2022年数据库系统基础教程第四章答案.pdf

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1、数据库系统基础教程第四章答案Solutions Chapter 4 4、1、1 4、1、2 a) b) c) In c we assume that a phone and address can only belong to a single customer (1-m relationship represented by arrow into customer)、精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 1 页，共 35 页 - - - - - - - - - - 数据库系统基础教程第四章答案

2、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 part of the composite key of the entity set、In c&d, we conve

3、rt attributes phones and addresses to entity sets、 Since entity sets often become relations in relational design, we must consider more efficient alternatives、Instead of querying multiple tables where key values are duplicated, we can also modify attributes: (i) Phones attribute can be converted int

4、o HomePhone, OfficePhone and CellPhone、(ii) A multivalued attribute such as alias can be kept as an attribute where a single column can be used in relational design i、e、 concatenate all values、SQL allows a query like %Junius% to search the multiple values in a column alias、精品资料 - - - 欢迎下载 - - - - -

5、- - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 2 页，共 35 页 - - - - - - - - - - 数据库系统基础教程第四章答案4、1、3 4、1、4 a) 精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 3 页，共 35 页 - - - - - - - - - - 数据库系统基础教程第四章答案b) c) The relationship played between Teams and Players is similar to relationship

6、plays between Teams and Players、4、1、5 精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 4 页，共 35 页 - - - - - - - - - - 数据库系统基础教程第四章答案4、1、6 The information about children can be ascertained from motherOf and fatherOf relationships、 Attribute ssNo is required since names are not uni

7、que、4、1、7 精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 5 页，共 35 页 - - - - - - - - - - 数据库系统基础教程第四章答案4、1、8 a) (b) 精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 6 页，共 35 页 - - - - - - - - - - 数据库系统基础教程第四章答案4、1、9 Assumptions A Professor only works in at mo

8、st one department、A course has at most one TA、A course is only taught by one professor and offered by one department、Students and professors have been assigned unique email ids、A course is uniquely identified by the course no, section no, and semester (e、g、 cs157-3 spring 09)、4、1、10 Given that for e

9、ach movie, a unique studio exists that produces the movie、 Each star is contracted to at most one studio、精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 7 页，共 35 页 - - - - - - - - - - 数据库系统基础教程第四章答案But stars could be unemployed at a given time、 Thus the four-way relationship in

10、fig 4、6 can be easily into converted equivalent relationships、4、2、1 Redundancy: The owner address is repeated in AccSets and Addresses entity sets、Simplicity: AccSets does not serve any useful purpose and the design can be more simply represented by creating many-to-many relationship between Custome

11、rs and Accounts、Right kind of element: The entity set Addresses has a single attribute address、A customer cannot have more than one address、Hence address should be an attribute of entity set Customers、精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 8 页，共 35 页 - - - - - - - - - -

12、 数据库系统基础教程第四章答案Faithfulness: Customers cannot be uniquely identified by their names、 In real world Customers would have a unique attribute such as ssNo or customerNo 4、2、2 Studios and Presidents can be combined into one entity set Studios with Presidents becoming an attribute of Studios under follow

13、ing circumstances: 1、 The Presidents entity set only contains a simple attribute viz、presidentName、 Additional attributes specific to Presidents might justify making Presidents into an entity set、4、2、3 4、2、4 The entity sets should have single attribute、a) Stars: starName b) Movies: movieName c) Stud

14、ios: studioName、 However there exists a many-to-many relationship between Studios and Contracts、 Hence, in addition, we need more information about studios involved、 If a contract always involves two studios, two attributes such as producingStudio and starStudio can replace the Studios entity set、 I

15、f a contact can be associated with at most five studios, it may be possible to replace the Studios entity set by five attributes viz、studio1, studio2, studio3, studio4, and studio5、 Alternately, a composite attribute containing concatenation of all studio names in a contact can be considered、 A sepa

16、rator character such as $ can be used、 SQL allows searching of such an attribute using query like %keyword% 4、2、5 From Augmentation rule of Functional Dependency, given B - M (B=Baby, M=Mother) then BND - M (N=Nurse, D=Doctor) Hence we can just put an arrow entering mother、a) Put an arrow entering e

17、ntity set Mothers for the simplest solution (As in fig、 4 、4, where a multi-way relationship was allowed, even though Movies alone could identify the Studio)、 However, we can display more accurate information with below figure、精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 9 页，

18、共 35 页 - - - - - - - - - - 数据库系统基础教程第四章答案b) c) Again from Augmentation rule of Functional Dependency, given BM - D then BMN - D 精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 10 页，共 35 页 - - - - - - - - - - 数据库系统基础教程第四章答案Thus we can just add an arrow entering Doctors to fig 4、1

19、5 、 Below figure represents more accurate information however、4、2、6 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、精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载

