组合数学【学习笔记(二)】

Combinatorial Mathematics
(TsinghuaX: 60240013x)

Linear Homogeneous Recurrence Relation

Fibonacci Rabbits

Fibonacci number
Fibonacci prime

$$F_1^2+F_2^2+\cdots +F_n^2=F_n{F}_{n-1}$$
Area of the rectangle = Sum of multiple quadrates

Expressions of Fibonacci Numbers

$$F_n=\frac{1}{\sqrt{5}}((\frac{1+\sqrt{5}}{2}{)}^n-(\frac{1-\sqrt{5}}{2}{)}^n)$$

Linear Homogeneous Recurrence Relation

Def If sequence { a n } \{a_n\} {an​} satisifies:
$$a_n+C_1{a}_{n-1}+C_2{a}_{n-2}+\cdots +C_k{a}_{n-k}=0$$$$a_0=d_0,a_0=d_1,\dots ,a_{k-1}=d_{k-1}$$$C_1,C_2,\dots ,C_k$ and $d_0,d_1,\dots ,d_{k-1}$ are constants, $C_k\ne 0$, so this expression is called a $k^{th}$-order linear homogeneous recurrence relation of { a n } \{a_n\} {an​}.
Fibonacci Recurrence
Characteristic Polynomial
$$C(x)=x^k+C_1{x}^{k-1}+\cdots +C_{k-1}x+C_k$$Characteristic polynomial has $k$

1. distinct real roots
2. Characteristic polynomial $C(x)$ has multiple roots
3. Characteristic polynomial $C(x)$ has conjugate complex roots
   Summary
   

Applications

• There’s a point $P$ on the plane. It’s the cross of $n$ fields $D_1,D_2,\dots D_n$. Color these $n$ fields with $k$ colors. We require the color of two adjacent areas to be different.
  Calculate the number of arrangements.

Let $a_n$ be the number of arrangement to color these areas. There are 2 situations:

1. $D_1$ and $D_{n-1}$ have the same color
   $D_n$ has $k-1$ choices, which is all colors except the one used by $D_1$ and $D_{n-1}$;
   the arrangements for $D_{n-2}$ to $D_1$ are one-to-one correspondent to the arrangements for $n-2$ areas.
   $$(k-1)a_{n-2}$$
2. $D_1$ and $D_{n-1}$ have different colors.
   $D_n$ has $k-2$ choices; the arrangements from $D_1$ to $D_{n-1}$ are one-to-one correspondent to the arrangements for $n-1$ areas.
   $$(k-2)a_{n-1}$$

$\therefore a_n=(k-2)a_{n-1}+(k-1)a_{n-2},a_2=k(k-1),a_3=k(k-1)(k-2).$
$a_1=0,a_0=k$
$x^2-(k-2)x-(k-1)=0$
$x_1=k-1,x_2=-1.$
$a_n=A(k-1{)}^n+B(-1{)}^n$
$$\begin{array}{c}\{\begin{array}{l}A+B=k,\\ (k-1)A-B=0.\end{array}\end{array}\begin{array}{c}\{\begin{array}{l}A=1,\\ B=k-1.\end{array}\end{array}$$
$\therefore a_n=(k-1{)}^n+(k-1)(-1{)}^n,n\ge 2.$
$a_1=k.$

Magical Sequences

Catalan Numbers

• One stack (of infinite size) has the “push” sequence: $1,2,3,...n$. How many ways can the numbers be popped out?

Partition the $n$ sequences into two sub-sequences.
$f(n)=f(k-1)\ast f(n-k)$

• Binary Tree
  How many different shapes can a binary tree with n nodes have?

The root will obviously contain one node.
Assume $T(i,j)$ stands the left subtree of the root containing $I$ nodes, and right subtree with $j$ nodes.
Not counting the root, there are $n-1$ nodes that could be structured as:

$C(n)=C(0)\ast C(n-1)+C(1)\ast C(n-2)+\cdots +C(n-2)\ast C(1)+C(n-1)\ast C(0)$
$C(n)={\sum }_{i=0}^{n-1}{C}_i{C}_{n-i-1}$

Actually,
$$C_{(}n)=C(2n,n))-C(2n-1)=\frac{1}{n+1}\left(\genfrac{}{}{0ex}{}{2n}{n}\right)=\frac{(2n)!}{(n+1)!n!}=\prod _{k=2}^n\frac{n+k}{k}$$Proof using generating functions,

Dyck Language
Hands across the table

Exponential Generating Functions

Binomial Theorem
$(1+x{)}^n={\sum }_{k=0}^{\infty }C(n,k)x^k={\sum }_{k=0}^{\infty }\frac{C(n,k)k!}{k!}{x}^k={\sum }_{k=0}^{\infty }P(n,k)\frac{x^k}{k!}$

