第五章 关系数据库理论 ( 8 学时)

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数据库原理和语言. 第五章 关系数据库理论 ( 8 学时). 2005 年 8 月 6 日. 主讲:谢益武 副教授 大连海事大学计算机科学与技术学院 研究方向:数据库与信息系统 电话: 84729625 Email : [email protected]. 章节目录. §5.1 问题的提出 §5.2 规范化( Normalization ) § 5.3 数据依赖的公理系统 § 5.4 关系模式的分解. §5.1 问题的提出. 在现实世界中,进行 数据处理 或进行数据库应用系统开发的 关键 是: - PowerPoint PPT Presentation

Transcript of 第五章 关系数据库理论 ( 8 学时)

  • 8 200586 [email protected]

  • 5.1 5.2 Normalization5.3 5.4

  • 5.1

  • RUDdomF UFR(UF)UFUDdom

  • ()P170Functional DependencyFDMultivalued DependencyMVDxf(x)P171-

  • (P171)(Student)SNoSDeptMNCNameGU={ SNoSNameSDeptMNCNameG }

  • UF={ SNOSDept,SDept MN,(SNO,CName) G } 5.1 Student

  • S123

  • ,3S (SNoSDeptSNoSDept)SG (SNoCNameG(SNoCName) G)D (SDeptMNSDeptMN)

  • 5.2 Normalization 1971E.F.Codd 5.2.1 1NF--BCNF5.2.2 5.2.3 Normal FormNF5.2.4 4NF5.2.5 4NF

  • 5.2.1 1R(U)UXYUR(U)rrt,sXt[X]=s[X]Yt[Y]s[Y]XYYXXY R(U)U X YU R(U)rts tX=sX tY=sY XY YXX Y

  • XY XY xf(x)=y

  • (1) R Rr Rr R (2) (3)

  • 2 YXXYX

  • 1 l SNo SName lSNo SDept l SNo SAge

  • 2

  • 3PJTP(** )

  • 3 R(U)XY(Y X)Y X Y Z X Z ZXY XY X XY R={A, B, C, D} F={AB, BC, AC, ABD} AC

  • 5.2.2 4KR(UF) KRCandidate KeyPrimary KeyPrime attributeNon Prime attributeNon-Key-attributeall key5 RXRXXRForeign Key

  • 5.2.3 Normal FormNF 19711972E.F.Codd1NF,2NF,3NF1974CoddBoyceBCNF1976Fagin4NF5NFRR xNF

  • 1NF 1NF

  • ------ 1A 2B

  • 2NF 2NF6 R 1NFR 2NF P175 2NF 2NF(Second Normal Form)(2NF)12NF 22NF

  • Sno SnameSex Sdept SL Spec (SC) Cno Cname T CreditG RUF U={Sno, Sname, Sex, ID, Sdept, SL, Spec, SC, Cno, Cname, T, Credit, G} F={SnoSname, SnoSex, SnoSdept, SnoSL, SnoSpec, SnoSC, CnoCname , CnoT, CnoCredit, (Sno, Cno)G} Sno,Cno--F R1NF--- R2NF---

  • R (1) (2) Cno=1011 Cname= =3 R (3) C5 C5 C5 C5

  • R2NF---RR1U1F1 ----Sno 2NFU1={Sno, Sname, Sex, ID, Sdept, SL, Spec, SC}F1={SnoSname, SnoSex, SnoSdept, SnoSL, SnoSpec, SnoSC}R2U2F2 ----Cno 2NFU2={Cno, Cname, T, Credit}F2={CnoCname , CnoT, CnoCredit}R3U3F3 ----(Sno,Cno) 2NFU3={Sno CnoG}F3={(Sno, Cno)G}

  • -- 1NF+ + + + 2NF* *

  • ()

  • 2NF --2NF

  • 3NF 7 RXYZ(Z Y ) X Y (Y X ) Y ZR 3NFA BCRRCBBACACARAB BC AC3NF(Third Normal Form)3NF)

  • 1

  • 2 * * * 3NF

  • QQ

  • BCNF( Boyce Codd Normal Form)Boyce Codd 19743NF8R 1NFXY Y!X X R BCNFRBCNFBCNF ()

  • BCNF3NFBCNFBCNF3NF

  • BCNFSTJ STJ S T J SJ T ST J T J5.6

  • SJ ST STJ3NF STJBCNFTTBCNFSTSTTJTJ

  • R3NF,R2NF,R2NF ()R3NF R3NFR2NF AX A, X RRAX

  • 5.2.4 (P178-P179)R(U)UXYUR(U)rXxYYUXYzYXXY Y1 XYXY

  • R(U)rt,st[X]s[X]w,vrw,vs,tw[X]=v[X]=t[X]w[Y]=t[Y]w[Z]=s[Z]v[Y]=s[Y]v[Z]=t[Z] s,tYr XYUZ=UXYYXXYXY Z=UXY= XY