20、 名师归纳 - - - - - - - - - -第 11 页，共 35 页 - - - - - - - - - - 数据库系统基础教程第四章答案c) Design flaws in abc above 1、 As suggested above, using Transitivity and Augmentation rules of Functional Dependency, much simpler design is possible、4、2、7 In below figure there exists a many-to-one relationship between Babie

21、s and Births and another many-to-one relationship between Births and Mothers、 From transitivity of relationships, there is a many-to-one relationship between Babies and Mothers、 Hence a baby has a unique mother while a birth can allow more than one baby、精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归

22、纳 - - - - - - - - - -第 12 页，共 35 页 - - - - - - - - - - 数据库系统基础教程第四章答案4、3、1 a) b) A captain cannot exist without a team、 However a player can (free agent)、 A recently formed (or defunct) team can exist without players or colors、c) Children can exist without mother and father (unknown)、精品资料 - - - 欢迎下载

23、 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 13 页，共 35 页 - - - - - - - - - - 数据库系统基础教程第四章答案4、3、2 a) The keys of both E1 and E2 are required for uniquely identifying tuples in R b) The key of E1 c) The key of E2 d) The key of either E1 or E2 4、3、3 Special Case: All entity sets have arrows go

24、ing into them i、e、 all relationships are 1-to-1 Any Ki Otherwise: Combination of all Kis where there does not exist an arrow going from R to Ei、4、4、1 No, grade is not part of the key for enrollments、 The keys of Students and Courses become keys of the weak entity set Enrollments、精品资料 - - - 欢迎下载 - -

25、- - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 14 页，共 35 页 - - - - - - - - - - 数据库系统基础教程第四章答案4、4、2 It is possible to make assignment number a weak key of Enrollments but this is not good design (redundancy since multiple assignments correspond to a course)、A new entity set Assignment is created an

26、d it is also a weak entity set、 Hence the key attributes of Assignment will come from the strong entity sets to which Enrollments is connected i、e、 studentID, dept, and CourseNo、精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 15 页，共 35 页 - - - - - - - - - - 数据库系统基础教程第四章答案4、4、3 a

27、) b) 精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 16 页，共 35 页 - - - - - - - - - - 数据库系统基础教程第四章答案c) 4、4、4 a) 精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 17 页，共 35 页 - - - - - - - - - - 数据库系统基础教程第四章答案b) 4、5、1 Customers(SSNo,name,addr,phone) Flights(numb

28、er,day,aircraft) Bookings(custSSNo,flightNo,flightDay,row,seat) Relations for toCust and toFlt relationships are not required since the weak entity set Bookings already contains the keys of Customers and Flights、4、5、2 精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 18 页，共 35 页 -

29、 - - - - - - - - - 数据库系统基础教程第四章答案(a) (b) Schema is changed、 Since toCust is no longer an identifying relationship, SSNo is no longer a part of Bookings relation、Bookings(flightNo,flightDay,row,seat) ToCust(custSSNO,flightNo,flightDay,row,seat) The above relations are merged into Bookings(flightNo,fl

30、ightDay,row,seat,custSSNo) However custSSNo is no longer a key of Bookings relation、 It becomes a foreign key instead、4、5、3 Ships(name, yearLaunched) SisterOf(name, sisterName) 4、5、4 (a) Stars(name,addr) 精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 19 页，共 35 页 - - - - - - - -

31、 - - 数据库系统基础教程第四章答案Studios(name,addr) Movies(title,year,length,genre) Contracts(starName,movieTitle,movieYear,studioName,salary) Depending on other relationships not shown in ER diagram, studioName may not be required as a key of Contracts (or not even required as an attribute of Contracts)、(b) Stud

32、ents(studentID) Courses(dept,courseNo) Enrollments(studentID,dept,courseNo,grade) (c) Departments(name) Courses(deptName,number) (d) Leagues(name) Teams(leagueName,teamName) Players(leagueName,teamName,playerName) 4、6、1 The weak relation Courses has the key from Depts along with number、 Hence there

33、is no relation for GivenBy relationship、(a) Depts(name, chair) Courses(number, deptName, room) LabCourses(number, deptName, allocation) (b) LabCourses has all the attributes of Courses、 Depts(name, chair) Courses(number, deptName, room) LabCourses(number, deptName, room, allocation) (c) Courses and

34、LabCourses are combined into one relation、 Depts(name, chair) Courses(number, deptName, room, allocation) 4、6、2 (a) Person(name,address) ChildOf(personName,personAddress,childName,childAddress) Child(name,address,fatherName,fatherAddress,motherName,motherAddresss) Father(name,address,wifeName,wifeAd

