# HG changeset patch # User Chengsong # Date 1688776568 -3600 # Node ID d8f82c690b32294a1d6e332f1c6d07cfb6579f5f # Parent 2ad20ba5b1787ff9667ec29afc38118cc992964d updated 4.2 diagram diff -r 2ad20ba5b178 -r d8f82c690b32 ChengsongTanPhdThesis/Chapters/Bitcoded2.tex --- a/ChengsongTanPhdThesis/Chapters/Bitcoded2.tex Fri Jul 07 20:03:05 2023 +0100 +++ b/ChengsongTanPhdThesis/Chapters/Bitcoded2.tex Sat Jul 08 01:36:08 2023 +0100 @@ -8,12 +8,8 @@ %simplifications and therefore introduce our version of the bitcoded algorithm and %its correctness proof in %Chapter 3\ref{Chapter3}. -{\color{red} \rule{\linewidth}{0.5mm}} -New content starts. -{\color{red} \rule{\linewidth}{0.5mm}} -\marginpar{\em Added a completely new overview section, highlighting the contributions.} - \section{Overview} +\marginpar{\em Added a completely new \\overview section, \\highlighting contributions.} This chapter is the point from which novel contributions of this PhD project are introduced @@ -27,7 +23,7 @@ %are about to describe. The fundamental reason is we cannot extend the correctness proof of theorem 4 because lemma 13 does not hold anymore when simplifications are involved. -\marginpar{\em rephrased things so they make better sense.} +\marginpar{\em rephrased things \\so why new \\proof makes sense.} %The proof details are necessary materials for this thesis %because it provides necessary context to explain why we need a %new framework for the proof of $\blexersimp$, which involves @@ -155,24 +151,40 @@ } } -\begin{figure}[H] +\begin{center} \begin{tikzpicture}[->, >=stealth', shorten >= 1pt, auto, thick] - \node [rectangle, draw] (1) at (-7, 2) {$\ldots$}; - \node [rectangle, draw] (2) at (-4, 2) {$_{bs'}(_Za+_Saa)^*$}; - \node [rectangle, draw] (3) at (4, 2) {$_{bs'}(_Z(_Z\ONE + _S(\ONE \cdot a)))\cdot(_0a+_1aa)^*$}; - \node [rectangle, draw] (4) at (7, 2) {$\ldots$}; - \node [rectangle, draw] (5) at (-7, -2) {$\ldots$}; - \node [rectangle, draw] (6) at (-4, -2) {$\Stars \; [\Left (a)]$}; - \node [rectangle, draw] (7) at ( 4, -2) {$\Seq (\Alt (\Left \; \Empty)) \; \Stars \, []$}; - \node [rectangle, draw] (8) at ( 7, -2) {$\ldots$}; - \node [rectangle, draw] (9) at (-7, -6) {$\ldots$}; - \node [rectangle, draw] (10) at (-4, -6) {$\textit{bits} = \retrieve \; r_i\;v_i$}; - \node [rectangle, draw] (11) at (4, -6) {$\textit{bits} = \retrieve \; r_{i+1}\;v_{i+1}$}; - \node [rectangle, draw] (12) at (7, -6) {$\ldots$}; + \node [rectangle ] (1) at (-7, 2) {$\ldots$}; + \node [rectangle, draw] (2) at (-4, 2) {$r_i = _{bs'}(_Za+_Saa)^*$}; + \node [rectangle, draw] (3) at (4, 2) {$r_{i+1} = _{bs'}(_Z(_Z\ONE + _S(\ONE \cdot a)))\cdot(_0a+_1aa)^*$}; + \node [rectangle] (4) at (9, 2) {$\ldots$}; + \node [rectangle] (5) at (-7, -2) {$\ldots$}; + \node [rectangle, draw] (6) at (-4, -2) {$v_i = \Stars \; [\Left (a)]$}; + \node [rectangle, draw] (7) at ( 4, -2) {$v_{i+1} = \Seq (\Alt (\Left \; \Empty)) \; \Stars \, []$}; + \node [rectangle] (8) at ( 9, -2) {$\ldots$}; + \node [rectangle] (9) at (-7, -6) {$\ldots$}; + \node [rectangle, draw] (10) at (-4, -6) {$\textit{bits}_{i} = \retrieve \; r_i\;v_i$}; + \node [rectangle, draw] (11) at (4, -6) {$\textit{bits}_{i+1} = \retrieve \; r_{i+1}\;v_{i+1}$}; + \node [rectangle] (12) at (9, -6) {$\ldots$}; \path (1) edge [] node {} (2); + \path (5) edge [] node {} (6); + \path (9) edge [] node {} (10); + + \path (11) edge [] node {} (12); + \path (7) edge [] node {} (8); + \path (3) edge [] node {} (4); + + \path (2) edge [] node {$\backslash a$} (3); + \path (2) edge [dashed, <->] node {$\vdash v_i : r_i$} (6); + \path (3) edge [dashed, <->] node {$\vdash v_{i+1} : r_{i+1}$} (7); + %\path (6) edge [] node {$\vdash v_i : r_i$} (10); + %\path (7) edge [dashed, <->] node {$\vdash v_i : r_i$} (11); + + \path (10) edge [dashed, <->] node {$=$} (11); + + \path (7) edge [] node {$\inj \; r_{i+1} \; a \; v_i$} (6); % \node [rectangle, draw] (r) at (-6, -1) {$(aa)^*(b+c)$}; % \node [rectangle, draw] (a) at (-6, 4) {$(aa)^*(_{Z}b + _{S}c)$}; @@ -200,8 +212,8 @@ \end{tikzpicture} - \caption{$\blexer$ with the regular expression $(aa)^*(b+c)$ and $aab$} -\end{figure} + %\caption{$\blexer$ with the regular expression $(aa)^*(b+c)$ and $aab$} +\end{center} When simplifications are added, the inhabitation relation no longer holds, causing the above diagram to break.