  • 1. XY XZ Z=UXY2.XYYZ XZ-Y 3. XY XY 4.XY XZ XYZ 5. XY XZXYZ 6. XYXZXY -ZX Y Z

  • XYUW(XYWU) XYW(WU)UXYR(U)Y Y XY XYR(U)Y YXY

  • 5.2.5 4NF R1NFRXY Y !XXR 4NF4NF XYXXY 4NFBCNF4NF

  • 4NFWSC(WSC)WS WC UWS=CUWC = SW,S,CW WSR 4NFWS(WS) WC(WC)4NF

  • 1

  • 2

  • R

  • FD

  • 5.3 5.3.1 Armstrong----ArmstrongR(UF) X Y U FR XYR Rr F rXY FXY XYF XYFFRrXY(rt,st[X]=s[X]t[Y]= s [Y] )FXY (XYF )

  • 5.3.1 ArmstrongArmstrong1974ArmstrongUFURRu A1ReflexivityYX U XYFuA2XYFZ U XZYZFu A3XYYZFXZF

  • 5.3.1 Armstrong5.1 Armstrong rR(U) t1 t2r X Y Z U t1X=t2X Xt1, t2 Y X t1Y=t2Y XY t1XZ=t2XZ t1XZ=t2XZ t1X=t2X t1Z=t2Z XY t1Y=t2Y t1YZ= t1Yt1Z=t2Yt2Z=t2YZ XZYZ XY YZ t1X=t2X t1Y=t2Y t1Y=t2Y t1Z=t2Z t1X=t2X t1Z=t2Z XZ

  • 5.3.1 ArmstrongA1A2A3u XY XZXYZ u XY WYZ WXZ u XYZ YXZ 5.1XA1,A2,,Ak XAii=1,2, , k

  • FRFFF + R(U) U={A, B, C}, F={ABC, CB}R(U) F+={AA, ABA, ACA, ABCA, BB, ABB, BCB, ABCB, CC, ACC, BCC, ABCC, ABAB, ABCAB, ACAC, ABCAC, BCBC, ABCBC, ABCABC, ABC, ABAC, ABBC, ABABC, CB, CBC, ACB}

  • XFFUX U XF+ ={A|X AArmstrong} XF+XF 2FR(U)XY U XYF ArmstrongY X+ R(U) U={A, B, C, D, E} F={ABC, CDE, BD, EA} B+F={B, D} A+F={A, B, C, D, E} 5.1 XXF+ XF FXX+ X(0)=X i=0; X(i+1)=X(i){A|V X(i) VWFAW}; X(i+1)= X(i)X(i+1)= X(i) X(i+1)=U i=i+1 (2)

  • R(U)F, U={A, B, C, D, E, I}; F={AD, ABE, BIE, CDI, EC} (AE)F+ X(0)=AE i=0; FAE AD, EC X(1)=AEDC X(0) U AEDC CDI, X(3)=AEDCI; VWF X(3+1)X(3) XF+=X(3), (AE)F+ =ACDEI

  • 5.2 Armstrong

    ArmstrongFF F+FArmstrong

  • 5.3.2F

  • 1. FGFG: RFG F+=G+ FG FG FG FG GF 5.3: GF FGF G+G F+

  • 2. F FF,Fm1F2FX A FF{X A }---F3FX A XZ F{X A }{ZA}F---F

    FFm

  • 5.4: FFm (1) FXY Y=A1A2Amm2 {XAi|i=1, 2, , m}XY (2) FXA G=F-{XA} AX+G, FXAFGAX+G F (3) FXA X=B1B2Bn Bi(i=1, 2, , n) A(X-Bi)+F (X-Bi)AXAF(F-{XA}){(X-Bi)A}A(X-Bi)+F F

    FF

  • R(U)F, U={A, B, C, D, E, G, }; F={ABC, BCD, ACDB, DEG, CA, BEC, CGBD, CEAG} F 1 F F1={ABC, BCD, ACDB, DE, DG, CA, BEC, CGB, CGD, CEA, CEG}2 CEA, CA, E ACDB,CD F1 +=ABCDEGCD B A F2={ABC, BCD, CDB, DE, DG, CA, BEC, CGB, CGD, CEG}

  • (3) F2 CGB, G= F2- CG B = {ABC, BCD, CDB, DE, DG, CA, BEC, CGD, CEG} (CG)G+=ABCDEG, CGB Fm Fm={ABC, BCD, CDB, DE, DG, CA, BEC, CGD, CEG}

  • SU={ SNoSDeptMNCNameG }F={ SNoSDeptSDept MN(SNo,CName) G }F F = { SNoSDept, SNoMN, SDept MN, (SNoCName) G , (SNoSDept) SDept } 5.15F 1 2F {SNoMN}F 3F {(SNo,SDept) SDept }F2

  • 5.4 FFmFmFFFmFDiXAF={ABBABCACCA}Fm1={ABBCCA}Fm2= {ABBAACCA}FFm1 Fm2

  • 5.3.3 (1) 1NF (2) 3NF

  • 1. R(U) U={A1, A2, , An}, FR(U) ={R1, R2, , Rn}R R1R2Rn=R, Ri(Ui), U=U1U2Un FRiUiFi={XY|XYF+, X, YUi}2.

