『Discrete Mathematics and Its Applications』离散数学及其应用学习笔记

本文深入探讨了离散数学的基础概念,包括逻辑与证明、谓词与量词、命题范式、集合及其运算、函数特性与组合数学习题。解析了逻辑等价、命题规则、全称与存在量化,以及集合的性质和操作。同时,涵盖了函数的定义域、值域、映射特性,如单射、满射、双射,并讨论了无限集基数的概念。

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教材是Discrete Mathematics and Its Applications 7th,做了一些简单的翻译和简单的笔记。

1. The Foundations: Logic and Proofs

ENGCHNDenote
negation否定¬p\neg p¬p
conjunction (and)合取p∧qp \land qpq
disjunction (or)析取p∨qp \lor qpq
exlusive or异或p⊕qp \oplus qpq
conditional statement/
implication
蕴含p→qp \to qpq
biconditional statement等价p↔qp \leftrightarrow qpq
tautology永真式≡1\equiv 11
contingency可能式0/10/10/1
contradiction矛盾式≡0\equiv 00
proposition命题p→qp \to qpq
converse proposition逆命题q→pq \to pqp
inverse proposition否命题¬p→¬q\neg p \to \neg q¬p¬q
contrapositive proposition逆否命题¬q→¬p\neg q \to \neg p¬q¬p

 Precedence of Logical Operators:
¬ > ∧ > ∨> → > ↔\neg\ >\ \land\ >\ \lor>\ \to\ >\ \leftrightarrow¬ >  > >  > 
 Implication Law:
p→q≡¬p∨q  ⟺  p \to q \equiv \neg p \lor q \iffpq¬pq if ppp then q  ⟺  qq \iff qqq if/when p  ⟺  pp \iff ppp only if q  ⟺  qq \iff qqq unless ¬p\neg p¬p
 Equivalence Law:
p↔q≡(p→q)∧(q→p)  ⟺  pp \leftrightarrow q \equiv (p \to q) \land (q \to p) \iff ppq(pq)(qp)p if and only if q  ⟺  pq \iff pqp iff qqq

Logical Equivalences:
EquivalenceENGCHN
p∧T≡pp \land T \equiv ppTp
p∨F≡pp \lor F \equiv ppFp
Identity Laws同一律
p∨T≡Tp \lor T \equiv TpTT
p∧F≡Fp \land F \equiv FpFF
Domination Laws零律
p∨p≡pp \lor p \equiv pppp
p∧p≡pp \land p \equiv pppp
Idempotent Laws幂等律
¬(¬p)≡p\neg (\neg p) \equiv p¬(¬p)pDouble Negation Law双重否定律
p∨q≡q∨pp \lor q \equiv q \lor ppqqp
p∧q≡q∧pp \land q \equiv q \land ppqqp
Comutative Laws交换律
(p∨q)∨r≡p∨(q∨r)(p \lor q) \lor r \equiv p \lor (q \lor r)(pq)rp(qr)
(p∧q)∧r≡p∧(q∧r)(p \land q) \land r \equiv p \land (q \land r)(pq)rp(qr)
Associative Laws结合律
p∨(q∧r)≡(p∨q)∧(p∨r)p \lor (q \land r) \equiv (p \lor q) \land (p \lor r)p(qr)(pq)(pr)
p∧(q∨r)≡(p∧q)∨(p∧r)p \land (q \lor r) \equiv (p \land q) \lor (p \land r)p(qr)(pq)(pr)
Distributive Laws分配律
¬(p∧q)≡¬p∨¬q\neg (p \land q) \equiv \neg p \lor \neg q¬(pq)¬p¬q
¬(p∨q)≡¬p∧¬q\neg (p \lor q) \equiv \neg p \land \neg q¬(pq)¬p¬q
De Morgan’s Laws德摩根定律
p∨(p∧q)≡pp \lor (p \land q) \equiv pp(pq)p
p∧(p∨q)≡pp \land (p \lor q) \equiv pp(pq)p
Absorption Laws吸收律
p∨¬p≡Tp \lor \neg p \equiv Tp¬pT
p∧¬p≡Fp \land \neg p \equiv Fp¬pF
Negation Laws排中律
矛盾律

Show p≡qp \equiv qpq :
 a. Using truth tables
 b. Using already-proved equivalences (Recommended)

Predicates and Quantifiers:

 Universal Quantifier ∀\forall :
∀xP(x)  ⟺  P(x)\forall xP(x) \iff P(x)xP(x)P(x) for all values of xxx in the (restricted) domain
 Existential Quantifier ∃\exist :
∃xP(x)  ⟺  \exist xP(x) \iffxP(x) There exists (at least) an element xxx in the domain such that P(x)P(x)P(x)
 De Morgan’s Laws for quantifiers:
¬∀xP(x)≡∃x¬P(x)\neg \forall xP(x) \equiv \exist x \neg P(x)¬xP(x)x¬P(x)
¬∃xQ(x)≡∀x¬Q(x)\neg \exist xQ(x) \equiv \forall x \neg Q(x)¬xQ(x)x¬Q(x)
Notice: ∀\forall and ∃\exist have higher precedence than any logical operators
 Remark:
 All P(x)P(x)P(x) are Q(x)  ⟺  ∀x(P(x)→Q(x))Q(x) \iff \forall x (P(x) \to Q(x))Q(x)x(P(x)Q(x))
 No P(x)P(x)P(x) are Q(x)  ⟺  ∀x(P(x)→¬Q(x))Q(x) \iff \forall x (P(x) \to \neg Q(x))Q(x)x(P(x)¬Q(x))
 Some P(x)P(x)P(x) are Q(x)  ⟺  ∃x(P(x)∧Q(x))Q(x) \iff \exist x (P(x) \land Q(x))Q(x)x(P(x)Q(x))
 Some P(x)P(x)P(x) are not Q(x)  ⟺  ∃x(P(x)∧¬Q(x))Q(x) \iff \exist x (P(x) \land \neg Q(x))Q(x)x(P(x)¬Q(x))
eg. Recall that the definition of lim⁡x→af(x)=L{\lim_{x \to a}f(x)=L}xalimf(x)=L  is ∀ϵ>0 ∃δ>0 ∀x(0<∣x−a∣<δ→∣f(x)−L∣<ϵ){\forall \epsilon>0\ \exist \delta>0\ \forall x(0<|x-a|<\delta \to |f(x)-L|<\epsilon)}ϵ>0 δ>0 x(0<xa<δf(x)L<ϵ)  And what is the inverse proposition of this statement?

Propositional Normal Forms:
ENGCHNDenote
Maxterm极大项M=a1∨a2∨⋯∧ai∨⋯M=a_1 \lor a_2 \lor \cdots \land a_i \lor \cdotsM=a1a2ai
(there exists unique value of each aia_iai so that M=0M=0M=0)
Minterm极小项m=b1∧b2∧⋯∧bi∧⋯m=b_1 \land b_2 \land \cdots \land b_i \land \cdotsm=b1b2bi
(there exists unique value of each bib_ibi so that m=1m=1m=1)
Conjunctive Normal Form (CNF)合取范式C=A1∧A2∧⋯∧Ai∧⋯C=A_1 \land A_2 \land \cdots \land A_i \land \cdotsC=A1A2Ai, AiA_iAi are propositions
Disjunctive Normal Form (DNF)析取范式D=B1∨B2∨⋯∨Bi∨⋯D=B_1 \lor B_2 \lor \cdots \lor B_i \lor \cdotsD=B1B2Bi, BiB_iBi are propositions
Full Conjunctive Normal Form主合取范式X=M1∧M2∧⋯∧Mi∧⋯X=M_1 \land M_2 \land \cdots \land M_i \land \cdotsX=M1M2Mi, MiM_iMi are maxterms
Full Disjunctive Normal Form主析取范式Y=m1∨m2∨⋯∨mi∨⋯Y=m_1 \lor m_2 \lor \cdots \lor m_i \lor \cdotsY=m1m2mi, mim_imi are minterms
Prenex Normal Form前约束范式O=Q1z1Q2z2⋯QnznEO=Q_1z_1Q_2z_2 \cdots Q_nz_nEO=Q1z1Q2z2QnznE, Qi=∀Q_i=\forallQi= or ∃\exist,
EEE is a quantifier-free proposition

