Question

Let, R=(A,B,C,D,E,G) and let F be {A→BDG, BG→DE, B→D, D→A}. Argue that R is not in BCNF by finding one functional dependency in F that violates the definition of BCNF. Add one more non-trivial dependency to F so that R is in BCNF with respect to the new set of dependencies.

Answer 1

The decomposition is lossless.

Explanation:

Lossless Join Decomposition:

The decomposition of a table or relation in DBMS is done to normalize the date if the table or relation is not following the rule of a normal form. We decompose the table into two or more tables if it preserves the dependency and lossless join.

By using a lossless join decomposition method we remove the redundancy from the database and the original data is preserved as it is.

If we can reconstruct a table from the decomposed tables using joins and join results in the same original table, then decomposition is lossless otherwise decomposition is lossy.

We can use three methods, to check the decomposition is lossless or lossy:

1. Matrix method
2. Binary method
3. Join method

The given relation is:

R = (A, B, C) is decomposed into

R1= (A, B) and

R2 = (B, C)

The given functional dependency is given below:

B → C

The decomposition is lossless because there is one column that is common.

The dependency will preserve because the functional dependency will remain in the second table.