离散数学英文版PPT
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离散数学课件(英文版)----Equivalence(II)
Ex. (x,y)R (y,z)R (x,z)R (x,x)R etc.
A1
A It is straight to prove that R is reflexible, symmetric and transitive, so, it is an equivalence relation.
Symmetry
Let A={1,2,3}, RAA {(1,1),(1,2),(1,3),(2,1),(3,1),(3,3)} symmetric. {(1,2),(2,3),(2,2),(3,1)} antisymmetric. {(1,2),(2,3),(3,1)} antisymmetric and asymmetric. {(11),(2,2)} symmetric and antisymmetric. symmetric and antisymmetric, and asymmetric!
• R is reflexive relation on A if and only if IAR
Visualized Reflexivity
A={a,b,c} a
1 0 0 1 1 1 MR 0 1 1
b
c
Symmetry
Relation R on A is Symmetric if whenever (a,b)R, then (b,a)R Antisymmetric if whenever (a,b)R and (b,a)R then a=b. Asymmetric if whenever (a,b)R then (b,a)R (Note: neither anti- nor a-symmetry is the negative of symmetry)
A1
A It is straight to prove that R is reflexible, symmetric and transitive, so, it is an equivalence relation.
Symmetry
Let A={1,2,3}, RAA {(1,1),(1,2),(1,3),(2,1),(3,1),(3,3)} symmetric. {(1,2),(2,3),(2,2),(3,1)} antisymmetric. {(1,2),(2,3),(3,1)} antisymmetric and asymmetric. {(11),(2,2)} symmetric and antisymmetric. symmetric and antisymmetric, and asymmetric!
• R is reflexive relation on A if and only if IAR
Visualized Reflexivity
A={a,b,c} a
1 0 0 1 1 1 MR 0 1 1
b
c
Symmetry
Relation R on A is Symmetric if whenever (a,b)R, then (b,a)R Antisymmetric if whenever (a,b)R and (b,a)R then a=b. Asymmetric if whenever (a,b)R then (b,a)R (Note: neither anti- nor a-symmetry is the negative of symmetry)
235-Ch5-离散数学英文版PPT
• Recursive Step: Provide rules for forming new elements in the set from those already known to be in the set
• Example:
– Basis Step: 3 S – Recursive Step: if x S and y S then x + y S – Then, S = {3, 6, 9, 12, 15, 18, 21, …}
For n = 4, f4 = 3 > 2 = (3 + 5)/2 • Why we need to prove n = 3 and n = 4?
Fibonacci Example
• Inductive step:
• Assume fj > j-2, for all j, 3 j k, k ≥ 4. We must show that fk+1 > k-1. Note that fk+1 = fk + fk-1 By induction assumption, fk > k-2 and fk-1 > k-3 So, fk+1 = fk + fk-1 > k-2 + k-3 We now only need to show that k-2 + k-3 = k-1
• We denote Fibonacci numbers as fn = f(n), n = 0, 1, 2, …
Fibonacci Numbers
Suppose a newly-born pair of rabbits, one male, one female, are put in a field. Rabbits are able to mate at the age of one month so that at the end of its second month a female can produce another pair of rabbits. Suppose that our rabbits never die and that the female always produces one new pair (one male, one female) every month
• Example:
– Basis Step: 3 S – Recursive Step: if x S and y S then x + y S – Then, S = {3, 6, 9, 12, 15, 18, 21, …}
For n = 4, f4 = 3 > 2 = (3 + 5)/2 • Why we need to prove n = 3 and n = 4?
Fibonacci Example
• Inductive step:
• Assume fj > j-2, for all j, 3 j k, k ≥ 4. We must show that fk+1 > k-1. Note that fk+1 = fk + fk-1 By induction assumption, fk > k-2 and fk-1 > k-3 So, fk+1 = fk + fk-1 > k-2 + k-3 We now only need to show that k-2 + k-3 = k-1
• We denote Fibonacci numbers as fn = f(n), n = 0, 1, 2, …
Fibonacci Numbers
Suppose a newly-born pair of rabbits, one male, one female, are put in a field. Rabbits are able to mate at the age of one month so that at the end of its second month a female can produce another pair of rabbits. Suppose that our rabbits never die and that the female always produces one new pair (one male, one female) every month
离散数学英文7.1-4
• Many experiments do not yield exactly the same results when performed repeatedly.