 Transforming to Prenex Normal Form:
  1.eliminate →\to and ↔\leftrightarrow by:
  p↔q≡(p→q)∧(q→p)≡(¬p∨q)∧(¬q∨p)≡(p∧q)∨(¬p∧¬q)p \leftrightarrow q \equiv (p \to q) \land (q \to p) \equiv (\neg p \lor q) \land (\neg q \lor p) \equiv (p \land q) \lor (\neg p \land \neg q)pq(pq)(qp)(¬pq)(¬qp)(pq)(¬p¬q)
  2.move all ¬\neg¬ inward by applying De Morgan’s Laws
  3.rename variables to avoid ambiguity if necessary
  4.move all quantifiers to the front by:
  QxA(x)∧P≡Qx(A(x)∧P)Q xA(x) \land P \equiv Q x(A(x) \land P)QxA(x)PQx(A(x)P),
  QxA(x)∨P≡Qx(A(x)∨P)Q xA(x) \lor P \equiv Q x(A(x) \lor P)QxA(x)PQx(A(x)P),
  Q′xA(x)∧Q′′yB(y)≡Q′xQ′′y(A(x)∧B(y))Q'xA(x) \land Q''yB(y) \equiv Q'xQ''y(A(x) \land B(y))QxA(x)QyB(y)QxQy(A(x)B(y)),
  Q′xA(x)∨Q′′yB(y)≡Q′xQ′′y(A(x)∨B(y))Q'xA(x) \lor Q''yB(y) \equiv Q'xQ''y(A(x) \lor B(y))QxA(x)QyB(y)QxQy(A(x)B(y)),
  QxQyC(x,y)≡QyQxC(x,y)QxQyC(x,y) \equiv QyQxC(x,y)QxQyC(x,y)QyQxC(x,y),
  where we have:
  Q,Q′,Q′′=∀Q,Q',Q''=\forallQ,Q,Q= or ∃\exist,
  PPP is a quantifier-free proposition,
  A(x),B(y),C(x,y)A(x),B(y),C(x,y)A(x),B(y),C(x,y) are propositions.
 *5.transform to Prenex CNF/DNF

Rules of Inference:
RuleTautologyENGCHN
pp→q∴  q\begin{aligned} &p \\ &p \to q \\ \hline \therefore \; &q \end{aligned}ppqq(p∧(p→q))→q(p \land (p \to q)) \to q(p(pq))qModus ponens假言推理式
¬qp→q∴  ¬p\begin{aligned} &\neg q \\ &p \to q \\ \hline \therefore \; &\neg p \end{aligned}¬qpq¬p(¬q∧(p→q))→¬p(\neg q \land (p \to q)) \to \neg p(¬q(pq))¬pModus tollens否定拒取式
p→qq→r∴  p→r\begin{aligned}&p \to q \\ &q \to r \\ \hline \therefore \; &p \to r \end{aligned}pqqrpr((p→q)∧(q→r))→(p→r)((p \to q) \land (q \to r)) \to (p \to r)((pq)(qr))(pr)Hypothetical syllogism假言三段论
p∨q¬p∴  q\begin{aligned}&p \lor q \\ &\neg p\\ \hline \therefore \; &q \end{aligned}pq¬pq((p∨q)∧¬p)→q((p \lor q) \land \neg p) \to q((pq)¬p)qDisjunctive syllogism析取三段论
p∴  p∨q\begin{aligned}&p \\ \hline \therefore \; &p \lor q \end{aligned}ppqp→(p∨q)p \to (p \lor q)p(pq)Addition附加律
p∧q∴  q\begin{aligned}&p \land q \\ \hline \therefore \; &q \end{aligned}pqq(p∧q)→q(p \land q) \to q(pq)qSimplification消去律
pq∴  p∧q\begin{aligned}&p \\ &q \\ \hline \therefore \; &p \land q \end{aligned}pqpq((p)∧(q))→(p∧q)((p) \land (q)) \to (p \land q)((p)(q))(pq)Conjunction合取式
p∨q¬p∨r∴  q∨r\begin{aligned}&p \lor q \\ &\neg p \lor r \\ \hline \therefore \; &q \lor r \end{aligned}pq¬prqr((p∨q)∧(¬p∨r))→(q∨r)((p \lor q) \land (\neg p \lor r)) \to (q \lor r)((pq)(¬pr))(qr)Resolution消解规则
p→rq→sp∨q∴  r∨s\begin{aligned}&p \to r \\ &q \to s \\ &p \lor q \\ \hline \therefore \; &r \lor s \end{aligned}prqspqrs((p→r)∧(q→r)∧(p∨q))→(r∨s)((p \to r) \land (q \to r) \land (p \lor q)) \to (r \lor s)((pr)(qr)(pq))(rs)Constructive dilemma构造性两难

Remark: (P1∧P2∧⋯∧Pn∧P)→Q≡(P1∧P2∧⋯∧Pn)→(P→Q)(P_1 \land P_2 \land \cdots \land P_n \land P) \to Q \equiv (P_1 \land P_2 \land \cdots \land P_n) \to (P \to Q)(P1P2PnP)Q(P1P2Pn)(PQ)

RuleName
∀xP(x)∴  P(c)\begin{aligned}&\forall xP(x) \\ \hline \therefore \; &P(c) \end{aligned}xP(x)P(c) , for an arbitary cccUniversal instantiation (UI)
P(c)∴  ∀xP(x)\begin{aligned}&P(c) \\ \hline \therefore \; &\forall xP(x) \end{aligned}P(c)xP(x) , for an arbitary cccUniversal generalization (UG)
∃xP(x)∴  P(c)\begin{aligned}&\exist xP(x) \\ \hline \therefore \; &P(c) \end{aligned}xP(x)P(c) , for some element cccExistential instantiation (EI)
P(c)∴  ∃xP(x)\begin{aligned}&P(c) \\ \hline \therefore \; &\exist xP(x) \end{aligned}P(c)xP(x) , for some element cccExistential generalization (EG)

Remark: ∃!x  s.t.  P(x)  ⟺  ∃x(P(x)∧∀y(x≠y→¬P(y)))\exist!x \; \mathrm{s.t.} \; P(x) \iff \exist x(P(x) \land \forall y(x \ne y \to \neg P(y)))!xs.t.P(x)x(P(x)y(x=y¬P(y)))

2. Basic Structures: Sets, Functions, Sequences, Sums, and Matrices

DenoteENG
N={0,1,2,… }\mathbb{N}=\{0, 1, 2, \dots\}N={0,1,2,}Natural numbers
Z={…,−1,0,1,… }\mathbb{Z}=\{\dots, -1, 0, 1, \dots\}Z={,1,0,1,}Intergers
Z+={1,2,3,… }\mathbb{Z^+}=\{1, 2, 3, \dots\}Z+={1,2,3,}Positive integers
Q={pq ∣ p∈Z+∪Z−,q∈Z}\mathbb{Q}=\left\{\dfrac{p}{q} \, \Big \vert \, p \in \mathbb{Z}^+ \cup \mathbb{Z}^-, q \in \mathbb{Z} \right\}Q={qppZ+Z,qZ}Rational numbers
R\mathbb{R}RReal numbers
R+\mathbb{R^+}R+Positve real numbers
C\mathbb{C}CComplex numbers
ENGCHNDenotei.e.
Subset子集A⊆BA \subseteq BAB∀x(x∈A→x∈B)\forall x(x \in A \to x \in B)x(xAxB)
Proper subset真子集A⊂BA \subset BAB∀x(x∈A→x∈B)∧∃x(x∈B∧x∉A)\forall x(x \in A \to x \in B) \land \exist x(x \in B \land x \notin A)x(xAxB)x(xBx/A)
Equal等集A=BA=BA=BA⊆B∧B⊆AA \subseteq B \land B \subseteq AABBA
∀x(x∈A↔x∈B)\forall x(x \in A \leftrightarrow x \in B)x(xAxB)
Cardinality基数∣S∣Card⁡(S)\vert S \vert \\ \operatorname{Card}(S)SCard(S)∑x∈S1\sum_{x \in S}1xS1
Power set幂集P(S)\mathcal{P}(S)P(S)⋃A⊆S{A}\bigcup_{A \subseteq S}\{A\}AS{A}

