From 76643e6b4a47f56392c330d018e11c070081f422 Mon Sep 17 00:00:00 2001 From: Antti H S Laaksonen Date: Fri, 10 Feb 2017 00:10:07 +0200 Subject: [PATCH] Corrections --- luku22.tex | 212 ++++++++++++++++++++++++++--------------------------- 1 file changed, 106 insertions(+), 106 deletions(-) diff --git a/luku22.tex b/luku22.tex index 96f6ccb..2dcf6ab 100644 --- a/luku22.tex +++ b/luku22.tex @@ -8,9 +8,9 @@ Usually, the goal is to find a way to count the combinations efficiently without generating each combination separately. -As an example, let's consider a problem where -our task is to calculate the number of representations -for an integer $n$ as a sum of positive integers. +As an example, let us 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 $4$: \begin{multicols}{2} @@ -28,10 +28,11 @@ for the number $4$: A combinatorial problem can often be solved using a recursive function. -In this case, we can define a function $f(n)$ +In this problem, we can define a function $f(n)$ that counts the number of representations for $n$. For example, $f(4)=8$ according to the above example. -The function can be recursively calculated as follows: +The values of the function +can be recursively calculated as follows: \begin{equation*} f(n) = \begin{cases} 1 & n = 1\\ @@ -40,8 +41,8 @@ The function can be recursively calculated as follows: \end{equation*} The base case is $f(1)=1$, because there is only one way to represent the number 1. -Otherwise, we go through all possibilities for -the last number in the sum. +When $n>1$, we go through all ways to +select the last number in the sum. For example, in when $n=4$, the sum can end with $+1$, $+2$ or $+3$. In addition, we also count the representation @@ -69,8 +70,8 @@ and we can choose any subset of them. \index{binomial coefficient} -A \key{binomial coefficient} ${n \choose k}$ -is the number of ways we can choose a subset +The \key{binomial coefficient} ${n \choose k}$ +equals the number of ways we can choose a subset of $k$ elements from a set of $n$ elements. For example, ${5 \choose 3}=10$, because the set $\{1,2,3,4,5\}$ @@ -87,22 +88,20 @@ recursively calculated as follows: {n \choose k} = {n-1 \choose k-1} + {n-1 \choose k} \] -The idea is to consider a fixed -element $x$ in the set. +The idea is to fix an element $x$ in the set. If $x$ is included in the subset, -the remaining task is to choose $k-1$ +we have to choose $k-1$ elements from $n-1$ elements, -and otherwise -the remaining task is to choose $k$ elements from $n-1$ elements. +and if $x$ is not included in the subset, +we have to choose $k$ elements from $n-1$ elements. -The base cases for the recursion are as follows: +The base cases for the recursion are \[ -{n \choose 0} = {n \choose n} = 1 +{n \choose 0} = {n \choose n} = 1, \] - -The reason for this is that there is always -one way to construct an empty subset, -or a subset that contains all the elements. +because there is always exactly +one way to construct an empty subset +and a subset that contains all the elements. \subsubsection{Formula 2} @@ -111,12 +110,12 @@ Another way to calculate binomial coefficients is as follows: {n \choose k} = \frac{n!}{k!(n-k)!}. \] -There are $n!$ permutations for $n$ elements. -We go through all permutations and in each case +There are $n!$ permutations of $n$ elements. +We go through all permutations and always select the first $k$ elements of the permutation to the subset. Since the order of the elements in the subset -and outside the subset doesn't matter, +and outside the subset does not matter, the result is divided by $k!$ and $(n-k)!$ \subsubsection{Properties} @@ -126,9 +125,9 @@ For binomial coefficients, {n \choose k} = {n \choose n-k}, \] because we can either select $k$ -elements to the subset, -or select $n-k$ elements that -will be outside the subset. +elements that belong to the subset +or $n-k$ elements that +do not belong to the subset. The sum of binomial coefficients is \[ @@ -149,8 +148,7 @@ is that Binomial coefficients also appear in \key{Pascal's triangle} -whose border consists of 1's, -and each value is the sum of two +where each value equals the sum of two above values: \begin{center} \begin{tikzpicture}{0.9} @@ -179,12 +177,12 @@ above values: \subsubsection{Boxes and balls} -''Boxes and models'' is a useful model, +''Boxes and balls'' is a useful model, where we count the ways to place $k$ balls in $n$ boxes. -Let's consider three cases: +Let us consider three scenarios: -\textit{Case 1}: Each box can contain +\textit{Scenario 1}: Each box can contain at most one ball. For example, when $n=5$ and $k=2$, there are 10 solutions: @@ -220,10 +218,10 @@ there are 10 solutions: \end{tikzpicture} \end{center} -In this case, the answer is directly the +In this scenario, the answer is directly the binomial coefficient ${n \choose k}$. -\textit{Case 2}: A box can contain multiple balls. +\textit{Scenario 2}: A box can contain multiple balls. For example, when $n=5$ and $k=2$, there are 15 solutions: @@ -263,25 +261,26 @@ there are 15 solutions: \end{tikzpicture} \end{center} -This process can be represented as a string +The process of placing the balls in the boxes +can be represented as a string that consists of symbols ''o'' and ''$\rightarrow$''. -Initially, we are standing at the leftmost box. -The symbol ''o'' means we place a ball +Initially, assume that we are standing at the leftmost box. +The symbol ''o'' means that we place a ball in the current box, and the symbol ''$\rightarrow$'' means that we move to -the next box right. +the next box to the right. Using this notation, each solution is a string -that has $k$ times the symbol ''o'' and +that contains $k$ times the symbol ''o'' and $n-1$ times the symbol ''$\rightarrow$''. For example, the upper-right solution -corresponds to the string +in the above picture corresponds to the string ''$\rightarrow$ $\rightarrow$ o $\rightarrow$ o $\rightarrow$''. Thus, the number of solutions is ${k+n-1 \choose k}$. -\textit{Case 3}: Each box may contain at most one ball, +\textit{Scenario 3}: Each box may contain at most one ball, and in addition, no two adjacent boxes may both contain a ball. For example, when $n=5$ and $k=2$, there are 6 solutions: @@ -313,37 +312,37 @@ there are 6 solutions: \end{tikzpicture} \end{center} -In this case, we can think that -$k$ balls are initially placed in boxes. -and between each such box there is an empty box. +In this scenario, we can assume that +$k$ balls are initially placed in boxes +and there is an empty box between each +two such boxes. The remaining task is to choose the positions for $n-k-(k-1)=n-2k+1$ empty boxes. -There are $k+1$ positions, so as in case 2, -the number of solutions is +There are $k+1$ positions, +so the number of solutions is ${n-2k+1+k+1-1 \choose n-2k+1} = {n-k+1 \choose n-2k+1}$. \subsubsection{Multinomial coefficient} \index{multinomial coefficient} -A generalization for a binomial coefficient is -a \key{multinomial coefficient} - +The \key{multinomial coefficient} \[ {n \choose k_1,k_2,\ldots,k_m} = \frac{n!}{k_1! k_2! \cdots k_m!}, \] - -where $k_1+k_2+\cdots+k_m=n$. -A multinomial coefficient i the number of ways +equals the number of ways we can divide $n$ elements into subsets -whose sizes are $k_1,k_2,\ldots,k_m$. -If $m=2$, the formula +of sizes $k_1,k_2,\ldots,k_m$, +where $k_1+k_2+\cdots+k_m=n$. +Multinomial coefficients can be seen as a +generalization of binomial cofficients; +if $m=2$, the above formula corresponds to the binomial coefficient formula. \section{Catalan numbers} \index{Catalan number} -A \key{Catalan number} $C_n$ is the +The \key{Catalan number} $C_n$ equals the number of valid parenthesis expressions that consist of $n$ left parentheses and $n$ right parentheses. @@ -379,8 +378,8 @@ then also the expression $AB$ is valid. \end{itemize} Another way to characterize valid -paranthesis expressions is that if -we choose any prefix of the expression, +parenthesis expressions is that if +we choose any prefix of such an expression, it has to contain at least as many left parentheses as right parentheses. In addition, the complete expression has to @@ -423,7 +422,7 @@ There are a total of ${2n \choose n}$ ways to construct a (not necessarily valid) parenthesis expression that contains $n$ left parentheses and $n$ right parentheses. -Let's calculate the number of such +Let us calculate the number of such expressions that are \emph{not} valid. If a parenthesis expression is not valid, @@ -448,7 +447,7 @@ expressions can be calculated using the formula \subsubsection{Counting trees} -Catalan numbers are also related to rooted trees: +Catalan numbers are also related to trees: \begin{itemize} \item there are $C_n$ binary trees of $n$ nodes @@ -554,18 +553,18 @@ The formula can be illustrated as follows: \end{tikzpicture} \end{center} -In the above example, our goal is to calculate +Our goal is to calculate the size of the union $A \cup B$ that corresponds to the area of the region -that is inside at least one circle. +that belongs to at least one circle. The picture shows that we can calculate the area of $A \cup B$ by first summing the areas of $A$ and $B$, and then subtracting the area of $A \cap B$. -The same idea can be applied, when the number +The same idea can be applied when the number of sets is larger. -When there are three sets, the formula becomes +When there are three sets, the inclusio-exclusion formula is \[ |A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C| \] and the corresponding picture is @@ -589,14 +588,14 @@ and the corresponding picture is In the general case, the size of the union $X_1 \cup X_2 \cup \cdots \cup X_n$ -can be calculated by going through all ways to -construct an intersection for a collection of -sets $X_1,X_2,\ldots,X_n$. +can be calculated by going through all possible +intersections that contain some of the sets $X_1,X_2,\ldots,X_n$. If the intersection contains an odd number of sets, -its size will be added to the answer, -and otherwise subtracted from the answer. +its size is added to the answer, +and otherwise its size is subtracted from the answer. -Note that similar formulas also work when counting +Note that similar formulas can also be used +for calculating the size