Normalization

Question 1
Consider the relation R(P, Q, S, T, X, Y, Z, W) with the following functional dependencies

Consider the decomposition of the relation R into the consistent relations according to the following two decomposition schemes.               
D1: R=[(P,Q,S,T); (P,T,X); (Q,Y); (Y,Z,W)]               
D2: R=[(P,Q,S);(T,X);(Q,Y);(Y,Z,W)]
Which one of the following options is correct?
A
D1is a lossy decomposition, but D2is a lossless decomposition.
B
Both D1and D2are lossy decompositions.
C
Both D1and D2are lossless decompositions.
D
D1is a lossless decomposition, but D2is a lossy decomposition.
Question 1 Explanation: 

Given functional dependencies set:

PQ->X

P->YX

Q->Y

Y->ZW

  • While merging the tables there should be some common attribute(s) and it should be a candidate key of one of the tables.


  • R1 should be merged with R2 because PT is a key of R2.
  • R3 should be merged with PQSTX because Q is a key of R3.
  • R4 should be merged with PQSTXY because Y is a key of R4.

  • R1 should be merged with R3 because Q is a key of R3.
  • R4 should be merged with PQSY because Y is a key of R4.
  • Now, there is no common attribute in between R2(TX) and PQSYZW.
  • Hence, D2 is lossy decomposition.
Question 2

State True or False with reason
There is always a decomposition into Boyce-Codd normal form (BCNF) that is lossless and dependency preserving.

A
True
B
False
Question 2 Explanation: 
BCNF decomposition can always be lossless, but it may not be always possible to get a dependency preserving BCNF decomposition.
Question 3

(a) Consider the relation scheme R(A, B, C) with the following functional dependencies:
A, B → C, C → A
Show that the scheme R is the Third Normal Form (3NF) but not in Boyce-Code Normal Form (BCNF).
(b) Determine the minimal keys of relation R.

A
Theory Explanation.
Question 4

Consider a relational table R that is in 3NF, but not in BCNF. Which one of the following statements is TRUE?

A
A cell in R holds a set instead of an atomic value.
B
R has a nontrivial functional dependency X→A, where X is not a superkey and A is a non-prime attribute and X is not a proper subset of any key.
C
R has a nontrivial functional dependency X→A, where X is not a superkey and A is a non-prime attribute and X is a proper subset of some key.
D
R has a nontrivial functional dependency X→A, where X is not a superkey and A is a prime attribute.
Question 4 Explanation: 
R(ABCD)
FDs:
AB → C
BC → A
(BD)+ = BD ✖
(ABD)+ = ABDC ✔
(CBD)+ = CBDA ✔
Candidate keys = {ABD, CBD}
• The relation R is in 3NF, as there are no transitive dependencies.
• The relation R is not in BCNF, because the left side of both the FD’s are not Super keys.
• In R, BC → A is a non-trivial FD and in which BC is not a Super key and A is a prime attribute.
Question 5

Which of the following is TRUE?

A
Every relation in 3NF is also in BCNF
B
A relation R is in 3NF if every non-prime attribute of R is fully functionally dependent on every key of R
C
Every relation in BCNF is also in 3NF
D
No relation can be in both BCNF and 3NF
Question 5 Explanation: 
BCNF is a stronger version 3NF. So straight from definition of BCNF every relation in BCNF will also be in 3NF.
Question 6

For a database relation R(a,b,c,d), where the domains a, b, c, d include only atomic values, only the following functional dependencies and those that can be inferred from them hold:

   a → c 
   b → d  

This relation is

A
in first normal form but not in second normal form
B
in second normal form but not in third normal form
C
in third normal form
D
None of the above
Question 6 Explanation: 
Candidate key is ab.
Since all a, b, c, d are atomic. So the relation is in 1NF.
Checking the FD's
a → c
b → d
We can see that there is partial dependencies. So it is not 2NF.
So answer is option (A).
Question 7

Which normal form is considered adequate for normal relational database design?

