数据库系统基础教程第三章答案

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1、Exercise3.1.1Answersforthisexercisemayvarybecauseofdifferentinterpretations.SomepossibleFDs:SocialSecuritynumbernameAreacodestatezipcodezipcodeStreetaddress,city,statePossiblekeys:SocialSecuritynumber,streetaddress,city,state,areacode,phonenumberNeedstreetaddress,city,statetouniquelydeterminelocatio

2、n.Apersoncouldhavemultipleaddresses.Thesameistrueforphones.Thesedays,apersoncouldhaveaIandlineandacellularphoneExercise3.1.2AnswersforthisexercisemayvarybecauseofdifferentinterpretationsSomepossibleFDs:IDx-position,y-position,z-positionIDIDIDx-velocity,y-velocity,z-velocityx-position,y-position,z-po

3、sitionPossiblekeys:IDx-position,y-position,z-positionThereasonwhythepositionswouldbeakeyisnotwomoleculescanoccupythesamepoint.ThesuperkeysareanysubsetthatcontainsA1.Thus,thereare2(n-1)suchsubsets,sinceeachofthen-1attributesAthroughAnmayindependentlybechoseninorout.Exercise3.1.3bThesuperkeysareanysub

4、setthatcontainsAThesuperkeysareanysubsetthatcontainsA1 orA2.Thereare2(n-1)suchsubsetswhenconsideringA1andthen-1attributesAconsideringA2andthen-2attributesA2 throughAn.Thereare2(n-2)suchsubsetswhen3 throughAn.WedonotcountA1inthesesubsetsbecausetheyarealreadycountedinthefirstgroupofsubsets.Thetotalnum

5、berofsubsetsis2(n-1)+2(n-2)Exercise3.1.3cThesuperkeysareanysubsetthatcontainsA1,A2orA3,A4.Thereare2(n-2)suchsubsetswhenconsideringA1A2andthen-2attributesA3throughAn.Thereare2(n-2)2(n-4)suchsubsetswhenconsideringA3,A4andattributesA5throughAnalongwiththeindividualattributesA1andA2.Wegetthe2(n-4)termbe

6、causewehavetodiscardthesubsetsthatcontainthekeyA1,A2toavoiddoublecounting.Thetotalnumberofsubsetsis2+2(n-2)_2(n-4)Exercise3.1.3dsubsetsbecausetheyarealreadycountedinthefirstgroupofsubsets.Thetotalnumberofsubsetsis2(n-2)+2(n-3).ThesuperkeysareanysubsetthatcontainsAwhenconsideringA1,A2andthen-2attribu

7、tesAwhenconsideringA1,A3andthen-3attributesA1,A2orA1,A3.Thereare2(n-2)suchsubsets(n-3)3 throughAn.Thereare2suchsubsets4 throughAnWedonotcountA2intheseWecouldtryinferencerulestodeducenewdependenciesuntilwearesatisfiedwehavethemall.Amoresystematicwayistoconsidertheclosuresofall15nonemptysetsofattribut

8、es.ForthesingleattributeswehaveA+=A,B+=B,C+=ACD,andD+=AD.Thus,theA.A.onlynewdependencywegetwithasingleattributeontheleftisCNowconsiderpairsofattributes:D.AC+=ACD,andACDisnontrivial.ADAB+=ABCD,sowegetnewdependencyAB=AD,sonothingnew.BC+=ABCD,sowegetBCA,andBCD.BD+=ABCD,givingusBDAandBDC.CD+=ACD,givingC

9、DA.Forthetriplesofattributes,ACD+=ACD,buttheclosuresoftheothersetsareeachABCD.Thus,wegetnewdependenciesABCD,ABDC,andBCDA.SineeABCD+=ABCD,wegetnonewdependencies.Thecollectionof11newdependenciesmentionedaboveare:CA,ABD,ACD,BCA,BCD,BDA,BDC,CDA,ABCD,ABDC,andBCDA.Exercise3.2.1bFromtheanalysisofclosuresab

