\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 $P(n): N \to [0, 1]$ that returns the probability that our particle stops at $A$, where $N$ is the set of nodes in $G$. \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{} 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{} 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{} How can we use \ref{oneunweighted} to find $P(n)$ for any $n$? \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. \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{} 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