@ -44,7 +44,7 @@ you can get it to balance on the beveled edge, as seen in Figure
|
||||
\end{figure}
|
||||
|
||||
\problem{}
|
||||
See Figure \ref{soda filled}. Let's take the can to be massless and intially empty. Let's also assume that we live in two dimensions. We start slowly filling it up with soda to a vertical height $h$. What is $h$ just before the can tips over?
|
||||
See Figure \ref{soda filled}. Let's take the can to be massless and initially empty. Let's also assume that we live in two dimensions. We start slowly filling it up with soda to a vertical height $h$. What is $h$ just before the can tips over?
|
||||
|
||||
\vfill
|
||||
|
||||
|
||||
@ -35,7 +35,9 @@ Then, decode the following:
|
||||
|
||||
\begin{solution}
|
||||
|
||||
% spell:off
|
||||
\texttt{ABCD$\cdot$ABCD$\cdot$BABABA$\cdot$ABCD$\cdot$ABCD} becomes \texttt{[ABCD<4, 4> BA<2,4> ABCD<4,4>]}.
|
||||
% spell:on
|
||||
|
||||
\linehack{}
|
||||
|
||||
|
||||
@ -256,7 +256,8 @@ Now, do the opposite: draw a tree that encodes \texttt{DEACBDD} \textit{less} ef
|
||||
|
||||
\remark{}
|
||||
As we just saw, constructing a prefix-free code is fairly easy. \par
|
||||
Constucting the \textit{most efficient} prefix-free code for a given message is a bit more difficult. \par
|
||||
Constructing the \textit{most efficient} prefix-free code for a
|
||||
given message is a bit more difficult. \par
|
||||
\pagebreak
|
||||
|
||||
|
||||
|
||||
@ -101,7 +101,7 @@ Run the algorithm. What is the resulting shared secret?
|
||||
\begin{solution}
|
||||
$g^b = 5$\par
|
||||
$g^a = 6$\par
|
||||
$g^{ab} = g^{ba} = 9$
|
||||
$g^{ab} = g^{ba} = 9$ % spell:disable-line
|
||||
\end{solution}
|
||||
|
||||
|
||||
|
||||
@ -276,7 +276,7 @@ Attempt the above construction a few times. Is $w$ a minimal Sturmian word?
|
||||
|
||||
|
||||
\theorem{}<sturmanthm>
|
||||
We can construct a miminal Sturmian word of order $n \geq 3$ as follows:
|
||||
We can construct a minimal Sturmian word of order $n \geq 3$ as follows:
|
||||
\begin{itemize}
|
||||
\item Start with $G_2$, create $R_2$ by removing one edge.
|
||||
\item Construct $\mathcal{L}(G_2)$, remove an edge if necessary. \par
|
||||
@ -315,7 +315,7 @@ Construct a minimal Sturmain word of order 4.
|
||||
|
||||
$R_4 = \mathcal{L}(R_3)$ is then as shown below, producing the
|
||||
order $4$ minimal Sturman word \texttt{11110000}. Disconnected
|
||||
nodes are ommited.
|
||||
nodes are omitted.
|
||||
|
||||
\begin{center}
|
||||
\begin{tikzpicture}
|
||||
@ -345,7 +345,7 @@ Construct a minimal Sturmain word of order 5.
|
||||
|
||||
\begin{solution}
|
||||
Use $R_4$ from \ref{sturmianfour} to construct $R_5$, shown below. \par
|
||||
Disconnected nodes are ommited.
|
||||
Disconnected nodes are omitted.
|
||||
|
||||
\begin{center}
|
||||
\begin{tikzpicture}
|
||||
@ -375,7 +375,7 @@ Construct a minimal Sturmain word of order 5.
|
||||
|
||||
|
||||
\problem{}
|
||||
Argue that the words we get by \ref{sturmanthm} are mimimal Sturmain words. \par
|
||||
Argue that the words we get by \ref{sturmanthm} are minimal Sturmain words. \par
|
||||
That is, the word $w$ has length $2n$ and $\mathcal{S}_m(w) = m + 1$ for all $m \leq n$.
|
||||
|
||||
\begin{solution}
|
||||
|
||||
@ -1,11 +1,11 @@
|
||||
\section{Turing}
|
||||
|
||||
\definition{}
|
||||
An \textit{esoteric programming langauge} is a programming langauge made for fun. \par
|
||||
An \textit{esoteric programming language} is a programming language made for fun. \par
|
||||
We'll work with two such languages today: \textit{Turing} and \textit{Befunge}.
