分解如何保持依赖关系？
How decomposition is dependency preserving?
设 R（A、B、C、D）是具有以下函数依赖关系的关系模式：
A→B，B→C，
C → D and D → B. R分解为(A, B), (B, C), (B, D)
How decomposition is dependency preserving?
某处解释为"The given decomposition is also dependency preserving as the dependencies A->B is present in (A, B), B->C is present in (B, C), D->B is present in (B, D) and C->D is indirectly present via C->B in (B, C) and B->D in (B, D)."
But my question is, if B→C is given then C→B need not be, right?
Can you explain please.
在这种情况下，C → B 由所有其他依赖项决定 F = {A → B, B → C, C → D, D → B}。
要证明这一点，您必须查看 C 相对于 F 的闭包是否包含 B。给出闭包一组属性的算法中间步骤的结果：
C+ = C
C+ = CD (by using the third dependency)
C+ = CDB (by using the fourth dependency)
因此，由于C+包含B，因此C → B可以从其他依赖项中导出。
设 R（A、B、C、D）是具有以下函数依赖关系的关系模式： A→B，B→C， C → D and D → B. R分解为(A, B), (B, C), (B, D)
How decomposition is dependency preserving?
某处解释为"The given decomposition is also dependency preserving as the dependencies A->B is present in (A, B), B->C is present in (B, C), D->B is present in (B, D) and C->D is indirectly present via C->B in (B, C) and B->D in (B, D)."
But my question is, if B→C is given then C→B need not be, right? Can you explain please.
在这种情况下，C → B 由所有其他依赖项决定 F = {A → B, B → C, C → D, D → B}。
要证明这一点，您必须查看 C 相对于 F 的闭包是否包含 B。给出闭包一组属性的算法中间步骤的结果：
C+ = C
C+ = CD (by using the third dependency)
C+ = CDB (by using the fourth dependency)
因此，由于C+包含B，因此C → B可以从其他依赖项中导出。