A
2 NF
B
5 NF
C
4 NF
D
3 NF
Question 7 Explanation: 
3NF, is considered as adequate for normal relational database design, because we can have a 3NF decomposition which is dependency preserving and lossless (not possible for any higher forms).
Question 8

Consider the schema R = (S T U V) and the dependencies S → T, T → U, U → V and V → S. Let R = (R1 and R2) be a decomposition such that R1 ∩ R2 ≠ ∅ . The decomposition is

A
not in 2NF
B
in 2NF but not 3NF
C
in 3NF but not in 2NF
D
in both 2NF and 3NF
Question 8 Explanation: 
Since R1 ∩ R2 ≠ ∅, so the decomposition is lossless join. Now since all the attributes are keys, so R1 ∩ R2 will be a key of the decomposed relation.
And since every attribute is key so the decomposed relation will be in BCNF and hence in 3NF.
Question 9

R(A,B,C,D) is a relation. Which of the following does not have a lossless join, dependency preserving BCNF decomposition?

A
A → B, B → CD
B
A → B, B → C, C → D
C
AB → C, C → AD
D
A → BCD
Question 9 Explanation: 
We have, R (A, B, C, D) and the Functional Dependency set = {AB→C, C→AD}. We decompose it as R1(A, B, C) and R2(C, D). This preserves all dependencies and the join is lossless too, but the relation R1 is not in BCNF. In R1 we keep ABC together otherwise preserving {AB→C} will fail, but doing so also causes {C→A} to appear in R1. {C→A} violates the condition for R1 to be in BCNF as C is not a super key. Condition that all relations formed after decomposition should be in BCNF is not satisfied here.
Question 10

For relation R = (L, M, N , O, P), the following dependencies hold:

   M → O NO → P P → L and L → MN 

R is decomposed into R1 = (L, M, N , P) and R2 = (M, O).

    (a) Is the above decomposition a lossless-join decomposition? Explain.
    (b) Is the above decomposition dependency-preserving? If not, list all the dependencies that are not preserved.
    (c) What is the highest normal form satisfied by the above decomposition?
A
Theory Explanation is given below.
Question 11

Relation R is decomposed using a set of functional dependencies, F, and relation S is decomposed using another set of functional dependencies, G. One decomposition is definitely BCNF, the other is definitely 3NF, but it is not known which is which. To make a quaranteed identification, which one of the following tests should be used on the decompositions? (Assume that the closures of F and G are available).

A
Dependency-preservation
B
Lossless-join
C
BCNF definition
D
3NF definition
Question 11 Explanation: 
If it is BCNF then it is 3NF. But the relation is in 3NF then it is need not to be in BCNF.
Question 12

Relation R with an associated set of functional dependencies, F, is decomposed into BCNF. The redundancy (arising out of functional dependencies) in the resulting set of relations is

A
Zero
B
More than zero but less than that of an equivalent 3NF decomposition
C
Proportional to the size of F+
D
Indetermine
Question 12 Explanation: 
If a relation is in BCNF then there is no functional dependencies.
Question 13

Consider the following functional dependencies in a database:

  Data_of_Birth → Age
  Age → Eligibility
  Name → Roll_number
  Roll_number → Name
  Course_number → Course_name
  Course_number → Instructor
  (Roll_number, Course_number) → Grade 

The relation (Roll_number, Name, Date_of_birth, Age) is:

A
in second normal form but not in third normal form
B
in third normal form but not in BCNF
C
in BCNF
D
in none of the above
Question 13 Explanation: 
Three FD's are valid from the above set of FD's for the given relation.
Date_of_Birth → Age
Name → Roll_number
Roll_number → Name
Candidate keys for the above are:
(Date_of_Birth, Name) and (Date_of_Birth, Roll_number)
Clearly, there is a partial dependency,
Date_of_Birth → Age
So, it is only in 1NF.
Question 14

A table has fields Fl, F2, F3, F4, F5 with the following functional dependencies 

 F1 → F3,  F2→ F4,  (F1.F2) → F5 

In terms of Normalization, this table is in

A
1 NF
B
2 NF
C
3 NF
D
None
Question 14 Explanation: 
F1 → F3 ......(i)
F2 → F4 ......(ii)
(F1⋅F2) → F5 .....(iii)
F1F2 is the candidate key.
F1 and F2 are the prime key.
In (i) and (ii) we can observe that the relation from P → NP which is partial dependency. So this is in 1NF.
Question 15

