数据库系统基础教程第五章答案
矿产资源开发利用方案编写内容要求及审查大纲
矿产资源开发利用方案编写内容要求及《矿产资源开发利用方案》审查大纲一、概述
㈠矿区位置、隶属关系和企业性质。
如为改扩建矿山, 应说明矿山现状、
特点及存在的主要问题。
㈡编制依据
(1简述项目前期工作进展情况及与有关方面对项目的意向性协议情况。
(2 列出开发利用方案编制所依据的主要基础性资料的名称。
如经储量管理部门认定的矿区地质勘探报告、选矿试验报告、加工利用试验报告、工程地质初评资料、矿区水文资料和供水资料等。
对改、扩建矿山应有生产实际资料, 如矿山总平面现状图、矿床开拓系统图、采场现状图和主要采选设备清单等。
二、矿产品需求现状和预测
㈠该矿产在国内需求情况和市场供应情况
1、矿产品现状及加工利用趋向。
2、国内近、远期的需求量及主要销向预测。
㈡产品价格分析
1、国内矿产品价格现状。
2、矿产品价格稳定性及变化趋势。
三、矿产资源概况
㈠矿区总体概况
1、矿区总体规划情况。
2、矿区矿产资源概况。
3、该设计与矿区总体开发的关系。
㈡该设计项目的资源概况
1、矿床地质及构造特征。
2、矿床开采技术条件及水文地质条件。
数据库原理与应用教程第四版 第五章答案
免责声明：私人学习之余整理，如有错漏，概不负责1.视图的优点简化数据查询语句、使用户能从多角度看待同一数据、提高了数据的安全性、提供了一定程度的逻辑独立性2.使用视图可以加快数据的查询速度吗？为什么？不对。
其本质上还是执行视图内部的查询语句，通过视图查询数据时，都是转换为对基本表的查询，其简化了数据查询语句但是并不能加快数据查询速度。
3.写出创建满足以下要求的视图的SQL语句。
1)查询学生的学号、姓名、所在系、课程号、课程名、课程学分。
CREATE VIEW v1(Sno,Sname,Sdept,Cno,Cname,Credit)ASSELECT s.Sno,Sname,Sdept,o,Cname,CreditFROM Student s JOIN SC ON s.Sno = SC.Sno JOIN Course c ON o = o2)查询学生的学号、姓名、选修的课程名和考试成绩。
CREATE VIEW v2(Sno,Sname,Cname,Grade)ASSELECT s.Sno,Sname,Cname,GradeFROM Student s JOIN SC ON s.Sno = SC.Sno JOIN Course c ON o = o3)统计每个学生的选课门数，列出学生学号和选课门数。
CREATE VIEW v3(Sno,选课门数)ASSELECT s.Sno,COUNT(*)FROM Student s JOIN SC ON s.Sno = SC.SnoGROUP BY Sno4)统计每个学生的修课总学分，列出学生学号和总学分。
(成绩大于等于60)CREATE VIEW v4(Sno,总学分)ASSELECT s.Sno,SUM(Credit)FROM Student s JOIN SC ON s.Sno = SC.Sno JOIN Course c ON o = oWHERE Grade > 60GROUP BY Sno5)查询计算机系Java考试成绩最高的学生的学号、姓名和Java考试成绩。
数据库系统基础教程课后答案第五章
Exercise 5.1.1 As a set:Average = 2.37 As a bag:Average = 2.48 Exercise 5.1.2 As a set:Average = 218 As a bag:Average = 215 Exercise 5.1.3a As a set:As a bag:Exercise 5.1.3bπbore(Ships Classes)Exercise 5.1.4aFor bags:On the left-hand side:Given bags R and S where a tuple t appears n and m times respectively, the union of bags R and S will have tuple t appear n + m times. The further union of bag T with the tuple t appearing o times will have tuple t appear n + m + o times in the final result.On the right-hand side:Given bags S and T where a tuple t appears m and o times respectively, the union of bags R and S will have tuple t appear m + o times. The further union of bag R with the tuple t appearing n times will have tuple t appear m + o + n times in the final result.For sets:This is a similar case when dealing with bags except the tuple t can only appear at most once in each set. The tuple t only appears in the result if all the sets have the tuple t. Otherwise, the tuple t will not appear in the result. Since we cannot have duplicates, the result only has at most one copy of the tuple t.Exercise 5.1.4bFor bags:On the left-hand side:Given bags R and S where a tuple t appears n and m times respectively, the intersectionof bags R and S will have tuple t appear min( n, m ) times. The further intersection of bag T with the tuple t appearing o times will produce tuple t min( o, min( n, m ) ) times in the final result.On the right-hand side:Given bags S and T where a tuple t appears m and o times respectively, the intersection of bags R and S will have tuple t appear min( m, o ) times. The further intersection of bag R with the tuple t appearing n times will produce tuple t min( n, min( m, o ) ) times in thefinal result.The intersection of bags R,S and T will yield a result where tuple t appears min( n,m,o ) times. For sets:This is a similar case when dealing with bags except the tuple t can only appear at most once in each set. The tuple t only appears in the result if all the sets have the tuple t. Otherwise, the tuple t will not appear in the result.Exercise 5.1.4cFor bags:On the left-hand side:Given that tuple r in R, which appears m times, can successfully join with tuple s in S,which appears n times, we expect the result to contain mn copies. Also given that tuple tin T, which appears o times, can successfully join with the joined tuples of r and s, weexpect the final result to have mno copies.On the right-hand side:Given that tuple s in S, which appears n times, can successfully join with tuple t in T,which appears o times, we expect the result to contain no copies. Also given that tuple rin R, which appears m times, can successfully join with the joined tuples of s and t, weexpect the final result to have nom copies.The order in which we perform the natural join does not matter for bags.For sets:This is a similar case when dealing with bags except the joined tuples can only appear at most once in each result. If there are tuples r,s,t in relations R,S,T that can successfully join, then the result will contain a tuple with the schema of their joined attributes.Exercise 5.1.4dFor bags:Suppose a tuple t occurs n and m times in bags R and S respectively. In the union of these two bags R ⋃ S, tuple t would appear n + m times. Likewise, in the union of these two bags S ⋃ R, tuple t would appear m + n times. Both sides of the relation yield the same result.For sets:A tuple t can only appear at most one time. Tuple t might appear each in sets R and S one or zero times. The combinations of number of occurrences for tuple t in R and S respectively are (0,0), (0,1), (1,0), and (1,1). Only when tuple t appears in both sets R and S will the union R ⋃ S have the tuple t. The same reasoning holds when we take the union S ⋃ R.Therefore the commutative law for union holds.Exercise 5.1.4eFor bags:Suppose a tuple t occurs n and m times in bags R and S respectively. In the intersection of these two bags R ∩ S, tuple t would appear min( n,m ) times. Likewise in the intersection of these two bags S ∩ R, tuple t would appear min( m,n ) times. Both sides of the