10、ove,wefindthatAB,BC,andBDarekeys.AllothersetseitherdonothaveABCDastheclosureorcontainoneofthesethreesets.Exercise3.2.1cThesuperkeysareallthosethatcontainoneofthosethreekeys.Thatis,asuperkeythatisnotakeymustcontainBandmorethanoneofA,C,andD.Thus,the(proper)superkeysareABC,ABD,BCD,andABCD.i)Forthesingl

11、eattributeswehaveA=ABCD,B=BCD,C=C,andD=D.Thus,thenewdependenciesareACandAD.Nowconsiderpairsofattributes:AB+=ABCD,AC+=ABCD,AD+=ABCD,BC+=BCD,BD+=BCD,CD+=CD.ThusthenewdependenciesareABC,ABD,ACB,ACD,ADB,ADC,BCDandBDC.Forthetriplesofattributes,BCD=BCD,buttheclosuresoftheothersetsareeachABCD.Thus,wegetnew

12、dependenciesABCD,ABDC,andACDB.SineeABCD+=ABCD,wegetnonewdependencies.Thecollectionof13newdependenciesmentionedaboveare:AC,AD,ABC,ABD,ACB,ACD,ADB,ADC,BCD,BDC,ABCD,ABDCandACDB.ii)ForthesingleattributeswehaveA=A,B=B,C=C,andD+=D.Thus,therearenonewdependencies.AB+=ABCD,AC+=AC,ADNowconsiderpairsofattribut

13、es:=ABCD,BC=ABCD,BD=BD,CD=ABCD.ThusthenewC,BCAandCDB.dependenciesareABD,ADForthetriplesofattributes,alltheclosuresofthesetsareeachABCD.Thus,wegetnewdependenciesABCD,ABDC,ACDBandBCDA.+SineeABCD=ABCD,wegetnonewdependencies.Thecollectionof8newdependenciesmentionedaboveare:ABD,ADC,BCA,CDB,ABCD,ABDC,ACDB

14、andBCDA.iii)ForthesingleattributeswehaveA=ABCD,B=ABCD,C=ABCD,andDABCD.Thus,thenewdependenciesareAC,AD,BD,BA,CA,CB,DBandDC.SinceallthesingleattributesclosuresareABCD,anysupersetofthesingleattributeswillalsoleadtoaclosureofABCD.Knowingthis,wecanenumeratetherestofthenewdependencies.Thecollectionof24new

15、dependenciesmentionedaboveare:AC,AD,BD,BA,CA,CB,DB,DC,ABC,ABD,ACB,ACD,ADB,ADC,BCA,BCD,BDA,BDC,CDA,CDB,ABCD,ABDC,ACDBandBCDA.Exercise3.2.2bFromtheanalysisofclosuresin3.2.2a(i),wefindthattheonlykeyisA.AllothersetseitherdonothaveABCDastheclosureorcontainA.i) Fromtheanalysisofclosures3.2.2a(ii),wefindth

16、atAB,AD,BC,andCDarekeys.AllothersetseitherdonothaveABCDastheclosureorcontainoneofthesefoursets.ii) Fromtheanalysisofclosures3.2.2a(iii),wefindthatA,B,CandDarekeys.AllothersetseitherdonothaveABCDastheclosureorcontainoneofthesefoursets.Exercise3.2.2cThesuperkeysareallthosesetsthatcontainoneofthekeysin

17、3.2.2b(i).ThesuperkeysareAB,AC,AD,ABC,ABD,ACD,BCDandABCD.i) Thesuperkeysareallthosesetsthatcontainoneofthekeysin3.2.2b(ii).ThesuperkeysareABC,ABD,ACD,BCDandABCD.ii) Thesuperkeysareallthosesetsthatcontainoneofthekeysin3.2.2b(iii).ThesuperkeysareAB,AC,AD,BC,BD,CD,ABC,ABD,ACD,BCDandABCD.Exercise3.2.3aS