• $3a_1,2a_2,3a_3$
  How many combinations of length $4$?
  Key: $70$
  Hint: $4!(\frac{4}{3!}+\frac{3}{2!2!}+\frac{3}{2!})=70$

For a seq KaTeX parse error: Unexpected character: '' at position 19: …, a_, a_2 \dots̲, construct:
$$G_e(x)=a_0+\frac{a_1}{1!}x+\frac{a_2}{2!}{x}^2+\frac{a_3}{3!}{x}^3+\cdots +\frac{a_k}{k!}{x}^k+\dots$$ $G(x)$ is known as the exponential generating function of $a_0,a_1,a_2,\dots$

• Using the digits 1, 2, 3 and 4, construct a five digit number where 1 doesn’t appear more than 2 times, but has to appear at least once; 2 can’t appear more than once; 3 can appear up to 3 times but can also not appear in the number; 4 can only appear an even number of times. How many numbers can satisfy these conditions?

Hint:
An $r$ digit number that satisfies the condition is $a_r$; the exponential generating function for the seq $a_0,a_1,\dots ,a_{10}$ is:

There are $215$ 5-digit numbers that satisfy the conditions.

• With digits $1,3,5,7,9$, how many n-digit numbers are there, where $3$ and $7$ appear an even number of times, and $1,5,9$ don’t have any conditions.

  Hint:
  Assume an r-digit number that satisfies the conditions is $a_r$. Then, the exponential generating function of the seq of $a_1,a_2,...,a_3$ is:

Derangements

之前浅显的学习笔记
A derangement is a permutation of the elements of a set such that none of the elements appear in their original position.

Suppose the derangement of n elements in a sequence $1,2,\dots n$ is $D_n$, for every $i^{th}$ number, it has to switch with the other $n-1$ numbers, followed by a derangement of the other $n-2$ elements. We will have a total of $(n-1)D_{n-2}$ number of derangements.
For the other part, not considering the number $i$, there is a derangement of the other $n-1$ elements to be done, followed by switching $i$ with the other numbers yielding a total of $(n-1)D_{n-1}$ number of derangements.
$D(n)=(n-1)(D_{n-1}+D_{n-2}),D_0=0,D_2=1$
$\therefore D_n=1$
$$D(n)=(n-1)(D_{n-1}+D_{n-2}),n\ge 3$$

Stirling Numbers

Stirling numbers of the first kind

$n$ people group dancing, divided into m disjoint circular groups.
$s(n,0)=0,s(1,1)=1$
$s(n+1,m)=s(n,m-1)+n\ast s(n,m)$

• The $n+1^{st}$ person can dance alone, and the others form $m-1$ groups.
• The $n+1^{st}$ person joins a group, and will have $n$ choices for groups, while the other $n$ people have $s(n,m)$ ways of forming groups.

Stirling numbers of the second kind

Definition: The number of ways $n$ non-identical balls in m identical boxes where none of the boxes are empty is known as the Stirling number of the second kind.

Theorem: Stirling number of the second kind $S(n,m)$ has the following properties:

• $S(n,0)=0$
• $S(n,1)=1$
• KaTeX parse error: Unexpected character: '' at position 16: S(n, 2) = 2^{n-̲1}-1
  Proof:
  Assume there are $n$ non-identical balls $b_1,b_2,...,b_n$, from which we pick $b_1$. Now for the other $n-1$ balls, there are two possibilities, either they can be in the same box as $b_1$ or a box that doesn’t contain $b_1$. But it can only satisfy one of these conditions, therefore there are $2^{n-1}-1$ ways of placing the balls.
• $S(n,n)=1$

Theorem: Stirling number of the second kind satisfies the following recurrence relation.
KaTeX parse error: Unexpected character: '' at position 23: … = m*S(n-1, m)+̲S(n-1, m-1) (n …

Proof: Suppose there are $n$ non-identical balls $b_1,b_2,\dots ,b_n$, from which we pick a ball $b_1$.
Placing n balls into m boxes where no box is empty can be classified into two possibilities.

1. $b_1$ is placed on its own in a box, therefore, the count is $S(n-1,m-1)$
2. $b_1$ is not the only ball in its box. This is equivalent to placing $n-1$ balls in $m$ boxes, where no box is empty. This yields the count $S(n-1,m).$ Out of these boxes, $b_1$ can be placed in any of these $m$ boxes, so the total number of ways in this possibility equals to $m\ast S(n-1,m)$.

A generating function is a clothesline on which we hang up a sequence of numbers.
— — Herbert Wilf

Summary of placing balls into boxes problem