Advanced Database Management System - Tutorials and Notes: Normalization in DBMS - Multiple Choice Questions with Answers

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# Normalization process in RDBMS, multiple choice questions with answers in RDBMS, normal forms and functional dependencies MCQs.

## Database Management Systems Quiz - Normalization Process in DBMS

Assume a relation R(A, B, C, D)  with set of functional dependencies F = { C → D, C → A, B → C}. Use this setup to answer the following questions;

### 1. Which of the following is the candidate keys of R?

a) C
b) BC
c) B
d) Both (b) and (c)

 Answer: (c) B Only left hand side (LHS) attributes of the given set of FDs are C and B. B determines C uniquely. Hence, we can find the closure of B to check whether it forms the candidate key or not as follows; B+ = BCDA through FDs B → C, C → D, and C → A. And, B+ = R. So, B is the only candidate key for R.

### 2. Which is the normal form that the relation R is currently complies with?

a) First Normal Form (1NF)
b) Second Normal Form (2NF)
c) Third Normal Form (3NF)
d) All of the above

 Answer: (b) Second Normal Form (2NF) C → D, and C → A are non-trivial FDs. C is not a super key and D is not part of any candidate key. Hence, R is not in 3NF. No partial key dependencies present in R since B is the only key (and is not composite key). Hence, R is currently in 2NF.

### 3. R is not in BCNF. Which of the following shows the correct decomposition of R into BCNF relations?

a) R1(CDA), R2(BC)
b) R1(BD), R2(CA)
c) R1(BC), R2(CA), R3(CD)
d) None of the above

 Answer: (c) R1(BC), R2(CA), R3(CD) BCNF relation – “The LHS of a functional dependency should be a key for the relation if the relation is in BCNF”. The FDs C → D and C → A violates the properties of BCNF because C is not a candidate key for R. (B → C does not violate because B is the candidate key for R). Let us decompose R using one of the violating FD C → D. To decompose, let us create separate relation for violating FD. We will get R1(A, B, C) and R2(C, D). Is the decomposition in BCNF? R2 is in BCNF due to the FD C → D whereas R1 is not due to the FD C → A. So, let us further decompose R1 by creating a relation for violating FD. We will get R11(B, C) and R12(C, A). Both are in BCNF. Hence, BCNF decomposition of R is R1(B, C), R2(C, A) and R3(C, D).

### 4. Which among the following is the canonical cover (minimal cover Fc) of the relation R?

a) Fc = {C → DA, B → C}
b) Fc = {BC → A, BC → D}
c) Fc = {C → A, B → C, D → A}
d) All of the above

 Answer: (a) Fc = {C → DA, B → C} A canonical cover (minimal cover) of a set of FDs F is a minimal set of functional dependencies Fmin that is equivalent to F. To understand the process of finding minimal cover, please refer here. For the relation R and set of FDs F, the set of functional dependencies {C → DA, B → C} is the minimal set.

5. R can be decomposed into set of 3NF relations R1(C, D, A) and R2(B, C). This decomposition is a _____.
a) Lossless join decomposition
b) Lossy join decomposition
c) Dependency preserving decomposition
d) Both lossless and dependency preserving decomposition

 Answer: (d) Both lossless and dependency preserving decomposition Decomposition is lossless if either (R1 ∩ R2) → R1 or (R1 ∩ R2) → R2 holds. According to the question, (R1 ∩ R2) → R1 is TRUE [How? (CDA) ∩ (BC) → (CDA) ⇒ C → CDA]. Hence, the decomposition is lossless. Decomposition is dependency preserving if (F1 U F2 U … Fn)+ = F+. that is, if the closure of union of individual relations functional dependencies is equal to the closure of FDs of original relation. This is also true here. Hence, the decomposition is a dependency preserving decomposition.

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