was the first column, is followed by some (possibly 0) number of   \tilecool\bhwire\hwire{1}{1}{\es}{1}{1}{\es} columns
	\item the rightmost column may be the single \tilecol\sea\are column, it may be \tilecool\bhwire\hwire{1}{1}{\es}{1}{1}{\es}, it may be \tilecol\vwire\are, it may be \tilecool\en\hwire{1}{1}{1}{\es}{1}{\es}, or if the $C\atop B$ section is only one column wide it may be \tilecol\sea\bump.
\end{enumerate}

We say a $B\atop C$ section is \defn{obstructive} if it is of the following form:
\begin{enumerate}
	\item its leftmost column may be \tilecol\jay\en, \tilecool\hwire\bhwire{1}{1}{\es}{1}{1}{\es}, \tilecol\jay\vwire, or \tilecool\hwire\sea{\es}{1}{\es}{1}{1}{1}, and if the width of the $B\atop C$ section is 1 then the first (and last) column is also allowed to be \tilecol\bump\en.
	\item after (and possibly including) the first column there may be some (possibly zero) sequence of \tilecol\hwire\bhwire columns 
	\item after the possibly empty sequence of \tilecool\hwire\bhwire{1}{1}{\es}{1}{1}{\es} columns there will be either a single \tilecol\jay\en column or there will be a single \tilecol\bump\en column and it will be the last column
	\item after the \tilecol\jay\en column, there may be a (possibly empty) string of \tilecol\nowire\bnowire columns
	\item the last column will be \tilecol\nowire\bnowire, \tilecol\bump\en, \tilecol\are\bnowire, or \tilecol\jay\en.  
\end{enumerate}}%end junk
We say that a $C\atop B$ section is \defn{obstructive} if it can be
made with the following non-deterministic finite state machine (on the left) by
starting at one of the leftmost 5 nodes and ending at one of the
rightmost 5 modes. In the finite state machine $j$ and $k$ mean to
repeat that node $j$ or $k$ times respectively where $j$ and $k$ may
be any non-negative integer including 0.

$$
\begin{tikzcd}[ row sep=tiny, column sep=small,
  cells={nodes={inner xsep=3pt}},
  every arrow/.append style = {-stealth, shorten &gt; = 2pt, shorten &lt;=2pt},
  ]
  &amp;&amp;          \tilecol \vwire \are \\
  \tilecol \bnowire \jay  \ar[r] &amp;
  {\tilecol \bnowire \nowire}^j  \ar[r] \ar[ru] &amp;
  \tilecol \sea \are  \ar[r] &amp; 
  {\tilecol \bhwire \hwire}^k  \ar[r] &amp;
  \tilecol \en \hwire \\
  &amp;&amp;\tilecol \sea \bump \ar[ru] 
\end{tikzcd}
% \input{diag.tex}
\qquad\qquad
%\input{diag2.tex}
\begin{tikzcd}[row sep = tiny, column sep=small,
	cells={nodes={inner xsep=3pt}},
	every arrow/.append style = {-stealth, shorten &gt; = 2pt, shorten &lt;=2pt},
	]
	&amp;&amp;          \tilecol \jay \vwire \ar[dr] \\
	\tilecol \hwire \sea  \ar[r] &amp;
	{\tilecol \hwire \bhwire}^k  \ar[r] \ar[rd] &amp;
	\tilecol \jay \en  \ar[r] &amp; 
	{\tilecol \nowire \bnowire}^j  \ar[r] &amp;
	\tilecol \are \bnowire \\
	&amp;&amp;\tilecol \bump \en 
\end{tikzcd}
$$
The state machine diagram defining \defn{obstructive} $B\atop C$ sections is
the one on the right.


\end{Definition}

\begin{Theorem}\label{obthm}
	There is a unique weight preserving bijection between decorated obstructive $C\atop B$ sections and decorated obstructive $B\atop C$ sections such that the labels along the top and bottom edges are preserved. %and $\phi$ seperates each obstructive $O\atop B$ section $D$ of width $k$ at $0$ and $k$.
\end{Theorem}
\begin{proof}
	Every obstructive $C\atop B$ section ($B\atop C$ section) begins with a possibly empty sequence of columns with weighty tiles in the bumpless (classic) row and non-weighty tiles in the classic (bumpless) row. After this, there will be a single column with no weighty tiles. Finally, the $C\atop B$ section ($B\atop C$ section) ends with a possibly empty  sequence of columns with weighty tiles in the classic (bumpless) row and non-weighty tiles in the bumpless (classic) row. An obstructive $C\atop B$ section ($B\atop C$ section) is completely determined by the labels on its top and bottom edges together with the position of its single non-weighty column. Further, for each possible boundary of an obstructive $C\atop B$ section ($B\atop C$ section) and each $i$ between $1$ and the width of the $C\atop B$ section ($B\atop C$ section), there exists an obstructive $C\atop B$ section ($B\atop C$ section) with those boundary labels that has a non-weighty $i$th column. 
	
