Is decomposition dependency preserving?
R is decomposed or divided into R1 with FD { f1 } and R2 with { f2 }, then there can be three cases: f1 U f2 = F —–> Decomposition is dependency preserving.
Which of the following decomposition is dependency preserving?
The relational R is decomposed into R1(ABC) and R2(AD) which is dependency preserving because FD A->BC is a part of relation R1(ABC).
How is decomposition related to functional dependency?
The closure of the combined FDs in the decomposed tables is the same, so the decomposition is dependency preserving. The resultant relation contains more rows (spurious tuples) than existed in the original R1. This is called a “lossy decomposition” as it causes constraint information to be lost.
Is 4NF decomposition dependency preserving?
Figure 6.4: Projection of relation onto a 4NF decomposition of . This decomposition is not dependency preserving as it fails to preserve .
Does the decomposition preserve dependencies Why or why not?
The dependencies are preserved because each dependency in F represents a constraint on the database. If decomposition is not dependency-preserving, some dependency is lost in the decomposition.
What is meant by dependency preserving?
The dependency preservation decomposition is another property of decomposed relational database schema D in which each functional dependency X -> Y specified in F either appeared directly in one of the relation schemas Ri in the decomposed D or could be inferred from the dependencies that appear in some Ri.
Is 3NF dependency preserving?
It is always possible to find a dependency-preserving lossless-join decomposition that is in 3NF. A database design is in 3NF if each member of the set of relation schemas is in 3NF. We now allow functional dependencies satisfying only the third condition.
What is a non trivial functional dependency?
Non-trivial Functional Dependency In Non-trivial functional dependency, the dependent is strictly not a subset of the determinant. i.e. If X → Y and Y is not a subset of X, then it is called Non-trivial functional dependency.
What is the importance of dependency preservation during decomposition?
Why is 4NF useful?
The benefit of conforming to BCNF or 4NF is very similar to the benefit of conforming to 2NF or 3NF. It eliminates some harmful redundancy, and thereby prevents certain cases where the database contradicts itself. It’s a rare case where a table can be in 3NF but fail to be in BCNF or 4NF.
Is R in 4NF?
A database design is in 4NF if each member of the set of relation schemas is in 4NF. The definition of 4NF differs from the BCNF definition only in the use of multivalued dependencies. Every 4NF schema is also in BCNF. , by the replication rule, R cannot be in 4NF.
Which is the decomposition for dependency preserving decomposition?
A Decomposition D = { R1, R2, R3….Rn } of R is dependency preserving wrt a set F of Functional dependency if (F1 ∪ F2 ∪ … ∪ Fm)+ = F+. Consider a relation R R —> F {…with some functional dependency (FD)….}
When to use lossless join and dependency preserving decomposition?
Lossless Join and Dependency Preserving Decomposition. Decomposition of a relation is done when a relation in relational model is not in appropriate normal form. Relation R is decomposed into two or more relations if decomposition is lossless join as well as dependency preserving.
Is the decomposition of FD1 a dependency preserving decomposition?
But, there is no follow up in Functional Dependency S holds P (S –> P). FD1 U FD2 is a subset of FD. So as a consequence, given decomposition is not dependency preserving.
Which is closure of functional dependency FD of R?
FD+ -> Closure of set of functional dependency FD of R. With FD (FD1) R is decomposed or divided into R1 and with FD (FD2) into R2, then the possibility of three cases arise, FD1 ∪ FD2 = FD -> Decomposition is dependency preserving.