数据库第五章答案
数据库第五章习题及答案
数据库第五章习题及答案本文档为数据库第五章的习题及答案，帮助读者巩固数据库相关知识。
习题1. 数据库的优点有哪些？数据库具有以下优点： - 数据共享：多个用户可以同时访问和共享数据库中的数据。
- 数据一致性：数据库提供事务管理能力，保证了数据的一致性。
- 数据持久性：数据在数据库中是永久存储的，不会因为系统关机或程序结束而丢失。
- 数据冗余度低：数据库通过规范化设计，减少了数据的冗余性，提高了数据的存储效率。
- 数据独立性：数据库支持数据与应用程序的独立性，提高了系统的灵活性和维护性。
- 数据安全性：数据库提供了用户权限管理和数据备份机制，保证了数据的安全性。
2. 数据库的三级模式结构是什么？数据库的三级模式结构包括： - 外模式（视图层）：外模式是用户所看到的数据库的子集，用于描述用户对数据库的逻辑视图。
每个用户可以有不同的外模式来满足自己的需求。
- 概念模式（逻辑层）：概念模式是全局数据库的逻辑结构和组织方式，描述了数据的总体逻辑视图。
概念模式独立于具体的应用程序，是数据库管理员的角度来看待数据库的。
- 内模式（物理层）：内模式是数据库的存储结构和物理组织方式，描述了数据在存储介质上的实际存储方式。
3. 数据库的完整性约束有哪些？数据库的完整性约束包括： - 实体完整性约束：确保表的主键不为空，每个实体都能够唯一标识。
- 参照完整性约束：确保外键的引用关系是有效的，即外键值必须等于被引用表中的主键值或者为空。
- 用户定义完整性约束：用户可以自定义额外的完整性约束，如检查约束、唯一约束、默认约束等。
4. 数据库的关系模型有哪些特点？数据库的关系模型具有以下特点： - 数据用二维表的形式进行组织，表由行和列组成，每一行表示一个实体，每一列表示一个属性。
- 表与表之间通过主键和外键建立关联关系，形成关系。
- 关系模型提供了一种数据独立性的设计方法，使得应用程序与数据的逻辑结构相分离，提高了系统的灵活性和可维护性。
数据库系统概论第五章完整性
一、选择题1．实体完整性要求主属性不能取空值，这一点可通过（ ）来保证。
A ．定义外部键B ．定义主键C ．用户定义的完整性D ．由关系系统自动答案：B2. （ ）定义了对参照关系的外部属性值域的约束。
A ．实体完整性规则B ．用户定义的完整性规则C ．参照完整性规则D ．以上均不是答案：C3.在如下2个数据库的表中，若雇员信息表EMP 的主键是雇员号，部门信息表DEPT 的主键是部门号。
若执行所列出的操作，哪个操作不能执行（ ）EMP DEPTA ．从雇员信息表EMP 中删除行（’010’,’’王利,’01’,’1200’）B ．在雇员信息表EMP 中插入行（’102’,’赵丽’,’01’,’1500’）C ．将雇员信息表EMP 中雇员号=’010’的工资改为1600元D ．将雇员信息表EMP 中雇员号=’101’的部门号改为’05’答案：D4.关系数据库中，实现主码标识元组的作用是通过（ ）A ．实体完整性规则B ．参照完整性规则C ．用户自定义的完整性D ．属性的值域答案：A5.在关系数据库中，实现“表中任意两行不能相同”的约束是靠（ ）A ．外码B ．主码C ．属性D ．列答案：B6.关系数据库中，实现表与表之间的联系是通过（ ）A ．实体完整性规则B ．参照完整性规则C ．用户自定义的完整性D ．值域答案：B雇员号 雇员名 部门号 工资001 张红 02 2000 010 王利 01 1200056 马明 02 1000101 赵丽 04 1500 部门号 部门名 主任 01 业务部 李林02 销售部 江平 03 服务部 周明 04 财务部 陈胜7.根据关系模式的完整性规则，一个关系中的“主键”（）A．不能有两个B．不能成为另一个关系的外部键C．不允许为空D．可以取空值答案：C8.在关系模型中，实现“关系中不允许发现相同的元组”的约束是通过（）A．候选键B．主键C．外键D．一般键答案：B二、填空题1.为了维护数据库中数据的完整性，在对关系数据库执行插入时首先应检查__________规则。
