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% use [nosolutions] flag to hide solutions.
% use [solutions] flag to show solutions.
\documentclass[
singlenumbering,
solutions
]{../../resources/ormc_handout}
\usepackage{../../resources/macros}
\usepackage{pdfpages}
\usepackage{sliderule}
\usepackage{changepage}
\usepackage{listings}
% Args:
% x, top scale y, label
\newcommand{\slideruleind}[3]{
\draw[
line width=1mm,
draw=black,
opacity=0.3,
text opacity=1
]
({#1}, {#2 + 1})
--
({#1}, {#2 - 1.1})
node [below] {#3};
}
\uptitlel{Advanced 2}
\uptitler{\smallurl{}}
\title{Fast Inverse Square Root}
\subtitle{Prepared by Mark on \today}
\begin{document}
\maketitle
\begin{center}
\begin{minipage}{6cm}
Dad says that anyone who can't use
a slide rule is a cultural illiterate
and should not be allowed to vote.
\vspace{1ex}
\textit{Have Space Suit --- Will Travel, 1958}
\end{minipage}
\end{center}
\hfill
%\input{parts/0 logarithms.tex}
%\input{parts/1 intro.tex}
%\input{parts/2 multiplication.tex}
%\input{parts/3 division.tex}
\pagebreak
\input{parts/4 float.tex}
\input{parts/5 approximate.tex}
\input{parts/6 quake.tex}
% Make sure the slide rule is on an odd page,
% so that double-sided prints won't require
% students to tear off problems.
\checkoddpage
\ifoddpage\else
\vspace*{\fill}
\begin{center}
{
\Large
\textbf{This page unintentionally left blank.}
}
\end{center}
\vspace{\fill}
\pagebreak
\fi
%\includepdf[
% pages=1,
% fitpaper=true
%]{resources/rule.pdf}
\end{document}

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src/Advanced/Fast Inverse Root/parts/0 logarithms.tex Normal file
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\section{Logarithms}
\definition{}<logdef>
The \textit{logarithm} is the inverse of the exponent. That is, if $b^p = c$, then $\log_b{c} = p$. \par
In other words, $\log_b{c}$ asks the question ``what power do I need to raise $b$ to to get $c$?''
\vspace{2mm}
In both $b^p$ and $\log_b{c}$, the number $b$ is called the \textit{base}.
\problem{}
Evaluate the following by hand:
\begin{enumerate}
\item $\log_{10}{(1000)}$
\vfill
\item $\log_2{(64)}$
\vfill
\item $\log_2{(\frac{1}{4})}$
\vfill
\item $\log_x{(x)}$ for any $x$
\vfill
\item $log_x{(1)}$ for any $x$
\vfill
\end{enumerate}
\pagebreak
\problem{}<logids>
Prove the following identities:
\begin{enumerate}[itemsep=2mm]
\item $\log_b{(b^x)} = x$
\item $b^{\log_b{x}} = x$
\item $\log_b{(xy)} = \log_b{(x)} + \log_b{(y)}$
\item $\log_b{(\frac{x}{y})} = \log_b{(x)} - \log_b{(y)}$
\item $\log_b{(x^y)} = y \log_b{(x)}$
\end{enumerate}
\vfill
\begin{instructornote}
A good intro to the following sections is the linear slide rule:
\begin{center}
\begin{tikzpicture}[scale=1]
\linearscale{2}{1}{}
\linearscale{0}{0}{}
\slideruleind
{5}
{1}
{2 + 3 = 5}
\end{tikzpicture}
\end{center}
Take two linear rulers, offset one, and you add. \par
If you do the same with a log scale, you multiply!
\vspace{2mm}
Note that the slide rules above start at 0.
\linehack{}
After assembling the paper slide rule, you can make a visor with some transparent tape. Wrap a strip around the slide rule, sticky side out, and stick it to itself to form a ring. Cover the sticky side with another layer of tape, and trim the edges to make them straight. Use the edge of the visor to read your slide rule!
