数据库系统基本教学教程第三章答案解析

**Exercise3.1.1Answersforthisexercisemayvarybecauseofdifferentinterpretations.感谢阅读SomepossibleFDs:SocialSecuritynumbername谢谢阅读AreacodestateStreetaddress,city,statezipcode谢谢阅读Possiblekeys:{SocialSecuritynumber,streetaddress,city,state,areacode,phonenumber}谢谢阅读Needstreetaddress,city,statetouniquelydeterminelocation.Apersoncouldhavemultipleaddresses.Thesameistrueforphones.Thesedays,apersoncouldhavealandlineandacellularphone精品文档放心下载Exercise3.1.2Answersforthisexercisemayvarybecauseofdifferentinterpretations精品文档放心下载SomepossibleFDs:IDx-position,y-position,z-position感谢阅读IDx-velocity,y-velocity,z-velocity感谢阅读x-position,y-position,z-positionID感谢阅读Possiblekeys:{ID}{x-position,y-position,z-position}感谢阅读Thereasonwhythepositionswouldbeakeyisnotwomoleculescanoccupythesamepoint.精品文档放心下载Exercise3.1.3aThesuperkeysareanysubsetthatcontainsA1.Thus,thereare2(n-1)suchsubsets,sinceeachofthen-1attributesA2throughAnmayindependentlybechoseninorout.谢谢阅读**Exercise3.1.3bThesuperkeysareanysubsetthatcontainsA1orA2.Thereare2(n-1)suchsubsetswhenconsideringA1andthen-1attributesA2throughAn.Thereare2(n-2)suchsubsetswhenconsideringA2andthen-2attributesA3throughAn.WedonotcountA1inthesesubsetsbecausetheyarealreadycountedinthefirstgroupofsubsets.Thetotalnumberofsubsets谢谢阅读is2(n-1)+2(n-2).Exercise3.1.3cThesuperkeysareanysubsetthatcontains{A1,A2}or{A3,A4}.Thereare2(n-2)suchsubsetswhenconsidering{A1,A2}andthen-2attributesA3throughAn.Thereare2(n-2)–2(n-4)suchsubsetswhenconsidering{A3,A4}andattributesA5throughAnalongwiththeindividualattributesA1andA2.Wegetthe2(n-4)termbecausewehavetodiscardthesubsetsthatcontainthekey{A1,A2}toavoiddoublecounting.Thetotalnumberofsubsetsis2(n-2)+2(n-2)精品文档放心下载–2(n-4).Exercise3.1.3dThesuperkeysareanysubsetthatcontains{A1,A2}or{A1,A3}.Thereare2(n-2)suchsubsetswhenconsidering{A1,A2}andthen-2attributesA3throughAn.Thereare2(n-3)suchsubsetswhenconsidering{A1,A3}andthen-3attributesA4throughAnWedonotcountA2inthesesubsetsbecausetheyarealreadycountedinthefirstgroupofsubsets.Thetotalnumberofsubsetsis2(n-2)+2(n-3).谢谢阅读Exercise3.2.1aWecouldtryinferencerulestodeducenewdependenciesuntilwearesatisfiedwehavethemall.Amoresystematicwayistoconsidertheclosuresofall15nonemptysetsofattributes.精品文档放心下载Forthesingleattributeswehave{A}+=A,{B}+=B,{C}+=ACD,and{D}+=AD.Thus,theonlynewdependencywegetwithasingleattributeontheleftisCA.感谢阅读**Nowconsiderpairsofattributes:精品文档放心下载{AB}+=ABCD,sowegetnewdependencyABD.{AC}+=ACD,andACDisnontrivial.{AD}+=AD,sonothingnew.{BC}+=ABCD,sowegetBCA,andBCD.{BD}+=ABCD,givingusBDAandBDC.{CD}+=ACD,givingCDA.感谢阅读Forthetriplesofattributes,{ACD}+=ACD,buttheclosuresoftheothersetsareeachABCD.谢谢阅读Thus,wegetnewdependenciesABCD,ABDC,andBCDA.感谢阅读Since{ABCD}+=ABCD,wegetnonewdependencies.感谢阅读Thecollectionof11newdependenciesmentionedaboveare:精品文档放心下载CA,ABD,ACD,BCA,BCD,BDA,BDC,CDA,ABCD,ABDC,andBCDA.