数据库系统原理课件：作业讲评 4NF的判定

1、4NF的判定3.6.2, 3.6.313.6.2R=(U, F)U=n, s, b, cn, cs, cb, as, amF=sn, sb, cscn, cscb, ass, asam1=(s,n,b), (cs,cn,cb),(as,s,am)1保持函数依赖并且满足4NF1不能够无损连接得到R，丢失了s与cs之间的亲子关系23.6.2R=(U, F)U=n, s, b, cn, cs, cb, as, amF=sn, sb, cscn, cscb, ass, asamKEY=cs,as3 U=n, s, b, cn, cs, cb, as, am F=sn, sb, cscn, cscb,

2、ass, asam KEY=cs,as计算: (s)F+ = s, n,bR1满足4NFR只满足1NF“X”“X”“A” U=s, cn, cs, cb, as, am KEY= cs,asF=cscn, cscb, ass, asamU去掉“A”（ s, cn, cs, cb, as, am ） U1=s,n,bKEY=s“XA”F1=sn, sb4 U=s, cn, cs, cb, as, am F=cscn, cscb, ass, asam KEY=cs,as计算: (cs)F+ = cs, cn,cbR2满足4NFR”只满足1NF“X”“X”“A” U”=s, cs, as, am K

3、EY= cs,as F”=ass, asamU去掉“A”（ s, cs, as, am ） U2=cs, cn,cbKEY=cs“XA”F2=cscn, cscb5 U”=s, cs, as, am F”=ass, asam KEY=cs,as计算: (as)F+ = as, s,amR3满足4NFR4满足4NF“X”“X”“A” U4=cs, asKEY= cs,as F4=U去掉“A”（cs, as） U3=as, s,amKEY=as“XA”F3=ass, asam63.6.2R=(U, F)U=n, s, b, cn, cs, cb, as, amF=sn, sb, cscn, csc

4、b, ass, asamKEY=cs,as2=(s,n,b), (cs,cn,cb),(as,s,am),(cs,as)2无损连接并且保持函数依赖2满足4NF73.6.2R=(U, F)U=n, s, b, cn, cs, cb, as, amF=sn, sb, cscn, cscb, ass, asamKEY=cs,as3=(s,n,b),(as,s,am), (s,cs)3从实体联系的语义出发进行分解3满足4NF83.6.3.aR(A, B, C, D)F=MVD=AB, AC=D不属于任一多值依赖KEY=ABCDR最高满足BCNF93.6.3.aR(A, B, C, D)F=MVD=AB

5、, ACKEY=ABCD=ABD, ACD，满足4NFR3(AB)=ABR1(ABD) =ABR(ABCD)R4(AC)=ACR2(ACD) =ACR(ABCD)103.6.3.bR(A, B, C, D)F=MVD=AB, BCD = ACDKEY=ABCDR最高满足BCNF=AB, BCD, ACD满足4NF113.6.3.cR(A, B, C, D)F=BCMVD=ABC = ABDKEY=ABDR只满足1NF123.6.3.cR(A, B, C, D)F=BC KEY=ABDMVD=ABC = ABD=ABC, ABD = A, BC, ABD =AB, BC, ABDR1(AB)=ABR3(ABD) =ABR(ABCD)=BC, ABD满足4NF，保持函数依赖，无损133.6.3.cR(A, B, C, D)F=BC KEY=ABDMVD=ABC = ABD保持函数依赖进行模式分解=BC, ABD满足4NF，保持函数依赖，无损143.6.3.dR(A, B, C, D, E)F=AD, ABEMVD=AB, AC=D, E不属于任一多值依赖KEY=ABCR最高满足1NF153.6.3.dR(A, B, C, D, E)F=AD, A