diff --git a/Intermediate/Instant Insanity/4Dcube2.jpg b/Intermediate/Instant Insanity/4Dcube2.jpg new file mode 100755 index 0000000..d2ca423 Binary files /dev/null and b/Intermediate/Instant Insanity/4Dcube2.jpg differ diff --git a/Intermediate/Instant Insanity/II.jpg b/Intermediate/Instant Insanity/II.jpg new file mode 100755 index 0000000..db587fa Binary files /dev/null and b/Intermediate/Instant Insanity/II.jpg differ diff --git a/Intermediate/Instant Insanity/dodecahedron.jpg b/Intermediate/Instant Insanity/dodecahedron.jpg new file mode 100755 index 0000000..c137f88 Binary files /dev/null and b/Intermediate/Instant Insanity/dodecahedron.jpg differ diff --git a/Intermediate/Instant Insanity/main.tex b/Intermediate/Instant Insanity/main.tex new file mode 100755 index 0000000..716d0db --- /dev/null +++ b/Intermediate/Instant Insanity/main.tex @@ -0,0 +1,1127 @@ +% https://git.betalupi.com/Mark/latex-packages +% use [nosolutions] flag to hide solutions. +% use [solutions] flag to show solutions. +% Last built with version 1.1.0 +\documentclass[ + solutions +]{ormc_handout} + + +\usepackage{tkz-graph} + + +\begin{document} + + \maketitle + + + {Graph Theory and Instant Insanity} + { + Prepared by Mark on \today \\ + Based on a handout by Oleg Gleizer + } + + + \section{Instant Insanity} + + The puzzle you have in front of you is called {\it Instant Insanity}. + + It consists of four cubes, with faces colored with four colors: + red, blue, green, and white. The objective is to put the cubes in a row + so that each side, front, back, upper, and lower, + of the row shows each of the four colors. \\ + + \begin{center} + \includegraphics[width=2.2in] + {II.jpg} + \end{center} + + There are 41,472 different arrangements + of the cubes. Only one is a solution. + Finding it by trial and error is quite difficult, + but we have witnessed a few students + do just that. + + However, that rarely happens. + We'd like to solve this puzzle today, and + to do that, we'll need a few tools. + + \section{Cubic Nets} + + A {\it cubic net} is a 2D picture + simultaneously showing all the six sides + (a.k.a.~faces) of a 3D cube, + please take a look at the examples below. \\ + + \begin{center} + \begin{tikzpicture} [scale = .3] + \draw [line width = 1.5pt] (0,0) -- (12,0) -- + (12,3) -- (0,3) -- (0,0); + \draw [line width = 1.5pt] (6,-3) -- + (9,-3) -- (9,6) -- (6,6) -- (6,-3); + \draw [line width = 1.5pt] (3,0) -- (3,3); + + \draw [line width = 1.5pt] (21,0) -- (33,0) -- + (33,3) -- (21,3) -- (21,0); + \draw [line width = 1.5pt] (24,0) -- + (24,6) -- (27,6) -- (27,0); + \draw [line width = 1.5pt] (27,0) -- + (27,-3) -- (30,-3) -- (30,3); + \end{tikzpicture} + \end{center} + + \problem{} + Draw a cubic net different from the two above. + + \vfill + \pagebreak + + + \problem{} + An ant wants to crawl from point $A$ of a cubic room + to the opposite point $B$, as in the picture below. + + \begin{center} \begin{small} + \begin{tikzpicture} + \draw (1,1) -- (4,1) -- (4,4) -- (1,4) -- (1,1); + \draw (0,0) -- (3,0) -- (3,3) -- (0,3) -- (0,0); + \draw (0,0) -- (1,1); + \draw (3,0) -- (4,1); + \draw (0,3) -- (1,4); + \draw (3,3) -- (4,4); + \filldraw (0,0) circle (3pt); + \filldraw (4,4) circle (3pt); + \coordinate [label=below left:{A}] (a) at (0,0); + \coordinate [label=above right:{B}] (b) at (4,4); + \end{tikzpicture} + \end{small} \end{center} + + The insect can crawl on any surface, + a floor, ceiling, or wall, but cannot fly through the air. + Find at least two different shortest paths for the ant + (there is more than one). + + Let's look at the nets of the puzzle's cubes. \\ + + \begin{center} \begin{small} + \begin{tikzpicture} [scale = .3] \label{pic:ii_cubes} + \filldraw [red] (0,0) -- (9,0) -- (9,3) -- (0,3) -- (0,0); + \filldraw [blue] (9,0) -- (12,0) -- (12,3) -- (9,3) -- (9,0); + \filldraw [green] (6,3) -- (9,3) -- (9,6) -- (6,6) -- (6,3); + \draw [line width = 1.5pt] (0,0) -- (12,0) -- + (12,3) -- (0,3) -- (0,0); + \draw [line width = 1.5pt] (6,-3) -- + (9,-3) -- (9,6) -- (6,6) -- (6,-3); + \draw [line width = 1.5pt] (3,0) -- (3,3); + \coordinate [label=below:{Cube 1}] (c1) at (7.6,-3.5); + + \node[text = white] at (0 + 1.5, 0 + 1.5) {\textbf{Red}}; + \node[text = white] at (0 + 4.5, 0 + 1.5) {\textbf{Red}}; + \node[text = white] at (0 + 7.5, 0 + 1.5) {\textbf{Red}}; + \node[text = white] at (0 + 10.5, 0 + 1.5) {\textbf{Blue}}; + \node[text = black] at (0 + 7.5, 0 + 4.5) {\textbf{Grn}}; + \node[text = black] at (0 + 7.5, 0 - 1.5) {\textbf{Wht}}; + + \filldraw [red] (21,0) -- (27,0) -- (27,3) -- + (21,3) -- (21,0); + \filldraw [green] (27,3) -- (27,0) -- (30,0) -- + (30,3) -- (27,3); + \filldraw [blue] (27,0) -- (27,-3) -- (30,-3) -- + (30,0) -- (27,0); + \draw [line width = 1.5pt] (21,0) -- (33,0) -- + (33,3) -- (21,3) -- (21,0); + \draw [line width = 1.5pt] (27,-3) -- + (30,-3) -- (30,6) -- (27,6) -- (27,-3); + \draw [line width = 1.5pt] (24,0) -- (24,3); + \coordinate [label=below:{Cube 2}] (c2) at (28.6,-3.5); + + \node[text = white] at (21 + 1.5, 0 + 1.5) {\textbf{Red}}; + \node[text = white] at (21 + 4.5, 0 + 1.5) {\textbf{Red}}; + \node[text = black] at (21 + 7.5, 0 + 1.5) {\textbf{Grn}}; + \node[text = black] at (21 + 10.5, 0 + 1.5) {\textbf{Wht}}; + \node[text = black] at (21 + 7.5, 0 + 4.5) {\textbf{Wht}}; + \node[text = white] at (21 + 7.5, 0 - 1.5) {\textbf{Blue}}; + + \filldraw [red] (0,-15) -- (3,-15) -- (3,-12) + -- (0,-12) -- (0,-15); + \filldraw [green] (3,-15) -- (6,-15) -- (6,-12) + -- (3,-12) -- (3,-15); + \filldraw [blue] (6,-18) -- (9,-18) -- (9,-12) + -- (6,-12) -- (6,-15); + \draw [line width = 1.5pt] (0,-15) -- (12,-15) -- + (12,-12) -- (0,-12) -- (0,-15); + \draw [line width = 1.5pt] (6,-18) -- + (9,-18) -- (9,-9) -- (6,-9) -- (6,-18); + \draw [line width = 