From 946752ef1b99b617ec6dee6697aef76870a325ac Mon Sep 17 00:00:00 2001 From: Mark Date: Tue, 7 Mar 2023 21:57:37 -0800 Subject: [PATCH] Started Mockingbird handout --- Advanced/Mock a Mockingbird/main.tex | 281 +++++++++++++++++++++++++++ 1 file changed, 281 insertions(+) create mode 100755 Advanced/Mock a Mockingbird/main.tex diff --git a/Advanced/Mock a Mockingbird/main.tex b/Advanced/Mock a Mockingbird/main.tex new file mode 100755 index 0000000..4e81ab3 --- /dev/null +++ b/Advanced/Mock a Mockingbird/main.tex @@ -0,0 +1,281 @@ +% use [nosolutions] flag to hide solutions. +% use [solutions] flag to show solutions. +\documentclass[ + solutions, + singlenumbering +]{../../resources/ormc_handout} + +\usepackage{mathtools} % for \coloneqq + +\usepackage{alltt} + + +\newenvironment*{helpbox}[1][0.5]{ + \begin{center} + \begin{tcolorbox}[ + colback=white!90!black, + colframe=white!90!black, + coltitle=black, + center title, + width = #1\textwidth, + leftrule = 0mm, + rightrule = 0mm, + toprule = 0mm, + bottomrule = 0mm, + left = 1mm, + right = 1mm, + top = 1mm, + bottom = 1mm, + toptitle = 1mm, + lefttitle = 1mm, + titlerule = 1pt, + title={\textbf{Things you will need:}} + ] +}{ + \end{tcolorbox} + \end{center} +} + +% Logic block comment +\newcommand{\cmnt}[1]{ + \textcolor{gray}{\# #1} +} + + +\newcommand{\thus}{\(\Rightarrow\)} +\newcommand{\qed}{\(\blacksquare\)} + + +\begin{document} + + \maketitle + + + {To Mock a Mockingbird} + { + Prepared by Mark on \today \\ + Based on a book of the same name. + } + + \section{Introduction} + + A certain enchanted forest is inhabited by talking birds. Each of these birds has a name, and will respond whenever it hears the name of another. Suppose you are exploring this forest and come across the bird $A$. You call the name of bird $B$. $A$ hears you and responds with the name of some other bird, which we will designate $AB$. + + Bird $AB$ is, by definition, $A$'s response to $B$. + + \vspace{2mm} + + As you wander around this forest, you quickly discover two interesting facts: + \begin{enumerate}[itemsep = 1mm] + \item $A$'s responds to $B$ mustn't be the same as $B$'s response to $A$. + \item Given three birds $A$, $B$, and $C$, $(AB)C$ and $A(BC)$ are not necessarily the same bird. \\ + Bird $A(BC)$ is $A$'s response to bird $BC$, while $(AB)C$ is $AB$'s response to $C$. \\ + Thus, $ABC$ is ambiguous. Parenthesis are mandatory. + \end{enumerate} + + \vspace{2mm} + + You also find that this forest has two laws: + \begin{enumerate}[itemsep = 1mm] + \item $L_1$, \textit{The Law of Composition}: \\ + For any two birds $A$ and $B$, there must be a bird $C$ so that $Cx = A(Bx)$ + \item $L_2$, \textit{The Law of the Mockingbird}: \\ + The forest must contain the Mockingbird $M$, which always satisfies $Mx = xx$. \\ + In other words, the Mockingbird's response to any bird $x$ is the same as $x$'s response to itself. + \end{enumerate} + + \vfill + + \definition{} + We say a bird $A$ is fond of a bird $B$ if $A$ responds to $B$ with $B$. \\ + In other words, $A$ is fond of $B$ if $AB = B$. + + \vfill + + \definition{} + We say a bird $C$ \textit{composes} $A$ with $B$ if for any bird $x$, + $$ + Cx = A(Bx) + $$ + In other words, this means that $C$'s response to $x$ is the same as $A$'s response to $B$'s response to $x$. + + \vfill + \pagebreak + + \section{To Mock a Mockingbird} + + \problem{} + The bear, a lifelong resident of the forest, tells you that any bird $A$ is fond of at least one other bird. \\ + Complete his proof. + \begin{alltt} + let A \cmnt{Let A be any any bird.} + let Cx := A(Mx) \cmnt{Define C as the composition of A and M} + + \cmnt{The rest is up to you.