北京邮电大学计算机学院 离散数学 数学结构 群论 chap9-2

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北京邮电大学 计算机学院 离散数学 第十章补充 传输+网络流

北京邮电大学  计算机学院  离散数学   第十章补充  传输+网络流

2015-2-5
College of Computer Science & Technology, BUPT
12
A Maximum Flow Algorithm



The algorithm we present is due to Ford and Fulkerson and is often called the labeling algorithm(标记算法). The labeling referred to is an additional labeling of nodes. We have used integer capacities for simplicity, but Ford and Fulkerson show that this algorithm will stop in a finite number of steps if the capacities are rational numbers.
2015-2-5
College of Computer Science & Technology, BUPT
10
Virtual Flow
2015-2-5
College of Computer Science & Technology, BUPT
11

Let N be a network and let G be the symmetric closure of N.
Content

Application of digraph Concept




Capacity (容量) Transport network (传输网) Flow (流量) Virtual flow (虚流量) Maximal flow (最大流量)

(精华版)国家开放大学电大本科《离散数学》网络课形考网考作业及答案

(精华版)国家开放大学电大本科《离散数学》网络课形考网考作业及答案

(精华版)国家开放大学电大本科《离散数学》网络课形考网考作业及答案(精华版)国家开放大学电大本科《离散数学》网络课形考网考作业及答案 100%通过考试说明:2020年秋期电大把该网络课纳入到“国开平台”进行考核,该课程共有5个形考任务,针对该门课程,本人汇总了该科所有的题,形成一个完整的标准题库,并且以后会不断更新,对考生的复习、作业和考试起着非常重要的作用,会给您节省大量的时间。

