Discrete Mathematics Midterm

逻辑和推理

命题逻辑

Operators

• $x\to y$, material implication (conditional statement)
  • $(x\to y)=1⟺(x=0)\vee (y=1)$
• $x\leftrightarrow y$, bi-conditional statement
  • $(x\leftrightarrow y)=1⟺(x=y=0)\vee (x=y=1)$

逻辑等价式

Distributive laws

$$\begin{array}{c}p\vee (q\wedge r)\equiv (p\vee q)\wedge (p\vee r)\\ p\wedge (q\vee r)\equiv (p\wedge q)\vee (p\wedge r)\end{array}$$

De Morgan's laws

$$\begin{array}{c}\neg (p\wedge q)\equiv \neg p\vee \neg q\\ \neg (p\vee q)\equiv \neg p\wedge \neg q\end{array}$$

Absorption laws

$$\begin{array}{c}p\vee (p\wedge q)\equiv p\\ p\wedge (p\vee q)\equiv p\end{array}$$

Conditional distributive laws

$$\begin{array}{c}(p\vee q)\to r\equiv (p\to r)\wedge (q\to r)\\ (p\wedge q)\to r\equiv (p\to r)\vee (q\to r)\end{array}$$

Like De Morgan's laws, ha?

Normal forms

Disjunction normal Form

$$F(x_1,x_2,\cdots ,x_n)=Q_1\vee Q_2\vee \cdots \vee Q_m$$

Conjunction Normal Form

$$F(x_1,x_2,\cdots ,x_n)=Q_1\wedge Q_2\wedge \cdots \wedge Q_m$$

Minimal Operators to Express Statements

One could use and only use

1. AND, NOT
2. OR, NOT
3. IMPLICATION

谓词和量词

Unique Quantifier

The uniqueness quantifier is denoted as $!\exists$, the statement $!\exists xP(x)$ states "There exists a unique $x$ such that $P(x)$ is true."

Quantifiers with Restricted Domains

$$\forall P(x)Q(x)\equiv \forall x\left(P(x)\to Q(x)\right)$$

and

$$\exists P(x)Q(x)\equiv \exists x\left(P(x)\wedge Q(x)\right)$$

Tip

Note that for existential quantification, here is $P(x)\wedge Q(x)$ but not $P(x)\to Q(x)$

推理规则

Tautology	Name
$(p\wedge (p\to q))\to q$	Modus ponens
$(\neg q\wedge (p\to q)\to \neg p)$	Modus tollens
$((p\to q)\wedge (q\to r))\to (p\to r)$	Hypothetical syllogism
$((p\vee q)\wedge \neg p)\to q$	Disjunctive syllogism
$p\to (p\vee q)$	Addition
$(p\vee q)\to p$	Simplification
$((p)\wedge (q))\to (p\wedge q)$	Conjunction
$((p\vee q)\wedge (\neg p\vee r))\to (q\vee r)$	Resolution
潘天麟 概率论与数理统计 2024年10月26日（周六）下午14:00-16:00 17 阶一6
潘天麟 离散数学 2024年10月26日（周六）上午09:00-10:30 18 阶一6
潘天麟 信号与系统 2024年10月27日（周日）下午14:00-15:40 21 阶二3

证明方法

Proof by Contraposition

Proofs by contraposition make use of the fact that the conditional statement $p\to q$ is equivalent to tis contrapositive, $\neg q\to \neg p$. This means that the conditional statement $p\to q$ can be proved by showing that its contrapositive, $\neg q\to \neg p$, is true. In a proof by contraposition of $p\to q$, we take $\neg q$ as a premise, and using axioms, definitions, and previously proven theorems, together with rules of inference, we show that $\neg p$ must follow.

Proof by Contradiction

Suppose we want to prove that a statement $p$ is true. Furthermore, suppose that we can find a contradiction $q$ such that $\neg p\to q$ is true. Because $q$ is false, but $\neg p\to q$ is true, we can conclude that $\neg p$ is false, which means that $p$ is true.

To find a contradiction $q$, we consider the statement $r\wedge \neg r$, which is a contradiction whenever $r$ is a proposition. We can prove that $p$ is true if we can show that $\neg p\to (r\wedge \neg r)$ is true for some proposition $r$.