– For example, if we toss a coin, we are not sure if we will get heads or tails. – If we toss a die, we have no way of knowing which of the six possible numbers will turn up.
• • • • • • 1. four blue balls and five red balls 2. sum of two dice is 7 3. lottery four digits 4. picks the correct six number out of 40 5. four cards of one kind 同花大顺(Royal Flush)，同花顺 6. a full house
• 7. 50 labeled balls
子(Flush Straight)，四条(Four of a Kind)，富卢(Full House， 即三条+一对)，同花(Flush)， 顺子(Straight)，三条(Three of a Kind)，两对(Two Pairs)，一 对 (One Pair)，无赖(Nothing)
• • • • • • • • • • Introduction Assigning Probabilities Combinations of Events Conditional Probability Independence Bernoulli Trials and the Binomial Distribution Random Variables The Birthday Problem Monte Carlo Algorithms The Probabilistic Method
– For example, if we toss a coin, we are not sure if we will get heads or tails. – If we toss a die, we have no way of knowing which of the six possible numbers will turn up.
• • • • • • 1. four blue balls and five red balls 2. sum of two dice is 7 3. lottery four digits 4. picks the correct six number out of 40 5. four cards of one kind 同花大顺(Royal Flush)，同花顺 6. a full house
• 7. 50 labeled balls
子(Flush Straight)，四条(Four of a Kind)，富卢(Full House， 即三条+一对)，同花(Flush)， 顺子(Straight)，三条(Three of a Kind)，两对(Two Pairs)，一 对 (One Pair)，无赖(Nothing)
• • • • • • • • • • Introduction Assigning Probabilities Combinations of Events Conditional Probability Independence Bernoulli Trials and the Binomial Distribution Random Variables The Birthday Problem Monte Carlo Algorithms The Probabilistic Method
离散数学英文课件 群论 Groups (II)
Cyclic group
Corollary 6 The generators of Zn are the integers r such that 1=<r < n and gcd(r, n) = 1. Example Let us examine the group Z16. The numbers 1, 3, 5, 7, 9, 11,13, and 15 are the elements of Z16 that are relatively prime to 16. Each of these elements generates Z16. For example, 1*9 = 9 2*9 = 2 3*9 = 11 4*9 = 4 5*9 = 13 6*9 = 6 7*9 = 15 8*9 = 8 9*9 = 1 10*9 = 10 11*9 = 3 12*9 = 12 13*9 = 5 14*9 = 14 15*9 = 7:
Subgroups of Cyclic Groups
Theorem 2: Every subgroup of a cyclic group is cyclic.
Proof. Let G be a cyclic group generated by a and suppose that H is a subgroup of G. If H = {e}, then trivially H is cyclic. Suppose that H contains some other element g distinct from the identity. Then g can be written as an for some integer n. We can assume that n > 0. Let m be the smallest natural number such that am ∈H. Such an m exists by the Principle of Well-Ordering. We claim that h = am is a generator for H. We must show that every h’ ∈ H can be written as a power of h. Since h’ ∈ H and H is a subgroup of G, h’ = ak for some positive integer k. Using the division algorithm, we can find numbers q and r such that k = mq + r where 0 =<r < m; hence, ak = amq+r = (am)qar = hqar: So ar = akh-q. Since ak and h-q are in H, ar must also be in H. However, m was the smallest positive number such that am was in H; consequently, r = 0 and so k = mq. Therefore, h’ = ak = amq = hq and H is generated by h.
离散数学英文课件：DM_lecture1_2Propositional Equivalence
Hui Gao
Discrete Mathematics
8
More Equivalence Laws
Distributive: p(qr) (pq)(pr) p(qr) (pq)(pr)
De Morgan’s: (p1p2…pn) (p1p2…pn) (p1p2…pn) (p1p2…pn)
Ex. p p [What is its truth table?] A contradiction is a compound proposition that
is false no matter what! Ex. p p [Truth table?] Other compound props. are contingencies.
Equivalence Laws - Examples
Identity:
pT p pF p
Domination: pT T pF F
Idempotent: pp p pp p
Double negation: p p
Commutative: pq qp pq qp
Associative: (pq)r p(qr) (pq)r p(qr)
Equivalent expressions can always be substituted for each other in a more complex expression - useful for simplification.