做考题时,利用本文档中的查找工具,把考题中的关键字输到查找工具的查找内容框内,就可迅速查找到该题答案。

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课程总成绩 = 形成性考核×30% + 终结性考试×70% 形考任务1 单项选择题题目1 若集合A={ a,{a},{1,2}},则下列表述正确的是().选择一项:题目2 若集合A={2,a,{ a },4},则下列表述正确的是( ).选择一项:题目3 设集合A={1 , 2 , 3 , 4}上的二元关系R={<1, 1>,<2, 2>,<2, 3>,<4, 4>},S={<1, 1>,<2, 2>,<2, 3>,<3, 2>,<4, 4>},则S是R的()闭包.选择一项:B. 对称题目4 设集合A={1, 2, 3},B={3, 4, 5},C={5, 6, 7},则A∪B–C=( ).选择一项:D. {1, 2, 3, 4} 题目5 如果R1和R2是A上的自反关系,则R1∪R2,R1∩R2,R1-R2中自反关系有()个.选择一项:C. 2 题目6 集合A={1, 2, 3, 4}上的关系R={<x,y>|x=y且x, y∈A},则R的性质为().选择一项:D. 传递的题目7 若集合A={1,2},B={1,2,{1,2}},则下列表述正确的是( ).选择一项:题目8 设A={a,b,c},B={1,2},作f:A→B,则不同的函数个数为().选择一项:C. 8 题目9 设A={1, 2, 3, 4, 5, 6, 7, 8},R是A上的整除关系,B={2, 4, 6},则集合B的最大元、最小元、上界、下界依次为 ( ).选择一项:B. 无、2、无、2 题目10 设集合A ={1 , 2, 3}上的函数分别为:f = {<1, 2>,<2, 1>,<3, 3>},g = {<1, 3>,<2, 2>,<3, 2>},h = {<1, 3>,<2,1>,<3, 1>},则h =().选择一项:D. f◦g 判断题题目11 设A={1, 2}上的二元关系为R={<x, y>|xA,yA, x+y =10},则R的自反闭包为{<1, 1>, <2, 2>}.()选择一项:对题目12 空集的幂集是空集.()选择一项:错题目13 设A={a, b},B={1, 2},C={a, b},从A到B的函数f={<a, 1>, <b, 2>},从B到C的函数g={<1, b>, <2, a >},则g° f ={<1,2 >, <2,1 >}.()选择一项:错题目14 设集合A={1, 2, 3, 4},B={2, 4, 6, 8},下列关系f = {<1, 8>, <2, 6>,<3, 4>, <4, 2,>}可以构成函数f:.()选择一项:对题目15 设集合A={1, 2, 3},B={2, 3, 4},C={3, 4, 5},则A∩(C-B )= {1, 2, 3, 5}.()选择一项:错题目16 如果R1和R2是A上的自反关系,则、R1∪R2、R1∩R2是自反的.()选择一项:对题目17 设集合A={a, b, c, d},A上的二元关系R={<a, b>, <b, a>, <b, c>, <c, d>},则R具有反自反性质.()选择一项:对题目18 设集合A={1, 2, 3},B={1, 2},则P(A)-P(B )={{3},{1,3},{2,3},{1,2,3}}.()选择一项:对题目19 若集合A = {1,2,3}上的二元关系R={<1, 1>,<1, 2>,<3, 3>},则R是对称的关系.()选择一项:错题目20 设集合A={1, 2, 3, 4 },B={6, 8, 12}, A到B的二元关系R=那么R-1={<6, 3>,<8,4>}.()选择一项:对形考任务2 单项选择题题目1 无向完全图K4是().选择一项:C. 汉密尔顿图题目2 已知一棵无向树T中有8个顶点,4度、3度、2度的分支点各一个,T的树叶数为( ).选择一项:D. 5 题目3 设无向图G的邻接矩阵为则G的边数为( ).选择一项:A. 7 题目4 如图一所示,以下说法正确的是 ( ) .选择一项:C. {(d, e)}是边割集题目5 以下结论正确的是( ).选择一项:C. 树的每条边都是割边题目6 若G是一个欧拉图,则G一定是( ).选择一项:B. 连通图题目7 设图G=<V, E>,v∈V,则下列结论成立的是 ( ) .选择一项:题目8 图G如图三所示,以下说法正确的是 ( ).选择一项:C. {b, c}是点割集题目9 设有向图(a)、(b)、(c)与(d)如图五所示,则下列结论成立的是( ).选择一项:A. (a)是强连通的题目10 设有向图(a)、(b)、(c)与(d)如图六所示,则下列结论成立的是( ).选择一项:D. (d)只是弱连通的判断题题目11 设图G是有6个结点的连通图,结点的总度数为18,则可从G中删去4条边后使之变成树.( ) 选择一项:对题目12 汉密尔顿图一定是欧拉图.( ) 选择一项:错题目13 设连通平面图G的结点数为5,边数为6,则面数为4.( ) 选择一项:错题目14 设G是一个有7个结点16条边的连通图,则G为平面图.( ) 选择一项:错题目15 如图八所示的图G存在一条欧拉回路.( ) 选择一项:错题目16 设图G如图七所示,则图G的点割集是{f}.( ) 选择一项:错题目17 设G是一个图,结点集合为V,边集合为E,则( ) 选择一项:对题目18 设图G是有5个结点的连通图,结点度数总和为10,则可从G中删去6条边后使之变成树.( ) 选择一项:错题目19 如图九所示的图G不是欧拉图而是汉密尔顿图.( ) 选择一项:对题目20 若图G=<V, E>,其中V={ a, b, c, d },E={ (a, b), (a, d),(b, c), (b, d)},则该图中的割边为(b, c).( ) 选择一项:对形考任务3 单项选择题题目1 命题公式的主合取范式是( ).选择一项:题目2 设P:我将去打球,Q:我有时间.命题“我将去打球,仅当我有时间时”符号化为( ).选择一项:题目3 命题公式的主析取范式是( ).选择一项:题目4 下列公式成立的为( ).选择一项:题目5 设A(x):x是书,B(x):x是数学书,则命题“不是所有书都是数学书”可符号化为().选择一项:题目6 前提条件的有效结论是( ).选择一项:B. ┐Q 题目7 命题公式(P∨Q)→R的析取范式是 ( ).选择一项:D. (┐P∧┐Q)∨R 题目8 下列等价公式成立的为( ).选择一项:题目9 下列等价公式成立的为( ).选择一项:题目10 下列公式中 ( )为永真式.选择一项:C. ┐A∧┐B ↔ ┐(A∨B) 判断题题目11 设个体域D={1, 2, 3},A(x)为“x小于3”,则谓词公式(∃x)A(x) 的真值为T.( ) 选择一项:对题目12 设P:小王来学校, Q:他会参加比赛.那么命题“如果小王来学校,则他会参加比赛”符号化的结果为P→Q.( ) 选择一项:对题目13 下面的推理是否正确.( ) (1) (∀x)A(x)→B(x) 前提引入(2) A(y)→B(y) US (1) 选择一项:错题目14 含有三个命题变项P,Q,R的命题公式P∧Q的主析取范式(P∧Q∧R)∨(P∧Q∧┐R).( ) 选择一项:对题目15 命题公式P→(Q∨P)的真值是T.( ) 选择一项:对题目16 命题公式┐P∧P的真值是T.( ) 选择一项:错题目17 谓词公式┐(∀x)P(x)(∃x)┐P(x)成立.( ) 选择一项:对题目18 命题公式┐(P→Q)的主析取范式是P∨┐Q.( ) 选择一项:错题目19 设个体域D={a, b},则谓词公式(∀x)(A(x)∧B(x))消去量词后的等值式为(A(a)∧B(a))∧(A(b)∧B(b)).( ) 选择一项:对题目20 设个体域D={a, b},那么谓词公式(∃x)A(x)∨(∀y)B(y)消去量词后的等值式为A(a)∨B(b).( ) 选择一项:错形考任务4 要求:学生提交作业有以下三种方式可供选择:1. 可将此次作业用A4纸打印出来,手工书写答题,字迹工整,解答题要有解答过程,完成作业后交给辅导教师批阅. 2. 在线提交word文档. 3. 自备答题纸张,将答题过程手工书写,并拍照上传形考任务 5 网上学习行为(学生无需提交作业,占形考总分的10%)附:元宇宙(新兴概念、新型虚实相融的互联网应用和社会形态)元宇宙(Metaverse)是整合了多种新技术而产生的新型虚实相融的互联网应用和社会形态,通过利用科技手段进行链接与创造的,与现实世界映射与交互的虚拟世界,具备新型社会体系的数字生活空间。

北京邮电大学计算机学院 离散数学 9.1~9.3-relations

北京邮电大学计算机学院  离散数学  9.1~9.3-relations

A B = {(a, b) | a A and b B}
2015-2-5
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A1 A2 Am


The Cartesian product A1 A2 Am of the nonempty sets A1, A2, , Am is the set of all ordered m-tuples (m元组) (al, a2, ... , am), where ai Ai, i = 1, 2, . . . , m Thus
9.1 Relations and Their Properties 关系及关系性质 9.2 n-ary Relations and Their Applications n元关系及应用 9.3 Representing Relations 关系的表示 9.4 Closures of Relations 关系闭包 9.5 Equivalence Relations 等价关系 9.6 Partial Orderings 偏序关系

R(A1) = {y B | x R y for some x in A1}
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Theorem

Let R be a relation from A to B, and let A1 and A2 be subsets of A. Then
College of Computer Science & Technology, BUPT

北京邮电大学计算机学院 离散数学 10.1~10.2 graphs

北京邮电大学计算机学院  离散数学 10.1~10.2 graphs

Leonard Euler 1736 (father of graph theory)
College of Computer Science & Technology, BUPT -- © Copyright Yang Juan
Kö nigsberg Bridge problem

Picture only what is essential to the problem.
College of Computer Science & Technology, BUPT -- © Copyright Yang Juan
p643
College of Computer Science & Technology, BUPT -- © Copyright Yang Juan
Directed Multigraphs
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P642
College of Computer Science & Technology, BUPT -- © Copyright Yang Juan
Pseudographs
A particular class of discrete structures (to be defined) that is useful for representing relations and has a convenient webby-looking graphical representation.