集合

幂集(Power Sets)

Given a set $S$, the power set of $S$ is the set of all subsets of the set $S$. The power set of $S$ is denoted as $\mathcal{P}(S)$

For example

$$\begin{array}{c}\mathcal{P}\left(\left\{0,1,2\right\}\right)=\left\{\varnothing ,\left\{0\right\},\left\{1\right\},\left\{2\right\},\left\{0,1\right\},\left\{0,2\right\},\left\{1,2\right\},\left\{0,1,2\right\}\right\}\\ \mathcal{P}(\varnothing )=\left\{\varnothing \right\}\\ \mathcal{P}(\left\{\varnothing \right\})=\left\{\varnothing ,\left\{\varnothing \right\}\right\}\end{array}$$

If $\left|S\right|=n$, than we have $\left|\mathcal{P}(S)\right|=2^n$

Cartesian积

$$A\times B=\left\{(a,b)\mid a\in A\wedge b\in B\right\}$$ $$A_1\times A_2\times \cdots \times A_n=\left\{(a_1,a_2,\dots ,a_n)\mid a_i\in A_i\text{ for }i=1,2,\dots ,n\right\}$$

Relation

A subset $R$ of the Cartesian Product $A\times B$ is called a relation from set $A$ to the set $B$. A relation from $A$ to itself is called a relation on $A$

重集(Multisets)

The multiset denoted by $\left\{a,a,a,b,b\right\}$ is the multiset that contains the element $a$ thrice and $b$ twice. There is another notation $\left\{m_1\cdot a_1,m_2\cdot a_2,\dots ,m_r\cdot a_r\right\}$ denotes the multiset with $a_i$ occurs $m_i$ times.

The numbers $m_i$ are called the multiplicities of the elements $a_i$. (Elements not in a multiset are assigned $0$ as their multiplicities in this set)

势(基数, Cardinality)

The sets $A$ and $B$ have the same cardinality if and only if there is one-to-one correspondence from $A$ to $B$. When $A$ and $B$ have the same cardinality, we write $\left|A\right|=\left|B\right|$

If there is a one-to-one function from $A$ to $B$, the cardinality of $A$ is less than or the same as the cardinality of $B$ and we write $\left|A\right|\le \left|B\right|$.

If $A$ and $B$ are sets with $\left|A\right|\le \left|B\right|$ and $\left|B\right|\le \left|A\right|$, then $\left|A\right|=\left|B\right|$

Countable Sets

A set that is either finite or has the same cardinality as the set of positive integers is called countable. A set that is not countable is called uncountable. When an infinite set $S$ is countable, we denote the cardinality of $S$ by ${\aleph }_0$. We write $\left|S\right|={\aleph }_0$ and say that $S$ has cardinality "aleph null"

If $A$ and $B$ are countable sets, then $A\cup B$ is also countable.

The Continuum Hypothesis

Cantor states that the cardinality of a set is always less than the cardinality of its power set. Hence, $\left|{\mathbb{Z}}^{+}\right|<\left|\mathcal{P}({\mathbb{Z}}^{+})\right|$. We can rewrite this as ${\aleph }_0<2^{{\aleph }_0}$. We denote $2^{{\aleph }_0}={\aleph }_1$

We have $\left|\mathbb{R}\right|={\aleph }_1$

The continuum hypothesis is that there is no cardinal number $X$ between these numbers:

$${\aleph }_0<{\aleph }_1<{\aleph }_2<\cdots$$

where ${\aleph }_{n+1}=2^{{\aleph }_n}$

矩阵

01矩阵

A 0-1 Matrix is a matrix whose elements are either 0 or 1.

