数据库系统基础教程答案.docx
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数据库系统基础教程答案.docx
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数据库系统基础教程答案
Exercise3.1.1
Answersforthisexercisemayvarybecauseofdifferentinterpretations.
SomepossibleFDs:
SocialSecuritynumber→name
Areacode→state
Streetaddress,city,state→zipcode
Possiblekeys:
{SocialSecuritynumber,streetaddress,city,state,areacode,phonenumber}
Needstreetaddress,city,statetouniquelydeterminelocation.Apersoncouldhavemultipleaddresses.Thesameistrueforphones.Thesedays,apersoncouldhavealandlineandacellularphone
Exercise3.1.2
Answersforthisexercisemayvarybecauseofdifferentinterpretations
SomepossibleFDs:
ID→x-position,y-position,z-position
ID→x-velocity,y-velocity,z-velocity
x-position,y-position,z-position→ID
Possiblekeys:
{ID}
{x-position,y-position,z-position}
Thereasonwhythepositionswouldbeakeyisnotwomoleculescanoccupythesamepoint.
Exercise3.1.3a
ThesuperkeysareanysubsetthatcontainsA1.Thus,thereare2(n-1)suchsubsets,sinceeachofthen-1attributesA2throughAnmayindependentlybechoseninorout.
Exercise3.1.3b
ThesuperkeysareanysubsetthatcontainsA1orA2.Thereare2(n-1)suchsubsetswhenconsideringA1andthen-1attributesA2throughAn.Thereare2(n-2)suchsubsetswhenconsideringA2andthen-2attributesA3throughAn.WedonotcountA1inthesesubsetsbecausetheyarealreadycountedinthefirstgroupofsubsets.Thetotalnumberofsubsetsis2(n-1)+2(n-2).
Exercise3.1.3c
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).
Exercise3.1.3d
Thesuperkeysareanysubsetthatcontains{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).
Exercise3.2.1a
Wecouldtryinferencerulestodeducenewdependenciesuntilwearesatisfiedwehavethemall.Amoresystematicwayistoconsidertheclosuresofall15nonemptysetsofattributes.
Forthesingleattributeswehave{A}+=A,{B}+=B,{C}+=ACD,and{D}+=AD.Thus,theonlynewdependencywegetwithasingleattributeontheleftisC→A.
Nowconsiderpairsofattributes:
{AB}+=ABCD,sowegetnewdependencyAB→D.{AC}+=ACD,andAC→Disnontrivial.{AD}+=AD,sonothingnew.{BC}+=ABCD,sowegetBC→A,andBC→D.{BD}+=ABCD,givingusBD→AandBD→C.{CD}+=ACD,givingCD→A.
Forthetriplesofattributes,{ACD}+=ACD,buttheclosuresoftheothersetsareeachABCD.Thus,wegetnewdependenciesABC→D,ABD→C,andBCD→A.
Since{ABCD}+=ABCD,wegetnonewdependencies.
Thecollectionof11newdependenciesmentionedaboveare:
C→A,AB→D,AC→D,BC→A,BC→D,BD→A,BD→C,CD→A,ABC→D,ABD→C,andBCD→A.
Exercise3.2.1b
Fromtheanalysisofclosuresabove,wefindthatAB,BC,andBDarekeys.AllothersetseitherdonothaveABCDastheclosureorcontainoneofthesethreesets.
Exercise3.2.1c
Thesuperkeysareallthosethatcontainoneofthosethreekeys.Thatis,asuperkeythatisnotakeymustcontainBandmorethanoneofA,C,andD.Thus,the(proper)superkeysareABC,ABD,BCD,andABCD.
Exercise3.2.2a
i)Forthesingleattributeswehave{A}+=ABCD,{B}+=BCD,{C}+=C,and{D}+=D.Thus,thenewdependenciesareA→CandA→D.
Nowconsiderpairsofattributes:
{AB}+=ABCD,{AC}+=ABCD,{AD}+=ABCD,{BC}+=BCD,{BD}+=BCD,{CD}+=CD.ThusthenewdependenciesareAB→C,AB→D,AC→B,AC→D,AD→B,AD→C,BC→DandBD→C.
Forthetriplesofattributes,{BCD}+=BCD,buttheclosuresoftheothersetsareeachABCD.Thus,wegetnewdependenciesABC→D,ABD→C,andACD→B.
Since{ABCD}+=ABCD,wegetnonewdependencies.
Thecollectionof13newdependenciesmentionedaboveare:
A→C,A→D,AB→C,AB→D,AC→B,AC→D,AD→B,AD→C,BC→D,BD→C,ABC→D,ABD→CandACD→B.
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.ThusthenewdependenciesareAB→D,AD→C,BC→AandCD→B.
Forthetriplesofattributes,alltheclosuresofthesetsareeachABCD.Thus,wegetnewdependenciesABC→D,ABD→C,ACD→BandBCD→A.
Since{ABCD}+=ABCD,wegetnonewdependencies.
Thecollectionof8newdependenciesmentionedaboveare:
AB→D,AD→C,BC→A,CD→B,ABC→D,ABD→C,ACD→BandBCD→A.
iii)Forthesingleattributeswehave{A}+=ABCD,{B}+=ABCD,{C}+=ABCD,and{D}+=ABCD.Thus,thenewdependenciesareA→C,A→D,B→D,B→A,C→A,C→B,D→BandD→C.