	Let $D$ be an obstructive $C\atop B$ section ($B\atop C$ section) with width $k$ and let $f$ be some decoration on it. \junk{For $f$ a decoration of an obstructive $C\atop B$ section ($B\atop C$ section) $D$,} Then,  $\mathrm{mon}(D,f)=x_c^a x_b^{k-1-a-m} \prod_{j=1}^m -y_{i_j}$ for some $0\leq m\leq k-1$, some $0\leq a\leq k-1-m$,and some sequence $i_1, i_2, \dots, i_m \in \{1,2,\dots, k\}$. 
	We claim that $\mathrm{mon}$ restricts to a bijection from $C\atop B$ sections ($B\atop C$ sections) with fixed labels along their top and bottom edges to monomials of the form in the previous sentence. Once proven, this claim implies that there is a unique weight preserving bijection between decorated obstructive $C\atop B$ sections and decorated obstructive $B\atop C$ sections.
	% given by $(\mathrm{mon}\mid_{\text{decorated } BC\text{-dreams with prescribed labels on top and bottom edges}})^{-1}\mathrm{mon}.$   
	\junk{We show the bijection between obstructive $C\atop B$ sections and obstructive $B\atop C$ sections indirectly by constructing bijections between each of them and polynomials of the above form.}
	 Suppose we have some $i_1,\dots, i_m \in [k]$ with $0\leq m\leq k-1$, and some $0\leq a \leq k-1-m$. Given labels along the top and bottom edges we want to show there is a unique obstructive decorated $C\atop B$ section $(D,f)$ and a unique obstructive decorated $B\atop C$ section $(D',f')$ with those labels and with $\mathrm{mon}(D,f)=\mathrm{mon}(D',f')=x_c^ax_b^{k-1-a-m} \prod_{j=1}^m -y_{i_j}$. To achieve that monomial, there needs to be a weighty tile in both $D$ and $D'$ decorated with $(-y)$ in the $(i_j)$th column for each $j=1,\dots, m$. The $a$ factors of $x_c$ in the monomial must come from the rightmost (leftmost) $a$ columns of $D$ ($D'$) which are not already reserved for giving $-y_{i_j} $ weights. The $k-1-a-m$ factors of $x_b$ must come from the leftmost (rightmost) $k-1-a-m$ columns of $D$  ($D'$) which are not already reserved for giving $-y_{i_j}$ weights. We have thus determined the weights we'll need to have coming from  $m+a+(k-1-a-m)=k-1$ of the $k$ columns of $D$ ($D'$). Hence, the unique remaining column of $D$ ($D'$) will be the unique non-weighty column of $D$ (D'). Setting a column as the non-weighty column uniquely determines what $D$ and $D'$ should each be based on the labels along the top and bottom edges. In the process of determining where the non-weighty column of $D$ and $D'$ had to be to achieve the monomial $x_c^ax_b^{k-1-a-m} \prod_{j=1}^m -y_{i_j}$, we specified which columns would contribute which weights. This tells us exactly how to decorate $D$ and $D'$ because each column of an obstructive $C\atop B$ section ($B\atop C$ section) has at most one weighty tile. We have thus shown that $\mathrm{mon}$ gives a weight preserving bijection between obstructive decorated $C\atop B$ sections ($B\atop C$ sections) with set labels on the top and bottom edges, and monomial weights of the form $x_c^a x_b^{k-1-a-m}\prod_{j=1}^m -y_{i_j}$ for appropriate $a,m,i_j$. We conclude that for any decorated $C\atop B$ section $(D,f)$, there is a unique decorated $B\atop C$ section $(D',f')$ such that the the top and bottom edge labels of $D$ match those of $D'$ and $\mathrm{mon}(D,f)=\mathrm{mon}(D',f').$ \qedhere
	
	
	
	\junk{
	 We have thus constructed a weight preserving bijection between obstructive decorated $C\atop B$ sections ($B\atop C$ sections) with set labels on the top and bottom edges, and monomial weights of the given form. This gives us a bijective map $\phi$ from obstructive decorated $C\atop B$ sections with certain labels on the top and bottom and obstructive decorated $B\atop C$ sections with the same labels on the top and bottom edges.
	
	The labels along the top and bottom edges of an obstructive $C\atop B$ section or $B\atop C$ section are very simple. There may be a label on the top edge of the first column and/or on the bottom edge of the final column, and there cannot be labels anywhere else.} %From the definitions of obstructive $O\atop B$ sections and obstructive $B\atop O$ sections (or from looking at representative examples of them) the following statements hold: an obstructive $O\atop B$ section $D$ has blank labels on its two vertical left edges if it does not have a label on the top of its first column and otherwise $C_0(D)=(1,\emptyset)$, an obstructive $B\atop O$ section $D'$ has both of its vertical left edges labeled $0$ iff it does not have a label on the top of its first column and otherwise $C_0(D')=(1,0)$. This proves that $\phi$ seperates every obstructive decorated $O\atop B$ section $D$ at $0$. Also from the definitions we can see that an obstructive $O\atop B$ section $D$ with width $k$ has $C_k(D)=(x,x)$ for some $x$ if the label below the final column is blank and has $C_k(D)=(x,\emptyset)$ for $x$ the label below the final column otherwise; on the other hand an obstructive $B\atop O$ section $D'$ with width $k$ has $C_k(D')=(\emptyset,\emptyset)$ if the label below the final column is blank and $C_k(D')=(x,\emptyset)$ with $x$ the label below the final column otherwise. This proves that $\phi$ separates every obstructive decorated $O\atop B$ section $D$ at $k$.
\end{proof}

