如何找到给定关系的最高范式
我已经浏览过互联网和书籍，但对于如何确定这种关系的正常形式仍然存在一些困难
R(a, b, c, d, e, f, g, h, i)
FDs =
B→G
BI→CD
EH→AG
G→DE
到目前为止，我已经知道唯一的候选键是BHI（如果我应该用F数，那么BFHI）。 由于属性 F 根本没有被使用。完全独立于给定的 FD。
- 那么我应该如何处理属性 F 呢？
- 如何确定关系R的最高范式？
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
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.
- What am I supposed to do with the attribute F then?
- How to determine the highest normal form for the realation R?
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那么我应该如何处理属性 F 呢？
您可以观察到这样一个事实：唯一提到 F 的 FD 是琐碎的 F->F。之所以没有明确提到它，是因为它很微不足道。尽管如此，阿姆斯特朗的所有公理同样适用于琐碎的公理。因此，您可以使用这个简单的方法，例如应用增强，从 B->G 到 BF->GF；
如何确定关系R的最高范式？
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。
注意：
“检查第一范式的条件”在正式流程中是有点奇怪的事情，因为不存在该条件的正式定义，除非您按照日期进行，但我几乎没有怀疑您的课程不遵循该定义。
提示：
假设唯一的密钥是 BFHI，它是“密钥、整个密钥，除了密钥之外什么都没有”的第一个子句，例如 B->G 违反了？
What am I supposed to do with the attribute F then?
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;
How to determine the highest normal form for the relation R?
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.
Note :
"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.
Hint :
Given that the sole key is BFHI, which is the first clause of "the key, the whole key, and nothing but the key" that gets violated by, say, B->G ?