Let $A=(a_{ij}{)}_{m\times n},B=(b_{ij}{)}_{m\times n}$ be 0-1 matrices, we define

$$A\vee B=(a_{ij}\vee b_{ij}{)}_{m\times n},A\wedge B=(a_{ij}\wedge b_{ij}{)}_{m\times n}$$

and

$$A\odot B=(c_{ij}{)}_{m\times n},\text{where}{c}_{ij}=(a_{i1}\wedge b_{1j})\vee (a_{i2}\wedge b_{2j})\vee \cdots \vee (a_{ik}\wedge b_{kj})$$

which is called the Boolean Product of $A$ and $B$

Tip

We can see Boolean Product as a normal matrix multiplication with $+$ replaced by $\vee$ and $\times$ replaced by $\wedge$

Based on this, we can also define the $r$-th Boolean power of $A$ as

$$A^{[r]}=\underset{A\text{ occurs }r\text{ times}}{\underbrace{A\odot A\odot \cdots \odot A}}$$

算法

性质

• 输入
• 输出
• 确定性: Well-defined, without a second interpretation
• 正确性: Generate correct output for every possible input
• 有限性: Finishing in finite steps
• 有效性: Each step finishing in finite time
• 通用性: Can be generalized

Note

The following example demonstrate correctness for specific cases but lack generality

def sort_five_numbers(a, b, c, d, e): 
	return sorted([a, b, c, d, e])

时间复杂度

Ranking

$1\le \log n\le n\le n\log n\le n^2\le 2^n\le n!$

记号

Big $\Theta$

For function $f(n),g(n)$, $f(n)=\Theta (g(n))$ if and only if $\exists c_1,c_2,n_0>0$ s.t.

$$\forall n\ge n_0,0\le c_1\cdot g(n)\le f(n)\le c_2\cdot g(n)$$

Big $\Theta$ symbol indicates both the upper and lower bound of a time function

Big $O$

We say $f(n)=O(g(n))$ if and only if $\exists c,n_0$ s.t.

$$\forall n\ge n_0,0\le f(n)\le c\cdot g(n)$$

This gives us the upper bound of a time function.

Big $\Omega$

We say $f(n)=\Omega (g(n))$ if and only if $\exists c,n_0$, s.t.

$$\forall n\ge n_0,0<c\cdot g(n)\le f(n)$$

This gives us the lower bound of a time function.

数论

模算数

$({\mathbb{Z}}_m,+,\cdot )$ is a commutative ring:

1. Definition

$$\begin{array}{r}a\equiv b(\mathrm{mod}m)⟺a\mathrm{mod}m=b\mathrm{mod}m\end{array}$$

2. If $a\equiv b(\mathrm{mod}m)$ and $c\equiv d(\mathrm{mod}m)$, then

$$a+c\equiv b+d(\mathrm{mod}m),ac\equiv bd(\mathrm{mod}m)$$

$$\begin{array}{rl}(a+b)\mathrm{mod}m& =(a\mathrm{mod}m+b\mathrm{mod}m)\mathrm{mod}m\\ ab\mathrm{mod}m& =\left[(a\mathrm{mod}m)(b\mathrm{mod}m)\right]\mathrm{mod}m\end{array}$$

${\mathbb{Z}}_m$中的乘法逆元

The modular inverse of $a$ in the ring of integers modulo $m$ is an integer $x$ such that

$$ax\equiv 1(\mathrm{mod}m)$$

Euler定理

i if $n$ is a positive integer, and let $a$ be an integer that is relatively prime to $n$. Then

$$a^{\varphi (n)}\equiv 1(\mathrm{mod}n)$$

where $\varphi (n)$ is Euler's totient function, which counts the number of positive integers less than $n$ that is also relative prime to $n$

Fermat小定理(Euler定理的一个特例)

If $p$ is a prime, then $\varphi (p)=p$, then Euler's Theorem becomes Fermat's Little Theorem

$$a^{p-1}\equiv 1(\mathrm{mod}p)$$

伪素数

Let $b$ be a positive integer. If $n$ is a composite positive integer, and $b^{n-1}\equiv 1(\mathrm{mod}n)$, then $n$ is called a pseudo-prime to be base $b$

Among the positive integer not exceeding $x$, where $x$ is a positive real number, compared to primes there are relatively few pseudo-primes to the base $b$. So determining whether $b^{n-1}\equiv 1(\mathrm{mod}n)$ is a useful test that provides some evidence concerning whether $n$ is prime.

However, there are some composite $n$ that satisfies the congruence $b^{n-1}\equiv 1(\mathrm{mod}n)$ for all positive integers $b$ with $\gcd (b,n)=1$ is called a Carmichael number. For example, the integer $561$ is a Carmichael number.

