I have not understood how the following problem must be approached. Any help in learning how to solve this question will be much appreciated!
我不明白必须如何处理以下问题。任何帮助学习如何解决这个问题将不胜感激!
Consider Relation Schema R = {ABCDEFG} with a set of Functional Depenedencies
F = {GA -> D, DC -> E, GF -> A, CA -> GB, AF -> D, F -> G}
Identify any redundant Functional Dependencies.
1 个解决方案
#1
5
Consider GA -> D. To check whether it is redundant we need to check whether we can infer D from GA by using dependencies other than GA -> D. However, no other dependency is applicable to GA so GA -> D is not redundant.
考虑GA - > D.要检查它是否是冗余的,我们需要检查我们是否可以通过使用GA以外的依赖关系从GA推断D.但是,没有其他依赖关系适用于GA因此GA - > D不是多余的。
Consider AF -> D. If we know AF, then we also know AFG since F -> G. Moreover, since GA -> D we also know D. Hence, we have inferred D from AF without using the dependency AF -> D meaning that AF -> D is redundant.
考虑AF - > D.如果我们知道AF,那么我们也知道AFG,因为F - > G.此外,由于GA - > D我们也知道D.因此,我们从AF推断D而不使用依赖AF - > D意味着AF - > D是多余的。
If your lecture slides also discuss redundant (extraneous) attributes you can check that G is an extraneous attribute in GF -> A since F -> G.
如果您的演讲幻灯片还讨论冗余(无关)属性,您可以检查G是GF中的无关属性 - > A,因为F - > G.
#1
5
Consider GA -> D. To check whether it is redundant we need to check whether we can infer D from GA by using dependencies other than GA -> D. However, no other dependency is applicable to GA so GA -> D is not redundant.
考虑GA - > D.要检查它是否是冗余的,我们需要检查我们是否可以通过使用GA以外的依赖关系从GA推断D.但是,没有其他依赖关系适用于GA因此GA - > D不是多余的。
Consider AF -> D. If we know AF, then we also know AFG since F -> G. Moreover, since GA -> D we also know D. Hence, we have inferred D from AF without using the dependency AF -> D meaning that AF -> D is redundant.
考虑AF - > D.如果我们知道AF,那么我们也知道AFG,因为F - > G.此外,由于GA - > D我们也知道D.因此,我们从AF推断D而不使用依赖AF - > D意味着AF - > D是多余的。
If your lecture slides also discuss redundant (extraneous) attributes you can check that G is an extraneous attribute in GF -> A since F -> G.
如果您的演讲幻灯片还讨论冗余(无关)属性,您可以检查G是GF中的无关属性 - > A,因为F - > G.