选择的串接律分解σL.LN=S.LNANDL.BN=B.BNσL.LN=S.LN和σL.BN=B.BN再利用选择对笛卡儿积的分配律得到σL.BN=B.BN(BσL.LN=S.LN(SσD<20020101(L)))利用选择与投影的串接律πN,T(σL.BN=B.BN(…)) πN,T(σL.BN=B.BN(πN,T,L.BN,B.BN(…)))……(取其中二步即可)八、(每小题4分,共8分)?1)σNOT(SA>20ANDSD=‘计算机系’)(S)?2)先做等价变换σSA≤20ORSD≠‘计算机系’(S) 用两个规则: a)S1(a,b,c,d)S(a,b,c,d)ANDc≤20 b)S1(a,b,c,d)S(a,b,c,d)ANDd≠‘计算机系’九、(每小题6分,共12分)键码:{}和{SNo,TN}函数依赖:SNoSN,SA(BC范式违例)(BC范式违例)TN,Ga)SN,SAP?SNo,TNGPb)?SNo,c)?SNo,TNSN,SA (a,b,c为部分依赖,可不写)?2) STC1(SNo,SN,SA) STC2() STC3(SNo,TN,G)十、(每小题6分,共12分)interfaceMovie(extentMovies)?{attributestringtitle; attributeintegeryear; relationshipSet<Act>atcs inverseAct::starredIn;?};?intefaceAct (extentActs){attributestringname;?attributecharsex;?relationshipSet<Movie>starredIn inverseMovie::acts;};?2)?SELECTm.title FROMActsa,a.starredInm WHEREa.name=‘孙小华’;
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