Let R (A, B, C, D, E, P, G) be a relational schema in which the following functional dependencies are known to hold:

AB → CD, DE → P, C → E, P → C and B → G.
The relational schema R is
A
in BCNF
B
in 3NF, but not in BCNF
C
in 2NF, but not in 3NF
D
not in 2NF
Question 15 Explanation: 
Candidate key is AB.
Since there is a partial dependency B→G.
So the relational schema R is Not in 2NF.
Question 16
In a relational data model, which one of the following statements is TRUE?
A
A relation with only two attributes is always in BCNF.
B
If all attributes of a relation are prime attributes, then the relation is in BCNF.
C
Every relation has at least one non-prime attribute.
D
BCNF decompositions preserve functional dependencies.
Question 16 Explanation: 
A relation with only two attributes will always be in BCNF.
Example:
R(A, B).
Two functional dependencies possible for the relation: (1) A->B and (2) B->A
If there is no functional dependency, we can assume trivial functional dependencies like AB->A and AB->B.
In all cases, functional dependencies like A->B, A must be a key.
So they all will be in BCNF irrespective of the functional depencies set.
Question 17

Choose the correct alternatives (More than one may be correct).

Indicate which of the following statements are true: A relational database which is in 3NF may still have undesirable data redundancy because there may exist:

A
Transitive functional dependencies.
B
Non-trivial functional dependencies involving prime attributes on the right-side.
C
Non-trivial functional dependencies involving prime attributes only on the left-side.
D
Non-trivial functional dependencies involving only prime attributes.
E
Both (B) and (D).
Question 17 Explanation: 
A) Transitive functional dependency, so not in 3NF.
B) 3NF because right side is prime attribute.
C) Not in 3NF, because lets suppose ABC is a candidate key. Now consider
AB → Non-prime attribute
which show it is not in 3NF
D) Involves only prime attribute, so right side should definitely contain only prime attribute. So in 3NF.
Question 18

The relation scheme Student Performance (name, courseNo, rollNo, grade) has the following functional dependencies:

   name, courseNo → grade
   rollNo, courseNo → grade
               name → rollNo
             rollNo → name 

The highest normal form of this relation scheme is

A
2 NF
B
3 NF
C
BCNF
D
4NF
Question 18 Explanation: 
Student Performance (name, courseNo, rollNo, grade)
name, courseNo  → grade →(I)
rollNo, courseNo → grade →(II)
name → rollNo →(III)
rollNo → name →(IV)
Candidate keys: name, courseNo (or) rollNo
Its is not BCNF, because the relation III, there is no relationship from super key.
name → rollNo
It is not BCNF, name is not super key.
It belongs to 3NF, because if X→Y, Y is prime then it is in 3NF.
Question 19

Which of the following is NOT a superkey in a relational schema with attributes V, W, X, Y, Z and primary key VY?

A
VXYZ
B
VWXZ
C
VWXY
D
VWXYZ
Question 19 Explanation: 
It is given that “VY” is a primary key of the relational schema.
Any superset of “VY” is a super key. So, option (B) does not contain “Y”.
Question 20
A database of research articles in a journal uses the following schema.(VOLUME, NUMBER, STARTPAGE, ENDPAGE, TITLE, YEAR, PRICE)
The primary key is (VOLUME, NUMBER, STARTPAGE, ENDPAGE) and the following functional dependencies exist in the schema. (VOLUME, NUMBER, STARTPAGE, ENDPAGE) → TITLE (VOLUME, NUMBER) → YEAR (VOLUME, NUMBER, STARTPAGE, ENDPAGE) → PRICE The database is redesigned to use the following schemas. (VOLUME, NUMBER, STARTPAGE, ENDPAGE, TITLE, PRICE) (VOLUME, NUMBER, YEAR) Which is the weakest normal form that the new database satisfies, but the old one does not?
A
1NF
B
2NF
C
3NF
D
BCNF
Question 20 Explanation: 
Journal (V, N, S, E, T, Y, P)
V – VOLUME
N – NUMBER
S – STARTPAGE
E – ENDPAGE
T – TITLE
Y – YEAR
P – PRICE
Primary key: (V, N, S, E)
FD set:
(V, N, S, E) → T
(V, N) → Y
(V, N, S, E) → P
In (V, N) → Y; V, N is a part of the key and Y is non-prime attribute.
So, it is a partial dependency.
Now, the schema “Journal” is in 1NF but not in 2NF.
The database is redesigned as follows:

