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# Answer the following questions: Consider the relation schema R = (N. Y, P. M. and assume... Answer the following questions:

Consider the relation schema R = (N. Y, P. M. and assume that the following set of functional dependencies holds on R:

The letters can be interpreted as follows: R=(Model_Number. Year, Price, Manufacturing Plant Color).

1. [25 points] Give a lossless.join decomposition of Rinto Boyce-Codd normal form. Make sure to use the algorithm studied in class (Figure 7.11, page 331 of the book) and to show all details. 2. [25 points] Does your decomposition preserve functional dependencies? Justify your answer.

3. [25 points] Is Rin 3NF? Justify your answer.

4. [25 points] In the previous Module's HW Assignment, you showed that Fis already in canonical cover form. Use the algorithm we studied in class (Figure 7.12, page 334 of the book) to find a lossless-join and dependency preserving decomposition of Rinto 3NF. Make sure to show all details

#### Homework Answers

Answer #1 ✔ Recommended Answer

The relation given is R ( N, Y, P, M, C ) .

The set of functional dependencies given is :

N -> M

NY -> P

M -> C

The candidate key of the given relation is NY because the attribute closure of NY is {N, Y, M, P, C}.

1.

A relation is said to be in BCNF if the left side attribute of every functional dependency is a superkey. Every candidate key is also a superkey.

The following points are considered for a lossless join decomposition of R into Boyce-Codd normal form :

• Include all the attributes in table T1.
• The first functional dependency is N -> M and this violates the BCNF definition because the left side attribute N is not a superkey. So, a separate table T2 has to be created for the attributes N and M with N being common to both T1 and T2.
• The second functional dependency is NY -> P and this follows the BCNF decomposition. So, no separate tables are created.
• The third functional dependency is M -> C and this violates the BCNF definition because the left side attribute M is not a superkey. So, a separate table T3 has to be created for the attributes M and C with M being common to both T1 and T3.

So, the tables formed as a result of lossless join decomposition of R into Boyce-Codd normal form are :

• T1 ( N, Y, P ) with NY as the primary key.
• T2 ( N, M ) with N as the primary key.
• T3 ( M, C ) with M as the primary key.

2.

In T1, the set of functional dependency F1 is :

NY -> P

In T2, the set of functional dependency F2 is :

N -> M

In T3, the set of functional depedency F3 is :

M -> C

Closure of ( F1 U F2 U F3 ) is same as the closure of the original set of functional dependency.

Hence, the decomposition is depedency preserving.

3.

A relation is in 3NF if either the left side attribute in every functional dependency is a superkey or the right side attribute of the same functional dependency is a prime attribute.

The given relation is not in 3NF because in the functional dependency M -> C, neither attribute M is a superkey nor attribute C is a prime attribute. Moreover, in the functional dependency N -> M, neither attribute N is a superkey nor attribute M is a prime attribute.

Hence, the relation is not in 3NF.

4.

The following points are considered a lossless decomposition of R into 3NF :

• Include all the attributes in table T4.
• The first functional dependency is N -> M and this violates the 3NF definition because neither attribute N is a superkey nor attribute M is a prime attribute. So, a separate table T5 has to be created for the attributes N and M with N being common to both T4 and T5.
• The second functional dependency is NY -> P and this follows the 3NF decomposition. So, no separate tables are created.
• The third functional dependency is M -> C and this violates the 3NF definition because neither attribute M is a superkey nor attribute C is a prime attribute. So, a separate table T6 has to be created for the attributes M and C with M being common to both T5 and T6.

So, the tables formed as a result of lossless join decomposition of R into 3NF are :

• T4 ( N, Y, P ) with NY as the primary key.
• T5 ( N, M ) with N as the primary key.
• T6 ( M, C ) with M as the primary key.
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