18、ineeAiAACcontainsAA,thentheclosureofAAnCcontainsB.ThusitfollowsthatA1A2AnCB.Exercise3.2.3bFrom3.2.3a,weknowthatA1A2AnCB.Usingtheconceptoftrivialdependencies,wecanshowthatA1A2AnCC.ThusA1A2AnCBC.Exercise3.2.3c1B2BmbecauseoftheFDA1A2A1B2BmbecauseoftheFDA1A2A1C2Ck.ThustheclosureofFromA1A2几曰匕E,weknowthat

19、theclosurecontainsBBB2BmTheB1B2BmandtheE1E2EcombinetoformtheC1A2AE1E2EjD.AAAnEEEjcontainsDaswell.Thus,AExercise3.2.3dFromA1A2AnCQC,weknowthattheclosurecontainsB1B2BmbecauseoftheFDA1A2ABB2BmTheC1C2GalsotellusthattheclosureofA1A2ACC2CcontainsD1D2D.Thus,AA2AnCQCkB1B2BkDDD.Exercise3.2.4aIfattributeArepr

20、esentedSocialSecurityNumberandBrepresentedapersonsname,thenwewouldassumeABbutBAwouldnotbevalidbecausetheremaybemanypeoplewiththesamenameanddifferentSocialSecurityNumbers.Exercise3.2.4bLetattributeArepresentSocialSecurityNumber,BrepresentgenderandCrepresentname.sname.ABC).SurelySocialSecurityNumberan

21、dgendercanuniquelyidentifyapersonsname.AC).However,ASocialSecurityNumbercanalsouniquelyidentifyapersongenderdoesnotuniquelydetermineaname.BCisnotvalid).Exercise3.2.4cLetattributeArepresentlatitudeandBrepresentIongitude.Together,bothattributescanuniquelydetermineC,apointontheworldmap.ABC).However,nei

22、therAnorBcanuniquelyidentifyapoint.ACandBCarenotvalid).Exercise3.2.5GivenarelationwithattributesAAn,wearetoldthattherearenofunctionaldependenciesoftheformB1B2B-1CwhereB1B2B-1isn-1oftheattributesfromA1AAandCistheremainingattributefromA1A2An.Inthiscase,thesetB1庄B1andanysubsetdonotfunctionallydetermine

23、C.ThustheonlyfunctionaldependenciesthatwecanmakeareoneswhereCisonboththeleftandrighthandsides.AllofthesefunctionaldependencieswouldbetrivialandthustherelationhasnonontrivialFDs.Exercise3.2.6Letsprovethisbyusingthecontrapositive.WewishtoshowthatifX+isnotasubset+ofY,thenitmustbethatXisnotasubsetofY.If

24、X+isnotasubsetofY+,theremustbeattributesA1AAinX+thatarenotinY+.IfanyoftheseattributeswereoriginallyinX,thenwearedonebecauseYdoesnotcontainanyoftheA1A2A.However,iftheAiAAwereaddedbytheclosure,thenwemustexaminethecasefurther.AssumethattherewassomeFDC1C2CmA1A2AwhereA1A2AissomesubsetofA1A2A.Itmustbethen

25、thatC1C2CmorsomesubsetofCiC2CmisinX.However,theattributesC1C2CmcannotbeinYbecauseweassumedthatattributesA1A2An+areonlyinXandarenotinY.Thus,XisnotasubsetofY.?Y,thenX+?Y+.Byprovingthecontrapositive,wehavealsoprovedifXExercise3.2.7ThealgorithmtofindX+isoutlinedonpg.76.Usingthatalgorithm,wecanprovethat+

26、(X)=X.Wewilldothisbyusingaproofbycontradiction.Supposethat(X+)+丰X+.Thenfor(X+)+,itmustbethatsomeFDallowedadditional+attributestobeaddedtotheoriginalsetX.Forexample,XAwhereAissomeattributenotinX+.However,ifthiswerethecase,thenX+wouldnotbetheclosureofX.TheclosureofXwouldhavetoincludeAaswell.Thiscontra

