數(shù)據(jù)庫(kù)系統(tǒng)原理課件：作業(yè)講評(píng) 4NF的判定

1、1,4NF的判定,3.6.2, 3.6.3,2,3.6.2,R=(U, F) U=n, s, b, cn, cs, cb, as, am F=sn, sb, cscn, cscb, ass, asam 1=(s,n,b), (cs,cn,cb),(as,s,am) 1保持函數(shù)依賴并且滿足4NF 1不能夠無(wú)損連接得到R，丟失了s與cs之間的親子關(guān)系,3,3.6.2,R=(U, F) U=n, s, b, cn, cs, cb, as, am F=sn, sb, cscn, cscb, ass, asam KEY=cs,as,4,U=n, s, b, cn, cs, cb, as, am F=sn

2、, sb, cscn, cscb, ass, asam KEY=cs,as,計(jì)算: (s)F+ = s, n,b,R1滿足4NF,R只滿足1NF,5,U=s, cn, cs, cb, as, am F=cscn, cscb, ass, asam KEY=cs,as,計(jì)算: (cs)F+ = cs, cn,cb,R2滿足4NF,R”只滿足1NF,6,U”=s, cs, as, am F”=ass, asam KEY=cs,as,計(jì)算: (as)F+ = as, s,am,R3滿足4NF,R4滿足4NF,7,3.6.2,R=(U, F) U=n, s, b, cn, cs, cb, as, am

3、F=sn, sb, cscn, cscb, ass, asam KEY=cs,as 2=(s,n,b), (cs,cn,cb),(as,s,am),(cs,as) 2無(wú)損連接并且保持函數(shù)依賴 2滿足4NF,8,3.6.2,R=(U, F) U=n, s, b, cn, cs, cb, as, am F=sn, sb, cscn, cscb, ass, asam KEY=cs,as 3=(s,n,b),(as,s,am), (s,cs) 3從實(shí)體聯(lián)系的語(yǔ)義出發(fā)進(jìn)行分解 3滿足4NF,9,3.6.3.a,R(A, B, C, D) F= MVD=AB, AC =D不屬于任一多值依賴 KEY=ABC

4、D R最高滿足BCNF,10,3.6.3.a,R(A, B, C, D) F= MVD=AB, AC KEY=ABCD =ABD, ACD，滿足4NF R3(AB)=ABR1(ABD) =ABR(ABCD) R4(AC)=ACR2(ACD) =ACR(ABCD),11,3.6.3.b,R(A, B, C, D) F= MVD=AB, BCD = ACD KEY=ABCD R最高滿足BCNF =AB, BCD, ACD 滿足4NF,12,3.6.3.c,R(A, B, C, D) F=BC MVD=ABC = ABD KEY=ABD R只滿足1NF,13,3.6.3.c,R(A, B, C, D

5、) F=BC KEY=ABD MVD=ABC = ABD =ABC, ABD = A, BC, ABD =AB, BC, ABD R1(AB)=ABR3(ABD) =ABR(ABCD) =BC, ABD 滿足4NF，保持函數(shù)依賴，無(wú)損,14,3.6.3.c,R(A, B, C, D) F=BC KEY=ABD MVD=ABC = ABD 保持函數(shù)依賴進(jìn)行模式分解 =BC, ABD 滿足4NF，保持函數(shù)依賴，無(wú)損,15,3.6.3.d,R(A, B, C, D, E) F=AD, ABE MVD=AB, AC =D, E不屬于任一多值依賴 KEY=ABC R最高滿足1NF,16,3.6.3.d,R(A, B, C, D, E) F=AD, ABE MVD=AB, AC K