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。
《MySQL数据库原理、设计与应用》第5章课后习题答案
第五章一、填空题1.逗号或,2. 33.FLOOR(3+RAND()*(11-3+1))或FLOOR(3+RAND()*9)4.NULL5.ON DUPLICATE KEY二、判断题1.错2.对3.错4.对5.对三、选择题1. D2. B3. D4. A5. C四、简答题1.请简述DELETE与TRUNCA TE的区别。
答：①实现方式不同：TRUNCATE本质上先执行删除（DROP）数据表的操作，然后再根据有效的表结构文件（.frm）重新创建数据表的方式来实现数据清空操作。
而DELETE语句则是逐条的删除数据表中保存的记录。
②执行效率不同：在针对大型数据表（如千万级的数据记录）时，TRUNCATE清空数据的实现方式，决定了它比DELETE语句删除数据的方式执行效率更高。
③对AUTO_INCREMENT的字段影响不同，TRUNCATE清空数据后，再次向表中添加数据，自动增长字段会从默认的初始值重新开始，而使用DELETE语句删除表中的记录时，则不影响自动增长值。
④删除数据的范围不同：TRUNCATE语句只能用于清空表中的所有记录，而DELETE语句可通过WHERE指定删除满足条件的部分记录。
⑤返回值含义不同：TRUNCATE操作的返回值一般是无意义的，而DELETE语句则会返回符合条件被删除的记录数。
⑥所属SQL语言的不同组成部分：DELETE语句属于DML数据操作语句，而TRUNCA TE通常被认为是DDL数据定义语句。
2.请简述WHERE与HA VING之间的区别。
1答：①WHERE操作是从数据表中获取数据，用于将数据从磁盘存储到内存中，而HA VING是对已存放到内存中的数据进行操作。
②HA VING位于GROUP BY子句后，而WHERE位于GROUP BY 子句之前。
③HA VING关键字后可以跟聚合函数，而WHERE则不可以。
通常情况下，HA VING关键字与GROUPBY一起使用，对分组后的结果进行过滤。
数据库原理及应用智慧树知到答案章节测试2023年山东建筑大学
第一章测试1数据库(DB)、数据库系统(DBS)、数据库管理系统(DBMS)之间的关系是（）。
A:DB包含DBS和DBMSB:DBMS包含DB和DBSC:DBS包含DB和DBMSD: 没有任何关系答案：C2.数据库系统的核心和基础是()。
A: 逻辑模型B: 数据模型C: 概念模型D: 物理模型答案：B3.单个用户使用的数据视图的描述称为（）A: 外模式B: 存储模式C: 概念模式D: 内模式答案：A4.数据模型的三要素是()。
A: 关系模型、层次模型和网状模型B: 外模式、模式和内模式C: 实体、属性和联系D: 数据结构、数据操作和完整性约束答案：D5. 数据库管理系统能实现对数据库中数据的查询、插入、修改和删除，这类功能称为（）。
A: 数据管理功能B: 数据操纵功能C: 数据定义功能D: 数据控制功能答案：B6在数据库的三级模式结构中，描述数据库中全体数据的逻辑结构和特性的是（）。
A: 模式B: 内模式C: 外模式D: 存储模式答案：A7.在数据库的体系结构中，数据库存储结构的改变会引起内模式的改变。
为了使数据库的模式保持不变，必须改变模式与内模式之间的映象。
使数据库具有（）。
A: 操作独立性B: 逻辑独立性C: 物理独立性D: 数据独立性答案：C8.数据库系统的三级模式结构中，下列属于三级模式的是()。
A: 内模式B: 抽象模式C: 模式D: 外模式答案：ACD9.一个数据库系统的外模式只能有一个。
（）A: 对B: 错答案：B10在数据库中，数据的物理独立性是指应用程序与数据库中数据的逻辑结果相互独立。
（）A: 对B: 错答案：B第二章测试1一个关系只有一个（）。