27、dictsthefactthatweweregiventheclosureofX,X+.Therefore,itmustbethat(X+)+=X+orelseX+isnottheclosureofX.Exercise3.2.8aIfallsetsofattributesareclosed,thentherecannotbeanynontrivialfunctional1A2.An+containsBdependencies.SupposeAA.AnBisanontrivialdependency.ThenAandthusA1A.Anisnotclosed.Exercise3.2.8bIfth

28、eonlyclosedsetsareIftheonlyclosedsetsareandA,B,C,D,thenthefollowingFDshold:ABACBABCCACBDADBABCABDACBACDADBADCBCABCDBDABDCADBDCDDCCDACDBCDACDBABCDABDCACDBBCDAExercise3.2.8cIftheonlyclosedsetsareIftheonlyclosedsetsare,A,BandA,B,C,D,thenthefollowingFDshold:ABBACACBCDDADBDCACBACDADBADCBCABCDBDABDCCDACDB

29、ABCDABDCACDBBCDAExercise3.2.9WecanthinkofthisproblemasasituationwheretheattributesA,B,Crepresentcitiesandthefunctionaldependenciesrepresentonewaypathsbetweenthecities.Theminimalbasesaretheminimalnumberofpathwaysthatareneededtoconnectthecities.Wedonotwanttocreateanotherroadwayifthetwocitiesarealready

30、connected.Thesystematicwaytodothiswouldbetocheckallpossiblesetsofthepathways.However,wecansimplifythesituationbynotingthatittakesmorethantwopathwaystovisitthetwoothercitiesandcomeback.Also,ifwefindasetofpathwaysthatisminimal,addingadditionalpathwayswillnotcreateanotherminimalset.Thetwosetsofminimalb

31、asesthatweregiveninexampleare:AB,BC,CAAB,BA,BC,CBTheadditionalsetsofminimalbasesare:CB,BA,ACAB,AC,BA,CAAC,BC,CA,CBExercise3.2.10aWeneedtocomputetheclosuresofallsubsetsofABC,althoughthereisnoneedtothinkabouttheemptysetorthesetofallthreeattributes.Herearethecalculationsfortheremainingsixsets:A+=AB+=BC

32、+=ACE+AB=ABCDEAC+=ACEBC+=ABCDEWeignoreDandE,soabasisfortheresultingfunctionaldependenciesforABCis:CAandABC.NotethatBC-Aistrue,butfollowslogicallyfromC-A,andthereforemaybeomittedfromourlist.Exercise3.2.10bWeneedtocomputetheclosuresofallsubsetsofABC,althoughthereisnoneedtothinkabouttheemptysetortheset

33、ofallthreeattributes.Herearethecalculationsfortheremainingsixsets:A+=ADB+=BC+=cAB+=ABDEAC+=ABCDEBC+=BCB.WeignoreDandE,soabasisfortheresultingfunctionaldependenciesforABCis:ACExercise3.2.10cWeneedtocomputetheclosuresofallsubsetsofABC,althoughthereisnoneedtothinkabouttheemptysetorthesetofallthreeattri

34、butes.Herearethecalculationsfortheremainingsixsets:A+=A+B=BC+=c+AB=ABDAC+=ABCDE+BC=ABCDEWeignoreDandE,soabasisfortheresultingfunctionaldependenciesforABCis:ACandBCA.Exercise3.2.10dWeneedtocomputetheclosuresofallsubsetsofABC,althoughthereisnoneedtothinkabouttheemptysetorthesetofallthreeattributes.Her

35、earethecalculationsfortheremainingsixsets:A+=ABCDEB+=ABCDEC+=ABCDEAB+=ABCDEAC+=ABCDEBC+=ABCDEWeignoreDandE,soabasisfortheresultingfunctionaldependenciesforABCis:AB,BCandCA.Exercise3.2.11ForsteponeofAlgorithm,supposewehavetheFDABCDE.WewanttouseArmstrongsAxiomstoshowthatABCDandABCEfollow.Surelythefunc

