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Showing posts with label Normalization. Show all posts
Showing posts with label Normalization. Show all posts

## Sunday, November 22, 2020

### Normalization MCQ with answers in DBMS 19

Normalization in DBMS, solved exercises in DBMS, lossless join decomposition, dependency preserving decomposition, bcnf decomposition

RDBMS MCQ solved quiz and answers

1. Let us assume that a relation R (A, B, C, D, E) with set of functional dependencies F = {A BC, C D} is decomposed into relations R1 (A, B, C) and R2 (A, D, E). This decomposition is ___________.

a) Lossless join decomposition

b) Dependency preserving decomposition

c) Not a dependency preserving decomposition

d) Lossy decomposition

 Answer: (a) and (c) Common attribute between R1 and R2 is A, and, attribute A determines all attributes of R1. Hence, it is a lossless decomposition. It is not a dependency preserving decomposition because the FD C → D is lost. Lossless join decomposition Decomposition of relation R into R1 and R2 is said to be lossless join decomposition if one of the following holds; (R1 ∩ R2) → R1 (R1 ∩ R2) → R2 Dependency preserving decomposition If a relation R with set F of functional dependencies is decomposed into relations R1, R2, R3, …, Ri then the closure of set of functional dependencies for these relations should satisfy the following;  (F1 U F2 U F3 U … U Fi)+ = F+ That is the closure of union of set of functional dependencies of relations R1, R2, …, Ri should be equal to the closure of set of functional dependencies F of R. In other words, all the functional dependencies in (F1 U F2 U F3 U … U Fi)+ should be in F+ also.

2. Let us assume that a relation R (A, B, C, D, E, F, G, H) with set of functional dependencies F = {AB E, C D, D E, FG A} is decomposed into relations R1(ABE), R2(CD), R3(FGA) and R4(BCFGH). The decomposition _____.

a) is resulted in all BCNF relations

b) is dependency preserving decomposition

c) is not a dependency preserving decomposition

d) is lossless decomposition

 Answer: (a) and (c) R1(ABE), R2(CD), R3(FGA) and R4(BCFGH). Keys are underlined. All relations are in BCNF. The functional dependency D → E is lost. Hence, decomposition of R into R1, R2, R3 and R4 is not a dependency preserving decomposition.

3. Consider a relation R(A, B, C, D, E) with the set of functional dependencies F = {A B, B E, E A}. Relation R is in ____.

a) Un-normalized form

b) Third Normal Form

c) Boyce-Codd Normal Form

d) Domain Key Normal Form

 Answer: (b) 3NF The candidate keys for R are ACD, BCD, and ECD. No non-key dependencies found in R. Hence, relation R is in third normal form.

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Related posts:

# 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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Related posts:

# How to prove that two sets of functional dependencies are equal or not - Proving equality of two sets of functional dependencies - How to find whether two sets of functional dependencies are equal or not? - Equivalent sets of functional dependencies

Exercise:
Let F = {A → C, AC → D, E → AD}, and let G = {A → CD, E → AHE}. Are they equivalent?

Solution:
To check whether two sets of functional dependency are equivalent, we have to verify every functional dependency (FD) of F is in G+ and every FD of G is in F+.
Theorem: If F G+ and G F+, then F and G are equivalent.

To check for F G+
Let us take the first FD of F, A → C and find the closure using FDs from G as follows;
• A+ = ACD through the FD A → CD of G. The result includes the RHS attribute of FD A → C. Hence, A → C is in G+.
Let us do the same for other FDs of F;
• AC+ = ACD through the FD A → CD of G. The result includes the RHS attribute of FD AC → D. Hence, AC → D is in G+.
• E+ = AHECD through the FDs E → AHE and A → CD of G. The result includes the RHS attribute of FD E → AD. Hence, E → AD is in G+.
As all the FDs of F can be derived using G, F G+ is true.

To check for G F+
• A+ = ACD through the FDs A → C and AC → D of F. The result includes the RHS attribute of FD A → CD. Hence, A → CD is in F+.
• E+ = ACDE through the FDs E → AD, A → C and AC → D of F. The result does not include the RHS attribute of FD E → AHE. Hence, E → AHE is in F+.
As all the FDs of G cannot be derived using F, G F+ is false.

Result:
We found that F G+ is true but not G F+. Hence, the given set of functional dependencies F and G are not equivalent (F ≠ G).

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## How to find extraneous attribute?

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