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@ -38,7 +38,7 @@ Using the two theorems above, detail an algorithm for finding $\gcd(a, b)$. \par
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Then, compute $\gcd(1610, 207)$ by hand. \par
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\begin{solution}
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Using \ref{gcd_abc} and the division algorthm,
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Using \ref{gcd_abc} and the division algorithm,
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% Minipage prevents column breaks inside body
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\begin{multicols}{2}
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@ -36,7 +36,7 @@ Using the two theorems above, detail an algorithm for finding $\gcd(a, b)$. \par
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Then, compute $\gcd(1610, 207)$ by hand. \par
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\begin{solution}
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Using \ref{gcd_abc} and the division algorthm,
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Using \ref{gcd_abc} and the division algorithm,
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% Minipage prevents column breaks inside body
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\begin{multicols}{2}
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@ -16,7 +16,7 @@ This is the \textit{discrete logarithm problem}, often abbreviated \textit{DLP}.
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\problem{}
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Does the discrete log function even exist? \par
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Show that $\exp$ is a bijection, which will guarantee the existence of $\log$. \par
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\note[Note]{Why does this guarantee the existence of log? Recall our lesson on funtions.}
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\note[Note]{Why does this guarantee the existence of log? Recall our lesson on functions.}
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\vfill
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@ -510,7 +510,7 @@ Draw a state diagram for a DFA over an alphabet of your choice that accepts exac
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\end{tikzpicture}
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\end{center}
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This automaton rejects all strings with three \texttt{'a'}s in a row. If we count accepted strings, we get the Tribonacci numbers with an offest: $f(0) = 1$, $f(1) = 2$, $f(2)=4$, ... \par
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This automaton rejects all strings with three \texttt{'a'}s in a row. If we count accepted strings, we get the Tribonacci numbers with an offset: $f(0) = 1$, $f(1) = 2$, $f(2)=4$, ... \par
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\pagebreak
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@ -112,7 +112,7 @@ Define $\{-2, 2\}$ in $S$.
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\problem{}
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Let $P$ be the set of all subsets of $\mathbb{Z}^+_0$. This is called a \textit{power set}. \par
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Let $S$ be the stucture $( P ~|~ \{\subseteq\})$ \par
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Let $S$ be the structure $( P ~|~ \{\subseteq\})$ \par
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\problempart{}
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Show that the empty set is definable in $S$. \par
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@ -99,7 +99,7 @@ Show that if a sequence $a_n$ has a limit, that limit is unique. \par
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Let $N = \max(N_A, N_B)$. \par
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Then, $|a_n - A| + |a_n - B| < 2\epsilon\ \forall n > N$, \par
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which can be writen as $|a_n - A| + |B - a_n| < 2\epsilon\ \forall n > N$. \par
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which can be written as $|a_n - A| + |B - a_n| < 2\epsilon\ \forall n > N$. \par
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By the triangle inequality, we have \par
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$|a_n - A + B - a_n| \leq |a_n - A| + |B - a_n|$, \par
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@ -5,7 +5,7 @@ Say have a network: a sequence of pipes, a set of cities and highways, an electr
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\vspace{1ex}
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We can draw this network as a directed weighted graph. If we take a transporation network, for example, edges will represent highways and nodes will be cities. There are a few conditions for a valid network graph:
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We can draw this network as a directed weighted graph. If we take a transportation network, for example, edges will represent highways and nodes will be cities. There are a few conditions for a valid network graph:
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\begin{itemize}
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\item The weight of each edge represents its capacity, e.g, the number of lanes in the highway.
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@ -33,7 +33,7 @@
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% Lazy evaluation (alternate Y)
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% Add a few theorems
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% Better ending -> applications?
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% - nix, comparison to imperitive
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% - nix, comparison to imperative
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@ -332,7 +332,7 @@ If we look closely, we'll find that $C$ pretends to take three arguments.
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\vspace{1mm}
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What does $C$ do? Evaluate $(C~a~b~x)$ for arbitary expressions $a, b,$ and $x$. \par
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What does $C$ do? Evaluate $(C~a~b~x)$ for arbitrary expressions $a, b,$ and $x$. \par
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\hint{Evaluate $(C~a)$ first. Remember, function application is left-associative.}
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\vfill
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@ -1,6 +1,9 @@
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\section{Polya's Orchard Problem}
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You are standing in the center of a circular orchard of integer radius $R$. A tree of raduis $r$ has been planted at every integer point in the circle. If $r$ is small, you will have a clear line of sight through the orchard. If $r$ is large, there will be no clear line of sight through in any direction:
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You are standing in the center of a circular orchard of integer radius $R$.
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A tree of radius $r$ has been planted at every integer point in the circle.
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If $r$ is small, you will have a clear line of sight through the orchard.
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If $r$ is large, there will be no clear line of sight through in any direction:
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\begin{center}
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\hfill
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@ -58,7 +58,7 @@ Can you develop geometric intuition for their sum and difference?