A: 超码B: 候选码C: 外码D: 主码答案：D2参照完整性规则是对（）的约束。
A: 候选码B: 外码C元组D: 列答案：B3设关系R和关系S的元数分别是3和4,关系T是R与S的厂义笛卡尔积，即：T=R S,则关系T的元数是（）A:12B:7C:16D:9答案：A4.设关系R和关系S具有相同的元数，且对应的属性取自相同的域。
第五章 数据库的维护和应用
第五章数据库的维护和应用一、例题解析【例题1】设数据表已经打开，为了在表尾增加一条空记录，应使用的命令是（）。
A．APPEND B．APPEND BLANK C．INSERT D．INSERT BLANK【解析】答案A，APPEND命令打开一个输入记录的浏览窗口；答案D，INSERT BLANK是插入一条空记录，但必须将记录指针移动到文件尾；答案B是在文件尾追加一条空记录，这个命令常常配合REPLACE命令用于程序中添加记录。
答案：B【例题2】在打开的职工表在有字符型字段“职称”和数值型字段“工资”等，若要求先按职称的升序，职称相同再按工资降序建立排序好的zcgz．dbf文件，应使用的命令是（）。
A.SORT ON职称,工资/D Tozcgz.dbfB.SORT ON工资/D,职称/A Tozcgz.dbfC.SORT ON职称+工资Tozcgz.dbfD.SORT ON职称+工资/D Tozcgz.dbf【解析】在SORT命令中,排序的依据只能是关键字段名,而不能使用关键字表达式,因此,答案C和D都是错误的。
答案B是先按工资排序,工资相同再按职称排序,与题意不符，所以正确的答案是Ａ。
答案：Ａ【例题３】在打开的学生档案表中有字符字段“性别”和日期型字段“出生日期”等若要先按性别排序，性别相同时再按出生日期排序创建单索引文件，应使用的命令是（）。
A．INDEX ON性别，出生日期TO Xbrq.idxB．INDEX ON性别＋出生日期TO xbrq.idxC．INDEX ON性别＋STR(出生日期)TO xbrq.idxD．INDEX ON性别＋DTOC（出生日期）TO xbrq.idx【解析】对于多重索引，索引表达式中的各字段数据类型必须是一致的。
本题中，“性别”和“出生日期”的数据类型不一致，为构成一个索引表达式，通常是使用转换函数把非字符型的数据转换成字符型的数据。
这里要用DTOC（）函数将日期型转换成字符型的，因而答案B、C是错误的；答案A中，索引表达式不能是用逗号隔开的式子。
(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)。
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4．设有关系模式R（U，F），其中，U={A，B，C，D，E}，F={A →BC，CD→E，B→D，E→A}
（1）计算B+；E+
（2）求出R的所有候选键；
(3) 求F的极小函数依赖集
（4）判断关系模式最高达到第几范式。
答:（1）B+=BD , E+=ABCDE
（2）R的所有候选键为：E，A，CD，BC
(3) {A→B，A→C, CD→E，B→D，E→A}
（4）没有非主属性，B不是候选键，所以为3NF。
7.设有R(A,B,C,D)F={A—>C,C—>A,B—>AC,D—>AC},计算(AD)+,Fmin,CK,将R分解使其满足BCNF且无损连接性 ,将R 分解使其满足3NF并且有无损连接性与保持依赖性
答:
(AD)+=ACD
Fmin之一{A—>C,C—>A,B—>A,D—>A }
候选键:BD
分解使其满足BCNF且无损连接性{A,CAB,BD}
R分解使其满足3NF并且有无损连接性与保持依赖性{AC,BA,DA,BD}
8.现在R(A,B,C),F={A—>C,B—>C},分解为{AB,AC}{AB,BC}是否具有无损连接性和依赖保持性.
答:
{AB,AC}:无损,没有函数依赖{AC,BC}:不具有无损,依赖保持。