数据库管理系统|依赖保留分解

依赖保护

R 的分解 D = { R1, R2, R3….Rn } 是依赖保留 wrt 函数依赖的集合 F 如果

(F1 ∪ F2 ∪ … ∪ Fm)+ = F+.
Consider a relation R
R ---> F{...with some functional dependency(FD)....}

R is decomposed or divided into R1 with FD { f1 } and R2 with { f2 }, then
there can be three cases:

f1 U f2 = F -----> Decomposition is dependency preserving. 
f1 U f2 is a subset of F -----> Not Dependency preserving.
f1 U f2 is a super set of F -----> This case is not possible.

问题：让关系 R (A, B, C, D ) 和函数依赖 {AB –> C, C –> D, D –> A}。关系R被分解为R1(A,B,C)和R2(C,D)。检查分解是否保留依赖关系。
解决方案：

R1(A, B, C) and R2(C, D)

Let us find closure of F1 and F2
To find closure of F1, consider all combination of
ABC. i.e., find closure of A, B, C, AB, BC and AC
Note ABC is not considered as it is always ABC 

closure(A) = { A }  // Trivial
closure(B) = { B }  // Trivial
closure(C) = {C, A, D} but D can't be in closure as D is not present R1.
           = {C, A}
C--> A   // Removing C from right side as it is trivial attribute

closure(AB) = {A, B, C, D}
            = {A, B, C}
AB --> C  // Removing AB from right side as these are trivial attributes

closure(BC) = {B, C, D, A}
            = {A, B, C}
BC --> A  // Removing BC from right side as these are trivial attributes

closure(AC) = {A, C, D}
AC --> D  // Removing AC from right side as these are trivial attributes             


F1 {C--> A, AB --> C, BC --> A}.
Similarly F2 { C--> D }

In the original Relation Dependency { AB --> C , C --> D , D --> A}.
AB --> C is present in F1.
C --> D is present in F2.
D --> A is not preserved.

F1 U F2 is a subset of F. So given decomposition is not dependency preservingg.

问题 1：
令 R (A, B, C, D) 是具有以下功能依赖关系的关系模式：

A → B, B → C,
C → D and D → B. 

The decomposition of R into 
(A, B), (B, C), (B, D)

(A)给出无损连接，并且是依赖保留
(B)提供无损连接，但不是依赖保留
(C)不提供无损连接，但保留依赖关系
(D)不提供无损连接并且不是依赖保留

请参阅此解决方案。

问题2
R(A,B,C,D) 是一个关系。下列哪项没有无损连接、保留依赖的 BCNF 分解?
(A) A->B, B->CD
(B) A->B、B->C、C->D
(C) AB->C, C->AD
(D) A -> BCD

请参阅此解决方案。

以下是上一年 GATE 问题的测验。

https://www.geeksforgeeks.org/dbms-gq/database-design-normal-forms-gq/