relation yield the same result.For sets:A tuple t can only appear at most one time. Tuple t might appear each in sets R and S one or zero times. The combinations of number of occurrences for tuple t in R and S respectively are (0,0), (0,1), (1,0), and (1,1). Only when tuple t appears in at least one of the sets R and S will the intersection R ∩ S have the tuple t. The same reasoning holds when we take the intersection S ∩ R.Therefore the commutative law for intersection holds.Exercise 5.1.4fFor bags:Suppose a tuple t occurs n times in bag R and tuple u occurs m times in bag S. Suppose also that the two tuples t,u can successfully join. Then in the natural join of these two bags R S, the joined tuple would appear nm times. Likewise in the natural join of these two bags S R, the joined tuple would appear mn times. Both sides of the relation yield the same result.For sets:An arbitrary tuple t can only appear at most one time in any set. Tuples u,v might appear respectively in sets R and S one or zero times. The combinations of number of occurrences for tuples u,v in R and S respectively are (0,0), (0,1), (1,0), and (1,1). Only when tuple u exists in Rand tuple v exists in S will the natural join R S have the joined tuple. The same reasoning holds when we take the natural join S R.Therefore the commutative law for natural join holds.Exercise 5.1.4gFor bags:Suppose tuple t appears m times in R and n times in S. If we take the union of R and S first, we will get a relation where tuple t appears m + n times. Taking the projection of a list of attributes L will yield a resulting relation where the projected attributes from tuple t appear m + n times. If we take the projection of the attributes in list L first, then the projected attributes from tuple t would appear m times from R and n times from S. The union of these resulting relations would have the projected attributes of tuple t appear m + n times.For sets:An arbitrary tuple t can only appear at most one time in any set. Tuple t might appear in sets R and S one or zero times. The combinations of number of occurrences for tuple t in R and S respectively are (0,0), (0,1), (1,0), and (1,1). Only when tuple t exists in R or S (or both R and S) will the projected attributes of tuple t appear in the result.Therefore the law holds.Exercise 5.1.4hFor bags:Suppose tuple t appears u times in R, v times in S and w times in T. On the left hand side, the intersection of S and T would produce a result where tuple t would appear min(v , w) times. With the addition of the union of R, the overall result would have u + min(v , w) copies of tuple t. On the right hand side, we would get a result of min(u + v, u + w) copies of tuple t. The expressions on both the left and right sides are equivalent.For sets:An arbitrary tuple t can only appear at most one time in any set. Tuple t might appear in sets R,S and T one or zero times. The combinations of number of occurrences for tuple t in R, S and T respectively are (0,0,0), (0,0,1), (0,1,0), (0,1,1), (1,0,0), (1,0,1), (1,1,0) and (1,1,1). Only when tuple t appears in R or in both S and T will the result have tuple t.Therefore the distributive law of union over intersection holds.Exercise 5.1.4iSuppose that in relation R, u tuples satisfy condition C and v tuples satisfy condition D. Suppose also that w tuples satisfy both conditions C and D where w≤ min(v , w). Then the left hand side will return those w tuples. On the right hand side, σC(R) produces u tuples and σD(R) produces v tuples. However, we know the intersection will produce the same w tuples in the result.When considering bags and sets, the only difference is bags allow duplicate tuples while sets only allow one copy of the tuple. The example above applies to both cases.Therefore the law holds.Exercise 5.1.5aFor sets, an arbitrary tuple t appears on the left hand side if it appears in both R,S and not in T. The same is true for the right hand side.As an example for bags, suppose that tuple t appears one time each in both R,T and two times in S. The result of the left hand side would have zero copies of tuple t while the right hand side would have one copy of tuple t.Therefore the law holds for sets but not for bags.Exercise 5.1.5bFor sets, an arbitrary tuple t appears on the left hand side if it appears in R and either S or T. This is equivalent to saying tuple t only appears when it is in at least R and S or in R and T. The equivalence is exactly the right side’s expression.As an example for bags, suppose that tuple t appears one time in R and two times each in S and T. Then the left hand side would have one copy of tuple t in the result while the right hand side would have two copies of tuple t.Therefore the law holds for sets but not for bags.Exercise 5.1.5cFor sets, an arbitrary tuple t appears on the left hand side if it satisfies condition C, condition Dor both condition C and D. On the right hand side, σC(R) selects those tuples that satisfy condition C while σD(R) selects those tuples that satisfy condition D. However, the union operator will eliminate duplicate tuples, namely those tuples that satisfy both condition C and D. Thus we are ensured that both sides are equivalent.As an example for bags, we only need to look at the union operator. If there are indeed tuples that satisfy both conditions C and D, then the right hand side will contain duplicate copies of those tuples. The left hand side, however, will only have one copy for each tuple of the original set of tuples.Exercise 5.2.1bExercise 