%An important property of obstructive $O\atop B$ sections and $B\atop O$ sections is that they can be identified as sub-$O\atop B$ sections or sub-$B\atop O$ sections of some larger $O\atop B$ section or $B\atop O$ section entirely by the boundary data of the larger $B\atop O$ section. That is, without


	%The pipe entering through the top of the first column is either blank or it exits through the left edge of the ordinary row (and its label does not appear anywhere else on the boundary of $D$). When that label above the first column is blank, the left edges of the $D$ have blank labels on them if $D$ is an $O\atop B$ section and they have the same label (which wont appear anywhere else on $\partial D$) on them otherwise 


%Although there is no completely local weight preserving bijection between decorated $O\atop B$ sections and decorated $B\atop O$ sections, there is a weight preserving completely local bijection between $O\atop B$ sections decorated only with $y$ and $BO$ dreams decorated only with $y$. We intro

%
%
%
%
%%


%In the absence of a completely local bijection, we  construct a bijection which is as local as it can be in a certain sense. To compute the bijection on a full $O\atop B$ section, we will break up the $O\atop B$ section into smaller pieces (each of which an $OB$ dream itself), describe what to do with these smaller pieces, and then put the results back together. This will be somewhat local in the sense that what the bijection does to a single one of these pieces will depend only on that piece. It turns out that all of the pieces we have to deal with which have width greater than $1$ but cannot be broken down any further will look something like the example above. \textcolor{red}{TODO: properly number example}. 

%\begin{lemma}
	%Let $C$ be the list of labels around the outside of some $O\atop B$ section. Label internal vertical %edges of an $O\atop B$ section $If every $O\atop B$ section with boundary data $C$ has the same pipes (or %lac) 
%\end{lemma}


%We'll soon break up a full $O\atop B$ section into many pieces, but we can start
%by thinking about breaking it into two pieces. If $D$ is an $O\atop B$ section
%we can cut it into two smaller $O\atop B$ sections $D_1,D_2$ where $D_1$ is
%the first $i$ columns of $D$ and $D_2$ is the last $i$ columns of
%$D$. Once we compute $\phi(D_1)$ and $\phi(D_2)$ we can {\bf Should
%  that be ``We might hope to"?} set $\phi(D)$
%to be the $B\atop O$ section that has $\phi(D_1)$ as its first $i$ columns and
%$\phi(D_2)$ as its last $i$ columns. We need to know (1) that the
%labels along the vertical edge of $D_1$ match the labels on the
%vertical edge of $D_2$ and (2) that our map is surjective. We make
%these tasks significantly easier by only cutting at $i$ such that the
%boundary data of $D$ uniquely determines the boundary data of $D_1$
%and $D_2$. Denoting the labels on the boundary of an $O\atop B$ section
%($B\atop O$ section) $D$ by $\partial D$, this can be rephrased as: we cut at
%$i$ if for all $D'$, such that $\partial D'=\partial D,$ we have
%$\partial D_1' = \partial D_1$ where $D_1'$ is the first $i$ columns
%of $D'$. One reason this works out especially well is that for fixed
%boundary data, the indices $i$ that have this property for $O\atop B$ sections
%are the same as the indices that have this property for
%$B\atop O$ sections. \textcolor{red}{I think this may be poorly explained.}












 %The row-swapping bijection we seek only needs to be defined on full $O\atop B$ sections; however, we define our bijection by defining it for a subset of small $O\atop B$ sections and describing how to break a full $O\atop B$ section into these pieces. We show that our bijection between full $O\atop B$ sections and full $B\atop O$ sections is the only possible bijection which can be computed by breaking up an $O\atop B$ section into these smaller pieces and computing the bijection on these pieces. Unfortunately, the map we construct is not a bijection between all $OB$ dreams and all $B\atop O$ sections. Our goal is to 


 %\textcolor{red}{Some care needs to be taken and I need to think carefully about what exactly is true and how to say it.}

	
%	The major distinction between full $O\atop B$ sections ($B\atop O$ sections) and $O\atop B$ sections ($B\atop O$ sections) that may not be full is that a full $O\atop B$ section ($B\atop O$ section) has blank labels on the right edge of its top row and the left edge of its bottom row (left edge of its top row and right edge of its bottom row) and blank labels on its other vertical edges, whereas an arbitrary $O\atop B$ section ($B\atop O$ section) may have any combination of blank and non-blank labels on its vertical edges. These labels tell us where pipes are and how they connect, but they are not a completely essential part of the data encoding an $O\atop B$ section ($B\atop O$ section). That is, the data of an $OB$-deam ($B\atop O$ section) is the $2\times k$ grid of tiles making it up, and we will say that two $O\atop B$ sections are the same if they have the same tiles, even if we think about them as subrectangles of different hybrid pipe dreams (that may have given different labels to the pipes). \textcolor{red}{this is blatantly false but I'm leaving it here for now so I can hopefully get to other stuff.} Although the precise labels on an $O\atop B$ section aren't important, we still want to be able to describe the connectivity of them. To do this, we will apply labels in a consistent manner. In particular,  we label the pipes from $0,1,\dots, p,p+1$ based on where they exit the $O\atop B$ section ($B\atop O$ section) so that a pipe that exits on the left edge of the ordinary row is labeled $1$ (unless it is in a $B\atop O$ section and originally entered through the left edge of the bumpless row, in which case we label it $0$), labels of pipes are increasing left to right across the bottom edge of the $O\atop B$ section ($B\atop O$ section), a pipe exiting the $O\atop B$ section ($B\atop O$ section) through the right edge of the bumpless row is labeled $p$ (unless it is in an $OB$ dream and initially entered through the right edge of the ordinary row, in which case we label it $p+1$), and pipes have the same label iff they are the same pipe, and all labels between $1$ and $p$ are used. The final condition of the previous sentence is equivalent to saying that for both $O\atop B$ sections and $B\atop O$ sections, $p$ will be the number of pipes that either enter through the top of the $O\atop B$ section or exit through the bottom. \textcolor{red}{This is kind of a lot of explaining to do for something that is just notational and not all that  important.} \textcolor{red}{Here are some examples (TODO):} 
	