36、tionaldependenciesDEDandDEEholdbecausetheyaretrivialandfollowthereflexivityproperty.Usingthetransitivityrule,wecanderivetheFDABCDfromtheFDsABCDEandDED.Likewise,wecandothesameforABCDEandDEEandderivetheFDABCE.ForstepstwothroughfourofAlgorithm,supposewehavetheinitialsetofattributesoftheclosureasABC.Sup

37、posealsothatwehaveFDsCDandDE.AccordingtoAlgorithm,theclosureshouldbecomeABCDE.TakingtheFDCDandaugmentingbothsideswithattributesABwegettheFDABCABD.WecanusethesplittingmethodinsteponetogettheFDABCD.SinceDisnotintheclosure,wecanaddattributeD.TakingtheFDDEandaugmentingbothsideswithattributesABCwegettheF

38、DABCDABCDE.UsingagainthesplittingmethodinsteponewegettheFDABCDE.SineeEisnotintheclosure,wecanaddattributeE.GivenasetofFDs,wecanprovethataFDFfollowsbytakingtheclosureoftheleftsideofFDF.ThestepstocomputetheclosureinAlgorithmcanbemimickedbyArmstrongsaxiomsandthuswecanproveFfromthegivensetofFDsusingArms

39、trongsaxioms.Exercise3.3.1aInthesolutiontoExercise3.2.1wefoundthatthereare14nontrivialdependencies,includingthethreegivenonesandelevenderiveddependencies.Theyare:CA,CD,DA,ABD,ABC,ACD,BCA,BCD,BDA,BDC,CDA,ABCD,ABDC,andBCDA.WealsolearnedthatthethreekeyswereAB,BC,andBD.Thus,anydependencyabovethatdoesnot

40、haveoneofthesepairsontheleftisaBCNFviolation.Theseare:CA,CD,DA,ACD,andCDA.OnechoiceistodecomposeusingtheviolationCD.UsingtheaboveFDs,wegetACDandBCasdecomposedrelations.BCissurelyinBCNF,sinceanytwo-attributerelationis.UsingAlgorithmtodiscovertheprojectionofFDsonrelationACD,wediscoverthatACDisnotinBCN

41、FsineeCisitsonlykey.However,DAisadependencythatholdsinABCDandthereforeholdsinACD.WemustfurtherdecomposeACDintoADandCD.Thus,thethreerelationsofthedecompositionareBC,AD,andCD.Exercise3.3.1bBycomputingtheclosuresofall15nonemptysubsetsofABCD,wecanfindallthenontrivialFDs.TheyareBC,BD,ABC,ABD,BCD,BDC,ABCD

42、andABDC.FromtheclosureswecanalsodeducethattheonlykeyisAB.Thus,anydependencyabovethatdoesnotcontainABontheleftisaBCNFviolation.Theseare:BC,BD,BCDandBDC.D,CDA,B,BC,B,ACD,BandBCDA.C.UsingtheaboveFDs,wegetBCDandOnechoiceistodecomposeusingtheviolationBABasdecomposedrelations.ABissurelyinBCNF,sinceanytwo-

43、attributerelationis.UsingAlgorithmtodiscovertheprojectionofFDsonrelationBCD,wediscoverthatBCDisinBCNFsinceBisitsonlykeyandtheprojectedFDsallhaveBontheleftside.ThusthetworelationsofthedecompositionareABandBCD.Exercise3.3.1cInthesolutiontoExercise3.2.2(ii),wefoundthatthereare12nontrivialdependencies,i

44、ncludingthefourgivenonesandtheeightderivedones.TheyareABC,BCADB,ABD,ADC,BCA,CDB,ABCD,ABDC,ACDBandBCDA.WealsofoundoutthatthekeysareAB,AD,BC,andCD.Thus,anydependencyabovethatdoesnothaveoneofthesepairsontheleftisaBCNFviolation.However,alloftheFDscontainakeyontheleftsotherearenoBCNFviolations.Nodecompos