35、dresss) Mother(name,address) Since FatherOf and MotherOf are many-one relationships from Child, there is no need for a separate relation for them、 Similarly the one-one relationship Married can be included in Father (or Mother)、 ChildOf is a many-many relationship and needs a separate relation、Howev

36、er the ChildOf relation is not required since the relationship can be deduced from FatherOf and MotherOf relationships contained in Child relation、(b) A person cannot be both Mother and Father、Person(name,address) PersonChild(name,address) PersonChildFather(name,address) PersonChildMother(name,addre

37、ss) PersonFather(name,address) PersonMother(name,address) ChildOf(personName,personAddress,childName,childAddress) FatherOf(childName,childAddress,fatherName,fatherAddress) MotherOf(childName,childAddress,motherName,motherAddress) Married(husbandName,husbandAddress,wifeName,wifeAddress) The many-man

38、y ChildOf relationship again requires a relation、精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 20 页，共 35 页 - - - - - - - - - - 数据库系统基础教程第四章答案An entity belongs to one and only one class when using object-oriented approach、 Hence, the many-one relations MotherOf and FatherOf cou

39、ld be added as attributes to PersonChild,PersonChildFather, and PersonChildMother relations、Similarly the Married relation can be added as attributes to PersonChildMother and PersonMother (or the corresponding father relations)、(c) For the Person relation at least one of husband and wife attributes

40、will be null、Person(personName,personAddress,fatherName,fatherAddress,motherName,motherAddresss,wifeName,wifeAddresss,husbandName,husbandAddress) ChildOf(personName,personAddress,childName,childAddress) 4、6、3 (a) People(name,fatherName,motherName) Males(name) Females(name) Fathers(name) Mothers(name

41、) ChildOf(personName,childName) (b) People(name) PeopleMale(name) PeopleMaleFathers(name) PeopleFemale(name) PeopleFemaleMothers(name) ChildOf(personName,childName) FatherOf(childName,fatherName) MotherOf(childName,motherName) People cannot belong to both male and female branch of the ER diagram、Mor

42、eover since an entity belongs to one and only one class when using object-oriented approach, no entity belongs to People relation、Again we could replace MotherOf and FatherOf relations by adding as attributes to PeopleMale,PeopleMaleFathers,PeopleFemale, and PeopleFemaleMothers relations、(c) People(

43、name,fatherName,motherName) ChildOf(personName,childName) 4、6、4 (a) Each entity set results in one relation、 Thus both the minimum and maximum number of relations is e、The root relation has a attributes including k keys、 Thus the minimum number of attributes is a、 All other relations include the k k

44、eys from root along with their a attributes、 Thus the maximum number of attributes is a+k、(b) The relation for root will have a attributes、 The relation representing the whole tree will have e*a attributes、The number of relations will depend on the shape of the tree、 A tree of e entities where only

45、one child exists(say left child only) would have the minimum number of relations、 Thus below figure will only contain 4 subtrees that contain root E1,E1E2,E1E2E3, and E1E2E3E4、 With e entity sets, minimum e relations are possible、精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 2

46、1 页，共 35 页 - - - - - - - - - - 数据库系统基础教程第四章答案The maximum number of subtrees result when all the entities(except root) are at depth 1、 Thus below figure will contain 8 subtrees that contain root E1,E1E2,E1E3,E1E4,E1E2E3,E1E3E4,E1E2E4,and E1E2E3E4、 With e entity sets, maximum 2(e-1) relations are poss

47、ible、(c) The nulls method always results in one relation and contains attributes from all e entities i、e、 e*a attributes、Summarizing for a,b, and c above; #Components #Relations Min Max Min Max Method straight-E/R a a e e object-oriented a e*a e 2(e-1) 精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳

48、 - - - - - - - - - -第 22 页，共 35 页 - - - - - - - - - - 数据库系统基础教程第四章答案nulls e*a e*a 1 1 4、7、1 4、7、2 a) b) c) 精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 23 页，共 35 页 - - - - - - - - - - 数据库系统基础教程第四章答案d) 4、7、3 精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第

49、24 页，共 35 页 - - - - - - - - - - 数据库系统基础教程第四章答案4、7、4 4、7、5 Males and Females subclasses are complete、 Mothers and Fathers are partial、 All subclasses are disjoint、精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 25 页，共 35 页 - - - - - - - - - - 数据库系统基础教程第四章答案4、7、6 精品资料 - - - 欢迎下载 -

50、 - - - - - - - - - - 欢迎下载 名师归纳 - - - - - - - - - -第 26 页，共 35 页 - - - - - - - - - - 数据库系统基础教程第四章答案4、7、7 4、7、8 We convert the ternary relationship Contracts into three binary relationships between a new entity set Contracts and existing entity sets、精品资料 - - - 欢迎下载 - - - - - - - - - - - 欢迎下载 名师归纳 - -