NIELIT 2021 Dec Scientist A - Section B: 57

Question

Consider a relation $\text{R}$ with attributes $\{ \text{A, B, C} \}$ and functional dependency set $\text {S= \{ A} \rightarrow \text{B, } \text{A} \rightarrow \text{C} \; \}.$ Then relation $\text{R}$ can be decomposed into two relations :

• A. $ \text{R1} \{ \text{A, B} \} \; \text{AND} \; \text{R2} \{ \text{A, C} \} $
• B. $ \text{R1} \{ \text{A, B} \} \; \text{AND} \; \text{R2} \{ \text{B, C} \} $
• C. $ \text{R1} \{ \text{A, B, C} \} \; \text{AND} \; \text{R2} \{ \text{A, C} \} $
• D. None of the above

Answer 1

The given relation is R{A,B,C}

Functional dependency {A->C,A->B}

Candidate key is A.

option B:- the decomposed relation is R1{A,B} ,R2{B,C}

the common attribute is B.

now B is not key in any of the table.

so this decomposition is lossy.

The given decomposition is binary so we can apply these method otherwise go with chase algorithm.

so option B is false.

Option C: – R1{A,B,C} ,R2{A,C}

they told to decompose and gave the same table along with another table R2{A,C}.

It only create redundancy makes no sense .

so this option is also false.

option A .

It is true.

the decomposed relation is R1{A,B} ,R2{A,C}

Common attribute is A which is key in Key in both the table.

so correct answer is A.