%	These labelings are not especially important, but they are helpful for distinguishing $O\atop B$ sections ($B\atop O$ sections) with different connectivity. We define the \textbf{configuration} of an $O\atop B$ section ($B\atop O$ section) as a list  $C=(C_1,C_2,x_1,x_2,y_1,y_2)$ where $C_1$ is the list of labels across the top edges (including blank labels), $C_2$ is the list of labels across the bottom edges (including blank labels), $x_1$ is the label on the right edge of the bumpless row, $x_2$ is the label on the left edge of the ordinary row, $y_1$ is the label on the left edge of the ordinary row, and $y_2$ is the label on the right edge of the bumpless row. 
	% If there are blanks on both left edges or on both right edges of an $O\atop B$ section ($B\atop O$ section) then
		 
	%The weighty tiles in an $OB$ dream ($BO$) dream are $\inlinehwire$ and $\inlinecross$ in the ordinary row and $\inlinenowire$ in the bumpless row. We attach decorations $f$ to $O\atop B$ sections ($B\atop O$ sections) in the same way that we did so with hybrid pipe dreams. To assign weights to these in the context of a complete hybrid pipe dream we need to know the labels on the two rows. However, for the purpose of verifying that our bijection preserves weights, we can pretend the rows are labeled $1$ and $2$ and attach weights of $x_1$ to weighty tiles decorated with $x$ in the ordinary row and $x_2$ to weighty tiles decorated with $y$ in the bumpless row.  
	
	
%We can roughly describe obstructive $O\atop B$ sections as $O\atop B$ sections where some pipe starts at the rightmost column, moves left for a while, and then doubles back and returns to the rightmost column. 

%There can be many $O\atop B$ sections ($B\atop O$ sections) with the same labels in the same positions around the boundary. In particular, for an $O\atop B$ section ($B\atop O$ section) $D$ of width $k$, there will be $k$ different $O\atop B$ sections ($B\atop O$ sections) that have the same boundary labels. In contrast, it turns out that a column of a full $O\atop B$ section ($B\atop O$ section) which is not contained in an obstructive $O\atop B$ section ($B\atop O$ section) will be completely determined by the connectivity of the full $O\atop B$ section ($B\atop O$ section).  


Columns of a full $C\atop B$ section ($B\atop C$ section) which aren't contained in some obstructive $C\atop B$ section ($B\atop C$ section) turn out to be completely determined by the connectivity of the full $C\atop B$ section ($B\atop C$ section).  To aid in the discussion of the connectivity of hybrid pipe dreams we introduce the following definitions. 


%A full $O\atop B$ section ($B\atop O$ section) $D$ may contain several obstructive $O\atop B$ sections ($B\atop O$ sections) within it. Each of these obstructive $O\atop B$ section ($B\atop O$ section) could be swapped out with some other one that has the same boundary to give a new full $O\atop B$ section ($B\atop O$ section) with the same boundary as the original one. I



\begin{Definition}
	Define the $i$th edge content $C_i(D)$ of either a $C\atop B$ section or $B\atop C$ section $D$ to be $(x,y)$ where $x$ is the label of the pipe passing from the tile in the $(i+1)$st column of the classic row of $D$ to the $i$th column of the classic row of $D$ (or $\emptyset$ if there is no such pipe) and $y$ is the label of the pipe passing from the $i$th column of the bumpless row of $D$ to the $(i+1)$st column of the bumpless row of $D$ (and $\emptyset$ if there is no such pipe). 
	
	Define the boundary $\partial D$ of $D$ to be the list of labels along the top, bottom, left, and right boundary edges of $D$ where $D$ may be either a $C\atop B$ section or a $B\atop C$ section.  
	