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\definition{Euclidean Norm}
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A \textit{norm} on $\mathbb{R}^n$ is a map from $\mathbb{R}^n$ to $\mathbb{R}^+_0$ \\
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Usually, one thinks of a norm as a way of mesuring \say{length} in a vector space. \\
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Usually, one thinks of a norm as a way of measuring \say{length} in a vector space. \\
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The norm of a vector $v$ is written $||v||$. \\
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\vspace{2mm}
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@ -28,7 +28,7 @@ Show that the dot product is
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\begin{itemize}
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\item Commutative
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\item Distributive $a \cdot (b + c) = a \cdot b + a \cdot c$
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\item Homogenous: $x(a \cdot b) = xa \cdot b = a \cdot xb$ \\
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\item Homogeneous: $x(a \cdot b) = xa \cdot b = a \cdot xb$ \\
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\note{$x \in \mathbb{R}$, and $a, b$ are vectors.}
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\item Positive definite: $a \cdot a \geq 0$, with equality iff $a = 0$ \\
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\note{$a \in \mathbb{R}^n$, and $0$ is the zero vector.}
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@ -75,7 +75,7 @@ This is a trick we often use when showing that two quantities are equal.
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\problem{}
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Although $A \iff B$ looks like a single statement, we often need to prove each direction seperately. \par
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Although $A \iff B$ looks like a single statement, we often need to prove each direction separately. \par
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Show that $x \in \mathbb{Z}$ iff $\lfloor x \rfloor = \lceil x \rceil$
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\begin{solution}
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@ -97,7 +97,7 @@ Show that $x \in \mathbb{Z}$ iff $\lfloor x \rfloor = \lceil x \rceil$
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\pagebreak
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\problem{}
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We don't always need to prove each direction of an iff statement seperately. \par
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We don't always need to prove each direction of an iff statement separately. \par
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\begin{itemize}[itemsep = 1mm]
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\item Convince yourself that we can \say{chain} iffs together: \par
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@ -1,8 +1,8 @@
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\section{Proofs by Contradiction}
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\definition{}
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A very common (and somewhat contraversial) proof technique is
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\textit{proof by contradiction}. It works as follows:
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A very common proof technique is \textit{proof by contradiction}.
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It works as follows:
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\vspace{2mm}
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@ -18,7 +18,7 @@ Show that the set of integers has no maximum using a proof by contradiction.
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\begin{solution}
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Assume there is a maximal integer $x$. \par
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$x + 1$ is also an integer. \par
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$x + 1$ is larger than $x$, which contradicts our original assumtion!
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$x + 1$ is larger than $x$, which contradicts our original assumption!
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\vspace{2mm}
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@ -229,7 +229,7 @@ For example, see the proof of the statement in \ref{binomsum} on the next page.
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\textbf{Solution 2:}\par
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We could also observe that there are $x - 1$ places to put a \say{bar} in
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the array of ones. This corresponds to $x - 1$ binary positions, and thus
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$2^{x-1}$ ways to seperate our array of $1$s with bars.
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$2^{x-1}$ ways to separate our array of $1$s with bars.
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\linehack{}
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@ -103,7 +103,7 @@ Show that every element of $A$ is in \textit{exactly one} equivalence class\foot
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We now have a proper definition of \say{mod $n$:} \par
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it is the equivalence relation $a \equiv_n b$, which is usually written as $a \equiv b \pmod{n}$. \par
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We will use this definition thoughout this handout.
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We will use this definition throughout this handout.
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\note[Note]{
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This is different than the \say{mod} operator $a ~\%~ b $,
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@ -23,7 +23,7 @@ In the definitions below, let $X$ be the set of nodes in a circuit.
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Note that this is different than current and resistance, which aren't functions
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of two arbitrary nodes --- rather, they are functions of \textit{edges}
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(i.e, two adjecent nodes).
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(i.e, two adjacent nodes).
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}
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@ -136,7 +136,7 @@ It exists only to create a potential difference between the two nodes.
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\end{center}
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\problem{}<onecurrents>
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From the circuit diagram above, we immediatly know that $V(A) = 1$ and $V(B) = 0$. \par
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From the circuit diagram above, we immediately know that $V(A) = 1$ and $V(B) = 0$. \par
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What equations related to the currents out of $x$ and $y$ does Kirchoff's law give us? \par
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\hint{Current into $x$ = current out of $x$}
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@ -168,7 +168,7 @@ Using Ohm's law and Kirchoff's law, calculate the effective resistance $R_{\text
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\pagebreak
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We can now use effective resistance to simplify complicated circuits. Whenever we see the above constructions
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(resistors in parellel or in series) in a graph, we can replace them with a single resistor of appropriate value.
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(resistors in parallel or in series) in a graph, we can replace them with a single resistor of appropriate value.
