update pairing function

latex/lafi2024/acmart.tex

@@ -107,7 +107,7 @@
   \end{footnotesize}
 \end{equation*}
 
-\noindent This initializes the superdiagonal entries, enabling us to compute the fixpoint $M_\infty$ by redefining $\oplus, \otimes: \mathbb{T}_3 \times \mathbb{T}_3 \rightarrow \mathbb{T}_3$ as:
+\noindent This initializes the superdiagonal, enabling us to compute the fixpoint $M_\infty$ by redefining $\oplus, \otimes: \mathbb{T}_3 \times \mathbb{T}_3 \rightarrow \mathbb{T}_3$ as:
 
 \begin{equation*}
   \begin{footnotesize}
@@ -210,7 +210,7 @@ \subsection{Sampling trees with replacement}\label{sec:replacement}
 \caption{A partial $\mathbb{T}_2$ for the grammar with productions $P=\{S \rightarrow BC \mid \ldots \mid AB, B\rightarrow RD \mid \ldots, A\rightarrow QC \mid \ldots\}$.}
 \end{figure}
 
-The number of binary trees inhabiting a single instance of $\mathbb{T}_2$ is sensititive to the number of nonterminals and rule expansions in the grammar. To obtain the total number of trees with breadth $n$, we abstractly parse the porous string using the algebra defined in \S~\ref{sec:method}, letting $T=\Lambda_{\underline\sigma}^* \circ S$, and compute the total number of trees using the recurrence:
+To obtain the total number of trees with breadth $n$, we abstractly parse the porous string using the algebra defined in \S~\ref{sec:method}, and compute the total number of trees using:
 
 \begin{equation*}
   |T: \mathbb{T}_2| \mapsto \begin{cases}
@@ -223,18 +223,35 @@ \subsection{Sampling trees with replacement}\label{sec:replacement}
 To sample all trees in a given $T: \mathbb{T}_2$ uniformly without replacement, we then construct a modular pairing function $\varphi: \mathbb{T}_2 \rightarrow \mathbb{Z}_{|T|} \rightarrow \texttt{BTree}$, which is defined as follows:
 
 \begin{small}
-\begin{equation*}
-  \varphi(T: \mathbb{T}_2, i: \mathbb{Z}_{|T|}) \mapsto \begin{cases}
-  \Big\langle\texttt{BTree}\big(\texttt{root}(T)\big), i\Big\rangle & \text{if $T$ is a leaf,} \vspace{5pt}\\
-  \text{Let } b = |\texttt{children}(T)|,\\
-  \phantom{\text{Let }} q_1, r=\big\langle\lfloor\frac{i}{b}\rfloor, i \pmod{b}\big\rangle,\\
-  \phantom{\text{Let }} lb, rb = \texttt{children}[r],\\
-  \phantom{\text{Let }} T_1, q_2 = \varphi(lb, q_1),\\
-  \phantom{\text{Let }} T_2, q_3 = \varphi(rb, q_2) \text{ in } \\
-  \Big\langle\texttt{BTree}\big(\texttt{root}(T), T_1, T_2\big), q_3\Big\rangle & \text{otherwise.} \\
-  \end{cases}
+\begin{equation*}\label{eq:pairing}
+\varphi(T: \mathbb{T}_2, i: \mathbb{Z}_{|T|}) \mapsto \begin{cases}
+              \texttt{BTree}\big(\texttt{root}(T)\big) \text{ if $T$ is a leaf,} \vspace{5pt}\\
+              \text{Let } r = |\texttt{children}(T)|,\\
+              \phantom{\text{Let }} F(n) = \sum_{\langle l, r\rangle \in \texttt{children}[0 \ldots n]}}|l|\cdot|r|,\\
+\phantom{\text{Let }} F^{-1}(u)=\inf \big\{x \mid u \leq F(x)\big\},\\
+\phantom{\text{Let }} q=i - F(F^{-1}(i)),\\
+\phantom{\text{Let }} l, r = \texttt{children}[t],\\
+\phantom{\text{Let }} q_1, q_2 =\big\langle\lfloor\frac{q}{|r|}\rfloor, q \pmod{|r|}\big\rangle,\\
+\phantom{\text{Let }} T_1 = \varphi(l, q_1),\\
+\phantom{\text{Let }} T_2 = \varphi(r, q_2) \text{ in } \\
+\texttt{BTree}\big(\texttt{root}(T), T_1, T_2\big) \text{ otherwise.} \\
+\end{cases}
 \end{equation*}
-\end{small}
+\begin{small}
+
+%\begin{small}
+%\begin{equation*}
+%  \varphi(T: \mathbb{T}_2, i: \mathbb{Z}_{|T|}) \mapsto \begin{cases}
+%  \Big\langle\texttt{BTree}\big(\texttt{root}(T)\big), i\Big\rangle & \text{if $T$ is a leaf,} \vspace{5pt}\\
+%  \text{Let } b = |\texttt{children}(T)|,\\
+%  \phantom{\text{Let }} q_1, r=\big\langle\lfloor\frac{i}{b}\rfloor, i \pmod{b}\big\rangle,\\
+%  \phantom{\text{Let }} lb, rb = \texttt{children}[r],\\
+%  \phantom{\text{Let }} T_1, q_2 = \varphi(lb, q_1),\\
+%  \phantom{\text{Let }} T_2, q_3 = \varphi(rb, q_2) \text{ in } \\
+%  \Big\langle\texttt{BTree}\big(\texttt{root}(T), T_1, T_2\big), q_3\Big\rangle & \text{otherwise.} \\
+%  \end{cases}
+%\end{equation*}
+%\end{small}
 
 Then, instead of sampling trees, we can simply sample integers uniformly without replacement from $\mathbb{Z}_{|T|}$ using the construction defined in \S~\ref{sec:method}, and lazily decode them into trees.
 

latex/splash2024/splash.tex

@@ -806,7 +806,7 @@
     L(p) = 1 + p L(p) \phantom{addspace} P(a) = \Sigma + V + V L\big(V^2P(a)^2\big)
   \end{equation}
 
-  Depicted in Fig.~\ref{fig:ptree} is a partial $\mathbb{T}_2$, where red nodes are \texttt{root}s and blue nodes are \texttt{children}. The shape of type $\mathbb{T}_2$ is congruent with an acyclic CFG in Chomsky Normal Form, i.e., $\mathbb{T}_2\cong\mathcal{G}'$, so assuming the CFG recognizes a finite language, as is the case for $G_\cap'$, we can translate it directly. If the language is infinite, we slice the CFL, $\mathcal{L}(G)\cap \Sigma^n$, and compute the fixpoint for each slice.
+  Depicted in Fig.~\ref{fig:ptree} is a partial $\mathbb{T}_2$, where red nodes are \texttt{root}s and blue nodes are \texttt{children}. The shape of type $\mathbb{T}_2$ is congruent with an acyclic CFG in Chomsky Normal Form, i.e., $\mathbb{T}_2\cong\mathcal{G}'$, so assuming the CFG recognizes a finite language, as is the case for $G_\cap'$, we can translate it directly. If the language is infinite, we slice the CFL, $\mathcal{L}(G)\cap \Sigma^n$, and solve the fixpoint for each slice $n \in [2, n]$.
 
   Given a porous string $\sigma: \underline\Sigma^n$ representing the slice, we can construct $\mathbb{T}_2$ from the bottom-up, and read off structures from the top-down. We construct the first upper diagonal $\hat\sigma_r = \Lambda(\sigma_r)$ as follows: