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Co-authored-by: Mark <mark@betalupi.com>
Co-committed-by: Mark <mark@betalupi.com>
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src/Advanced/Random Walks/main.tex Executable file
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% use [nosolutions] flag to hide solutions.
% use [solutions] flag to show solutions.
\documentclass[
solutions,
singlenumbering
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}
\input{tikxset.tex}
% For \nicefrac
\usepackage{units}
\usepackage{circuitikz}
\uptitlel{Advanced 2}
\uptitler{\smallurl{}}
\title{Random Walks and Resistance}
\subtitle{
Prepared by Mark on \today{} \\
Based on a handout by Aaron Anderson
}
\begin{document}
\maketitle
\input{parts/0 random.tex}
\input{parts/1 circuits.tex}
\input{parts/2 equivalence.tex}
\input{parts/3 effective.tex}
\end{document}

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src/Advanced/Random Walks/meta.toml Normal file
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[metadata]
title = "Random Walks"
[publish]
handout = true
solutions = true

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src/Advanced/Random Walks/parts/0 random.tex Normal file
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\section{Random Walks}
Consider the graph below. A particle sits on some node $n$. Every second, this particle moves left or
right with equal probability. Once it reaches node $A$ or $B$, it stops. \par
We would like to compute the probability of our particle stopping at node $A$. \par
\vspace{2mm}
In other words, we want a function $P: \text{Nodes} \to [0, 1]$ that maps each node of the graph
to the probability that our particle stops at $A$.
\begin{center}
\begin{tikzpicture}
\begin{scope}[layer = nodes]
\node[accept] (a) at (0, 0) {$A$};
\node[main] (x) at (2, 0) {$x$};
\node[main] (y) at (4, 0) {$y$};
\node[reject] (b) at (6, 0) {$B$};
\end{scope}
\draw[-]
(a) edge (x)
(x) edge (y)
(y) edge (b)
;
\end{tikzpicture}
\end{center}
\problem{}<firstgraph>
What are $P(A)$ and $P(B)$ in the graph above? \par
\note{Note that these values hold for all graphs.}
\begin{solution}
$P(A) = 1$ and $P(B) = 0$
\end{solution}
\vfill
\problem{}
Find an expression for $P(x)$ in terms of $P(y)$ and $P(A)$. \par
Find an expression for $P(y)$ in terms of $P(x)$ and $P(B)$. \par
\begin{solution}
$P(x) = \frac{P(A) + P(y)}{2}$
\vspace{2mm}
$P(y) = \frac{P(B) + P(x)}{2}$
\end{solution}
\vfill
\problem{}
Use the previous problems to find $P(x)$ and $P(y)$.
\begin{solution}
$P(x) = \nicefrac{2}{3}$
\vspace{2mm}
$P(y) = \nicefrac{1}{3}$
\end{solution}
\vfill
\pagebreak
\problem{}<oneunweighted>
Say we have a graph $G$ and a particle on node $x$ with neighbors $v_1, v_2, ..., v_n$. \par
Assume that our particle is equally likely to travel to each neighbor. \par
Find $P(x)$ in terms of $P(v_1), P(v_2), ..., P(v_n)$.
\begin{solution}
We have
$$
P(x) = \frac{P(v_1) + P(v_2) + ... + P(v_n)}{n}
$$
\end{solution}
\vfill
\problem{}
In general, how do we find $P(n)$ for any node $n$?
\begin{solution}
If we write an equation for each node other than $A$ and $B$, we have a system of $|N| - 2$
linear equations in $|N| - 2$ variables.
\vspace{2mm}
We still need to show that this system is nonsingular, but
that's outside the scope of this handout. This could
be offered as a bonus problem.
\end{solution}
\vfill
\pagebreak
\problem{}
Find $P(n)$ for all nodes in the graph below.
\begin{center}
\begin{tikzpicture}
\begin{scope}[layer = nodes]
\node[accept] (a) at (0, 0) {$A$};
\node[main] (x) at (2, 0) {$x$};
\node[main] (y) at (0, -2) {$y$};
\node[reject] (b) at (2, -2) {$B$};
\end{scope}
\draw[-]
(a) edge (x)
(x) edge (b)
(b) edge (y)
(y) edge (a)
;
\end{tikzpicture}
\end{center}
\begin{solution}
$P(x) = \nicefrac{1}{2}$ for both $x$ and $y$.
\end{solution}
\vfill
\problem{}
Find $P(n)$ for all nodes in the graph below. \par
\note{Note that this is the graph of a cube with $A$ and $B$ on opposing vertices.}
\begin{center}
\begin{tikzpicture}
\begin{scope}[layer = nodes]
\node[main] (q) at (0, 0) {$q$};
\node[main] (r) at (2, 0) {$r$};
\node[main] (s) at (0, -2) {$s$};
\node[reject] (b) at (2, -2) {$B$};
\node[accept] (a) at (-1, 1) {$A$};
\node[main] (z) at (3, 1) {$z$};
\node[main] (x) at (-1, -3) {$x$};
\node[main] (y) at (3, -3) {$y$};
\end{scope}
\draw[-]
(a) edge (z)
(z) edge (y)
(y) edge (x)
(x) edge (a)
(q) edge (r)
(r) edge (b)
(b) edge (s)
(s) edge (q)
(a) edge (q)
(z) edge (r)
(y) edge (b)
(x) edge (s)
;
\end{tikzpicture}
\end{center}
\begin{solution}
$P(z,q, \text{ and } x) = \nicefrac{3}{5}$ \par
$P(s,r, \text{ and } y) = \nicefrac{2}{5}$
\end{solution}
\vfill
\pagebreak
\definition{}<weightedgraph>
Let us now take a look at weighted graphs. The problem remains the same: we want to compute the probability that
our particle stops at node $A$, but our graphs will now feature weighted edges. The probability of our particle
taking a certain edge is proportional to that edge's weight.
\vspace{2mm}
For example, if our particle is on node $y$ of the graph below, it has a $\frac{3}{8}$ probability of moving
to $x$ and a $\frac{1}{8}$ probability of moving to $z$. \par
\note{Note that $3 + 3 + 1 + 1 = 8$.}
\begin{center}
\begin{tikzpicture}
\begin{scope}[layer = nodes]
\node[reject] (b) at (0, 0) {$B$};
\node[main] (x) at (0, 2) {$x$};
\node[main] (y) at (2, 0) {$y$};
\node[main] (z) at (4, 0) {$z$};
\node[accept] (a) at (3, -2) {$A$};
\end{scope}
\draw[-]
(a) edge node[label] {$3$} (y)
(y) edge node[label] {$1$} (z)
(b) edge node[label] {$2$} (x)
(x) edge[bend left] node[label] {$3$} (y)
(a) edge[bend right] node[label] {$2$} (z)
(y) edge node[label] {$1$} (b)
;
\end{tikzpicture}
\end{center}
\problem{}
Say a particle on node $x$ has neighbors $v_1, v_2, ..., v_n$ with weights $w_1, w_2, ..., w_n$. \par
The edge $(x, v_1)$ has weight $w_1$. Find $P(x)$ in terms of $P(v_1), P(v_2), ..., P(v_n)$.
\begin{solution}
$$
P(x) = \frac{w_1 P(v_1) + w_2 P(v_2) + ... + w_n P(v_n)}{w_1 + w_2 + ... + w_n}
$$
\end{solution}
\vfill
\pagebreak
\problem{}
Consider the following graph. Find $P(x)$, $P(y)$, and $P(z)$.
\begin{center}
\begin{tikzpicture}
\begin{scope}[layer = nodes]
\node[reject] (b) at (3, 2) {$B$};
\node[main] (x) at (0, 0) {$x$};
\node[main] (y) at (2, 0) {$y$};
\node[main] (z) at (1, 2) {$z$};
\node[accept] (a) at (-2, 0) {$A$};
\end{scope}
\draw[-]
(x) edge node[label] {$1$} (y)
(y) edge node[label] {$1$} (z)
(x) edge[bend left] node[label] {$2$} (z)
(a) edge node[label] {$1$} (x)
(z) edge node[label] {$1$} (b)
;
\end{tikzpicture}
\end{center}
\begin{solution}
$P(x) = \nicefrac{7}{12}$ \par
$P(y) = \nicefrac{6}{12}$ \par
$P(z) = \nicefrac{5}{12}$
\end{solution}
\vfill
\problem{}
Consider the following graph. \par
What expressions can you find for $P(w)$, $P(x)$, $P(y)$, and $P(z)$?
\begin{center}
\begin{tikzpicture}
\begin{scope}[layer = nodes]
\node[accept] (a) at (0, 0) {$A$};
\node[main] (w) at (2, 1) {$w$};
\node[main] (x) at (4, 1) {$x$};
\node[main] (y) at (2, -1) {$y$};
\node[main] (z) at (4, -1) {$z$};
\node[accept] (b) at (6, 0) {$B$};
\end{scope}
\draw[-]
(a) edge node[label] {$2$} (w)
(a) edge node[label] {$1$} (y)
(w) edge node[label] {$2$} (x)
(y) edge node[label] {$2$} (z)
(x) edge node[label] {$1$} (y)
(x) edge node[label] {$1$} (b)
(z) edge node[label] {$2$} (b)
;
\end{tikzpicture}
\end{center}
Solve this system of equations. \par
\hint{Use symmetry. $P(w) = 1 - P(z)$ and $P(x) = 1 - P(y)$. Why?}
\begin{solution}
$P(w) = \nicefrac{3}{4}$ \par
$P(x) = \nicefrac{2}{4}$ \par
$P(y) = \nicefrac{2}{4}$ \par
$P(z) = \nicefrac{1}{4}$
\end{solution}
\vfill
\pagebreak

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\section{Circuits}
An \textit{electrical circuit} is a graph with a few extra properties,
called \textit{current}, \textit{voltage}, and \textit{resistance}. \par
In the definitions below, let $X$ be the set of nodes in a circuit.
\begin{itemize}[itemsep=3mm]
\item \textbf{Voltage} is a function $V: X \to \mathbb{R}$ that assigns a number to each node of our graph. \par
In any circuit, we pick a \say{ground} node, and define the voltage\footnotemark{} there as 0. \par
We also select a \say{source} node, and define its voltage as 1. \par
\vspace{1mm}
Intuitively, you could say we're connecting the ends of a 1-volt battery to our source and ground nodes.
\footnotetext{
In the real world, voltage is always measured \textit{between two points} on a circuit.
Voltage is defined as the \textit{difference} in electrical charge between two points.
Hence, voltage is a function of two nodes.
\vspace{2mm}
Note that this is different than current and resistance, which aren't functions
of two arbitrary nodes --- rather, they are functions of \textit{edges}
(i.e, two adjacent nodes).
}
\item \textbf{Current} is a function $I: X^2 \to \mathbb{R}$ that assigns a number to each
\textit{oriented edge} in our graph. An \say{oriented edge} is just an ordered pair of nodes $(n_1, n_2)$. \par
\vspace{1mm}
Current through an edge $(a, b)$ is a measure of the flow of charge from $a$ to $b$. \par
Naturally, $I(a, b) = -I(b, a)$.
\item \textbf{Resistance} is a function $R: X^2 \to \mathbb{R}^+_0$ that represents a certain edge's
resistance to the flow of current through it. \par
Resistance is a property of each \textit{link} between nodes, so order doesn't matter: $R(a, b) = R(b, a)$.
\end{itemize}
\vspace{2mm}
It is often convenient to compare electrical circuits to systems of pipes. Say we have a pipe from point $A$ to point $B$.
The size of this pipe represents resistance (smaller pipe $\implies$ more resistance), the pressure between $A$ and $B$
is voltage, and the speed water flows through it is to current.
\definition{Ohm's law}
With this \say{pipe} analogy in mind, you may expect that voltage, current, and resistance are related:
if we make our pipe bigger (and change no other parameters), we'd expect to see more current. This is indeed
the case! Any circuit obeys \textit{Ohm's law}, stated below:
$$
V(a, b) = I(a,b) \times R(a,b)
$$
This handout uses two notations for voltage: two-variable $V(a, b)$ and one-variable $V(a)$. \par
The first represents the voltage between points $a$ and $b$, better reflecting reality (see the footnote below).