\end{instructornote}
\pagebreak

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\section{Slide Rules}
Mathematicians, physicists, and engineers needed to quickly solve complex equations even before computers were invented.
\vspace{2mm}
The \textit{slide rule} is an instrument that uses the logarithm to solve this problem. \par
Before you continue, tear off the last page of this handout and assemble your slide rule.
\vspace{2mm}
There are four scales on your slide rule, each labeled with a letter on the left side:
\def\sliderulewidth{13}
\begin{center}
\begin{tikzpicture}[scale=1]
\tscale{0}{9}{T}
\kscale{0}{8}{K}
\abscale{0}{7}{A}
\abscale{0}{5.5}{B}
\ciscale{0}{4.5}{CI}
\cdscale{0}{3.5}{C}
\cdscale{0}{2}{D}
\lscale{0}{1}{L}
\sscale{0}{0}{S}
\end{tikzpicture}
\end{center}
Each scale's ``generating function'' is on the right:
\begin{itemize}
\item T: $\tan$
\item K: $x^3$
\item A,B: $x^2$
\item CI: $\frac{1}{x}$
\item C, D: $x$
\item L: $\log_{10}(x)$
\item S: $\sin$
\end{itemize}
Once you understand the layout of your slide rule, move on to the next page.
\pagebreak

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src/Advanced/Fast Inverse Root/parts/2 multiplication.tex Normal file
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\section{Multiplication}
We'll use the C and D scales of your slide rule to multiply. \par
Say we want to multiply $2 \times 3$. First, move the \textit{left-hand index}
of the C scale over the smaller number, $2$:
\def\sliderulewidth{10}
\begin{center}
\begin{tikzpicture}[scale=1]
\cdscale{\cdscalefn(2)}{1}{C}
\cdscale{0}{0}{D}
\end{tikzpicture}
\end{center}
Then we'll find the second number, $3$ on the C scale, and read the D scale under it:
\begin{center}
\begin{tikzpicture}[scale=1]
\cdscale{\cdscalefn(2)}{1}{C}
\cdscale{0}{0}{D}
\slideruleind
{\cdscalefn(6)}
{1}
{6}
\end{tikzpicture}
\end{center}
Of course, our answer is 6.
\problem{}
What is $1.15 \times 2.1$? \par
Use your slide rule.
\begin{solution}
\begin{center}
\begin{tikzpicture}[scale=1]
\cdscale{\cdscalefn(1.15)}{1}{C}
\cdscale{0}{0}{D}
\slideruleind
{\cdscalefn(1.15)}
{1}
{1.15}
\slideruleind
{\cdscalefn(1.15) + \cdscalefn(2.1)}
{1}
{2.415}
\end{tikzpicture}
\end{center}
\end{solution}
\vfill
Note that your answer isn't exact. $1.15 \times 2.1 = 2.415$, but an answer accurate within
two decimal places is close enough for most practical applications.
\pagebreak
Look at your C and D scales again. They contain every number between 1 and 10, but no more than that.
What should we do if we want to calculate $32 \times 210$? \par
\problem{}
Using your slide rule, calculate $32 \times 210$.
\begin{solution}
\begin{center}
\begin{tikzpicture}[scale=1]
\cdscale{\cdscalefn(2.1)}{1}{C}
\cdscale{0}{0}{D}
\slideruleind
{\cdscalefn(2.1)}
{1}
{2.1}
\slideruleind
{\cdscalefn(2.1) + \cdscalefn(3.2)}
{1}
{6.72}
\end{tikzpicture}
\end{center}
Placing the decimal point correctly is your job. \par
$10^1 \times 10^2 = 10^3$, so our final answer is $6.72 \times 10^3 = 672$.
\end{solution}
\vfill
\problem{}
Compute the following:
\begin{enumerate}
\item $1.44 \times 52$
\item $0.38 \times 1.24$
\item $\pi \times 2.35$
\end{enumerate}
\begin{solution}
\begin{enumerate}
\item $1.44 \times 52 = 74.88$
\item $0.38 \times 1.24 = 0.4712$
\item $\pi \times 2.35 = 7.382$
\end{enumerate}
\end{solution}
\vfill
\problem{}<provemult>
Note that the numbers on your C and D scales are logarithmically spaced.
