數(shù)據(jù)庫系統(tǒng)基礎(chǔ)教學(xué)教材第三章答案_第1頁
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1、_ExerciseExercise3.1.13.1.1Answersfthisexercisemayvarybecauseofdifferentinterpretations.SomepossibleFDs:SocialSecuritynumber?nameAreacode?stateStreetaddresscitystate?zipcodePossiblekeys:SocialSecuritynumberstreetaddressc

2、itystateareacodephonenumberNeedstreetaddresscitystatetouniquelydeterminelocation.Apersoncouldhavemultipleaddresses.Thesameistruefphones.ThesedaysapersoncouldhaveallineacellularphoneExerciseExercise3.1.23.1.2Answersfthise

3、xercisemayvarybecauseofdifferentinterpretationsSomepossibleFDs:ID?xpositionypositionzpositionID?xvelocityyvelocityzvelocityxpositionypositionzposition?IDPossiblekeys:IDxpositionypositionzpositionThereasonwhythepositionsw

4、ouldbeakeyisnotwomoleculescanoccupythesamepoint.ExerciseExercise3.1.3a3.1.3aThesuperkeysareanysubsetthatcontainsA1.Thusthereare2(n1)suchsubsetssinceeachofthen1attributesA2throughAnmayindependentlybechoseninout.ExerciseEx

5、ercise3.1.3b3.1.3bThesuperkeysareanysubsetthatcontainsA1A2.Thereare2(n1)suchsubsetswhenconsideringA1then1attributesA2throughAn.Thereare2(n2)suchsubsetswhenconsideringA2then2attributesA3throughAn.WedonotcountA1inthesesubs

6、etsbecausetheyarealreadycountedinthefirstgroupofsubsets.Thetotalnumberofsubsetsis2(n1)2(n2).ExerciseExercise3.1.3c3.1.3c_Thesuperkeysareallthosethatcontainoneofthosethreekeys.ThatisasuperkeythatisnotakeymustcontainBmetha

7、noneofACD.Thusthe(proper)superkeysareABCABDBCDABCD.ExerciseExercise3.2.2a3.2.2ai)FthesingleattributeswehaveA=ABCDB=BCDC=CD=D.ThusthenewdependenciesareA?CA?D.Nowconsiderpairsofattributes:AB=ABCDAC=ABCDAD=ABCDBC=BCDBD=BCDC

8、D=CD.ThusthenewdependenciesareAB?CAB?DAC?BAC?DAD?BAD?CBC?DBD?C.FthetriplesofattributesBCD=BCDbuttheclosuresoftheothersetsareeachABCD.ThuswegetnewdependenciesABC?DABD?CACD?B.SinceABCD=ABCDwegetnonewdependencies.Thecollect

9、ionof13newdependenciesmentionedaboveare:A?CA?DAB?CAB?DAC?BAC?DAD?BAD?CBC?DBD?CABC?DABD?CACD?B.ii)FthesingleattributeswehaveA=AB=BC=CD=D.Thustherearenonewdependencies.Nowconsiderpairsofattributes:AB=ABCDAC=ACAD=ABCDBC=ABC

10、DBD=BDCD=ABCD.ThusthenewdependenciesareAB?DAD?CBC?ACD?B.FthetriplesofattributesalltheclosuresofthesetsareeachABCD.ThuswegetnewdependenciesABC?DABD?CACD?BBCD?A.SinceABCD=ABCDwegetnonewdependencies.Thecollectionof8newdepen

11、denciesmentionedaboveare:AB?DAD?CBC?ACD?BABC?DABD?CACD?BBCD?A.iii)FthesingleattributeswehaveA=ABCDB=ABCDC=ABCDD=ABCD.ThusthenewdependenciesareA?CA?DB?DB?AC?AC?BD?BD?C.Sinceallthesingleattributes’closuresareABCDanysuperse

12、tofthesingleattributeswillalsoleadtoaclosureofABCD.Knowingthiswecanenumeratetherestofthenewdependencies.Thecollectionof24newdependenciesmentionedaboveare:A?CA?DB?DB?AC?AC?BD?BD?CAB?CAB?DAC?BAC?DAD?BAD?CBC?ABC?DBD?ABD?CCD

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