如何从函数依赖中获取最小键

如何从函数依赖中获取最小键

本文介绍了如何从函数依赖中获取最小键?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我需要一些帮助和指导。

I need some help and guidelines.

我有以下关系: R = {A,B,C,D,E, F} 和函数依赖性集合

I have the following relation: R = {A, B, C, D, E, F} and the set of functional dependencies


F = {
  {AB -> C};
  {A  -> D};
  {D  -> AE};
  {E  -> F};
}

R的主键是什么?

如果我应用推理规则,我得到这些额外的函数依赖:

If i apply inference rules i get these additional Function dependencies:

D -> A
D -> E
D -> F

D -> AEF

A -> E
A -> F
A -> DEF

如何继续?

推荐答案

有一个众所周知的算法来做到这一点。我不记得了,但是练习似乎很简单,不能使用它。

There is a well known algorithm to do this. I don't remember it, but the excercise seems to be simple enough not to use it.

我认为这一切都是关于传递性:

I think this is all about transitivity:

CurrentKey = {A, B, C, D, E, F}

你知道D决定E和E决定F.因此,D通过传递性确定F.由于F不确定任何东西,我们可以删除它,并且E可以从D中获得,我们也可以删除它:

You know D determines E and E determines F. Hence, D determines F by transitivity. As F doesn't determine anything, we can remove it and as E can be obtained from D we can remove it as well:

CurrentKey = {A, B, C, D}

由于AB决定C和C确定我们知道它不能作为密钥的一部分,所以我们删除它:

As AB determines C and C doesn't determine anything we know it can't be part of the key, so we remove it:

CurrentKey = {A, B, D}

最后,我们知道A决定D,所以我们可以从密钥中删除后者:

Finally we know A determines D so we can remove the latter from the key:

CurrentKey = {A, B}

如果一旦你有这个可能的键,你可以重新创建所有的功能依赖它是一个可能的键。

If once you have this possible key, you can recreate all functional dependencies it is a possible key.

PS:如果你碰巧有算法,请张贴它,因为我很高兴重新学习:)

PS: If you happen to have the algorithm handy, please post it as I'd be glad to re-learn that :)

这篇关于如何从函数依赖中获取最小键?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!

09-05 01:56