Correspond to symmetric, irreflexive binary relations R. A simple graph G=(V,E) Visual Representation of a Simple Graph consists of:

北京邮电大学计算机学院 离散数学 数学结构 群论 chap9-3

北京邮电大学计算机学院  离散数学  数学结构 群论 chap9-3

16
Example 5

Let B = {0, l }, and let + be the operation defined on B as follows:

Then B is a group.
2015-2-5
College of Computer Science & Technology, BUPT

a*a' = a*4/a = a(4/a)/2 = 2 = (4/a)(a)/2 = (4/a)*a = a' *a. a*b = ab/2 = ba/2 = b*a

Abelian


So, G is an Abelian group.
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2015-2-5
Theorem 1


Let G be a group. Each element a in G has only one inverse in G. Proof

Let

a' and a" be inverses of a. a' = a'e = a'(aa") = (a'a)a" = ea" = a".

Then

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Theorem 2

Let

G be a group and a, b, and c be elements of G. left cancellation – 左消去律

北京邮电大学 计算机学院 离散数学 3.2-Growth of Functions

北京邮电大学  计算机学院  离散数学 3.2-Growth of Functions
However, you should prove that the values you choose do work.
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2015-2-6
little-o of g

In calculus

If

Then

Theorem 1

Let f(x) = anxn + an-1xn-1 +…+ a1x+a0 ,where a0 , a1 , … an-1, an are real numbers. Then f(x) is O(xn) .



n! is O(nn) log n! is O(nlogn) log n is O(n) 1, log n, n, nlogn, n2, 2n, n!
The Growth of Functions


We quantify the concept that g grows at least as fast as f. What really matters in comparing the complexity of algorithms?

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2015-2-6
Visualizing Orders of Growth

On a graph, as you go to the right, the fastergrowing function always eventually becomes the larger one...

离散数学高等里离散数学课件-CHAP

离散数学高等里离散数学课件-CHAP
图论
图的基本概念

连接两个节点的线段称为边。
简单图与多重图
只含一条边的图称为简单图, 含有相同端点的多条边称为多 重边。
节点
图中的顶点称为节点。
定向图与无向图
如果边有方向,则称为定向图; 如果边无方向,则称为无向图。
有限图与无限图
节点和边都有限的图称为有限 图,节点或边至少有一个为无 限的图称为无限图。
发展
随着计算机科学的快速发展,离散数学也得到了迅速的发展 。许多新的分支如组合数学、离散概率论等不断涌现,并广 泛应用于计算机科学、工程学、物理学等领域。
离散数学的应用领域
计算机科学
离散数学在计算机科学中有着广泛的 应用,如算法设计、数据结构、计算 机图形学、数据库系统等。
工程学
离散数学在工程学中也有着广泛的应 用,如电子工程、通信工程、机械工 程等。
要点二
详细描述
集合可以用列举法、描述法、图示法等多种方法来表示。 列举法是将集合中的所有元素一一列举出来,适用于元素 数量较少的集合。描述法是用数学符号和逻辑表达式来描 述集合中的元素,适用于元素数量较多且具有共同特征的 集合。图示法则是用图形来表示集合,直观易懂,适用于 具有明显包含关系的集合。
03
如果图中任意两个节点之间都存在一 条路径,则称该图为连通图。
路径与回路
欧拉回路与哈密顿回路
如果一条回路恰好经过图中的每条边 一次,则称为欧拉回路;如果一条回 路恰好经过图中的每个节点一次,则 称为哈密顿回路。
连接两个节点的序列称为路径,如果 路径的起点和终点是同一点,则称为 回路。
04
离散概率论
离散概率的基本概念
图的表示方法
邻接矩阵
用矩阵表示图中节点之 间的关系,如果节点i与 节点j之间存在一条边, 则矩阵中第i行第j列的 元素为1,否则为0。

离散数学北京邮电大学

离散数学北京邮电大学
– Analysis using order-ofgrowth notation.
• §3.5: Primes and Greatest Common Divisors • §3.6: Integers & Algorithms
– Alternate bases, algorithms for basic arithmetic
Algorithm Characteristics
Some important general features of algorithms: • Input. Information or data that comes in. • Output. Information or data that goes out. • Definiteness. Algorithm is precisely defined. • Correctness. Outputs correctly relate to inputs. • Finiteness. Won‟t take forever to describe or run. • Effectiveness. Individual steps are all do-able. • Generality. Works for many possible inputs. • Efficiency. Takes little time & memory to run.
– Example assignment statement: v := 3x+7 (If x is 2, changes v to 13.)
• In pseudocode (but not real code), the expression might be informally stated:

北京邮电大学 计算机学院 离散数学 3.3-Complexity of algorithm

北京邮电大学  计算机学院  离散数学 3.3-Complexity of algorithm

§3.3: Algorithmic omplexity


The algorithmic complexity of a computation is, most generally, a measure of how difficult it is to perform the computation. That is, it measures some aspect of the cost of computation (in a general sense of “cost”).
Constant Logarithmic (same order c) (With c Polylogarithmica constant.) Linear Polynomial (for any c) Exponential (for c>1) Factorial
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Complexity analysis, cont.
procedure max(a1, a2, …, an: integers) v := a1 t1 Times for for i := 2 to n t2 each execution if ai > v then v := ai t3 of each return v t4 line. w.c.t.c.: n


Characterize complexity as a function of input size: Worst-case, best-case, or average-case. Use orders-of-growth notation to concisely summarize the growth properties of complexity functions.