5.2.1cExercise 5.2.1dExercise 5.2.1fExercise 5.2.1gExercise 5.2.1hExercise 5.2.1iExercise 5.2.1jExercise 5.2.1kExercise 5.2.1lExercise 5.2.1mExercise 5.2.1nExercise 5.2.2aApplying the δ operator on a relation with no duplicates will yield the same relation. Thus δ is idempotent.Exercise 5.2.2bThe result of πL is a relation over the list of attributes L. Performing the projection again will return the same relation because the relation only contains the list of attributes L. Thus πL is idempotent.Exercise 5.2.2cThe result of σC is a relation where condition C is satisfied by every tuple. Performing the selection again will return the same relation because the relation only contains tuples that satisfy the condition C. Thus σC is idempotent.Exercise 5.2.2dThe result of γL is a relation whose schema consists of the grouping attributes and the aggregated attributes. If we perform the same grouping operation, there is no guarantee that the expression would make sense. The grouping attributes will still appear in the new result. However, the aggregated attributes may or may not appear correctly. If the aggregated attribute is given a different name than the original attribute, then performing γL would not make sense because it contains an aggregation for an attribute name that does not exist. In this case, the resultingrelation would, according to the definition, only contain the grouping attributes. Thus, γL is not idempotent.Exercise 5.2.2eThe result of τ is a sorted list of tuples based on some attributes L. If L is not the entire schema of relation R, then there are attributes that are not sorted on. If in relation R there are two tuples that agree in all attributes L and disagree in some of the remaining attributes not in L, then it is arbitrary as to which order these two tuples appear in the result. Thus, performing the operation τ multiple times can yield a different relation where these two tuples are swapped. Thus, τ is not idempotent.Exercise 5.2.3If we only consider sets, then it is possible. We can take πA(R) and do a product with itself. From this product, we take the tuples where the two columns are equal to each other.If we consider bags as well, then it is not possible. Take the case where we have the two tuples (1,0) and (1,0). We wish to produce a relation that contains tuples (1,1) and (1,1). If we use the classical operations of relational algebra, we can either get a result where there are no tuples or four copies of the tuple (1,1). It is not possible to get the desired relation because no operation can distinguish between the original tuples and the duplicated tuples. Thus it is not possible to get the relation with the two tuples (1,1) and (1,1).Exercise 5.3.1a)Answer(model) ← PC(model,speed,_,_,_) AND speed ≥ 3.00b)Answer(maker) ← Laptop(model,_,_,hd,_,_) AND Product(maker,model,_) AND hd ≥100c)Answer(model,price) ← PC(model,_,_,_,price) AND Product(maker,model,_) ANDmaker=’B’Answer(model,price) ← Laptop(mode l,_,_,_,_,price) AND Product(maker,model,_)AND maker=’B’Answer(model,price) ← Printer(model,_,_,price) AND Product(maker,model,_) ANDmaker=’B’d)Answer(model) ← Printer(model,color,type,_) AND color=’true’ AND type=’laser’e)PCMaker(maker) ← Product(maker,_,type) AND type=’pc’LaptopMaker(maker) ← Product(maker,_,type) AND type=’laptop’Answer(maker) ← LaptopMaker(maker) AND NOT PCMaker(maker)f)Answer(hd) ← PC(model1,_,_,hd,_) AND PC(model2,_,_,hd,_) AND model1 <>model2g)Answer(model1,model2) ← PC(model1,speed, ram,_,_) ANDPC(model2,_speed,ram,_,_) AND model1 < model2h)FastComputer(model) ← PC(model,speed,_,_,_) AND speed ≥ 2.80FastComputer(model) ← Laptop(model,speed,_,_,_,_) AND speed ≥ 2.80Answer(maker) ← Product(maker,model1,_) AND Product(maker,mod el2,_) ANDFastComputer(model1) AND FastComputer(model2) AND model1 <> model2i)Computers(model,speed) ← PC(model,speed,_,_,_)Computers(model,speed) ← Laptop(model,speed,_,_,_,_)SlowComputers(model) ← Computers(model,speed) AND Computers(model1,speed1)AND speed < speed1FastestComputers(model) ← Computers(model,_) AND NOT SlowComputers(model)Answer(maker) ← Fast estComputers(model) AND Product(maker,model,_)j)PCs(maker,speed) ← PC(model,speed,_,_,_) AND Product(maker,model,_) Answer(maker) ← PCs(maker,spe ed) AND PCs(maker,speed1) AND PCs(maker,speed2) AND speed <> speed1 AND speed <> speed2 AND speed1 <> speed2k)PCs(maker,model) ← Product(maker,model,type) AND type=’pc’Answer(maker) ← PCs(maker,model) AND PCs(maker,model1) ANDPCs(maker,model2) AND PCs(maker,model3) AND model <> model1 AND model <>model2 AND model1 <> model2 AND (model3 = model OR model3 = model1 ORmodel3 = model2)Exercise 5.3.2a)Answer(class,country) ← Classes(class,_,country,_,bore,_) AND bore ≥ 16b)Answer(name) ← Ships(name,_,launch ed) AND launched < 1921c)Answer(ship) ← Outcomes(ship,battle,result) AND battle=’Denmark Strait’ AND result= ‘sunk’d)Answer(name) ← Classes(class,_,_,_,_,displacement) AND Ships(name,class,launched)AND displacement > 35000 AND launched > 1921e)Answer(nam e,displacement,numGuns) ← Classes(class,_,_,numGuns,_,displacement)AND Ships(name,class,_) AND Outcomes (ship,battle,_) AND battle=’Guadalcanal’AND ship=namef)Answer(name) ← Ships(name,_,_)Answer(name) ← Outcomes(name,_,_) AND NOT Answer(name)g)MoreThan One(class) ← Ships(name,class,_) AND Ships(name1,class,_) AND name <>name1Answer(class) ← Classes(class,_,_,_,_,_) AND NOT MoreThanOne(class)h)Battleship(country) ← Classes(_,type,country,_,_,_) AND type=’bb’Battlecruiser(country) ← Classes(_,type,country,_,_,_) AND