Advanced Database Management System - Tutorials and Notes: Normalization - TRUE/FALSE Questions with Proofs

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Monday, 15 September 2014

Normalization - TRUE/FALSE Questions with Proofs

Normalization TRUE/FALSE Questions with explained proofs for understanding / TRUE/FALSE Questions with answer discussion on Normalization / TRUE/FALSE Questions on Normal Forms Discussion

TRUE/FALSE Questions – Normalization and Normal Forms

1. For a relation schema R(A, B, C, D), if AB C and C D, then AB is a key.
Proof: Try to find (AB)+, ie., closure of (AB).
Step 1: result = AB
Step 2: If you know AB, then using AB C, result = ABC
Step 3: If you know AB and C, and using C D, the result = ABCD
At this stage, result included all the attributes of R. Hence, (AB) is a key for R.

2. If a relation is in 3NF, then it is also in BCNF.
Proof: 3NF permits Non-key Key dependency. That is, it accepts a dependency where a non-key attribute can determine the value of a key attribute. But in BCNF, it expects only candidate keys to be determiner, not non-key attribute. Hence, a relation that is in 3NF, not in BCNF too.

3. From A B, we can derive AC BC, which can further lead to A BC.
Proof: From the question, A determines the value of B. According to Armstrong’s Augmentation rule AC BC is true for any attribute C. this is because C C is a trivial functional dependency. But, this augmentation cannot lead to the dependency A BC where A alone determine both B and C.

4. BCNF decomposition of a relation R(A, B, C, D) with functional dependencies A B and C D leads to three relations, R1(A, B), R2(C, D), and R3(C, A).
Proof: According to the rule of BCNF, all the functional dependencies should have keys as determiners (keys as left hand side attributes of any FDs). For the given relation R and according to the set of FDs given, the key is AC which is composite. Hence, we break the relation into a relation R1 with attributes A and B [from FD A B], a relation R2 with attributes C and D [from FD C D], and a relation R3 with attributes C and A [to link R1 and R2].

5. A relation R(A, B, C) with the functional dependencies AB C and C B is in BCNF.
Proof: From AB C, it is clear that AB is the key. But, one of the key attributes is determined by a non-key attribute C according to C B. Hence, the relation R is not in BCNF.

6. A relation R which is in 3NF is also in 4NF.
Proof: 3NF uses the functional dependencies of Single valued facts. That is, they are about one or more attributes uniquely determine the values of one or more other attributes. But 4NF is about a different kind of functional dependency called Multi-valued functional dependencies. That is, functional dependencies about multi-valued facts.

7. For a relation R with attribute ABCDE and set of functional dependencies F = {A B, BC D, D E}, the closure for attribute combination {AC} is ABCDE.
Proof: Find the closure of (AC)
Step 1: result = AC
Step 2: from A B, A can determine the value of B. hence, result = AC U B = ACB.
Step 3: through BC D, result becomes ABCD.
Step 4: as we know D, then through D E, result = ABCDE.
Hence, the closure of (AC) is ABCDE.

8. For a relation R with attribute ABCDE and set of functional dependencies F = {A B, BC E, ED A}, the attribute combination {ACE} is one of the keys.
Proof: Let us find the closure of (ACE)
Step 1: result = ACE
Step 2: from A B, result = ABCE.
Step 3: from BC E, result = ABCE, does not change.
The result is not ABCDE (all the attributes). Hence, (ACE) is not a key for R.

9. In a relation R(A, B, C, D), if AC CB then A B.
Proof: As per the given FD AC CB, we can understand both A and C together identifies the value of B. We cannot say, by removing C from both sides, A alone can determine B. We would say AC B.

10. In a relation R with attributes ABC, if a FD A B holds, then AC B also holds.
Proof: Attribute A alone can determine B. Adding another attribute C would not change the determination of B by A.

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