数据库系统基础教程第三章答案解析.docx
《数据库系统基础教程第三章答案解析.docx》由会员分享,可在线阅读,更多相关《数据库系统基础教程第三章答案解析.docx(38页珍藏版)》请在冰豆网上搜索。
![数据库系统基础教程第三章答案解析.docx](https://file1.bdocx.com/fileroot1/2023-5/21/eda1a67f-7161-4815-aa77-8fd68dd96d25/eda1a67f-7161-4815-aa77-8fd68dd96d251.gif)
数据库系统基础教程第三章答案解析
Exercise3.1.1
Answersforthisexercisemayvarybecauseofdifferentinterpretations.
SomepossibleFDs:
SocialSecuritynumbername
Areacodestate
Streetaddress,city,statezipcode
Possiblekeys:
{SocialSecuritynumber,streetaddress,city,state,areacode,phonenumber}
Needstreetaddress,city,statetouniquelydeterminelocation.Apersoncouldhavemultipleaddresses.Thesameistrueforphones.Thesedays,apersoncouldhavealandlineandacellularphone
Exercise3.1.2
Answersforthisexercisemayvarybecauseofdifferentinterpretations
SomepossibleFDs:
IDx-position,y-position,z-position
IDx-velocity,y-velocity,z-velocity
x-position,y-position,z-positionID
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,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.
Exercise3.2.1b
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,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.
Exercise3.2.2b
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…AnCB.
Exercise3.2.3b
From3.2.3a,weknowthatA1A2…AnCB.Usingtheconceptoftrivialdependencies,wecanshowthatA1A2…AnCC.ThusA1A2…AnCBC.
Exercise3.2.3c
FromA1A2…AnE1E2…Ej,weknowthattheclosurecontainsB1B2…BmbecauseoftheFDA1A2…AnB1B2…Bm.TheB1B2…BmandtheE1E2…EjcombinetoformtheC1C2…Ck.ThustheclosureofA1A2…AnE1E2…EjcontainsDaswell.Thus,A1A2…AnE1E2…EjD.
Exercise3.2.3d
FromA1A2…AnC1C2…Ck,weknowthattheclosurecontainsB1B2…BmbecauseoftheFDA1A2…AnB1B2…Bm.TheC1C2…CkalsotellusthattheclosureofA1A2…AnC1C2…CkcontainsD1D2…Dj.Thus,A1A2…AnC1C2…CkB1B2…BkD1D2…Dj.
Exercise3.2.4a
IfattributeArepresentedSocialSecurityNumberandBrepresentedaperson’sname,thenwewouldassumeABbutBAwouldnotbevalidbecausetheremaybemanypeoplewiththesamenameanddifferentSocialSecurityNumbers.
Exercise3.2.4b
LetattributeArepresentSocialSecurityNumber,BrepresentgenderandCrepresentname.SurelySocialSecurityNumberandgendercanuniquelyidentifyaperson’sname(i.e.ABC).ASocialSecurityNumbercanalsouniquelyidentifyaperson’sname(i.e.AC).However,genderdoesnotuniquelydetermineaname(i.e.BCisnotvalid).
Exercise3.2.4c
LetattributeArepresentlatitudeandBrepresentlongitude.Together,bothattributescanuniquelydetermineC,apointontheworldmap(i.e.ABC).However,neitherAnorBcanuniquelyidentifyapoint(i.e.ACandBCarenotvalid).
Exercise3.2.5
GivenarelationwithattributesA1A2…An,wearetoldthattherearenofunctionaldependenciesoftheformB1B2…Bn-1CwhereB1B2…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…CmA1A2…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.TheclosureofXwouldhavetoincludeAaswell.ThiscontradictsthefactthatweweregiventheclosureofX,X+.Therefore,i