|
||||
|
||||
\definition{}
|
||||
\textit{Turing} is one of the most famous esoteric langauges, and is extremely minimalist. \par
|
||||
\textit{Turing} is one of the most famous esoteric languages, and is extremely minimalist. \par
|
||||
It consists only of eight symbols, a data pointer, and an instruction pointer.
|
||||
|
||||
Turing's eight symbols are as follows:
|
||||
|
||||
@ -1,7 +1,7 @@
|
||||
\section{Befunge}
|
||||
|
||||
\definition{}
|
||||
\textit{Befunge} is another esoteric programming langauge, designed to be very difficult to compile. \par
|
||||
\textit{Befunge} is another esoteric programming language, designed to be very difficult to compile. \par
|
||||
It consists of a \say{field} of instructions, a two-dimensional program counter, and a stack of values. \par
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
@ -79,7 +79,7 @@ Using generating functions, find two six-sided dice whose sum has the same
|
||||
distribution as the sum of two standard six-sided dice? \par
|
||||
|
||||
That is, for any integer $k$, the number if ways that the sum of the two
|
||||
nonstandard dice rolls as $k$ is equal to the numer of ways the sum of
|
||||
nonstandard dice rolls as $k$ is equal to the number of ways the sum of
|
||||
two standard dice rolls as $k$.
|
||||
\hint{factor polynomials.}
|
||||
\begin{solution}
|
||||
|
||||
@ -170,7 +170,7 @@
|
||||
\problem{}
|
||||
Draw a circle, then draw two distinct tangents $\ell_1$ and $\ell_2$ that intersect at point $A$. \par
|
||||
Let $P$ be a point on the circle between the tangents, and $BC$ be the tangent at that point.
|
||||
Describe how $P$ shoud be selected in order to minimize the perimeter of triangle $ABC$.
|
||||
Describe how $P$ should be selected in order to minimize the perimeter of triangle $ABC$.
|
||||
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale = 2]
|
||||
|
||||
@ -44,7 +44,11 @@ you can get it to balance on the beveled edge, as seen in Figure
|
||||
\end{figure}
|
||||
|
||||
\problem{}
|
||||
See Figure \ref{soda filled}. Let's take the can to be massless and intially empty. Let's also assume that we live in two dimensions. We start slowly filling it up with soda to a vertical height $h$. What is $h$ just before the can tips over?
|
||||
See Figure \ref{soda filled}.
|
||||
Let's take the can to be massless and initially empty.
|
||||
Let's also assume that we live in two dimensions.
|
||||
We start slowly filling it up with soda to a vertical height $h$.
|
||||
What is $h$ just before the can tips over?
|
||||
|
||||
\vfill
|
||||
|
||||
|
||||
@ -305,7 +305,7 @@
|
||||
\problem{}
|
||||
Consider a rectangular chocolate bar of arbitrary size. \par
|
||||
What is the minimum number of breaks you need to make to
|
||||
seperate all its pieces?
|
||||
separate all its pieces?
|
||||
|
||||
\begin{solution}
|
||||
number of squares minus one, simple proof by induction.
|
||||
|
||||
@ -32,7 +32,7 @@
|
||||
\def\bra#1{\left\langle#1\right|}
|
||||
|
||||
|
||||
% TODO: spend more time on probabalistic bits.
|
||||
% TODO: spend more time on probabalastic bits.
|
||||
% This could even be its own handout, especially
|
||||
% for younger classes!
|
||||
|
||||
|
||||
@ -498,7 +498,7 @@ Compute the following product:
|
||||
|
||||
\generic{Remark:}
|
||||
Also, recall that every matrix is linear map, and that every linear map may be written as a matrix. \par
|
||||
We often use the terms \textit{matrix}, \textit{transformation}, and \textit{linear map} interchangably.
|
||||
We often use the terms \textit{matrix}, \textit{transformation}, and \textit{linear map} interchangeably.
|
||||
|
||||
\pagebreak
|
||||
|
||||
|
||||
@ -18,7 +18,7 @@ $\ket{0}$ is pronounced \say{ket zero,} and $\ket{1}$ is pronounced \say{ket one
|
||||
\note[Note]{$\bra{0}$ is called a \say{bra,} but we won't worry about that for now.}
|
||||
|
||||
\vspace{2mm}
|
||||
This is very similiar to the \say{box} $[~]$ notation we used for probabilistic bits. \par
|
||||
This is very similar to the \say{box} $[~]$ notation we used for probabilistic bits. \par
|
||||
As before, we will write $\ket{0} = \left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$
|
||||
and $\ket{1} = \left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$.