  • 3. (1) (2) (3)

  • 1. R(U) FR ={R1, R2, , Rn}R RFr

  • 2. R(A1, A2, , An) R={R1, R2, ,Rn} RF

  • (1) kn i Ri jAj AjRi ijai bij (2) F FXY X Y aj, bijaj; bij(bij ) (3) a1, a2, , an a F

  • R(U) U={A, B, C, D, E} RF={AC, BC, CD, DEC, CEA} ={R1({A, D}), R2({A, B}), R3({B, E}), R4({C, D, E}), R5({A, E}), (1) 5-1

  • ={R1({A, D}), R2({A, B}), R3({B, E}), R4({C, D, E}), R5({A, E})

    5-1

    ABCDER1a1b12b13a4b15R2a1a2b23b24b25R3b31a2b33b34a5R4b41b42a3a4a5R5a1b52b53b54a5

  • (2) AC R1A=R2A=R5A, b13, b23, b53 b13 5-2

    5-2

    ABCDER1a1b12b13a4b15R2a1a2b13b24b25R3b31a2b33b34a5R4b41b42a3a4a5R5a1b52b13b54a5

  • (3) BC R2B=R3B, b13, b33 b13 5-35-3

    ABCDER1a1b12b13a4b15R2a1a2b13b24b25R3b31a2b13b34a5R4b41b42a3a4a5R5a1b52b13b54a5

  • (4) CD R1C=R2C=R3C=R5C, b24, b34, b54 a4 5-4

    ABCDER1a1b12b13a4b15R2a1a2b13a4b25R3b31a2b13a4a5R4b41b42a3a4a5R5a1b52b13a4a5

  • (5) DEC R3DE=R4DE=R5DE, R3C, R5C, a3 5-5

    5-5

    ABCDER1a1b12b13a4b15R2a1a2b13a4b25R3b31a2a3a4a5R4b41b42a3a4a5R5a1b52a3a4a5

  • (6) CEA R3CE=R4CE=R5CE, R3A, R4A, a1 5-6

    5-6

    ABCDER1a1b12b13a4b15R2a1a2b13a4b25R3b31-a1a2a3a4a5R4b41-a1b42a3a4a5R5a1b52a3a4a5

  • 1. R(U) ={R1, R2, , Rn}R UiRi FiFUi FF1F2Fn,

  • 5.4 R(A1,A2,An)FLFRFNFLRF

  • 1RFXXRLXR1RABCDF={DBBDADBACD}RFACL1ACRACF+={ABCD}ACR1RFXXRLX F+RXR

  • 2RFXXRRX3RFXXRNXR

  • 2RABCDEPF={ADEDDBBCD DCA}RFECL1ECRPNPRCEPF+={ABCDEP}CEPR2RFXXRLNX F+RXR

  • 51 (1) R{A, B, C} F={ABC} (2) R{A, B, C, D} F={BD, ABC} (3) R{A, B, C, D, E, G} F={CG, EA, CED, AB} (4) R{A, B, C} F={BA, CB, AB} (5) R{A, B, C} F={ACB, BC} (6) R{A, B, C, D} F={ABC, CDA, BCD, ADB} (7) R{A, B, C, D, E} F={ACB, BD, DC, DE}

  • 2R{A, B, C}, F={AB, AC, BA, BC, CA, CB, ABC, ACB, BCA} F 3R{A, B, C, D, E} F={ABC, CDE, BD, EA} (1) B+, E+ (2) F (3) R 4S(S, SN, SD, MN, C, GR) RF={SSD SSN, SDMN (S, C)GR } S1(S, SN, SD, MN) SC(S, C, GR) S1S11(S, SN, MN) S12(SD, MN), S S1SC S11 S12

  • 5 R{A, B, C, D, E} F={ AD, ED, BCD, DCA, DB} ={R1(A, B), R2(A, E), R3(C, E), R4(B, C, D), R5(A, C)}6R{C, D, M, N, } F={MC, DCM, NCM, CM} (1) R (2) F (3) R (4) R 3NF6 BCNF3NF

  • Thank You