The second measures the voltage between $a$ and ground, and is more convenient to use in equations.
\textbf{Try to use the single-variable notation in your equations.}
Convince yourself that $V(a, b) = V(a) - V(b)$.
\vfill
\definition{Kirchoff's law}
The second axiom of electrical circuits is also fairly simple. \textit{Kirchoff's law} states that the sum of all currents connected to
a given edge is zero. You can think of this as \say{conservation of mass}: nodes in our circuit do not create or
destroy electrons, they simply pass them around to other nodes.\par
Formally, we can state this as follows:
\vspace{2mm}
Let $x$ be a node in our circuit and $N_x$ the set of its neighbors. We than have
$$
\sum_{b \in N_x} I(x, b) = 0
$$
which must hold at every node \textbf{except the source and ground vertices.} \par
\hint{Keep this exception in mind, it is used in a few problems later on.}
\vfill
\pagebreak
\begin{instructornote}
Be aware that some students may not be comfortable with these concepts from physics,
nor with the circuit notation on the next page.
\vspace{2mm}
It may be a good idea to give the class a quick lecture on this topic,
explaining the basics of electronic circuits and circuit diagrams.
\vspace{2mm}
Things to cover:
\begin{itemize}
\item All the definitions on the previous page, in detail.
\item What's an Ohm, an Amp, a Volt?
\item Measuring voltage. Why is $V(a, b) = V(a) - V(b)$?
\item What does the $\Omega$ in the picture below mean?
\item Circuit symbols in the diagram below.
\end{itemize}
\vspace{2mm}
You could also draw connections to the graph flow handout,
if the class covered it before.
\end{instructornote}
Consider the circuit below. \textbf{This the graph from \ref{firstgraph}}, turned into a circuit by:
\begin{itemize}
\item Replacing all edges with $1\Omega$ resistors
\item Attaching a 1 volt battery between $A$ and $B$
\end{itemize}
\vspace{2mm}
Note that the battery between $A$ and $B$ isn't really an edge.
It exists only to create a potential difference between the two nodes.
\begin{center}
\begin{circuitikz}[american voltages]
\draw
(0,0) node[above left] {$A$ (source)}
to[R, l=$1\Omega$, *-*] (2,0) node[above] {$x$}
to[R, l=$1\Omega$, *-*] (4,0) node[above] {$y$}
to[R, l=$1\Omega$, *-*] (6,0) node[above right] {$B$ (ground)}
to[short] (6, -1) node[below] {$-$}
to[battery,invert,l={1 Volt}] (0, -1) node[below] {$+$}
to[short] (0, 0)
;
\end{circuitikz}
\end{center}
\problem{}<onecurrents>
From the circuit diagram above, we immediately know that $V(A) = 1$ and $V(B) = 0$. \par
What equations related to the currents out of $x$ and $y$ does Kirchoff's law give us? \par
\hint{Current into $x$ = current out of $x$}
\vfill
\problem{}
Use Ohm's law to turn the equations from \ref{onecurrents} into equations about voltage and resistance. \par
Find an expression for $V(x)$ and $V(y)$ in terms of other voltages, then solve the resulting system of equations.
Does your result look familiar?
\begin{solution}
\setlength{\abovedisplayskip}{0pt} % Fix spacing on top
\begin{flalign*}
V(x) &= \frac{V(A) - V(y)}{2} &&\\
V(y) &= \frac{V(x) - V(B)}{2} &&
\end{flalign*}
\end{solution}
\vfill
\pagebreak

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\section{The Equivalence}
In the last problem, we found that the equations for $V(x)$ were the same as the equations for $P(x)$ on the same graph.
It turns out that this is true in general: problems about voltage in circuits directly correspond to problems about probability
in graphs. We'll spend the next section proving this fact.