Hui Gao
Discrete Mathematics
7
Assume (p q) F, then (p q) T, then p q T, then p=F and q=F, then pq =F.
离散数学英文DMAv7_10.1_2_3_4 (1)
Roadmaps
This is a GPS trajectory dataset collected in (Microsoft Research Asia) GeoLife project by 182 users in a period of over two years (from April 2007 to August 2012). This trajectory dataset can be used in many research fields, such as mobility pattern mining, user activity recognition, location-based social networks, location privacy, and location recommendation. The following heat maps visualize its distribution in Beijing.
Graph Theory
Rosen 7th ed., ch. 10
Chapter 10
Graphs
图
图/Graph：
论
可直观地表示离散对象之间的相 互关系，研究它们的共性和特性，以 便解决具体问题。
10.1 图的概念/Introduction of Graph 10.2 图的术语/Graph Terminology 10.3 图的表示与同构/ Representing Graph and Graph Isomorphism 10.4 连通性/Connectivity
10.2 图的术语/Graph Terminology
[定义]相邻和关联：
在无向图G中，若e=（ａ，ｂ） ∈Ｅ，则称ａ 与/connect。a、b称为边e的端点或结束顶点 /endpoint. 在有向图G中，若e=（ａ，ｂ）∈Ｅ，即箭头由 ａ到ｂ，称ａ相邻到ｂ，或a关联或联结b。a称为e 的起点/initial vertex，b称为e的终点/terminal or end vertex。
ChLogic离散数学英文实用PPT课件
Discrete Structures: Terminology
• Algorithm (Algorithmic Analysis, Algorithmic Thinking) • Universal, Existential Quantification, Rule of Inference • Probability, Statistics, Event, Independent, Dependent • Operation, Operator, Product, Sum, Division, Integer, Real Number,
• Steps to convert an English sentence to a statement in propositional logic • Identify propositions and using propositional variables • Determine appropriate logical operations
• “If I go to iShow (爱秀) or to CHIC (东京秀客 ), t hen I will no t go sho pping at Golden Ea gle .” • p: I go to iShow • q: I go to CHIC • r: I go shopping at GE
Solution: Let p be A is a knight, and q be B is a
kniቤተ መጻሕፍቲ ባይዱht
• If A is a knight, then p True. Since knights tell
the truth, then
If p or q then not r.
• Algorithm (Algorithmic Analysis, Algorithmic Thinking) • Universal, Existential Quantification, Rule of Inference • Probability, Statistics, Event, Independent, Dependent • Operation, Operator, Product, Sum, Division, Integer, Real Number,
• Steps to convert an English sentence to a statement in propositional logic • Identify propositions and using propositional variables • Determine appropriate logical operations
• “If I go to iShow (爱秀) or to CHIC (东京秀客 ), t hen I will no t go sho pping at Golden Ea gle .” • p: I go to iShow • q: I go to CHIC • r: I go shopping at GE
Solution: Let p be A is a knight, and q be B is a
kniቤተ መጻሕፍቲ ባይዱht
• If A is a knight, then p True. Since knights tell
the truth, then
If p or q then not r.
离散数学英文课件：DM_lecture3_4_5 The Integers and Division
Hui Gao
Discrete Mathematics
2
Facts re: the Divides Relation
Theorem: a,b,c Z:
1. a|0 2. (a|b a|c) a | (b + c) 3. a|b a|bc 4. (a|b b|c) a|c
Proof of (2): a|b means there is an s such
1
Divides, Factor, Multiple
Let a,bZ with a0. Def.: a|b “a divides b” : (cZ: b=ac)
“There is an integer c such that c times a equals b.” Example: 312 True, but 37 False. Iff a divides b, then we say a is a factor or a divisor of b, and b is a multiple of a. Ex.: “b is even” :≡ 2|b.
So, the security of your day-to-day web transactions depends critically on the fact that all known factoring algorithms are intractable!
Hui Gao
Discrete Mathematics
Hui Gao
Discrete Mathematics
4
Fundamental Tpositive integer has a unique representation as the product of a nondecreasing series of zero or more primes.
离散数学课件(英文版)----Counting
– p(A) = 2n. Why?