type=’bc’Answer(country) ← Battleship(country) AND Battlecruiser(country)i)Results(ship,result,date) ← Battles(name,date) AND Outcomes(ship,battle,result) ANDbattle=nameAnswer(ship) ← Results(ship,result,date) AND Results(ship,_,date1) ANDresult=’damaged’ AND date < date1Exercise 5.3.3A nswer(x,y) ← R(x,y) AND z = zExercise 5.4.1aAnswer(a,b,c) ← R(a,b,c)Answer(a,b,c) ← S(a,b,c)Exercise 5.4.1bAnswer(a,b,c) ← R(a,b,c) AND S(a,b,c)Exercise 5.4.1cAnswer(a,b,c) ← R(a,b,c) AND NOT S(a,b,c)Exercise 5.4.1dUnion(a,b,c) ← R(a,b,c)Union(a,b,c) ← S(a,b,c)Answer(a,b,c) ← Union(a,b,c) AND NOT T(a,b,c)Exercise 5.4.1eJ(a,b,c) ← R(a,b,c) AND NOT S(a,b,c)K(a,b,c) ← R(,a,b,c) AND NOT T(a,b,c)Answer(a,b,c) ← J(a,b,c) AND K(a,b,c)Exercise 5.4.1fAnswer(a,b) ← R(a,b,_)Exercise 5.4.1gJ(a,b) ← R(a,b,_)K(a,b) ← S(_,a,b)Answer(a,b) ← J(a,b) AND K(a,b)Exercise 5.4.2aAnswer(x,y,z) ← R(x,y,z) AND x = yExercise 5.4.2bAnswer(x,y,z) ← R(x,y,z) AND x < y AND y < z Exercise 5.4.2cAnswer(x,y,z) ← R(x,y,z) AND x < yAnswer(x,y,z) ← R(x,y,z) AND y < zExercise 5.4.2dChange: NOT(x < y OR x > y)To: x ≥ y AND x ≤ yThe above simplifies to x = yAnswer(x,y,z) ← R(x,y,z) AND x = yExercise 5.4.2eChange: NOT((x < y OR x > y) AND y < z)NOT(x < y OR x > y) OR y ≥ z(x ≥ y AND x ≤ y) OR y ≥ zTo: x = y OR y ≥ zAnswer(x,y,z) ← R(x,y,z) AND x = yAnswer(x,y,z) ← R(x,y,z) AND y ≥ zExercise 5.4.2fChange: NOT((x < y OR x < z) AND y < z)NOT(x < y OR x < z) OR y ≥ z To: (x ≥ y AND x ≥ z) OR y ≥ zAnswer(x,y,z) ← R(x,y,z) AND x ≥ y AND x ≥ zAnswer(x,y,z) ← R(x,y,z) AND y ≥zExercise 5.4.3aAnswer(a,b,c,d) ← R(a,b,c) AND S(b,c,d)Exercise 5.4.3bAnswer(b,c,d,e) ← S(b,c,d) AND T(d,e)Exercise 5.4.3cAnswer(a,b,c,d,e) ← R(a,b,c) AND S(b,c,d) AND T(d,e)Exercise 5.4.4a)Answer(rx,ry,rz,sx,sy,sz) ← R(rx,ry,rz) AND S(sx,sy,sz) AND rx = syb)Answer(rx,ry,rz,sx,sy,sz) ← R(rx,ry,rz) AND S(sx,sy,sz) AND rx < sy AND ry < szc)Answer(rx,ry,rz,sx,sy,s z) ← R(rx,ry,rz) AND S(sx,sy,sz) AND rx < syAnswer(rx,ry,rz,sx,sy,sz) ← R(rx,ry,rz) AND S(sx,sy,sz) AND ry < szd)Answer(rx,ry,rz,sx,sy,sz) ← R(rx,ry,rz) AND S(sx,sy,sz) AND rx = sye)Answer(rx,ry,rz,sx,sy,sz) ← R(rx,ry,rz) AND S(sx,sy,sz) AND rx = syAnswe r(rx,ry,rz,sx,sy,sz) ← R(rx,ry,rz) AND S(sx,sy,sz) AND ry ≥ szf)Answer(rx,ry,rz,sx,sy,sz) ← R(rx,ry,rz) AND S(sx,sy,sz) AND rx ≥ sy AND rx ≥ szAnswer(rx,ry,rz,sx,sy,sz) ← R(rx,ry,rz) AND S(sx,sy,sz) AND ry ≥ szExercise 5.4.5aR1 := πx,y(Q R)Exercise 5.4.5bR1 := ρR1(x,z)(Q)R2 := ρR2(z,y)(Q)R3 := πx,y(R1 (R1.z = R2.z) R2)Exercise 5.4.5cR1 := πx,y(Q R)R2 := σx < y(R1)。
数据库系统原理课后答案 第五章
5.1 名词解释(1)SQL模式：SQL模式是表和授权的静态定义。
一个SQL模式定义为基本表的集合。
一个由模式名和模式拥有者的用户名或账号来确定,并包含模式中每一个元素(基本表、视图、索引等)的定义。
(2)SQL数据库：SQL(Structured Query Language)，即‘结构式查询语言’，采用英语单词表示和结构式的语法规则。
一个SQL数据库是表的汇集，它用一个或多个SQL模式定义。
(3)基本表：在SQL中，把传统的关系模型中的关系模式称为基本表(Base Table)。
基本表是实际存储在数据库中的表，对应一个关系。
(4)存储文件：在SQL中，把传统的关系模型中的存储模式称为存储文件(Stored File)。
每个存储文件与外部存储器上一个物理文件对应。
(5)视图：在SQL中，把传统的关系模型中的子模式称为视图(View)，视图是从若干基本表和（或）其他视图构造出来的表。
(6)行：在SQL中，把传统的关系模型中的元组称为行(row)。
(7)列：在SQL中，把传统的关系模型中的属性称为列(coloumn)。
(8)实表：基本表被称为“实表”，它是实际存放在数据库中的表。
(9)虚表：视图被称为“虚表”，创建一个视图时，只把视图的定义存储在数据词典中，而不存储视图所对应的数据。
(10)相关子查询：在嵌套查询中出现的符合以下特征的子查询：子查询中查询条件依赖于外层查询中的某个值，所以子查询的处理不只一次，要反复求值，以供外层查询使用。
(11)联接查询：查询时先对表进行笛卡尔积操作，然后再做等值联接、选择、投影等操作。
联接查询的效率比嵌套查询低。
(12)交互式SQL：在终端交互方式下使用的SQL语言称为交互式SQL。
(13)嵌入式SQL：嵌入在高级语言的程序中使用的SQL语言称为嵌入式SQL。
(14)共享变量：SQL和宿主语言的接口。
共享变量有宿主语言程序定义，再用SQL 的DECLARE语句说明， SQL语句就可引用这些变量传递数据库信息。
南京理工大学《数据库系统基础教程》试题和标准答案
一、选择题60(选择一个最合适的答案，在答题纸上涂黑)１.一个事务中的一组更新操作是一个整体,要么全部执行,要么全部不执行。
这是事务的：A．原子性B.一致性 C.隔离性 D.持久性２．在数据库的三级模式结构中,描述一个数据库中全体数据的全局逻辑结构和特性的是：Ａ．外模式 B．内模式 C.存储模式D．模式3．关于联系的多重性,下面哪种说法不正确？A.一个多对多的联系中允许多对一的情形。
B.一个多对多的联系中允许一对一的情形。
C．一个多对一的联系中允许一对一的情形。
Ｄ．一个多对一的联系中允许多对多的情形。
4.考虑学校里的"学生"和"课程＂之间的联系，该联系的多重性应该是:A. 一对一 B．多对一 C．一对多 D. 多对多5．下面哪种约束要求一组属性在同一实体集任意两个不同实体上的取值不同。
A. 键(key）约束。
Ｂ．单值约束。
C．参照完整性。
D．域(domａin)约束6．关系模型要求各元组的每个分量的值必须是原子性的。
对原子性,下面哪种解释不正确: A.每个属性都没有内部结构。
ﻩＢ.每个属性都不可再分解。
Ｃ.各属性值应属于某种基本数据类型。
ﻩD.属性值不允许为ＮUＬL。
７．对于一个关系的属性(列)集合和元组（行）集合，下面哪种说法不正确:A.改变属性的排列次序不影响该关系。
B．改变元组的排列次序不影响该关系。
C.改变元组的排列次序会改变该关系。
D.关系的模式包括其名称及其属性集合。
８．若Ｒ是实体集R1与R２间的一个多对多联系,将其转换为关系R'，哪种说法不正确：Ａ.R'属性应包括R１与R2的所有属性。
B.R'属性应包括R1与R2的键属性。
C.R1与R２的键属性共同构成R'的键。
D.R'的属性应包括R自身定义的属性。
9.关于函数依赖的判断,下面哪种说法不正确？A.若任意两元组在属性A上一致，在Ｂ上也一致，则有A →B成立。
B.若任意两元组在属性A上一致,在B上不一致,则A → B不成立。
数据库系统教程课后答案(施伯乐)(第二版)
目录第1部分课程的教与学第2部分各章习题解答及自测题第1章数据库概论1.1 基本内容分析1.2 教材中习题1的解答1.3 自测题1.4 自测题答案第2章关系模型和关系运算理论2.1基本内容分析2.2 教材中习题2的解答2.3 自测题2.4 自测题答案第3章关系数据库语言SQL3.1基本内容分析3.2 教材中习题3的解答3.3 自测题3.4 自测题答案第4章关系数据库的规范化设计4.1基本内容分析4.2 教材中习题4的解答4.3 自测题4.4 自测题答案第5章数据库设计与ER模型5.1基本内容分析5.2 教材中习题5的解答5.3 自测题5.4 自测题答案第6章数据库的存储结构6.1基本内容分析6.2 教材中习题6的解答第7章系统实现技术7.1基本内容分析7.2 教材中习题7的解答7.3 自测题7.4 自测题答案第8章对象数据库系统8.1基本内容分析8.2 教材中习题8的解答8.3 自测题8.4 自测题答案第9章分布式数据库系统9.1基本内容分析9.2 教材中习题9的解答9.3 自测题9.4 自测题答案第10章中间件技术10.1基本内容分析10.2 教材中习题10的解答10.3 自测题及答案第11章数据库与WWW11.1基本内容分析11.2 教材中习题11的解答第12章 XML技术12.1基本内容分析12.2 教材中习题12的解答学习推荐书目1．国内出版的数据库教材（1）施伯乐,丁宝康,汪卫. 数据库系统教程(第2版). 北京:高等教育出版社,2003（2）丁宝康,董健全. 数据库实用教程(第2版). 北京:清华大学出版社,2003（3）施伯乐,丁宝康. 数据库技术. 北京:科学出版社,2002（4）王能斌. 数据库系统教程(上、下册). 北京:电子工业出版社,2002（5）闪四清. 数据库系统原理与应用教程. 北京:清华大学出版社,2001（6）萨师煊,王珊. 数据库系统概论(第3版). 北京:高等教育出版社,2000（7）庄成三,洪玫,杨秋辉. 数据库系统原理及其应用. 北京:电子工业出版社,20002．出版的国外数据库教材（中文版或影印版）（1）Silberschatz A，Korth H F，Sudarshan S. 数据库系统概念(第4版). 杨冬青,唐世渭等译. 北京:机械工业出版社,2003（2）Elmasri R A，Navathe S B. 数据库系统基础(第3版). 邵佩英,张坤龙等译. 北京:人民邮电出版社,2002（3）Lewis P M,Bernstein A,Kifer M. Databases and Transaction Processing:An Application-Oriented Approach, Addison-Wesley, 2002（影印版, 北京:高等教育出版社；中文版，施伯乐等译，即将由电子工业出版社出版）（4）Hoffer J A,Prescott M B,McFadden F R. Modern Database Management. 6th ed. Prentice Hall, 2002（中文版，施伯乐等译，即将由电子工业出版社出版）3．上机实习教材（1）廖疆星,张艳钗,肖金星. PowerBuilder 8.0 & SQL Server 2000数据库管理系统管理与实现. 北京:冶金工业出版社,2002（2）伍俊良. PowerBuilder课程设计与系统开发案例. 北京:清华大学出版社,20034．学习指导书（1）丁宝康,董健全,汪卫,曾宇昆. 数据库系统教程习题解答及上机指导. 北京:高等教育出版社,2003（2）丁宝康,张守志,严勇. 数据库技术学习指导书. 北京:科学出版社,2003（3）丁宝康,董健全,曾宇昆. 数据库实用教程习题解答. 北京:清华大学出版社,2003 （4）丁宝康. 数据库原理题典. 长春:吉林大学出版社,2002（5）丁宝康,陈坚,许建军,楼晓鸿. 数据库原理辅导与练习. 北京:经济科学出版社,2001第1部分课程的教与学1．课程性质与设置目的现在，数据库已是信息化社会中信息资源与开发利用的基础，因而数据库是计算机教育的一门重要课程，是高等院校计算机和信息类专业的一门专业基础课。
