数据库系统基础教程第二章答案解析

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1、Exercise 2.2.1aFor relation Accounts, the attributes are:acctNo, type, balanceFor relation Customers, the attributes are:firstName, lastName, idNo, accountExercise 2.2.1bFor relation Accounts, the tuples are:(12345, savings, 12000),(23456, checking, 1000),(34567, savings, 25)For relation Customers,

2、the tuples are:(Robbie, Banks, 901-222, 12345),(Lena, Hand, 805-333, 12345),(Lena, Hand, 805-333, 23456)Exercise 2.2.1cFor relation Accounts and the first tuple, the components are:123456 acctNosavings type12000 balanceFor relation Customers and the first tuple, the components are:Robbie firstNameBa

3、nks lastName901-222 idNo12345 accountExercise 2.2.1dFor relation Accounts, a relation schema is:Accounts(acctNo, type, balance)For relation Customers, a relation schema is:Customers(firstName, lastName, idNo, account)Exercise 2.2.1eAn example database schema is:Accounts (acctNo,type,balance)Customer

4、s (firstName,lastName,idNo,account)Exercise 2.2.1fA suitable domain for each attribute:acctNo Integertype Stringbalance IntegerfirstName StringlastName StringidNo String (because there is a hyphen we cannot use Integer)account IntegerExercise 2.2.1gAnother equivalent way to present the Account relat

5、ion:acctNobalancetype3456725savings234561000checking1234512000savingsAnother equivalent way to present the Customers relation:idNofirstNamelastNameaccount805-333LenaHand23456805-333LenaHand12345901-222RobbieBanks12345Exercise 2.2.2Examples of attributes that are created for primarily serving as keys

6、 in a relation:Universal Product Code (UPC) used widely in United States and Canada to track products in stores.Serial Numbers on a wide variety of products to allow the manufacturer to individually track each product.Vehicle Identification Numbers (VIN), a unique serial number used by the automotiv

7、e industry to identify vehicles.Exercise 2.2.3aWe can order the three tuples in any of 3! = 6 ways. Also, the columns can be ordered in any of 3! = 6 ways. Thus, the number of presentations is 6*6 = 36.Exercise 2.2.3bWe can order the three tuples in any of 5! = 120 ways. Also, the columns can be ord

8、ered in any of 4! = 24 ways. Thus, the number of presentations is 120*24 = 2880Exercise 2.2.3cWe can order the three tuples in any of m! ways. Also, the columns can be ordered in any of n! ways. Thus, the number of presentations is n!m!Exercise 2.3.1aCREATE TABLE Product (makerCHAR(30),modelCHAR(10)

9、 PRIMARY KEY,type CHAR(15);Exercise 2.3.1bCREATE TABLE PC (model CHAR(30),speed DECIMAL(4,2),ram INTEGER,hd INTEGER,price DECIMAL(7,2);Exercise 2.3.1cCREATE TABLE Laptop (model CHAR(30),speed DECIMAL(4,2),ram INTEGER,hd INTEGER,screen DECIMAL(3,1),price DECIMAL(7,2);Exercise 2.3.1dCREATE TABLE Print

10、er (model CHAR(30),color BOOLEAN,type CHAR (10),price DECIMAL(7,2);Exercise 2.3.1eALTER TABLE Printer DROP color;Exercise 2.3.1fALTER TABLE Laptop ADD od CHAR (10) DEFAULT none;Exercise 2.3.2aCREATE TABLE Classes (class CHAR(20),type CHAR(5),country CHAR(20),numGuns INTEGER,bore DECIMAL(3,1),displac

11、ement INTEGER);Exercise 2.3.2bCREATE TABLE Ships (name CHAR(30),class CHAR(20),launched INTEGER);Exercise 2.3.2cCREATE TABLE Battles (name CHAR(30),date DATE);Exercise 2.3.2dCREATE TABLE Outcomes (ship CHAR(30),battle CHAR(30),result CHAR(10);Exercise 2.3.2eALTER TABLE Classes DROP bore;Exercise 2.3

