written 8.2 years ago by | • modified 2.8 years ago |

**Mumbai University > Computer Engineering > Sem 4 > Database Management System**

**Marks:** 5M

**Year:** May 2014

**1 Answer**

0

28kviews

Explain lossless join decomposition and dependency preserving decomposition.

written 8.2 years ago by | • modified 2.8 years ago |

**Mumbai University > Computer Engineering > Sem 4 > Database Management System**

**Marks:** 5M

**Year:** May 2014

ADD COMMENT
EDIT

3

360views

written 8.2 years ago by |

**Lossless Join Decomposition:**

The lossless join property is a feature of decomposition supported by normalization. It is the ability to ensure that any instance of the original relation can be identified from corresponding instances in the smaller relations.

R : relation, F : set of functional dependencies on R,

X,Y : decomposition of R

- A decomposition {R1, R2,…, Rn} of a relation R is called a lossless decomposition for R if the natural join of R1, R2,…, Rn produces exactly the relation R.
A decomposition is lossless if we can recover:

R(A, B, C) -> Decompose -> R1(A, B) R2(A, C) -> Recover -> R’(A, B, C)

Thus,R’ = R

Decomposition is lossles if :

X ∩ Y -> X, that is: all attributes common to both X and Y functionally determine ALL the attributes in X.

X ∩ Y -> Y, that is: all attributes common to both X and Y functionally determine ALL the attributes in Y

If X ∩ Y forms a superkey of either X or Y, the decomposition of R is a lossless decomposition.

**Dependency Preserving Decomposition:**

A decomposition D = {R1, R2, ..., Rn} of R is dependency-preserving with respect to F if the union of the projections of F on each Ri in D is equivalent to F;

if (F1∪ F2 ∪ …∪Fn)+ = F +

Example-

R= (A, B, C )

F = {A ->B, B->C}

Key = {A}

Ris not in BCNF

Decomposition R1 = (A, B), R2 = (B, C)

R1 and R2 are in BCNF, Lossless-join decomposition, Dependency preserving

- Each Functional Dependency specified in F either appears directly in one of the relations in the decomposition.
- It is not necessary that all dependencies from the relation R appear in some relation Ri.
- It is sufficient that the union of the dependencies on all the relations Ri be equivalent to the dependencies on R.
- is lost in the decomposition

ADD COMMENT
EDIT

Please log in to add an answer.