• 加法原则
– 一件事情有两种做法，第一种做法有n种方式，第二种 做法有m种方式，则完成这件事情共有m＋n种方法 – 定义标识符：由字符开头的8位字符数字串或者一位字 符。共有多少个合法标识符？ – 含数字1的小于10000的正整数个数
Permutations
• Problem 1:
• In General:
– Order matters => permutation – Order does not matter =>combination
Pigeonhole Principle
• If n pigeons are assigned to m pigeonholes, and m<n, then at least one pigeonhole contains two or more pigeons.
– Proof by contradiction: Suppose each pigeonhole contains at most 1 pigeon. Then at most m pigeons have been assigned. Since m<n, so n-m>0, there are (n-m) pigeons have not been assigned. It’s a contradiction.
集合的等势关系
• 等势关系的定义：
– 如果存在从集合A到集合B的双射，则称集合A 与B等势。 – 集合A与B等势记为：AB, 否则A≉B – AB意味着：A，B中的元素可以“一一对应”。 – 要证明AB，找出任意一个从A到B的双射即可。
• “等势”的集合就被认为是“一样大”
• 加法原则
– 一件事情有两种做法，第一种做法有n种方式，第二种 做法有m种方式，则完成这件事情共有m＋n种方法 – 定义标识符：由字符开头的8位字符数字串或者一位字 符。共有多少个合法标识符？ – 含数字1的小于10000的正整数个数
Permutations
• Problem 1:
• In General:
– Order matters => permutation – Order does not matter =>combination
Pigeonhole Principle
• If n pigeons are assigned to m pigeonholes, and m<n, then at least one pigeonhole contains two or more pigeons.
– Proof by contradiction: Suppose each pigeonhole contains at most 1 pigeon. Then at most m pigeons have been assigned. Since m<n, so n-m>0, there are (n-m) pigeons have not been assigned. It’s a contradiction.
集合的等势关系
• 等势关系的定义：
– 如果存在从集合A到集合B的双射，则称集合A 与B等势。 – 集合A与B等势记为：AB, 否则A≉B – AB意味着：A，B中的元素可以“一一对应”。 – 要证明AB，找出任意一个从A到B的双射即可。
• “等势”的集合就被认为是“一样大”
离散数学课件 离散6.3-6.4节PPT
Example: Let S = {1, 2, 3}. The sequence 3,1,2 is a permutation of S. The sequence 3,1 is a 2-permutation of S.
Theorem: Let P (n, r) denote the number of r-permutations
5 / 18
Corollaries
Corollary 1: ∑nj=0 C(n, j) = 2n combinatorial proof
Corollary 2: ∑nj=0(−1)jC(n, j) = 0 Corollary 3: ∑nj=0 2jC(n, j) = 3n
Example: How many permutations of the letters ABCDEFGH contain the string ABC?
2 / 18
Combinations (|Ü)
Definition: An unordered selection of r elements of a set is called an r-combination.
Assignment 1 due in two weeks
1/4
Review of last time
Basic counting principles: the product, sum, substraction and division rules Tree diagrams The pigeonhole principle and its generalized version
How many bit strings of length n contain exactly r 1s?
离散数学课件(英文版)----Counting精选 课件
... <0,1> <1,1> <2,1> <3,1> ... <0,2> <1,2> <2,2> ... <0,3> <1,3>
Hale Waihona Puke <0,4><0,0>, <0,1>, <1,0>, <0,2>, <1,1>, <2,0>, <0,3>, ......
So, the set of rational numbers is countable.
as a sequence, { r1, r2, r3, ... } • (3) Assume, for example, that the decimal expansions of
the beginning of the sequence are as follows. r1 = 0 . 0 1 0 5 1 1 0 ... r2 = 0 . 4 1 3 2 0 4 3 ... r3 = 0 . 8 2 4 5 0 2 6 ... r4 = 0 . 2 3 3 0 1 2 6 ... r5 = 0 . 4 1 0 7 2 4 6 ... r6 = 0 . 9 9 3 7 8 3 8 ... r7 = 0 . 0 1 0 5 1 3 0 ...
– If the list goes on forever, it is infinite.
Proof of Countability
• The set of all integers is countable.
– We can arrange all integer in a linear list as follows: 0,-1,1,-2,2,-3,3,... that is: positive k is the (2k+1)th element, and negative k is the 2kth element in the list.
Hale Waihona Puke <0,4><0,0>, <0,1>, <1,0>, <0,2>, <1,1>, <2,0>, <0,3>, ......