Remark: for every set SSS, we have ∅∈S\varnothing \in SS and S∈SS \in SSS
Cartesian products(笛卡尔积):
A×B={(a,b) ∣ a∈A∧b∈B}A \times B=\{(a,b) \, \vert \, a \in A \land b \in B\}A×B={(a,b)aAbB}
A1×A2×⋯×An={(a1,a2,…,an) ∣ ∀i∈{1,2,…,n}(ai∈An)}A_1 \times A_2 \times \cdots \times A_n=\{(a_1, a_2, \dots, a_n) \, \vert \, \forall i \in \{1, 2, \dots, n\}(a_i \in A_n)\}A1×A2××An={(a1,a2,,an)i{1,2,,n}(aiAn)}

Set Operations:
OperationCHNDenotei.e.
Union并集A∪BA \cup BAB{x ∣ x∈A∨x∈B}\{x \, \vert \, x \in A \lor x \in B\}{xxAxB}
Intersection交集A∩BA \cap BAB{x ∣ x∈A∧x∈B}\{x \, \vert \, x \in A \land x \in B\}{xxAxB}
Complement补集A‾\overline{A}A{x∈U ∣ x∉A}\{x \in U \, \vert \, x \notin A\}{xUx/A}
Difference差集A−BA∖BA - B \\ A \setminus BABAB{x ∣ x∈A∧x∉B}\{x \, \vert \, x \in A \land x \notin B\}{xxAx/B}

Remark:
A−B=A∩B‾A-B=A \cap \overline{B}AB=AB
A∩B=∅  ⟺  AA \cap B=\varnothing \iff AAB=A and BBB are disjoint
Principle of Inclusion–Exclusion(容斥原理):
∣A∪B∣=∣A∣+∣B∣−∣A∩B∣|A \cup B|=|A|+|B|-|A \cap B|AB=A+BAB
Symmetric difference(对称差):
A⊕B=(A−B)∪(B−A)=(A∪B)−(A∩B)A \oplus B=(A-B) \cup (B-A)=(A \cup B)-(A \cap B)AB=(AB)(BA)=(AB)(AB)

Set Identities:
IdentityENGCHN
A∪∅=AA∩U=AA \cup \varnothing = A \\ A \cap U =AA=AAU=AIdentity Laws同一律
A∪U=UA∩∅=∅A \cup U = U \\ A \cap \varnothing = \varnothingAU=UA=Domination Laws支配律
A∪A=AA∩A=AA \cup A = A \\ A \cap A = AAA=AAA=AIdempotent Laws幂等律
(A‾)‾=A\overline{(\overline{A})}=A(A)=AComplementation Law双重互补律
A∪B=B∪AA∩B=B∩AA \cup B = B \cup A \\ A \cap B = B \cap AAB=BAAB=BAComutative Laws交换律
(A∪B)∪C=A∪(B∪C)(A∩B)∩C=A∩(B∩C)(A \cup B) \cup C = A \cup (B \cup C) \\ (A \cap B) \cap C = A \cap (B \cap C)(AB)C=A(BC)(AB)C=A(BC)Associative Laws结合律
A∪(B∩C)=(A∪B)∩(A∪C)A∩(B∪C)=(A∩B)∪(A∩C)A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \\ A \cap (B \cup C) = (A \cap B) \cup (A \cap C)A(BC)=(AB)(AC)A(BC)=(AB)(AC)Distributive Laws分配律
A∩B‾=A‾∪B‾A∪B‾=A‾∩B‾\overline{A \cap B} = \overline{A} \cup \overline{B} \\ \overline{A \cup B} = \overline{A} \cap \overline{B}AB=ABAB=ABDe Morgan’s Laws德摩根定律
A∪(A∩B)=AA∩(A∪B)=AA \cup (A \cap B) = A \\ A \cap (A \cup B) = AA(AB)=AA(AB)=AAbsorption Laws吸收律
A∪A‾=UA∩A‾=∅A \cup \overline{A} = U \\ A \cap \overline{A} = \varnothingAA=UAA=Complement Laws互补律
Functions:

 Let A,B≠∅A,B \ne \varnothingA,B=, we define a mapping fff from AAA to BBB:f:A↦Bf:A \mapsto Bf:AB i.e.∀a(a∈A→∃!b(b∈B∧f(a)=b))\forall a(a \in A \to \exist !b(b \in B \land f(a)=b))a(aA!b(bBf(a)=b))

DenoteENGCHN
AAAdomain定义域
BBBcodomain上域
f(A)f(A)f(A)range值域
bbbimage
aaapreimage原像

Remark:f(A)={f(x) ∣ ∀x∈A}⊆Bf(A)=\{f(x) \, \vert \, \forall x \in A\} \subseteq Bf(A)={f(x)xA}B

ENGCHNDenote
Injective/
One-to-one
单射∀a∈A ∀b∈A(f(a)=f(b)→a=b)  ⟺  ∀a∈A ∀b∈A(a≠b→f(a)≠f(b))\forall a \in A \ \forall b \in A(f(a)=f(b) \to a=b) \\ \iff \forall a \in A \ \forall b \in A(a \ne b \to f(a) \ne f(b))aA bA(f(a)=f(b)a=b)aA bA(a=bf(a)=f(b))
Surjective/
Onto
满射∀b∈B ∃a∈A(f(a)=b)\forall b \in B \ \exist a \in A(f(a)=b)bB aA(f(a)=b)
Bijective/
One-to-one correspondent
双射both injective and surjective
(strictly) Increasing(严格)单调递增∀x∈D ∀y∈D(x<y→f(x)≤f(y))∀x∈D ∀y∈D(x<y→f(x)<f(y))\forall x \in D \ \forall y \in D(x < y \to f(x) \leq f(y)) \\ \forall x \in D \ \forall y \in D(x < y \to f(x) < f(y))xD yD(x<yf(x)f(y))xD yD(x<yf(x)<f(y))
(strictly) Decreasing(严格)单调递减∀x∈D ∀y∈D(x<y→f(x)≥f(y))∀x∈D ∀y∈D(x<y→f(x)>f(y))\forall x \in D \ \forall y \in D(x < y \to f(x) \ge f(y)) \\ \forall x \in D \ \forall y \in D(x < y \to f(x) > f(y))xD yD(x<yf(x)f(y))xD yD(x<yf(x)>f(y))
Inverse function反函数∀y∈B,∃!x∈A s.t.f(x)=y  ⟺  f−1(y)=x\forall y \in B , \exist !x \in A \ \mathrm{s.t.} f(x)=y \\ \iff f^{-1}(y)=xyB,!xA s.t.f(x)=yf1(y)=x

Remark: f:A↦Bf:A \mapsto Bf:AB is a bijection ⟹∣A∣=∣B∣\Longrightarrow \vert A \vert = \vert B \vertA=B
Addition of Functions:
(f+g)(x)=f(x)+g(x)(f+g)(x)=f(x)+g(x)(f+g)(x)=f(x)+g(x)
Mutiplication of Functions:
(f⋅g)(x)=f(x)⋅g(x)(f \cdot g)(x)=f(x) \cdot g(x)(fg)(x)=f(x)g(x)
Composition of Functions:
(f∘g)(x)=f(g(x))(f \circ g)(x)=f(g(x))(fg)(x)=f(g(x))
 Useful properties of Floor Function f(x)=⌊x⌋f(x)=\lfloor x \rfloorf(x)=x and Ceiling Function g(x)=⌈x⌉g(x)=\lceil x \rceilg(x)=x:
⌊−x⌋=−⌈x⌉\lfloor -x \rfloor =- \lceil x \rceilx=x , ⌈−x⌉=−⌊x⌋\lceil -x \rceil =- \lfloor x \rfloorx=x
x−1<⌊x⌋≤x≤⌈x⌉<x+1x-1<\lfloor x \rfloor \leq x \leq \lceil x \rceil <x+1x1<xxx<x+1

