如何找到给定关系的最高范式 [英] how to find the highest normal form for a given relation
问题描述
我浏览了互联网和书籍，但在如何确定这种关系的范式上仍有一些困难
I've gone through internet and books and still have some difficulties on how to determine the normal form of this relation
R(a, b, c, d, e, f, g, h, i)
FDs =
B→G
BI→CD
EH→AG
G→DE
到目前为止，我知道唯一的候选键是 BHI(如果我应该用 F 来计算，那么 BFHI).因为根本没有使用属性 F.完全独立于给定的 FD.
So far I've got that the only candidate key is BHI (If I should count with F, then BFHI). Since the attribute F is not in use at all. Totally independent from the given FDs.
- 那么我应该如何处理属性 F?
- 如何确定关系 R 的最高范式?
推荐答案
那么我应该如何处理 F 属性?
What am I supposed to do with the attribute F then?
您可以观察到，唯一提到 F 的 FD 是微不足道的 F-> F.之所以没有明确提及，正是因为它微不足道.尽管如此，阿姆斯特朗的所有公理都同样适用于琐碎的公理.所以，你可以使用这个微不足道的，例如应用增强，从 B->G 到 BF->GF;
You could observe the fact that the only FD in which F gets mentioned, is the trivial one F->F. It's not explicitly mentioned precisely because it is trivial. Nonetheless, all of Armstrong's axioms apply to trivial ones equally well. So, you can use this trivial one, e.g. applying augmentation, to go from B->G to BF->GF;
如何确定关系R的最高范式?
How to determine the highest normal form for the relation R?
首先，测试第一范式的条件.如果满足，NF 至少为 1.检查第二范式的条件.如果满足，NF 至少为 2.检查第三范式的条件.如果满足，NF 至少为 3.
first, test the condition of first normal form. If satisfied, NF is at least 1. Check the condition of second normal form. If satisfied, NF is at least 2. Check the condition of third normal form. If satisfied, NF is at least three.
注意:
检查第一范式的条件"，在正式过程中做的事情有点奇怪，因为不存在该条件的正式定义这样的东西，除非你通过日期，但我有毫无疑问，您的课程不遵循该定义.
"checking the condition of first normal form", is a bit of a weird thing to do in a formal process, because there exists no such thing as a formal definition of that condition, unless you go by Date's, but I have little doubt that your course does not follow that definition.
提示:
鉴于唯一的密钥是 BFHI，这是密钥，整个密钥，除了密钥什么都没有"的第一个子句，它被 B->G 违反?
Given that the sole key is BF which is the first clause of "the key, the whole key, and nothing but the key" that gets violated by, say, B->G ?
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