So, the set of rational numbers is countable.
as a sequence, { r1, r2, r3, ... } • (3) Assume, for example, that the decimal expansions of
the beginning of the sequence are as follows. r1 = 0 . 0 1 0 5 1 1 0 ... r2 = 0 . 4 1 3 2 0 4 3 ... r3 = 0 . 8 2 4 5 0 2 6 ... r4 = 0 . 2 3 3 0 1 2 6 ... r5 = 0 . 4 1 0 7 2 4 6 ... r6 = 0 . 9 9 3 7 8 3 8 ... r7 = 0 . 0 1 0 5 1 3 0 ...
– If the list goes on forever, it is infinite.
Proof of Countability
• The set of all integers is countable.
– We can arrange all integer in a linear list as follows: 0,-1,1,-2,2,-3,3,... that is: positive k is the (2k+1)th element, and negative k is the 2kth element in the list.
离散数学英文版PPT
– I is given in the semester and not made up by the end of the next semester
How I Can Help You
• If you have a question, you can ask me face-to-face
Prerequisite and Description
• Prerequisite: MATH 170 Calculus I and CSCI 185 Programming II • Description: An introduction to discrete structures with applications to computing problems. Topics include logic, sets, functions, relations, proof techniques and algorithmic analysis. Graph theory and trees may be studied as well
Course Objectives
1. Relate practical examples to the appropriate set, function, or relation model, and interpret the associated operations and terminology in context 2. Apply proof techniques, including logic, to problems 3. Differentiate between dependent and independent events 4. Apply the binomial theorem to independent events and Bayes’ theorem to dependent events, and solve problems such as Hashing 5. Relate ideas of mathematical induction(归纳) to recursion and apply it to problems in computer science setting 6. Apply the basic counting principles, permutations and combinations to problems in computer science setting 7. Implementing algorithms in C or C++ programs
How I Can Help You
• If you have a question, you can ask me face-to-face
Prerequisite and Description
• Prerequisite: MATH 170 Calculus I and CSCI 185 Programming II • Description: An introduction to discrete structures with applications to computing problems. Topics include logic, sets, functions, relations, proof techniques and algorithmic analysis. Graph theory and trees may be studied as well
Course Objectives
1. Relate practical examples to the appropriate set, function, or relation model, and interpret the associated operations and terminology in context 2. Apply proof techniques, including logic, to problems 3. Differentiate between dependent and independent events 4. Apply the binomial theorem to independent events and Bayes’ theorem to dependent events, and solve problems such as Hashing 5. Relate ideas of mathematical induction(归纳) to recursion and apply it to problems in computer science setting 6. Apply the basic counting principles, permutations and combinations to problems in computer science setting 7. Implementing algorithms in C or C++ programs
离散数学课件(英文版)----Semigroup
离散数学课件英文版semigroup离散数学课件离散数学离散数学及其应用离散数学符号离散数学复习资料离散数学试题离散数学pdf离散数学吧离散数学知识点总结离散数学重点
ห้องสมุดไป่ตู้Algebraic Systems and Groups
Lecture 13 Discrete Mathematical Structures
If “” is associative, then x1x2x3… xn can be computed by any order of among the (n-1) operations, with the constraint that the order of all operands are not changed.
If S has a left identity and a right identity as well, then they must be equal, and this element is also an identity of the system: e l = e l e r= e r If existing, the identity of an algebraic system is unique: e 1= e 1e 2= e 2
Association
What a pity!
Semigroup
Axiom of semigroup
– Association
An example ({1,2},*), * defined as follows:
For any x,y∈{1,2}, x*y=y Proof: it should be proved that for any x,y,z in {1,2}, (x*y)*z = x* (y*z)
ห้องสมุดไป่ตู้Algebraic Systems and Groups
Lecture 13 Discrete Mathematical Structures
If “” is associative, then x1x2x3… xn can be computed by any order of among the (n-1) operations, with the constraint that the order of all operands are not changed.
If S has a left identity and a right identity as well, then they must be equal, and this element is also an identity of the system: e l = e l e r= e r If existing, the identity of an algebraic system is unique: e 1= e 1e 2= e 2
Association
What a pity!