Useful Summation Formulea:
SumClosed Form
∑k=0nark,r≠0,1\sum\limits_{k=0}^{n}{ar^k},r \ne 0,1k=0nark,r=0,1a(1−rn+1)1−r\dfrac{a(1-r^{n+1})}{1-r}1ra(1rn+1)
∑k=1nk\sum\limits_{k=1}^{n}{k}k=1nkn(n+1)2\dfrac{n(n+1)}{2}2n(n+1)
∑k=1nk2\sum\limits_{k=1}^{n}{k^2}k=1nk2n(n+1)(2n+1)6\dfrac{n(n+1)(2n+1)}{6}6n(n+1)(2n+1)
∑k=1nk3\sum\limits_{k=1}^{n}{k^3}k=1nk3n2(n+1)24\dfrac{n^2(n+1)^2}{4}4n2(n+1)2
∑k=0∞xk,∣x∣<1\sum\limits_{k=0}^{\infty}{x^k},\vert x \vert <1k=0xk,x<111−x\dfrac{1}{1-x}1x1
∑k=1∞kxk−1,∣x∣<1\sum\limits_{k=1}^{\infty}{kx^{k-1}},\vert x \vert <1k=1kxk1,x<11(1−x)2\dfrac{1}{(1-x)^2}(1x)21
Cardinality of infinite sets:

Definition: infinite set AAA is countable   ⟺  ∣A∣=∣Z+∣=ℵ0\iff |A|=|\mathbb{Z}^+|=\aleph_0A=Z+=0
Remark:
  ∣Q+∣=∣Z+×Z+∣=ℵ0⟸∣Q+∣≤∣Z+×Z+∣=∣Z+∣≤∣Q+∣|\mathbb{Q}^+|=|\mathbb{Z}^+\times\mathbb{Z}^+|=\aleph_0 \Longleftarrow|\mathbb{Q}^+|\leq|\mathbb{Z}^+\times\mathbb{Z}^+|=|\mathbb{Z}^+|\leq|\mathbb{Q}^+|Q+=Z+×Z+=0Q+Z+×Z+=Z+Q+
  ∣R∣=∣(0,1)∣=ℵ1⟸f:(−π2,π2)↦R,f(x)=tan⁡(xπ+12)|\mathbb{R}|=|(0,1)|=\aleph_1\Longleftarrow f:(-\frac{\pi}{2},\frac{\pi}{2}) \mapsto \mathbb{R},f(x)=\tan(\frac{x}{\pi}+\frac12)R=(0,1)=1f:(2π,2π)R,f(x)=tan(πx+21)
  ∣[0,1]∣=∣(0,1)∣=ℵ1⟸∣(0,1)∣≤∣[0,1]∣=∣[14,34]∣≤∣(0,1)∣|[0,1]|=|(0,1)|=\aleph_1 \Longleftarrow |(0,1)|\leq|[0,1]|=|[\frac14,\frac34]|\leq|(0,1)|[0,1]=(0,1)=1(0,1)[0,1]=[41,43](0,1)

Cantor’s Theorem(康托尔定理) and proof: ∣P(A)∣>∣A∣|\mathcal{P}(A)|>|A|P(A)>A
  Let f:A↦P(A)f:A \mapsto \mathcal{P}(A)f:AP(A) be an arbitary function.
  Consider B={x∈A:x∉f(x)}B=\{x \in A : x \notin f(x)\}B={xA:x/f(x)}.
  Assume that there exists y∈Ay \in AyA so that f(y)=Bf(y)=Bf(y)=B.
  If y∈By \in ByB then y∈f(y)y \in f(y)yf(y), conflicting with the definition of BBB.
  If y∉By \notin By/B then y∉f(y)y \notin f(y)y/f(y), so we have y∈By \in ByB conflicting with y∉By \notin By/B.
  Thus there is no such yyy, showing B∈P(A)B \in \mathcal{P}(A)BP(A) does not have preimage in AAA.
  Hence, fff is not a surjection.
  Q.E.D.\mathrm{Q.E.D.}Q.E.D.
Schrőder-Bernstein Theorem(伯恩斯坦定理): ∣A∣≤∣B∣∧∣B∣≤∣A∣⟺∣A∣=∣B∣|A|\leq|B|\land|B|\leq|A| \Longleftrightarrow |A|=|B|ABBAA=B
The Continuum Hypothesis(连续统假设): ∄a s.t. ℵ0<a<ℵ1\not\exists a \ \mathrm{s.t.} \ \aleph_0<a<\aleph_1a s.t. 0<a<1

6. Counting

The Generalized Pigeonhole Principle(鸽巢原理/抽屉原理):
  If NNN objects are placed into kkk boxes, then there is at least one box containing at least ⌈Nk⌉\lceil \frac{N}{k} \rceilkN objects.

Permutations and Combinations:

P(n,r)=n(n−1)(n−2)⋯(n−r+1)=n!(n−r)!P(n,r)=n(n-1)(n-2)\cdots(n-r+1)=\dfrac{n!}{(n-r)!}P(n,r)=n(n1)(n2)(nr+1)=(nr)!n!
C(n,r)=(nr)=P(n,r)r!=n(n−1)(n−2)⋯(n−r+1)r!=n!r!(n−r)!C(n,r)=\dbinom{n}{r}=\dfrac{P(n,r)}{r!}=\dfrac{n(n-1)(n-2)\cdots(n-r+1)}{r!}=\dfrac{n!}{r!(n-r)!}C(n,r)=(rn)=r!P(n,r)=r!n(n1)(n2)(nr+1)=r!(nr)!n!
Remark: (nr)=(nn−r)\dbinom{n}{r}=\dbinom{n}{n-r}(rn)=(nrn)
The Binomial Theorem(二项式定理): (x+y)n=∑k=0n(nk)xn−kyk(x+y)^n=\sum\limits_{k=0}^{n}{\dbinom{n}{k}x^{n-k}y^k}(x+y)n=k=0n(kn)xnkyk
  ∑k=0n(nk)=2n⟸x=y=1\sum\limits_{k=0}^{n}{\dbinom{n}{k}}=2^n \Longleftarrow x=y=1k=0n(kn)=2nx=y=1
  ∑k=0n(−1)k(nk)=0⟸x=−y=1\sum\limits_{k=0}^{n}{(-1)^k\dbinom{n}{k}}=0 \Longleftarrow x=-y=1k=0n(1)k(kn)=0x=y=1
  ∑k=0n2k(nk)=3n⟸x+1=y=2\sum\limits_{k=0}^{n}{2^k\dbinom{n}{k}}=3^n \Longleftarrow x+1=y=2k=0n2k(kn)=3nx+1=y=2
Pascal’s Identity: (n+1k)=(nk)+(nk−1)\dbinom{n+1}{k}=\dbinom{n}{k}+\dbinom{n}{k-1}(kn+1)=(kn)+(k1n)
Vandermonde’s Identity and proof: (m+nr)=∑k=0r(mr−k)(nk)\dbinom{m+n}{r}=\sum\limits_{k=0}^r{\dbinom{m}{r-k}\dbinom{n}{k}}(rm+n)=k=0r(rkm)(kn)
  (x+y)m+n=⋯+(m+nr)xm+n−ryr+⋯(x+y)^{m+n}=\cdots+\dbinom{m+n}{r}x^{m+n-r}y^r+\cdots(x+y)m+n=+(rm+n)xm+nryr+
  (x+y)m(x+y)n=⋯+(∑k=0r(mr−k)xm−r+kyr−k)(∑k=0n(nk)xn−kyk)+⋯=⋯+∑k=0r((mr−k)xm−r+kyr−k⋅(nk)xn−kyk)+⋯=⋯+∑k=0r(mr−k)(nk)xm+n−ryr+⋯\begin{aligned}(x+y)^m(x+y)^n&=\cdots+\left(\sum\limits_{k=0}^{r}{\dbinom{m}{r-k}x^{m-r+k}y^{r-k}}\right)\left(\sum\limits_{k=0}^{n}{\dbinom{n}{k}x^{n-k}y^k}\right)+\cdots \\ &=\cdots+\sum\limits_{k=0}^{r}{\left(\dbinom{m}{r-k}x^{m-r+k}y^{r-k}\cdot\dbinom{n}{k}x^{n-k}y^k\right)}+\cdots \\ &=\cdots+\sum\limits_{k=0}^{r}{\dbinom{m}{r-k}\dbinom{n}{k}x^{m+n-r}y^r}+\cdots\end{aligned}(x+y)m(x+y)n=+(k=0r(rkm)xmr+kyrk)(k=0n(kn)xnkyk)+=+k=0r((rkm)xmr+kyrk(kn)xnkyk)+=+k=0r(rkm)(kn)xm+nryr+
  Corollary: (2nn)=∑k=0n(nn−k)(nk)=∑k=0n(nk)2⟸m=r=n\dbinom{2n}{n}=\sum\limits_{k=0}^{n}{\dbinom{n}{n-k}\dbinom{n}{k}}=\sum\limits_{k=0}^{n}{\dbinom{n}{k}^2} \Longleftarrow m=r=n(n2n)=k=0n(nkn)(kn)=k=0n(kn)2m=r=n
Remark: (n+1r+1)=∑k=rn(kr)\dbinom{n+1}{r+1}=\sum\limits_{k=r}^{n}{\dbinom{k}{r}}(r+1n+1)=k=rn(rk)
  (n+1r+1)=(nr+1)+(nr)=(n−1r+1)+(n−1r)+(nr)=⋯=(r+1r+1)+∑k=r+1n(kr)=(rr)+∑k=r+1n(kr)=∑k=rn(kr)\begin{aligned}\dbinom{n+1}{r+1}&=\dbinom{n}{r+1}+\dbinom{n}{r}=\dbinom{n-1}{r+1}+\dbinom{n-1}{r}+\dbinom{n}{r}=\cdots \\ &=\dbinom{r+1}{r+1}+\sum\limits_{k=r+1}^{n}{\dbinom{k}{r}}=\dbinom{r}{r}+\sum\limits_{k=r+1}^{n}{\dbinom{k}{r}}=\sum\limits_{k=r}^{n}{\dbinom{k}{r}}\end{aligned}(r+1n+1)=(r+1n)+(rn)=(r+1n1)+(rn1)+(rn)==(r+1r+1)+k=r+1n(rk)=(rr)+k=r+1n(rk)=k=rn(rk)