|
||||
|
||||
|
||||
@ -64,7 +64,7 @@ and like $X$ otherwise. The two circuits above illustrate this fact---take a loo
|
||||
|
||||
\vspace{2mm}
|
||||
Of course, we can give a gate multiple controls. \par
|
||||
An $X$ gate with multiplie controls behaves like an $X$ gate if...
|
||||
An $X$ gate with multiple controls behaves like an $X$ gate if...
|
||||
\begin{itemize}
|
||||
\item all non-inverted controls are $\ket{1}$, and
|
||||
\item all inverted controls are $\ket{0}$
|
||||
|
||||
@ -1,7 +1,8 @@
|
||||
\section{Quantum Teleportation}
|
||||
|
||||
Superdense coding lets us convert quantum bandwidth into classical bandwidth. \par
|
||||
Quantum teleporation does the opposite, using two classical bits and an entangled pair to transmit a quantum state.
|
||||
Quantum teleportation does the opposite, using two classical bits and an entangled pair
|
||||
to transmit a quantum state.
|
||||
|
||||
\generic{Setup:}
|
||||
Again, suppose Alice and Bob each have half of a $\ket{\Phi^+}$ state. \par
|
||||
@ -131,7 +132,7 @@ With an informal proof, show that it is not possible to use superdense coding to
|
||||
more than two classical bits through an entangled two-qubit quantum state.
|
||||
|
||||
\begin{solution}
|
||||
If superdense coding was any more efficient, we could repeatedly apply superdense coding and quantum teleporation,
|
||||
If superdense coding was any more efficient, we could repeatedly apply superdense coding and quantum teleportation,
|
||||
to compress an arbitrary number of bits into two \say{seed} bits.
|
||||
|
||||
\linehack{}
|
||||
|
||||
@ -52,7 +52,7 @@ An ordered field must satisfy the following properties:
|
||||
An ordered field that contains $\mathbb{R}$ is called an \textit{extension} of $\mathbb{R}$.
|
||||
|
||||
\definition{}
|
||||
The \textit{Archimedian property} states the following: \par
|
||||
The \textit{Archimedean property} states the following: \par
|
||||
For all positive $x, y$, there exists an $n$ so that $nx \geq y$.
|
||||
|
||||
\theorem{}
|
||||
@ -149,7 +149,9 @@ In an ordered field, the \textit{magnitude} of a number x is defined as follows:
|
||||
\end{equation*}
|
||||
|
||||
\definition{}
|
||||
We say an element $\delta$ of an ordered field is \textit{infinitesimal} if $|nd| < 1$ for all $n \in \mathbb{Z^+}$. \par
|
||||
We say an element $\delta$ of an ordered field is \textit{infinitesimal} if
|
||||
$|nd| < 1$ % spell:disable-line
|
||||
for all $n \in \mathbb{Z^+}$. \par
|
||||
\note{Note that $\mathbb{Z}^+$ is a subset of any nonarchimedian extension of $\mathbb{R}$.} \par
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
@ -380,7 +380,10 @@
|
||||
In a year, there are $364$ clean breaks. This leaves $499 - 364 = 135$ ``dirty'' breaks. \par
|
||||
We therefore have $135$ places to start a block on a dirty break, and $135$ places to end a block on a dirty break. This gives us a maximum of $270$ dirty blocks. \par
|
||||
|
||||
However, there are $401$ possible blocks, since we can start one at the $1^{\text{st}}, 2^{\text{nd}}, ..., 401^{\text{st}}$ espresso. \par
|
||||
% spell:off
|
||||
However, there are $401$ possible blocks,
|
||||
since we can start one at the $1^{\text{st}},2^{\text{nd}}, ..., 401^{\text{st}}$ espresso. \par
|
||||
% spell:on
|
||||
|
||||
Out of $401$ blocks, a maximum of $270$ can be dirty. We are therefore guaranteed at least $131$ clean blocks. This completes the problem---each clean block represents a set of consecutive, whole days during which exactly 100 espressos were consumed.
|
||||
|
||||
|
||||
@ -12,11 +12,13 @@ What was Black's last move, and what was White's last move? \par
|
||||
There a few empty boards at the end of this handout as well.