\definition{}
For the following problems, \textit{conductance} will be more convenient than resistance. \par
The definition of conductance is quite simple:
$$
C(a, b) = \frac{1}{R(a,b)}
$$
\note[Aside]{
Resistance is usually measured in Ohms, denoted $\Omega$. \\
A few good-natured physicists came up with the \say{mho} (denoted \reflectbox{\rotatebox[origin=c]{180}{$\Omega$}})
as a unit of conductance, which is equivalent to an inverse Ohm.
Unfortunately, NIST discourages the use of Mhos in favor of the equivalent (and less amusing) \say{Siemens.}
}
\problem{}
Let $x$ be a node in a graph. \par
Let $N_x$ be the set of $x$'s neighbors, $w(x, y)$ the weight of the edge between nodes $x$ and $y$, and $W_x$
the sum of the weights of all edges connected to $x$.
We saw earlier that the probability function $P$ satisfies the following sum:
$$
P(x) = \sum_{b \in N_x} \biggl( P(b) \times \frac{w(x, b)}{W_x} \biggr)
$$
\note{This was never explicitly stated, but is noted in \ref{weightedgraph}.}
\vspace{4mm}
Use Ohm's and Kirchoff's laws to show that the voltage function $V$ satisfies a similar sum:
$$
V(x) = \sum_{b \in N_x} \biggl( V(b) \times \frac{C(x, b)}{C_x} \biggr)
$$
where $C(x, b)$ is the conductance of edge $(x, b)$ and $C_x$ is the sum of the conductances of all edges connected to $x$.
\begin{solution}
First, we know that
$$
\sum_{b \in N_x} I(x, b) = 0
$$
for all nodes $x$. Now, substitute $I(x, b) = \frac{V(x) - V(b)}{R(x, y)}$ and pull out $V(x)$ terms to get
$$
V(x) \sum_{b \in N_x} \frac{1}{R(x, b)} - \sum_{b \in N_x} \frac{V(b)}{R(x, b)} = 0
$$
Rearranging and replacing $R(x, b)^{-1}$ with $C(x, b)$ and $\sum C(x, b)$ with $C_x$ gives us
$$
V(x) = \sum_{b \in N_x} V(b) \frac{C(x, b)}{C_x}
$$
\end{solution}
\vfill
\pagebreak
Thus, if $w(a, b) = C(a, b)$, $P$ and $V$ satisfy the same system of linear equations. To finish proving that
$P = V$, we now need to show that there can only be one solution to this system. We will do this in the next
two problems.
\problem{}<generaleq>
Let $q$ be a solution to the following equations, where $x \neq a, b$.
$$
q(x) = \sum_{b \in N_x} \biggl( q(b) \times \frac{w(x, b)}{W_x} \biggr)
$$
Show that the maximum and minimum of $q$ are $q(a)$ and $q(b)$ (not necessarily in this order).
\begin{solution}
The domain of $q$ is finite, so a maximum and minimum must exist.
\vspace{2mm}
Since $q(x)$ is a weighted average of all $q(b), ~b \in N_x$, there exist $y, z \in N_x$ satisfying
$q(y) \leq q(x) \leq q(z)$. Therefore, none of these can be an extreme point.
\vspace{2mm}
$A$ and $B$ are the only vertices for which this may not be true, so they must be the minimum and maximum.
\end{solution}
\vfill
\problem{}
Let $p$ and $q$ be functions that solve our linear system \par
and satisfy $p(A) = q(A) = 1$ and $p(B) = q(B) = 0$. \par
\vspace{1mm}
Show that the function $p - q$ satisfies the equations in \ref{generaleq}, \par
and that $p(x) - q(x) = 0$ for every $x$. \note{Note that $p(x) - q(x) = 0 ~ \forall x \implies p = q$}
\begin{solution}
The equations in \ref{generaleq} for $p$ and $q$ directly imply that
$$
[p - q](x) = \sum_{b \in N_x} \biggl( [p - q](b) \times \frac{w(x, b)}{W_x} \biggr)
$$
Which are the equations from \ref{generaleq} for $(p - q)$.
\vspace{2mm}
Hence, the minimum and maximum values of $p - q$ are $[p - q](a) = 1 - 1 = 0$
and $[p - q](b) = 1 - 1 = 0$.