离散数学北邮内部资料

离散数学北邮内部资料

景晓军
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等值演算(判断命题公式类型法Ⅱ)
(置换定理):设Φ(A)是含命题公式A的命题公式,Φ(B) 是命题公式B置换了Φ(A)中A之后得到的命题公式。如 果AB,则Φ(A)Φ(B)。
例如:P∧7(q∧r)P∧(7qV7r)
景晓军
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等值演算
设A,B是两个命题,若等价式A↔B是重言式, 则A与B是等值的,记为A<=>B。 注意和“<=>”、“=”的关系。 A↔B是重言式(说明只出现A与B 的值 同时为真或同时为假的两种情况),所 以肯定A <=> B 。 注意和“<=>”、“↔”的关系。如A <=> B 则A↔B必是重言式。若A↔B,未必A <=> B 因为A↔B有4种情况.
不是所有的“和”、“与”都可用“∧”表示。
李文和李武是兄弟: p
景晓军
15
析取联结词
设p,q为两个命题,复合命题“p或q” 称 作p与q的析取式,记作p ∨ q , ∨为 析取联结词。 p ∨ q为真当且仅当p与 q中至少一个为真。 析取联结词是逻辑“或”的意思。
王燕学过英语(p)或法语(q) p∨q
景晓军
简单命题与复合命题
简单命题:命题不能分成更简单的句子的命题,又称为 命题常项。 2是素数. 雪是黑色的. 2+3=5. 明年十月一日是晴天. 复合命题:由简单命题用联结词联结而成的命题。 3不是偶数; 2是素数和偶数; 林芳学过英语或日语; 如果角A和角B是对顶角,则角A等于角B.
景晓军
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景晓军
24
真值表
含n(n>1)个命题变项的命题公式,共有2n组赋值,将命题 公式A在所有的赋值之下取值的情况列成表,称为A的真 值表. 如计算 p ∧ (q ∨ ┐r ) 的真值表

