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Small fixes

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Antti H S Laaksonen 2017-05-30 19:59:15 +03:00
parent 6c694da7a8
commit e79e59a839
5 changed files with 29 additions and 26 deletions
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5
chapter21.tex
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@ -621,8 +621,7 @@ Let $x^{-1}_m$ be the inverse of $x$ modulo $m$, and
\[ X_k = \frac{m_1 m_2 \cdots m_n}{m_k}.\] \[ X_k = \frac{m_1 m_2 \cdots m_n}{m_k}.\]
Using this notation, a solution to the equations is Using this notation, a solution to the equations is
\[x = a_1 X_1 {X_1}^{-1}_{m_1} + a_2 X_2 {X_2}^{-1}_{m_2} + \cdots + a_n X_n {X_n}^{-1}_{m_n}.\] \[x = a_1 X_1 {X_1}^{-1}_{m_1} + a_2 X_2 {X_2}^{-1}_{m_2} + \cdots + a_n X_n {X_n}^{-1}_{m_n}.\]
In this solution, it holds for each number In this solution, for each $k=1,2,\ldots,n$,
$k=1,2,\ldots,n$ that
\[a_k X_k {X_k}^{-1}_{m_k} \bmod m_k = a_k,\] \[a_k X_k {X_k}^{-1}_{m_k} \bmod m_k = a_k,\]
because because
\[X_k {X_k}^{-1}_{m_k} \bmod m_k = 1.\] \[X_k {X_k}^{-1}_{m_k} \bmod m_k = 1.\]
@ -722,6 +721,6 @@ and the number 12 is not prime, because
Hence, Wilson's theorem can be used to find out Hence, Wilson's theorem can be used to find out
whether a number is prime. However, in practice, the theorem cannot be whether a number is prime. However, in practice, the theorem cannot be
applied to large values of $n$, because it is difficult applied to large values of $n$, because it is difficult
to calculate the value of $(n-1)!$ when $n$ is large. to calculate values of $(n-1)!$ when $n$ is large.

