Normalization
Question 1 
A cell in R holds a set instead of an atomic value.
 
R has a nontrivial functional dependency X>A, where X is not a superkey and A is a nonprime attribute and X is not a proper subset of any key.
 
R has a nontrivial functional dependency X>A, where X is not a superkey and A is a nonprime attribute and X is a proper subset of some key.
 
R has a nontrivial functional dependency X>A, where X is not a superkey and A is a prime attribute.

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 nontrivial FD and in which BC is not a Super key and A is a prime attribute.
Question 2 
Consider the following four relational schemas. For each schema, all nontrivial functional dependencies are listed. The underlined attributes are the respective primary keys.
Schema I: Registration(rollno, courses) Field ‘courses’ is a setvalued attribute containing the set of courses a student has registered for. Nontrivial functional dependency rollno → courses Schema II: Registration (rollno, coursid, email) Nontrivial functional dependencies: rollno, courseid → email email → rollno Schema III: Registration (rollno, courseid, marks, grade) Nontrivial functional dependencies: rollno, courseid, → marks, grade marks → grade Schema IV: Registration (rollno, courseid, credit) Nontrivial functional dependencies: rollno, courseid → credit courseid → credit
Which one of the relational schemas above is in 3NF but not in BCNF?
Schema I  
Schema II  
Schema III  
Schema IV 
Registration (rollno, courses) rollno → courses
For the given schema Registration ‘rollno’ is a primary key.
Leftside 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 nonprime 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 nonprime attributes.
So, the schema is in 2NF.
In the FD marks → grade, both the attributes marks and grade are nonprime.
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 nonprime.
So, it is a partial dependency.
The schema is violating 2NF.
Question 3 
Which of the following is NOT a superkey in a relational schema with attributes V, W, X, Y, Z and primary key VY?
VXYZ  
VWXZ  
VWXY  
VWXYZ 
Any superset of “VY” is a super key. So, option (B) does not contain “Y”.
Question 4 
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?
1NF  
2NF  
3NF  
BCNF 
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 nonprime 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 R_{1} and R_{2} are in BCNF.
Therefore, 2NF is the weakest normal form that the new database satisfies, but the old one does not.
Question 5 
S1: Every table with two singlevalued attributes is in 1NF, 2NF, 3NF and BCNF. S2: AB>C, D>E, E>C is a minimal cover for the set of functional dependencies AB>C, D>E, AB>E, E>C.Which one of the following is CORRECT?
S1 is TRUE and S2 is FALSE.  
Both S1 and S2 are TRUE.  
S1 is FALSE and S2 is TRUE.  
Both S1 and S2 are FALSE. 
If we can prove the relation is in BCNF then by default it would be in 1NF, 2NF, 3NF also.
Let R(AB) be a two attribute relation, then
If {A→B} exists then BCNF since {A}^{+} = AB = R
If {B→A} exists then BCNF since {B}^{+} = AB = R
If {A→B, B→A} exists then BCNF since A and B both are Super Key now.
If {No nontrivial Functional Dependency} then default BCNF.
Hence it’s proved that a Relation with two singlevalued attributes is in BCNF hence it’s also in 1NF, 2NF, 3NF.
S2: False
The canonical cover for the given FD set is {AB→C, D→E, AB→E, E→C}. As we can see AB→E is not covered in minimal cover since {AB}^{+} = ABC in the given cover {AB→C, D→E, E→C}
Question 6 
3  
4  
5  
6 
Now D+ = {D}.
Hence we have to add A,B,C,E,F,G,H to D and check which of them are Candidate keys of size 2.
AD^{+} = {ABCDEFGH}
BD^{+} = {ABCDEFGH}
ED^{+} = {ABCDEFGH}
FD^{+} = {ABCDEFGH}
But CD^{+}, GD^{+} and HD^{+} does not give all the attributes hence CD, GD and HD are not candidate keys.
Hence no. of candidate keys are 4: AD, BD, ED, FD.
Question 7 
in 1NF, but not in 2NF.  
in 2NF, but not in 3NF.  
in 3NF, but not in BCNF.  
in BCNF. 
Now D+ = {D}.
Hence we have to add A,B,C,E,F,G,H to D and check which of them are Candidate keys of size 2.
AD^{+} = {ABCDEFGH}
BD^{+} = {ABCDEFGH}
ED^{+} = {ABCDEFGH}
FD^{+}= {ABCDEFGH}
But CD^{+}, GD^{+} and HD^{+} does not give all the attributes hence CD, GD and HD are not candidate keys.
Here Candidate keys are AD, BD, ED and FD.
A → BC, B → CFH and F → EG etc are partial dependencies.
So given relation is in 1NF, but not in 2NF.
Question 8 
Every relation in 3NF is also in BCNF  
A relation R is in 3NF if every nonprime attribute of R is fully functionally dependent on every key of R  
Every relation in BCNF is also in 3NF  
No relation can be in both BCNF and 3NF 
Question 9 
The schema is in BCNF  
The schema is in 3NF but not in BCNF
 