Sinceallthesingleattributes’closuresareABCD,anysupersetofthesingleattributeswillalsoleadtoaclosureofABCD.Knowingthis,wecanenumeratetherestofthenewdependencies.
Thecollectionof24newdependenciesmentionedaboveare:
A→C,A→D,B→D,B→A,C→A,C→B,D→B,D→C,AB→C,AB→D,AC→B,AC→D,AD→B,AD→C,BC→A,BC→D,BD→A,BD→C,CD→A,CD→B,ABC→D,ABD→C,ACD→BandBCD→A.
Exercise3.2.2b
i)Fromtheanalysisofclosuresin3.2.2a(i),wefindthattheonlykeyisA.AllothersetseitherdonothaveABCDastheclosureorcontainA.
ii)Fromtheanalysisofclosures3.2.2a(ii),wefindthatAB,AD,BC,andCDarekeys.AllothersetseitherdonothaveABCDastheclosureorcontainoneofthesefoursets.
iii)Fromtheanalysisofclosures3.2.2a(iii),wefindthatA,B,CandDarekeys.AllothersetseitherdonothaveABCDastheclosureorcontainoneofthesefoursets.
Exercise3.2.2c
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.
Exercise3.2.3a
SinceA1A2…AnCcontainsA1A2…An,thentheclosureofA1A2…AnCcontainsB.ThusitfollowsthatA1A2…AnC→B.
Exercise3.2.3b
From3.2.3a,weknowthatA1A2…AnC→B.Usingtheconceptoftrivialdependencies,wecanshowthatA1A2…AnC→C.ThusA1A2…AnC→BC.
Exercise3.2.3c
FromA1A2…AnE1E2…Ej,weknowthattheclosurecontainsB1B2…BmbecauseoftheFDA1A2…An→B1B2…Bm.TheB1B2…BmandtheE1E2…EjcombinetoformtheC1C2…Ck.ThustheclosureofA1A2…AnE1E2…EjcontainsDaswell.Thus,A1A2…AnE1E2…Ej→D.
Exercise3.2.3d
FromA1A2…AnC1C2…Ck,weknowthattheclosurecontainsB1B2…BmbecauseoftheFDA1A2…An→B1B2…Bm.TheC1C2…CkalsotellusthattheclosureofA1A2…AnC1C2…CkcontainsD1D2…Dj.Thus,A1A2…AnC1C2…Ck→B1B2…BkD1D2…Dj.
Exercise3.2.4a
IfattributeArepresentedSocialSecurityNumberandBrepresentedaperson’sname,thenwewouldassumeA→BbutB→AwouldnotbevalidbecausetheremaybemanypeoplewiththesamenameanddifferentSocialSecurityNumbers.
Exercise3.2.4b
LetattributeArepresentSocialSecurityNumber,BrepresentgenderandCrepresentname.SurelySocialSecurityNumberandgendercanuniquelyidentifyaperson’sname(i.e.AB→C).ASocialSecurityNumbercanalsouniquelyidentifyaperson’sname(i.e.A→C).However,genderdoesnotuniquelydetermineaname(i.e.B→Cisnotvalid).
Exercise3.2.4c
LetattributeArepresentlatitudeandBrepresentlongitude.Together,bothattributescanuniquelydetermineC,apointontheworldmap(i.e.AB→C).However,neitherAnorBcanuniquelyidentifyapoint(i.e.A→CandB→Carenotvalid).
Exercise3.2.5
GivenarelationwithattributesA1A2…An,wearetoldthattherearenofunctionaldependenciesoftheformB1B2…Bn-1→CwhereB1B2…Bn-1isn-1oftheattributesfromA1A2…AnandCistheremainingattributefromA1A2…An.Inthiscase,thesetB1B2…Bn-1andanysubsetdonotfunctionallydetermineC.ThustheonlyfunctionaldependenciesthatwecanmakeareoneswhereCisonboththeleftandrighthandsides.AllofthesefunctionaldependencieswouldbetrivialandthustherelationhasnonontrivialFD’s.
Exercise3.2.6
Let’sprovethisbyusingthecontrapositive.WewishtoshowthatifX+isnotasubsetofY+,thenitmustbethatXisnotasubsetofY.
IfX+isnotasubsetofY+,theremustbeattributesA1A2…AninX+thatarenotinY+.IfanyoftheseattributeswereoriginallyinX,thenwearedonebecauseYdoesnotcontainanyoftheA1A2…An.However,iftheA1A2…Anwereaddedbytheclosure,thenwemustexaminethecasefurther.AssumethattherewassomeFDC1C2…Cm→A1A2…AjwhereA1A2…AjissomesubsetofA1A2…An.ItmustbethenthatC1C2…CmorsomesubsetofC1C2…CmisinX.However,theattributesC1C2…CmcannotbeinYbecauseweassumedthatattributesA1A2…AnareonlyinX+andarenotinY+.Thus,XisnotasubsetofY.
Byprovingthecontrapositive,wehavealsoprovedifX⊆Y,thenX+⊆Y+.
Exercise3.2.7
ThealgorithmtofindX+isoutlinedonpg.76.Usingthatalgorithm,wecanprovethat
(X+)+=X+.Wewilldothisbyusingaproofbycontradiction.
Supposethat(X+)+≠X+.Thenfor(X+)+,itmustbethatsomeFDallowedadditionalattributestobeaddedtotheoriginalsetX+.Forexample,X+→AwhereAissomeattributenotinX+.However,ifthiswerethecase,thenX+wouldnotbetheclosureofX.Theclosureo
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