GO Classes DPP | GATE CS, DA | DBMS | 3NF

Answer 1

First find the candidate key.

Attributes $A$ and $B$ do not appear on the RHS of any FD, so both must be present in every candidate key.

Compute $(AB)^+$.

From $AB \to C$, add $C$.

Now $B$ and $C$ are available, so $BC \to D$ gives $D$.

$\therefore (AB)^+=\{A,B,C,D\}$

Thus $AB$ is the candidate key.

Hence,

Prime attributes $=\{A,B\}$

Non-prime attributes $=\{C,D\}$.

Now examine the FDs.

For $AB \to C$, $AB$ is a candidate key and hence a superkey. 

Therefore, this FD satisfies $\text{3NF}$.

For $BC \to D$, $BC$ is not a superkey because $(BC)^+=\{B,C,D\}$, which does not contain $A$.

Also, $D$ is non-prime.

Thus neither $\text{3NF}$ condition is satisfied:

• LHS is not a superkey
   
• RHS is not prime

$\therefore BC \to D$ violates $\text{3NF}$.

Hence, Answer : B

Answer 2

R(A, B, C, D)

FDs :- AB -> C, BC -> D

AB -> C // AB is a Super Key because AB⁺ = {A, B, C, D} and A and B are prime attributes

BC -> D // Neither BC is a Super Key not D is a Prime Attribute

Therefore, Option B is correct

Answer 3

Step 1: Find the Candidate Key(s)

To find the candidate key, we compute the closure of attributes. Attributes $A$ and $B$ are not present on the right-hand side of any functional dependency, meaning they must be part of any candidate key.

Let's compute the closure of $AB$, denoted as $(AB)^+$:

• $(AB)^+ = \{A, B\}$ (trivial)
• $(AB)^+ = \{A, B, C\}$ (using the dependency $AB \rightarrow C$)
• $(AB)^+ = \{A, B, C, D\}$ (using the dependency $BC \rightarrow D$)

Since the closure of $AB$ contains all attributes of relation $R$, $AB$ is the only candidate key.

Step 2: Identify Prime and Non-Prime Attributes

• Prime attributes (part of candidate key): $A, B$
• Non-prime attributes (not part of candidate key): $C, D$

Step 3: Check 3NF Conditions

For a relation to be in 3NF, every non-trivial functional dependency $X \rightarrow Y$ must satisfy at least one of the following:

1. $X$ is a superkey.
2. $Y$ is a prime attribute.

Let's test both given dependencies:

• $AB \rightarrow C$: The left side $AB$ is a candidate key (superkey). This satisfies 3NF.
• $BC \rightarrow D$: Let's find $(BC)^+$. $(BC)^+ = \{B, C, D\}$. Since it doesn't contain all attributes of $R$, $BC$ is not a superkey. Furthermore, $D$ is a non-prime attribute. This dependency violates 3NF.

Conclusion:

The correct statement is B: $BC \rightarrow D$ violates 3NF because $BC$ is not a superkey and $D$ is non-prime.