北京邮电大学 离散数学下 群论 作业错题讲解

北京邮电大学 离散数学下 群论 作业错题讲解

群论9.1-9.2 (1)20,28 @323-324a. not commutative, a*b ≠b*a.b. a*(b*c)=a*c=b, (a*b)*c=c*c=b.c. not associative, (c*c)*c=b*c=c ≠a=c*b=c*(c*c).群论9.1-9.2 (1)20,28 @323-3241. ≤is reflexiveSince a= a*a, a≤a for all a in A.2. ≤is antisymmetricsuppose that a≤b and b≤a.Then, b = a*b= b*a= a, so a= b.3. ≤is transitiveIf a≤b and b≤c, then c= b*c= (a*b)*c= a*(b*c)= a*c, so a≤c.群论9.1-9.2 (1)28 @323-324show a*b= a∨b, for all a and b in A(1)a*b= a*(b*b) =(a*b)*b=b*(a*b), so b ≤a*b.In a similar way, a*b= (a*a)*b=a*(a*b), so a≤a*b.so a*b is a upper bound for a and b.(2)if a≤c and b≤c.then c= a*c and c= b*c by definition. Thus c= a*(b*c)= (a*b)*c. so a*b≤c.This shows that a*b is the leastest upper bound of a and b.群论9.1-9.2 (2)12,16@page 348;︒f1 f2 f3 f4 f5 f6closure;f1 f1 f2 f3 f4 f5 f6identity=f1;f2 f2 f1 f5 f6 f3 f4reverse:f2-1=f2,f3-1=f3, f3 f3 f4 f1 f2 f6 f5f4-1=f5,f6-1=f6.f4 f4 f3 f6 f5 f1 f2associate: ︒is associate.f5 f5 f6 f2 f1 f4 f3f6 f6 f5 f4 f3 f2 f1群论9.1-9.2 (2)12,16@page 348; If a*a=e, ∀a∈G. show commutative.(1)group propertise (a*b)-1=b-1*a-1 because a*a=e=a*a-1 so, b-1*a-1=b*a, (2) (a*b)*(a*b)=e => (a*b)-1=a*b, reverse be only one in Group.so b*a=a*b.群论9.1-9.2 Ex1: Let G be a group. Fora,b∈G,we say that b is conjugate to a,written by b~a, if there exist g∈G such that b=gag-1.show that ~ is a equivalence relation on G. The equivalence classes of are called the conjugacy classes of G.proof: (1)reflexive a=eae, so a~a;(2)symmetric If b~a, then b=gag-1,a=eae=g-1gag-1g=g-1bg.so a~b.(3)transitive suppose a~b and b~c,a=gbg-1 ,b=h ch-1; so a=ghch-1g-1=(gh)c(gh)-1群论9.1-9.2 Ex2: Let G be a group, and suppose that a and b are any elements of G.Show that if (ab)2=a2b2,,then ba=ab.proof:(ab)2=ab*ab=a*(b*a)*ba2b2=aa*bb=a*(a*b)*b,based left/right cancellation, b*a=a*b.群论9.1-9.2 Ex3: Let G ={x∈R|x>1} be the set of all real numbers greater than 1. For x,y∈G, definex*y=xy-x-y+2. Show that (G,*) is a group.proof: (1)closure; because x>1, xy>x+y,so x*y=xy-x-y+2>2, x*y∈G.(2)associate; (x*y)*z=(xy-x-y+2)*z=xyz-xz-yz+2z-(xy-x-y+2)-z+2=xyz-xy-xz-yz+x+y+z .x*(y*z)=x*(yz-y-z+2)=xyz-xy-xz+2x-x-yz+y+z-2+2=xyz-xy-xz-yz+x+y+z .(3)identity: e=2, 2*x=2x-x-2+2=x.(4)reverse: x*x-1=xx-1-x-x-1+2=e=2, x-1=x/(x-1).群论9.4 (1)22,28@page 331;22-Show S 1⋂S 2is a subsemigroup of S.proof: closure; ∀a,b ∈S 1⋂S 2, then a*b ∈S 1, a*b ∈S 2, so a*b ∈S 1⋂S 2.if S 1⋂S 2 =∅, still be subsemigroup.28-If f:S 1→S 2and g :S 2→S 3 be isomorphisms, show g ︒f :S 1→S 3is an isomorphism.proof: (g ︒f)(x *1y)= g(f(x *1y))=g(f(x) *2f(y))=g(f(x))*3g(f(y))=(g ︒f)(x) *3(g ︒f)(y).one-to-one: suppose (g ︒f)(a)=(g ︒f)(b), g(f(a))=g(f(b)), f(a)=f(b), a=b.onto: ∀z ∈S 3 有g(y)=z; ∀y ∈S 2 有f(x)=y;群论9.4 (2)28,32@page 349; 24@page338. 28-Show f:G→G defined by f(a)=a n is a homo morphism.proof: (数学归纳法证)1.because ab=ba, ab*ab=aabb=a2b2,2.(ab)n=a n b n,ab*(ab)n=ab*a n b n=aa n bb n=a n+1b n+1 Hence, f(ab)=(ab)n=a n b n=f(a)f(b).群论9.4 (2)28,32@page 349; 24@page338.32-Show f:G→G by f(a)=a-1 is a isomorphism iff Abelian.Proof: (1)Suppose f is isomorphism,f(xy)=(xy)-1=f(x)f(y)=x-1y-1so xy=((xy)-1)-1=(x-1y-1)-1=yx. G is Abelian. (2)Suppose G is Abelian.f(xy)=(xy)-1=x-1y-1=f(x)f(y), homomorphism. onto: ∀x∈G, f(x-1)=(x-1)-1=x.one-to-one: suppose f(x)=f(y), x-1=y-1,xx-1=e=yy-1, right cancelation x=y.群论9.4 (2)28,32@page 349; 24@page338.24-Let A={0,1} and consider (A*, ⋅), (N,+).(1)f(α)=length(α), show f:A*->N is homomorphism.f(α⋅β)=f(αβ)=length(α)+length(β)=f(α)+f(β)(2)R: f(α)=f(β),show R is congruence relation.First show R is equivalence relation.2. if f(α)=f(β) and f(χ)=f(δ),then f(α⋅χ) = f(β⋅δ)(3)A*/R and N is isomorphic.(A*/R,*) : [α]*[β]=[α⋅β].Let g([α])=length(α), (1)show g: A*/R->N is homomorphism. g([α] *[β])=g([α⋅β])=length(α)+length(β)=g([α])+g([β])(2) onto : ∀x∈N, let α=00…0(x factors) ,g([α])=x.(3) one-to-one : suppose g([α])=g([β]),length(α)=length(β),so f(α)=f(β), then αRβ, [α]=[β].4: Show f:G 1⨯G 2→G 1by f(a,b)=a is a homomorphism. f((a,b )*‘(c,d))=f(a*c,b*d)=a*c=f(a,b)*f(c,d)18: Prove N is normal subgroup iff a -1Na=N for all a ∈G.If N is normal subgroup, then ∀a ∈G, aN=Na, ande ∈N. so a -1Na=Na -1a=Ne=N.If a -1Na=N for all a ∈G, then N=a -1aN=a -1Na,left cancelation, aN=Na.29: Prove H is normal subgroup if H only two left coset.let a∉H, The left cosets of H are H and aH.The right coset are H and Ha.H⋂aH = ∅= H⋂Ha. and H⋃aH=G=H⋃Ha.Thus aH=Ha.let b∈H, bH=H=Hb. so ∀x∈G, xH=Hx.H is a normal subgroup.30: Prove that if N is a normal subgroup of G, then H⋂N is a normal subgroup of H.(1)show H⋂N is a subset of H;∀x∈H⋂N, x ∈H , H⋂N is a subset of H.(2)show H⋂N is a closure;∀x,y∈H⋂N, x∈H and y∈H, x*y∈H; x∈N and y∈N, x*y∈N; so x*y∈H⋂N.(3)Since H and N are subgroup of G, e∈N and e∈H,so e∈H⋂N.(4)∀x∈H⋂N, x ∈H , x-1∈H, and x ∈N, x-1∈N;so x-1∈H⋂N.30: Prove that if N is a normal subgroup of G, thenH ⋂N is a normal subgroup of H.(5) ∀x ∈H, x(H ⋂N) = {xn | n ∈H ⋂N}, let any n 1∈H ⋂N,because aN=Na, x*n 1=n 2*x for some n 2∈N,n 2= x*n 1*x -1,since x -1 ∈H, n 1∈H, then n 2∈H.n 2∈H ⋂N,Thus x(H ⋂N) = (H ⋂N)x.Hence, H ⋂N is a normal subgroup of H.群论9.5 Ex1: Let G be a group, and let N and H be subgroups of G such that N is normal in G. Prove that (1)HN is a subgroup of G.(2)N is normal subgroup of HN.(1) first show HN is closure;∀x,y ∈HN, x=an 1,y=bn 2 for some a,b ∈H.xy=an 1*bn 2, because bN=Nb => n 1*b=bn 3,so xy=abn 3n 2∈HN.e ∈H and e ∈N, so e*e=e ∈HN.∀x ∈HN, x=an 1, x -1=(an 1)-1= n1-1a -1 ,because Nb=bN,so n1-1a -1 =a -1n 4 ∈HN.群论9.5 Ex1: Let G be a group, and let N and H be subgroups of G such that N is normal in G. Prove that(1)HN is a subgroup of G.(2)N is normal subgroup of HN.(2) first show N is subset of HN;∀n∈N, because e∈H, so e*n=n∈HN.Since N is subgroup of G, so N is closed and have identity, and inverse of all elements, thus N is subgroup of HN.Second, show aN=Na, for ∀a∈HN;because (1), a∈G,N is mormal in G, so aN=Na,Hence N is normal subgroup of HN.。