Both R1 and R2 are in BCNF.
Therefore, 2NF is the weakest normal form that the new database satisfies, but the old one does not.
Question 21

Consider the following four relational schemas. For each schema, all non-trivial functional dependencies are listed. The underlined attributes are the respective primary keys.

Schema I: Registration(rollno, courses)
          Field ‘courses’ is a set-valued attribute containing the set of 
          courses a student has registered for.
          Non-trivial functional dependency
          rollno → courses

Schema II: Registration (rollno, coursid, email)
          Non-trivial functional dependencies:
          rollno, courseid → email
          email → rollno
 	
Schema III: Registration (rollno, courseid, marks, grade)
          Non-trivial functional dependencies:
          rollno, courseid, → marks, grade
          marks → grade
 	
Schema IV: Registration (rollno, courseid, credit)
          Non-trivial functional dependencies:
          rollno, courseid → credit
          courseid → credit

Which one of the relational schemas above is in 3NF but not in BCNF?

A
Schema I
B
Schema II
C
Schema III
D
Schema IV
Question 21 Explanation: 
Schema I:
Registration (rollno, courses) rollno → courses
For the given schema Registration ‘rollno’ is a primary key.
Left-side of the functional dependency is a superkey so, Registration is in BCNF.
Schema II:
Registrstion (rollno, courseid, email)
rollno, courseid → email
email → rollno
From the given schema the candidate key is (rollno + courseid).
There is no part of the key in the left hand of the FD’s so, it is in 2NF.
In the FD email→rollno, email is non-prime attribute but rollno is a prime attribute.
So, it is not a transitive dependency.
No transitive dependencies so, the schema is in 3NF.
But in the second FD email→rollno, email is not a superkey.
So, it is violating BCNF.
Hence, the schema Registration is in 3NF but not in BCNF.
Schema III:
Registration (rollno, courseid, marks, grade)
rollno, courseid → marks, grade
marks → grade
For the schema the candidate key is (rollno + courseid).
There are no part of the keys are determining non-prime attributes.
So, the schema is in 2NF.
In the FD marks → grade, both the attributes marks and grade are non-prime.
So, it is a transitive dependency.
The FD is violating 3NF.
The schema Registration is in 2NF but not in 3NF.
Schema IV:
Registration (rollno, courseid, credit)
rollno, courseid → credit
courseid → credit
The candidate key is (rollno + courseid).
In the FD, courseid → credit, courseid is part of the key (prime attribute) and credit is non-prime.
So, it is a partial dependency.
The schema is violating 2NF.
Question 22

Let the set of functional dependencies F = {QR → S, R → P, S → Q} hold on a relation schema X = (PQRS). X is not in BCNF. Suppose X is decomposed into two schemas Y and Z, where Y = (PR) and Z = (QRS).

Consider the two statements given below.

    • I. Both Y and Z are in BCNF
 
    II. Decomposition of X into Y and Z is dependency preserving and lossless

Which of the above statements is/are correct?

A
I only
B
Neither I nor II
C
II only
D
Both I and II
Question 22 Explanation: 
Y = (PR)
R → P
R+ = RP
* In R → P, 'R' is a super key. So, Y is in BCNF.
Z = (QRS)
QR → S
S → Q
CK's = QR, RS
* In, S → Q, 'S' is not a super key. So, Z is not in BCNF.
* Y is in BCNF and Z is not in BCNF.
* 'R' is common attribute in the relations Y and Z. and R is candidate key for Y. So, the decomposition is lossless.
* The FD, R → P is applicable on Y and QR → S, S → Q are applicablein 2.
So, the decomposition is dependency preserving.
* Hence, Statement II is correct.
There are 22 questions to complete.

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