數(shù)據(jù)庫(kù)系統(tǒng)基礎(chǔ)教程第三章答案

數(shù)據(jù)庫(kù)系統(tǒng)基礎(chǔ)教程第三章答案數(shù)據(jù)庫(kù)系統(tǒng)基礎(chǔ)教程第三章答案數(shù)據(jù)庫(kù)系統(tǒng)基礎(chǔ)教程第三章答案Exercise3、1、1Answersforthisexercisemayvarybecauseofdifferentinterpretations、SomepossibleFDs: SocialSecuritynumbername Areacodestate Streetaddress,city,statezipcodePossiblekeys: {SocialSecuritynumber,streetaddress,city,state,areacode,phonenumber}Needstreetaddress,city,statetouniquelydeterminelocation、Apersoncouldhavemultipleaddresses、Thesameistrueforphones、Thesedays,apersoncouldhavealandlineandacellularphoneExercise3、1、2AnswersforthisexercisemayvarybecauseofdifferentinterpretationsSomepossibleFDs: IDx-position,y-position,z-position IDx-velocity,y-velocity,z-velocity x-position,y-position,z-positionIDPossiblekeys: {ID} {x-position,y-position,z-position}Thereasonwhythepositionswouldbeakeyisnotwomoleculescanoccupythesamepoint、Exercise3、ThesuperkeysareanysubsetthatcontainsA1、Thus,thereare2(n-1)suchsubsets,sinceeachofthen-1attributesA2throughAnmayindependentlybechoseninorout、Exercise3、1、3bThesuperkeysareanysubsetthatcontainsA1orA2、Thereare2(n-1)suchsubsetswhenconsideringA1andthen-1attributesA2throughAn、Thereare2(n-2)suchsubsetswhenconsideringA2andthen-2attributesA3throughAn、WedonotcountA1inthesesubsetsbecausetheyarealreadycountedinthefirstgroupofsubsets、Thetotalnumberofsubsetsis2(n-1)+2(n-2)Exercise3、Thesuperkeysareanysubsetthatcontains{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)–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)Exercise3、Wecouldtryinferencerulestodeducenewdependenciesuntilwearesatisfiedwehavethemall、Amoresystematicwayistoconsidertheclosuresofall15nonemptysetsofattributes、Forthesingleattributeswehave{A}+=A,{B}+=B,{C}+=ACD,and{D}+=AD、Thus,theonlynewdependencywegetwithasingleattributeontheleftisCA、Nowconsiderpairsofattributes:{AB}+=ABCD,sowegetnewdependencyABD、{AC}+=ACD,andACDisnontrivial、{AD}+=AD,sonothingnew、{BC}+=ABCD,sowegetBCA,andBCD、{BD}+=ABCD,givingusBDAandBDC、{CD}+=ACD,givingCDA、Forthetriplesofattributes,{ACD}+=ACD,buttheclosuresoftheothersetsareeachABCD、Thus,wegetnewdependenciesABCD,ABDC,andBCDA、Since{ABCD}+=ABCD,wegetnonewdependencies、Thecollectionof11newdependenciesmentionedaboveare:CA,ABD,ACD,BCA,BCD,BDA,BDC,CDA,ABCD,ABDC,andBCDA、Exercise3、2、1bFromtheanalysisofclosuresabove,wefindthatAB,BC,andBDarekeys、AllothersetseitherdonothaveABCDastheclosureorcontainoneofthesethreesets、Exercise3、Thesuperkeysareallthosethatcontainoneofthosethreekeys、Thatis,asuperkeythatisnotakeymustcontainBandmorethanoneofA,C,andD、Thus,the(proper)superkeysareABC,ABD,BCD,andABCD、Exercise3、i)Forthesingleattributeswehave{A}+=ABCD,{B}+=BCD,{C}+=C,and{D}+=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,wegetnewdependenciesABCD,ABDC,andACDB、Since{ABCD}+=ABCD,wegetnonewdependencies、Thecollectionof13newdependenciesmentionedaboveare:AC,AD,ABC,ABD,ACB,ACD,ADB,ADC,BCD,BDC,ABCD,ABDCandACDB、ii)Forthesingleattributeswehave{A}+=A,{B}+=B,{C}+=C,and{