谢谢阅读Exercise3.2.1bFromtheanalysisofclosuresabove,wefindthatAB,BC,andBDarekeys.AllothersetseitherdonothaveABCDastheclosureorcontainoneofthesethreesets.谢谢阅读Exercise3.2.1cThesuperkeysareallthosethatcontainoneofthosethreekeys.Thatis,asuperkeythatisnotakeymustcontainBandmorethanoneofA,C,andD.Thus,the(proper)superkeysareABC,ABD,BCD,andABCD.谢谢阅读Exercise3.2.2aForthesingleattributeswehave{A}+=ABCD,{B}+=BCD,{C}+=C,and{D}+=D.Thus,thenewdependenciesareACandAD.感谢阅读Nowconsiderpairsofattributes:感谢阅读{AB}+=ABCD,{AC}+=ABCD,{AD}+=ABCD,{BC}+=BCD,{BD}+=BCD,{CD}+=CD.ThusthenewdependenciesareABC,ABD,ACB,ACD,ADB,ADC,BCDandBDC.感谢阅读**Forthetriplesofattributes,{BCD}+=BCD,buttheclosuresoftheothersetsareeachABCD.谢谢阅读Thus,wegetnewdependenciesABCD,ABDC,andACDB.精品文档放心下载Since{ABCD}+=ABCD,wegetnonewdependencies.谢谢阅读Thecollectionof13newdependenciesmentionedaboveare:谢谢阅读AC,AD,ABC,ABD,ACB,ACD,ADB,ADC,BCD,BDC,ABCD,ABDCandACDB.精品文档放心下载Forthesingleattributeswehave{A}+=A,{B}+=B,{C}+=C,and{D}+=D.Thus,therearenonewdependencies.感谢阅读Nowconsiderpairsofattributes:感谢阅读{AB}+=ABCD,{AC}+=AC,{AD}+=ABCD,{BC}+=ABCD,{BD}+=BD,{CD}+=ABCD.ThusthenewdependenciesareABD,ADC,BCAandCDB.谢谢阅读Forthetriplesofattributes,alltheclosuresofthesetsareeachABCD.Thus,wegetnewdependenciesABCD,ABDC,ACDBandBCDA.感谢阅读Since{ABCD}+=ABCD,wegetnonewdependencies.感谢阅读Thecollectionof8newdependenciesmentionedaboveare:谢谢阅读ABD,ADC,BCA,CDB,ABCD,ABDC,ACDBandBCDA.谢谢阅读Forthesingleattributeswehave{A}+=ABCD,{B}+=ABCD,{C}+=ABCD,and{D}+=ABCD.Thus,thenewdependenciesareAC,AD,BD,BA,CA,CB,DBandDC.谢谢阅读Sinceallthesingleattributes’closuresareABCD,anysupersetofthesingleattributeswillalsoleadtoaclosureofABCD.Knowingthis,wecanenumeratetherestofthenewdependencies.精品文档放心下载Thecollectionof24newdependenciesmentionedaboveare:谢谢阅读AC,AD,BD,BA,CA,CB,DB,DC,ABC,ABD,ACB,ACD,ADB,ADC,BCA,BCD,BDA,BDC,CDA,CDB,ABCD,ABDC,ACDBandBCDA.感谢阅读**Exercise3.2.2bFromtheanalysisofclosuresin3.2.2a(i),wefindthattheonlykeyisA.AllothersetseitherdonothaveABCDastheclosureorcontainA.感谢阅读Fromtheanalysisofclosures3.2.2a(ii),wefindthatAB,AD,BC,andCDarekeys.AllothersetseitherdonothaveABCDastheclosureorcontainoneofthesefoursets.谢谢阅读Fromtheanalysisofclosures3.2.2a(iii),wefindthatA,B,CandDarekeys.AllothersetseitherdonothaveABCDastheclosureorcontainoneofthesefoursets.精品文档放心下载Exercise3.2.2cThesuperkeysareallthosesetsthatcontainoneofthekeysin3.2.2b(i).ThesuperkeysareAB,AC,AD,ABC,ABD,ACD,BCDandABCD.谢谢阅读Thesuperkeysareallthosesetsthatcontainoneofthekeysin3.2.2b(ii).ThesuperkeysareABC,ABD,ACD,BCDandABCD.感谢阅读Thesuperkeysareallthosesetsthatcontainoneofthekeysin3.2.2b(iii).ThesuperkeysareAB,AC,AD,BC,BD,CD,ABC,ABD,ACD,BCDandABCD.精品文档放心下载Exercise3.2.3aSinceA1A2…AnCcontainsA1A2…An,thentheclosureofA1A2…AnCcontainsB.ThusitfollowsthatA1A2…AnCB.