} + CC = ?? + \end{alltt} + + + \begin{helpbox} + \texttt{Law:} There exists a Mockingbird, $Mx := xx$ \\ + \texttt{Def:} $A$ is fond of $B$ if $AB = B$ + \end{helpbox} + + + \begin{solution} + \begin{alltt} + let A \cmnt{Let A be any any bird.} + let Cx := A(Mx) \cmnt{Define C as the composition of A and M} + CC = A(MC) + = A(CC) \qed{} + \end{alltt} + \end{solution} + + \vfill + \problem{} + We say a bird $A$ is \textit{egocentric} if it is fond if itself. + + Show that the laws of the forest guarantee that at least one bird is egocentric. + + + \begin{helpbox} + \texttt{Law:} There exists a Mockingbird, $Mx := xx$ \\ + \texttt{Def:} $A$ is fond of $B$ if $AB = B$ \\ + \texttt{Lem:} Any bird is fond of at least one bird. + \end{helpbox} + + \begin{solution} + \begin{alltt} + \cmnt{We know M is fond of at least one bird.} + let E so that ME = E + + ME = E \cmnt{By definition of fondness} + ME = EE \cmnt{By definition of M} + \thus{} EE = E \qed{} + \end{alltt} + \end{solution} + + \vfill + \pagebreak + + + \problem{} + We say a bird $A$ is \textit{agreeable} if for all birds $B$, there is at least one bird $x$ on which $A$ and $B$ agree. \\ + This means that $Ax = Bx$. + + \begin{helpbox} + \texttt{Def:} $Mx := xx$ + \end{helpbox} + + \begin{solution} + We know that $Mx = xx$. \\ + + From this definition, we see that $M$ agrees with any $x$ on $x$ itself. + \end{solution} + + \vfill + + \problem{} + Take two birds $A$ and $B$. Let $C$ be their composition. \\ + Show that $A$ must be agreeable if $C$ is agreeable. \\ + The bear has again given you a hint. + \begin{alltt} + \cmnt{Given information} + let A, B + let Cx := A(Bx) + + let D \cmnt{Arbitrary bird} + let Ex := D(Bx) \cmnt{Define E as the composition of D and B} + Cy = ?? + \end{alltt} + + \begin{helpbox}[0.65] + \texttt{Def:} $A$ is agreeable if $Ax = Bx$ for all $B$ with some $x$. \\ + \texttt{Law:} For any $A, B$, there is C defined by Cx = A(Bx) + \end{helpbox} + + + \begin{solution} + \begin{alltt} + \cmnt{Given information} + let A, B + let Cx := A(Bx) + + let D \cmnt{Arbitrary bird} + let Ex := D(Bx) \cmnt{Define E as the composition of D and B} + Cy = Ey \cmnt{For some y, because C is agreeable} + \thus{} A(By) = Ey + \thus{} A(By) = D(By) \qed{} + \end{alltt} + \end{solution} + + \vfill + \pagebreak + + \problem{} + Given three arbitrary birds $A$, $B$, and $C$, show that there exists a bird $D$ defined by $Dx = A(B(Cx))$ + + \begin{solution} + \begin{alltt} + let A, B, C + + \cmnt{Invoke the Law of Composition:} + let Q := BC + let D := AQ + + D = AQ + = A(BC) \qed{} + \end{alltt} + \end{solution} + + \vfill + + + \problem{} + We say two birds $A$ and $B$ are \textit{compatible} if there are birds $x$ and $y$ so that $Ax = y$ and $By = x$. \\ + Note that $x$ and $y$ may be the same bird. \\ + Show that any two birds in this forest are compatible. \\ + \begin{alltt} + let A, B + let Cx := A(Bx) + \end{alltt} + + \begin{helpbox} + \texttt{Law:} Law of composition \\ + \texttt{Lem:} Any bird is fond of at least one bird. + \end{helpbox} + + \begin{solution} + \begin{alltt} + let A, B + + let Cx := A(Bx) \cmnt{Composition} + let y := Cy \cmnt{Let C be fond of y} + + Cy = y + \thus{} A(By) = y + + let x := By \cmnt{Rename By to x} + Ax = y \qed{} + \end{alltt} + \end{solution} + + \vfill + + \problem{} + Show that any bird that is fond of at least one bird is compatible with itself. + + \begin{solution} + \begin{alltt} + let A + let x so that Ax := x + Ax = x \qed{} + \end{alltt} + That's it. + \end{solution} + + + \vfill + \pagebreak + +\end{document} \ No newline at end of file