数据库系统教程课后答案
数据库系统教程课后答案数据库系统是一种用于管理和组织大量数据的软件系统，它通过数据模型、数据结构和数据操作等方式来实现数据的存储、访问、更新和管理等功能。
数据库系统广泛应用于各个领域，如企业管理、电子商务、科学研究等。
在数据库系统的学习过程中，会出现一些练习题目和问题，需要我们进行思考和解答。
下面我将结合一些常见的课后答案，对数据库系统进行详细的解析。
首先，我们需要了解数据库的基本概念和组成部分。
数据库是按照一定的数据模型组织、存储和管理数据的集合，它由数据库管理系统(DBMS)、数据库、数据库模式(或称为数据库结构)、数据库实例(或称为数据系统)等组成。
其中，数据库管理系统是进行数据库管理的软件系统，它负责数据库的创建、维护和管理等操作。
接着，我们需要了解数据库的设计和规范化。
数据库设计是指根据用户的需求和系统的要求，将现实世界的数据和关系转化为数据库模式的过程。
在设计数据库时，要符合一定的规范化原则，如第一范式、第二范式和第三范式等。
规范化可以提高数据库的性能、减少数据冗余，同时也方便数据库的操作和维护。
然后，我们需要了解数据库的查询和操作。
数据库查询是指根据用户的需求，从数据库中检索出符合条件的数据的过程。
查询语言是进行数据库查询的工具，如结构化查询语言(SQL)。
SQL语言包括数据定义语言(DDL)、数据操纵语言(DML)和数据控制语言(DCL)等。
DDL用于定义和管理数据库的结构，如创建表、定义键等；DML用于对数据库中的数据进行操作，如插入、更新和删除等；DCL用于控制数据库的安全性和权限，如授权和回收权限。
此外，我们还需要了解数据库的事务和并发控制。
事务是指一组对数据库的操作，要么全部执行成功，要么全部放弃。
事务具备ACID特性，即原子性、一致性、隔离性和持久性。
并发控制是指多个事务同时对数据库进行操作时，保证数据库的一致性和完整性的措施。
并发控制可以通过加锁、多版本并发控制(MVCC)、时间戳等方式来实现。
数据库原理第五章课后答案
5.4 对于教学数据库的三个基本表学生S(S#,SNAME,AGE,SEX)学习SC(S#,,GRADE)课程C(C#,CNAME,TEACHER)试用SQL的查询语句表达下列查询：（1）检索LIU老师所授课程的课程号和课程名。
SELECT C#，CNAMEFROM CWHERE TEACHER=‘LIU’（2）检索年龄大于23岁的男学生的学号和姓名。
SELECT S#，SNAMEFROM SWHERE (AGE>23) AND (SEX=‘M’)（3）检索至少选修LIU老师所授课程中一门课程的女学生姓名。
SELECT SNAMEFROM SWHERE SEX=‘F’ AND S# IN(SELECT S#FROM SCWHERE C# IN(SELECT C#FROM CWHERE TEACHER=‘LIU’)NOTICE：有多种写法，比如联接查询写法：SELECT SNAMEFROM S,SC,CWHERE SEX=‘F’ AND SC.S#=S.S#AND SC.C#=C.C#AND TEACHER='LIU'但上一种写法更好一些。
（4）检索WANG同学不学的课程的课程号。
SELECT C#FROM CWHERE C# NOT IN(SELECT C#FROM SCWHERE S# INFROM SWHERE SNAME='WANG'))（5）检索至少选修两门课程的学生学号。
SELECT DISTINCT X.SNOFROM SC X,SC YWHERE X.SNO=Y.SNO AND O<>ONotice:对表SC进行自连接，X，Y是SC的两个别名。
（6）检索全部学生都选修的课程的课程号与课程名。
SELECT C#,CNAMEFROM CWHERE NOT EXISTS(SELECT *FROM SWHERE S# NOT IN(SELECT *FROM SCWHERE SC.C#=C.C#))要从语义上分解：（1）选择课程的课程号与课程名，不存在不选这门课的同学。
(完整版)数据库系统基础教程第五章答案
Exercise 5.1.1 As a set:speed2.662.101.422.803.202.202.001.863.06 Average = 2.37 As a bag:speed2.662.101.422.803.203.202.202.202.002.801.862.803.06 Average = 2.48 Exercise 5.1.2 As a set:hd25080320200300160 Average = 218 As a bag:hd2502508025025032020025025030016016080 Average = 215 Exercise 5.1.3a As a set:bore15161418As a bag:bore1516141615151418Exercise 5.1.3bπbore(Ships Classes)Exercise 5.1.4aFor bags:On the left-hand side:Given bags R and S where a tuple t appears n and m times respectively, the union of bags R and S will have tuple t appear n + m times. The further union of bag T with the tuple t appearing o times will have tuple t appear n + m + o times in the final result.On the right-hand side:Given bags S and T where a tuple t appears m and o times respectively, the union of bags R and S will have tuple t appear m + o times. The further union of bag R with the tuple t appearing n times will have tuple t appear m + o + n times in the final result.For sets:This is a similar case when dealing with bags except the tuple t can only appear at most once in each set. The tuple t only appears in the result if all the sets have the tuple t. Otherwise, the tuple t will not appear in the result. Since we cannot have duplicates, the result only has at most one copy of the tuple t.Exercise 5.1.4bFor bags:On the left-hand side:Given bags R and S where a tuple t appears n and m times respectively, the intersectionof bags R and S will have tuple t appear min( n, m ) times. The further intersection of bag T with the tuple t appearing o times will produce tuple t min( o, min( n, m ) ) times in the final result.On the right-hand side:Given bags S and T where a tuple t appears m and o times respectively, the intersection of bags R and S will have tuple t appear min( m, o ) times. The further intersection of bag R with the tuple t appearing n times will produce tuple t min( n, min( m, o ) ) times in thefinal result.The intersection of bags R,S and T will yield a result where tuple t appears min( n,m,o ) times. For sets:This is a similar case when dealing with bags except the tuple t can only appear at most once in each set. The tuple t only appears in the result if all the sets have the tuple t. Otherwise, the tuple t will not appear in the result.Exercise 5.1.4cFor bags:On the left-hand side:Given that tuple r in R, which appears m times, can successfully join with tuple s in S,which appears n times, we expect the result to contain mn copies. Also given that tuple tin T, which appears o times, can successfully join with the joined tuples of r and s, weexpect the final result to have mno copies.On the right-hand side:Given that tuple s in S, which appears n times, can successfully join with tuple t in T,which appears o times, we expect the result to contain no copies. Also given that tuple rin R, which appears m times, can successfully join with the joined tuples of s and t, weexpect the final result to have nom copies.The order in which we perform the natural join does not matter for bags.For sets:This is a similar case when dealing with bags except the joined tuples can only appear at most once in each result. If there are tuples r,s,t in relations R,S,T that can successfully join, then the result will contain a tuple with the schema of their joined attributes.Exercise 