感谢阅读Exercise3.2.3bFrom3.2.3a,weknowthatA1A2…AnCB.Usingtheconceptoftrivialdependencies,wecanshowthatA1A2…AnCC.ThusA1A2…AnCBC.谢谢阅读Exercise3.2.3c**FromA1A2…AnE1E2…Ej,weknowthattheclosurecontainsB1B2…BmbecauseoftheFDA1A2…AnB1B2…Bm.TheB1B2…BmandtheE1E2…EjcombinetoformtheC1C2…Ck.ThustheclosureofA1A2…AnE1E2…EjcontainsDaswell.Thus,A1A2…AnE1E2…EjD.感谢阅读Exercise3.2.3dFromA1A2…AnC1C2…Ck,weknowthattheclosurecontainsB1B2…BmbecauseoftheFDA1A2…AnB1B2…Bm.TheC1C2…CkalsotellusthattheclosureofA1A2…AnC1C2…CkcontainsD1D2…Dj.Thus,A1A2…AnC1C2…CkB1B2…BkD1D2…Dj.谢谢阅读Exercise3.2.4aIfattributeArepresentedSocialSecurityNumberandBrepresentedaperson’sname,thenwewouldassumeABbutBAwouldnotbevalidbecausetheremaybemanypeoplewiththesamenameanddifferentSocialSecurityNumbers.谢谢阅读Exercise3.2.4bLetattributeArepresentSocialSecurityNumber,BrepresentgenderandCrepresentname.谢谢阅读SurelySocialSecurityNumberandgendercanuniquelyidentifyaperson’sname(i.e.谢谢阅读ABC).ASocialSecurityNumbercanalsouniquelyidentifyaperson’sname(i.e.AC).感谢阅读However,genderdoesnotuniquelydetermineaname(i.e.BCisnotvalid).精品文档放心下载Exercise3.2.4cLetattributeArepresentlatitudeandBrepresentlongitude.Together,bothattributescanuniquelydetermineC,apointontheworldmap(i.e.ABC).However,neitherAnorBcanuniquelyidentifyapoint(i.e.ACandBCarenotvalid).谢谢阅读Exercise3.2.5GivenarelationwithattributesA1A2…An,wearetoldthattherearenofunctionaldependenciesoftheformB1B2…Bn-1CwhereB1B2…Bn-1isn-1oftheattributesfromA1A2…AnandCistheremainingattributefromA1A2…An.Inthiscase,thesetB1B2…Bn-1andanysubsetdonotfunctionallydetermineC.Thustheonlyfunctionaldependenciesthatwecan感谢阅读**makeareoneswhereCisonboththeleftandrighthandsides.AllofthesefunctionaldependencieswouldbetrivialandthustherelationhasnonontrivialFD’s.谢谢阅读Exercise3.2.6Let’sprovethisbyusingthecontrapositive.WewishtoshowthatifX+isnotasubsetofY+,thenitmustbethatXisnotasubsetofY.谢谢阅读IfX+isnotasubsetofY+,theremustbeattributesA1A2…AninX+thatarenotinY+.IfanyoftheseattributeswereoriginallyinX,thenwearedonebecauseYdoesnotcontainanyoftheA1A2…An.However,iftheA1A2…Anwereaddedbytheclosure,thenwemustexaminethecasefurther.AssumethattherewassomeFDC1C2…CmA1A2…AjwhereA1A2…AjissomesubsetofA1A2…An.ItmustbethenthatC1C2…CmorsomesubsetofC1C2…CmisinX.However,theattributesC1C2…CmcannotbeinYbecauseweassumedthatattributesA1A2…AnareonlyinX+andarenotinY+.Thus,XisnotasubsetofY.谢谢阅读Byprovingthecontrapositive,wehavealsoprovedifX⊆Y,thenX+⊆Y+.感谢阅读Exercise3.2.7ThealgorithmtofindX+isoutlinedonpg.76.Usingthatalgorithm,wecanprovethat(X+)+=X+.Wewilldothisbyusingaproofbycontradiction.谢谢阅读Supposethat(X+)+≠X+.Thenfor(X+)+,itmustbethatsomeFDallowedadditionalattributestobeaddedtotheoriginalsetX+.Forexample,X+AwhereAissomeattributenotinX+.However,ifthiswerethecase,thenX+wouldnotbetheclosureofX.TheclosureofXwouldhavetoincludeAaswell.ThiscontradictsthefactthatweweregiventheclosureofX,X+.Therefore,itmustbethat(X+)+=X+orelseX+isnottheclosureofX.