# Decompose the Relation R till BCNF – Question

## Decompose the Relation R till BCNF – Question

```Question 2 :
R(ABDLPT)
FD : {B → PT, T → L, A → D}
Decompose the Relation R till BCNF.```
```Solution :
Step 1 : Find all the candidate keys of R.
Candidate Key : {AB}

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

All attributes of R not in Closure
(B)+ = {BPTL}
 BPTL
B → PT √
T → L  √
A → D ×
 Applying Decomposition Algorithm to FD: A → D BAD
Compute Closure of LHS Relationi =
All Attributes in Closure
Relationj =
All attributes on LHS of FD

All attributes of R not in Closure
 AB
A → D √ There is no FD for this relation.
But {AB} is a CK and is missing
in other’s decomposed relation,
So, the relation “AB” is added.
```Check the CK of R is preserved in the decomposed relations- yes,
it is preserved in "AB" relation.
Hence the decomposition in 2NF :```
 CK : B BPTL
 CK : AB AB
B → PT √
T → L  √
A → D √ AB : CK
```Step 3 : Checking For 3NF :
(a) FD which violates 3NF :
T → L
(b)```
 Applying Decomposition Algorithm to FD: T → L BPTL
Compute Closure of LHS Relationi =
All Attributes in Closure
Relationj =
All attributes on LHS of FD

All attributes of R not in Closure
(T)+ = {TL}
 TL
 TBP
T → L √  B → PT √
```Check the CK of R is preserved in all the decomposed relations- yes
it is preserved in "AB" relation.
Hence the decomposition in 3NF :```
 CK : AB AB
 CK : T TL
 CK : B TBP
A → D √ CK relation T → L √ B → PT √
```Step 4 : Checking For BCNF :
FD which violates BCNF : None
Hence the decomposition is already in BCNF also.```

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