## MCQ on Database Management Systems Normalization process, Normalization quiz on 1NF, 2NF, 3NF and BCNF

##
__Normalization Quiz Questions with Answers__

Consider the
attribute set

*R*=*ABCDEGH*and the FD set*F*=*{AB → C, AC → B*,*AD → E*,*B → D*,*BC → A*,*E → G}*to answer the following questions.
Q1. Which of the
following relations

**if decomposed from R**satisfies BCNF?**a) ABC**

b) ABCD

c) ABCEG

d) None of the
above

**View Answer**Answer:
(a)
Let
us say R
_{1} = ABC. If decomposed from R, R_{1} will have the
following set of functional dependencies F_{1};
F
_{1}
= {AB → C, AC → B, BC → A}
The
are AB, AC, and BC. candidate
keys for R1. Hence, R1 in BCNF.As per rules
governing BCNF, LHS of all the functional dependencies must be the candidate key
which is true for R1 |

Q2. Which of the
following relations

**if decomposed from R**satisfies BCNF?
a) ABCD

b) ABCEG

**c) AECH**

d) None of the
above

**View Answer**Answer:
(c)
Let
us say R
_{1} = AECH. If decomposed from R, . Hence, the key for RR_{1}
will have no functional dependencies_{1}
is AECH itself. So, R_{1} is in BCNF. |

Q3. Which of the
following relations

**if decomposed from R**does not satisfy 2NF?
a) ABC

b) AB

**c) ABCEG**

d) All of the above

**View Answer**Answer:
(c)
Let
us say R
_{1} = ABCEG. If decomposed from R, R_{1} will have
the following set of functional dependencies F_{1};
F
_{1}
= { AB → C, AC → B, BC → A, E → G}
Candidate
keys;
(AB)
^{+} = ABC ≠ R_{1}
(AC)
^{+} = ACB ≠ R_{1}
(BC)
^{+} = BCA ≠ R_{1}
(ABE)
^{+} = ABCEG = R_{1}.
Likewise, (ACE)^{+} = (BCE)^{+} = R.
Hence,
.candidate keys are ABE, ACE, and
BCEAs per the rules of 2NF, there shouldn’t be any partial key
dependencies.
But,
in R
_{1}, the FD E → G is a partial key
dependency. That is, E is not a candidate key but a key
attribute. And E alone determines another non-key attribute G uniquely
(the FD E → G). Hence, R_{1}(ABCEG) is not in 2NF. |

Q4. Let us suppose
that R

_{1 }= DCEGH is a relation which is decomposed from R satisfies the set F of functional dependencies F= {E → G}. What is the strongest normal form currently R_{1}is in?**a) 1NF**

b) 2NF

c) 3NF

d) BCNF

**View Answer**Answer:
(a)
The
key is DCEH.
R
_{1}
is not in BCNF because E on the LHS of FD E → G is not a candidate key.
R
_{1}
is not in 3NF and not even in 2NF because the FD E → G is a partial key dependency.
Hence, R
_{1} is in 1NF. |

Q5. If R

_{1 }= ABCD is decomposed from R, which of the following decomposition of R_{1}satisfies BCNF?
a) R

_{11}(AB), R_{12}(CD)**b) R**

_{11}(ABC), R_{12}(BD)
c) R

_{11}(AC), R_{12}(BD)
d) R

_{11}(BCD), R_{12}(AB)

**View Answer**Answer:
(b)
If
R
_{1}(ABCD) is decomposed from R, then the set of functional
dependencies holds by R1 is as follows;
F
_{1}
= { AB → C, AC
→ B, B → D, BC → A }
We
can take, for instance, the first FD AB
→ C to find the closure. The closure of AB = ABC. So, we can create a separate
table R
_{11}(ABC) with the first FD. The remaining is the FD B → D,
and B^{+} is BD. This can go as R_{12}(BD).
Hence, the decomposition results in
R_{11}(ABC), R_{12}(BD) |

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