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