\def\sliderulewidth{13}
\begin{center}
\begin{tikzpicture}[scale=1]
\cdscale{0}{1}{C}
\cdscale{0}{0}{D}
\end{tikzpicture}
\end{center}
Why does our multiplication procedure work?
\vfill
\pagebreak
Now we want to compute $7.2 \times 5.5$:
\def\sliderulewidth{10}
\begin{center}
\begin{tikzpicture}[scale=0.8]
\cdscale{\cdscalefn(5.5)}{1}{C}
\cdscale{0}{0}{D}
\slideruleind
{\cdscalefn(5.5)}
{1}
{5.5}
\slideruleind
{\cdscalefn(5.5) + \cdscalefn(7.2)}
{1}
{???}
\end{tikzpicture}
\end{center}
No matter what order we go in, the answer ends up off the scale. There must be another way.
\vspace{2mm}
Look at the far right of your C scale. There's an arrow pointing to the $10$ tick, labeled \textit{right-hand index}.
Move it over the \textit{larger} number, $7.2$:
\begin{center}
\begin{tikzpicture}[scale=1]
\cdscale{\cdscalefn(7.2) - \cdscalefn(10)}{1}{C}
\cdscale{0}{0}{D}
\slideruleind
{\cdscalefn(7.2)}
{1}
{7.2}
\end{tikzpicture}
\end{center}
Now find the smaller number, $5.5$, on the C scale, and read the D scale under it:
\begin{center}
\begin{tikzpicture}[scale=1]
\cdscale{\cdscalefn(7.2) - \cdscalefn(10)}{1}{C}
\cdscale{0}{0}{D}
\slideruleind
{\cdscalefn(7.2)}
{1}
{7.2}
\slideruleind
{\cdscalefn(3.96)}
{1}
{3.96}
\end{tikzpicture}
\end{center}
Our answer should be about $7 \times 5 = 35$, so let's move the decimal point: $5.5 \times 7.2 = 39.6$.
We can do this by hand to verify our answer.
\vspace{2mm}
\problem{}
Why does this work?
\begin{solution}
Consider the following picture, where we have two D scales next to each other:
\begin{center}
\begin{tikzpicture}[scale=0.7]
\cdscale{\cdscalefn(7.2) - \cdscalefn(10)}{1}{C}
\cdscale{0}{0}{}
\cdscale{-10}{0}{}
\draw[
draw=black,
]
(0, 0)
--
(0, -0.3)
node [below] {D};
\draw[
draw=black,
]
(-10, 0)
--
(-10, -0.3)
node [below] {D};
\slideruleind
{-10 + \cdscalefn(7.2)}
{1}
{7.2}
\slideruleind
{\cdscalefn(7.2)}
{1}
{7.2}
\slideruleind
{\cdscalefn(3.96)}
{1}
{3.96}
\end{tikzpicture}
\end{center}
\vspace{2mm}
The second D scale has been moved to the right by $(\log{10})$, so every value on it is $(\log{10})$
smaller than it should be.
\vspace{2mm}
\vspace{2mm}
In other words, the answer we get from reverse multiplication is $\log{a} + \log{b} - \log{10}$. \par
This reduces to $\log{(\frac{a \times b}{10})}$, and explains the misplaced decimal point in $7.2 \times 5.5$.
\end{solution}
\vfill
\pagebreak
\problem{}
Compute the following using your slide rule:
\begin{enumerate}
\item $9 \times 8$
\item $15 \times 35$
\item $42.1 \times 7.65$
\end{enumerate}
\begin{solution}
\begin{enumerate}
\item $9 \times 8 = 72$
\item $15 \times 35 = 525$
\item $42.1 \times 7.65 = 322.065$
\end{enumerate}
\end{solution}
\vfill
\pagebreak

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\section{Division}
Now that you can multiply, division should be easy. All you need to do is work backwards. \\
Let's look at our first example again: $3 \times 2 = 6$.
