**Decomposition of a Relation-**

The process of breaking up or dividing a single relation into two or more sub relations is called as decomposition of a relation. |

**Properties of Decomposition-**

The following two properties must be followed when decomposing a given relation-

**1. Lossless decomposition-**

Lossless decomposition ensures-

- No information is lost from the original relation during decomposition.
- When the sub relations are joined back, the same relation is obtained that was decomposed.

Every decomposition must always be lossless.

**2. Dependency Preservation-**

Dependency preservation ensures-

- None of the functional dependencies that holds on the original relation are lost.
- The sub relations still hold or satisfy the functional dependencies of the original relation.

**Types of Decomposition-**

Decomposition of a relation can be completed in the following two ways-

**1. Lossless Join Decomposition-**

- Consider there is a relation R which is decomposed into sub relations R
_{1}, R_{2}, …. , R_{n.} - This decomposition is called lossless join decomposition when the join of the sub relations results in the same relation R that was decomposed.
- For lossless join decomposition, we always have-

R_{1} ⋈ R_{2} ⋈ R_{3} ……. ⋈ R_{n} = R |

where ⋈ is a natural join operator

**Example-**

Consider the following relation R( A , B , C )-

A | B | C |

1 | 2 | 1 |

2 | 5 | 3 |

3 | 3 | 3 |

**R( A , B , C )**

Consider this relation is decomposed into two sub relations R_{1}( A , B ) and R_{2}( B , C )-

The two sub relations are-

A | B |

1 | 2 |

2 | 5 |

3 | 3 |

**R _{1}( A , B )**

B | C |

2 | 1 |

5 | 3 |

3 | 3 |

**R _{2}( B , C )**

Now, let us check whether this decomposition is lossless or not.

For lossless decomposition, we must have-

**R _{1} ⋈ R_{2} = R**

Now, if we perform the natural join ( ⋈ ) of the sub relations R_{1} and R_{2} ,_{}we get-

A | B | C |

1 | 2 | 1 |

2 | 5 | 3 |

3 | 3 | 3 |

This relation is same as the original relation R.

Thus, we conclude that the above decomposition is lossless join decomposition.

**NOTE-**

- Lossless join decomposition is also known as
**non-additive join decomposition.** - This is because the resultant relation after joining the sub relations is same as the decomposed relation.
- No extraneous tuples appear after joining of the sub-relations.

**2. Lossy Join Decomposition-**

- Consider there is a relation R which is decomposed into sub relations R
_{1}, R_{2}, …. , R_{n.} - This decomposition is called lossy join decomposition when the join of the sub relations does not result in the same relation R that was decomposed.
- The natural join of the sub relations is always found to have some extraneous tuples.
- For lossy join decomposition, we always have-

R_{1} ⋈ R_{2} ⋈ R_{3} ……. ⋈ R_{n} ⊃ R |

where ⋈ is a natural join operator

**Example-**

Consider the following relation R( A , B , C )-

A | B | C |

1 | 2 | 1 |

2 | 5 | 3 |

3 | 3 | 3 |

**R( A , B , C )**

Consider this relation is decomposed into two sub relations as R_{1}( A , C ) and R_{2}( B , C )-

The two sub relations are-

A | C |

1 | 1 |

2 | 3 |

3 | 3 |

**R _{1}( A , B )**

B | C |

2 | 1 |

5 | 3 |

3 | 3 |

**R _{2}( B , C )**

Now, let us check whether this decomposition is lossy or not.

For lossy decomposition, we must have-

**R _{1} ⋈ R_{2} ⊃ R**

Now, if we perform the natural join ( ⋈ ) of the sub relations R_{1} and R_{2 }we get-

A | B | C |

1 | 2 | 1 |

2 | 5 | 3 |

2 | 3 | 3 |

3 | 5 | 3 |

3 | 3 | 3 |

This relation is not same as the original relation R and contains some extraneous tuples.

Clearly, R_{1}⋈ R_{2} ⊃ R.

Thus, we conclude that the above decomposition is lossy join decomposition.

**NOTE-**

- Lossy join decomposition is also known as
**careless decomposition.** - This is because extraneous tuples get introduced in the natural join of the sub-relations.
- Extraneous tuples make the identification of the original tuples difficult.

**Next Article-****Rules to Determine Lossless and Lossy Decomposition**

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