Semigroup
Axiom of semigroup
– Association
An example ({1,2},*), * defined as follows:
For any x,y∈{1,2}, x*y=y Proof: it should be proved that for any x,y,z in {1,2}, (x*y)*z = x* (y*z)
235Ch5-1Induction离散数学英文版PPT
P(k+1) is correct, for all positive k. k < 2k (assumption) k+1 < 2k + 1 (from assumption)
< 2k + 2k = 2k+1
Inequality Example
• Prove that 2n < n! for integers n ≥ 4 • Basis step: P(4) = 24 = 16 < 4! = 4321 = 24 • Inductive step:
Mathematical Induction
• For a propositional function P(n) and any integers n ≥ b, b is a starting value of n in N
• (Basis Step) we prove: P(b) is true • (Inductive Step) and we prove P(k) P(k+1) for
P(n): a + ar1 + ar2 + … + arn = (arn+1 – a)/(r – 1) • Proof by induction (on blackboard)
Inequality Example
• Prove that n < 2n for positive integers n. • Basis step: P(1): 1 < 21 • Inductive step: assume P(k) is correct, prove
Assume 2k < k! for k ≥ 4, then 2k+1 = 22k < 2k! < (k + 1)k! = (k + 1)! • So, 2k+1 < (k + 1)!
< 2k + 2k = 2k+1
Inequality Example
• Prove that 2n < n! for integers n ≥ 4 • Basis step: P(4) = 24 = 16 < 4! = 4321 = 24 • Inductive step:
Mathematical Induction
• For a propositional function P(n) and any integers n ≥ b, b is a starting value of n in N
• (Basis Step) we prove: P(b) is true • (Inductive Step) and we prove P(k) P(k+1) for
P(n): a + ar1 + ar2 + … + arn = (arn+1 – a)/(r – 1) • Proof by induction (on blackboard)
Inequality Example
• Prove that n < 2n for positive integers n. • Basis step: P(1): 1 < 21 • Inductive step: assume P(k) is correct, prove
Assume 2k < k! for k ≥ 4, then 2k+1 = 22k < 2k! < (k + 1)k! = (k + 1)! • So, 2k+1 < (k + 1)!
离散数学课件(英文版)----Function
Try to prove it What is ({<x,y>|x,y∈R, y=x+1})? R×{-1}
x-y>,
:N×N→N, (<x,y>) = | x2-y2|
(N×{0}) ={ n2|n∈N}, -1({0}) ={<n,n>|n∈N}
Characteristic Function of Set
A’ B’ B A
f
Special Types of Functions
Surjection
:A→B is a surjection or “onto” iff. Ran()=B, iff. y∈B, x∈A, such that f(x)=y
Injection (one-to-one)
:A→B is one-to-one iff. y∈Ran(f), there is at most one x∈A, such that f(x)=y iff. x1,x2∈A, if x1≠x2， then (x1) ≠(x2) iff. x1,x2∈A, if (x1) =(x2)，then x1=x2
Note: if <x,y>∈f, then <x,y>∈f and <x,x>∈1A if <x,y>∈ f °1A，则<t,y>∈ f , 且<x,t>∈1A， 则t=x, 所以<x,y>∈f .
Invertible Function
The inverse relation of f :A→B is not necessarily a function, even though f is.
If |A|>|B| then 0 else |A|!*|A|C|B|
x-y>,
:N×N→N, (<x,y>) = | x2-y2|
(N×{0}) ={ n2|n∈N}, -1({0}) ={<n,n>|n∈N}
Characteristic Function of Set
A’ B’ B A
f
Special Types of Functions
Surjection
:A→B is a surjection or “onto” iff. Ran()=B, iff. y∈B, x∈A, such that f(x)=y
Injection (one-to-one)
:A→B is one-to-one iff. y∈Ran(f), there is at most one x∈A, such that f(x)=y iff. x1,x2∈A, if x1≠x2， then (x1) ≠(x2) iff. x1,x2∈A, if (x1) =(x2)，then x1=x2
Note: if <x,y>∈f, then <x,y>∈f and <x,x>∈1A if <x,y>∈ f °1A，则<t,y>∈ f , 且<x,t>∈1A， 则t=x, 所以<x,y>∈f .
Invertible Function
The inverse relation of f :A→B is not necessarily a function, even though f is.