\medskip
We can easily see that $6 \div 3 = 2$
\begin{center}
\begin{tikzpicture}[scale=1]
\cdscale{\cdscalefn(2)}{1}{C}
\cdscale{0}{0}{D}
\slideruleind
{\cdscalefn(6)}
{1}
{Align here}
\slideruleind
{\cdscalefn(2)}
{1}
{2}
\end{tikzpicture}
\end{center}
and that $6 \div 2 = 3$:
\begin{center}
\begin{tikzpicture}[scale=1]
\cdscale{\cdscalefn(3)}{-3}{C}
\cdscale{0}{-4}{D}
\slideruleind
{\cdscalefn(6)}
{-3}
{Align here}
\slideruleind
{\cdscalefn(3)}
{-3}
{3}
\end{tikzpicture}
\end{center}
If your left-hand index is off the scale, read the right-hand one. \\
Consider $42.25 \div 6.5 = 6.5$:
\begin{center}
\begin{tikzpicture}[scale=1]
\cdscale{\cdscalefn(6.5) - \cdscalefn(10)}{1}{C}
\cdscale{0}{0}{D}
\slideruleind
{\cdscalefn(4.225)}
{1}
{Align here}
\slideruleind
{\cdscalefn(6.5)}
{1}
{6.5}
\end{tikzpicture}
\end{center}
Place your decimal points carefully.
\vfill
\problem{}
Compute the following using your slide rule. \par
\begin{enumerate}
\item $135 \div 15$
\item $68.2 \div 0.575$
\item $(118 \times 0.51) \div 6.6$
\end{enumerate}
\begin{solution}
\begin{enumerate}
\item $135 \div 15 = 9$
\item $68.2 \div 0.575 = 118.609$
\item $(118 \times 0.51) \div 6.6 = 9.118$
\end{enumerate}
\end{solution}
\vfill
\pagebreak

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\section{\texttt{int}s and \texttt{float}s}
\definition{}
A \textit{signed 32-bit integer} (equivalently, a \texttt{long int}) consists of thirty-two binary digits, \par
and is used represent a subset of the integers.
\vspace{2mm}
The first bit of a \texttt{long} tells us its sign:
\begin{itemize}
\item if the first bit of a \texttt{long} is \texttt{1}, it represents a negative number;
\item if the first bit is \texttt{0}, it represents a positive number.
\end{itemize}
We do not need negative numbers today, so we will assume that the first bit is always zero. \par
\note{If you'd like to know how negative integers are written, look up \say{two's complement} after class.}
\vspace{2mm}
We'll denote binary strings with the prefix \texttt{0b}. \par
Underscores are added between every eight digits for readability, and have no meaning.
\vspace{2mm}
The value of a positive signed \texttt{long} is simply the value of its binary digits. \par
For example:
\begin{itemize}
\item $\texttt{0b00000000\_00000000\_00000000\_00000000} = 0$
\item $\texttt{0b00000000\_00000000\_00000000\_00000011} = 3$
\item $\texttt{0b00000000\_00000000\_00000000\_00100000} = 32$
\item $\texttt{0b00000000\_00000000\_00000000\_10000010} = 130$
\end{itemize}
Remember---we only need positive integers today. Assume the \say{sign} bit is always \texttt{0}.
\problem{}
What is the largest number that can be represented with a \texttt{long}?
\begin{solution}
$\texttt{0b01111111\_11111111\_11111111\_11111111} = 2^{31}$
\end{solution}
\vfill
\problem{}
What is the smallest possible number that can be represented with a \texttt{long}? \par
\hint{
You do not need to know \textit{how} negative numbers are represented. \par
Assume that we do not skip any integers, and don't forget about zero.
}
\begin{solution}
There are $2^{64}$ possible 32-bit patterns,
of which 1 represents zero and $2^{31}$ represent positive numbers.