1.5pt] (3,-15) -- (3,-12); + \coordinate [label=below:{Cube 3}] (c3) at (7.6,-18.5); + + \node[text = white] at (0 + 1.5, -15 + 1.5) {\textbf{Red}}; + \node[text = black] at (0 + 4.5, -15 + 1.5) {\textbf{Grn}}; + \node[text = white] at (0 + 7.5, -15 + 1.5) {\textbf{Blue}}; + \node[text = black] at (0 + 10.5, -15 + 1.5) {\textbf{Wht}}; + \node[text = black] at (0 + 7.5, -15 + 4.5) {\textbf{Wht}}; + \node[text = white] at (0 + 7.5, -15 - 1.5) {\textbf{Blue}}; + + \filldraw [red] (21,-15) -- (24,-15) -- + (24,-12) -- (21,-12) -- (21,-15); + \filldraw [blue] (24,-15) -- (27,-15) -- + (27,-12) -- (24,-12) -- (24,-15); + \filldraw [green] (27,-18) -- (30,-18) -- + (30,-12) -- (27,-12) -- (27,-18); + \filldraw [blue] (30,-15) -- (33,-15) -- + (33,-12) -- (30,-12) -- (30,-15); + \draw [line width = 1.5pt] (21,-15) -- (33,-15) -- + (33,-12) -- (21,-12) -- (21,-15); + \draw [line width = 1.5pt] (27,-18) -- + (30,-18) -- (30,-9) -- (27,-9) -- (27,-18); + \draw [line width = 1.5pt] (24,-15) -- (24,-12); + \coordinate [label=below:{Cube 4}] (c4) at (28.6,-18.5); + + \node[text = white] at (21 + 1.5, -15 + 1.5) {\textbf{Red}}; + \node[text = white] at (21 + 4.5, -15 + 1.5) {\textbf{Blue}}; + \node[text = black] at (21 + 7.5, -15 + 1.5) {\textbf{Grn}}; + \node[text = white] at (21 + 10.5, -15 + 1.5) {\textbf{Blue}}; + \node[text = black] at (21 + 7.5, -15 + 4.5) {\textbf{Wht}}; + \node[text = black] at (21 + 7.5, -15 - 1.5) {\textbf{Grn}}; + + \end{tikzpicture} + \end{small} \end{center} + \medskip + + Note that each cube is different. + + \vfill + \pagebreak + + \section{Graphs} + + \begin{tcolorbox}[ + colback=white, + colframe=gray!75!black, + title={Last week's lesson} + ] + A \textit{graph} is a collection of nodes (vertices) and connections between them (edges). If an edge $e$ connects the vertices $v_i$ and $v_j$, then we write $e = {v_i, v_j}$. An example is below. + + \begin{center} + \begin{tikzpicture} [scale = .6] \label{pic:1} + \SetGraphUnit{5} + \Vertex{B} + \WE(B){A} + \EA(B){C} + \Edge(B)(A) + \Edge(C)(B) + \tikzset{EdgeStyle/.append style = {bend left = 50}} + \Edge(A)(C) + \Edge(C)(A) + \coordinate [label=above:{$e_1$}] (e1) at (-2.1,.0); + \coordinate [label=above:{$e_2$}] (e2) at (0,2.45); + \coordinate [label=below:{$e_3$}] (e3) at (0,-2.5); + \coordinate [label=above:{$e_4$}] (e4) at (2.1,.0); + \end{tikzpicture} + \end{center} + + More formally, a graph is defined by a set of vertices $\{v_1, v_2, ...\}$, and a set of edges $\{\ \{v_1, v_2\}, \{v_1, v_3\}, ...\ \}$. + + \medskip + + If the order of the vertices in an edge does not matter, + a graph is called {\it undirected}. A graph is called + a {\it directed graph} if the order of the vertices does matter. + For example, the (undirected) graph above + has three vertices, $A$, $B$, and $C$, and four edges, + $e_1 =\{A,B\}$, $e_2 = \{A,C\}$, $e_3 = \{A,C\}$, + and $e_4 = \{B,C\}$. + + \end{tcolorbox} + + + Let's represent Cube 1 by a graph. \\ + The vertices will be the face colors: Blue, Green, Red, and White, so $V = \{B,G,R,W\}$. \\ + Two vertices are be connected by an edge if and only if the corresponding faces are opposing each other on the cube. \\ + Cube 1 has the following edges: $e_1 = \{B,R\}$, $e_2 = \{G,W\}$, and the loop $e_3 = \{R,R\}$. To emphasize that all the three edges represent the first cube, let us mark them with the number $1$. \\ + + \begin{center} \begin{small} + \begin{tikzpicture} + \filldraw [blue] (0,5) -- (1,5) -- (1,6) -- + (0,6) -- (0,5); + \draw [line width = 1.5pt] (0,5) -- + (1,5) -- (1,6) -- (0,6) -- (0,5); + \filldraw [green] (5,5) -- (6,5) -- (6,6) -- + (5,6) -- (5,5); + \draw [line width = 1.5pt] (5,5) -- + (6,5) -- (6,6) -- (5,6) -- (5,5); + \filldraw [red] (0,0) -- (1,0) -- (1,1) -- + (0,1) -- (0,0); + \draw [line width = 1.5pt] (0,0) -- + (1,0) -- (1,1) -- (0,1) -- (0,0); + \draw [line width = 1.5pt] (5,0) -- + (6,0) -- (6,1) -- (5,1) -- (5,0); + \draw (-.5,-.5) circle (.3) node {1}; + \draw [line width = 1.5pt] (0,.5) .. + controls (- .9,.4) and (-.65,-.2) .. (-.7,-.3); + \draw [line width = 1.5pt] (.5,0) .. + controls (.4,-.9) and (-.2,-.65) .. (-.3,-.7); + \draw (.5,2) circle (.3) node {1}; + \draw [line width = 1.5pt] (.5,1) -- (.5,1.7); + \draw [line width = 1.5pt] (.5,2.3) -- (.5,5); + \draw (5.5,4) circle (.3) node {1}; + \draw [line width = 1.5pt] (5.5,5) -- (5.5,4.3); + \draw [line width = 1.5pt] (5.5,3.7) -- (5.5,1); + + \node[text = white] at (0.5, 5.5) {\textbf{Blue}}; + \node[text = white] at (0.5, 0.5) {\textbf{Red}}; + \node[text = black] at (5.5, 0.5) {\textbf{Wht}}; + \node[text = black] at (5.5, 5.5) {\textbf{Grn}}; + + \end{tikzpicture} + \end{small} \end{center} + \bigskip + + Cube 2 has the following pairs + of opposing faces, $\{B,W\}$, $\{G,R\}$, + and $\{R,W\}$. Let us add them to the graph + as the edges $e_4$, $e_5$, and $e_6$. \\ + + \begin{center} \begin{small} + \begin{tikzpicture} + \filldraw [blue] (0,5) -- (1,5) -- (1,6) -- + (0,6) -- (0,5); + \draw [line width = 1.5pt] (0,5) -- + (1,5) -- (1,6) -- (0,6) -- (0,5); + \filldraw [green] (5,5) -- (6,5) -- (6,6) -- + (5,6) -- (5,5); + \draw [line width = 1.5pt] (5,5) -- + (6,5) -- (6,6) -- (5,6) -- (5,5); + \filldraw [red] (0,0) -- (1,0) -- (1,1) -- + (0,1) -- (0,0); + \draw [line width = 1.5pt] (0,0) -- + (1,0) -- (1,1) -- (0,1) -- (0,0); + \draw [line width = 1.5pt] (5,0) -- + (6,0) -- (6,1) -- (5,1) -- (5,0); + \draw (-.5,-.5) circle (.3) node {1}; + \draw [line width = 1.5pt] (0,.5) .. + controls (-.9,.4) and (-.65,-.2) .. (-.7,-.3); + \draw [line width = 1.5pt] (.5,0) .. + controls (.4,-.9) and (-.2,-.65) .. (-.3,-.7); + \draw (.5,2) circle (.3) node {1}; + \draw [line width = 1.5pt] (.5,1) -- (.5,1.7); + \draw [line width = 1.5pt] (.5,2.3) -- (.5,5); + \draw (5.5,4) circle (.3) node {1}; + \draw [line width = 1.5pt] (5.5,5) -- (5.5,4.3); + \draw [line width = 1.5pt] (5.5,3.7) -- (5.5,1); + \draw (2,4) circle (.3) node {2}; + \draw [line width = 1.5pt] (1,5) -- (1.8,4.2); + \draw [line width = 1.5pt] (5,1) -- (2.2,3.8); + \draw (4,4) circle (.3) node {2}; + \draw [line width = 1.5pt] (5,5) -- (4.2,4.2); + \draw [line width = 1.5pt] (1,1) -- (3.8,3.8); + \draw (4,.5) circle (.3) node {2}; + \draw [line width = 1.5pt] (1,.5) -- (3.7,.5); + \draw [line width = 1.5pt] (5,.5) -- (4.3,.5); + + \node[text = white] at (0.5, 5.5) {\textbf{Blue}}; + \node[text = white] at (0.5, 0.5) {\textbf{Red}}; + \node[text = black] at (5.5, 0.5) {\textbf{Wht}}; + \node[text = black] at (5.5, 5.5) {\textbf{Grn}}; + \end{tikzpicture} + \end{small} \end{center} + \bigskip + + Let us now make the graph + representing all four cubes. \\ + + \begin{center} \begin{small} + \begin{tikzpicture} \label{pic:II_comfiguration} + \filldraw [blue] (0,5) -- (1,5) -- (1,6) -- + (0,6) -- (0,5); + \draw [line width = 1.5pt] (0,5) -- + (1,5) -- (1,6) -- (0,6) -- (0,5); + \filldraw [green] (5,5) -- (6,5) -- (6,6) -- + (5,6) -- (5,5); + \draw [line width = 1.5pt] (5,5) -- + (6,5) -- (6,6) -- (5,6) -- (5,5); + \filldraw [red] (0,0) -- (1,0) -- (1,1) -- + (0,1) -- (0,0); + \draw [line width = 1.5pt] (0,0) -- + (1,0) -- (1,1) -- (0,1) -- (0,0); + \draw [line width = 1.5pt] (5,0) -- + (6,0) -- (6,1) -- (5,1) -- (5,0); + \draw (-.5,-.5) circle (.3) node {1}; + \draw [line width = 1.5pt] (0,.5) .. + controls (-.9,.4) and (-.65,-.2) .. (-.7,-.3); + \draw [line width = 1.5pt] (.5,0) .. + controls (.4,-.9) and (-.2,-.65) .. (-.3,-.7); + \draw (.5,2) circle (.3) node {1}; + \draw [line width = 1.5pt] (.5,1) -- (.5,1.7); + \draw [line width = 1.5pt] (.5,2.3) -- (.5,5); + \draw (5.5,4) circle (.3) node {1}; + \draw [line width = 1.5pt] (5.5,5) -- (5.5,4.3); + \draw [line width = 1.5pt] (5.5,3.7) -- (5.5,1); + \draw (1.5,3.5) circle (.3) node {2}; + \draw [line width = 1.5pt] (.8,5) .. + controls (.8,4.6) and (1,4.2) .. (1.3,3.7); + \draw [line width = 1.5pt] (5,.8) .. + controls (2.5,1.5) and (2.1,2.5) .. (1.6,3.22); + \draw (4,4) circle (.3) node {2}; + \draw [line width = 1.5pt] (5,5) -- (4.2,4.2); + \draw [line width = 1.5pt] (1,1) -- (3.8,3.8); + \draw (4,.5) circle (.3) node {2}; + \draw [line width = 1.5pt] (1,.5) -- (3.7,.5); + \draw [line width = 1.5pt] (5,.5) -- (4.3,.5); + \draw (-1,3) circle (.3) node {3}; + \draw [line width = 1.5pt] (0,5) .. + controls (-.7,4.3) and (-.9,3.6) .. (-1,3.3); + \draw [line width = 1.5pt] (0,1) .. + controls (-.7,1.7) and (-.9,2.3) .. (-1,2.7); + \draw (2.5,4) circle (.3) node {3}; + \draw [line width = 1.5pt] (1,5.2) .. + controls (1.4,5.2) and (2,4.5) .. (2.3,4.2); + \draw [line width = 1.5pt] (5.2,1) .. + controls (4.7,1.8) and (3.5,3) .. (2.7,3.8); + \draw (7,3) circle (.3) node {3}; + \draw [line width = 1.5pt] (6,5) .. + controls (6.7,4.3) and (6.9,3.6) .. (7,3.3); + \draw [line width = 1.5pt] (6,1) .. + controls (6.7,1.7) and (6.9,2.3) .. (7,2.7); + \draw (-.5,6.5) circle (.3) node {4}; + \draw [line width = 1.5pt] (0,5.5) .. + controls (-.9,5.4) and (-.65,6.2) .. (-.7,6.26); + \draw [line width = 1.5pt] (-.24,6.65) .. + controls (.1,6.7) and (.45,6.7) .. (.5,6); + \draw (8.5,3) circle (.3) node {4}; + \draw [line width = 1.5pt] (6,5.5) .. + controls (7.3,5) and (8.2,4) .. (8.5,3.3); + \draw [line width = 1.5pt] (6,.8) .. + controls (7.3,1) and (8.2,2) .. (8.5,2.7); + \draw (2.3,1.2) circle (.3) node {4}; + \draw [line width = 1.5pt] (1,.8) -- (2.04,1.1); + \draw [line width = 1.5pt] (2.6,1.3) .. + controls (4,2) and (5,4) .. (5.2,5); + + \node[text = white] at (0.5, 5.5) {\textbf{Blue}}; + \node[text = white] at (0.5, 0.5) {\textbf{Red}}; + \node[text = black] at (5.5, 0.5) {\textbf{Wht}}; + \node[text = black] at (5.5, 5.5) {\textbf{Grn}}; + \end{tikzpicture} + \end{small} \end{center} + + \problem{} + Check if the above representation + is correct for Cubes 3 and 4. + \vfill + \pagebreak + + With the help of the above graph, + solving the puzzle becomes as easy + as a walk in the park, literally. + Imagine that the vertices of the above graph + are the clearings and the edges are the paths. + An edge marked by the number $i$ represents + two opposing faces of the $i$-th cube. + Let us try to find a closed walk, a.k.a.~a cycle, + in the graph that visits each clearing once + and uses the paths marked by the different numbers, + $i=1,2,3,4$. If we order the front and rear sides + of the cubes accordingly, + then the front and rear of the stack + will show all the four different colors + in the order prescribed by our walk. \\ + + For example, here is such an (oriented) cycle, + represented by the magenta arrows + on the picture below. \\ + + \begin{center} \begin{small} + \begin{tikzpicture} + \draw [line width = 3pt, color = magenta, <-] + (.5,1) -- (.5,1.7); + \draw [line width = 3pt, color = magenta, <-] + (.5,2.3) -- (.5,5); + \draw [line width = 3pt, color = magenta, <-] + (5,5) -- (4.2,4.2); + \draw [line width = 3pt, color = magenta, ->] + (1,1) -- (3.8,3.8); + \draw [line width = 3pt, color = magenta, ->] + (6,5.5) .. controls (7.3,5) and (8.2,4) .. (8.5,3.3); + \draw [line width = 3pt, color = magenta, <-] + (6,.8) .. controls (7.3,1) and (8.2,2) .. (8.5,2.7); + \draw [line width = 3pt, color = magenta, <-] + (1,5.2) .. controls (1.4,5.2) and (2,4.5) .. (2.3,4.2); + \draw [line width = 3pt, color = magenta, ->] + (5.2,1) .. controls (4.7,1.8) and (3.5,3) .. (2.7,3.8); + + + \filldraw [blue] (0,5) -- (1,5) -- (1,6) -- + (0,6) -- (0,5); + \draw [line width = 1.5pt] (0,5) -- + (1,5) -- (1,6) -- (0,6) -- (0,5); + \filldraw [green] (5,5) -- (6,5) -- (6,6) -- + (5,6) -- (5,5); + \draw [line width = 1.5pt] (5,5) -- + (6,5) -- (6,6) -- (5,6) -- (5,5); + \filldraw [red] (0,0) -- (1,0) -- (1,1) -- + (0,1) -- (0,0); + \draw [line width = 1.5pt] (0,0) -- + (1,0) -- (1,1) -- (0,1) -- (0,0); + \draw [line width = 1.5pt] (5,0) -- + (6,0) -- (6,1) -- (5,1) -- (5,0); + \draw (-.5,-.5) circle (.3) node {1}; + \draw [line width = 1.5pt] (0,.5) .. + controls (-.9,.4) and (-.65,-.2) .. (-.7,-.3); + \draw [line width = 1.5pt] (.5,0) .. + controls (.4,-.9) and (-.2,-.65) .. (-.3,-.7); + \draw (.5,2) circle (.3) node {1}; + \draw (5.5,4) circle (.3) node {1}; + \draw [line width = 1.5pt] (5.5,5) -- (5.5,4.3); + \draw [line width = 1.5pt] (5.5,3.7) -- (5.5,1); + \draw (1.5,3.5) circle (.3) node {2}; + \draw [line width = 1.5pt] (.8,5) .. + controls (.8,4.6) and (1,4.2) .. (1.3,3.7); + \draw [line width = 1.5pt] (5,.8) .. + controls (2.5,1.5) and (2.1,2.5) .. (1.6,3.22); + \draw (4,4) circle (.3) node {2}; + \draw (4,.5) circle (.3) node {2}; + \draw [line width = 1.5pt] (1,.5) -- (3.7,.5); + \draw [line width = 1.5pt] (5,.5) -- (4.3,.5); + \draw (-1,3) circle (.3) node {3}; + \draw [line width = 1.5pt] (0,5) .. + controls (-.7,4.3) and (-.9,3.6) .. (-1,3.3); + \draw [line width = 1.5pt] (0,1) .. + controls (-.7,1.7) and (-.9,2.3) .. (-1,2.7); + \draw (2.5,4) circle (.3) node {3}; + \draw (7,3) circle (.3) node {3}; + \draw [line width = 1.5pt] (6,5) .. + controls (6.7,4.3) and (6.9,3.6) .. (7,3.3); + \draw [line width = 1.5pt] (6,1) .. + controls (6.7,1.7) and (6.9,2.3) .. (7,2.7); + \draw (-.5,6.5) circle (.3) node {4}; + \draw [line width = 1.5pt] (0,5.5) .. + controls (-.9,5.4) and (-.65,6.2) .. (-.7,6.26); + \draw [line width = 1.5pt] (-.24,6.65) .. + controls (.1,6.7) and (.45,6.7) .. (.5,6); + \draw (8.5,3) circle (.3) node {4}; + \draw (2.3,1.2) circle (.3) node {4}; + \draw [line width = 1.5pt] (1,.8) -- (2.04,1.1); + \draw [line width = 1.5pt] (2.6,1.3) .. + controls (4,2) and (5,4) .. (5.2,5); + + + \node[text = white] at (0.5, 5.5) {\textbf{Blue}}; + \node[text = white] at (0.5, 0.5) {\textbf{Red}}; + \node[text = black] at (5.5, 0.5) {\textbf{Wht}}; + \node[text = black] at (5.5, 5.5) {\textbf{Grn}}; + \end{tikzpicture} + \end{small} \end{center} + \bigskip + + The first leg of the walk tells us + to take Cube 1 and to make sure + that its blue side is facing forward. + Then the red side, opposite to the blue one, + will face the rear. + + \begin{center} + \begin{tikzpicture} + \coordinate [label=center:{Front:}] (f) at (0,.5); + \filldraw [blue] (1,0) -- (2,0) -- (2,1) -- (1,1) -- (1,0); + \draw [line width = 1.5pt] (1,0) -- (2,0) -- + (2,1) -- (1,1) -- (1,0); + + \coordinate [label=center:{Rear:}] (f) at (0,-1); + \filldraw [red] (1,-1.5) -- (2,-1.5) -- (2,-.5) -- + (1,-.5) -- (1,-1.5); + \draw [line width = 1.5pt] (1,-1.5) -- (2,-1.5) -- (2,-.5) -- + (1,-.5) -- (1,-1.5); + + \node[text = white] at (1.5, 0.5) {\textbf{Blue}}; + \node[text = white] at (1.5, -1) {\textbf{Red}}; + \end{tikzpicture} + \end{center} + \bigskip + + The next leg of the walk tells us + to take Cube 2 and to place it in such a way + that its red side faces + us while the opposing green side faces the rear. + Since we go in a cycle that visits all the colors + one-by-one, neither color repeats the ones + already used on their sides of the stack. \\ + + \begin{center} + \begin{tikzpicture} + \coordinate [label=center:{Front:}] (f) at (0,.5); + \filldraw [blue] (1,0) -- (2,0) -- (2,1) -- (1,1) -- (1,0); + \filldraw [red] (2,0) -- (3,0) -- (3,1) -- (2,1) -- (2,0); + \draw [line width = 1.5pt] (1,0) -- (2,0) -- + (2,1) -- (1,1) -- (1,0); + \draw [line width = 1.5pt] (2,0) -- (3,0) -- + (3,1) -- (2,1); + + \coordinate [label=center:{Rear:}] (f) at (0,-1); + \filldraw [red] (1,-1.5) -- (2,-1.5) -- (2,-.5) -- + (1,-.5) -- (1,-1.5); + \filldraw [green] (2,-1.5) -- (3,-1.5) -- (3,-.5) -- + (2,-.5) -- (2,-1.5); + \draw [line width = 1.5pt] (1,-1.5) -- (2,-1.5) -- (2,-.5) -- + (1,-.5) -- (1,-1.5); + \draw [line width = 1.5pt] (2,-1.5) -- (3,-1.5) -- + (3,-.5) -- (2,-.5); + + \node[text = white] at (1.5, 0.5) {\textbf{Blue}}; + \node[text = black] at (2.5, 0.5) {\textbf{Red}}; + \node[text = white] at (1.5, -1) {\textbf{Red}}; + \node[text = black] at (2.5, -1) {\textbf{Grn}}; + + \end{tikzpicture} + \end{center} + \bigskip + + The third leg of the walk tells us + to take Cube 4, not Cube 3, and to place it + green side forward, white side facing the rear. \\ + + \begin{center} + \begin{tikzpicture} + \coordinate [label=center:{Front:}] (f) at (0,.5); + \filldraw [blue] (1,0) -- (2,0) -- (2,1) -- (1,1) -- (1,0); + \filldraw [red] (2,0) -- (3,0) -- (3,1) -- (2,1) -- (2,0); + \filldraw [green] (4,0) -- (5,0) -- (5,1) -- (4,1) -- (4,0); + \draw [line width = 1.5pt] (1,0) -- (2,0) -- + (2,1) -- (1,1) -- (1,0); + \draw [line width = 1.5pt] (2,0) -- (3,0) -- + (3,1) -- (2,1); + \draw [line width = 1.5pt] (4,0) -- (5,0) -- + (5,1) -- (4,1) -- (4,0); + + \coordinate [label=center:{Rear:}] (f) at (0,-1); + \filldraw [red] (1,-1.5) -- (2,-1.5) -- (2,-.5) -- + (1,-.5) -- (1,-1.5); + \filldraw [green] (2,-1.5) -- (3,-1.5) -- (3,-.5) -- + (2,-.5) -- (2,-1.5); + \draw [line width = 1.5pt] (1,-1.5) -- (2,-1.5) -- (2,-.5) -- + (1,-.5) -- (1,-1.5); + \draw [line width = 1.5pt] (2,-1.5) -- (3,-1.5) -- + (3,-.5) -- (2,-.5); + \draw [line width = 1.5pt] (4,-1.5) -- (5,-1.5) -- + (5,-.5) -- (4,-.5) -- (4,-1.5); + + \node[text = white] at (1.5, 0.5) {\textbf{Blue}}; + \node[text = black] at (2.5, 0.5) {\textbf{Red}}; + \node[text = white] at (1.5, -1) {\textbf{Red}}; + \node[text = black] at (2.5, -1) {\textbf{Grn}}; + + \node[text = black] at (4.5, 0.5) {\textbf{Grn}}; + \node[text = black] at (4.5, -1) {\textbf{Wht}}; + + \end{tikzpicture} + \end{center} + \bigskip + + Finally, the last leg of the walk + tells us to take Cube 3 and to place it + the white side facing forward, + the opposite blue side facing the