Questions on Normal Forms – Part 1

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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Normal Form Shortcuts for Gate Students Next Question 1 : Decompose the relation R, till BCNF
     

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