精品文档放心下载Exercise3.2.8aIfallsetsofattributesareclosed,thentherecannotbeanynontrivialfunctionaldependencies.SupposeA1A2...AnBisanontrivialdependency.Then{A1A2...An}+containsBandthusA1A2...Anisnotclosed.精品文档放心下载**Exercise3.2.8bIftheonlyclosedsetsareøand{A,B,C,D},thenthefollowingFDshold:感谢阅读ABBACADAABCABDACBACDADBADCBCABCDBDABDCCDACDBABCDABDCACDBBCDA感谢阅读

AC ADBC BDCB CDDB DCExercise3.2.8cIftheonlyclosedsetsareø,{A,B}and{A,B,C,D},thenthefollowingFDshold:精品文档放心下载ABBACA CB CDDA DB DCACBACDADBADCBCABCDBDABDCCDACDBABCDABDCACDB**BCDAExercise3.2.9WecanthinkofthisproblemasasituationwheretheattributesA,B,Crepresentcitiesandthefunctionaldependenciesrepresentonewaypathsbetweenthecities.Theminimalbasesaretheminimalnumberofpathwaysthatareneededtoconnectthecities.Wedonotwanttocreateanotherroadwayifthetwocitiesarealreadyconnected.谢谢阅读Thesystematicwaytodothiswouldbetocheckallpossiblesetsofthepathways.However,wecansimplifythesituationbynotingthatittakesmorethantwopathwaystovisitthetwoothercitiesandcomeback.Also,ifwefindasetofpathwaysthatisminimal,addingadditionalpathwayswillnotcreateanotherminimalset.精品文档放心下载Thetwosetsofminimalbasesthatweregiveninexample3.11are:感谢阅读{AB,BC,CA}{AB,BA,BC,CB}Theadditionalsetsofminimalbasesare:感谢阅读{CB,BA,AC}{AB,AC,BA,CA}{AC,BC,CA,CB}Exercise3.2.10aWeneedtocomputetheclosuresofallsubsetsof{ABC},althoughthereisnoneedtothink精品文档放心下载abouttheemptysetorthesetofallthreeattributes.Herearethecalculationsforthe感谢阅读remainingsixsets:{A}+=A{B}+=B{C}+=ACE{AB}+=ABCDE{AC}+=ACE**{BC}+=ABCDEWeignoreDandE,soabasisfortheresultingfunctionaldependenciesforABCis:CAandABC.NotethatBC->Aistrue,butfollowslogicallyfromC->A,andthereforemaybeomittedfromourlist.谢谢阅读Exercise3.2.10bWeneedtocomputetheclosuresofallsubsetsof{ABC},althoughthereisnoneedtothink感谢阅读abouttheemptysetorthesetofallthreeattributes.Herearethecalculationsforthe谢谢阅读remainingsixsets:{A}+=AD{B}+=B{C}+=C{AB}+=ABDE{AC}+=ABCDE{BC}+=BCWeignoreDandE,soabasisfortheresultingfunctionaldependenciesforABCis:ACB.感谢阅读Exercise3.2.10cWeneedtocomputetheclosuresofallsubsetsof{ABC},althoughthereisnoneedtothink谢谢阅读abouttheemptysetorthesetofallthreeattributes.Herearethecalculationsforthe谢谢阅读remainingsixsets:{A}+=A{B}+=B{C}+=C{AB}+=ABD{AC}+=ABCDE{BC}+=ABCDEWeignoreDandE,soabasisfortheresultingfunctionaldependenciesforABCis:ACBandBCA.感谢阅读**Exercise3.2.10dWeneedtocomputetheclosuresofallsubsetsof{ABC},althoughthereisnoneedtothink谢谢阅读abouttheemptysetorthesetofallthreeattributes.Herearethecalculationsforthe感谢阅读remainingsixsets:{A}+=ABCDE{B}+=ABCDE{C}+=ABCDE{AB}+=ABCDE{AC}+=ABCDE{BC}+=ABCDEWeignoreDandE,soabasisfortheresultingfunctionaldependenciesforABCis:AB,BCandCA.