Stirling Number(II):

nnn distinguishable objects into kkk indistinguishable boxes: ∑j=1kS(n,j)=∑j=1k[1j!∑i=0j−1(−1)i(ji)(j−i)n]\sum\limits_{j=1}^{k}{S(n, j)}=\sum\limits_{j=1}^{k}{\left[\dfrac{1}{j!}\sum\limits_{i=0}^{j-1}{(-1)^i\dbinom{j}{i}(j-i)^n}\right]}j=1kS(n,j)=j=1k[j!1i=0j1(1)i(ij)(ji)n]

8. Advanced Counting Techniques

Solving Linear Recurrence Relations:

an=c1an−1+c2an−2+⋯+ckan−k+F(n)a_n=c_1a_{n-1}+c_2a_{n-2}+\cdots+c_ka_{n-k}+F(n)an=c1an1+c2an2++ckank+F(n), F(n)=(btnt+bt−1nt−1+⋯+b1n+b0)snF(n)=(b_tn^t+b_{t-1}n^{t-1}+\cdots+b_1n+b_0)s^nF(n)=(btnt+bt1nt1++b1n+b0)sn.
 Let the form of general solutions be an(p)+an(h)a_n^{(p)}+a_n^{(h)}an(p)+an(h)
an(h)a_n^{(h)}an(h) is the solution of homogeneous equation an=c1an−1+c2an−2+⋯+ckan−ka_n=c_1a_{n-1}+c_2a_{n-2}+\cdots+c_ka_{n-k}an=c1an1+c2an2++ckank
 Assume that sss is a root of multiplicity mmm of the equation
 Then an(p)a_n^{(p)}an(p)is the form as nm(dtnt+dt−1nt−1+⋯+d1n+d0)snn^m(d_tn^t+d_{t-1}n^{t-1}+\cdots+d_1n+d_0)s^nnm(dtnt+dt1nt1++d1n+d0)sn.
 Particularly, when sss is not a root of the homogeneous equation, m=0m=0m=0.
 Now we can solve the coefficients eie_iei of an=e1x1n+e2x2n+⋯+ekxkn+an(p)a_n=e_1x_1^n+e_2x_2^n+\cdots+e_kx_k^n+a_n^{(p)}an=e1x1n+e2x2n++ekxkn+an(p).
 If xix_ixi is of multiplicity lll, the form is as (einl+ei+1nl−1+⋯+ei+l−2n+ei+l−1)xin(e_in^l+e_{i+1}n^{l-1}+\cdots+e_{i+l-2}n+e_{i+l-1})x_i^n(einl+ei+1nl1++ei+l2n+ei+l1)xin.

Generating Functions:

G(x)=a0+a1x+a2x2+⋯+anxn+⋯=∑k=0∞akxkG(x)=a_0+a_1x+a_2x^2+\cdots+a_nx^n+\cdots=\sum\limits_{k=0}^{\infty}{a_kx^k}G(x)=a0+a1x+a2x2++anxn+=k=0akxk
Remark:
  f(x)+g(x)=∑k=0∞(ak+bk)xkf(x)+g(x)=\sum\limits_{k=0}^{\infty}{(a_k+b_k)x^k}f(x)+g(x)=k=0(ak+bk)xk
  α⋅f(x)=∑k=0∞α⋅akxk\alpha \cdot f(x)=\sum\limits_{k=0}^{\infty}{\alpha \cdot a_kx^k}αf(x)=k=0αakxk
  x⋅f′(x)=∑k=0∞k⋅akxkx \cdot f'(x)=\sum\limits_{k=0}^{\infty}{k \cdot a_kx^k}xf(x)=k=0kakxk
  f(αx)=∑k=0∞αk⋅akxkf(\alpha x)=\sum\limits_{k=0}^{\infty}{\alpha^k \cdot a_kx^k}f(αx)=k=0αkakxk
  f(x)g(x)=∑k=0∞(∑j=0kajbk−j)xkf(x)g(x)=\sum\limits_{k=0}^{\infty}{(\sum\limits_{j=0}^{k}{a_jb_{k-j}})x^k}f(x)g(x)=k=0(j=0kajbkj)xk
eg.
  bk=∑i=0kaib_k=\sum\limits_{i=0}^{k}{a_i}bk=i=0kai
  ⟹F(x)=∑k=0∞bkxk=∑k=0∞(∑i=0kai⋅1)xk=G(x)⋅11−x\Longrightarrow F(x)=\sum\limits_{k=0}^{\infty}{b_kx^k}=\sum\limits_{k=0}^{\infty}{\left(\sum\limits_{i=0}^{k}{a_i \cdot 1}\right)x^k}=G(x) \cdot \dfrac{1}{1-x}F(x)=k=0bkxk=k=0(i=0kai1)xk=G(x)1x1
  ak=k2a_k=k^2ak=k2
  ⟹G(x)=∑k=0∞k2xk=x(∑k=0∞kxk)′=x[x(∑k=0∞xk)′]′=x[x(11−x)′]′=x(1+x)(1−x)3\Longrightarrow G(x)=\sum\limits_{k=0}^{\infty}{k^2x^k}=x\left(\sum\limits_{k=0}^{\infty}{kx^k}\right)'=x\left[x\left(\sum\limits_{k=0}^{\infty}{x^k}\right)'\right]'=x\left[x\left(\dfrac{1}{1-x}\right)'\right]'=\dfrac{x(1+x)}{(1-x)^3}G(x)=k=0k2xk=x(k=0kxk)=x[x(k=0xk)]=x[x(1x1)]=(1x)3x(1+x)
  ak=∑i=0ki2a_k=\sum\limits_{i=0}^{k}{i^2}ak=i=0ki2
  ⟹G(x)=∑k=0∞(∑i=0ki2⋅1)xk=x(1+x)(1−x)3⋅11−x=x(1+x)(1−x)4\Longrightarrow G(x)=\sum\limits_{k=0}^{\infty}{(\sum\limits_{i=0}^{k}{i^2 \cdot 1})x^k}=\dfrac{x(1+x)}{(1-x)^3} \cdot \dfrac{1}{1-x}=\dfrac{x(1+x)}{(1-x)^4}G(x)=k=0(i=0ki21)xk=(1x)3x(1+x)1x1=(1x)4x(1+x)
  f(x)=11−4x2=1(1−2x)(1+2x)=12(11−2x+11+2x)=∑k=0∞akxkf(x)=\dfrac{1}{1-4x^2}=\dfrac{1}{(1-2x)(1+2x)}=\dfrac12\left(\dfrac{1}{1-2x}+\dfrac{1}{1+2x}\right)=\sum\limits_{k=0}^{\infty}{a_kx^k}f(x)=14x21=(12x)(1+2x)1=21(12x1+1+2x1)=k=0akxk
  ⟹ak=2k+(−2)k2\Longrightarrow a_k=\dfrac{2^k+(-2)^k}{2}ak=22k+(2)k
The Extended Binomial Therom:
  (1+x)u=∑k=0∞(uk)xk,∣x∣<1,u∈R(1+x)^u=\sum\limits_{k=0}^{\infty}{\dbinom{u}{k}x^k},|x|<1,u \in \mathbb{R}(1+x)u=k=0(ku)xk,x<1,uR
  Collary:
  (1+x)−n=∑k=0∞(−nk)xk=∑k=0∞(−1)k(n+k−1k)xk(1+x)^{-n}=\sum\limits_{k=0}^{\infty}{\dbinom{-n}{k}x^k}=\sum\limits_{k=0}^{\infty}{(-1)^k\dbinom{n+k-1}{k}x^k}(1+x)n=k=0(kn)xk=k=0(1)k(kn+k1)xk
  (1−x)−n=∑k=0∞(−nk)(−x)k=∑k=0∞(n+k−1k)xk(1-x)^{-n}=\sum\limits_{k=0}^{\infty}{\dbinom{-n}{k}(-x)^k}=\sum\limits_{k=0}^{\infty}{\dbinom{n+k-1}{k}x^k}(1x)n=k=0(kn)(x)k=k=0(kn+k1)xk