\vspace{2mm}
Therefore $p(x) - q(x) = 0$ for all $x$, so $p(x) = q(x)$.
\end{solution}
\vfill
\pagebreak

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\section{Effective Resistance}
As we have seen, calculating the properties of a circuit by creating an equation for each vertex is
a fairly time-consuming ordeal. Fortunately, there is a better strategy we can use.
\vspace{2mm}
Consider a graph (or a circuit) with source and ground vertices. All parts of the circuit that aren't these
two vertices are hidden inside a box, as shown below:
\begin{center}
\begin{circuitikz}[american voltages]
\draw
(0,0) node[above left] {$A$ (source)}
to[short, *-] (1,0)
(5, 0) to[short, -*] (6,0) node[above right] {$B$ (ground)}
to[short] (6, -2) node[below] {$-$}
to[battery,invert,l={1 Volt}] (0, -2) node[below] {$+$}
to[short] (0, 0)
;
\node[
draw,
minimum width = 4cm,
minimum height = 2cm,
anchor = south west
] at (1, -1) {Unknown circuit};
\end{circuitikz}
\end{center}
What do we know about this box? If this was a physical system, we'd expect that the current flowing
out of $A$ is equal to the current flowing into $B$.
\problem{}
Using Kirchoff's law, show that the following equality holds. \par
Remember that we assumed Kirchoff's law holds only at nodes other than $A$ and $B$. \par
\note[Note]{As before, $N_x$ is the set of neighbors of $x$.}
$$
\sum_{b \in N_A} I(A, b) = \sum_{b \in N_B} I(b, B)
$$
\begin{solution}
Add Kirchoff's law for all vertices $x \neq A$ to get
$$
\sum_{\forall x} \biggl( ~ \sum_{b \in N_x } I(x, b) \biggr) = 0
$$
This sum counts both $I(x, y)$ and $I(x, y)$ for all edges $x, y$, except $I(x, y)$ when $x$ is
$A$ or $B$. Since $I(a, b) + I(b, a) = 0$, these cancel out, leaving us with
$$
\sum_{b \in N_A} I(A, b) + \sum_{b \in N_B} I(B, b) = 0
$$
\vspace{2mm}
Rearrange and use the fact that $I(a, b) = -I(b, a)$ to get the final equation.
\end{solution}
\vfill
If we call this current $I_A = \sum_{b \in N_A} I(A, b)$, we can pretend that the box contains only one resistor,
carrying $I_A$ units of current. Using this information and Ohm's law, we can calculate the
\textit{effective resistance} of the box.
\pagebreak
\problem{Resistors in parallel}<parallelresistors>
Using Ohm's law and Kirchoff's law, calculate the effective resistance $R_\text{eff}$ of the circuit below.
\begin{center}
\begin{circuitikz}[american voltages]
\draw
(0,0) node[above left] {$A$ (source)}
to[short, *-*] (1, 0)
(1, 0) to[short] (2, 1)
to[R, l=$R_1$, o-o] (4, 1)
to[short] (5, 0)
(1, 0) to[short] (2, -1)
to[R, l=$R_n$, o-o] (4, -1)
to[short] (5, 0)
(1, 0) to[short, -o] (2, 0.5)
(2, 0.5) to (2.3, 0.5)
(4, 0.5) to (3.7, 0.5)
(4, 0.5) to[short, o-] (5, 0)
(1, 0) to[short, -o] (2, -0.5)
(2, -0.5) to (2.3, -0.5)
(4, -0.5) to (3.7, -0.5)
(4, -0.5) to[short, o-] (5, 0)
(1, 0) to[short] (1.7, 0)
(4.3, 0) to[short] (5, 0)
(5, 0) to[short, *-*] (6, 0) node[above right] {$B$ (ground)}
to[short] (6, -2) node[below] {$-$}
to[battery,invert,l={1 Volt}] (0, -2) node[below] {$+$}
to[short] (0, 0)
;
\node at (3, 0) {$...$ a few more $...$};
\end{circuitikz}
\end{center}
\begin{solution}
Let $I_i$ be the current across resistor $R_i$, from left to right. \par
By Ohm's law, $I_i = \frac{V}{R_i}$ (Note that $V = 1$ in this problem). \par
\vspace{2mm}
The source current is then $I_A = \sum_{i=1}^n = \Bigl( V \Bigr) \Bigl( \frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n} \Bigr)$.