rear. \\ + + \begin{center} + \begin{tikzpicture} + \coordinate [label=center:{Front:}] (f) at (0,.5); + \filldraw [blue] (1,0) -- (2,0) -- (2,1) -- (1,1) -- (1,0); + \filldraw [red] (2,0) -- (3,0) -- (3,1) -- (2,1) -- (2,0); + \filldraw [green] (4,0) -- (5,0) -- (5,1) -- (4,1) -- (4,0); + \draw [line width = 1.5pt] (1,0) -- (2,0) -- + (2,1) -- (1,1) -- (1,0); + \draw [line width = 1.5pt] (2,0) -- (3,0) -- + (3,1) -- (2,1); + \draw [line width = 1.5pt] (4,0) -- (5,0) -- + (5,1) -- (4,1) -- (4,0); + \draw [line width = 1.5pt] (3,0) -- (4,0); + \draw [line width = 1.5pt] (3,1) -- (4,1); + + \coordinate [label=center:{Rear:}] (f) at (0,-1); + \filldraw [red] (1,-1.5) -- (2,-1.5) -- (2,-.5) -- + (1,-.5) -- (1,-1.5); + \filldraw [green] (2,-1.5) -- (3,-1.5) -- (3,-.5) -- + (2,-.5) -- (2,-1.5); + \filldraw [blue] (3,-1.5) -- (4,-1.5) -- (4,-.5) -- + (3,-.5) -- (3,-1.5); + \draw [line width = 1.5pt] (1,-1.5) -- (2,-1.5) -- (2,-.5) -- + (1,-.5) -- (1,-1.5); + \draw [line width = 1.5pt] (2,-1.5) -- (3,-1.5) -- + (3,-.5) -- (2,-.5); + \draw [line width = 1.5pt] (4,-1.5) -- (5,-1.5) -- + (5,-.5) -- (4,-.5) -- (4,-1.5); + \draw [line width = 1.5pt] (3,-1.5) -- (4,-1.5); + \draw [line width = 1.5pt] (3,-.5) -- (4,-.5); + + \node[text = white] at (1.5, 0.5) {\textbf{Blue}}; + \node[text = black] at (2.5, 0.5) {\textbf{Red}}; + \node[text = white] at (1.5, -1) {\textbf{Red}}; + \node[text = black] at (2.5, -1) {\textbf{Grn}}; + \node[text = black] at (3.5, 0.5) {\textbf{Wht}}; + \node[text = white] at (3.5, -1) {\textbf{Blue}}; + \node[text = black] at (4.5, 0.5) {\textbf{Grn}}; + \node[text = black] at (4.5, -1) {\textbf{Wht}}; + \end{tikzpicture} + \end{center} + \bigskip + + Now the front and rear of the stack are done. + If we manage to find a second oriented cycle + in the original graph that has all the properties + of the first cycle, but uses none of its edges, + we would be able to do the upper and lower sides + of the stack and to complete the puzzle. + Using the edges we have already traversed + during our first walk will mess up + the front-rear configuration, but there are still + a plenty of the edges left! + + \problem{} + Complete the puzzle. + + \vfill + \pagebreak + + + \section{Traveling salesman problem} + + \problem{} \label{pr:tsp} + A salesman with the home office + in Albuquerque has to fly to Boston, + Chicago, and Denver, visiting each city once, + and then to come back to the home office. + The airfare prices, shown on the graph below, + do not depend on the direction of the travel. + Find the cheapest way. \\ + \note{This was on last week's handout, but not everyone had the chance to solve it.} + + \begin{center} + \begin{normalsize} + \begin{tikzpicture} + \tikzset{EdgeStyle/.append style = {-}} + \SetGraphUnit{3} + \Vertex{A} + \SOWE(A){B} + \SOEA(A){C} + \SO(A){D} + \Edge(A)(B) + \Edge(A)(C) + \Edge(A)(D) + \Edge(B)(D) + \Edge(C)(D) + \tikzset{EdgeStyle/.append style = {bend right = 70}} + \Edge(B)(C) + \coordinate [label=left:{\$1400}] (ab) at (-1.8,-1.75); + \coordinate [label=right:{\$1000}] (ac) at (1.8,-1.75); + \coordinate [label=right:{\$400}] (ad) at (-.1,-1.75); + \coordinate [label=below:{\$800}] (bc) at (0,-4.8); + \coordinate [label=below:{\$1200}] (bd) at (-1.5,-3); + \coordinate [label=below:{\$900}] (cd) at (1.5,-3); + \end{tikzpicture} + \end{normalsize} + \end{center} + + \vfill + \pagebreak + + \section{Planar graphs} + \label{sec:PG} + + A graph is called {\it planar}, + if it can be drawn in the plane in such a way + that no edges cross one another. + \problem{} + Show that the following graph is planar. + + \begin{center} + \tikzset{EdgeStyle/.append style = {-}} + \begin{tikzpicture} [scale = .8] + \SetGraphUnit{3.5} + \draw [color = white] (0,0) -- (0,-5); + \Vertex{A} + \EA(A){B} + \SO(B){C} + \SO(A){D} + \Edge(A)(B) + \Edge(A)(C) + \Edge(A)(D) + \Edge(B)(C) + \Edge(B)(D) + \Edge(C)(D) + \end{tikzpicture} + \end{center} + \vfill + + \problem{} + Is it possible to connect three houses, + A, B, and C, to three utility sources, + water (W), gas (G), and electricity (E), + without using the third dimension, + either on the plane or sphere, + so that the utility lines do not intersect? \\ + + \begin{center} + \begin{tikzpicture} [scale = .8] + \draw (0,0) circle (.5); + \coordinate [label=center:{W}] (w) at (-.05,-.03); + \draw (4,0) circle (.5); + \coordinate [label=center:{G}] (g) at (3.95,-.03); + \draw (8,0) circle (.5); + \coordinate [label=center:{E}] (e) at (7.95,-.03); + \draw (0,3) circle (.5); + \coordinate [label=center:{A}] (a) at (-.05,2.97); + \draw (4,3) circle (.5); + \coordinate [label=center:{B}] (b) at (3.95,2.97); + \draw (8,3) circle (.5); + \coordinate [label=center:{C}] (c) at (7.95,2.97); + \end{tikzpicture} + \end{center} + + \vfill + \pagebreak + + A {\it subdivision} of a graph $G$ + is a graph resulting from the subdivision + of the edges of $G$. The subdivision + of an edge $e = (v_1,v_2)$ is a graph + containing one new vertex $v_3$, + with the edges $e_1 = (v_1,v_3)$ and + $e_2 = (v_3,v_2)$ replacing the edge $e$. \\ + + \begin{center} + \begin{tikzpicture} [scale = .8] + \draw (0,0) circle (.5); + \coordinate [label=center:{$v_1$}] (v12) at (0,-.03); + \draw (4,0) circle (.5); + \coordinate [label=center:{$v_3$}] (v3) at (4,-.03); + \draw (8,0) circle (.5); + \draw (.5,0) -- (3.5,0); + \draw (4.5,0) -- (7.5,0); + \draw (.5,3) -- (7.5,3); + \coordinate [label=center:{$v_2$}] (v22) at (8,-.03); + \draw (0,3) circle (.5); + \coordinate [label=center:{$v_1$}] (v11) at (0,2.97); + \draw (8,3) circle (.5); + \coordinate [label=center:{$v_2$}] (v21) at (8,2.97); + \coordinate [label=above:{$e$}] (e) at (4,3); + \coordinate [label=above:{$e_1$}] (e1) at (2,0); + \coordinate [label=above:{$e_2$}] (e2) at (6,0); + \end{tikzpicture} + \end{center} + + \problem{} + What is the degree of a subdivision vertex? + + \vfill + + A graph $H$ is called a {\it subgraph} + of a graph $G$ if the sets of vertices + and edges of $H$ are subsets of the sets + of vertices and edges of $G$. \\ + + The following graphs are known as $K_{3,3}$ + and $K_5$. \\ + + \begin{center} + \begin{tikzpicture} %[scale = .8] + \filldraw (0,0) circle (3pt); + \filldraw (2,0) circle (3pt); + \filldraw (4,0) circle (3pt); + \filldraw (0,2) circle (3pt); + \filldraw (2,2) circle (3pt); + \filldraw (4,2) circle (3pt); + \draw (0,0) -- (0,2); + \draw (0,0) -- (2,2); + \draw (0,0) -- (4,2); + \draw (2,0) -- (0,2); + \draw (2,0) -- (2,2); + \draw (2,0) -- (4,2); + \draw (4,0) -- (0,2); + \draw (4,0) -- (2,2); + \draw (4,0) -- (4,2); + \coordinate [label=below:{$K_{3,3}$}] (l) at (2,-.2); + \end{tikzpicture} \hspace{60pt} + \begin{tikzpicture} %[scale = .8] + \filldraw (90:2) circle (3pt); + \filldraw (162:2) circle (3pt); + \filldraw (234:2) circle (3pt); + \filldraw (306:2) circle (3pt); + \filldraw (18:2) circle (3pt); + \draw (90:2) -- (162:2); + \draw (90:2) -- (234:2); + \draw (90:2) -- (306:2); + \draw (90:2) -- (18:2); + \draw (162:2) -- (234:2); + \draw (162:2) -- (306:2); + \draw (162:2) -- (18:2); + \draw (234:2) -- (306:2); + \draw (234:2) -- (18:2); + \draw (306:2) -- (18:2); + \coordinate [label=below:{$K_5$}] (l) at (0,-1.7); + \end{tikzpicture} + \end{center} + \vspace{15pt} + + Let $H$ be a graph that is a subdivision + of either $K_{3,3}$ or $K_5$. If $H$ is + a subgraph of a graph $G$, then $H$ is called + a {\it Kuratowski subgraph}, + after a famous Polish mathematician + Kazimierz Kuratowski (1896-1980). \\ + + \pagebreak + + \theorem{} + A graph is planar if and only if + it has no Kuratowski subgraph. + + \problem{} + Is the following graph planar? + Why or why not? \\ + + \begin{center} + \begin{tikzpicture} [scale = .8] + \filldraw (90:2) circle (3pt); + \filldraw (150:2) circle (3pt); + \filldraw (210:2) circle (3pt); + \filldraw (270:2) circle (3pt); + \filldraw (330:2) circle (3pt); + \filldraw (30:2) circle (3pt); + \draw (90:2) -- (150:2); + \draw (90:2) -- (210:2); + \draw (90:2) -- (270:2); + \draw (90:2) -- (330:2); + \draw (90:2) -- (30:2); + \draw (150:2) -- (210:2); + \draw (150:2) -- (270:2); + \draw (150:2) -- (330:2); + \draw (150:2) -- (30:2); + \draw (210:2) -- (270:2); + \draw (210:2) -- (330:2); + \draw (210:2) -- (330:2); + \draw (210:2) -- (30:2); + \draw (270:2) -- (330:2); + \draw (270:2) -- (30:2); + \draw (330:2) -- (30:2); + \end{tikzpicture} + \end{center} + + \vfill + + \problem{} + Is the following graph planar? + Why or why not? + + \begin{center} + \includegraphics[width=2.2in]{4Dcube2.jpg} + \end{center} + + \vfill + \pagebreak + + + \section{Euler characteristic} + + Let $G$ be a planar graph, + drawn with no edge intersections. + The edges of $G$ divide the plane into regions, + called {\it faces}. The regions + enclosed by the graph are called + the {\it interior faces}. + The region surrounding the graph is called + the {\it exterior (or infinite) face}. + The faces of $G$ include both the interior faces + and the exterior one. For example, + the following graph has two interior faces, + $F_1$, bounded by the edges $e_1$, $e_2$, $e_4$; + and $F_2$, bounded by the edges $e_1$, $e_3$, $e_4$. + Its exterior face, $F_3$, is bounded by the edges + $e_2$, $e_3$. + + \begin{center} \label{pic:nsgrap} + \begin{tikzpicture} [scale = .6] + \SetGraphUnit{5} + \Vertex{B} + \WE(B){A} + \EA(B){C} + \Edge(B)(A) + \Edge(C)(B) + \tikzset{EdgeStyle/.append style = {bend left = 50}} + \Edge(A)(C) + \Edge(C)(A) + \coordinate [label=above:{$e_1$}] (e1) at (-2.1,.0); + \coordinate [label=above:{$e_2$}] (e2) at (0,2.45); + \coordinate [label=below:{$e_3$}] (e3) at (0,-2.5); + \coordinate [label=above:{$e_4$}] (e4) at (2.1,.0); + \end{tikzpicture} + \end{center} + + The {\it Euler characteristic} of a graph + is the number of the graph's vertices minus + the number of the edges plus the number of the faces. + \begin{equation} + \chi = V - E + F + \end{equation} + + \problem{} + Compute the Euler characteristic + of the graph above. + \vfill + + \problem{} + Compute the Euler characteristic + of the following graph. + + + \begin{center} + \begin{tikzpicture} [scale = .8] + \draw (.5,3) -- (7.5,3); + \draw (0,3) circle (.5); + \coordinate [label=center:{$v_1$}] (v11) at (0,2.97); + \draw (8,3) circle (.5); + \coordinate [label=center:{$v_2$}] (v21) at (8,2.97); + \coordinate [label=above:{$e$}] (e) at (4,3); + \end{tikzpicture} + \end{center} + + \vfill + \pagebreak + + \problem{}<3Dcube> + Is the following graph planar? + If you think it is, please re-draw + the graph so that it has no intersecting edges. + If you think the graph is not planar, + please explain why. \\ + + \begin{center} + \begin{tikzpicture} + \draw (1,1) -- (4,1) -- (4,4) -- (1,4) -- (1,1); + \draw (0,0) -- (3,0) -- (3,3) -- (0,3) -- (0,0); + \draw (0,0) -- (1,1); + \draw (3,0) -- (4,1); + \draw (0,3) -- (1,4); + \draw (3,3) -- (4,4); + \filldraw (0,0) circle (3pt); + \filldraw (3,0) circle (3pt); + \filldraw (0,3) circle (3pt); + \filldraw (3,3) circle (3pt); + \filldraw (4,1) circle (3pt); + \filldraw (1,1) circle (3pt); + \filldraw (1,4) circle (3pt); + \filldraw (4,4) circle (3pt); + \end{tikzpicture} + \end{center} + + \vfill + + \problem{} + Compute the Euler characteristic of the graph + from Problem \ref{3Dcube}. + + \vfill + \pagebreak + + Let us consider the below picture + of a {\it regular dodecahedron} as a graph, + the vertices representing those of the graph, + and the edges, both solid and dashed, + representing the edges of the graph. + + \problem{} + Is the graph planar? + If you think it is planar, + please