If |A|>|B| then 0 else |A|!*|A|C|B|
离散数学课件(英文版)----Graph
A D C A D
C
B
B
Graph and Diagram
Graph G is a triple: G =〈VG, EG, 〉
VG and EG are sets，satisgying VGEG=φ, :EG {{vi, vj}| vi, vj∈VG} Note: {vi, vj}={vj, vi} A graph can be represented conveniently by some diagram: : each element of VG as a dot, the vertex, and each element of EG as a line segment, the edge, between vertices. So, VG is called the set of vertices, and EG, the set of edges.
i i =1
The number of vertices with odd degree must be even.
Complete Graph
A graph is a complete graph if and only if any two of its vertices are adjacent.
Subgraph
Let G=<V,E>, G’=<V’,E’>, if V’V, E’E, then G’is called a subgraph of G. If V’V, or E’E, then G’ is a proper subgraph. If V’=V, then G’ is a spanning subgraph.
Determination of Euler Graph
C
B
B
Graph and Diagram
Graph G is a triple: G =〈VG, EG, 〉
VG and EG are sets，satisgying VGEG=φ, :EG {{vi, vj}| vi, vj∈VG} Note: {vi, vj}={vj, vi} A graph can be represented conveniently by some diagram: : each element of VG as a dot, the vertex, and each element of EG as a line segment, the edge, between vertices. So, VG is called the set of vertices, and EG, the set of edges.
i i =1
The number of vertices with odd degree must be even.
Complete Graph
A graph is a complete graph if and only if any two of its vertices are adjacent.
Subgraph
Let G=<V,E>, G’=<V’,E’>, if V’V, E’E, then G’is called a subgraph of G. If V’V, or E’E, then G’ is a proper subgraph. If V’=V, then G’ is a spanning subgraph.
Determination of Euler Graph
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• Prerequisite: MATH 170 Calculus I and CSCI 185 Programming II • Description: An introduction to discrete structures with applications to computing problems. Topics include logic, sets, functions, relations, proof techniques and algorithmic analysis. Graph theory and trees may be studied as well
Course Objectives
1. Relate practical examples to the appropriate set, function, or relation model, and interpret the associated operations and terminology in context 2. Apply proof techniques, including logic, to problems 3. Differentiate between dependent and independent events 4. Apply the binomial theorem to independent events and Bayes’ theorem to dependent events, and solve problems such as Hashing 5. Relate ideas of mathematical induction(归纳) to recursion and apply it to problems in computer science setting 6. Apply the basic counting principles, permutations and combinations to problems in computer science setting 7. Implementing algorithms in C or C++ programs
Attendance Policy
• A student is expected to attend each class session on a regular and punctual basis • Students will be allowed to be late OR absent during the semester no more than three (3) times. Students who exceed these limits may be withdrawn from the course, or given an F grade
– Homework and attendance: 20% – Midterm exams 35% – Final exam 35% – Project 10%
Letter Grade A AB+ B BC+ Point Grade Letter Grade 95 - 100 C 90 - 94 C85 - 89 D+ 80 - 84 D 75 - 79 70 - 74 F Point Grade 65 – 69 60 – 64 55 – 59 50 - 54 Below 50
• Absence from a scheduled test is generally disallowed. Approval from the Department or College is required under extreme circumstances
Academic Honesty
• Cheating or copying on an examination or an assignment will result in an F grade on the exam or assignment • If it occurs the second time, the grade for the whole course will be an F • If it occurs the third time, the case will be brought to the university and the ultimate punishment could be a dismissal from the universities • Please refer to the Cheating Policy being enforced by both NYIT and NUPT on the website
• Exams are always open book!
Importance of Homework and Attendance
• Homework and attending the classes help you learn the course material and improve your grades in the exams • Your final grade is calculated based on the formula HW+Attn (20) + Midterm (35) + Final (35) + Project (10) • Example: Assume midterm grade is 20 out of 35 and your final is 20 out of 35. If you did not do any homework and failed to attend the class for three (not four!) times, your homework and attendance grade is 0, your project grade is 8, then your final grade is 0+20+20+8= 48, which is a Fail (F). If you did all the homework and attended all the classes (easy!), your final grade is 20+20+20+8 = 68, which is a C • Missing one or two homework assignments sometimes makes a big difference!