谢谢阅读Exercise3.2.11ForsteponeofAlgorithm3.7,supposewehavetheFDABCDE.WewanttouseArmstrong’sAxiomstoshowthatABCDandABCEfollow.SurelythefunctionaldependenciesDEDandDEEholdbecausetheyaretrivialandfollowthereflexivityproperty.Usingthetransitivityrule,wecanderivetheFDABCDfromtheFDsABCDEandDED.Likewise,wecandothesameforABCDEandDEEandderivetheFDABCE.精品文档放心下载ForstepstwothroughfourofAlgorithm3.7,supposewehavetheinitialsetofattributesoftheclosureasABC.SupposealsothatwehaveFDsCDandDE.AccordingtoAlgorithm3.7,theclosureshouldbecomeABCDE.TakingtheFDCDandaugmentingbothsideswithattributesABwegettheFDABCABD.WecanusethesplittingmethodinsteponetogettheFDABCD.SinceDisnotintheclosure,wecanaddattributeD.TakingtheFDDEandaugmentingbothsideswithattributesABCwegettheFDABCDABCDE.UsingagainthesplittingmethodinsteponewegettheFDABCDE.SinceEisnotintheclosure,wecanaddattributeE.精品文档放心下载GivenasetofFDs,wecanprovethataFDFfollowsbytakingtheclosureoftheleftsideofFDF.ThestepstocomputetheclosureinAlgorithm3.7canbemimickedbyArmstrong’saxiomsandthuswecanproveFfromthegivensetofFDsusingArmstrong’saxioms.精品文档放心下载**Exercise3.3.1aInthesolutiontoExercise3.2.1wefoundthatthereare14nontrivialdependencies,includingthethreegivenonesandelevenderiveddependencies.Theyare:CA,CD,DA,ABD,ABC,ACD,BCA,BCD,BDA,BDC,CDA,ABCD,ABDC,andBCDA.精品文档放心下载WealsolearnedthatthethreekeyswereAB,BC,andBD.Thus,anydependencyabovethatdoesnothaveoneofthesepairsontheleftisaBCNFviolation.Theseare:CA,CD,DA,ACD,andCDA.谢谢阅读OnechoiceistodecomposeusingtheviolationCD.UsingtheaboveFDs,wegetACDandBCasdecomposedrelations.BCissurelyinBCNF,sinceanytwo-attributerelationis.UsingAlgorithm3.12todiscovertheprojectionofFDsonrelationACD,wediscoverthatACDisnotinBCNFsinceCisitsonlykey.However,DAisadependencythatholdsinABCDandthereforeholdsinACD.WemustfurtherdecomposeACDintoADandCD.Thus,thethreerelationsofthedecompositionareBC,AD,andCD.感谢阅读Exercise3.3.1bBycomputingtheclosuresofall15nonemptysubsetsofABCD,wecanfindallthenontrivialFDs.TheyareBC,BD,ABC,ABD,BCD,BDC,ABCDandABDC.FromtheclosureswecanalsodeducethattheonlykeyisAB.Thus,anydependencyabovethatdoesnotcontainABontheleftisaBCNFviolation.Theseare:BC,BD,BCDandBDC.精品文档放心下载OnechoiceistodecomposeusingtheviolationBC.UsingtheaboveFDs,wegetBCDandABasdecomposedrelations.ABissurelyinBCNF,sinceanytwo-attributerelationis.UsingAlgorithm3.12todiscovertheprojectionofFDsonrelationBCD,wediscoverthatBCDisinBCNFsinceBisitsonlykeyandtheprojectedFDsallhaveBontheleftside.ThusthetworelationsofthedecompositionareABandBCD.精品文档放心下载Exercise3.3.1c**InthesolutiontoExercise3.2.2(ii),wefoundthatthereare12nontrivialdependencies,includingthefourgivenonesandtheeightderivedones.TheyareABC,BCD,CDA,ADB,ABD,ADC,BCA,CDB,ABCD,ABDC,ACDBandBCDA.谢谢阅读WealsofoundoutthatthekeysareAB,AD,BC,andCD.Thus,anydependencyabovethatdoesnothaveoneofthesepairsontheleftisaBCNFviolation.However,alloftheFDscontainakeyontheleftsotherearenoBCNFviolations.