	If $D$ is a $C\atop B$ section ($B\atop C$ section) such that for all $C\atop B$ sections ($B\atop C$ sections) $D'$ such that $\partial D' = \partial D,$ $C_i(D')=(x,y)$ we call $i$ \defn{separating} for $D$.
\end{Definition}	


\junk{\begin{Lemma}\label{compose}
	Suppose $i$ is separating for a $C\atop B$ section ($B\atop C$ section) $D$ with width $n$. Let $D_1$ be the $C\atop B$ section ($B\atop C$ section) consisting of the first $i$ columns of $D$ and $D_2$ be the $C\atop B$ section ($B\atop C$ section) consisting of the columns of $D$ right of $i$. For any $D'$ define $D_1'$ and $D_2'$ analogously. A $C\atop B$ section ($B\atop C$ section) $D'$ satisfies $\partial D'=\partial D$ if and only if $\partial D_1'=\partial D_1$ and $\partial D_2'=\partial D_2$. Note that  $\partial D_1'=\partial D_1$ if and only if $\partial D_2'=\partial D_2$. 
\end{Lemma}
}

	

		
\begin{Lemma}\label{distinctEdge} If $D$ is a $C\atop B$ section ($B\atop C$ section) and
$C_i(D)=(x,y)$ with $x\neq y$ then $i$ is separating for $D$ and in fact
$C_i(D')=(x,y)$ for any $C\atop B$ section ($B\atop C$ section) $D'$ with $\partial D=\partial D'$.  In particular, if we
introduce the following conditions:
\begin{enumerate}
\item[(i)] the label $x$ appears on the top
  edge of $D$ in a column right of the $i$th column (or on the right
  edge of the classic row)
\item[(ii)] the label $x$ appears on the
  bottom edge of $D$ in a column left of the $i$th column (or on the
  right edge of the classic row)
\item[(iii)] the label $y$ appears on the top
  edge of $D$ in the $i$th column or to the left of it (or on the left
  edge of the bumpless row)
\item[(iv)] the label $y$ appears on the
  bottom edge of $D$ in a column right of the $i$th column (or on the
  right edge of the bumpless row),
\end{enumerate} then
\begin{enumerate}
\item $C_i(D)=(x,y)$ with $x\neq y$ and $x\neq
  \emptyset \neq y$ iff conditions (i),(ii),(iii), and (iv) all hold.
\item $C_i(D)=(x,\emptyset)$ with $x\neq
  \emptyset$ iff conditions (i) and (ii) hold and for all $y$, either
  condition (iii) does not hold or condition (iv) does not hold.
\item $C_i(D)=(\emptyset, y)$ with $y\neq \emptyset$
  iff conditions (iii) and (iv) hold, and for all $x$, either condition
  (i) does not hold or condition (ii) does not hold.
\end{enumerate}
\end{Lemma}

\begin{Lemma}\label{lem:noright} If the label $x\neq \emptyset$ appears on 
  the top edge of the $i$th column of $D$ and the bottom edge of the $j$th column of $D$ for some $j\geq i$ or the right edge of the bumpless row of $D$ when $D$ has width $j$, then $i-1, i, \dots, j$ are all separating for $D$. Additionally, if the label $x\neq \emptyset$ appears on the bottom edge of the $i$th column and the left edge of the bumpless row of $D$ then $1,2,\dots,i$ are all separating for $D$.
%  $C_i(D)=\begin{cases} (\emptyset,\emptyset) &amp; D \text{ is an
%\(OB\)-dream}\\ (x,x) &amp; D \text{ is a \(BO\)-dream}
 %         \end{cases}$ and $C_{i-1}(D)=\begin{cases}
%(x,x) &amp; D \text{ is an \(OB\)-dream}\\ (\emptyset, \emptyset) &amp; D
%\text{ is a \(BO\)-dream}
%		\end{cases}.$
%	Consequently, whenever $x\neq \emptyset$ appears on both the top and bottom edge of the right column 
\end{Lemma}
	
	
	

\begin{Definition}
	Let $D$ be a $C\atop B$ section ($B\atop C$ section) of width $n$. The interval $[a,b]$ is irreducible in $D$ if no $i\in [a,b-1]$ is separating for $D$. An irreducible interval $[a,b]$ in $D$ is called maximal if $a-1$ and $b$ are separating for $D$. 
 \end{Definition}

%\begin{Lemma}\label{lem:dontEndEarly}
	%A maximal irreducible interval of a full $O\atop B$ section cannot end with $\tilecol \bnowire \jay$ or $ \tilecol \bnowire \nowire$. 
%	Let $F$ be a full $O\atop B$ section ($B\atop O$ section). Let $D$ be an $O\atop B$ section ($B\atop O$ section) contained in $F$. If $D$ can be extended to an obstructive $O\atop B$ section ($B\atop O$ section) by adding columns to the left and right, then there is some obstructive $O\atop B$ section 
%\end{Lemma}
%\begin{proof}
%	The $O\atop B$ section columns that can be immediately to the right of  $\tilecol \vwire \are$ or  $\tilecol \bnowire \jay$ are  $\tilecol \bnowire \nowire$, $\tilecol \sea \are,$ and $\tilecol \bhwire \hwire.$ After some number of blank columns there must be either  
	
%	Since a full $O\atop B$ section can't end with $\tilecol \bnowire \nowire$, if we see $\tilecol \vwire \are$ or  $\tilecol \bnowire \jay$, 
%\end{proof}
	
	

	
	
	
\begin{Lemma}\label{obstruct} %If $D$ is an $O\atop B$ section ($B\atop O$ section) such that no $i\in [1,\mathrm{width}(w)-1]$ is separating for $D$ then we call $D$ \defn{irreducible}. Every irreducible $O\atop B$ section ($B\atop O$ section) with width greater than 1 is obstructive. 
	
