# Normal Forms Questions – Part 2

## Normal Forms Questions – Part 2

#### Decompose the relation R, till BCNF.

```Question 1 :
R(ABCDEFGHIJ)
FD : {AB → C, A → DE, B → F, F → GH, D → IJ}```
```Solution :
Step 1 : Find all the candidate keys of R.
Candidate Key : {AB}

Step 2 : Checking For 2NF :
(a) FD which violates 2NF :
A → DE
B → F
(b)```
 Applying Decomposition Algorithm to FD: A → DE ABCDEFGHIJ
Compute Closure of LHS Relationi =
All Attributes in Closure
Relationj =
All attributes on LHS of FD

All attributes of R not in Closure
 ABCFGH
A → DE √
D → IJ  √
AB → C √
B → F    ×
F → GH √
 Applying Decomposition Algorithm to FD: B → F ABCFGH
Compute Closure of LHS Relationi =
All Attributes in Closure
Relationj =
All attributes on LHS of FD

All attributes of R not in Closure
(B)+ = {BFGH}
 BFGH
 BAC
B → F √
F → GH √
AB → C √
```Check the CK of R is preserved in the decomposed relations- yes,
it is preserved in "BAC" relation.
Hence the decomposition in 2NF :```
 CK : B BFGH
 CK : AB BAC
A → DE √
D → IJ  √
B → F √
F → GH √
AB → C √
```Step 3 : Checking For 3NF :
(a) FD which violates 3NF :
D → IJ
F → GH
(b)```
 Applying Decomposition Algorithm to FD: D → IJ ADEIJ
Compute Closure of LHS Relationi =
All Attributes in Closure
Relationj =
All attributes on LHS of FD

All attributes of R not in Closure
(D)+ = {DIJ}
 DIJ
 DAE
D → IJ √  A → DE √
 Applying Decomposition Algorithm to FD: F → GH BFGH
Compute Closure of LHS Relationi =
All Attributes in Closure
Relationj =
All attributes on LHS of FD

All attributes of R not in Closure
(F)+ = {FGH}
 FGH
 FB
F → GH √  B → F √
```Check the CK of R is preserved in the decomposed relations- yes
it is preserved in "BAC" relation.
Hence the decomposition in 3NF :```
 CK : D DIJ
 CK : A DAE
 CK : F FGH
 CK : B BF
 CK : AB BAC
D → IJ √ A → DE √ F → GH √ B → F √ AB → C √
```Step 4 : Checking For BCNF :
FD which violates BCNF : None
Hence the decomposition is already in BCNF also.```