# 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,
`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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