Let $F$ be a full $C\atop B$ section ($B\atop C$ section) and let $D$ be the $C\atop B$ section ($B\atop C$ section) sitting in columns $a$ through $b$ of $F$ where $b&gt;a$. If $[a,b]$ is a maximal irreducible interval for $F$ then $D$ is an obstructive $C\atop B$ section ($B\atop C$ section).
	
	 %If $D$ is an irreducible $O\atop B$ section then the label on
 % the top of the first column of $D$ (which may be blank) is the same
  %as the label on the left edge of the ordinary row, the label on the
 % right edge of the ordinary row is equal to either the label on the
%  right edge of the bumpless row or the label on the bottom edge of
 % the final column.
\end{Lemma}
\begin{proof}


	
	
Suppose that $[a,b]$ is a maximal irreducible interval for $F$. 

 Say a pipe $p$ enters $D$ from the top edge of the $i$th column (with $i\in [a,b]$ of course). If $p$ exits $D$ from the bottom edge of a column right of $i$ (or exits from the bottom of $i$) then $i-1$ and $i$ are both separating for $D$ by Lemma \ref{lem:noright}. In that case, irreducibility implies that $i-1=a-1$ and $i=b$ which would then contradict the assumption that $b&gt;a$.  We conclude that $p$ must exit $D$ either to the left of the $i$th column or through a vertical edge of $D$. If $p$ exits through the right edge of the bumpless row of $D$ then $b-1$ is separating for $D$ by Lemma \ref{lem:noright} so either $D$ is not irreducible or $b=a+1$. Either case is a contradiction so we conclude that $p$ exits through the left edge of the classic row of $D$ or somewhere left of the $i$th column. If $p$ exits $D$ in the $j$th column for some $a\leq j&lt;i$ then $i-1$ will be separating for $D$ by Lemma \ref{distinctEdge} parts 1 and 2. Irreducibility then implies $i-1=a-1$ which contradicts our assumption that $a\leq j &lt;i.$ Since $p$ can't exit $D$ in the bottom of column $i$ and it can't exit $D$ in a column $j&lt;i$ or a column $j&gt;i$, the only remaining possibility is that $p$ exits $D$ through the left edge of the classic row of $D$ and $i=a$. In summary, if there is a pipe entering $D$ from a top edge, it must enter from the top edge of column $a$ and exit through the left edge of the classic row of $D$. 
 
 
 Suppose a pipe $p$ exits $D$ from the bottom edge of the $i$th column of $D$. By the preceding paragraph, $p$ cannot have entered $D$ from a top edge, so $p$ must have entered $D$ from the right edge of the classic row of $D$ or the left edge of the bumpless row of $D$. Suppose towards a contradiction that $p$ entered through the left edge of the bumpless row. Then, $a$ will be separating for $D$ by Lemma \ref{lem:noright}. Since $b-1&gt;a,$ that contradicts irreducibility. Therefore, we conclude that $p$ can only have entered $D$ from the right edge of the classic row. If $i$ is less than $b$, then Lemma \ref{distinctEdge} shows $b-1$ to be separating for $D$. Thus, we conclude $i=b$ by irreducibility. Our takeaway is that if a pipe $p$ exits $D$ from a bottom edge, it must be entering from the right edge of the classic row and exiting from the bottom edge of column $b$. 
 
In addition to what we know already, Lemma \ref{distinctEdge} tells us that for all $i=a,\dots, b-1$, $C_i(D)=(\emptyset,\emptyset)$ or $C_i(D)=(x,x)$ for some $x$. With all the restrictions on the contents of $D$ we can now go through the list of $2\times 1$ $C\atop B$ section columns and and find that if $D$ is a $C\atop B$ section, the possible columns are: 
 $
 \tilecol \vwire \are,  \tilecol \bnowire \jay , \tilecol \bnowire \nowire, \tilecol \sea \are, \tilecol \bhwire \hwire, \tilecol \en \hwire,$ and $\tilecol \sea \bump.$   
 If $D$ is a $B\atop C$ section, the possible tiles will be:  
 $\tilecol \jay \vwire, 
 \tilecol \hwire \sea ,
 \tilecol \hwire \bhwire, 
\tilecol \jay \en, 
\tilecol \nowire \bnowire,
\tilecol \are \bnowire,$ and 
$\tilecol \bump \en. $ 
 This does not yet tell us that $D$ is an obstructive $C\atop B$ section ($B\atop C$ section). The problem is that $D$ could start with a column which is not a start state of the finite state machine or end with a column that is not an end state of the finite state machine.  Call a column of a $C\atop B$ section ($B\atop C$ section) problematic if it is included in the finite state machine for obstructive $C\atop B$ sections ($B\atop C$ sections). In the finite state machine for $C\atop B$ sections ($B\atop C$ sections), the valid start states are exactly the columns which can be preceded by some non-problematic tile and the valid end states are exactly the columns that can be followed by some non-problematic tile. The columns which are not valid start states cannot appear at the beginning of a full $C\atop B$ section ($B\atop C$ section). The columns which are not valid end states cannot appear at the end of a full $C\atop B$ section ($B\atop C$ section). Hence, if $D$ starts with an invalid start state/ends with an invalid end state, it will be preceded/followed by a sequence of problematic tiles beginning/ending with a valid start/end column. This larger $C\atop B$ section ($B\atop C$ section) will be obstructive. If this containment is proper, then either $a-1$ or $b$ won't be separating. That would contradict maximality, so we conclude that $D$ is an obstructive $C\atop B$ section ($B\atop C$ section). \qedhere