5.1.4dFor bags:Suppose a tuple t occurs n and m times in bags R and S respectively. In the union of these two bags R ⋃ S, tuple t would appear n + m times. Likewise, in the union of these two bags S ⋃ R, tuple t would appear m + n times. Both sides of the relation yield the same result.For sets:A tuple t can only appear at most one time. Tuple t might appear each in sets R and S one or zero times. The combinations of number of occurrences for tuple t in R and S respectively are (0,0), (0,1), (1,0), and (1,1). Only when tuple t appears in both sets R and S will the union R ⋃ S have the tuple t. The same reasoning holds when we take the union S ⋃ R.Therefore the commutative law for union holds.Exercise 5.1.4eFor bags:Suppose a tuple t occurs n and m times in bags R and S respectively. In the intersection of these two bags R ∩ S, tuple t would appear min( n,m ) times. Likewise in the intersection of these two bags S ∩ R, tuple t would appear min( m,n ) times. Both sides of the relation yield the same result.For sets:A tuple t can only appear at most one time. Tuple t might appear each in sets R and S one or zero times. The combinations of number of occurrences for tuple t in R and S respectively are (0,0), (0,1), (1,0), and (1,1). Only when tuple t appears in at least one of the sets R and S will the intersection R ∩ S have the tuple t. The same reasoning holds when we take the intersection S ∩R.Therefore the commutative law for intersection holds.Exercise 5.1.4fFor bags:Suppose a tuple t occurs n times in bag R and tuple u occurs m times in bag S. Suppose also that the two tuples t,u can successfully join. Then in the natural join of these two bags R S, the joined tuple would appear nm times. Likewise in the natural join of these two bags S R, the joined tuple would appear mn times. Both sides of the relation yield the same result.For sets:An arbitrary tuple t can only appear at most one time in any set. Tuples u,v might appear respectively in sets R and S one or zero times. The combinations of number of occurrences for tuples u,v in R and S respectively are (0,0), (0,1), (1,0), and (1,1). Only when tuple u exists in Rand tuple v exists in S will the natural join R S have the joined tuple. The same reasoning holds when we take the natural join S R.Therefore the commutative law for natural join holds.Exercise 5.1.4gFor bags:Suppose tuple t appears m times in R and n times in S. If we take the union of R and S first, we will get a relation where tuple t appears m + n times. Taking the projection of a list of attributes L will yield a resulting relation where the projected attributes from tuple t appear m + n times. If we take the projection of the attributes in list L first, then the projected attributes from tuple t would appear m times from R and n times from S. The union of these resulting relations would have the projected attributes of tuple t appear m + n times.For sets:An arbitrary tuple t can only appear at most one time in any set. Tuple t might appear in sets R and S one or zero times. The combinations of number of occurrences for tuple t in R and S respectively are (0,0), (0,1), (1,0), and (1,1). Only when tuple t exists in R or S (or both R and S) will the projected attributes of tuple t appear in the result.Therefore the law holds.Exercise 5.1.4hFor bags:Suppose tuple t appears u times in R, v times in S and w times in T. On the left hand side, the intersection of S and T would produce a result where tuple t would appear min(v , w) times. With the addition of the union of R, the overall result would have u + min(v , w) copies of tuple t. On the right hand side, we would get a result of min(u + v, u + w) copies of tuple t. The expressions on both the left and right sides are equivalent.For sets:An arbitrary tuple t can only appear at most one time in any set. Tuple t might appear in sets R,S and T one or zero times. The combinations of number of occurrences for tuple t in R, S and T respectively are (0,0,0), (0,0,1), (0,1,0), (0,1,1), (1,0,0), (1,0,1), (1,1,0) and (1,1,1). Only when tuple t appears in R or in both S and T will the result have tuple t.Therefore the distributive law of union over intersection holds.Exercise 5.1.4iSuppose that in relation R, u tuples satisfy condition C and v tuples satisfy condition D. Suppose also that w tuples satisfy both conditions C and D where w≤ min(v , w). Then the left hand side will return those w tuples. On the right hand side, σC(R) produces u tuples and σD(R) produces v tuples. However, we know the intersection will produce the same w tuples in the result.When considering bags and sets, the only difference is bags allow duplicate tuples while sets only allow one copy of the tuple. The example above applies to both cases.Therefore the law holds.Exercise 5.1.5aFor sets, an arbitrary tuple t appears on the left hand side if it appears in both R,S and not in T. The same is true for the right hand side.As an example for bags, suppose that tuple t appears one time each in both R,T and two times in S. The result of the left hand side would have zero copies of tuple t while the right hand side would have one copy of tuple t.Therefore the law holds for sets but not for bags.Exercise 5.1.5bFor sets, an arbitrary tuple t appears on the left hand side if it appears in R and either S or T. This is equivalent to saying tuple t only appears when it is in at least R and S or in R and T. The equivalence is exactly the right side’s expression.As an example for bags, suppose that tuple t appears one