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}
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