Generating Functions of Common Sequences:
SequenceGenerating Function
11111−x\dfrac{1}{1-x}1x1
αk\alpha^kαk11−αx\dfrac{1}{1-\alpha x}1αx1
k+1k+1k+11(1−x)2\dfrac{1}{(1-x)^2}(1x)21
(nk)\dbinom{n}{k}(kn)(1+x)n(1+x)^n(1+x)n
(nk)αk\dbinom{n}{k} \alpha^k(kn)αk(1+αx)n(1+\alpha x)^n(1+αx)n
(n+k−1k)\dbinom{n+k-1}{k}(kn+k1)(1−x)−n(1-x)^{-n}(1x)n
(−1)k(n+k−1k)(-1)^k\dbinom{n+k-1}{k}(1)k(kn+k1)(1+x)−n(1+x)^{-n}(1+x)n
1k!\dfrac{1}{k!}k!1exe^xex
(−1)k+1k!\dfrac{(-1)^{k+1}}{k!}k!(1)k+1ln⁡(1+x)\ln(1+x)ln(1+x)

9. Relations

Definition: A binary relation RRR from set AAA to set BBB is a subset of A×BA \times BA×B.
Remark: R={(a,b) ∣ a∈A∧b∈B∧aRb}⊆A×BR=\{(a,b) \, | \, a \in A \land b \in B \land aRb\} \subseteq A \times BR={(a,b)aAbBaRb}A×B
Definition: A relation on set AAA is a binary relation from set AAA to set AAA.
eg. How many binary relations are there on a set AAA with nnn elements?
  ∣A∣=n⟹∣A×A∣=n⋅n=n2⟹∣R∣=2n2|A|=n \Longrightarrow |A \times A|=n \cdot n=n^2 \Longrightarrow |R|=2^{n^2}A=nA×A=nn=n2R=2n2

Connection Matrices

 Let RRR be a relation from A={a1,a2,…,an}A=\{a_1, a_2, \dots , a_n\}A={a1,a2,,an} to B={b1,b2,…,bm}B=\{b_1, b_2, \dots , b_m\}B={b1,b2,,bm}.
 The n×mn \times mn×m connection matrix MR=[mij]M_R=[m_{ij}]MR=[mij] for RRR is defined by:
mij={1,(ai,bj)∈R0,(ai,bj)∉Rm_{ij}=\begin{cases} 1, (a_i, b_j) \in R \\ 0, (a_i, b_j) \notin R \end{cases}mij={1,(ai,bj)R0,(ai,bj)/R

Special Properties of Binary Relations
TypeCHNDenote
Reflexive自反性∀x(x∈A→(x,x)∈R)\forall x(x \in A \to (x,x) \in R)x(xA(x,x)R)
Irreflexive非自反性∀x(x∈A→(x,x)∉R)\forall x(x \in A \to (x,x) \notin R)x(xA(x,x)/R)
Symmetric对称性∀x∀y((x,y)∈R→(y,x)∈R)\forall x \forall y((x,y) \in R \to (y,x) \in R)xy((x,y)R(y,x)R)
Antisymmetric反对称性∀x∀y((x,y)∈R∧(y,x)∈R→x=y)\forall x \forall y((x,y) \in R \land (y,x) \in R \to x=y)xy((x,y)R(y,x)Rx=y)
∀x∀y((x,y)∈R∧x≠y→(y,x)∉R)\forall x \forall y((x,y) \in R \land x \neq y \to (y,x) \notin R)xy((x,y)Rx=y(y,x)/R)
Asymmetric非对称性∀x∀y(x≠y→(x,y)∈R∧(y,x)∉R∨(x,y)∉R∧(y,x)∈R)\forall x \forall y(x \neq y \to (x,y) \in R \land (y,x) \notin R \lor (x,y) \notin R \land (y,x) \in R)xy(x=y(x,y)R(y,x)/R(x,y)/R(y,x)R)
Transitive传递性∀x∀y∀z((x,y)∈R∧(y,z)∈R→(x,z)∈R)\forall x \forall y \forall z((x,y) \in R \land (y,z) \in R \to (x,z) \in R)xyz((x,y)R(y,z)R(x,z)R)

Remark: (mij∧mjk)‾∨mik=1\overline{(m_{ij} \land m_{jk})} \lor m_{ik}=1(mijmjk)mik=1
eg.
  How many relations on a set with nnn elements that are reflexive? 2n2−n2^{n^2-n}2n2n
  How many relations on a set with nnn elements that are symmetric? 2n⋅2n2−n2=2n2+n22^n \cdot 2^\frac{n^2-n}{2}=2^\frac{n^2+n}{2}2n22n2n=22n2+n
  How many relations on a set with nnn elements that are antisymmetric? 2n⋅3n2−n22^n \cdot 3^\frac{n^2-n}{2}2n32n2n
  How many relations on a set with nnn elements that are reflexive and symmetric? 2n2−n22^\frac{n^2-n}{2}22n2n
  How many relations on a set with nnn elements that are transitive? ∑j=1nS(n,j)\sum\limits_{j=1}^{n}{S(n,j)}j=1nS(n,j)

Combining Relations

 Let A={a1,a2,…,an}A=\{a_1, a_2, \dots , a_n\}A={a1,a2,,an}, B={b1,b2,…,bm}B=\{b_1, b_2, \dots , b_m\}B={b1,b2,,bm}, MR1=[cij]M_{R_1}=[c_{ij}]MR1=[cij], MR2=[dij]M_{R_2}=[d_{ij}]MR2=[dij].