12、.2fALTER TABLE Ships ADD yard CHAR(30);Exercise 2.4.1aR1 := speed 3.00 (PC)R2 := model(R1)model100510061013Exercise 2.4.1bR1 := hd 100 (Laptop)R2 := Product (R1)R3 := maker (R2)makerEABFGExercise 2.4.1cR1 := maker=B (Product PC)R2 := maker=B (Product Laptop)R3 := maker=B (Product Printer)R4 := model

13、,price (R1)R5 := model,price (R2)R6: = model,price (R3)R7 := R4 R5 R6modelprice100464910056301006104920071429Exercise 2.4.1dR1 := color = true AND type = laser (Printer)R2 := model (R1)model30033007Exercise 2.4.1eR1 := type=laptop (Product)R2 := type=PC(Product)R3 := maker(R1)R4 := maker(R2)R5 := R3

14、 R4makerFGExercise 2.4.1fR1 := PC1(PC)R2 := PC2(PC)R3 := R1 (PC1.hd = PC2.hd AND PC1.model PC2.model) R2R4 := hd(R3)hd25080160Exercise 2.4.1gR1 := PC1(PC)R2 := PC2(PC)R3 := R1 (PC1.speed = PC2.speed AND PC1.ram = PC2.ram AND PC1.model PC2.model) R2R4 := PC1.model,PC2.model(R3)PC1.modelPC2.model10041

15、012Exercise 2.4.1hR1 := model(speed 2.80(PC) model(speed 2.80(Laptop)R2 := maker,model(R1 Product)R3 := R3(maker2,model2)(R2)R4 := R2 (maker = maker2 AND model model2) R3R5 := maker(R4)makerBEExercise 2.4.1iR1 := model,speed(PC)R2 := model,speed(Laptop)R3 := R1 R2R4 := R4(model2,speed2)(R3)R5 := mod

16、el,speed (R3 (speed speed2 ) R4)R6 := R3 R5R7 := maker(R6 Product)makerBExercise 2.4.1jR1 := maker,speed(Product PC)R2 := R2(maker2,speed2)(R1)R3 := R3(maker3,speed3)(R1)R4 := R1 (maker = maker2 AND speed speed2) R2R5 := R4 (maker3 = maker AND speed3 speed2 AND speed3 speed) R3R6 := maker(R5)makerAD

17、E Exercise 2.4.1kR1 := maker,model(Product PC)R2 := R2(maker2,model2)(R1)R3 := R3(maker3,model3)(R1)R4 := R4(maker4,model4)(R1)R5 := R1 (maker = maker2 AND model model2) R2R6 := R3 (maker3 = maker AND model3 model2 AND model3 model) R5R7 := R4 (maker4 = maker AND (model4=model OR model4=model2 OR mo

18、del4=model3) R6R8 := maker(R7)makerABDEExercise 2.4.2aExercise 2.4.2bExercise 2.4.2cExercise 2.4.2dExercise 2.4.2eExercise 2.4.2fExercise 2.4.2gExercise 2.4.2hExercise 2.4.2iExercise 2.4.2jExercise 2.4.2kExercise 2.4.3aR1 := bore 16 (Classes)R2 := class,country (R1)classcountryIowaUSANorth CarolinaU

19、SAYamatoJapanExercise 2.4.3bR1 := launched 1921 AND displacement 35000 (R1)R3 := name (R2)nameIowaMissouriMusashiNew JerseyNorth CarolinaWashingtonWisconsinYamatoExercise 2.4.3eR1 := battle=Guadalcanal(Outcomes)R2 := Ships (ship=name) R1R3 := Classes R2R4 := name,displacement,numGuns(R3)namedisplace

20、mentnumGunsKirishima320008Washington370009Exercise 2.4.3fR1 := name(Ships)R2 := ship(Outcomes)R3 := R3(name)(R2)R4 := R1 R3nameCaliforniaHarunaHieiIowaKirishimaKongoMissouriMusashiNew JerseyNorth CarolinaRamilliesRenownRepulseResolutionRevengeRoyal OakRoyal SovereignTennesseeWashingtonWisconsinYamat