re-draw the graph so that it has + no intersecting edges. If you think the graph + is not planar, please explain why. \\ + + \begin{center} + \includegraphics[width=2.5in] + {dodecahedron.jpg} + \end{center} + + \vfill + + \problem{} + Compute the Euler characteristic of the graph + from Problem \ref{dodec}. + Can you conjecture what the Euler characteristic + of every planar graph is equal to? + + \vfill + \pagebreak + + A graph is called a {\it tree} + if it is connected and has no cycles. + Here is an example. \\ + + \begin{center} + \begin{tikzpicture} [scale = .6] + \SetGraphUnit{5} + \Vertex{A} + \SOWE(A){B} + \SO(A){C} + \SOEA(A){D} + \SOWE(B){E} + \SO(B){F} + \SOEA(D){G} + \Edge(A)(B) + \Edge(A)(C) + \Edge(A)(D) + \Edge(B)(E) + \Edge(B)(F) + \Edge(D)(G) + %\tikzset{EdgeStyle/.append style = {bend left = 50}} + \end{tikzpicture} + \end{center} + \bigskip + + A path is called {\it simple} + if it does not include any of its edges + more than once. + + \problem{} + Prove that a graph in which any two vertices are connected by one + and only one simple path is a tree. + + \problem{} + What is the Euler characteristic + of a finite tree? + + \theorem{} + Let a finite connected planar graph have + $V$ vertices, $E$ edges, and $F$ faces. + Then $V - E + F = 2$. + + \problem{} + Prove Theorem \ref{eu_char}. + Hint: removing an edge from a cycle + does not change the number of vertices + and reduces the number of edges and faces + by one. + + \problem{} + There are three ponds in a botanical garden, + connected by ten non-intersecting brooks + so that the ducks can sweem from any pond to any other. + How many islands are there in the garden? + + \problem{} + All the vertices of a finite graph + have degree three. + Prove that the graph has a cycle. + + \problem{} + Draw an infinite tree with every vertex + of degree three. + + \problem{} + Prove that a connected finite graph is a tree + if and only if $V = E + 1$. + + \problem{} + Give an example of a finite graph + that is not a tree, + but satisfies the relation $V = E + 1$. + +\end{document} \ No newline at end of file diff --git a/Intermediate/Slide Rules/main.tex b/Intermediate/Slide Rules/main.tex new file mode 100755 index 0000000..60a1016 --- /dev/null +++ b/Intermediate/Slide Rules/main.tex @@ -0,0 +1,794 @@ +% https://git.betalupi.com/Mark/latex-packages +% use [nosolutions] flag to hide solutions. +% use [solutions] flag to show solutions. +% Last built with version 1.1.0 +\documentclass[ + solutions +]{ormc_handout} + +\usepackage{pdfpages} +\usepackage{sliderule} + +% 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}; +} + +\begin{document} + + \maketitle + + + {Slide Rules} + { + Prepared by Mark on \today + } + + \vspace{1cm} + + \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} + \hfill + + \section{Logarithms} + + \definition{} + The \textit{logarithm} is the inverse of the exponent. That is, if $b^p = c$, then $\log_b{c} = p$. \\ + In other words, $\log_b{c}$ asks the question ``what power do I need to raise $b$ to to get $c$?'' \\ + + \medskip + + 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 + + + \definition{} + There are a few ways to write logarithms: + \begin{itemize} + \item[] $\log{x} = \log_{10}{x}$ + \item[] $\lg{x} = \log_{10}{x}$ + \item[] $\ln{x} = \log_e{x}$ + \end{itemize} + + \definition{} + The \textit{domain} of a function is the set of values it can take as inputs. \\ + The \textit{range} of a function is the set of values it can produce. + + \medskip + + For example, the domain and range of $f(x) = x$ is $\mathbb{R}$, all real numbers. \\ + The domain of $f(x) = |x|$ is $\mathbb{R}$, and its range is $\mathbb{R}^+ \cup \{0\}$, all positive real numbers and 0. \\ + + \medskip + + Note that the domain and range of a function are not always equal. + + \problem{} + What is the domain of $f(x) = 5^x$? \\ + What is the range of $f(x) = 5^x$? + \vfill + + \problem{} + What is the domain of $f(x) = \log{x}$? \\ + What is the range of $f(x) = \log{x}$? + \vfill + + \pagebreak + + + \problem{} + 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. \\ + If you do the same with a log scale, you multiply! \\ + \vspace{1ex} + 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 + + \section{Introduction} + + Mathematicians, physicists, and engineers needed to quickly solve complex equations even before computers were invented. + + \medskip + + The \textit{slide rule} is an instrument that uses the logarithm to solve this problem. Before you continue, cut out and assemble your slide rule. + + \medskip + + 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 + + \section{Multiplication} + + We'll use the C and D scales of your slide rule to multiply. \\ + + 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 $1.15 \times 2.1$? \\ + 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 $32 \times 210$? \\ + + \problem{} + Using your slide rule, calculate $32 \times 210$. \\ + %\hint{$32 = 3.2 \times 10^1$} + + \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. \\ + $10^1 \times 10^2 = 10^3$, so our final answer is $6.72 \times 10^3 = 672$. + \end{solution} + + \vfill + + %This method of writing numbers is called \textit{scientific notation}. In the form $a \times 10^b$, $a$ is called the \textit{mantissa}, and $b$, the \textit{exponent}. \\ + + %You may also see expressions like $4.3\text{e}2$. This is equivalent to $4.3 \times 10^2$, but is more compact. + + + \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 + \pagebreak + + \problem{} + 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? \\ + %\hint{See \ref{logids}} + + \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. \\ + + \medskip + + 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. \\ + + \medskip + + \iftrue + \problem{} + Why does this work? + + \else + Why does this work? \\ + + \medskip + + Consider the following picture, where I've put 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} + + \medskip + + 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. + + \medskip + + \medskip + In other words, the answer we get from reverse multiplication is the following: $\log{a} + \log{b} - \log{10}$. \\ + This reduces to $\log{(\frac{a \times b}{10})}$, which explains the misplaced decimal point in $7.2 \times 5.5$. + \fi + + \vfill + \pagebreak + + \problem{} + Compute the following using your slide rule: + \begin{enumerate} + \item $9 \times 8$ + \item $15 \times 35$ + \item $42.1 \times 7.65$ + \item $6.5^2$ + \end{enumerate} + + \begin{solution} + \begin{enumerate} + \item $9 \times 8 = 72$ + \item $15 \times 35 = 525$ + \item $42.1 \times 7.65 = 322.065$ + \item $6.5^2 = 42.25$ + \end{enumerate} + \end{solution} + + \vfill + \pagebreak + + \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 + \pagebreak + + \problem{} + Compute the following using your slide rule. \\ + + \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 + + + + \section{Squares, Cubes, and Roots} + + Now, take a look at scales A and B, and note the label on the right: $x^2$. If C, D are $x$, A and B are $x^2$, and K is $x^3$. + + \medskip + + Finding squares of numbers up to ten is straightforward: just read the scale. \\ + Square roots are also easy: find your number on B and read its pair on C. \\ + + + \def\sliderulewidth{13} + \begin{center} + \begin{tikzpicture}[scale=1] + \abscale{0}{1}{B} + \cdscale{0}{0}{C} + \end{tikzpicture} + \end{center} + + \problem{} + Compute the following. + \begin{enumerate} + \item $1.5^2$ + \item $3.1^2$ + \item $7^3$ + \item $\sqrt{14}$ + \item $\sqrt[3]{150}$ + \end{enumerate} + + \begin{solution} + \begin{enumerate} + \item $1.5^2 = 2.25$ + \item $3.1^2 = 9.61$ + \item $7^3 = 343$ + \item $\sqrt{14} = 3.74$ + \item $\sqrt[3]{150} = 5.313$ + \end{enumerate} + \end{solution} + + \vfill + \problem{} + Compute the following. + \begin{enumerate} + \item $42^2$ + \item $\sqrt{200}$ + \item $\sqrt{2000}$ + \item $\sqrt{0.9}$ + \item $\sqrt[3]{0.12}$ + \end{enumerate} + + \begin{solution} + \begin{enumerate} + \item $42^2 = 1,764$ + \item $\sqrt{200} = 14.14$ + \item $\sqrt{2000} = 44.72$ + \item $\sqrt{0.9} = 0.948$ + \item $\sqrt[3]{0.12} = 0.493$ + \end{enumerate} + \end{solution} + + + \vfill + \pagebreak + + \section{Inverses} + + Try finding $1 \div 32$ using your slide rule. \\ + The procedure we learned before doesn't work! + + \medskip + + This is why we have the CI scale, or the ``C Inverse'' scale. + + \problem{} + Figure out how the CI scale works and compute the following: + \begin{enumerate}[itemsep=1mm] + \item $\frac{1}{7}$ + \item $\frac{1}{120}$ + \item $\frac{1}{\pi}$ + \end{enumerate} + + \vfill + \pagebreak + + \section{Logarithms Base 10} + + When we take a logarithm, the resulting number has two parts: the \textit{characteristic} and the \textit{mantissa}. \\ + The characteristic is the integral (whole-numbered) part of the answer, and the mantissa is the fractional part (what comes after the decimal). \\ + + \medskip + + For example, $\log_{10}{18} = 1.255$, so in this case the characteristic is $1$ and the mantissa is $0.255$. + + \problem{} + Approximate the following logs without a slide rule. Find the exact characteristic, and approximate the mantissa. + \begin{enumerate} + \item $\log_{10}{20}$ + \item $\log_{2}{18}$ + \end{enumerate} + + \begin{solution} + \begin{enumerate} + \item $\log_{10}{20} = 1.30$ + \item $\log_{2}{18} = 4.17$ + \end{enumerate} + \end{solution} + + \vfill + + Now, find the L scale on your slide rule. As you can see on the right, its generating function is $\log_{10}{x}$. + + \problem{} + Compute the following logarithms using your slide rule. \\ + You'll have to find the characteristic yourself, but your L scale will give you the mantissa. \\ + Don't forget your log identities! + + \begin{enumerate} + \item $\log_{10}{20}$ + \item $\log_{10}{15}$ + \item $\log_{10}{150}$ + \item $\log_{10}{0.024}$ + \end{enumerate} + + \begin{solution} + Careful with number 4. + + \begin{enumerate} + \item $\log_{10}{20} = 1.30$ + \item $\log_{10}{15} = 1.176$ + \item $\log_{10}{150} = 2.176$ + \item $\log_{10}{0.024} = -1.6197$ + \end{enumerate} + \end{solution} + + \vfill + \pagebreak + + %\problem{} + %Find the following. + %\begin{enumerate}[itemsep=2mm] + % \item $\frac{118 \times 0.51}{6.6}$ + % \item $\sqrt{33.8} \times \sqrt[3]{226}$ + % \item $\frac{\sqrt{152}}{\sqrt[3]{4.95}}$ + % \item $\frac{\sqrt{96 \times 250}}{\sqrt{7 \times 0.88}}$ + % \item The area of a circle with radius $1.47$ + % \item The circumference of a circle with radius $31.4$ + % \item The radius of a circle with area $6\pi$ + % \item $\log_{10}{17.38}$ + %\end{enumerate} + %\vfill + %\pagebreak + + \section{Logarithms in Any Base} + + Our slide rule easily computes logarithms in base 10, but we can also use it to find logarithms in \textit{any} base. + + \proposition{} + This is usually called the \textit{change-of-base} formula: + + \[ + \log_{b}{a} = \frac{\log_c{a}}{\log_c{b}} + \] + + \problem{} + Using log identities, prove \ref{logcob}. + + \vfill + + \problem{} + Approximate the following: + \begin{enumerate} + \item $\log_{2}{56}$ + \item $\log_{5.2}{26}$ + \item $\log_{12}{500}$ + \item $\log_{43}{134}$ + \end{enumerate} + + \begin{solution} + \begin{enumerate} + \item $\log_{2}{56} = 5.81$ + \item $\log_{5.2}{26} = 1.97$ + \item $\log_{12}{500} = 2.50$ + \item $\log_{43}{134} = 1.30$ + \end{enumerate} + \end{solution} + + + + + % Make sure the slide rule is on an odd page, + % so that 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