考虑通用关系R = {A，B，C，D，E，F，G，H，I，J}。 R的关键是什么？将R分解为2NF，然后分解为3NF关系 [英] Consider the universal relation R = {A, B, C, D, E, F, G, H, I, J}. What is the key for R? Decompose R into 2NF and then 3NF relations

问题描述

考虑通用关系R = {A，B，C，D，E，F，G，H，I，J}和
函数依赖项集合F = {{{A，B}→{ C}，{A}→{D，E}，{B}→{F}，{F}→{G，
H}，{D}→{I，J}}。 R的关键是什么？将R分解为2NF，然后分解为
3NF关系。

Consider the universal relation R = {A, B, C, D, E, F, G, H, I, J} and the set of functional dependencies F = { {A, B}→{C}, {A}→{D, E}, {B}→{F}, {F}→{G, H}, {D}→{I, J} }. What is the key for R? Decompose R into 2NF and then 3NF relations.

我尝试了Internet上给出的所有解决方案，但即使我的讲师没有回答我，也仍然无法理解答案令人满意。有人可以向我解释一下吗？

I tried every solution given on internet but still not able to understand the answer even my instructor is not answering me satisfactorily. Can someone please explain me this?

是的，这是在作业中问到的，但它已被标记为错误，我只想学习这个概念。谢谢。

Yes this was asked in homework but it has already been marked wrong and i just want to learn this concept. Thank you.

推荐答案

有一个6个步骤的过程将引导您找到答案，但是在很多情况下，关键是要弄清楚哪一个属性或一组属性只有外向关系而没有传入关系。在此，除了A，B以外，其他所有属性都直接或间接依赖于A，B。因此，A，B是这种关系的关键。这不是完美的答案，但是在大多数情况下，它将带您实现目标。
到达密钥后，请使用该密钥并检查是否可以直接或传递到达所有属性。如果是，那么宾果游戏，您就有了钥匙。对于您使用A，B的情况，我们可以访问所有属性。

There is a 6 step process that will lead you to the answer but in many cases the key is to figure out which attribute or set of attributes have only out going relations and no incoming. Here except for A,B all other attributes have dependencies on A,B directly or indirectly. Hence A,B is the key for this relation. This is not the perfect answer but will lead you to the goal in most of the cases. Once you arrive at the key, use that key and check if you can reach all the attributes directly or transitively. If yes then bingo, you have your key. In your case with A,B we can get to all the attributes.

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