of an intersection from the sizes of unions. For example, \[ |A \cap B| = |A| + |B| - |A \cup B|\] @@ -607,16 +606,16 @@ and \index{derangement} -As an example, let's count the number of \key{derangements} -of numbers $\{1,2,\ldots,n\}$, i.e., permutations +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: $(2,3,1)$ ja $(3,1,2)$. +two possible derangements: $(2,3,1)$ and $(3,1,2)$. -One approach for the problem is to use +One approach for solving the problem is to use inclusion-exclusion. Let $X_k$ be the set of permutations -that contain the number $k$ at index $k$. +that contain the element $k$ at position $k$. For example, when $n=3$, the sets are as follows: \[ \begin{array}{lcl} @@ -625,7 +624,7 @@ X_2 & = & \{(1,2,3),(3,2,1)\} \\ X_3 & = & \{(1,2,3),(2,1,3)\} \\ \end{array} \] -Using these sets the number of derangements is +Using these sets, the number of derangements equals \[ n! - |X_1 \cup X_2 \cup \cdots \cup X_n|, \] so it suffices to calculate the size of the union. Using inclusion-exclusion, this reduces to @@ -642,8 +641,8 @@ $|X_1 \cup X_2 \cup X_3|$ is \] so the number of solutions is $3!-4=2$. -It turns out that there is also another way for -solving the problem without inclusion-exclusion. +It turns out that the problem can also be solved +without using inclusion-exclusion. Let $f(n)$ denote the number of derangements for $\{1,2,\ldots,n\}$. We can use the following recursive formula: @@ -657,31 +656,32 @@ recursive formula: \end{equation*} The formula can be derived by going through -the possibilities how the number 1 changes +the possibilities how the element 1 changes in the derangement. -There are $n-1$ ways to choose a number $x$ -that will replace the number 1. +There are $n-1$ ways to choose an element $x$ +that replaces the element 1. In each such choice, there are two options: -\textit{Option 1:} We also replace the number $x$ -by the number 1. +\textit{Option 1:} We also replace the element $x$ +with the element 1. After this, the remaining task is to construct -a derangement for $n-2$ numbers. +a derangement of $n-2$ elements. -\textit{Option 2:} We replace the number $x$ -by some other number than 1. -Now we should construct a derangement -for $n-1$ numbers, because we can't replace -the number $x$ with number $1$, and all other -numbers should be changed. +\textit{Option 2:} We replace the element $x$ +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. \section{Burnside's lemma} \index{Burnside's lemma} \key{Burnside's lemma} counts the number of -combinations so that for each group of -symmetric combinations, only one representative is counted. +combinations so that +only one representative is counted +for each group of symmetric combinations, Burnside's lemma states that the number of combinations is \[\sum_{k=1}^n \frac{c(k)}{n},\] @@ -690,7 +690,7 @@ position of a combination, and there are $c(k)$ combinations that remain unchanged when the $k$th way is applied. -As an example, let's calculate the number of +As an example, let us calculate the number of necklaces of $n$ pearls, where the color of each pearl is one of $1,2,\ldots,m$. @@ -750,12 +750,12 @@ a total of necklaces remain the same, where $\textrm{gcd}(k,n)$ is the greatest common divisor of $k$ and $n$. -The reason for this is that sequences -of pearls of size $\textrm{syt}(k,n)$ +The reason for this is that blocks +of pearls of size $\textrm{gcd}(k,n)$ will replace each other. Thus, according to Burnside's lemma, the number of necklaces is -\[\sum_{i=0}^{n-1} \frac{m^{\textrm{syt}(i,n)}}{n}. \] +\[\sum_{i=0}^{n-1} \frac{m^{\textrm{gcd}(i,n)}}{n}. \] For example, the number of necklaces of length 4 with 3 colors is \[\frac{3^4+3+3^2+3}{4} = 24. \] @@ -769,7 +769,7 @@ there are $n^{n-2}$ labeled trees that contain $n$ nodes. The nodes are labeled $1,2,\ldots,n$, and two trees are different -if either their structure or their +if either their structure or labeling is different. \begin{samepage} @@ -829,11 +829,10 @@ be derived using Prüfer codes. A \key{Prüfer code} is a sequence of $n-2$ numbers that describes a labeled tree. -The code is calculated by removing $n-2$ -leaves from the tree. -At each step, we remove the leaf whose number -is the smallest, and simultaneously -add the number of its only neighbor to the code. +The code is constructed by following a process +that removes $n-2$ leaves from the tree. +At each step, the leaf with the smallest label is removed, +and the label of its only neighbor is added to the code. For example, the Prüfer code for \begin{center} @@ -856,10 +855,11 @@ For example, the Prüfer code for is $[4,4,2]$, because we first remove node 1, then node 3 and finally node 5. -We can calculate a Prüfer code for any tree, +We can construct a Prüfer code for any tree, and more importantly, -the original tree can be constructed -from the Prüfer code. -Hence, the number of labeled trees equals -the number of Prüfer codes that is $n^{n-2}$. - +the original tree can be reconstructed +from a Prüfer code. +Hence, the number of labeled trees +of $n$ nodes equals +$n^{n-2}$, the number of Prüfer codes +of size $n$.