|
||||
}
|
||||
|
||||
% spell:off
|
||||
\manyboards{
|
||||
ka8,Kc8,
|
||||
Ph2,
|
||||
Bg1
|
||||
}
|
||||
% spell:on
|
||||
|
||||
\begin{solution}
|
||||
It's pretty clear that Black just moved out of check from A7.
|
||||
@ -59,12 +61,14 @@ In the game below, no pieces have moved from a black square to a white square, o
|
||||
There is a pawn at G3. What color is it? \par
|
||||
As before, White started on the bottom of the board.
|
||||
|
||||
% spell:off
|
||||
\manyboards{
|
||||
ke8,
|
||||
Kb4,
|
||||
Ug3,
|
||||
Pd2,Pf2
|
||||
}
|
||||
% spell:on
|
||||
|
||||
\begin{solution}
|
||||
The white king is the key to this solution. How did it get off of E1? \par
|
||||
@ -99,11 +103,13 @@ As before, White started on the bottom of the board.
|
||||
The black king has turned himself invisible. Unfortunately, his position is hopeless. \par
|
||||
Mate the king in one move. \par
|
||||
|
||||
% spell:off
|
||||
\manyboards{
|
||||
Ra8,rb8,Kf8,
|
||||
Nb7,Pc7,
|
||||
Pa6,Rc6
|
||||
}
|
||||
% spell:on
|
||||
|
||||
\begin{solution}
|
||||
Since it is White's move, Black cannot be in check. \par
|
||||
@ -132,11 +138,13 @@ Mate the king in one move. \par
|
||||
In the game below, no pieces have moved from a black square to a white square, or from a white square to a black square.
|
||||
There is one more piece on the board, which isn't shown. What color square does it stand on? \par
|
||||
|
||||
% spell:off
|
||||
\manyboards{
|
||||
ke8,
|
||||
Pd2,Pf2,
|
||||
Ke1
|
||||
}
|
||||
% spell:on
|
||||
|
||||
\begin{solution}
|
||||
|
||||
@ -161,11 +169,13 @@ In the game below, no pieces have moved from a black square to a white square, o
|
||||
The white king has made less than fourteen moves. \par
|
||||
Use this information to show that a pawn was promoted. \par
|
||||
|
||||
% spell:off
|
||||
\manyboards{
|
||||
ke8,
|
||||
Pb2,Pd2,
|
||||
Ke1
|
||||
}
|
||||
% spell:on
|
||||
|
||||
\begin{solution}
|
||||
Knights always move to a different colored square, so all four missing knights must have been captured on their home square.
|
||||
@ -196,10 +206,12 @@ Use this information to show that a pawn was promoted. \par
|
||||
|
||||
It is White's move. Have there been any promotions this game? \par
|
||||
|
||||
% spell:off
|
||||
\manyboards{
|
||||
Pb2,Pe2,kf2,Pg2,Ph2,
|
||||
Bc1,Kd1,Rh1
|
||||
}
|
||||
% spell:on
|
||||
|
||||
\begin{solution}
|
||||
|
||||
@ -231,6 +243,7 @@ It is White's move. Have there been any promotions this game? \par
|
||||
It is Black's move. Can Black castle? \par
|
||||
\hint{Remember the rules of chess: you may not castle if you've moved your rook.}
|
||||
|
||||
% spell:off
|
||||
\manyboards{
|
||||
ra8,bc8,ke8,rh8,
|
||||
pa7,pc7,pe7,pg7,
|
||||
@ -239,6 +252,7 @@ It is Black's move. Can Black castle? \par
|
||||
Pb2,Pc2,Pd2,Pe2,Pf2,Pg2,Ph2,
|
||||
Bc1,Qd1,Ke1,Bf1
|
||||
}
|
||||
% spell:on
|
||||
|
||||
\begin{solution}
|
||||
White's last move was with the pawn. \par
|
||||
|
||||
@ -7,6 +7,8 @@
|
||||
|
||||
The results of a game of chess are shown below. \par
|
||||
Did White start on the north or south side of the board? \par
|
||||
|
||||
% spell:off
|
||||
\manyboards{
|
||||
ka8,Kc8,
|
||||
Qe7,
|
||||
@ -15,6 +17,7 @@ Did White start on the north or south side of the board? \par
|
||||
Ph3,
|
||||
Bh1
|
||||
}
|
||||
% spell:on
|
||||
|
||||
\begin{solution}
|
||||
Let us first find White's last move. It wasn't with the pawns on D4 and E5, since Black wouldn't have a move before that.