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\problem{}
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@ -243,13 +243,13 @@ If we place $A$ and $B$ at opposing vertices, what is the effective resistance o
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and $B$ at \texttt{111...1}. We can divide our cube into $n+1$ layers based on how many ones are in each
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node's binary string, with the $k^\text{th}$ layer having $k$ ones. By symmetry, all the nodes in each layer
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have the same voltage. This means we can think of the layers as connected in series, with the resistors
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inside each layer connected in parellel.
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inside each layer connected in parallel.
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\vspace{2mm}
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There are $\binom{n}{k}$ nodes in the $k^\text{th}$ layer. Each node in this layer has $k$ ones, so
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there are $n - k$ ways to flip a zero to get to the $(k + 1)^\text{th}$ layer. In total, there are
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$\binom{n}{k}(n - k)$ parellel connections from the $k^\text{th}$ layer to the $(k + 1)^\text{th}$
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$\binom{n}{k}(n - k)$ parallel connections from the $k^\text{th}$ layer to the $(k + 1)^\text{th}$
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layer, creating an effective resistance of
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$$
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\frac{1}{\binom{n}{k}(n - k)}
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@ -197,7 +197,7 @@ There is one more piece on the board, which isn't shown. What color square does
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\begin{solution}
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Which piece performed the last capture on a black square? It couldn't have been a white pawn, which haven't moved.
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It couldn't have been the white king, which is trapped; or the black king, which is restriced to white squares.
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It couldn't have been the white king, which is trapped; or the black king, which is restricted to white squares.
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\vspace{2mm}
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@ -76,7 +76,7 @@ Show that he must be on C7.
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The pawn on A6 came from B7 by capturing one piece, and the pawn on B5 came from D7 by capturing two.
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(Note that this may not be true if we don't assume the pawn on C7 is real.)
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We've counted three captures, all on white squares, so the white black-square bishop must have been captured seperately.
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We've counted three captures, all on white squares, so the white black-square bishop must have been captured separately.
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\vspace{2mm}
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@ -7,7 +7,7 @@ This means that the sets $\{1, 2, 3\}$ and $\{3, 2, 1\}$ are identical.
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\definition{}
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A \textit{graph} $G = (N, E)$ consists of two sets: a set of \textit{vertices} $V$, and a set of \textit{edges} $E$. \par
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Verticies are simply named \say{points,} and edges are connections between pairs of vertices. \par
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Vertices are simply named \say{points,} and edges are connections between pairs of vertices. \par
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In the graph below, $V = \{a, b, c, d\}$ and $E = \{~ (a,b),~ (a,c),~ (a,d),~ (c,d) ~\}$.
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\begin{center}
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@ -252,7 +252,7 @@
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\problem{}
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A pharmaceutical study shows that a new drug causes negative side effects in 3 of every 100 patients.
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To check the number, a researcher chooses 5 random people to survey.
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Assuming the study is accurate, what is the probabilty of the following? \\
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Assuming the study is accurate, what is the probability of the following? \\
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\begin{enumerate}
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\item None of the five patients experience side effects.
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@ -2,7 +2,7 @@
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\begin{document}
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\problem{One Hundred and Fourty-Five Doors}
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\problem{One Hundred and Forty-Five Doors}
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A prisoner was thrown into a medieval dungeon with 145 doors. Nine, shown by black bars, are locked, but each one will open if before you reach it you pass through exactly 8 open doors. \\
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You don't have to go through every open door but you do have to go through every cell and all 9 locked doors. If you enter a cell or go through a door a second time, the doors clang shut, trapping you. \\
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\begin{document}
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\problem{One Hundred and Fourty-Five Doors}
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\problem{The Air Parade}
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Volodya asked, ``What plane did you fly during the air parade?'' His father sketched a formation of 9 planes.
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@ -151,7 +151,7 @@
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\problem{}
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If you'd like to know more, check out \url{https://regexr.com}. It offers an interative regex prompt, as well as a cheatsheet that explains every other regex token there is. \par
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If you'd like to know more, check out \url{https://regexr.com}. It offers an interactive regex prompt, as well as a cheatsheet that explains every other regex token there is. \par
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You will find a nice set of challenges at \url{https://alf.nu/RegexGolf}.
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I especially encourage you to look into this if you are interested in computer science.
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\end{document}
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\ifnum #1 > 0
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% The $$ are required around \bigstar.
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% the special \odot chess piece breaks
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% star sizing if they are ommited.
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% star sizing if they are omitted.
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\loop
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{\oncolor $\bigstar$}%
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\ifnum\count255 < #1
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\IfNoValueF{#2}{\@customlabel{#2}{#1}}
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}
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% Make a new secion type.
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% Make a new section type.
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% Args: command, counter, title.
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\newcommand\@newobj[3]{
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\NewDocumentCommand{#1}{ m d<> }{
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