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 :)
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