感谢阅读NodecompositionisnecessarysincealltheFDsdonotviolateBCNF.谢谢阅读Exercise3.3.1dInthesolutiontoExercise3.2.2(iii),wefoundthatthereare28nontrivialdependencies,精品文档放心下载includingthefourgivenonesandthe24derivedones.TheyareAB,BC,CD,DA,感谢阅读AC,AD,BD,BA,CA,CB,DB,DC,ABC,ABD,ACB,ACD,ADB,ADC,感谢阅读BCA,BCD,BDA,BDC,CDA,CDB,ABCD,ABDC,ACDBandBCDA.精品文档放心下载WealsofoundoutthatthekeysareA,B,C,D.Thus,anydependencyabovethatdoesnothaveoneoftheseattributesontheleftisaBCNFviolation.However,alloftheFDscontainakeyontheleftsotherearenoBCNFviolations.精品文档放心下载NodecompositionisnecessarysincealltheFDsdonotviolateBCNF.感谢阅读Exercise3.3.1eBycomputingtheclosuresofall31nonemptysubsetsofABCDE,wecanfindallthenontrivialFDs.TheyareABC,DEC,BD,ABD,BCD,BEC,BED,ABCD,ABDC,ABEC,ABED,ADEC,BCED,BDEC,ABCED,andABDEC.FromtheclosureswecanalsodeducethattheonlykeyisABE.Thus,anydependencyabovethatdoesnotcontainABEontheleftisaBCNFviolation.Theseare:ABC,DEC,BD,ABD,BCD,BEC,BED,ABCD,ABDC,ADEC,BCEDandBDEC.精品文档放心下载OnechoiceistodecomposeusingtheviolationABC.UsingtheaboveFDs,wegetABCDandABEasdecomposedrelations.UsingAlgorithm3.12todiscovertheprojectionofFDsonrelationABCD,wediscoverthatABCDisnotinBCNFsinceABisitsonlykeyandtheFD感谢阅读**BDfollowsforABCD.UsingviolationBDtofurtherdecompose,wegetBDandABCasdecomposedrelations.BDisinBCNFbecauseitisatwo-attributerelation.UsingAlgorithm3.12again,wediscoverthatABCisinBCNFsinceABistheonlykeyandABCistheonlynontrivialFD.GoingbacktorelationABE,followingAlgorithm3.12tellsusthatABEisinBCNFbecausetherearenokeysandnonontrivialFDs.ThusthethreerelationsofthedecompositionareABC,BDandABE.精品文档放心下载Exercise3.3.1fBycomputingtheclosuresofall31nonemptysubsetsofABCDE,wecanfindallthe精品文档放心下载nontrivialFDs.Theyare:CB,CD,CE,DB,DE,ABC,ABD,ABE,ACB,ACD,感谢阅读ACE,ADB,ADC,ADE,BCD,BCE,BDE,CDB,CDE,CEB,CED,DEB,感谢阅读ABCD,ABCE,ABDC,ABDE,ABEC,ABED,ACDB,ACDE,ACEB,ACED,精品文档放心下载ADEB,ADEC,BCDE,BCED,CDEB,ABCDE,ABCED,ABDECandACDEB.From精品文档放心下载theclosureswecanalsodeducethatthekeysareAB,ACandAD.Thus,anydependency精品文档放心下载abovethatdoesnotcontainoneoftheabovepairsontheleftisaBCNFviolation.Theseare:谢谢阅读CB,CD,CE,DB,DE,BCD,BCE,BDE,CDB,CDE,CEB,CED,DEB,精品文档放心下载BCDE,BCEDandCDEB.OnechoiceistodecomposeusingtheviolationDB.UsingtheaboveFDs,wegetBDEandABCasdecomposedrelations.UsingAlgorithm3.12todiscovertheprojectionofFDsonrelationBDE,wediscoverthatBDEisinBCNFsinceD,BD,DEaretheonlykeysandalltheprojectedFDscontainD,BD,orDEintheleftside.GoingbacktorelationABC,followingAlgorithm3.12tellsusthatABCisnotinBCNFbecausesinceABandACareitsonlykeysandtheFDCBfollowsforABC.UsingviolationCBtofurtherdecompose,wegetBCandACasdecomposedrelations.BothBCandACareinBCNFbecausetheyaretwo-attributerelations.ThusthethreerelationsofthedecompositionareBDE,BCandAC.