time in R and two times each in S and T. Then the left hand side would have one copy of tuple t in the result while the right hand side would have two copies of tuple t.Therefore the law holds for sets but not for bags.Exercise 5.1.5cFor sets, an arbitrary tuple t appears on the left hand side if it satisfies condition C, condition D or both condition C and D. On the right hand side, σC(R) selects those tuples that satisfy condition C while σD(R) selects those tuples that satisfy condition D. However, the union operator will eliminate duplicate tuples, namely those tuples that satisfy both condition C and D. Thus we are ensured that both sides are equivalent.As an example for bags, we only need to look at the union operator. If there are indeed tuples that satisfy both conditions C and D, then the right hand side will contain duplicate copies of those tuples. The left hand side, however, will only have one copy for each tuple of the original set of tuples.A+B A2B210154910164167916 Exercise 5.2.1bB+1C-1103334431143 Exercise 5.2.1cA B0101232434 Exercise 5.2.1dB C010224253434A B01232434 Exercise 5.2.1fB C0124253402 Exercise 5.2.1gA SUM(B)022734 Exercise 5.2.1hB AVG(C)0 1.52 4.534 Exercise 5.2.1iA23Exercise 5.2.1jA MAX(C)24 Exercise 5.2.1kA B C23423401┴01┴24┴34┴Exercise 5.2.1lA B C234234┴01┴24┴25┴02 Exercise 5.2.1mA B C23423401┴01┴24┴34┴┴01┴24┴25┴02Exercise 5.2.1nA R.B S.B C0124012501340134012401250134013423┴┴24┴┴34┴┴┴┴01┴┴02Exercise 5.2.2aApplying the δ operator on a relation with no duplicates will yield the same relation. Thus δ is idempotent.Exercise 5.2.2bThe result of πL is a relation over the list of attributes L. Performing the projection again will return the same relation because the relation only contains the list of attributes L. Thus πL is idempotent.Exercise 5.2.2cThe result of σC is a relation where condition C is satisfied by every tuple. Performing the selection again will return the same relation because the relation only contains tuples that satisfy the condition C. Thus σC is idempotent.Exercise 5.2.2dThe result of γL is a relation whose schema consists of the grouping attributes and the aggregated attributes. If we perform the same grouping operation, there is no guarantee that the expression would make sense. The grouping attributes will still appear in the new result. However, the aggregated attributes may or may not appear correctly. If the aggregated attribute is given a different name than the original attribute, then performing γL would not make sense because it contains an aggregation for an attribute name that does not exist. In this case, the resultingrelation would, according to the definition, only contain the grouping attributes. Thus, γL is not idempotent.Exercise 5.2.2eThe result of τ is a sorted list of tuples based on some attributes L. If L is not the entire schema of relation R, then there are attributes that are not sorted on. If in relation R there are two tuples that agree in all attributes L and disagree in some of the remaining attributes not in L, then it is arbitrary as to which order these two tuples appear in the result. Thus, performing the operation τmultiple times can yield a different relation where these two tuples are swapped. Thus, τ is not idempotent.Exercise 5.2.3If we only consider sets, then it is possible. We can take πA(R) and do a product with itself. From this product, we take the tuples where the two columns are equal to each other.If we consider bags as well, then it is not possible. Take the case where we have the two tuples (1,0) and (1,0). We wish to produce a relation that contains tuples (1,1) and (1,1). If we use the classical operations of relational algebra, we can either get a result where there are no tuples or four copies of the tuple (1,1). It is not possible to get the desired relation because no operation can distinguish between the original tuples and the duplicated tuples. Thus it is not possible to get the relation with the two tuples (1,1) and (1,1).Exercise 5.3.1a)Answer(model) ← PC(model,speed,_,_,_) AND speed ≥ 3.00b)Answer(maker) ← Laptop(model,_,_,hd,_,_) AND Product(maker,model,_) AND hd ≥100c)Answer(model,price) ← PC(model,_,_,_,price) AND Product(maker,model,_) ANDmaker=’B’Answer(model,price) ← Laptop(model,_,_,_,_,price) AND Product(maker,model,_)AND maker=’B’Answer(model,price) ← Printer(model,_,_,price) AND Product(maker,model,_) ANDmaker=’B’d)Answer(model) ← Printer(model,color,type,_) AND color=’true’ AND type=’laser’e)PCMaker(maker) ← Product(maker,_,type) AND type=’pc’LaptopMaker(maker) ← Product(maker,_,type) AND type=’laptop’Answer(maker) ← LaptopMaker(maker) AND NOT PCMaker(maker)f)Answer(hd) ← PC(model1,_,_,hd,_) AND PC(model2,_,_,hd,_) AND model1 <>model2g)Answer(model1,model2) ← PC(model1,speed, ram,_,_) ANDPC(model2,_speed,ram,_,_) AND model1 < model2h)FastComputer(model) ← PC(model,speed,_,_,_) AND speed ≥ 2.80FastComputer(model) ← Laptop(model,speed,_,_,_,_) AND speed ≥ 2.80Answer(maker) ← Product(maker,model1,_) AND Product(maker,model2,_) ANDFastComputer(model1) AND FastComputer(model2) AND model1 <> model2i)Computers(model,speed) ← PC(model,speed,_,_,_)Computers(model,speed) ← Laptop(model,speed,_,_,_,_)SlowComputers(model) ← Computers(model,speed) AND Computers(model1,speed1) AND speed < speed1FastestComputers(model) ← Computers(model,_) AND NOT SlowComputers(model)Answer(maker) ← FastestComputers(model) AND Product(maker,model,_) j)PCs(maker,speed) ← PC(model,speed,_,_,_) AND Product(maker,model,_) Answer(maker) ← PCs(maker,speed) AND