D}+=D、Thus,therearenonewdependencies、Nowconsiderpairsofattributes:{AB}+=ABCD,{AC}+=AC,{AD}+=ABCD,{BC}+=ABCD,{BD}+=BD,{CD}+=ABCD、ThusthenewdependenciesareABD,ADC,BCAandCDB、Forthetriplesofattributes,alltheclosuresofthesetsareeachABCD、Thus,wegetnewdependenciesABCD,ABDC,ACDBandBCDA、Since{ABCD}+=ABCD,wegetnonewdependencies、Thecollectionof8newdependenciesmentionedaboveare:ABD,ADC,BCA,CDB,ABCD,ABDC,ACDBandBCDA、iii)Forthesingleattributeswehave{A}+=ABCD,{B}+=ABCD,{C}+=ABCD,and{D}+=ABCD、Thus,thenewdependenciesareAC,AD,BD,BA,CA,CB,DBandDC、Sinceallthesingleattributes’closuresareABCD,anysupersetofthesingleattributeswillalsoleadtoaclosureofABCD、Knowingthis,wecanenumeratetherestofthenewdependencies、Thecollectionof24newdependenciesmentionedaboveare: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、2bi)Fromtheanalysisofclosuresin3、2、2aii)Fromtheanalysisofclosures3、2、2aiii)Fromtheanalysisofclosures3、2、2aExercise3、i)Thesuperkeysareallthosesetsthatcontainoneofthekeysin3、2、2b(i)、ThesuperkeysareAB,AC,AD,ABC,ABD,ACD,BCDandABCD、ii)Thesuperkeysareallthosesetsthatcontainoneofthekeysin3、2、2b(ii)、ThesuperkeysareABC,ABD,ACD,BCDandABCD、iii)Thesuperkeysareallthosesetsthatcontainoneofthekeysin3、2、2b(iii)、ThesuperkeysareAB,AC,AD,BC,BD,CD,ABC,ABD,ACD,BCDandABCD、Exercise3、SinceA1A2…AnCcontainsA1A2…An,thentheclosureofA1A2…AnCcontainsB、ThusitfollowsthatA1A2…AExercise3、2、3bFrom3、2、3a,weknowthatA1A2…AnCB、Usingtheconceptoftrivialdependencies,wecanshowthatA1A2…AnCC、ThusA1AExercise3、FromA1A2…AnE1E2…Ej,weknowthattheclosurecontainsB1B2…BmbecauseoftheFDA1A2…AnB1B2…Bm、TheB1B2…BmandtheE1E2…EjcombinetoformtheC1C2…Ck、ThustheclosureofA1A2…AnE1E2…EjcontainsDaswell、Thus,A1A2…AnE1E2Exercise3、2、3dFromA1A2…AnC1C2…Ck,weknowthattheclosurecontainsB1B2…BmbecauseoftheFDA1A2…AnB1B2…Bm、TheC1C2…CkalsotellusthattheclosureofA1A2…AnC1C2…CkcontainsD1D2…Dj、Thus,A1A2…AnC1C2…CkB1B2…BkExercise3、IfattributeArepresentedSocialSecurityNumberandBrepresentedaperson’sname,thenwewouldassumeABbutBAwouldnotbevalidbecausetheremaybemanypeoplewiththesamenameanddifferentSocialSecurityNumbers、Exercise3、2、4bLetattributeArepresentSocialSecurityNumber,BrepresentgenderandCrepresentname、SurelySocialSecurityNumberandgendercanuniquelyidentifyaperson’sname(i、e、ABC)、ASocialSecurityNumbercanalsouniquelyidentifyaperson’sname(i、e、AC)、However,genderdoesnotuniquelydetermineaname(i、e、BCisnotvalid)、Exercise3、LetattributeArepresentlatitudeandBrepresentlongitude、Together,bothattributescanuniquelydetermineC,apointontheworldmap(i、e、ABC)、However,neitherAnorBcanuniquelyidentifyapoint(i、e、ACandBCarenotvalid)、Exercise3、2、5GivenarelationwithattributesA1A2…An,wearetoldthattherearenofunctionaldependenciesoftheformB1B2…Bn-1CwhereB1B2…Bn-1isn-1oftheattributesfromA1A2…AnandCistheremainingattributefromA1A2…An、Inthiscase,thesetB1B2…Bn-1andanysubsetdonotfunctionallydetermineC、ThusExercise3、2、6Let’sprovethisbyusingthecontrapositive、WewishtoshowthatifX+isnotasubsetofY+,thenitmustbethatXisnotasubsetofY、IfX+isnotasubsetofY+,theremustbeattributesA1A2…AninX+thatarenotinY+、IfanyoftheseattributeswereoriginallyinX,thenwearedonebecauseYdoesnotcontainanyoftheA1A2…An、However,iftheA1A2…Anwereaddedbytheclosure,thenwemustexaminethecasefurther、AssumethattherewassomeFDC1C2…CmA1A2…AjwhereA1A2…AjissomesubsetofA1A2…An、ItmustbethenthatC1C2…CmorsomesubsetofC1C2…CmisinX、However,theattributesC1C2…CmcannotbeinYbecauseweassumedthatattributesAByprovingthecontrapositive,wehavealsoprovedifX?Y,thenX+?Y+、Exercise3、2、7ThealgorithmtofindX+isoutlinedonpg、76、Usingthatalgorithm,wecanprovethat(X+)+=X+、Wewilldothisbyusingaproofbycontradiction、Supposethat(X+)+≠X+、Thenfor(X+)+,itmustbethatsomeFDallowedadditionalattributestobeaddedtotheoriginalsetX+、Forexample,X+AwhereAissomeattributenotinX+、However,ifthiswerethecase,thenX+wouldnotbetheclosureofX、TheclosureofXwouldhavetoincludeAaswell、ThiscontradictsthefactthatweweregiventheclosureofX,X+、Therefore,itmustbethat(X+)+=X+orelseX+isnottheclosureofX、Exercise3、Ifallsetsofattributesareclosed,thentherecannotbeanynontrivialfunctionaldependencies、SupposeA1A2、、、AnBisanontrivialdependency、Then{A1A2、、、An}+containsBandthusA1A2Exercise3、2、8bIftheonlyclosedsetsare?and{A,B,C,D},thenthefollowingFDshold:AB AC ADBA BC BDCA CB CDDA DB DCABC ABDACB ACDADB ADCBCA BCDBDA BDCCDA CDBABCDABDCACDBBCDAExercise3、Iftheonlyclosedsetsare?,{A,B}and{A,B,C,D},thenthefollowingFDshold:ABBACA CB CDDA DB DCACB ACDADB ADCBCA BCDBDA BDCCDA CDBABCDABDCACDBBCDAExercise3、2、9WecanthinkofthisproblemasasituationwheretheattributesA,B,Crepresentcitiesandthefunctionaldependenciesrepresentonewaypathsbetweenthecities、Theminimalbasesaretheminimalnumberofpathwaysthatareneededtoconnectthecities、Wedonotwanttocreateanotherroadwayifthetwocitiesarealreadyconnected、Thesystematicwaytodothiswouldbetocheckallpossiblesetsofthepathways、However,wecansimplifythesituationbynotingthatittakesmorethantwopathwaystovisitthetwoothercitiesandcomeback、Also,ifwefindasetofpathwaysthatisminimal,addingadditionalpathwayswillnotcreateanotherminimalset、Thetwosetsofminimalbasesthatweregiveninexample3、11are:{AB,BC,CA}{AB,BA,BC,CB}Theadditionalsetsofminimalbasesare:{CB,BA,AC}{AB,AC,BA,CA}{AC,BC,CA,CB}Exercise3、Weneedtocomputetheclosuresofallsubsetsof{ABC},althoughthereisnoneedtothinkabouttheemptysetorthesetofallthreeattributes、Herearethecalculationsfortheremainingsixsets:{A}+=A{B}+=B{C}+=ACE{AB}+=ABCDE{AC}+=ACE{BC}+=ABCDEWeignoreDandE,soabasisfortheresultingfunctionaldependenciesforABCis:CAandABC、NotethatBC->Aistrue,butfollowslogicallyfromC->A,andthereforemaybeomittedfromourlist、Exercise3、2、10bWeneedtocomputetheclosuresofallsubsetsof{ABC},althoughthereisnoneedtothinkabouttheemptysetorthesetofallthreeattributes、Herearethecalculationsfortheremainingsixsets:{A}+=AD{B}+=B{C}+=C{AB}+=ABDE{AC}+=ABCDE{BC}+=BCWeignoreDandE,soabasisfortheresultingfunctionaldependenciesforABCis:ACB、Exercise3、Weneedtocomputetheclosuresofallsubsetsof{ABC},althoughthereisnoneedtothinkabouttheemptysetorthesetofallthreeattributes、Herearethecalculationsfortheremainingsixsets:{A}+=A{B}+=B{C}+=C{AB}+=ABD{AC}+=ABCDE{BC}+=ABCDEWeignoreDandE,soabasisfortheresultingfunctionaldependenciesforABCis:ACBandBCA、Exercise3、2、10dWeneedtocomputetheclosuresofallsubsetsof{ABC},althoughthereisnoneedtothinkabouttheemptysetorthesetofallthreeattributes、Herearethecalculationsfortheremainingsixsets:{A}+=ABCDE{B}+=ABCDE{C}+=ABCDE{AB}+=ABCDE{AC}+=ABCDE{BC}+=ABCDEWeignoreDandE,soabasisfortheresultingfunctionaldependenciesforABCis:AB,BCandCA、Exercise3、2、11ForsteponeofAlgorithm3、7,supposewehavetheFDABCDE、WewanttouseArmstrong’sAxiomstoshowthatABCDandABCEfollow、SurelythefunctionaldependenciesDEDandDEEholdbecausetheyaretrivialandfollowthereflexivityproperty、Usingthetransitivityrule,wecanderivetheFDABCDfromtheFDsABCDEandDED、Likewise,wecandothesameforABCForstepstwothroughfourofAlgorithm3、7,supposewehavetheinitialsetofattributesoftheclosureasABC、SupposealsothatwehaveFDsCDandDE、AccordingtoAlgorithm3、7,theclosureshouldbecomeABCDE、TakingtheFDCDandaugmentingbothsideswithattributesABwegettheFDABCABD、WecanusethesplittingmethodinsteponetogettheFDABCD、SinceDisnotintheclosure,wecanaddattributeD、TakingtheFDDEandaugmentingbothsideswithattributesABCwegettheFDABCDABCDE、UsingagainthesplittingmethodinsteponewegettheFDABCDE、SinceEisnotintheclosure,wecanaddattributeE、GivenasetofFDs,wecanprovethataFDFfollowsbytakingtheclosureoftheleftsideofFDF、ThestepstocomputetheclosureinAlgorithm3、7canbemimickedbyArmstrong’saxiomsandthuswecanproveFfromthegivensetofFDsusingArmstrong’saxioms、Exercise3、InthesolutiontoExercise3、2、1wefoundthatthereare14nontrivialdependencies,includingthethreegivenonesandelevenderiveddependencies、Theyare:CA,CD,DA,ABD,ABC,ACD,BCA,BCD,BDA,BDC,CDA,ABCD,ABDC,andBCDA、WealsolearnedthatthethreekeyswereAB,BC,andBD、Thus,anydependencyabovethatdoesnothaveoneofthesepairsontheleftisaBCNFviolation、Theseare:CA,CD,DA,ACD,andCDA、OnechoiceistodecomposeusingtheviolationCD、UsingtheaboveFDs,wegetACDandBCasdecomposedrelations、BCissurelyinBCNF,sinceanytwo-attributerelationis、UsingAlgorithm3、12todiscovertheprojectionofFDsonrelationACD,wediscoverthatACDisnotinBCNFsinceCisitsonlykey、However,DAisadependencythatholdsinABCDandthereforeholdsinACD、WemustfurtherdecomposeACDintoADandCD、Thus,thethreerelationsofthedecompositionareBC,AD,andCD、Exercise3、3、1bBycomputingtheclosuresofall15nonemptysubsetsofABCD,wecanfindallthenontrivialFDs、TheyareBC,BD,ABC,ABD,BCD,BDC,ABCDandABDC、FromtheclosureswecanalsodeducethattheonlykeyisAB、Thus,anydependencyabovethatdoesnotcontainABontheleftisaBCNFviolation、Theseare:BC,BD,BCDandBDC、OnechoiceistodecomposeusingtheviolationBC、UsingtheaboveFDs,wegetBCDandABasdecomposedrelations、ABissurelyinBCNF,sinceanytwo-attributerelationis、UsingAlgorithm3、12todiscovertheprojectionofFDsonrelationBCD,wediscoverthatBCDisinBCNFsinceBisitsonlykeyandtheprojectedFDsallhaveBontheleftside、ThusthetworelationsofthedecompositionareABandBCD、Exercise3、InthesolutiontoExercise3、2、2(ii),wefoundthatthereare12nontrivialdependencies,includingthefourgivenonesandtheeightderivedones、TheyareABC,BCD,CDA,ADB,ABD,ADC,BCA,CDB,ABCD,ABDC,ACDBandBCDA、WealsofoundoutthatthekeysareAB,AD,BC,andCD、Thus,anydependencyabovethatdoesnothaveoneofthesepairsontheleftisaBCNFviolation、However,alloftheFDscontainakeyontheleftsotherearenoBCNFviolations、NodecompositionisnecessarysincealltheFDsdonotviolateBCNF、Exercise3、3、1dInthesolutiontoExercise3、2、2(iii),wefoundthatthereare28nontrivialdependencies,includingthefourgivenonesandthe24derivedones、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、WealsofoundoutthatthekeysareA,B,C,D、Thus,anydependencyabovethatdoesnothaveoneoftheseattributesontheleftisaBCNFviolation、However,alloftheFDscontainakeyontheleftsotherearenoBCNFviolations、NodecompositionisnecessarysincealltheFDsdonotviolateBCNF、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、OnechoiceistodecomposeusingtheviolationABC、UsingtheaboveFDs,wegetABCDandABEasdecomposedrelations、UsingAlgorithm3、12todiscovertheprojectionofFDsonrelationABCD,wediscoverthatABCDisnotinBCNFsinceABisitsonlykeyandtheFDBDfollowsforABCD、UsingviolationBDtofurtherdecompose,wegetBDandABCasdecomposedrelations、BDisinBCNFbecauseitisatwo-attributerelation、UsingAlgorithm3、12again,wediscoverthatABCisinBCNFsinceABistheonlykeyandABCistheonlynontrivialFD、GoingbacktorelationABE,followingAlgorithm3、12tellsusthatABEisinBCNFbecausetherearenokeysandnonontrivialFDs、ThusthethreerelationsofthedecompositionareABC,BDandABE、Exercise3、Bycomputingtheclosuresofall31nonemptysubsetsofABCDE,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、FromtheclosureswecanalsodeducethatthekeysareAB,ACandAD、Thus,anydependencyabovethatdoesnotcontainoneoftheabovepairsontheleftisaBCNFviolation、Theseare:CB,CD,CE,DB,DE,BCD,BCE,BDE,CDB,CDE,CEB,CED,DEB,BCDE,BCEDandCDEB、OnechoiceistodecomposeusingtheviolationDB、UsingtheaboveFDs,wegetBDEandABCasdecomposedrelations、UsingAlgorithm3、12todiscovertheprojectionofFDsonrelationBDE,wediscoverthatBDEisinBCNFsinceD,BD,DEaretheonlykeysandalltheprojectedFDscontainD,BD,orDEintheleftside、GoingbacktorelationABC,followingAlgorithm3、12tellsusthatABCisnotinBCNFbecausesinceABandACareitsonlykeysandtheFDCBfollowsforABC、UsingviolationCBtofurtherdecompose,wegetBCandACasdecomposedrelations、BothBCandACareinBCNFbecausetheyaretwo-attributerelations、ThusthethreerelationsofthedecompositionareBDE,Exercise3、3、2Yes,wewillgetthesameresult、BothABandABChaveAontheleftsideandpartoftheprocessofdecompositioninvolvesfinding{A}+toformonedecomposedrelationandAplustherestoftheattributesnotin{A}+asthesecondrelation、Bothcasesyieldthesamedecomposedrelations、Exercise3、3、3Yes,wewillstillgetthesameresult、BothABandABChaveAontheleftsideandpartoftheprocessofdecompositioninvolvesfinding{A}+toformonedecomposedrelationandAplustherestoftheattributesnotin{A}+asthesecondrelation、Bothcasesyieldthesamedecomposedrelations、Exercise3、3、4ThisistakenfromExample3、21pg、95、SupposethataninstanceofrelationRonlycontainstwotuples、ABC123425TheprojectionsofRontotherelationswithschemas{A,B}and{B,C}are:AB1242BC2325Ifwedoanaturaljoinonthetwoprojections,wewillget:ABC123125423425TheresultofthenaturaljoinisnotequaltotheoriginalrelationR、Exercise3、Thisistheinitialtableau:ABCDEabcd1e1a1bcde1ab1cd1eThisisthefinaltableauafterapplyingFDsBEandCEA、ABCDEabcd1e1abcde1ab1cd1eSincethereisnotanunsubscriptedrow,thedecompositionforRisnotlosslessforthissetofFDs、WecanusethefinaltableauasaninstanceofRasanexampleforwhythejoinisnotlossless、Theprojectedrelationsare:ABCabcab1cBCDbcd1bcdb1cd1ACEace1aceThejoinedrelationis:ABCDEabcd1e1abcde1ab1cd1e1abcd1eabcdeab1cd1eThejoinedrelationhasthreemoretuplesthantheoriginaltableau、Exercise3、4、1bThisistheinitialtableau:ABCDEabcd1e1a1bcde1ab1cd1eThisisthefinaltableauafterapplyingFDsACEandBCDABCDEabcdea1bcde1ab1cd1eSincethereisanunsubscriptedrow,thedecompositionforRislosslessforthissetofFDs、Exercise3、Thisistheinitialtableau:ABCDEabcd1e1a1bcde1ab1cd1eThisisthefinaltableauafterapplyingFDsAD,DEandBD、ABCDEabcdea1bcdeab1cdeSincethereisanunsubscriptedrow,thedecompositionforRislosslessforthissetofFDs、Exercise3、4、1dThisistheinitialtableau:ABCDEabcd1e1a1bcde1ab1cd1eThisisthefinaltableauafterapplyingFDsAD,CDEandEDABCDEabcdea1bcdeab1cdeSincethereisanunsubscriptedrow,thedecompositionforRislosslessforthissetofFDs、Exercise3、4、2WhenwedecomposearelationintoBCNF,wewillprojecttheFDsontothedecomposedrelationstogetnewsetsofFDs、ThesedependenciesarepreservediftheunionofthesenewsetsisequivalenttotheoriginalsetofFDs、FortheFDsof3、4、1a,thedependenciesarenotpreserved、TheunionofthenewsetsofFDsisCEA、However,theFDBEisnotintheunionandcannotbederived、ThusthetwosetsofFDsarenotequivalent、FortheFDsof3、4、1b,thedependenciesarepreserved、TheunionofthenewsetsofFDsisACEandBCD、ThisispreciselythesameastheoriginalsetofFDsandthusthetwosetsofFDsareequivalent、FortheFDsof3、4、1c,thedependenciesarenotpreserved、TheunionofthenewsetsofFDsisBDandAE、TheFDsADandDEarenotintheunionandcannotbederived、ThusthetwosetsofFDsarenotequivalent、FortheFDsof3、4、1d,thedependenciesarenotpreserved、TheunionofthenewsetsofFDsisACE、However,theFDsAD,CDEandEDarenotintheunionandcannotbederived、ThusthetwosetsofFDsarenotequivalent、Exercise3、5、1aInthesolutiontoExercise3、3、1awefoundthatthereare14nontrivialdependencies、Theyare:CA,CD,DA,ABD,ABC,ACD,BCA,BCD,BDA,BDC,CDA,ABCD,ABDC,andBCDA、WealsolearnedthatthethreekeyswereAB,BC,andBD、SincealltheattributesontherightsidesoftheFDsareprime,thereareno3NFviolations、Sincethereareno3NFviolations,itisnotnecessarytodecomposetherelation、Exercise3、5、1bInthesolutiontoExercise3、3、1bwefoundthatthereare8nontrivialdependencies、TheyareBC,BD,ABC,ABD,BCD,BDC,ABCDandABDC、WealsofoundoutthattheonlykeyisAB、FDswheretheleftsideisnotasuperkeyortheattributesontherightarenotpartofsomekeyare3NFviolations、The3NFviolationsareBC,BD,BCDandBDC、Usingalgorithm3、26,wecandecomposeintorelationsusingtheminimalbasisBCandBD、TheresultingdecomposedrelationswouldbeBCandBD、However,noneofthesetwosetsofattributesisasuperkey、ThusweaddrelationABtotheresult、ThefinalsetofdecomposedrelationsisBC,BDandAB、Exercise3、5、1cInthesolutiontoExercise3、3、1cwefoundthatthereare12nontrivialdependencies、TheyareABC,BCD,CDA,ADB,ABD,ADC,BCA,CDB,ABCD,ABDC,ACDBandBCDA、WealsofoundoutthatthekeysareAB,AD,BC,andCD、SincealltheattributesontherightsidesoftheFDsareprime,thereareno3NFviolations、Sincethereareno3NFviolations,itisnotnecessarytodecomposetherelation、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、WealsofoundoutthatthekeysareA,B,C,D、SincealltheattributesontherightsidesoftheFDsareprime,thereareno3NFviolations、Sincethereareno3NFviolations,itisnotnecessarytodecomposetherelation、Exercise3、5、1eInthesolutiontoExercise3、3、1ewefoundthatthereare16nontrivialdependencies、TheyareABC,DEC,BD,ABD,BCD,BEC,BED,ABCD,ABDC,ABEC,ABED,ADEC,BCED,BDEC,ABCED,andABDEC、WealsofoundoutthattheonlykeyisABE、FDswheretheleftsideisnotasuperkeyortheattributesontherightarenotpartofsomekeyare3NFviolations、The3NFviolationsareABC,DEC,BD,ABD,BCD,BEC,BED,ABCD,ABDC,ADEC,BCEDandBDEC、Usingalgorithm3、26,wecandecomposeintorelationsusingtheminimalbasisABC,DECandBD、TheresultingdecomposedrelationswouldbeABC,CDEandBD、However,noneofthesethreesetsofattributesisasuperkey、ThusweaddrelationABEtotheresult、ThefinalsetofdecomposedrelationsisABC,CDE,BDandABE、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、WealsofoundoutthatthekeysareAB,ACandAD、FDswheretheleftsideisnotasuperkeyortheattributesontherightarenotpartofsomekeyare3NFviolations、The3NFviolationsareCE,DE,BCE,BDE,CDUsingalgorithm3、26,wecandecomposeintorelationsusingtheminimalbasisABC,CD,DBandDE、TheresultingdecomposedrelationswouldbeABC,CD,BDandDE、SincerelationABCcontainsakey,wecanstopwiththedecomposition、ThefinalsetofdecomposedrelationsisABC,CD,BDandDE、Exercise3、5、2aTheusualproceduretofindthekeyswouldbetotaketheclosureofall63nonemptysubsets、However,ifwenoticethatnoneoftherightsidesoftheFDscontainsattributesHandS、ThusweknowthatattributesHandSmustbepartofanykey、WeeventuallywillfindoutthatHSistheonlykeyfortheCoursesrelation、Exercise3、5、2bThefirststeptoverifythatthegivenFDsaretheirownminimalbasisistochecktoseeifanyoftheFDscanberemoved、However,ifweremoveanyoneofthefiveFDs,theremainingfourFDsdonotimplytheremovedFD、ThesecondsteptoverifythatthegivenFDsaretheirownminimalbasisistochecktoseeifanyoftheleftsidesofanFDcanhaveoneormoreattributesremovedwithoutlosingthedependencies、However,thisisnotthecaseforthefourFDsthatcontaintwoattributesontheleftside、Thus,thegivensetofFDshasbeenverifiedtobetheminimalbasis、Exercise3、5、2cSincetheonlykeyisHS,thegivensetofFDshassomedependenciesthatviolate3NF、WealsoknowthatthegivensetofFDsisaminimalbasis、ThusthedecomposedrelationsareCT,HRC,HTR,HSRandCSG、SincetherelationHSRcontainsakey,wedonotneedtoaddanadditionalrelation、ThefinalsetofdecomposedrelationsisCT,HRC,HTR,HSRandCSG、NoneofthedecomposedrelationsviolateBCNF、Thiscanbeverifiedbyprojecting