感谢阅读Exercise3.3.2Yes,wewillgetthesameresult.BothABandABChaveAontheleftsideandpartoftheprocessofdecompositioninvolvesfinding{A}+toformonedecomposedrelationandAplustherestoftheattributesnotin{A}+asthesecondrelation.Bothcasesyieldthesamedecomposedrelations.精品文档放心下载**Exercise3.3.3Yes,wewillstillgetthesameresult.BothABandABChaveAontheleftsideandpartoftheprocessofdecompositioninvolvesfinding{A}+toformonedecomposedrelationandAplustherestoftheattributesnotin{A}+asthesecondrelation.Bothcasesyieldthesamedecomposedrelations.感谢阅读Exercise3.3.4ThisistakenfromExample3.21pg.95.感谢阅读SupposethataninstanceofrelationRonlycontainstwotuples.谢谢阅读ABC123425TheprojectionsofRontotherelationswithschemas{A,B}and{B,C}are:谢谢阅读ABBC12234225Ifwedoanaturaljoinonthetwoprojections,wewillget:精品文档放心下载ABC123125423425TheresultofthenaturaljoinisnotequaltotheoriginalrelationR.谢谢阅读Exercise3.4.1a**Thisistheinitialtableau:感谢阅读ABCDEabcd1e1a1bcde1ab1cd1eThisisthefinaltableauafterapplyingFDsBEandCEA.谢谢阅读ABCDEabcd1e1abcde1ab1cd1eSincethereisnotanunsubscriptedrow,thedecompositionforRisnotlosslessforthissetofFDs.感谢阅读WecanusethefinaltableauasaninstanceofRasanexampleforwhythejoinisnotlossless.谢谢阅读Theprojectedrelationsare:感谢阅读ABCabcab1cBCDbcd1bcdb1cd1ACEace1aceThejoinedrelationis:**ABCDEabcd1e1abcde1ab1cd1e1abcd1eabcdeab1cd1eThejoinedrelationhasthreemoretuplesthantheoriginaltableau.精品文档放心下载Exercise3.4.1bThisistheinitialtableau:感谢阅读ABCDEabcd1e1a1bcde1ab1cd1eThisisthefinaltableauafterapplyingFDsACEandBCD感谢阅读ABCDEabcdea1bcde1ab1cd1eSincethereisanunsubscriptedrow,thedecompositionforRislosslessforthissetofFDs.感谢阅读Exercise3.4.1cThisistheinitialtableau:感谢阅读ABCDEabcd1e1a1bcde1**a b1 c d1 eThisisthefinaltableauafterapplyingFDsAD,DEandBD.精品文档放心下载ABCDEabcdea1bcdeab1cdeSincethereisanunsubscriptedrow,thedecompositionforRislosslessforthissetofFDs.谢谢阅读Exercise3.4.1dThisistheinitialtableau:精品文档放心下载ABCDEabcd1e1a1bcde1ab1cd1eThisisthefinaltableauafterapplyingFDsAD,CDEandED谢谢阅读ABCDEabcdea1bcdeab1cdeSincethereisanunsubscriptedrow,thedecompositionforRislosslessforthissetofFDs.感谢阅读Exercise3.4.2WhenwedecomposearelationintoBCNF,wewillprojecttheFDsontothedecomposedrelationstogetnewsetsofFDs.ThesedependenciesarepreservediftheunionofthesenewsetsisequivalenttotheoriginalsetofFDs.谢谢阅读**FortheFDsof3.4.1a,thedependenciesarenotpreserved.TheunionofthenewsetsofFDsisCEA.However,theFDBEisnotintheunionandcannotbederived.ThusthetwosetsofFDsarenotequivalent.谢谢阅读FortheFDsof3.4.1b,thedependenciesarepreserved.TheunionofthenewsetsofFDsisACEandBCD.ThisispreciselythesameastheoriginalsetofFDsandthusthetwosetsofFDsareequivalent.谢谢阅读FortheFDsof3.4.1c,thedependenciesarenotpreserved.TheunionofthenewsetsofFDsisBDandAE.TheFDsADandDEarenotintheunionandcannotbederived.ThusthetwosetsofFDsarenotequivalent.精品文档放心下载FortheFDsof3.4.1d,thedependenciesarenotpreserved.TheunionofthenewsetsofFDsisACE.However,theFDsAD,CDEandEDarenotintheunionandcannotbederived.ThusthetwosetsofFDsarenotequivalent.