|
||||
@ -52,11 +55,13 @@ Did White start on the north or south side of the board? \par
|
||||
The white king has again become invisible. Find him. \par
|
||||
\hint{White started on the bottom. En passant.} \par
|
||||
|
||||
% spell:off
|
||||
\manyboards{
|
||||
rb5,bd5,
|
||||
Ba4,
|
||||
kd1
|
||||
}
|
||||
% spell:on
|
||||
|
||||
|
||||
\begin{solution}
|
||||
@ -77,6 +82,7 @@ The white king has again become invisible. Find him. \par
|
||||
|
||||
\begin{minipage}{0.5\linewidth}
|
||||
\begin{center}
|
||||
% spell:off
|
||||
\chessboard[
|
||||
setpieces = {
|
||||
rb5,
|
||||
@ -86,6 +92,7 @@ The white king has again become invisible. Find him. \par
|
||||
kd1
|
||||
}
|
||||
]
|
||||
% spell:on
|
||||
\end{center}
|
||||
\end{minipage}
|
||||
\hfill
|
||||
@ -123,11 +130,13 @@ The white king has again become invisible. Find him. \par
|
||||
White to move. Which side of the board did each color start on? \par
|
||||
\hint{What was Black's last move? }
|
||||
|
||||
% spell:off
|
||||
\manyboards{
|
||||
Re3,
|
||||
Nc2,Rd2,
|
||||
Nd1,kf1,Kh1
|
||||
}
|
||||
% spell:on
|
||||
|
||||
\begin{solution}
|
||||
Black's last move was from F2, where his king was in double-check from both a rook and a knight.
|
||||
@ -175,6 +184,7 @@ White to move. Which side of the board did each color start on? \par
|
||||
There is a piece at G4, marked with a $\odot$. \par
|
||||
What is it, and what is its color? \par
|
||||
|
||||
% spell:off
|
||||
\manyboards{
|
||||
ra8,ke8,rh8,
|
||||
pc7,pd7,
|
||||
@ -185,6 +195,7 @@ What is it, and what is its color? \par
|
||||
ba2,Pb2,Pc2,Pd2,Pf2,qg2,bh2,
|
||||
Kc1,Rd1,nf1,Bh1
|
||||
}
|
||||
% spell:on
|
||||
|
||||
|
||||
\makeatletter
|
||||
|
||||
@ -7,6 +7,7 @@
|
||||
There is a white castle hidden on this board. Where is it? \par
|
||||
None of the royalty has moved or been under attack. \par
|
||||
|
||||
% spell:off
|
||||
\manyboards{
|
||||
nb8,qd8,ke8,ng8,rh8,
|
||||
pa7,pb7,pc7,pf7,pg7,
|
||||
@ -16,6 +17,7 @@ None of the royalty has moved or been under attack. \par
|
||||
Pb2,Pd2,Pf2,Pg2,
|
||||
Qd1,Ke1
|
||||
}
|
||||
% spell:on
|
||||
|
||||
\begin{solution}
|
||||
See \say{The Hidden Castle} in \textit{The Chess Mysteries of the Arabian Knights}.
|
||||
@ -33,6 +35,7 @@ None of the royalty has moved or been under attack. \par
|
||||
After many moves of chess, the board looks as follows. \par
|
||||
Who moved last? \par
|
||||
|
||||
% spell:off
|
||||
\manyboards{
|
||||
ka8,Kc8,bf8,rh8,
|
||||
pb7,pc7,pf7,pg7,
|
||||
@ -41,6 +44,7 @@ Who moved last? \par
|
||||
Pa2,Pb2,Pd2,Pg2,Ph2,
|
||||
Ra1,Nb1,Bc1,Qd1,Rh1
|
||||
}
|
||||
% spell:on
|
||||
|
||||
\begin{solution}
|
||||
See \say{A Vital Decision} in \textit{The Chess Mysteries of the Arabian Knights}.
|
||||
@ -59,6 +63,7 @@ Who moved last? \par
|
||||
The white king is exploring his kingdom under a disguise. He could look like any piece of any color.\par
|
||||
Show that he must be on C7.
|
||||
|
||||
% spell:off
|
||||
\manyboards{
|
||||
qa8,nb8,be8,Qg8,kh8,
|
||||
pa7,Pb7,pc7,Nd7,pe7,Pf7,ph7,
|
||||
@ -66,6 +71,7 @@ Show that he must be on C7.
