无损加入分解 [英] Lossless Join Decomposition

问题描述

我正在学习一门考试，这是在学习指南上。

关系模式R =（A，B，C，D，E）

功能依赖性=（AB-> E，C-> AD，D-> B，E-> C）

是r1 = ，C，D）r2 =（B，C，E）OR

p x1 =（A，C，D）x2 =连接分解？为什么？

解决方案

我的关系代数是可怕的生锈，但这里是我记得它去了

如果 r1∩r2 - > r1-r2 或 r1∩r2 - > r2 - r1 ，然后你有无损分解。

  r1∩r2 = C 
 r1  -  r2 = AD 
  

C-> AD 在函数依赖性=> lossless

为x1和x2

  x1∩x2 = A 
 x1  -  x2 = CD 
  

A-> CD 不在 FD s
现在检查x2 - x1

  x2  -  x1 = BE 
  

code> A-> BE 不在 FD 中，因此有损

引用此处，请检查我可能犯下的可怕错误

I am studying for a test, and this is on the study guide sheet. This is not homework, and will not be graded.

Relation Schema R = (A,B,C,D,E)

Functional Dependencies = (AB->E, C->AD, D->B, E->C)

Is r1 = (A,C,D) r2 = (B,C,E) OR

x1 = (A,C,D) x2 = (A,B,E) a lossless join decomposition? and why?

解决方案

My relational algebra is horribly rusty, but here is how I remember it to go

If r1 ∩ r2 -> r1 - r2 or r1 ∩ r2 -> r2 - r1 in FDs then you have lossless decomposition.

r1 ∩ r2 = C
r1 - r2 = AD

C->AD is in functional dependencies => lossless

for x1 and x2

x1 ∩ x2 = A
x1 - x2 = CD

A->CD is not in FDs now check x2 - x1

x2 - x1 = BE

A->BE is not in FDs either, therefore lossy

references here, please check for horrible mistakes that I might have committed

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