\vspace{2mm}
We therefore have access to $2^{64} - 1 - 2^{31}$ negative numbers,
giving us a minimum representable value of $-2^{31} + 1$.
\end{solution}
\problem{}
What is the value of the following longs?
\begin{itemize}
\item \texttt{0b00000000\_00000000\_00000101\_00111001}
\item \texttt{0b00000000\_00000000\_00000001\_00101100}
\item \texttt{0b00000000\_00000000\_00000100\_10110000}
\end{itemize}
\hint{The third conversion is easy---look carefully at the second.}
\begin{solution}
\begin{itemize}
\item $\texttt{0b00000000\_00000000\_00000101\_00111001} = 1337$
\item $\texttt{0b00000000\_00000000\_00000001\_00101100} = 300$
\item $\texttt{0b00000000\_00000000\_00000010\_01011000} = 1200$
\end{itemize}
Notice that the third long is the second shifted left twice (i.e, multiplied by 4)
\end{solution}
\vfill
\pagebreak
\definition{}
A \textit{signed 32-bit floating-point decimal} (equivalently, a \textit{float})
consists of 32 binary digits, and is used to represent a subset of the real numbers.
These 32 bits are interpreted as follows:
\begin{center}
\begin{tikzpicture}
\node[anchor=south west] at (0, 0) {\texttt{\texttt{0}}};
\node[anchor=south west] at (0.25, 0) {\texttt{\texttt{b}}};
\node[anchor=south west] at (0.50, 0) {\texttt{\texttt{0}}};
\node[anchor=south west] at (0.75, 0) {\texttt{\texttt{\_}}};
\node[anchor=south west] at (1.00, 0) {\texttt{\texttt{0}}};
\node[anchor=south west] at (1.25, 0) {\texttt{\texttt{0}}};
\node[anchor=south west] at (1.50, 0) {\texttt{\texttt{0}}};
\node[anchor=south west] at (1.75, 0) {\texttt{\texttt{0}}};
\node[anchor=south west] at (2.00, 0) {\texttt{\texttt{0}}};
\node[anchor=south west] at (2.25, 0) {\texttt{\texttt{0}}};
\node[anchor=south west] at (2.50, 0) {\texttt{\texttt{0}}};
\node[anchor=south west] at (2.75, 0) {\texttt{\texttt{0}}};
\node[anchor=south west] at (3.00, 0) {\texttt{\texttt{\_}}};
\node[anchor=south west] at (3.25, 0) {\texttt{\texttt{0}}};
\node[anchor=south west] at (3.50, 0) {\texttt{\texttt{0}}};
\node[anchor=south west] at (3.75, 0) {\texttt{\texttt{0}}};
\node[anchor=south west] at (4.00, 0) {\texttt{\texttt{0}}};
\node[anchor=south west] at (4.25, 0) {\texttt{\texttt{0}}};
\node[anchor=south west] at (4.50, 0) {\texttt{\texttt{0}}};
\node[anchor=south west] at (4.75, 0) {\texttt{\texttt{0}}};
\node[anchor=south west] at (5.00, 0) {\texttt{\texttt{\_}}};
\node[anchor=south west] at (5.25, 0) {\texttt{\texttt{0}}};
\node[anchor=south west] at (5.50, 0) {\texttt{\texttt{0}}};
\node[anchor=south west] at (5.75, 0) {\texttt{\texttt{0}}};
\node[anchor=south west] at (6.00, 0) {\texttt{\texttt{0}}};
\node[anchor=south west] at (6.25, 0) {\texttt{\texttt{0}}};
\node[anchor=south west] at (6.50, 0) {\texttt{\texttt{0}}};
\node[anchor=south west] at (6.75, 0) {\texttt{\texttt{0}}};
\node[anchor=south west] at (7.00, 0) {\texttt{\texttt{0}}};
\node[anchor=south west] at (7.25, 0) {\texttt{\texttt{\_}}};
\node[anchor=south west] at (7.50, 0) {\texttt{\texttt{0}}};
\node[anchor=south west] at (7.75, 0) {\texttt{\texttt{0}}};
\node[anchor=south west] at (8.00, 0) {\texttt{\texttt{0}}};
\node[anchor=south west] at (8.25, 0) {\texttt{\texttt{0}}};
\node[anchor=south west] at (8.50, 0) {\texttt{\texttt{0}}};
\node[anchor=south west] at (8.75, 0) {\texttt{\texttt{0}}};
\node[anchor=south west] at (9.00, 0) {\texttt{\texttt{0}}};
\node[anchor=south west] at (9.25, 0) {\texttt{\texttt{0}}};
\draw (0.50, 0) -- (0.95, 0) node [midway, below=1mm] {sign};
\draw (1.05, 0) -- (3.15, 0) node [midway, below=1mm] {exponent};
\draw (3.30, 0) -- (9.70, 0) node [midway, below=1mm] {fraction};
\end{tikzpicture}
\end{center}
In other words:
\begin{itemize}[itemsep = 1mm]
\item The first bit denotes the sign of the float's value. We'll label it $s$. \par
If $s = 1$, this float is negative; if $s = 0$, it is positive.
