从函数依赖性检查第三范式
我有一个关于根据函数依赖关系确定关系是否处于第三范式的问题。
R = {A,B,C,D,E}
<代码>A - > B
<代码>BC - > E <代码>ED - > A
由此，我确定候选键为： {ACD},{BCD},{CDE}
维基百科表示，如果对于每个函数依赖 X->Y，关系至少满足以下要求之一，则该关系处于第三范式： 1. Y是X的子集 2. X 是超级键 Y 是某个键 K 的 K 子集
我的工作：A ->由于键 满足 3，由于 {BCD}，B{CDE}，BC->E 满足 3，并且CD->由于 满足3。{ACD}，A
这是解释这些规则的正确方式吗？
I have a question about determining whether a relation is in 3rd Normal Form based on functional dependencies.
R = {A,B,C,D,E}A -> BBC -> EED -> A
From this, I determined the candidate keys to be:{ACD},{BCD},{CDE}
Wikipedia says that a relation is in 3rd Normal Form if, for each functional dependency X->Y, it meets at least one of the following requirements:
1. Y is a subset of X
2. X is a superkey
Y is a subset of K for some key K
My work: A -> B satisfies 3 because of the key {BCD}, BC->E satisfies 3 because of {CDE}, and CD -> A satisfies 3 because of {ACD}.
Is this the correct way of interpreting those rules?
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<代码>A - > B 表示函数依赖。但是这些
R1 = {A,B}
R2 = {B,C,E}
R3 = {E,D,A}
表示关系。 （我假设您的意思是我们将这些视为您分解 R 得到的关系。）
范式适用于关系；它们不适用于函数依赖。说“
A -> Bis in 3NF”或“A 决定 B is in 3NF”是没有任何意义的。所以 R1 = {A,B} 是一个关系。它不“由于密钥 {B,C,D} 而满足规则 3”。不能，因为 {B,C,D} 不在 R1 中。
这些规则与分解 R 所产生的关系的候选键有关，而不是与 R 本身的候选键有关。
稍后。 。 .我发现我做出了错误的假设。您正在尝试评估 R，而不是分解它。在这种情况下，你的推理是正确的，R 属于 3NF。您还可以得出结论，R 属于 3NF，因为不存在非素数属性。由于不存在非素数属性，因此每个 FD 的右侧必须是候选键的一部分。
但R不在BCNF中。 （或者 4NF，或者 5NF。）你能解决这个问题吗？
A -> Bexpresses a functional dependency. But theseR1 = {A,B}
R2 = {B,C,E}
R3 = {E,D,A}
express relations. (I'm assuming you meant us to take these as the relations you got from decomposing R.)
Normal forms apply to relations; they don't apply to functional dependencies. It doesn't make any sense to say "
A -> Bis in 3NF", or "A determines B is in 3NF".So R1 = {A,B} is a relation. It doesn't "satisfy rule 3 because of the key {B,C,D}". It can't, because {B,C,D} isn't in R1.
Those rules have to do with the candidate keys of the relations that result from decomposing R, not with the candidate keys of R itself.
Later . . . I see I made an incorrect assumption. You're trying to evaluate R, not decompose it. In that case your reasoning is correct, and R is in 3NF. You could also conclude that R is in 3NF because there are no non-prime attributes. Since there are no non-prime attributes, the right-hand side of every FD must be part of a candidate key.
But R is not in BCNF. (Or 4NF, or 5NF.) Can you fix that?