Normalization solved exercise bcnf 1

Normalize up to BCNF, Boyce-codd normal form normalization, steps in normalizing a relation to bcnf relation

Question:

Normalize the following table. Show all work and clearly indicate the primary and foreign keys.

R(elevator_no, building_no, building_name, capacity, staff_no, first_name, last_name, date_examined) with the following functional dependencies:

1. elevator_no → building_no,capacity

2. building_no → building_name

3. staff_no → first_name,last_name

4. elevator_no,staff_no → date_examined

Normalize table R up to BCNF.

Solution:

1NF – The table R is already in first normal form because of atomic attributes. All the attributes of R are atomic.

2NF – Second normal form is about elimination of partial key functional dependency. To proceed further, we need to find the key for R. For that we need to find the how to find attribute closure for all the left hand side attributes of the given functional dependencies.

elevator_no⁺

Result = building_no, capacity – from the FD elevator_no → building_no,capacity

           = building_no, capacity, building_name - from building_no → building_name

Result of elevator_no⁺ does not include all attributes of R. Hence, elevator_no cannot be the key of R.

staff_no⁺

Result = first_name, last_name – from the FD staff_no → first_name,last_name

Result of staff_no⁺ does not include all attributes of R. hence, staff_no cannot be the key of R.

(Elevator_no, staff_no)⁺

Result = date_examined

from the FD elevator_no,staff_no → date_examined

            = date_examined, building_no, capacity

from elevator_no → building_no,capacity

            = date_examined, building_no, capacity, building_name

            from building_no → building_name

            = elevator_no, building_no, building_name, capacity, staff_no, first_name, last_name, date_examined

            from staff_no → first_name,last_name

The result includes all the attributes of R for (Elevator_no, staff_no)⁺. Hence, (Elevator_no, staff_no) is the key (composite key) of R.

Do we have partial key dependencies in R?

Yes. The following are the partial key dependencies.

elevator_no → building_no,capacity

staff_no → first_name,last_name

The reason is, both elevator_no and staff_no are parts of the key of R. They can uniquely identify some of the non-key attributes. Because of these partial key dependencies, R is not in 2NF.

To convert R into 2NF relations, we need to decompose R on partial key dependencies. Thus, we get the following relations; (primary keys are underlined)

R1 (elevator_no, building_no, capacity, building_name)

          From FDs elevator_no → building_no,capacity and building_no → building_name

R2 (staff_no, first_name, last_name)

From FD staff_no → first_name,last_name

R3 (elevator_no, staff_no, date_examined)

From FD elevator_no,staff_no → date_examined

 – in R3 elevator_no is one foreign key referencing R1, and staff_no is another foreign key referencing the relation R2.

Are R1, R2, and R3 in 2NF?

Yes. In all the relations, all non-key attributes are fully-functionally dependent on the primary keys.

3NF – Third normal form is about elimination of non-key dependencies, that is, a functional dependency with a non-key attribute is dependent on another non-key attribute.

Do we have non-key dependencies in R1?

Yes. See below;

R1 (elevator_no, building_no, capacity, building_name)

          F = {elevator_no → building_no,capacity and building_no → building_name}

In R1 we have a non-key dependency building_no → building_name, that is the non-key attribute building_no uniquely determines the other non-key attribute building_name.

R1 is in 2NF because of no partial key dependencies and not in 3NF because of non-key dependencies.

To convert R1 into 3NF relations, decompose R1 on non-key dependencies. Thus, we get the following relations;

R11 (elevator_no, building_no, capacity)

From FD elevator_no → building_no, capacity

R12 (building_no, building_name)

From FD building_no → building_name a non-key dependency.

– in R12 building_no is the primary key and in R11 building_no is the foreign key referencing the relation R12.

Do we have non-key dependencies in R11 and R12?

No. Hence, R11 and R12 are in 3NF.

Do we have non-key dependencies in R2 and R?

No. Hence, R2 and R3 are in 3NF.

BCNF – The determinants in a FD should be the key for the relation.

Are R11, R12, R2, and R3 are in BCNF?

Yes. Each relation have only one functional dependency. And the determinants of the functional dependencies are the primary keys. Hence the relations R11, R12, R2, and R3 are in BCNF.

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