作业：
考虑关系（,,,,,,）及其FD集= {→，→，→，→，→，→}。 如果它不在BCNF中，则将其分解为BCNF关系的集合。确保你的分解是无损连接。
解释
嗨，我正在处理我的数据库作业（关系设计章节）。 我想我已经根据课堂示例指导了基本流程。 然而，这里棘手的部分是我们有一个属性＆＃39; H＆＃39;与他人无关，这让我深感困惑。我该怎么处理呢？
尝试回答
• We start from a schema: ABCDEGH, since H has no relations with any attribute, we decompose it into tables: H and ABCDEG
• The FDs for ABCDEG remains the same, therefore key is E.
• The FD D →AG violates BCNF (FD with non-key on LHS).
• To fix, we need to decompose into tables: ADG and BCDE
• FDs for ADG are { D → AG }, therefore key is D, therefore BCNF.
• FDs for BCDE are { B → CD, E → D, BC → DE, E → B, CD → BE }
• Key for BCDE is also E, and FD B → CD violates BCNF (FD with non-key on LHS).
• To fix, we need to decompose into tables: BCD and BE
• FDs for BCD are { B → CD } therefore key is B, therefore BCNF.
• FDs for BE are { E → B } therefore key is E, therefore BCNF.
• Final schema: H, ADG, BCD, BE
您对我的错误有何看法？
答案 0 :(得分：1)
你的第一颗子弹错了。如果您使用具有正确BCNF分解算法的已发布学术教科书的参考文献，那么它就不会有这一步骤。