26
chapter22.tex
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@ -12,7 +12,7 @@ As an example, consider the problem
of counting the number of ways to of counting the number of ways to
represent an integer $n$ as a sum of positive integers. represent an integer $n$ as a sum of positive integers.
For example, there are 8 representations For example, there are 8 representations
for the number $4$: for $4$:
\begin{multicols}{2} \begin{multicols}{2}
\begin{itemize} \begin{itemize}
\item $1+1+1+1$ \item $1+1+1+1$
@ -59,7 +59,10 @@ f(3) & = & 4 \\
f(4) & = & 8 \\ f(4) & = & 8 \\
\end{array} \end{array}
\] \]
It turns out that the function also has a closed-form formula
Sometimes, a recursive formula can be replaced
with a closed-form formula.
In this problem,
\[ \[
f(n)=2^{n-1}, f(n)=2^{n-1},
\] \]
@ -316,7 +319,7 @@ there are 6 solutions:
In this scenario, we can assume that In this scenario, we can assume that
$k$ balls are initially placed in boxes $k$ balls are initially placed in boxes
and there is an empty box between each and there is an empty box between each
pair of two adjacent boxes. two adjacent boxes.
The remaining task is to choose the The remaining task is to choose the
positions for the remaining empty boxes. positions for the remaining empty boxes.
There are $n-2k+1$ such boxes and There are $n-2k+1$ such boxes and
@ -354,7 +357,7 @@ $n$ left parentheses and $n$ right parentheses.
For example, $C_3=5$, because For example, $C_3=5$, because
we can construct the following parenthesis we can construct the following parenthesis
expressions using three expressions using three
left parentheses and three right parentheses: left and right parentheses:
\begin{itemize}[noitemsep] \begin{itemize}[noitemsep]
\item \texttt{()()()} \item \texttt{()()()}
@ -411,7 +414,8 @@ not counting the outermost parentheses
\item $C_{n-i-1}$: the number of ways to construct an \item $C_{n-i-1}$: the number of ways to construct an
expression using the parentheses of the second part expression using the parentheses of the second part
\end{itemize} \end{itemize}
In addition, the base case is $C_0=1$,
The base case is $C_0=1$,
because we can construct an empty parenthesis because we can construct an empty parenthesis
expression using zero pairs of parentheses. expression using zero pairs of parentheses.
@ -442,10 +446,10 @@ the expression becomes \texttt{)((()(}.
The resulting expression consists of $n+1$ The resulting expression consists of $n+1$
left parentheses and $n-1$ right parentheses. left parentheses and $n-1$ right parentheses.
The number of such expressions is ${2n \choose n+1}$ The number of such expressions is ${2n \choose n+1}$,
that equals the number of non-valid which equals the number of non-valid
parenthesis expressions. parenthesis expressions.
Thus the number of valid parenthesis Thus, the number of valid parenthesis
expressions can be calculated using the formula expressions can be calculated using the formula
\[{2n \choose n}-{2n \choose n+1} = {2n \choose n} - \frac{n}{n+1} {2n \choose n} = \frac{1}{n+1} {2n \choose n}.\] \[{2n \choose n}-{2n \choose n+1} = {2n \choose n} - \frac{n}{n+1} {2n \choose n} = \frac{1}{n+1} {2n \choose n}.\]
@ -541,7 +545,7 @@ the intersections are known, and vice versa.
A simple example of the technique is the formula A simple example of the technique is the formula
\[ |A \cup B| = |A| + |B| - |A \cap B|,\] \[ |A \cup B| = |A| + |B| - |A \cap B|,\]
where $A$ and $B$ are sets and $|X|$ where $A$ and $B$ are sets and $|X|$
is the size of a set $X$. denotes the size of $X$.
The formula can be illustrated as follows: The formula can be illustrated as follows:
\begin{center} \begin{center}
@ -614,7 +618,7 @@ As an example, let us count the number of \key{derangements}
of elements $\{1,2,\ldots,n\}$, i.e., permutations of elements $\{1,2,\ldots,n\}$, i.e., permutations
where no element remains in its original place. where no element remains in its original place.
For example, when $n=3$, there are For example, when $n=3$, there are
two possible derangements: $(2,3,1)$ and $(3,1,2)$. two derangements: $(2,3,1)$ and $(3,1,2)$.
One approach for solving the problem is to use One approach for solving the problem is to use
inclusion-exclusion. inclusion-exclusion.
@ -676,7 +680,7 @@ with some other element than 1.
Now we have to construct a derangement Now we have to construct a derangement
of $n-1$ element, because we cannot replace of $n-1$ element, because we cannot replace
the element $x$ with the element $1$, and all other the element $x$ with the element $1$, and all other
elements should be changed. elements must be changed.
\section{Burnside's lemma} \section{Burnside's lemma}

6
chapter23.tex
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@ -39,7 +39,7 @@ is a vector that contains three elements.
\index{transpose} \index{transpose}
The \key{transpose} $A^T$ of a matrix $A$ The \key{transpose} $A^T$ of a matrix $A$
is obtained when the rows and columns in $A$ is obtained when the rows and columns of $A$
are swapped, i.e., $A^T[i,j]=A[j,i]$: are swapped, i.e., $A^T[i,j]=A[j,i]$:
\[ \[
A^T = A^T =
@ -541,9 +541,9 @@ X^n \cdot
\end{bmatrix}. \end{bmatrix}.
\] \]
The value of $X^n$ can be calculated in The value of $X^n$ can be calculated in
$O(k^3 \log n)$ time, $O(\log n)$ time,
so the value of $f(n)$ can also be calculated so the value of $f(n)$ can also be calculated
in $O(k^3 \log n)$ time. in $O(\log n)$ time.
\subsubsection{General case} \subsubsection{General case}