感谢阅读Exercise3.5.1aInthesolutiontoExercise3.3.1awefoundthatthereare14nontrivialdependencies.Theyare:CA,CD,DA,ABD,ABC,ACD,BCA,BCD,BDA,BDC,CDA,ABCD,ABDC,andBCDA.精品文档放心下载WealsolearnedthatthethreekeyswereAB,BC,andBD.SincealltheattributesontherightsidesoftheFDsareprime,thereareno3NFviolations.谢谢阅读Sincethereareno3NFviolations,itisnotnecessarytodecomposetherelation.感谢阅读Exercise3.5.1bInthesolutiontoExercise3.3.1bwefoundthatthereare8nontrivialdependencies.TheyareBC,BD,ABC,ABD,BCD,BDC,ABCDandABDC.精品文档放心下载WealsofoundoutthattheonlykeyisAB.FDswheretheleftsideisnotasuperkeyortheattributesontherightarenotpartofsomekeyare3NFviolations.The3NFviolationsareBC,BD,BCDandBDC.精品文档放心下载**Usingalgorithm3.26,wecandecomposeintorelationsusingtheminimalbasisBCand谢谢阅读BD.TheresultingdecomposedrelationswouldbeBCandBD.However,noneofthesetwosetsofattributesisasuperkey.ThusweaddrelationABtotheresult.ThefinalsetofdecomposedrelationsisBC,BDandAB.精品文档放心下载Exercise3.5.1cInthesolutiontoExercise3.3.1cwefoundthatthereare12nontrivialdependencies.TheyareABC,BCD,CDA,ADB,ABD,ADC,BCA,CDB,ABCD,ABDC,ACDBandBCDA.感谢阅读WealsofoundoutthatthekeysareAB,AD,BC,andCD.SincealltheattributesontherightsidesoftheFDsareprime,thereareno3NFviolations.精品文档放心下载Sincethereareno3NFviolations,itisnotnecessarytodecomposetherelation.感谢阅读Exercise3.5.1dInthesolutiontoExercise3.3.1dwefoundthatthereare28nontrivialdependencies.TheyareAB,BC,CD,DA,AC,AD,BD,BA,CA,CB,DB,DC,ABC,ABD,ACB,ACD,ADB,ADC,BCA,BCD,BDA,BDC,CDA,CDB,ABCD,ABDC,ACDBandBCDA.谢谢阅读WealsofoundoutthatthekeysareA,B,C,D.SincealltheattributesontherightsidesoftheFDsareprime,thereareno3NFviolations.精品文档放心下载Sincethereareno3NFviolations,itisnotnecessarytodecomposetherelation.谢谢阅读Exercise3.5.1eInthesolutiontoExercise3.3.1ewefoundthatthereare16nontrivialdependencies.TheyareABC,DEC,BD,ABD,BCD,BEC,BED,ABCD,ABDC,ABEC,ABED,ADEC,BCED,BDEC,ABCED,andABDEC.谢谢阅读**WealsofoundoutthattheonlykeyisABE.FDswheretheleftsideisnotasuperkeyortheattributesontherightarenotpartofsomekeyare3NFviolations.The3NFviolationsareABC,DEC,BD,ABD,BCD,BEC,BED,ABCD,ABDC,ADEC,BCEDandBDEC.谢谢阅读Usingalgorithm3.26,wecandecomposeintorelationsusingtheminimalbasisABC,感谢阅读DECandBD.TheresultingdecomposedrelationswouldbeABC,CDEandBD.However,noneofthesethreesetsofattributesisasuperkey.ThusweaddrelationABEtotheresult.ThefinalsetofdecomposedrelationsisABC,CDE,BDandABE.精品文档放心下载Exercise3.5.1fInthesolutiontoExercise3.3.1fwefoundthatthereare41nontrivialdependencies.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.精品文档放心下载WealsofoundoutthatthekeysareAB,ACandAD.FDswheretheleftsideisnotasuperkeyortheattributesontherightarenotpartofsomekeyare3NFviolations.The3NFviolationsareCE,DE,BCE,BDE,CDEandBCDE.谢谢阅读Usingalgorithm3.26,wecandecomposeintorelationsusingtheminimalbasisABC,CD,