|
||||
ra5,pb5,Rd5,Ph5,
|
||||
Pa4,Nc4,Pe4,Bg4
|
||||
}
|
||||
% spell:on
|
||||
|
||||
\begin{solution}
|
||||
Black is in check, so we know that it is Black's move and White is not in check.\par
|
||||
@ -118,6 +124,7 @@ Show that he must be on C7.
|
||||
The white king is again exploring his kingdom, now under a different disguise. Where is he? \par
|
||||
\hint{\say{different disguise} implies that the white king looks like a different piece!}
|
||||
|
||||
% spell:off
|
||||
\manyboards{
|
||||
nb8,be8,Qg8,kh8,
|
||||
pa7,Pb7,pc7,Nd7,pe7,Pf7,ph7,
|
||||
@ -125,6 +132,7 @@ The white king is again exploring his kingdom, now under a different disguise. W
|
||||
ra5,pb5,Rd5,Ph5,
|
||||
Pa4,Nc4,Pe4,Bg4
|
||||
}
|
||||
% spell:on
|
||||
|
||||
\begin{solution}
|
||||
Use the same arguments as before, but now assume that the king isn't a black pawn.
|
||||
|
||||
@ -46,17 +46,17 @@ a \textit{random variable} is a function from $\Omega$ to a specified output set
|
||||
|
||||
For example, given the three-coin-toss sample space
|
||||
$\Omega = \{
|
||||
\texttt{TTT},~ \texttt{TTH},~ \texttt{THT},~
|
||||
\texttt{THH},~ \texttt{HTT},~ \texttt{HTH},~
|
||||
\texttt{HHT},~ \texttt{HHH}
|
||||
\texttt{TTT},~ \texttt{TTH},~ \texttt{THT},~ % spell:disable-line
|
||||
\texttt{THH},~ \texttt{HTT},~ \texttt{HTH},~ % spell:disable-line
|
||||
\texttt{HHT},~ \texttt{HHH} % spell:disable-line
|
||||
\}$,
|
||||
We can define a random variable $\mathcal{H}$ as \say{the number of heads in a throw of three coins}. \par
|
||||
As a function, $\mathcal{H}$ maps values in $\Omega$ to values in $\mathbb{Z}^+_0$ and is defined as:
|
||||
\begin{itemize}
|
||||
\item $\mathcal{H}(\texttt{TTT}) = 0$
|
||||
\item $\mathcal{H}(\texttt{TTH}) = 1$
|
||||
\item $\mathcal{H}(\texttt{THT}) = 1$
|
||||
\item $\mathcal{H}(\texttt{THH}) = 2$
|
||||
\item $\mathcal{H}(\texttt{TTT}) = 0$ % spell:disable-line
|
||||
\item $\mathcal{H}(\texttt{TTH}) = 1$ % spell:disable-line
|
||||
\item $\mathcal{H}(\texttt{THT}) = 1$ % spell:disable-line
|
||||
\item $\mathcal{H}(\texttt{THH}) = 2$ % spell:disable-line
|
||||
\item ...and so on.
|
||||
\end{itemize}
|
||||
|
||||
@ -70,7 +70,7 @@ the set of outcomes that produce that value. \par
|
||||
\vspace{2mm}
|
||||
|
||||
For example, if we wanted to compute $\mathcal{P}(\mathcal{H} = 2)$, we would find
|
||||
$\mathcal{P}\bigl(\{\texttt{THH}, \texttt{HTH}, \texttt{HHT}\}\bigr)$.
|
||||
$\mathcal{P}\bigl(\{\texttt{THH}, \texttt{HTH}, \texttt{HHT}\}\bigr)$. % spell:disable-line
|
||||
|
||||
|
||||
\problem{}
|
||||
|
||||
@ -191,7 +191,7 @@ what is the probability that we select the best candidate? \par
|
||||
Call this probability $\phi_n(k)$.
|
||||
|
||||
\begin{solution}
|
||||
Using \ref{seca} and \ref{secb}, this is straightfoward:
|
||||
Using \ref{seca} and \ref{secb}, this is straightforward:
|
||||
\[
|
||||
\phi_n(k)
|
||||
= \sum_{x = k}^{n}\left( \frac{1}{n} \times \frac{k-1}{x-1} \right)
|
||||
|
||||
Reference in New Issue
Block a user