PCs(maker,speed1) ANDPCs(maker,speed2) AND speed <> speed1 AND speed <> speed2 AND speed1 <>speed2k)PCs(maker,model) ← Product(maker,model,type) AND type=’pc’Answer(maker) ← PCs(maker,model) AND PCs(maker,model1) ANDPCs(maker,model2) AND PCs(maker,model3) AND model <> model1 AND model <> model2 AND model1 <> model2 AND (model3 = model OR model3 = model1 ORmodel3 = model2)Exercise 5.3.2a)Answer(class,country) ← Classes(class,_,country,_,bore,_) AND bore ≥ 16b)Answer(name) ← Ships(name,_,launched) AND launched < 1921c)Answer(ship) ← Outcomes(ship,battle,result) AND battle=’Denmark Strait’ AND result= ‘sunk’d)Answer(name) ← Classes(class,_,_,_,_,displacement) AND Ships(name,class,launched)AND displacement > 35000 AND launched > 1921e)Answer(name,displacement,numGuns) ← Classes(class,_,_,numGuns,_,displacement)AND Ships(name,class,_) AND Outcomes (ship,battle,_) AND battle=’Guadalcanal’AND ship=namef)Answer(name) ← Ships(name,_,_)Answer(name) ← Outcomes(name,_,_) AND NOT Answer(name)g)MoreThanOne(class) ← Ships(name,class,_) AND Ships(name1,class,_) AND name <>name1Answer(class) ← Classes(class,_,_,_,_,_) AND NOT MoreThanOne(class)h)Battleship(country) ← Classes(_,type,country,_,_,_) AND type=’bb’Battlecruiser(country) ← Classes(_,type,country,_,_,_) AND type=’bc’Answer(country) ← Battleship(country) AND Battlecruiser(country)i)Results(ship,result,date) ← Battles(name,date) AND Outcomes(ship,battle,result) ANDbattle=nameAnswer(ship) ← Results(ship,result,date) AND Results(ship,_,date1) ANDresult=’damaged’ AND date < date1Exercise 5.3.3Answer(x,y) ← R(x,y) AND z = zExercise 5.4.1aAnswer(a,b,c) ← R(a,b,c)Answer(a,b,c) ← S(a,b,c)Exercise 5.4.1bAnswer(a,b,c) ← R(a,b,c) AND S(a,b,c)Exercise 5.4.1cAnswer(a,b,c) ← R(a,b,c) AND NOT S(a,b,c)Exercise 5.4.1dUnion(a,b,c) ← R(a,b,c)Union(a,b,c) ← S(a,b,c)Answer(a,b,c) ← Union(a,b,c) AND NOT T(a,b,c)Exercise 5.4.1eJ(a,b,c) ← R(a,b,c) AND NOT S(a,b,c)K(a,b,c) ← R(,a,b,c) AND NOT T(a,b,c)Answer(a,b,c) ← J(a,b,c) AND K(a,b,c)Exercise 5.4.1fAnswer(a,b) ← R(a,b,_)Exercise 5.4.1gJ(a,b) ← R(a,b,_)K(a,b) ← S(_,a,b)Answer(a,b) ← J(a,b) AND K(a,b)Exercise 5.4.2aAnswer(x,y,z) ← R(x,y,z) AND x = yExercise 5.4.2bAnswer(x,y,z) ← R(x,y,z) AND x < y AND y < z Exercise 5.4.2cAnswer(x,y,z) ← R(x,y,z) AND x < yAnswer(x,y,z) ← R(x,y,z) AND y < zExercise 5.4.2dChange:NOT(x < y OR x > y)To:x ≥ y AND x ≤ yThe above simplifies to x = yAnswer(x,y,z) ← R(x,y,z) AND x = yExercise 5.4.2eChange:NOT((x < y OR x > y) AND y < z)NOT(x < y OR x > y) OR y ≥ z(x ≥ y AND x ≤ y) OR y ≥ zTo:x = y OR y ≥ zAnswer(x,y,z) ← R(x,y,z) AND x = yAnswer(x,y,z) ← R(x,y,z) AND y ≥ zExercise 5.4.2fChange:NOT((x < y OR x < z) AND y < z)NOT(x < y OR x < z) OR y ≥ z To:(x ≥ y AND x ≥ z) OR y ≥ zAnswer(x,y,z) ← R(x,y,z) AND x ≥ y AND x ≥ zAnswer(x,y,z) ← R(x,y,z) AND y ≥zExercise 5.4.3aAnswer(a,b,c,d) ← R(a,b,c) AND S(b,c,d)Exercise 5.4.3bAnswer(b,c,d,e) ← S(b,c,d) AND T(d,e)Exercise 5.4.3cAnswer(a,b,c,d,e) ← R(a,b,c) AND S(b,c,d) AND T(d,e)Exercise 5.4.4a)Answer(rx,ry,rz,sx,sy,sz) ← R(rx,ry,rz) AND S(sx,sy,sz) AND rx = syb)Answer(rx,ry,rz,sx,sy,sz) ← R(rx,ry,rz) AND S(sx,sy,sz) AND rx < sy AND ry < szc)Answer(rx,ry,rz,sx,sy,sz) ← R(rx,ry,rz) AND S(sx,sy,sz) AND rx < syAnswer(rx,ry,rz,sx,sy,sz) ← R(rx,ry,rz) AND S(sx,sy,sz) AND ry < szd)Answer(rx,ry,rz,sx,sy,sz) ← R(rx,ry,rz) AND S(sx,sy,sz) AND rx = sye)Answer(rx,ry,rz,sx,sy,sz) ← R(rx,ry,rz) AND S(sx,sy,sz) AND rx = syAnswer(rx,ry,rz,sx,sy,sz) ← R(rx,ry,rz) AND S(sx,sy,sz) AND ry ≥ szf)Answer(rx,ry,rz,sx,sy,sz) ← R(rx,ry,rz) AND S(sx,sy,sz) AND rx ≥ sy AND rx ≥ szAnswer(rx,ry,rz,sx,sy,sz) ← R(rx,ry,rz) AND S(sx,sy,sz) AND ry ≥ szExercise 5.4.5aR1 := πx,y(Q R)Exercise 5.4.5bR1 := ρR1(x,z)(Q)R2 := ρR2(z,y)(Q)R3 := πx,y(R1 (R1.z = R2.z) R2)Exercise 5.4.5cR1 := πx,y(Q R)R2 := σx < y(R1)。
数据库系统教程课后习题答案(部分)--何玉洁 李宝安
第一部分基础理论第1章数据库概述1．试说明数据、数据库、数据库管理系统和数据库系统的概念。
数据：描述事务的符号记录数据库：存储数据的仓库数据库管理系统：用于管理和维护数据的系统软件数据库系统：计算机中引入数据库后的系统，包括数据库，数据库管理系统，应用程序，数据库管理员2．数据管理技术的发展主要经历了哪几个阶段？两个阶段，文件管理和数据库管理9．数据独立性指的是什么？应用程序不因数据的物理表示方式和访问技术改变而改变，分为逻辑独立性和物理独立性。
物理独立性是指当数据的存储结构或存储位置发生变化时，不影响应用程序的特性；逻辑独立性是指当表达现实世界的信息内容发生变化时，不影响应用程序的特性。
10．数据库系统由哪几部分组成？由数据库、数据库管理系统、应用程序、数据库管理员组成。
第2章数据模型与数据库系统的结构4．说明实体一联系模型中的实体、属性和联系的概念。
实体是具有公共性质的并可相互区分的现实世界对象的集合。
属性是实体所具有的特征或性质。
联系是实体之间的关联关系。
6．数据库系统包含哪三级模式？试分别说明每一级模式的作用。
外模式、模式和内模式。
外模式：是对现实系统中用户感兴趣的整体数据结构的局部描述，用于满足不同用户对数据的需求，保证数据安全。
模式：是数据库中全体数据的逻辑结构和特征的描述，它满足所有用户对数据的需求。
内模式：是对整个数据库的底层表示，它描述了数据的存储结构。
7．数据库管理系统提供的两级映像的作用是什么？它带来了哪些功能？两级映像是外模式/模式映像和模式/内模式映像。
外模式/模式映像保证了当模式发生变化时可以保证外模式不变，从而使用户的应用程序不需要修改，保证了程序与数据的逻辑独立性。
模式/内模式映像保证了当内模式发生变化，比如存储位置或存储文件名改变，可以保持模式不变，保证了程序与数据的物理独立性。
两级印象保证了应用程序的稳定性。
第3章关系数据库1．试述关系模型的三个组成部分。
数据结构、关系操作集合、关系完整性约束2．解释下列术语的含义：(3)候选码当一个属性或属性集的值能够唯一标识一个关系的元组，而又不包含多余的元素，则称该属性或属性集为候选码。
数据库系统基础教程答案
a)
b) Transitivity and Augmentation rules of Functional Dependency allow arrow entering Mothers from Births. However, a new relationship in below figure represents more accurate information.
d)
In d we assume that an address can only belong to one customer and a phone can exist at only one address.
If the multiplicity of above relationships were m-to-n, the entity set becomes weak and the key ssNo of customers will be needed as partof the composite key of the entity set.
4.2.7
In below figure there exists a many-to-one relationship between Babies and Births and another many-to-one relationship between Births andMothers. From transitivity of relationships, there is a many-to-one relationship between Babies and Mothers. Hence a baby has a uniquemother while a birth can allow more than one baby.