Test by Bikram | Databases | Test 2 | Question: 5

Question

Consider the following set of FDs on a relation $R(ABCDE)$:

$F1 = \{ A \rightarrow B, \ AB \rightarrow C, \ D \rightarrow AC, \ D \rightarrow E \}$
$F2 = \{ A \rightarrow BC, \ D \rightarrow AE \}$

Which of the following is TRUE regarding $F1$ and $F2$?

• A. $F2$ covers $F1$, but $F1$ doesn’t cover $F2$.
• B. $F1$ covers $F2$, but $F2$ doesn’t cover $F1$.
• C. $F1$ is equivalent to $F2$.
• D. $F1$ is not equivalent to $F2$.

Comments

Sir, what it means that F1 "covers" F2 ??

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F1 covers F2 means all the FD's of F2 can be determined by F1

Answer 1

F1 covers F2 means all the FD's of F2 can be determined by F1

FD's are :

F1 = { A --> B, AB --> C, D --> AC, D--> E }
F2 = { A --> BC, D --> AE }

Considering F1 covers F2:

FD of  F2 A->B is directly determined in F1 by A->B

FD of F2 A->C is Determined in F1 by  AB->C where in AB B is redundant attribute as A->B so we can say A->C

FD of F2 D->A is determined in F1 by D->A

FD of F2 D->E is determined in F1 by D->E

therefore F1 covers F2--------------------------------(1)

Now considering F2 covers F1:

FD of  F1 A->B is directly determined in F2 by A->B

FD of F1 AB->C is determined in F2 by  A->B and A->C therefore AB->C

FD of F1 D->A is determined in F2 by D->A

FD of F1 D->C is determined in F2 by D->A and A->C therefore D->C

FD of F1 D->E is determined in F2 by D->E

therefore F2 covers F1--------------------------(2)

From (1) and (2) F1 is equivalent to F2