北京邮电大学计算机学院离散数学下半学期

北京邮电大学计算机学院离散数学下半学期

Semigroups and Groups
• • • • • Binary Operations Revisited Semigroups Products and Quotients of Semigroups Groups Products and Quotients of Groups
Mathematical Structures
Inverse(逆元)
• If a binary operation has an identity e, we say y is a -inverse of x if x y=y x=e.
Theorem 2
• If is an associative operation and x has a -inverse y, then y is unique. • Proof
Example 2
• The collection of 3 3 matrices with the operations of addition, multiplication, and transpose is a mathematical structure denoted by [ 3 3 matrices, +, , T ].
• The structure [5 5 matrices, +, *, T] is closed with respect to addition because the sum of two 5 5 matrices is another 5 5 matrix. • The structure [odd integers, +, *] is not closed with respect to addition.

北京邮电大学离散数学下群论作业错题讲解

北京邮电大学离散数学下群论作业错题讲解

北京邮电⼤学离散数学下群论作业错题讲解群论9.1-9.2 (1)20,28 @323-324a. not commutative, a*b ≠b*a.b. a*(b*c)=a*c=b, (a*b)*c=c*c=b.c. not associative, (c*c)*c=b*c=c ≠a=c*b=c*(c*c).群论9.1-9.2 (1)20,28 @323-3241. ≤is reflexiveSince a= a*a, a≤a for all a in A.2. ≤is antisymmetricsuppose that a≤b and b≤a.Then, b = a*b= b*a= a, so a= b.3. ≤is transitiveIf a≤b and b≤c, then c= b*c= (a*b)*c= a*(b*c)= a*c, so a≤c.群论9.1-9.2 (1)28 @323-324show a*b= a∨b, for all a and b in A(1)a*b= a*(b*b) =(a*b)*b=b*(a*b), so b ≤a*b.In a similar way, a*b= (a*a)*b=a*(a*b), so a≤a*b.so a*b is a upper bound for a and b.(2)if a≤c and b≤c.then c= a*c and c= b*c by definition. Thus c= a*(b*c)= (a*b)*c. so a*b≤c.This shows that a*b is the leastest upper bound of a and b.群论9.1-9.2 (2)12,16@page 348;f1 f2 f3 f4 f5 f6closure;f1 f1 f2 f3 f4 f5 f6identity=f1;f2 f2 f1 f5 f6 f3 f4reverse:f2-1=f2,f3-1=f3, f3 f3 f4 f1 f2 f6 f5f4-1=f5,f6-1=f6.f4 f4 f3 f6 f5 f1 f2associate: ?is associate.f5 f5 f6 f2 f1 f4 f3f6 f6 f5 f4 f3 f2 f1群论9.1-9.2 (2)12,16@page 348; If a*a=e, ?a∈G. show commutative.(1)group propertise (a*b)-1=b-1*a-1 because a*a=e=a*a-1 so, b-1*a-1=b*a, (2) (a*b)*(a*b)=e => (a*b)-1=a*b, reverse be only one in Group.so b*a=a*b.群论9.1-9.2 Ex1: Let G be a group. Fora,b∈G,we say that b is conjugate to a,written by b~a, if there exist g∈G such that b=gag-1.show that ~ is a equivalence relation on G. The equivalence classes of are called the conjugacy classes of G. proof: (1)reflexive a=eae, so a~a;(2)symmetric If b~a, then b=gag-1,a=eae=g-1gag-1g=g-1bg.so a~b.(3)transitive suppose a~b and b~c,a=gbg-1 ,b=h ch-1; so a=ghch-1g-1=(gh)c(gh)-1群论9.1-9.2 Ex2: Let G be a group, and suppose that a and b are any elements of G.Show that if (ab)2=a2b2,,then ba=ab.proof:(ab)2=ab*ab=a*(b*a)*ba2b2=aa*bb=a*(a*b)*b,based left/right cancellation, b*a=a*b.群论9.1-9.2 Ex3: Let G ={x∈R|x>1} be the set of all real numbers greater than 1. For x,y∈G, definex*y=xy-x-y+2. Show that (G,*) is a group.proof: (1)closure; because x>1, xy>x+y,so x*y=xy-x-y+2>2, x*y∈G.(2)associate; (x*y)*z=(xy-x-y+2)*z=xyz-xz-yz+2z-(xy-x-y+2)-z+2=xyz-xy-xz-yz+x+y+z .x*(y*z)=x*(yz-y-z+2)=xyz-xy-xz+2x-x-yz+y+z-2+2=xyz-xy-xz-yz+x+y+z .(3)identity: e=2, 2*x=2x-x-2+2=x.(4)reverse: x*x-1=xx-1-x-x-1+2=e=2, x-1=x/(x-1).群论9.4 (1)22,28@page 331;22-Show S 1?S 2is a subsemigroup of S.proof: closure; ?a,b ∈S 1?S 2, then a*b ∈S 1, a*b ∈S 2, so a*b ∈S 1?S 2.if S 1?S 2 =?, still be subsemigroup.28-If f:S 1→S 2and g :S 2→S 3 be isomorphisms, show g ?f :S 1→S 3is an isomorphism.proof: (g ?f)(x *1y)= g(f(x *1y))=g(f(x) *2f(y))=g(f(x))*3g(f(y))=(g ?f)(x) *3(g ?f)(y).one-to-one: suppose (g ?f)(a)=(g ?f)(b), g(f(a))=g(f(b)), f(a)=f(b), a=b.onto: ?z ∈S 3 有g(y)=z; ?y ∈S 2 有f(x)=y;群论9.4 (2)28,32@page 349; 24@page338. 