Exgcd

For example, calculate $\mathrm{exgcd}(108,68)$

$a(r_i)$	$b$	$x_i=x_{i-2}-x_{i-1}{q}_{i-1}$	$y_i=y_{i-2}-y_{i-1}{q}_{i-1}$
0	108	1	0
108	68	0	1
68	40	1	-1
40	28	-1	2
28	12	2	-3
12	4	-5	8
4	0
this gives

$$-5\times 108+8\times 68=\gcd (108,8)=4$$

中国剩余定理

Let $n_1,\dots ,n_k$ be integers greater than $1$, and $N={\prod }_{i=1}^k{n}_i$. If the $n_i$ are pair-wise co-prime, and if $a_1,\dots ,a_k$ are any integers, then the system

$$\begin{array}{rl}x& \equiv a_1(\mathrm{mod}{n}_1)\\ & ⋮\\ x& \equiv a_k(\mathrm{mod}{n}_k)\end{array}$$

has a solution, and any two solutions, say $x_1$ and $x_2$, satisfy $x_1\equiv x_2(\mathrm{mod}N)$

To construct a solution, let $N_i=N/n_i$ be the product of all moduli but one. As the $n_i$ are pair-wise co-prime, $N_i$ and $n_i$ are co-prime. Thus Bezout's identity applies, and there exist integers $M_i$ and $m_i$ such that

$$M_i{N}_i+m_i{n}_i=1$$

A solution of the system of congruences is

$$x=\sum _{i=1}^k{a}_i{M}_i{N}_i$$

In fact, as $N_j$ is a multiple of $n_i$ for $i\ne j$, we have

$$x\equiv a_i{M}_i{N}_i\equiv a_i(1-m_i{n}_i)\equiv a_i(\mathrm{mod}{n}_i)$$

原根和离散对数

A primitive root module a prime $p$ is an integer $r$ in ${\mathbb{Z}}_p$ such that every non-zero element of ${\mathbb{Z}}_p$ is a power of $r$.

We can say here $r$ can generate the space of ${\mathbb{Z}}_p$

An important fact is that there is a primitive root module $p$ for every prime $p$. Suppose this primitive root is $r$, and $a$ is an integer between $1$ and $p-1$ inclusive. If $r^e\mathrm{mod}p=a$ and $0\le e\le p-1$, we say that $e$ is the discrete logarithm of $a$ module $p$ to the base $r$ and we write ${\log }_{r}a=e$

例题

T1

Prove that $(0,1)$ and $[0,1]$ have the same cardinality

Using Hilbert's Hotel. Let $H=\left\{\frac{1}{n}:n\in \mathbb{N}\right\}$, then define $f:(0,1)\to [0,1]$ so that

$$\begin{array}{r}\frac{1}{2}\to 0,\frac{1}{3}\to 1,\frac{1}{n}\to \frac{1}{n-2}\end{array}$$

and

$$x\to x,x\notin H$$

Tip

We can also construct a bijection from $(0,1)$ to $[0,1)$ by letting $\frac{1}{2}\to 0$ and $\frac{1}{n}\to \frac{1}{n-1}$

T2

Let $S\subset \left\{1,2,3,\dots ,98,99,100\right\}$ where $\left|S\right|=62$. Prove that there exist $x,y\in S$ such that $x-y=23$

Let's divide $\left\{1,2,\dots ,100\right\}$ into threes sections

$$\left\{1,2,\dots ,100\right\}=\underset{A}{\underbrace{\left\{1,2,\dots ,46\right\}}}\cup \underset{B}{\underbrace{\left\{47,48,\dots ,69\right\}}}\cup \underset{C}{\underbrace{\left\{93,94,\dots ,100\right\}}}$$

First we prove that we can select $23$ elements at most from $A$ such that there are no $x,y$ in this selection satisfying $x-y=23$. To conclude this, we further divide $A$ into pairs

$$A=\left\{1,24\right\}\cup \left\{2,25\right\}\cup \cdots \cup \left\{23,46\right\}$$

We can select no more than $1$ element in each pair, so we can select no more than $23$ elements in $A$.

Similarly, we can select no more than $23$ elements in $B$. Therefore, we can select no more than $23+23+8=54$ elements in $\left\{1,2,\dots ,100\right\}$

T3