45、itionisnecessarysincealltheFDsdonotviolateBCNF.Exercise3.3.1dInthesolutiontoExercise3.2.2(iii),wefoundthatthereare28nontrivialdependencies,includingthefourgivenonesandthe24derivedones.TheyareACD,DA,AC,AD,BD,BA,CA,CB,DB,DC,ABC,ABD,ACADB,ADC,BCA,BCD,BDA,BDC,CDA,CDB,ABCD,ABDC,ACDWealsofoundoutthattheke

46、ysareA,B,C,D.Thus,anydependencyabovethatdoesnothaveoneoftheseattributesontheleftisaBCNFviolation.However,alloftheFDscontainakeyontheleftsotherearenoBCNFviolations.NodecompositionisnecessarysincealltheFDsdonotviolateBCNF.Exercise3.3.1eBycomputingtheclosuresofall31nonemptysubsetsofABCDE,wecanfindallth

47、enontrivialFDs.TheyareABC,DEC,BD,ABD,BCD,BEC,BED,ABCD,ABDC,ABEC,ABED,ADEC,BCED,BDEC,ABCED,andABDEC.FromtheclosureswecanalsodeducethattheonlykeyisABE.Thus,anydependencyabovethatdoesnotcontainABEontheleftisaBCNFviolation.Theseare:ABABEontheleftisaBCNFviolation.Theseare:ABC,DEC,BD,ABD,BCD,BEC,BED,ABCD,

48、ABDC,ADEC,BCEDandBDEC.OnechoiceistodecomposeusingtheviolationABC.UsingtheaboveFDs,wegetABCDandABEasdecomposedrelations.UsingAlgorithmtodiscovertheprojectionofFDsonrelationABCD,wediscoverthatABCDisnotinBCNFsinceABisitsonlykeyandtheFDBDfollowsforABCD.UsingviolationBDtofurtherdecompose,wegetBDandABCasd

49、ecomposedrelations.BDisinBCNFbecauseitisatwo-attributerelation.UsingAlgorithmagain,wediscoverthatABCisinBCNFsinceABistheonlykeyandABCistheonlynontrivialFD.GoingbacktorelationABE,followingAlgorithmtellsusthatABEisinBCNFbecausetherearenokeysandnonontrivialFDs.Thusthethreerelationsofthedecompositionare

50、ABC,BDandABE.Exercise3.3.1fBycomputingtheclosuresofall31nonemptysubsetsofABCDE,wecanfindallthenontrivialFDs.Theyare:CB,CD,CE,DB,DE,ABC,ABD,ABE,ACB,ACD,ACE,ADB,ADC,ADE,BCD,BCE,BDE,CDB,CDE,CEB,CED,DEB,ABCD,ABCE,ABDC,ABDE,ABEC,ABED,ACDB,ACDE,ACEB,ACED,ADEB,ADEC,BCDE,BCED,CDEB,ABCDE,ABCED,ABDECandACDEB.

51、FromtheclosureswecanalsodeducethatthekeysareAB,ACandAD.Thus,anydependencyabovethatdoesnotcontainoneoftheabovepairsontheleftisaBCNFviolation.Theseare:CB,CD,CE,DB,DE,BCD,BCE,BDE,CDB,CDE,CEB,CED,DEB,BCDE,BCEDandCDEB.B.UsingtheaboveFDs,wegetBDEandOnechoiceistodecomposeusingtheviolationDABCasdecomposedre

52、lations.UsingAlgorithmtodiscovertheprojectionofFDsonrelationBDE,wediscoverthatBDEisinBCNFsinceD,BD,DEaretheonlykeysandalltheprojectedFDscontainD,BD,orDEintheleftside.GoingbacktorelationABC,followingAlgorithmtellsusthatABCisnotinBCNFbecausesinceABandACareitsonlykeysandtheFDCBfollowsforABC.Usingviolat

