分解为第三范式 (3NF)

发布于 2024-10-06 04:10:20 字数 179 浏览 4 评论 0原文

Scheme (R) = (A,B,C,D,E,F,G,H)

Function Dependencies (F) = {A->CGH, AD->C, DE->F, G->H}

如何将模式 R 无损连接分解为第三范式 (3NF)？

任何帮助将不胜感激。

原文

Scheme (R) = (A,B,C,D,E,F,G,H)

Function Dependencies (F) = {A->CGH, AD->C, DE->F, G->H}

How would I perform a lossless-join decomposition of the schema R into Third Normal Form (3NF)?

Any help will be appreciated.

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如何视而不见 2024-10-13 04:10:20

由于 A→CGH 和 Ax→C 对于任何字母 x，我们可以忽略第二个函数依赖关系 (AD→C)，因为它没有告诉我们 A→CGH 没有告诉我们的任何信息。

没有什么可以决定B；由于G决定

H，A决定G和H，所以我们可以将G→H分成一个关系（存在传递依赖A→G和G→H）。

R1 = { G, H }       : PK = { G }

剩下 F' = { A→CG, DE→F } 和 R' = (A, B, C, D, E, F, G)。

剩下的两个函数依赖关系可以形成另外两个关系：

R2 = { A, C, G }    : PK = { A }
R3 = { D, E, F }    : PK = { D, E }

R'' = { A, B, D, E }

R4 = { A, B, D, E } : PK = { A, B, D, E }

R1、R2、R3 和 R4 的连接应该让您得到您开始时使用的任何起始值的 R R 的（满足给定函数依赖关系的约束）。

Since A→CGH and Ax→C for any letter x, we can ignore the second of the functional dependencies (AD→C) because it doesn't tell us anything that A→CGH doesn't also tell us.

There is nothing that determines B; there is nothing that determines D.

Since G determines H, and A determines both G and H, we can separate G→H into a relation (there is a transitive dependency A→G and G→H).

R1 = { G, H }       : PK = { G }

That leaves F' = { A→CG, DE→F } and R' = (A, B, C, D, E, F, G).

The two functional dependencies left can form two more relations:

R2 = { A, C, G }    : PK = { A }
R3 = { D, E, F }    : PK = { D, E }

That leaves R'' = { A, B, D, E }

R4 = { A, B, D, E } : PK = { A, B, D, E }

The join of R1, R2, R3, and R4 should leave you with the R you started with for any starting value of R (that satisfies the constraints of the given functional dependencies).

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