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在3NF中找到关系但在BCNF中找不到关系

[英]Finding a relation in 3NF but not in BCNF

原文 2014-05-15 14:39:02 0 2 database/ database-normalization/ functional-dependencies/ 3nf/ bcnf

I've been reading many different sources on how to differentiate relations that are in 3NF/BCNF. 我一直在阅读许多关于如何区分3NF / BCNF关系的不同来源。 And I've so far this is my understanding... 到目前为止，这是我的理解......

I will use this relation as an example... 我将以此关系为例......

R = {A, B, C, D, E}

and 和

F = {A -> B, BC - > E, ED -> A} . F = {A -> B, BC - > E, ED -> A} 。

Firstly we must find the keys of the relation. 首先，我们必须找到关系的关键。 I used this video to help me do that. 我用这个视频来帮助我做到这一点。 And I got 我得到了

Keys = {ACD, BCD, CDE}

Now to make sure R is in BCNF , we must make sure that the left hand side of every functional dependency in F is one of the Keys . 现在为了确保R在BCNF中 ，我们必须确保F中每个功能依赖的左侧是其中一个Keys 。 We instantly know this is not the case, because the first FD is A -> B and A is not one of the keys. 我们立即知道情况并非如此，因为第一个FD是A -> B而A不是其中一个键。 So it is not in BCNF. 所以它不在BCNF。

Now to make sure R is in 3NF , we must make sure that the left hand side of every functional dependency in F is one of the Keys OR the right hand side of every functional dependency in F is a subset of one of the Keys . 现在，以确保R是3NF，我们必须确保在每一个函数依赖的左边F是一个Keys 或每一个函数依赖的右边F是一个子集Keys 。 If you look at the right hand side of every FD, they are B , E and A . 如果你看看每个FD的右侧，它们是B ， E和A These are each a subset of a Key , so this means that it is in 3NF . 这些都是Key的子集，因此这意味着它在3NF中 。

So this is one of the rare cases (according to wiki) where a relation is in 3NF but not in BCNF . 所以这是罕见的情况之一（根据wiki），其中关系是3NF而不是 BCNF 。 Is this method correct? 这种方法是否正确？ Is it reliable? 它可靠吗？ Am I missing anything? 我错过了什么吗？

2 个解决方案

First you need to learn superkeys, candidate keys, and primary attributes. 首先，您需要学习超级键，候选键和主要属性。

However, this rule of thumb helps: 但是，这个经验法则有助于：

A 3NF table that does not have multiple overlapping candidate keys is guaranteed to be in BCNF. 不具有多个重叠候选键的3NF表保证在BCNF中。

In other words, if the candidate keys in a 3NF relation are 换句话说，如果3NF关系中的候选键是

all atomic, or 所有原子，或
non-atomic but non-overlapping, 非原子但不重叠，

it is guaranteed that the relation is in BCNF. 保证关系在BCNF中。

The simplest relation which violates BCNF but meets 3NF has the following functional dependencies: 违反BCNF但满足3NF的最简单关系具有以下功能依赖性：

A,B -> CC -> B

In this case, candidate keys are (A,B) and (A,C) . 在这种情况下，候选键是(A,B)和(A,C) 。
It meets 3NF because 它满足3NF因为

the right-hand-side of all functional dependencies is a primary attribute . 所有功能依赖项的右侧是主要属性 。

It violates BCNF because 它违反了BCNF，因为

C -> B , but the left-hand-side is not a superkey . C -> B ，但左侧不是超级键 。

BCNF: BCNF：

X->Y where Y can be prime or non-prime X-> Y，其中Y可以是素数 或非素数
Here,X must be a super key 这里，X 必须是超级密钥

3NF : 3NF ：

X->Y where Y is non-prime X-> Y，其中Y 是非素数
then, 然后，
X must be a super key X 必须是超级密钥
else 其他
X need not be super key X 不一定是超级密钥