Small fixes

chapter21.tex

@@ -621,8 +621,7 @@ Let $x^{-1}_m$ be the inverse of $x$ modulo $m$, and
 \[ X_k = \frac{m_1 m_2 \cdots m_n}{m_k}.\]
 Using this notation, a solution to the equations is
 \[x = a_1 X_1 {X_1}^{-1}_{m_1} + a_2 X_2 {X_2}^{-1}_{m_2} + \cdots + a_n X_n {X_n}^{-1}_{m_n}.\]
-In this solution, it holds for each number
-$k=1,2,\ldots,n$ that
+In this solution, for each $k=1,2,\ldots,n$,
 \[a_k X_k {X_k}^{-1}_{m_k} \bmod m_k = a_k,\]
 because
 \[X_k {X_k}^{-1}_{m_k} \bmod m_k = 1.\]
@@ -722,6 +721,6 @@ and the number 12 is not prime, because
 Hence, Wilson's theorem can be used to find out
 whether a number is prime. However, in practice, the theorem cannot be
 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.
 
 

chapter22.tex

@@ -12,7 +12,7 @@ As an example, consider the problem
 of counting the number of ways to
 represent an integer $n$ as a sum of positive integers.
 For example, there are 8 representations
-for the number $4$:
+for $4$:
 \begin{multicols}{2}
 \begin{itemize}
 \item $1+1+1+1$
@@ -59,7 +59,10 @@ f(3) & = & 4 \\
 f(4) & = & 8 \\
 \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,
 \[
 f(n)=2^{n-1},
 \]
@@ -316,7 +319,7 @@ there are 6 solutions:
 In this scenario, we can assume that
 $k$ balls are initially placed in boxes
 and there is an empty box between each
-pair of two adjacent boxes.
+two adjacent boxes.
 The remaining task is to choose the
 positions for the remaining empty boxes.
 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
 we can construct the following parenthesis
 expressions using three
-left parentheses and three right parentheses:
+left and right parentheses:
 
 \begin{itemize}[noitemsep]
 \item \texttt{()()()}
@@ -411,7 +414,8 @@ not counting the outermost parentheses
 \item $C_{n-i-1}$: the number of ways to construct an
 expression using the parentheses of the second part
 \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
 expression using zero pairs of parentheses.
 
@@ -442,10 +446,10 @@ the expression becomes \texttt{)((()(}.
 
 The resulting expression consists of $n+1$
 left parentheses and $n-1$ right parentheses.
-The number of such expressions is ${2n \choose n+1}$
-that equals the number of non-valid
+The number of such expressions is ${2n \choose n+1}$,
+which equals the number of non-valid
 parenthesis expressions.
-Thus the number of valid parenthesis
+Thus, the number of valid parenthesis
 expressions can be calculated using the formula
 \[{2n \choose n}-{2n \choose n+1} = {2n \choose n} - \frac{n}{n+1} {2n \choose n} = \frac{1}{n+1} {2n \choose n}.\]
 
@@ -541,7 +545,7 @@ the intersections are known, and vice versa.
 A simple example of the technique is the formula
 \[ |A \cup B| = |A| + |B| - |A \cap B|,\]
 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:
 
 \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
 where no element remains in its original place.
 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
 inclusion-exclusion.
@@ -676,7 +680,7 @@ with some other element than 1.
 Now we have to construct a derangement
 of $n-1$ element, because we cannot replace
 the element $x$ with the element $1$, and all other
-elements should be changed.
+elements must be changed.
 
 \section{Burnside's lemma}
 

chapter23.tex

@@ -39,7 +39,7 @@ is a vector that contains three elements.
 \index{transpose}
 
 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., $A^T[i,j]=A[j,i]$:
 \[
 A^T = 
@@ -541,9 +541,9 @@ X^n \cdot
  \end{bmatrix}.
 \]
 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
-in $O(k^3 \log n)$ time.
+in $O(\log n)$ time.
 
 \subsubsection{General case}
 

chapter24.tex

@@ -83,7 +83,7 @@ corresponds to the set
 \[A = \{2,4,6\}.\]
 
 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
 of probabilities of outcomes using the formula
 \[P(A) = \sum_{x \in A} p(x).\]
@@ -169,7 +169,7 @@ Hence, when calculating the
 probability of $A$, we only consider the outcomes
 that also belong to $B$.
 
-Using the above sets,
+Using the previous sets,
 \[P(A | B)= 1/3,\]
 because the outcomes of $B$ are
 $\{1,2,3\}$, 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 $k$.
 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.
 The next distribution is $[0,1,0,0,0]$,
 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
 represented as a matrix that updates the
 probability distribution.
-In this example, the matrix is
+In the above scenario, the matrix is
 
 \[ 
  \begin{bmatrix}
@@ -543,7 +543,7 @@ When each element $x$ is randomly chosen,
 the size of the array about halves at each step,
 so the time complexity for
 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,
 because it is possible that $x$ is always chosen

chapter25.tex

@@ -19,7 +19,7 @@ used in those games to other games.
 
 Let us consider a game where there is initially
 a heap of $n$ sticks.
-Players $A$ and $B$ move alternatively,
+Players $A$ and $B$ move alternately,
 and player $A$ begins.
 On each move, the player has to remove
 1, 2 or 3 sticks from the heap,
@@ -227,7 +227,7 @@ to other games.
 
 There are $n$ heaps in nim,
 and each heap contains some number of sticks.
-The players move alternatively,
+The players move alternately,
 and on each turn, the player chooses
 a heap that still contains sticks
 and removes any number of sticks from it.
@@ -367,7 +367,7 @@ strategy used in nim to all games that fulfil
 the following requirements:
 
 \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
 in a state do not depend on whose turn it is.
 \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.
 
 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
 states whose Grundy numbers are positive,
 and if the Grundy number is $x>0$, we can move