53、ionCBtofurtherdecompose,wegetBCandACasdecomposedrelations.BothBCandACareinBCNFbecausetheyaretwo-attributerelations.ThusthethreerelationsofthedecompositionareBDE,BCandAC.Exercise3.3.2Yes,wewillgetthesameresult.BothABandABChaveAontheleftsideandpartoftheprocessofdecompositioninvolvesfindingA+toformoned

54、ecomposedrelationandAplustherestoftheattributesnotinA+asthesecondrelation.Bothcasesyieldthesamedecomposedrelations.Exercise3.3.3BandABChaveAontheleftsideand+toformonedecomposed+asthesecondrelation.BothBandABChaveAontheleftsideand+toformonedecomposed+asthesecondrelation.BothYes,wewillstillgetthesamer

55、esult.BothApartoftheprocessofdecompositioninvolvesfindingArelationandAplustherestoftheattributesnotinAcasesyieldthesamedecomposedrelations.Exercise3.3.4ThisistakenfromExamplepg.95.SupposethataninstaneeofrelationRonlycontainstwotuples.ABC123425TheprojectionsofRontotherelationswithschemasA,BandB,Care:

56、AB1242BC2325Ifwedoanaturaljoinonthetwoprojections,wewillget:ABC123125423425TheresultofthenaturaljoinisnotequaltotheoriginalrelationR.Exercise3.4.1aA.A.Thisistheinitialtableau:ABCDEabcd1e1a1bcde1ab1cd1eThisisthefinaltableauafterapplyingFDsBEandCEABCDEabcd1e1abcde1ab1cd1eSincethereisnotanunsubscripted

57、row,thedecompositionforRisnotlosslessforthissetofFDs.WecanusethefinaltableauasaninstaneeofRasanexampleforwhythejoinisnotlossless.Theprojectedrelationsare:ABCabcab1cBCDbcd1bcdb1cd1ACEace1aceThejoinedrelationis:ABCDEabcd1e1abcde1ab1cd1e1abcd1eabcdeab1cd1eThejoinedrelationhasthreemoretuplesthantheorigi

58、naltableau.Exercise3.4.1bThisistheinitialtableau:ABCDEabcd1e1a1bcde1ab1cd1eThisisthefinaltableauafterapplyingFDsACEandBCDABCDEabcdea1bcde1ab1cd1eSincethereisanunsubscriptedrow,thedecompositionforRislosslessforthissetofFDs.Exercise3.4.1cThisistheinitialtableau:ABCDEabcd1e1a1bcde1ab1cd1eThisisthefinal

59、tableauafterapplyingFDsAD,DEandBD.ABCDEabcdea1bcdeab1cdeSincethereisanunsubscriptedrow,thedecompositionforRislosslessforthissetofFDs.Exerciseistheinitialtableau:ABCDEabcd1e1a1bcde1ab1cd1eThisisthefinaltableauafterapplyingFDsAD,CDEandEDABCDEabcdea1bcdeab1cdeSincethereisanunsubscriptedrow,thedecomposi

60、tionforRislosslessforthissetofFDs.ExercisewedecomposearelationintoBCNF,wewillprojecttheFDsontothedecomposedrelationstogetnewsetsofFDs.ThesedependenciesarepreservediftheunionofthesenewsetsisequivalenttotheoriginalsetofFDs.FortheFDsofthedependenciesarenotpreserved.TheunionofthenewsetsofFDsisCEA.However,theFDBEisnotintheunionandcannotbederived.ThusthetwosetsofFDsarenotequivalent.FortheFDsofthedependenciesarepreserved.TheunionofthenewsetsofFDsisACEandBCD.ThisispreciselythesameastheoriginalsetofFDsandthusthetwosetsofFDsareequivalent.FortheFDsofthedependenciesarenotpreserved.Theunionofthenewsetsof