# Decomposition

A Relation 'R' can be decomposed into a collection of relation schemes to eliminate some of the anomalies in original relation R.

LossLess Join decomposition: Let R is a relation and has set of FD's 'F' over R. The decomposition of R into R1 and R2 is lossless with respect to F if R1 ⛝ R2 = R.

Lossy decomposition: Lossy decomposition have contains extra tuples.

[Note :- R1 , R2. If (R1∩R2) forms a super key in either R1 or R2, then it is LossLess.]

R1∩R2 = B

Here, B is not a key in either R1 or R2, this means i.e Lossy decomposition.

(R1∩R2) = A

Here, A is a key in Relation R1, It means this is a LossLess decomposition.

The decomposition of relation 'R' with FDs F into R1 and R2 with FDs F1 and F2 respectively is said to be dependency preserving if.

(F) = (F1UF2)

**There is two type of decomposition :-**LossLess Join decomposition: Let R is a relation and has set of FD's 'F' over R. The decomposition of R into R1 and R2 is lossless with respect to F if R1 ⛝ R2 = R.

Lossy decomposition: Lossy decomposition have contains extra tuples.

[Note :- R1 , R2. If (R1∩R2) forms a super key in either R1 or R2, then it is LossLess.]

**Solved Examples on Decomposition :-***R(ABC) , decomposed in R1(AB) and R2(BC).check whether LossLess and Lossy.***(1)****Given :-**FD of R { A→B}R1∩R2 = B

Here, B is not a key in either R1 or R2, this means i.e Lossy decomposition.

**R(ABC) , F:{A→B}, decomposition of R1(AB), R2(AC) is LossLess or Lossy.***(2)***Given:-**(R1∩R2) = A

Here, A is a key in Relation R1, It means this is a LossLess decomposition.

**Dependency Preserving Decomposition**The decomposition of relation 'R' with FDs F into R1 and R2 with FDs F1 and F2 respectively is said to be dependency preserving if.

(F) = (F1UF2)

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