\item The next 8 bits represent the \textit{exponent} of this float. \par
We'll call the value of these eight bits $E$. \par
Naturally, $0 \leq E \leq 255$
\item The remaining 23 bits represent the \textit{fraction} of this float. \par
These 23 bits are interpreted as the fractional part of a binary decimal. \par
For example, the bits \texttt{0b1010000\_00000000\_00000000} represents $0.5 + 0.125 = 0.625$.
\end{itemize}
\vspace{2mm}
The final value of a float with sign $s$, exponent $E$, and fraction $F$ is
\begin{equation*}
(-1)^s ~\times~ 2^{E - 127} ~\times~ \left(1 + \frac{F}{2^{23}}\right)
\end{equation*}
Notice that this is very similar to decimal scientific notation, which is written as
\begin{equation*}
(\pm 1) ~\times~ 10^{e} ~\times~ (f)
\end{equation*}
\vfill
\pagebreak
\problem{}
What is the value of \texttt{0b01000001\_10101000\_00000000\_00000000} if it is interpreted as a float? \par
\hint{$21 \div 16 = 1.3125$}
\begin{solution}
This is 21:
\begin{align*}
&=~ 2^{131} \times \biggl(1 + \frac{2^{21} + 2^{19}}{2^{23}}\biggr) \\
&=~ 2^{4} \times (1 + 0.25 + 0.0625) \\
&=~ 16 \times (1.3125) \\
&=~ 21
\end{align*}
\end{solution}
\vfill
\problem{}
Encode $12.5$ as a float. \par
\hint{$12.5 \div 8 = 1.5625$}
\vspace{2mm}
What is the value of the resulting 32 bits if they are interpreted as a long? \par
\hint{A sum of powers of two is fine.}
\begin{solution}
\begin{align*}
12.5
&=~ 8 \times 1.5625 \\
&=~ 2^{3} \times \biggl(1 + (0.5 + 0.0625)\biggr) \\
&=~ 2^{130} \times \biggl(1 + \frac{2^{22} + 2^{19}}{2^{23}}\biggr)
\end{align*}
\linehack{}
This is \texttt{0b01000001\_01001000\_00000000\_00000000}, \par
which is $2^{30} + 2^{24} + 2^{22} + 2^{19} = 11,095,237,632$
\end{solution}
\vfill
\pagebreak

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\section{Approximation}
\definition{}
Say we have a bit string $x$. \par
Let $x_f$ denote the value of $x$ interpreted as a float, \par
and let $x_i$ denote the value of $x$ interpreted as an integer.
\problem{}
Let $x = \texttt{0b01000001\_01001000\_00000000\_00000000}$. \
What are $x_f$ and $x_i$?
\begin{solution}
$x_f = 12.5$ \par
\vspace{2mm}
$x_i = 2^{30} + 2^{24} + 2^{22} + 2^{19} = 11,095,237,632$
\end{solution}
\generic{Observation:}
For small values of $a$, $\log_2(1 + a)$ is approximately equal to $a$. \par
Note that this equality is exact for $a = 0$ and $a = 1$, since $\log_2(1) = 0$ and $\log_2(2) = 1$.