OperationDenote
R1∪R2R_1 \cup R_2R1R2MR1∪R2=[cij∨dij]M_{R_1 \cup R_2}=[c_{ij} \lor d_{ij}]MR1R2=[cijdij]
R1∩R2R_1 \cap R_2R1R2MR1∩R2=[cij∧dij]M_{R_1 \cap R_2}=[c_{ij} \land d_{ij}]MR1R2=[cijdij]
R1‾\overline{R_1}R1MR1‾=[cij‾]M_{\overline{R_1}}=[\overline{c_{ij}}]MR1=[cij]
R1−R2R_1-R_2R1R2
R1∩R2‾R_1 \cap \overline{R_2}R1R2
MR1−R2=MR1∩R2‾=[cij∧dij‾]M_{R_1-R_2}=M_{R_1 \cap \overline{R_2}}=[c_{ij} \land \overline{d_{ij}}]MR1R2=MR1R2=[cijdij]

 Let R={(a,b) ∣ a∈A∧b∈B∧aRb}R=\{(a,b) \, | \, a \in A \land b \in B \land aRb\}R={(a,b)aAbBaRb}, S={(b,c) ∣ b∈B∧c∈C∧bSc}S=\{(b,c) \, | \, b \in B \land c \in C \land bSc\}S={(b,c)bBcCbSc}.
 The composition of RRR and SSS: S∘R={(a,c) ∣ a∈A∧c∈C∧∃b∈B(aRb∧bSc)}S \circ R=\{(a,c) \, | \, a \in A \land c \in C \land \exists b \in B(aRb \land bSc)\}SR={(a,c)aAcCbB(aRbbSc)}
Remark: Rn=Rn−1∘RR^n=R^{n-1} \circ RRn=Rn1R
Theroem: The relation R on a set A is transitive   ⟺  Rn⊆R\iff R^n \subseteq RRnR
 Let R={(a,b) ∣ a∈A∧b∈B∧aRb}R=\{(a,b) \, | \, a \in A \land b \in B \land aRb\}R={(a,b)aAbBaRb}.
 The inverse relation of RRR: Rc=R−1={(b,a) ∣ a∈A∧b∈B∧aRb}R^c=R^{-1}=\{(b,a) \, | \, a \in A \land b \in B \land aRb\}Rc=R1={(b,a)aAbBaRb}
Remark: MR−1=(MR)TM_{R^{-1}}=(M_R)^TMR1=(MR)T

Properties of Relation Operations

(R∪S)−1=R−1∪S−1(R \cup S)^{-1}=R^{-1} \cup S^{-1}(RS)1=R1S1
(R∩S)−1=R−1∩S−1(R \cap S)^{-1}=R^{-1} \cap S^{-1}(RS)1=R1S1
(R‾)−1=R−1‾(\overline{R})^{-1}=\overline{R^{-1}}(R)1=R1
(R−S)−1=R−1−S−1(R-S)^{-1}=R^{-1}-S^{-1}(RS)1=R1S1
(A×B)−1=B×A(A \times B)^{-1}=B \times A(A×B)1=B×A
R‾=A×B−R\overline{R}=A \times B -RR=A×BR
(S∘T)−1=T−1∘S−1(S \circ T)^{-1}=T^{-1} \circ S^{-1}(ST)1=T1S1
(R∘T)∘P=R∘(T∘P)(R \circ T) \circ P=R \circ (T \circ P)(RT)P=R(TP)
(R∪S)∘T=(R∘T)∪(S∘T)(R \cup S) \circ T=(R \circ T) \cup (S \circ T)(RS)T=(RT)(ST)

Closures of Relations
ClosureCHNDenote
Reflexive Closure自反闭包r(R)=R∪IAr(R)=R \cup I_Ar(R)=RIA
Symmetric Closure对称闭包s(R)=R∪R−1s(R)=R \cup R^{-1}s(R)=RR1
Transitive Closure传递闭包t(R)=R∗t(R)=R^*t(R)=R

 The diagonal relation on AAA: IA={(x,x) ∣ x∈A}I_A=\{(x,x) \, | \, x \in A\}IA={(x,x)xA}
 The connectivity relation on AAA: R∗=⋃n=1∞RnR^*=\bigcup\limits_{n=1}^{\infty}{R^n}R=n=1Rn
Collary: ∣A∣=n⟹t(R)=R∪R2∪⋯∪Rn|A|=n \Longrightarrow t(R)=R \cup R^2 \cup \cdots \cup R^nA=nt(R)=RR2Rn

Equivalence Relations and Partitions

Definition:
  A relation RRR on set AAA is an equivalence relation iff RRR is reflexive, symmetric and transitive.
  The equivalence class of x∈Ax \in AxA via equivalence relation RRR is [x]R[x]_R[x]R, or [x][x][x] for short.
  aaa and bbb of set AAA are equivalent by equivalence relation RRR   ⟺  a∼b\iff a \sim bab
Remark:
  Congruence Modulo mmm: R={(a,b) ∣ a≡b(modm),a,b∈Z}R=\{(a,b) \, | \, a \equiv b \pmod{m}, a,b \in \mathbb{Z}\}R={(a,b)ab(modm),a,bZ}
  Congruence class Modulo mmm: [t]m={t+km ∣ k∈Z},t=0,1,2,…,m−1[t]_m=\{t+km \, | \, k \in \mathbb{Z}\}, t=0,1,2,\dots,m-1[t]m={t+kmkZ},t=0,1,2,,m1
Theroem:
  Let RRR be an equivalence relation on set AAA.
  aRb  ⟺  [a]=[b]  ⟺  [a]∩[b]≠∅aRb \iff [a]=[b] \iff [a] \cap [b] \neq \varnothingaRb[a]=[b][a][b]=
Theroem:
  Let R1R_1R1 and R2R_2R2 be equivalence relations on AAA.
  R1∩R2R_1 \cap R_2R1R2 is an equivalence relation, R1∪R2R_1 \cup R_2R1R2 is reflexive and symmetric.
  Collary: (R1∪R2)∗(R_1 \cup R_2)^*(R1R2) is an equivalence relation
Definition:
  A partition of set AAA is a collection of disjoint nonempty subsets of AAA that have AAA as their union.
  pr(A)={Ai ∣ i∈I}pr(A)=\{A_i \, | \, i \in I\}pr(A)={AiiI}, where III is an index set, Ai≠∅A_i \neq \varnothingAi=, Ai∩Aj=∅(i≠j),⋃i∈IAi=AA_i \cap A_j=\varnothing(i \neq j), \bigcup\limits_{i \in I}{A_i}=AAiAj=(i=j),iIAi=A.

Partial Orderings

Definition:
  A relation RRR on set SSS is a partial ordering iff RRR is reflexive, antisymmetric and transitive.
  For notation, (S,R)(S,R)(S,R) is a partially ordered set, or poset for short.
  The elements aaa and bbb of a poset (S,≼)(S,\preccurlyeq)(S,) are comparable if either a≼ba \preccurlyeq bab or b≼ab \preccurlyeq aba.
  When neither a≼ba \preccurlyeq bab or b≼ab \preccurlyeq aba, then aaa and bbb are called incomparable.
  If (S,≼)(S,\preccurlyeq)(S,) is a poset and ∀(a,b)∈S\forall (a,b) \in S(a,b)S are comparable, SSS is totally ordered or linearly ordered.
  In this case, ≼\preccurlyeq is called a total order or linear order.

ENGCHNDenote
Maximal Element(s)极大值∃a∈A(¬∃b∈A(a≺b))\exist a \in A(\neg \exist b \in A(a \prec b))aA(¬bA(ab))
Minimal Element(s)极小值∃a∈A(¬∃b∈A(b≺a))\exist a \in A(\neg \exist b \in A(b \prec a))aA(¬bA(ba))
Greatest Element最大值∃a∈A(∀b∈A(b≼a))\exist a \in A(\forall b \in A(b \preccurlyeq a))aA(bA(ba))
Least Element最小值∃a∈A(∀b∈A(a≼b))\exist a \in A(\forall b \in A(a \preccurlyeq b))aA(bA(ab))
Upper Bound上界∃a∈S(∀b∈A(b≺a))\exist a \in S(\forall b \in A(b \prec a))aS(bA(ba))
A⊆SA \subseteq SAS
Lower Bound下界∃a∈S(∀b∈A(a≺b))\exist a \in S(\forall b \in A(a \prec b))aS(bA(ab))
A⊆SA \subseteq SAS
Least Upper Bound上确界min⁡a\min{a}mina
Greatest Lower Bound下确界max⁡a\max{a}maxa

  A poset (A,R)(A,R)(A,R) is well-ordered iff every nonempty subset of AAA has a least element.
  A poset is called a lattice iff every pair of elements has a lub\mathrm{lub}lub and a glb\mathrm{glb}glb.