Applying Ohm's law again, we find that
$$
R_\text{eff} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n}}
$$
\end{solution}
\vfill
\problem{Resistors in series}
Using Ohm's law and Kirchoff's law, calculate the effective resistance $R_{\text{total}}$ of the circuit below.
\begin{center}
\begin{circuitikz}[american voltages]
\draw
(0,0) node[above left] {$A$ (source)}
to[R, l=$R_1$, *-*] (2,0)
to[short] (2.5, 0)
(5.5, 0) to[short] (6, 0)
to[R, l=$R_n$, *-*] (8,0) node[above right] {$B$ (ground)}
to[short] (8, -1) node[below] {$-$}
to[battery,invert,l={1 Volt}] (0, -1) node[below] {$+$}
to[short] (0, 0)
;
\node at (4, 0) {$...$ a few more $...$};
\end{circuitikz}
\end{center}
\begin{solution}
This solution uses the same notation as the solution for \ref{parallelresistors}.
\vspace{2mm}
By Kirchoff's law, all $I_i$ are equal in this circuit. Let's say $I = I_i$. \par
Let $V_i$ denote the voltage at the node to the left of $R_i$. \par
By Ohm's law, $V_i - V_{i+1} = IR_i$.
\vspace{2mm}
The sum of this over all $i$ telescopes, and we get $V(A) - V(B) = I(R_1 + R_2 + ... + R_n)$. \par
Dividing, we find that
$$
R_\text{eff} = R_1 + R_2 + ... + R_n
$$
\end{solution}
\vfill
\pagebreak
We can now use effective resistance to simplify complicated circuits. Whenever we see the above constructions
(resistors in parallel or in series) in a graph, we can replace them with a single resistor of appropriate value.
\problem{}
Consider the following circuits. Show that the triangle has the same effective resistance as the star if
\begin{itemize}
\item $x = R_1R_2 + R_1R_3 + R_2R_3$
\item $S_1 = \nicefrac{x}R_3$
\item $S_2 = \nicefrac{x}R_1$
\item $S_3 = \nicefrac{x}R_2$
\end{itemize}
\vspace{2mm}
\hfill
\begin{minipage}{0.48\textwidth}
\begin{center}
\textbf{The star:}\par
\begin{circuitikz}[american voltages]
\draw
(-1, 1.732) to[R, l=$R_1$, *-*] (0, 0)
(2, 0) to[R, l=$R_2$, *-*] (0, 0)
(-1, -1.732) to[R, l=$R_3$, *-*] (0, 0)
(-1, 1.732) to[short, *-o] (-1.5, 1.732)
(-1, -1.732) to[short, *-o] (-1.5, -1.732)
(2, 0) to[short, *-o] (2.5, 0)
;
\node[above] at (-1, 1.732) {a};
\node[above] at (2, 0) {b};
\node[below] at (-1, -1.732) {c};
\end{circuitikz}
\end{center}
\end{minipage}
\hfill
\begin{minipage}{0.48\textwidth}
\begin{center}
\textbf{The triangle:}\par
\begin{circuitikz}[american voltages]
\draw
(-1, 1.732)
to[R, l=$S_1$, *-*] (2, 0)
to[R, l=$S_2$, *-*] (-1, -1.732)
to[R, l=$S_3$, *-*] (-1, 1.732)
(-1, 1.732) to[short, *-o] (-1.5, 1.732)
(-1, -1.732) to[short, *-o] (-1.5, -1.732)
(2, 0) to[short, *-o] (2.5, 0)
;
\node[above] at (-1, 1.732) {a};
\node[above] at (2, 0) {b};
\node[below] at (-1, -1.732) {c};
\end{circuitikz}
\end{center}
\end{minipage}
\hfill
\vfill
\pagebreak
\problem{}
Suppose we construct a circuit by connecting the $2^n$ vertices of an $n$-dimensional cube with $1\Omega$ resistors.