\vspace{2mm}
We'll add a \say{correction term} $\varepsilon$ to this approximation, so that $\log_2(1 + a) \approx a + \varepsilon$.
TODO: why? Graphs.
\problem{}
Use the fact that $\log_2(1 + a) \approx a + \varepsilon$ to approximate $\log_2(x_f)$ in terms of $x_i$, \par
\begin{solution}
Let $E$ and $F$ be the exponent and float bits of $x_f$. \par
We then have:
\begin{align*}
\log_2(x_f)
&=~ \log_2 \left( 2^{E-127} \times \left(1 + \frac{F}{2^{23}}\right) \right) \\
&=~ E-127 + \log_2\left(1 + \frac{F}{2^{23}}\right) \\
&\approx~ E-127 + \frac{F}{2^{23}} + \varepsilon \\
&=~ \frac{1}{2^{23}}(2^{23}E + F) - 127 + \varepsilon \\
&=~ \frac{1}{2^{23}}(x_i) - 127 + \varepsilon
\end{align*}
\end{solution}
\vfill
\pagebreak

71
src/Advanced/Fast Inverse Root/parts/6 quake.tex Normal file
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\section{The Fast Inverse Square Root}
The following code is present in \textit{Quake III Arena} (1999):
\lstset{
breaklines=false,
numbersep=5pt,
xrightmargin=0in
}
\begin{lstlisting}[language=C]
float Q_rsqrt( float number ) {
long i = * ( long * ) &number;
i = 0x5f3759df - ( i >> 1 );
return * ( float * ) &i;
}
\end{lstlisting}
It defines a method \texttt{Q\_rsqrt} that consumes a float named \texttt{number} and quickly approximates its inverse
square root (in other words, \texttt{Q\_rsqrt} computes $1/\sqrt{\texttt{number}}$).
\vspace{8mm}
If we rewrite this using the notation we're familiar with, we get the following:
\begin{equation*}
\frac{1}{\sqrt{n_f}} \approx \kappa - (n_i \div 2)
\end{equation*}
Where $\kappa$ is the magic constant $6240089$ (which is \texttt{0x5f3759df} in hexadecimal)
\problem{}
Find the exact value of $\kappa$ in terms of $\varepsilon$. \par
\hint{Remember that $\varepsilon$ is the correction term in the approximation $\log_2(1 + a) = a + \varepsilon$.}
\problem{}
Rewrite $\log_2(1 / \sqrt{x})$ in terms of $\log_2{x}$.
\begin{solution}
Say $g_f = \frac{1}{\sqrt{n_f}}$---that is, $g_f$ is the value we want to compute. \par
Then:
\begin{align*}
\log_2(g_f)
&=\log_2(\frac{1}{\sqrt{n_f}}) \\
&=\frac{-1}{2}\log_2(n_f) \\
&=\frac{-1}{2}\left( \frac{n_i}{2^{23}} + \varepsilon - 127 \right)
\end{align*}
But we also know that
\begin{align*}
\log_2(g_f)
&=\frac{g_i}{2^{23}} + \varepsilon - 127
\end{align*}
So,
\begin{align*}
\frac{g_i}{2^{23}} + \varepsilon - 127
&=\frac{-1}{2}\left( \frac{n_i}{2^{23}} + \varepsilon - 127 \right) \\
\frac{g_i}{2^{23}}
&=\frac{-1}{2}\left( \frac{n_i}{2^{23}} \right) + \frac{3}{2}(\varepsilon - 127) \\
g_i
&=\frac{-1}{2}\left(n_i \right) + 2^{23}\frac{3}{2}(\varepsilon - 127) \\
&=2^{23}\frac{3}{2}(\varepsilon - 127) - \frac{n_i}{2}
\end{align*}
thus, $\kappa = 2^{23}\frac{3}{2}(\varepsilon - 127)$.