The schema is in 2NF but not in 3NF  
The schema is not in 2NF 
(Sid, Street) → Sname
As Sid is a primary key, then
(Sid, Street) will be super key.
Hence, it is in BCNF.
Question 10 
Both Book and Collection are in BCNF
 
Both Book and Collection are in 3NF only
 
Book is in 2NF and Collection is in 3NF
 
Both Book and Collection are in 2NF only

Book(Title, Author, Catalog_no, Publisher, Year, Price)
Collection(Title, Author, Catalog_no)
I) Title Author ⟶ Catalog_no ⟶ BCNR
II) Catalog_no ⟶ Title, Author, Publisher, Year ⟶ 3NF
III) Publisher Title Year ⟶Price ⟶ 2NF Book’s in 2NF
Collection is in 3NF.
Question 11 
in BCNF  
in 3NF, but not in BCNF  
in 2NF, but not in 3NF  
not in 2NF 
Since there is a partial dependency B→G.
So the relational schema R is Not in 2NF.
Question 12 
Any relation with two attributes is in BCNF  
A relation in which every key has only one attribute is in 2NF
 
A prime attribute can be transitively dependent on a key in a 3NF relation  
A prime attribute can be transitively dependent on a key in a BCNF relation 
i) It is in 3NF.
ii) For any dependency X→ Y
where X is a super key.
iii) Functional dependency has been removed.
Option D is false.
→ Because a prime attribute can’t be transitive dependent on a key in a BCNF relation.
Question 13 
BCNF is stricter than 3NF
 
Lossless, dependencypreserving decomposition into 3NF is always possible
 
Lossless, dependencypreserving decomposition into BCNF is always possible  
Any relation with two attributes is in BCNF

Option B: Lossless, dependency preserving decomposition into 3NF is always possible.
Option C: It is false.
It is not possible to have dependency preserving in BCNF decomposition.
→ Let take an example, 3NF can't be decomposed into BCNF.
Option D: It is true.
Let consider t wo attributes (X, Y).
If (X→Y), X is a candidate key. It is in BCNF and viceversa.
Question 14 
1 NF  
2 NF  
3 NF  
None 
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 
2 NF
 
3 NF
 
BCNF
 
4NF

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 16 
Data_of_Birth → Age Age → Eligibility Name → Roll_number Roll_number → Name Course_number → Course_name Course_number → Instructor (Roll_number, Course_number) → GradeThe relation (Roll_number, Name, Date_of_birth, Age) is:
in second normal form but not in third normal form  
in third normal form but not in BCNF  
in BCNF
 
in none of the above

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 17 
Zero  
More than zero but less than that of an equivalent 3NF decomposition  
Proportional to the size of F^{+}  
Indetermine 
Question 18 
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).
Dependencypreservation  
Losslessjoin  
BCNF definition  
3NF definition 
Question 20 
not in 2NF  
in 2NF but not 3NF  
in 3NF but not in 2NF  
in both 2NF and 3NF 
And since every attribute is key so the decomposed relation will be in BCNF and hence in 3NF.
Question 21 
2 NF  
5 NF  
4 NF  
3 NF 
Question 22 
in first normal form but not in second normal form  
in second normal form but not in third normal form  
in third normal form  
None of the above 
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 23 
A → B, B → CD  
A → B, B → C, C → D  
AB → C, C → AD  
A → BCD 
Question 24 
Transitive functional dependencies.  
Nontrivial functional dependencies involving prime attributes on the rightside.
 
Nontrivial functional dependencies involving prime attributes only on the leftside.
 
Nontrivial functional dependencies involving only prime attributes.  
Both (B) and (D). 
B) 3NF because right side is prime attribute.
C) Not in 3NF, because lets suppose ABC is a candidate key. Now consider
AB → Nonprime 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.