If we place $A$ and $B$ at opposing vertices, what is the effective resistance of this circuit? \par
\textbf{Bonus:} As $n \rightarrow \infty$, what happens to $R_\text{eff}$? \par
\note[Note]{Leave your answer as a sum.}
\begin{solution}
Think of the vertices of the $n$-dimensional cube as $n$-bit binary strings, with $A$ at \texttt{000...0}
and $B$ at \texttt{111...1}. We can divide our cube into $n+1$ layers based on how many ones are in each
node's binary string, with the $k^\text{th}$ layer having $k$ ones. By symmetry, all the nodes in each layer
have the same voltage. This means we can think of the layers as connected in series, with the resistors
inside each layer connected in parallel.
\vspace{2mm}
There are $\binom{n}{k}$ nodes in the $k^\text{th}$ layer. Each node in this layer has $k$ ones, so
there are $n - k$ ways to flip a zero to get to the $(k + 1)^\text{th}$ layer. In total, there are
$\binom{n}{k}(n - k)$ parallel connections from the $k^\text{th}$ layer to the $(k + 1)^\text{th}$
layer, creating an effective resistance of
$$
\frac{1}{\binom{n}{k}(n - k)}
$$
The total effective resistance is therefore
$$
\sum_{k = 0}^{n-1} \frac{1}{\binom{n}{k}(n - k)}
$$
\linehack{}
To calculate the limit as $n \rightarrow \infty$, note that
$$
\binom{n}{k}(n - k) = \frac{n!}{(n - k - 1)! \times k!} = n \binom{n-1}{k}
$$
So, the sum is
$$
\frac{1}{n} \sum_{k=0}^{n-1} \frac{1}{\binom{n - 1}{k}}
$$
\vspace{8mm}
Note that for $n \geq 4$, $\binom{n}{k} \geq \binom{n}{2}$ for $2 \leq k \leq n-2$, so
$$
\sum_{k = 0}^{n} \frac{1}{\binom{n}{k}} \leq 2 + \frac{2}{n} + \frac{n - 3}{\binom{n}{2}}
$$
which approaches $2$ as $n \rightarrow \infty$.
So, $R_\text{eff} \rightarrow 0$ as $n \rightarrow \infty$.
\end{solution}

88
src/Advanced/Random Walks/tikxset.tex Normal file
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@ -0,0 +1,88 @@
\usetikzlibrary{arrows.meta}
\usetikzlibrary{shapes.geometric}
\usetikzlibrary{patterns}
% We put nodes in a separate layer, so we can
% slightly overlap with paths for a perfect fit
\pgfdeclarelayer{nodes}
\pgfdeclarelayer{path}
\pgfsetlayers{main,nodes}
% Layer settings
\tikzset{
% Layer hack, lets us write
% later = * in scopes.
layer/.style = {
execute at begin scope={\pgfonlayer{#1}},
execute at end scope={\endpgfonlayer}
},
%
% Arrowhead tweak
>={Latex[ width=2mm, length=2mm ]},
%
% Labels inside edges
label/.style = {
rectangle,
% For automatic red background in solutions
fill = \ORMCbgcolor,
draw = none,
rounded corners = 0mm
},
%
% Nodes
main/.style = {
draw,
circle,
fill = white,
line width = 0.35mm
},
accept/.style = {
draw,
circle,
fill = white,
double,
double distance = 0.5mm,
line width = 0.35mm
},
reject/.style = {
draw,
rectangle,
fill = white,
text = black,
double,
double distance = 0.5mm,
line width = 0.35mm
},
start/.style = {
draw,
rectangle,
fill = white,
line width = 0.35mm
},
%
% Loop tweaks
loop above/.style = {
min distance = 2mm,
looseness = 8,
out = 45,
in = 135
},
loop below/.style = {
min distance = 5mm,
looseness = 10,
out = 315,
in = 225
},
loop right/.style = {
min distance = 5mm,
looseness = 10,
out = 45,
in = 315
},
loop left/.style = {
min distance = 5mm,
looseness = 10,
out = 135,
in = 215
}
}