\end{solution}
\pagebreak

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\NeedsTeXFormat{LaTeX2e}
\ProvidesPackage{sliderule}[2022/08/22 Slide rule tools]
\RequirePackage{tikz}
\RequirePackage{ifthen}
% Scale functions:
% See https://sliderulemuseum.com/SR_Scales.htm
%
% l: length of the rule
% n: the number on the rule
%
% A/B: (l/2) * log(n)
% C/D: l / log(n)
% CI: abs(l * log(10 / n) - l)
% K: (l/3) * log(n)
%
% L: n * l
% T: l * log(10 * tan(n))
% S: l * log(10 * sin(n))
\def\sliderulewidth{10}
\def\abscalefn(#1){(\sliderulewidth/2) * log10(#1)}
\def\cdscalefn(#1){(\sliderulewidth * log10(#1))}
\def\ciscalefn(#1){(\sliderulewidth - \cdscalefn(#1))}
\def\kscalefn(#1){(\sliderulewidth/3) * log10(#1)}
\def\lscalefn(#1){(\sliderulewidth * #1)}
\def\tscalefn(#1){(\sliderulewidth * log10(10 * tan(#1)))}
\def\sscalefn(#1){(\sliderulewidth * log10(10 * sin(#1)))}
% Arguments:
% Label
% x of start
% y of start
\newcommand{\linearscale}[3]{
\draw[black] ({#1}, #2) -- ({#1 + \sliderulewidth}, #2);
\draw[black] ({#1}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.9);
\draw[black] ({#1}, #2 + 0.9) -- ({#1}, #2 + 0.7);
\draw[black] ({#1 + \sliderulewidth}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.7);
\draw ({#1 - 0.1}, #2 + 0.5) node[left] {#3};
% Numbers and marks
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\draw[black]
({#1 + (\sliderulewidth / 10) * \i}, #2) --
({#1 + (\sliderulewidth / 10) * \i}, #2 + 0.3)
node[above] {\i};
}
% Submarks
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\foreach \i in {1,..., 9} {
\ifthenelse{\i=5}{
\draw[black]
({#1 + (\sliderulewidth / 10) * (\n + \i / 10)}, #2) --
({#1 + (\sliderulewidth / 10) * (\n + \i / 10)}, #2 + 0.2);
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% Arguments:
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% Numbers and marks 1 - 9
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% Submarks 1 - 9
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% Submarks 10 - 100
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\draw ({#1 - 0.1}, #2 + 0.5) node[left] {#3};
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% Submarks 1 - 9
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{
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% Submarks 1 - 9
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}
{
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\newcommand{\kscale}[3]{
\draw[black] ({#1}, #2) -- ({#1 + \sliderulewidth}, #2);
\draw[black] ({#1}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.9);
\draw[black] ({#1}, #2 + 0.9) -- ({#1}, #2 + 0.7);
\draw[black] ({#1 + \sliderulewidth}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.7);
\draw ({#1 - 0.1}, #2 + 0.5) node[left] {#3};
% Numbers and marks 1 - 9
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\draw[black]
({#1 + \kscalefn(\i)}, #2) --
({#1 + \kscalefn(\i)}, #2 + 0.3)
node[above] {\i};
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% Numbers and marks 10 - 90
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({#1 + \kscalefn(10 * \i)}, #2 + 0.3)
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% Numbers and marks 100 - 1000
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({#1 + \kscalefn(100 * \i)}, #2) --
({#1 + \kscalefn(100 * \i)}, #2 + 0.3)
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% Submarks 1 - 9
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({#1 + \kscalefn(\n + \i / 10)}, #2) --
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% Submarks 10 - 90
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{
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({#1 + \kscalefn(\n + \i)}, #2 + 0.2);
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\draw[black]
({#1 + \kscalefn(\n + \i)}, #2) --
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% Submarks 100 - 1000
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{
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\draw[black] ({#1}, #2) -- ({#1 + \sliderulewidth}, #2);
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% Submarks 15 - 45
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% First line
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