Questions on Normal Forms Part 1
Identify the Normal Forms of the relation R :
Question 1 :
R(ABCD)
FD : {A → B, B → C}
Solution :
(A)+ = {ABC}
As no dependency defined for D, hence R is in 1NF.
Question 2:
R(ABCD)
FD1 : A → B,
FD2 : B → C,
FD3 : AD → BC
Solution :
Step 1 : 
Find Candidate Key : (AD)^{+} = {ADBC} 
Step 2 : 
Functional
Dependency 
Find Highest
Normal Form 
Reason 
FD1 : A → B, 
1NF 
Partial Dependency as A is part of
Key 
FD2 : B → C, 
2NF 
Nonkey Nonkey allowed 
FD3 : AD → BC 
BCNF 
LHS is the SuperKey 

Question 3:
R(ABCDE)
FD1 : AB → C,
FD2 : C → D,
FD3 : D → E
FD4 : E → A
Solution :
Step 1 : 
Find Candidate Key : (AB)^{+} = {ABCDE}
Other Candidate Keys derived from AB
{AB,EB,DB,CB} 
Step 2 : 
Functional
Dependency 
Find Highest
Normal Form 
Reason 
FD1 : AB → C, 
BCNF 
LHS is the SuperKey 
FD2 : C → D, 
3NF 
Part of Key → Part of Key is allowed. 
FD3 : D → E 
3NF 
Part of Key → Part of Key is allowed. 
FD3 : E → A 
3NF 
Part of Key → Part of Key is allowed. 
R is in 3NF. 

Question 4:
R(ABCDEF)
FD1 : AB → C,
FD2 : C → D,
FD3 : B → E,
FD4 : B → F
Solution :
Step 1 : 
Find Candidate Key : (AB)^{+} = {ABCDEF} 
Step 2 : 
Functional
Dependency 
Find Highest
Normal Form 
Reason 
FD1 : AB → C, 
BCNF 
LHS is the SuperKey 
FD2 : C → D, 
2NF 
NonKey → NonKey is allowed. 
FD3 : B → E 
1NF 
Partial Dependency as B is part of
Key 
FD3 : B → F 
1NF 
Partial Dependency as B is part of
Key 
R is in 1NF. 

Question 5 :
a) If relation R consists of only simple candidate keys then R
should be in .......?
b) If relation R consists of only prime attributes, then R should
be in .......?
c) If relation R is in 3NF and every CK is simple CK, then
relation is in ....?
d) If relation R with no non trivial FD, then R is in ......?
Solution :
a) If relation R consists of only simple candidate keys then R
should be in 2NF but may or may not be in BCNF,3NF
b) If relation R consists of only prime attributes, then R should be
in 3NF but may or may not be in BCNF
X → Y {X: Candidate Key , Y : Prime Attribute }
Example of this type of relation is defined in Question 3, i.e.
R(ABCDE)
FD : {AB → C, C → D, D → E, E → A}
CK : {AB,EB,DB,CB}
c) If relation R is in 3NF and every CK is simple CK, then relation
is in BCNF.
Example of this type of relation is :
R(ABCD)
FD : {A → B, B → C, C → D, D → A}
d) If relation R with no non trivial FD, then R is in always BCNF.
Example of this type of relation is :
R(ABC)
FD : {A → A, B → B, C → C, AB → AB, AC → AC, BC → BC, ABC → ABC}
Incoming search terms:
 dbms questions on normal forms
 questions on normal forms
 question on normalization in dbms
 R={ABCD} normalisation questions
 questions for solving normal forms
 question solve fir 1nf and 2nf in dbms
 question of normal form in dbms
 numericals of norma forms in dbms
 examples to solve problems on normal forms
 what is the highest normal form of relational schema bank
Related