数据库概论第二次作业

题目

参考答案

\[E-\prod_{E.person\_name,E.street,E.city}{(\sigma_{E.person\_name=M.person\_name}{(E\times M)})} \]

或

\[E-\prod_{person\_name,street,city}{(E\Join M)} \]

2. 令 \(F = E\)，

\[\prod_{E.person\_name}{((E \Join M) \Join_{manager\_name=F.person\_name \wedge E.street=F.street \wedge E.city=F.city}F)} \]

或

\[\prod_{E.person\_name}{(\sigma_{E.person\_name=M.person\_name \wedge F.person\_name=M.manager\_name \wedge E.street=F.street \wedge E.city=F.city}(E\times F \times M))} \]

\[\prod_{person\_name}{(E)}-\prod_{person\_name}{(\sigma_{company\_name="First\ Bank\ Corp."}(W))} \]

4. 令 \(V=W\)，

\[\prod_{person\_name}{W}-\prod_{W.person\_name}{(\sigma_{W.salary \leq V.salary \wedge V.company\_name="Small\ Bank\ Corp."}{(W\times V)})} \]

或

\[\prod_{person\_name}{W}-\prod_{W.person\_name}{(W \Join_{W.salary \leq V.salary \wedge V.company\_name="Small\ Bank\ Corp."}{V} )} \]

5.令 \(N=M\)，

\[\prod_{M.manage\_name}{(\sigma_{M.manage\_name=N.manage\_name \wedge M.person\_name<>N.person\_name}(M\times N))} \]

或

\[\prod_{M.manage\_name}{(M \Join_{M.manage\_name=N.manage\_name \wedge M.person\_name<>N.person\_name}{\ N})} \]

\[\prod_{person\_name,company\_name}{(W)} \div \prod_{company\_name}{(\sigma_{city="Nanjing"}{(C)})} \]

后记

注：本答案为博主自己整理所写，可能会有部分错误，仅供参考，同时欢迎各位读者指出错误！