28-Show f:G→G defined by f(a)=a n is a homo morphism.proof: (数学归纳法证)1.because ab=ba, ab*ab=aabb=a2b2,2.(ab)n=a n b n,ab*(ab)n=ab*a n b n=aa n bb n=a n+1b n+1 Hence, f(ab)=(ab)n=a n b n=f(a)f(b).群论9.4 (2)28,32@page 349; 24@page338.32-Show f:G→G by f(a)=a-1 is a isomorphism iff Abelian.Proof: (1)Suppose f is isomorphism,f(xy)=(xy)-1=f(x)f(y)=x-1y-1so xy=((xy)-1)-1=(x-1y-1)-1=yx. G is Abelian. (2)Suppose G is Abelian.f(xy)=(xy)-1=x-1y-1=f(x)f(y), homomorphism. onto: ?x∈G, f(x-1)=(x-1)-1=x.one-to-one: suppose f(x)=f(y), x-1=y-1,xx-1=e=yy-1, right cancelation x=y.群论9.4 (2)28,32@page 349; 24@page338.24-Let A={0,1} and consider (A*, ?), (N,+).(1)f(α)=length(α), show f:A*->N is homomorphism.f(α?β)=f(αβ)=length(α)+length(β)=f(α)+f(β)(2)R: f(α)=f(β),show R is congruence relation.First show R is equivalence relation.2. if f(α)=f(β) and f(χ)=f(δ),then f(α?χ) = f(β?δ)(3)A*/R and N is isomorphic.(A*/R,*) : [α]*[β]=[α?β].Let g([α])=length(α), (1)show g: A*/R->N is homomorphism. g([α] *[β])=g([α?β])=length(α)+length(β)=g([α])+g([β])(2) onto : ?x∈N, let α=00…0(x factors) ,g([α])=x.(3) one-to-one : suppose g([α])=g([β]),length(α)=length(β),so f(α)=f(β), then αRβ, [α]=[β].4: Show f:G 1?G 2→G 1by f(a,b)=a is a homomorphism. f((a,b )*‘(c,d))=f(a*c,b*d)=a*c=f(a,b)*f(c,d)18: Prove N is normal subgroup iff a -1Na=N for all a ∈G.If N is normal subgroup, then ?a ∈G, aN=Na, ande ∈N. so a -1Na=Na -1a=Ne=N.If a -1Na=N for all a ∈G, then N=a -1aN=a -1Na,left cancelation, aN=Na.29: Prove H is normal subgroup if H only two left coset.let a?H, The left cosets of H are H and aH.The right coset are H and Ha.H?aH = ?= H?Ha. and H?aH=G=H?Ha.Thus aH=Ha.let b∈H, bH=H=Hb. so ?x∈G, xH=Hx.H is a normal subgroup.30: Prove that if N is a normal subgroup of G, then H?N is a normal subgroup of H.(1)show H?N is a subset of H;x∈H?N, x ∈H , H?N is a subset of H.(2)show H?N is a closure;x,y∈H?N, x∈H and y∈H, x*y∈H; x∈N and y∈N, x*y∈N; so x*y∈H?N.(3)Since H and N are subgroup of G, e∈N and e∈H,so e∈H?N.(4)?x∈H?N, x ∈H , x-1∈H, and x ∈N, x-1∈N;so x-1∈H?N.30: Prove that if N is a normal subgroup of G, thenH ?N is a normal subgroup of H.(5) ?x ∈H, x(H ?N) = {xn | n ∈H ?N}, let any n 1∈H ?N,because aN=Na, x*n 1=n 2*x for some n 2∈N,n 2= x*n 1*x -1,since x -1 ∈H, n 1∈H, then n 2∈H.n 2∈H ?N,Thus x(H ?N) = (H ?N)x.Hence, H ?N is a normal subgroup of H.群论9.5 Ex1: Let G be a group, and let N and H be subgroups of G such that N is normal in G. Prove that (1)HN is a subgroup of G.(2)N is normal subgroup of HN.(1) first show HN is closure;x,y ∈HN, x=an 1,y=bn 2 for some a,b ∈H.xy=an 1*bn 2, because bN=Nb => n 1*b=bn 3,so xy=abn 3n 2∈HN.e ∈H and e ∈N, so e*e=e ∈HN.x ∈HN, x=an 1, x -1=(an 1)-1= n1-1a -1 ,because Nb=bN,so n1-1a -1 =a -1n 4 ∈HN.群论9.5 Ex1: Let G be a group, and let N and H be subgroups of G such that N is normal in G. Prove that(1)HN is a subgroup of G.(2)N is normal subgroup of HN.(2) first show N is subset of HN;n∈N, because e∈H, so e*n=n∈HN.Since N is subgroup of G, so N is closed and have identity, and inverse of all elements, thus N is subgroup of HN.Second, show aN=Na, for ?a∈HN;because (1), a∈G,N is mormal in G, so aN=Na,Hence N is normal subgroup of HN.。

