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Relational Database Management System (RDBMS) and Normalization Forms – Part – 4

Relational Database Management System (RDBMS) and Normalization Forms – Part – 4

Irreducible Sets of Dependencies

Let’s assume S 1 as well as S 2 be two (2) sets of practical dependencies if every single practical dependencies inferred by means of S 1 is inferred through S 2 which is when S 1 + is a subclass of S 2 + an individual can state that S 2 is an equivalent set to S 1 + (the equivalent set is also termed as “Cover”). This depicts that if the Database Management System (DBMS) carry out the practical dependencies in S 2, at that moment it will spontaneously be put into the effect of practical dependencies in S 1.

Subsequently when the S 2 is equivalent for S 1 as well as S 1 is equivalent for S 2 which is – S 1 + = S 2 +, an individual can say that S 1 as well as S 2 are equal, obviously, when S 1 as well as S 2 are equal, then if the Database Management System (DBMS) carry out the practical dependencies in S 2 it will automatically be put into effect the practical dependencies in S 1, in addition to vice versa.

At this moment an individual can describe a set of practical dependencies to be complex (Generally known as negligible in the literature) if and solely if it contents the subsequent three (3) things

1. The right hand side (RHS) which is known as the dependent of every single practical dependency in S includes just one (1) column or attribute which is known as singleton set.

2. The left hand side (LHS) which is known as determinant of every single S is complicated in turn which depicts that no column or attribute can be rejected from the determinant without altering the closure S +, that is without changing S into certain set not corresponding to S. An individual can say that such a practical dependencies is left irreducible.

3. No practical dependencies in S can be rejected from S without altering the closure S +. That is, without changing S into certain set not equal to S.

The following is the discussion about the above mentioned three (3) things.

Table or Relation R { Z, Y, X, W, V, U } fulfils the subsequent practical dependencies

ZY → X

X → Z

YX → W

ZXW → Y

YW → X

XV → UZ

XV → VW

W → VU

Now what can be the irreducible equal set of these practical dependencies?

The explanation is simple. Let’s find the clarification for the above practical dependencies.

1. ZY → X

2. X → Z

3. YX → W

4. ZXW → Y

5. YV → X

6. XV → Z

7. XV → U

8. XV → Y

9. XU → W

10. W → V

11. W → U

At this point of time:

2 denotes 6, as a result an individual can drop 6.

8 denotes XU → YX (By means of augmentation), through which 3 denotes XU → W (By means of Transitivity), thus an individual can delete 10.

8 denotes ZXU → ZY (Via augmentation), as well as 11 denotes ZXW → ZXU (By means of augmentation), in addition to ZXW → ZY (By means of Transitivity), plus ZXW → Y (By means of Decomposition), as a result an individual can eliminate 4.

Definitely additional reductions are not at all possible, as well as an individual are left with the subsequent irreducible sets:

1. ZY → X

2. X → Z

3. YX → W

4. YV → X

5. XV → U

6. XU → Y

7. W → V

8. W → U

On the other hand:

2 denotes XW → ZXW (By means of Composition), next with 4 denotes XW → YV (By means of Transitivity), consequently an individual can substitute 4 XW → Y

2 denote 6, as a result an individual can drop 6 like before.

2 as well as 10 denotes XV → ZW (Through composition), which denotes XU → ZWX (Through Augmentation), which with the unique 4 denotes XU → Y (By means of Transitivity), accordingly an individual can drop 8.

Definitely additional reductions are not at all possible, as well as an individual are left with the subsequent irreducible sets:

1. ZY → X

2. X → Z

3. YX → W

4. XW → Y

5. YV → X

6. XV → U

7. XU → W

8. W → V

9. W → U

For that reason, there are two (2) discrete irreducible equivalence for the unique set of practical dependencies.