依存理论

发布于 2024-10-22 23:43:17 字数 421 浏览 6 评论 0原文

有谁知道有一个好的网站、书籍或任何其他资源可以很好地解释依赖理论吗？我陷入了与下面所示类似的问题：

鉴于

R   < A = {P,Q,R,S,T,U,Y },


gamma = {Y->S   …(1)
       Q->ST….(2)  

U-> Y……(3)
       S->R  …...(4)

RS->T…….(5) }>.
RTP U->T 成立

答案是：

U -> Y -> S -> RS -> T
aug (4) by S  S->R

原文

Does anyone know of a good website, book or any other resources that would explain dependency theory well? I am stuck on a similar question to the one shown below:

Given

R   < A = {P,Q,R,S,T,U,Y },


gamma = {Y->S   …(1)
       Q->ST….(2)  

U-> Y……(3)
       S->R  …...(4)

RS->T…….(5) }>.
RTP U->T  holds

Answer is:

U -> Y -> S -> RS -> T
aug (4) by S  S->R

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一页 2024-10-29 23:43:17

我认为您需要搜索功能依赖而不是依赖理论。维基百科有一篇关于函数依赖的介绍性文章。表达式“Y->S”表示

Y 决定 S，或者
如果您知道“Y”的一个值，则您可以
知道“S”的一个值（而不是
两个或三个或七个“S”值），或者
如果两个元组具有相同的“Y”值，它们也将具有相同的“S”值

我不熟悉您发布的所有符号。但我认为你会被要求从一个关系R和一组编号为1到4的函数依赖gamma开始，以供参考。

Relation R = {P,Q,R,S,T,U,Y }

FD gamma = {Y->S   (1)
            Q->ST  (2)  
            U-> Y  (3)
            S->R   (4) }

这似乎是几个问题的“设置”。然后你会被要求承担这个额外的功能依赖性。

RS->T  (5)

根据设置和附加 FD，您应该证明函数依赖性 U->T 成立。讲师的答案是“U -> Y -> S -> RS -> T”，我认为这是讲师希望你遵循的推论链。首先给你 U->Y 和 Y->S，所以具体的推理链是这样的。

U->Y 和 Y->S，因此 U->S。（传递性，讲师的 U->Y->S）
S->R，因此 S->RS。（增强，中间步骤）
U->S 和 S->RS，因此 U->RS。（传递性，讲师的 U->Y->S->RS）
U->RS 和 RS->T，因此 U->T。（传递性，讲师的 U->Y->S->RS->T）

I think you'll need to search for functional dependency instead of dependency theory. Wikipedia has an introductory article on functional dependency. The expression "Y->S" means

Y determines S, or
if you know one value for 'Y', you
know one value for 'S' (instead of
two or three or seven values for 'S'), or
if two tuples have the same value for 'Y', they'll also have the same value for 'S'

I'm not familiar with all the notation you posted. But I think you're asked to begin with a relation R and a set of functional dependencies gamma numbered 1 to 4 for reference.

Relation R = {P,Q,R,S,T,U,Y }

FD gamma = {Y->S   (1)
            Q->ST  (2)  
            U-> Y  (3)
            S->R   (4) }

This appears to be the "setup" for several problems. You're then asked to assume this additional functional dependency.

RS->T  (5)

Based on the setup and on that additional FD, you're supposed to prove that the functional dependency U->T holds. The lecturer's answer is "U -> Y -> S -> RS -> T", which I think is the chain of inferences the lecturer wants you to follow. You're given U->Y and Y->S to start with, so here's how that specific chain of inference goes.

U->Y and Y->S, therefore U->S. (transitivity, Lecturer's U->Y->S)
S->R, therefore S->RS. (augmentation, an intermediate step)
U->S and S->RS, therefore U->RS. (transitivity, Lecturer's U->Y->S->RS)
U->RS and RS->T, therefore U->T. (transitivity, Lecturer's U->Y->S->RS->T)