21、oArizonaBismarckDuke of YorkFusoHoodKing George VPrince of WalesRodneyScharnhorstSouth DakotaWest VirginiaYamashiroExercise 2.4.3gFrom 2.3.2, assuming that every class has one ship named after the class.R1 := class(Classes)R2 := class(name class(Ships)R3 := R1 R2classBismarckExercise 2.4.3hR1 := cou

22、ntry(type=bb(Classes)R2 := country(type=bc(Classes)R3 := R1 R2countryJapanGt. BritainExercise 2.4.3iR1 := ship,result,date(Battles (battle=name) Outcomes)R2 := R2(ship2,result2,date2)(R1)R3 := R1 (ship=ship2 AND result=damaged AND date date2) R2R4 := ship(R3)No results from sample data.Exercise 2.4.

23、4aExercise 2.4.4bExercise 2.4.4cExercise 2.4.4dExercise 2.4.4eExercise 2.4.4fExercise 2.4.4gExercise 2.4.4hExercise 2.4.4iExercise 2.4.5The result of the natural join has only one attribute from each pair of equated attributes. On the other hand, the result of the theta-join has both columns of the

24、attributes and their values are identical.Exercise 2.4.6UnionIf we add a tuple to the arguments of the union operator, we will get all of the tuples of the original result and maybe the added tuple. If the added tuple is a duplicate tuple, then the set behavior will eliminate that tuple. Thus the un

25、ion operator is monotone.IntersectionIf we add a tuple to the arguments of the intersection operator, we will get all of the tuples of the original result and maybe the added tuple. If the added tuple does not exist in the relation that it is added but does exist in the other relation, then the resu

26、lt set will include the added tuple. Thus the intersection operator is monotone.DifferenceIf we add a tuple to the arguments of the difference operator, we may not get all of the tuples of the original result. Suppose we have relations R and S and we are computing R S. Suppose also that tuple t is i

27、n R but not in S. The result of R S would include tuple t. However, if we add tuple t to S, then the new result will not have tuple t. Thus the difference operator is not monotone.ProjectionIf we add a tuple to the arguments of the projection operator, we will get all of the tuples of the original r

28、esult and the projection of the added tuple. The projection operator only selects columns from the relation and does not affect the rows that are selected. Thus the projection operator is monotone.SelectionIf we add a tuple to the arguments of the selection operator, we will get all of the tuples of

29、 the original result and maybe the added tuple. If the added tuple satisfies the select condition, then it will be added to the new result. The original tuples are included in the new result because they still satisfy the select condition. Thus the selection operator is monotone.Cartesian ProductIf

30、we add a tuple to the arguments of the Cartesian product operator, we will get all of the tuples of the original result and possibly additional tuples. The Cartesian product pairs the tuples of one relation with the tuples of another relation. Suppose that we are calculating R x S where R has m tupl

31、es and S has n tuples. If we add a tuple to R that is not already in R, then we expect the result of R x S to have (m + 1) * n tuples. Thus the Cartesian product operator is monotone.Natural JoinsIf we add a tuple to the arguments of a natural join operator, we will get all of the tuples of the orig

32、inal result and possibly additional tuples. The new tuple can only create additional successful joins, not less. If, however, the added tuple cannot successfully join with any of the existing tuples, then we will have zero additional successful joins. Thus the natural join operator is monotone.Theta

33、 JoinsIf we add a tuple to the arguments of a theta join operator, we will get all of the tuples of the original result and possibly additional tuples. The theta join can be modeled by a Cartesian product followed by a selection on some condition. The new tuple can only create additional tuples in t

34、he result, not less. If, however, the added tuple does not satisfy the select condition, then no additional tuples will be added to the result. Thus the theta join operator is monotone.RenamingIf we add a tuple to the arguments of a renaming operator, we will get all of the tuples of the original re