10. Graphs

 Let G=(V,E)G=(V,E)G=(V,E) be an undirected graph with eee edges, then ∑v∈Vdeg⁡(v)=2e\sum\limits_{v \in V}{\deg(v)}=2evVdeg(v)=2e.
 Let G=(V,E)G=(V,E)G=(V,E) be a digraph, then ∑v∈Vdeg⁡+(v)=∑v∈Vdeg⁡−(v)=∣E∣\sum\limits_{v \in V}{\deg^+(v)}=\sum\limits_{v \in V}{\deg^-(v)}=|E|vVdeg+(v)=vVdeg(v)=E.

Some Special Simple Graphs
Denotei.e.
KnK_nKnComplete Graph
Cn (n≥3)C_n \ (n \ge 3)Cn (n3)Cycle
Wn (n≥3)W_n \ (n \ge 3)Wn (n3)Wheel
QnQ_nQnnnn-Cube
Km,n (m=∣V1∣,n=∣V2∣)K_{m,n} \ (m=\vert V_1 \vert , n=\vert V_2 \vert)Km,n (m=V1,n=V2)Complete Bipartite Graph

Regular graph:
  A simply graph is called regular if every vertex of this graph has the same degree.
  A regular graph is called nnn-regular if every vertex in this graph has degree nnn.

New Graphs from Old

Subgraph:
  Let G=(V,E)G=(V,E)G=(V,E), H=(W,F)H=(W,F)H=(W,F).
  HHH is a subgraph of GGG if W⊆VW \subseteq VWV and F⊆EF \subseteq EFE.
  Subgraph HHH is a proper subgraph of GGG if H≠GH \neq GH=G.
  HHH is a spanning subgraph of GGG if W=VW=VW=V and F⊆EF \subseteq EFE.

Union:
  Let G1=(V1,E1)G_1=(V_1,E_1)G1=(V1,E1), G2=(V2,E2)G_2=(V_2,E_2)G2=(V2,E2), then G1∪G2=(V1∪V2,E1∪E2)G_1 \cup G_2=(V_1 \cup V_2, E_1 \cup E_2)G1G2=(V1V2,E1E2).

Representing Graphs

Adjacency matrix: AG=[aij]nA_G=[a_{ij}]_nAG=[aij]n, aij=1a_{ij}=1aij=1 if {vi,vj}\{v_i,v_j\}{vi,vj} is an edge of GGG, or aij=0a_{ij}=0aij=0 otherwise.
Incidence matrix: MG=[mij]n×mM_G=[m_{ij}]_{n \times m}MG=[mij]n×m, mij=1m_{ij}=1mij=1 if eje_jej is incident with viv_ivi, or mij=0m_{ij}=0mij=0 otherwise.
Isomorphism(同构):
  Formally, two simple graphs G1=(V1,E1)G_1=(V_1,E_1)G1=(V1,E1) and G2=(V2,E2)G_2=(V_2,E_2)G2=(V2,E2) are isomorphic.
  If and only if there is an bijection fff from V1V_1V1 to V2V_2V2 such that ∀a,b∈V1\forall a, b \in V_1a,bV1.
  aaa and bbb are adjacent in G1G_1G1 while f(a)f(a)f(a) and f(b)f(b)f(b) are adjacent in G2G_2G2.

Connectivity

 The number of different paths of length rrr from viv_ivi to vjv_jvj is equal to the (i,j)(i, j)(i,j)-th entry of ArA^rAr.
 The maximally connected subgraphs of GGG are called the connected components.
 A vertex is a cut vertex or articulation point, if removing it and its incident edges results in more connected components.
 Similarly if removal of an edge creates more components the edge is called a cut edge or bridge.
 A directed graph is strongly connected if there is a path from aaa to bbb and from bbb to aaa for all vertices aaa and bbb in the graph.
 The graph is weakly connected if the underlying undirected graph is connected.
 For directed graph, the maximal strongly connected subgraphs are called the strongly connected components.

Euler and Hamilton Paths

Definition:
  A connected multigraph has an Euler circuit if and only if each of its vertices has even degree.
  A connected multigraph has an Euler path but not an Euler circuit if and only if it has exactly two vertices of odd degree.
  A Hamilton path in a graph GGG is a path which visits every vertex in GGG exactly once.
  A Hamilton circuit visits every vertex exactly once, except for the first vertex, which is also the end of the cycle.
Sufficient Condition:
  Let GGG be a simple graph with nnn (n≥3)(n \ge 3)(n3) vertices.
  Dirac’s Theroem(狄拉克定理) If deg⁡v∈G(v)≥n2\deg\limits_{v \in G}(v)\ge \dfrac{n}{2}vGdeg(v)2n, then GGG has a Hamilton circuit.
  Ore’s Theroem(奥勒定理) If deg⁡(u)+deg⁡(v)≥n\deg(u)+\deg(v) \ge ndeg(u)+deg(v)n (∀u,v∈G,¬euv)(\forall u,v \in G, \neg e_{uv})(u,vG,¬euv), then GGG has a Hamilton circuit.

Planar Graphs

Euler’s formula:
  Let GGG be a connected planar simple graph with eee edges and vvv vertices.
  Let rrr be the number of regions in a planar representation of GGG.
  Then we have the formula r=e−v+2r=e-v+2r=ev+2.
Degree of Region:
  Suppose RRR is a region of a connected planar simple graph.
  The number of edges on the boundary of RRR is called the Degree of RRR, denoted by Deg(R)\mathrm{Deg}(R)Deg(R).
Collary:
  If GGG is a connected planar simple graph with eee edges and vvv (v≥3)(v \ge 3)(v3) vertices, then e≤3v−6e \leq3v-6e3v6.
   2e=∑Ri∈GDeg(Ri)≥3r⟹r=e−v+2≤23e⟹e≤3v−62e=\sum\limits_{R_i \in G}{\mathrm{Deg}(R_i)} \ge 3r \Longrightarrow r=e-v+2 \leq \dfrac23 e \Longrightarrow e \leq 3v-62e=RiGDeg(Ri)3rr=ev+232ee3v6
  If GGG is a connected planar simple graph, then GGG has a vertex viv_ivi with deg⁡(vi)≤5\deg(v_i) \leq 5deg(vi)5.
   (2e=∑vi∈Vdeg⁡(vi)≥6v)∧(e≤3v−6⇔2e≤6v−12)=0(2e=\sum\limits_{v_i \in V}{\deg(v_i)} \ge 6v) \land (e \leq 3v-6 \Leftrightarrow 2e \leq 6v-12)=0(2e=viVdeg(vi)6v)(e3v62e6v12)=0
  If G has eee edges and vvv (v≥3)(v \ge 3)(v3) vertices and no circuits of length 333,then e≤2v−4e \leq 2v-4e2v4.
   2e=∑Ri∈GDeg(Ri)≥4r⟹r=e−v+2≤12e⟹e≤2v−42e=\sum\limits_{R_i \in G}{\mathrm{Deg}(R_i)} \ge 4r \Longrightarrow r=e-v+2 \leq \dfrac12 e \Longrightarrow e \leq 2v-42e=RiGDeg(Ri)4rr=ev+221ee2v4
  Generally, if every region of G has at least kkk edges, then e≤k(v−2)k−2e \leq \dfrac{k(v-2)}{k-2}ek2k(v2).
Homeomorphism(同胚):
  The graph G1=(V1,E1)G_1=(V_1,E_1)G1=(V1,E1) and G2=(V2,E2)G_2=(V_2,E_2)G2=(V2,E2) are called homeomorphic.
  If and only if they can be obtained from the same graph by a sequence of elementary subdivision.
Kuratowski’s Theorem(库拉托斯基定理):
  A graph is nonplanar if and only if it contains a subgraph homeomorphic to K3,3K_{3,3}K3,3 or K5K_5K5.


(updated 2021.5.14)

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