 
 
 %To summarize, the first column is the only column that may have a label on its top edge, the last column is the only column that may have a label on its bottom edge, if the first column has a label on its top edge that label also appears on the left edge of the ordinary row, and if the last column has a label on its bottom edge that label also appears on the right edge of the ordinary row. By Lemma \ref{distinctEdge} we also get that $C_i(D)=(\emptyset,\emptyset)$ or $C_i(D)=(x,x)$ for some $x$ for all $i=1,\dots, w-1$. The restrictions these observations put on possible first columns, possible last columns, and possible intermediate columns lead us to find the potential first columns, intermediate columns, and last columns to be exactly the potential first columns, intermediate columns, and last columns respectively of an obstructing $O\atop B$ section ($B\atop O$ section) with width greater than 1. 
 
 
 
 
% The definition of obstructing $O\atop B$ sections ($B\atop O$ sections) includes every valid combination of columns that each individually sometimes appear in an obstructing $O\atop B$ section ($B\atop O$ section) and both begin and end with columns allowed to bookend an obstructive $O\atop B$ section ($B\atop O$ section). Thus, every irreducible $O\atop B$ section ($B\atop O$ section) is obstructive.
\end{proof}

\begin{Theorem}\label{bijection}
	There is a weight preserving bijection between decorated full $C\atop B$ sections and decorated full $B\atop C$ sections.
\end{Theorem}
\begin{proof}
To map a $C\atop B$ section $D$ to a $B\atop C$ section we first split it up into smaller pieces, splitting it up between columns $i$ and $i+1$ whenever $i$ is separating for $D$. This leaves us with maximal irreducible intervals which by Lemma \ref{obstruct} will either be single columns or obstructive $C\atop B$ sections. Obstructive $C\atop B$ sections (including ones which are single columns) are mapped to obstructive $B\atop C$ sections in the manner described in Theorem \ref{obthm} and all other single columns are mapped to single columns in the manner described in Theorem \ref{thm:justy}. The 23 replacement rules in Theorem \ref{thm:justy} are characterized by the property that $C_i(\phi_y(D))=\begin{cases} (\emptyset, \emptyset) &amp; C_i(D)=(x,x) \text{ for some } x\\ (x,x) \text{ for some } x &amp; C_i(D)=(\emptyset, \emptyset)\\ C_i(D) &amp; \text{else.}\end{cases}$ This property also holds on the leftmost and rightmost boundary edges for the bijection between decorated obstructive $C\atop B$ sections ($B\atop C$ sections). Thus, when we apply our construction on each maximal irreducible interval, the results will fit together. This construction sends maximal irreducible intervals to maximal irreducible intervals so it is indeed a bijection.
%Finally the results of these maps on the irreducible pieces are concatenated to produce a $B\atop O$ section. Computing the bijection on these pieces and concatenating the results makes sense because of Lemma \ref{compose}. 
\end{proof}

	
	%By Lemma \ref{distinctEdge}, if the two vertical edges to the left of a column of  
	
	

	
	
	% A \textbf{configuration} for an $O\atop B$ section or $B\atop O$ section which does not have blank labels on both vertical edges of the left or both vertical edges to the right is a tuple \textcolor{red}{TODO: figure out how to label these.}
	
As an example of Theorem \ref{bijection}, the following $C\atop B$ section and $B\atop C$ section are mapped to one another when all decorations are $x$.\\	
	\begin{tikzpicture}[scale=0.4]
	\node at (0,0)	{\tilecool{\bnowire}{\jay}{1}{\phantom 0}{\phantom 1}{\phantom 0}{\phantom 0}{1}};
	\node at (1,0) {\tilecool{\sea}{\are}{\phantom 0}{\phantom 0}{\phantom 1}{\phantom 0}{\phantom 0}{\phantom 1}};
	\node at (2,0) {\tilecool{\en}{\jay}{\phantom 0}{\phantom 0}{2}{\phantom 0}{\phantom 0}{2}};
	\node at (3,0) {\tilecool{\sea}{\are}{\phantom 0}{\phantom 0}{\phantom 1}{\phantom 0}{\phantom 0}{\phantom 1}};
	\node at (4,0) {\tilecool{\en}{\hwire}{\phantom 0}{\phantom 0}{3}{\phantom 0}{\phantom 0}{\phantom 1}};
	\node at (5,0) {\tilecool{\vwire}{\cross}{\phantom 0}{\phantom 0}{4}{\phantom 0}{\phantom 0}{4}};
	\node at (6,0) {\tilecool{\bnowire}{\jay}{\phantom 0}{\phantom 0}{\phantom 1}{\phantom 0}{\phantom 0}{3}};
	\node at (7,0) {\tilecool{\sea}{\are}{\phantom 0}{\phantom 0}{\phantom 1}{\phantom 0}{\phantom 0}{\phantom 1}};
	\node at (8,0) {\tilecool{\bhwire}{\hwire}{\phantom 0}{\phantom 0}{\phantom 1}{\phantom 0}{\phantom 0}{\phantom 1}};
	\node at (9,0) {\tilecool{\bcross}{\bump}{\phantom 0}{\phantom 0}{5}{\phantom 0}{\phantom 0}{7}};
	\node at (10,0) {\tilecool{\bcross}{\cross}{\phantom 0}{\phantom 0}{6}{\phantom 0}{\phantom 0}{6}};
	\node at (11,0) {\tilecool{\en}{\jay}{\phantom 0}{\phantom 0}{7}{\phantom 0}{\phantom 0}{5}};
	\node at (12,0) {\tilecool{\bnowire}{\nowire}{\phantom 0}{\phantom 0}{\phantom 1}{\phantom 0}{\phantom 0}{\phantom 1}};
	\node at (13,0) {\tilecool{\bnowire}{\nowire}{\phantom 0}{\phantom 0}{\phantom 1}{\phantom 0}{\phantom 0}{\phantom 1}};
	\node at (14,0) {\tilecool{\sea}{\are}{\phantom 0}{\phantom 0}{\phantom 1}{\phantom 0}{\phantom 0}{\phantom 1}};
	\node at (15,0) {\tilecool{\bhwire}{\hwire}{\phantom 0}{\phantom 0}{\phantom 1}{\phantom 0}{\phantom 0}{\phantom 1}};
	\node at (16,0) {\tilecool{\bhwire}{\hwire}{\phantom 0}{\phantom 0}{\phantom 1}{\phantom 0}{\phantom 0}{\phantom 1}};
	\node at (17,0) {\tilecool{\bhwire}{\jay}{\phantom 0}{\phantom 0}{\phantom 1}{8}{\phantom 0}{8}};
	\end{tikzpicture}
\begin{tikzpicture}[scale=0.4]
	\node at (0,0)	{\tilecool{\jay}{\vwire}{\phantom 1}{1}{\phantom 1}{\phantom 0}{\phantom 0}{1}};
	\node at (1,0) {\tilecool{\nowire}{\bnowire}{\phantom 0}{\phantom 0}{\phantom 1}{\phantom 0}{\phantom 0}{\phantom 1}};
	\node at (2,0) {\tilecool{\are}{\sea}{\phantom 0}{\phantom 0}{2}{\phantom 0}{\phantom 0}{2}};
	\node at (3,0) {\tilecool{\hwire}{\bhwire}{\phantom 0}{\phantom 0}{\phantom 1}{\phantom 0}{\phantom 0}{\phantom 1}};
	\node at (4,0) {\tilecool{\bump}{\en}{\phantom 0}{\phantom 0}{3}{\phantom 0}{\phantom 0}{\phantom 1}};
	\node at (5,0) {\tilecool{\cross}{\vwire}{\phantom 0}{\phantom 0}{4}{\phantom 0}{\phantom 0}{4}};
	\node at (6,0) {\tilecool{\hwire}{\sea}{\phantom 0}{\phantom 0}{\phantom 1}{\phantom 0}{\phantom 0}{3}};
	\node at (7,0) {\tilecool{\jay}{\en}{\phantom 0}{\phantom 0}{\phantom 1}{\phantom 0}{\phantom 0}{\phantom 1}};
	\node at (8,0) {\tilecool{\nowire}{\bnowire}{\phantom 0}{\phantom 0}{\phantom 1}{\phantom 0}{\phantom 0}{\phantom 1}};
	\node at (9,0) {\tilecool{\are}{\sea}{\phantom 0}{\phantom 0}{5}{\phantom 0}{\phantom 0}{7}};
	\node at (10,0) {\tilecool{\cross}{\bcross}{\phantom 0}{\phantom 0}{6}{\phantom 0}{\phantom 0}{6}};
	\node at (11,0) {\tilecool{\bump}{\bcross}{\phantom 0}{\phantom 0}{7}{\phantom 0}{\phantom 0}{5}};
	\node at (12,0) {\tilecool{\hwire}{\bhwire}{\phantom 0}{\phantom 0}{\phantom 1}{\phantom 0}{\phantom 0}{\phantom 1}};
	\node at (13,0) {\tilecool{\hwire}{\bhwire}{\phantom 0}{\phantom 0}{\phantom 1}{\phantom 0}{\phantom 0}{\phantom 1}};
	\node at (14,0) {\tilecool{\jay}{\en}{\phantom 0}{\phantom 0}{\phantom 1}{\phantom 0}{\phantom 0}{\phantom 1}};
	\node at (15,0) {\tilecool{\nowire}{\bnowire}{\phantom 0}{\phantom 0}{\phantom 1}{\phantom 0}{\phantom 0}{\phantom 1}};
	\node at (16,0) {\tilecool{\nowire}{\bnowire}{\phantom 0}{\phantom 0}{\phantom 1}{\phantom 0}{\phantom 0}{\phantom 1}};
	\node at (17,0) {\tilecool{\nowire}{\sea}{\phantom 0}{\phantom 0}{\phantom 1}{\phantom 8}{8}{8}};
\end{tikzpicture}
	


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