10
chapter24.tex
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@ -83,7 +83,7 @@ corresponds to the set
\[A = \{2,4,6\}.\] \[A = \{2,4,6\}.\]
Each outcome $x$ is assigned a probability $p(x)$. Each outcome $x$ is assigned a probability $p(x)$.
Furthermore, the probability $P(A)$ of an event Then, the probability $P(A)$ of an event
$A$ can be calculated as a sum $A$ can be calculated as a sum
of probabilities of outcomes using the formula of probabilities of outcomes using the formula
\[P(A) = \sum_{x \in A} p(x).\] \[P(A) = \sum_{x \in A} p(x).\]
@ -169,7 +169,7 @@ Hence, when calculating the
probability of $A$, we only consider the outcomes probability of $A$, we only consider the outcomes
that also belong to $B$. that also belong to $B$.
Using the above sets, Using the previous sets,
\[P(A | B)= 1/3,\] \[P(A | B)= 1/3,\]
because the outcomes of $B$ are because the outcomes of $B$ are
$\{1,2,3\}$, and one of them is even. $\{1,2,3\}$, and one of them is even.
@ -409,7 +409,7 @@ $[p_1,p_2,\ldots,p_n]$, where $p_k$ is the
probability that the current state is $k$. probability that the current state is $k$.
The formula $p_1+p_2+\cdots+p_n=1$ always holds. The formula $p_1+p_2+\cdots+p_n=1$ always holds.
In the example, the initial distribution is In the above scenario, the initial distribution is
$[1,0,0,0,0]$, because we always begin in floor 1. $[1,0,0,0,0]$, because we always begin in floor 1.
The next distribution is $[0,1,0,0,0]$, The next distribution is $[0,1,0,0,0]$,
because we can only move from floor 1 to floor 2. because we can only move from floor 1 to floor 2.
@ -428,7 +428,7 @@ a walk of $m$ steps in $O(n^2 m)$ time.
The transitions of a Markov chain can also be The transitions of a Markov chain can also be
represented as a matrix that updates the represented as a matrix that updates the
probability distribution. probability distribution.
In this example, the matrix is In the above scenario, the matrix is
\[ \[
\begin{bmatrix} \begin{bmatrix}
@ -543,7 +543,7 @@ When each element $x$ is randomly chosen,
the size of the array about halves at each step, the size of the array about halves at each step,
so the time complexity for so the time complexity for
finding the $k$th order statistic is about finding the $k$th order statistic is about
\[n+n/2+n/4+n/8+\cdots \le 2n = O(n).\] \[n+n/2+n/4+n/8+\cdots < 2n = O(n).\]
The worst case of the algorithm requires still $O(n^2)$ time, The worst case of the algorithm requires still $O(n^2)$ time,
because it is possible that $x$ is always chosen because it is possible that $x$ is always chosen

8
chapter25.tex
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@ -19,7 +19,7 @@ used in those games to other games.
Let us consider a game where there is initially Let us consider a game where there is initially
a heap of $n$ sticks. a heap of $n$ sticks.
Players $A$ and $B$ move alternatively, Players $A$ and $B$ move alternately,
and player $A$ begins. and player $A$ begins.
On each move, the player has to remove On each move, the player has to remove
1, 2 or 3 sticks from the heap, 1, 2 or 3 sticks from the heap,
@ -227,7 +227,7 @@ to other games.
There are $n$ heaps in nim, There are $n$ heaps in nim,
and each heap contains some number of sticks. and each heap contains some number of sticks.
The players move alternatively, The players move alternately,
and on each turn, the player chooses and on each turn, the player chooses
a heap that still contains sticks a heap that still contains sticks
and removes any number of sticks from it. and removes any number of sticks from it.
@ -367,7 +367,7 @@ strategy used in nim to all games that fulfil
the following requirements: the following requirements:
\begin{itemize}[noitemsep] \begin{itemize}[noitemsep]
\item There are two players who move alternatively. \item There are two players who move alternately.
\item The game consists of states, and the possible moves \item The game consists of states, and the possible moves
in a state do not depend on whose turn it is. in a state do not depend on whose turn it is.
\item The game ends when a player cannot make a move. \item The game ends when a player cannot make a move.
@ -442,7 +442,7 @@ and the Grundy number of a winning state is
a positive number. a positive number.
The Grundy number of a state corresponds to The Grundy number of a state corresponds to
a number of sticks in a nim heap. the number of sticks in a nim heap.
If the Grundy number is 0, we can only move to If the Grundy number is 0, we can only move to
states whose Grundy numbers are positive, states whose Grundy numbers are positive,
and if the Grundy number is $x>0$, we can move and if the Grundy number is $x>0$, we can move