Reading and Reviewing
• Read the section or chapter of the textbook “before” and after the lecture that covers it. It will help in learning the material, answering questions in the homework and the exams • Read and review the presentation slides (课件). It is most helpful in understanding the course contents and the focuses of the course • Slides can be obtained in class through USB drive
Incomplete Grade: I
• A grade of incomplete, I, is used when a student has been unable to complete all assigned work for the course because of some unavoidable circumstance. The instructor must certify that the student’s work is at least a pass at this point and the student must agree to complete the missing work without paying the fee again • A grade of I will become an F in the following situation:
Learning Outcomes
1. An ability to apply algorithmic concepts and constructs in problem analysis and design (Specific about algorithmic concepts and constructs) 2. An ability to apply concepts of discrete mathematics, probability, and statistics (General about discrete mathematics) 3. A knowledge of mathematics as it applies to solving problems in computer science (More general about mathematics)
– I is given in the semester and not made up by the end of the next semester
How I Can Help You
• If you have a question, you can ask me face-to-face
Exams and Grading Policy
• • • • There is a midterm exam in week 7 or 8 There is a non-comprehensive final exam (week 15) There is a small programming in Visual Basic project Grading
Course Objectives
1. Relate practical examples to the appropriate set, function, or relation model, and interpret the associated operations and terminology in context 2. Apply proof techniques, including logic, to problems 3. Differentiate between dependent and independent events 4. Apply the binomial theorem to independent events and Bayes’ theorem to dependent events, and solve problems such as Hashing 5. Relate ideas of mathematical induction(归纳) to recursion and apply it to problems in computer science setting 6. Apply the basic counting principles, permutations and combinations to problems in computer science setting 7. Implementing algorithms in C or C++ programs
Attendance Policy
• A student is expected to attend each class session on a regular and punctual basis • Students will be allowed to be late OR absent during the semester no more than three (3) times. Students who exceed these limits may be withdrawn from the course, or given an F grade
– Homework and attendance: 20% – Midterm exams 35% – Final exam 35% – Project 10%
Letter Grade A AB+ B BC+ Point Grade Letter Grade 95 - 100 C 90 - 94 C85 - 89 D+ 80 - 84 D 75 - 79 70 - 74 F Point Grade 65 – 69 60 – 64 55 – 59 50 - 54 Below 50
• Absence from a scheduled test is generally disallowed. Approval from the Department or College is required under extreme circumstances
Academic Honesty
• Cheating or copying on an examination or an assignment will result in an F grade on the exam or assignment • If it occurs the second time, the grade for the whole course will be an F • If it occurs the third time, the case will be brought to the university and the ultimate punishment could be a dismissal from the universities • Please refer to the Cheating Policy being enforced by both NYIT and NUPT on the website
• Exams are always open book!
Importance of Homework and Attendance
• Homework and attending the classes help you learn the course material and improve your grades in the exams • Your final grade is calculated based on the formula HW+Attn (20) + Midterm (35) + Final (35) + Project (10) • Example: Assume midterm grade is 20 out of 35 and your final is 20 out of 35. If you did not do any homework and failed to attend the class for three (not four!) times, your homework and attendance grade is 0, your project grade is 8, then your final grade is 0+20+20+8= 48, which is a Fail (F). If you did all the homework and attended all the classes (easy!), your final grade is 20+20+20+8 = 68, which is a C • Missing one or two homework assignments sometimes makes a big difference!
Reading and Reviewing
• Read the section or chapter of the textbook “before” and after the lecture that covers it. It will help in learning the material, answering questions in the homework and the exams • Read and review the presentation slides (课件). It is most helpful in understanding the course contents and the focuses of the course • Slides can be obtained in class through USB drive
Incomplete Grade: I
• A grade of incomplete, I, is used when a student has been unable to complete all assigned work for the course because of some unavoidable circumstance. The instructor must certify that the student’s work is at least a pass at this point and the student must agree to complete the missing work without paying the fee again • A grade of I will become an F in the following situation:
Learning Outcomes
1. An ability to apply algorithmic concepts and constructs in problem analysis and design (Specific about algorithmic concepts and constructs) 2. An ability to apply concepts of discrete mathematics, probability, and statistics (General about discrete mathematics) 3. A knowledge of mathematics as it applies to solving problems in computer science (More general about mathematics)
– I is given in the semester and not made up by the end of the next semester
How I Can Help You
• If you have a question, you can ask me face-to-face
Exams and Grading Policy
• • • • There is a midterm exam in week 7 or 8 There is a non-comprehensive final exam (week 15) There is a small programming in Visual Basic project Grading