35、sult and the added tuple. The renaming operator does not have any effect on whether a tuple is selected or not. In fact, the renaming operator will always return as many tuples as its argument. Thus the renaming operator is monotone.Exercise 2.4.7aIf all the tuples of R and S are different, then the

36、 union has n + m tuples, and this number is the maximum possible.The minimum number of tuples that can appear in the result occurs if every tuple of one relation also appears in the other. Then the union has max(m , n) tuples. Exercise 2.4.7bIf all the tuples in one relation can pair successfully wi

37、th all the tuples in the other relation, then the natural join has n * m tuples. This number would be the maximum possible.The minimum number of tuples that can appear in the result occurs if none of the tuples of one relation can pair successfully with all the tuples in the other relation. Then the

38、 natural join has zero tuples.Exercise 2.4.7cIf the condition C brings back all the tuples of R, then the cross product will contain n * m tuples. This number would be the maximum possible.The minimum number of tuples that can appear in the result occurs if the condition C brings back none of the tu

39、ples of R. Then the cross product has zero tuples.Exercise 2.4.7dAssuming that the list of attributes L makes the resulting relation L(R) and relation S schema compatible, then the maximum possible tuples is n. This happens when all of the tuples of L(R) are not in S.The minimum number of tuples tha

40、t can appear in the result occurs when all of the tuples in L(R) appear in S. Then the difference has max(n m , 0) tuples. Exercise 2.4.8Defining r as the schema of R and s as the schema of S:1. r(R S)2. R (rs(S)where is the duplicate-elimination operator in Section 5.2 pg. 2133. R (R r(R S)Exercise

41、 2.4.9Defining r as the schema of R1. R - r(R S)Exercise 2.4.10A1,A2An(R S)Exercise 2.5.1aspeed 500(PC) = Model 1011 violates this constraint.Exercise 2.5.1bscreen 15.4 AND hd PC.ram AND Laptop.price PC.price(PC Laptop) = Models 2002,2006,2008 violate the constraint.Exercise 2.5.2aclass(bore 16(Clas

42、ses) = The Yamato class violates the constraint.Exercise 2.5.2bclass(numGuns 9 AND bore 14(Classes) = No violations to the constraint.Exercise 2.5.2cThis complex expression is best seen as a sequence of steps in which we define temporary relations R1 through R5 that stand for nodes of expression tre

43、es. Here is the sequence: R1(class,name) := class,name(Classes Ships) R2(class2,name2) := R2(class2,name2)(R1) R3(class3,name3) := R3(class3,name3)(R1) R4(class,name,class2,name2) := R1 (class = class2 AND name name2) R2 R5(class,name,class2,name2,class3,name3) := R4 (class=class3 AND name name3 AND

44、 name2 name3) R3The constraint is R5 = The Kongo, Iowa and Revenge classes violate the constraint.Exercise 2.5.2dcountry(type = bb(Classes) country(type = bc(Classes) = Japan and Gt. Britain violate the constraint.Exercise 2.5.2eThis complex expression is best seen as a sequence of steps in which we

45、 define temporary relations R1 through R5 that stand for nodes of expression trees. Here is the sequence: R1(ship,battle,result,class) := ship,battle,result,class(Outcomes (ship = name) Ships) R2(ship,battle,result,numGuns) := ship,battle,result,numGuns(R1 Classes) R3(ship,battle) := ship,battle(num

46、Guns 9(R2) R5(ship2) := ship2(R3 (battle = battle2) R4)The constraint is R5 = No violations to the constraint. Since there are some ships in the Outcomes table that are not in the Ships table, we are unable to determine the number of guns on that ship.Exercise 2.5.3Defining r as the schema A1,A2,An

47、and s as the schema B1,B2,Bn:r(R) s(S) = where is the antisemijoinExercise 2.5.4The form of a constraint as E1 = E2 can be expressed as the other two constraints. Using the “equating an expression to the empty set” method, we can simply say:E1 E2 = As a containment, we can simply say:E1 E2 AND E2 E1Thus, the form E1 = E2 of a constraint cannot express more than the two other forms discussed in this section.