离散数学(第二版)第9章树

离散数学(第二版)第9章树

第九章 树
删去v0及其关联的边, 得到图T′, 由假设知T′无回路, 现将v0及其关联的边再加到T′, 则还原成T, 所以T没有回 路。
如果在连通图T中增加一条新边(vi, vj), 则(vi, vj)与 T中从vi 到vj 的一条初级路径构成一个初级回路, 且该回路 必定是唯一的, 否则当删去新边(vi, vj)时, T中必有回路, 产生矛盾。
e4}; e5}; e4, e5};
对应树枝e6, 对应树枝e7,
有 有基基本本割割集集{{ee• •67,,
e9}; e10};
对应树枝e8, 有基本割集{e• •8, e9, e10}。
第九章 树
同样对于图9.1.3中的G和T2, 对应树枝e1, e2, e4, e6, e7, e9, 分别有基本割集: {e1, e3, e5}, {e2, e5}, {e4, e3, e5}, {e6, e8, e10}, {e7,e10}, {e9, e8, e10}。
第九章 树
图9.1.2 不同构的七阶无向树
第九章 树
生成树 有一些图, 本身不是树, 但它的某些子图却是树, 其 中很重要的一类是生成树。
定义9.1.2 若无向图G的一个生成子图T是树, 则称T 是G的一棵生成树。
如果T是G的一棵生成树, 则称G在T中的边为T的树枝, G不在T中的边为T的弦, T的所有弦的集合的导出子图称 为T的余树。 易知, 余树不一定是树, 更不一定是生成树。
e10, 则分别产生初级回路e1e3e4, e1e4e5e2, e6e8e9,



e7e6e9e10。

第九章 树
这些初级回路有一个共同特点: 它们中均只含一条弦,
其余的边均是树枝, 我们称这样的回路为基本回路。 对于

北大离散数学chap9

北大离散数学chap9

例3、(2) 一棵树有2个4度顶点,3个3度顶点, 其余都是树叶,求这棵树共有多少个顶点?
解:设有 x片树叶,则顶点数 2 3 x,
边数 2 3 x 1 , 由握手定理, 2 4 33 x 2(2 3 x 1) , 解得 x 9 , 故顶点总数为 2 3 9 14 个。
二、生成树。
e1 )。
(2) 若 e2 不与 e1 构成回路,取 e2 在 T 中,否则弃 e2 ,再查 e3 ,继续这一过程,直到形成树为止。
例5、求以下连通图的最小生成树
5 b5
36
a
5
1
5e f
42
解:
b
3
T 及 W (T )。
a
1
5
e
f
42
c6 d
(1)
c
d
T1
W (T1) 15
例5、求以下连通图的最小生成树
例5、求以下连通图的最小生成树
T 及 W (T )。
a
6
b
1
9
7 f
3
28
10 c
4
d
5 e7
(2)
解: 1 a
b
f
3
2
7
c
e
5
d
T2 '
W (T2 ') 18 W (T )
注意: G的最小生成树可能不唯一, 但 G 的不同最小生成树权的值一样。
第二节 根树及其应用
内容:有向树,根树,最优二元树。
重点:1、有向树及根树的定义,
2、家族树,有序树, r 元树的概念,
3、最优2元树的概念及哈夫曼
(Huffman)
算法。
一、根树。
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* is a binary operation on Ss * is associative. (Ss, *) is a semigroup. The semigroup Ss is not commutative.
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2015-2-5
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Example 4

Let S be a fixed nonempty set, and let Ss be the set of al1 functions f : SS. If f and g are elements of Ss, we define f *g as fg, the composite function.


The associative property holds in any subset of a semigroup so that a subsemigroup (T, *) of a semigroup (S, *) is itself a semigroup. Similarly, a submonoid of a monoid is itself a monoid.
11
Identity – 单位元、幺元

An element e in a semigroup (S, *) is called an identity element if e*a = a*e = a for all a S.

an identity element must be unique.
13
Example 11,12,13

The semigroup Ss defined in Example 4

It has the identity ls, since ls*f = lsf = fls = f * ls for any element f Ss , we see that Ss is a monoid. It has the identity , the empty sequence, since = = for all A* .
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Examples



If (S, *) is a semigroup, then (S, *) is a subsemigroup of (S, *). Let (S,*) be a monoid, then (S, *) is a submonoid of (S, *). If T= {e}, then (T, *) is a submonoid of (S, *)


Denote the semigroup by (S, *) or S. a*b is referred as the product of a and b. The semigroup (S, *) is said to be commutative if * is a commutative operation.
Example 16Fra bibliotek2015-2-5
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Theorem 1

If a1, a2,..., an, n 3, are arbitrary elements of a semigroup, then all products of the elements al, a2,..., an that can be formed by inserting meaningful parentheses arbitrarily are equal.

if T is closed under the operation *, then

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Submonoid – 子独异点

Let

(S, *) be a monoid with identity e, and T be a nonempty subset of S.
6
2015-2-5
Example 5

Let (L, ) be a lattice. Define a binary operation on L by

a* b = ab.

Then L is a semigroup.
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The catenation is a binary operation on A*.

= a1a2...anb1b2…bk
( ) = ( ) .

if , , and are any elements of A*,

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If al, a2,..., an are elements in a semigroup (S, *), then the product can be written as

al*a2*... *an
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2015-2-5
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Content

Binary operations

Binary operation on a set A Properties of binary operations Semigroup Free semigroup generated by A Monoid(独异点) Subsemigroup(子半群)and submonoid(子独异点) Isomorphism (同构)

The semigroup A* defined in Example 6


The set of all relations on set A with the operation of composition

It is a monoid . The identity element is the equality relation .
7
Example 6

Let

A = {al, a2,..., an} be a nonempty set. A* is the set of all finite sequences of elements of A. and be elements of A*. if = a1a2...an and = b1b2…bk
8
Example 6


is an associative binary operation, and (A*, ) is a semigroup. The semigroup (A*, ) is called the free semigroup generated by A(由A生成的自由 半群).

If T is closed under the operation * and eT,then

(T, *) is called a submonoid of (S, *).
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16
Note
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2015-2-5
Subsemigroup – 子半群

Let

(S, *) be a semigroup and T be a subset of S. (T, *) is called a subsemigroup of (S, *).
Semigroups and Groups (半群与群)
Yang Juan
yangjuan@
College of Computer Science & Technology
Beijing University of Posts & Telecommunications
Content

Example 8,9


(Z, +) (Z+, +)
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Monoid – 独异点、含幺半群


A monoid is a semigroup (S, *) that has an identity. Example 10

Define the powers of an recursively as follows:


if m and n are nonnegative integers, then
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