To compute B + : To compute B +, the closure of an attribute “B” over a set of functional dependencies “F”, refer the algorithm in Figure 7.19 of the textbook. The result is initialized to {B}. For the first repetition, The condition is not satisfied in first and second functional dependency. The condition B is a subset of {B} is satisfied for the functional dependency B → D Hence “D” is added to the result. Therefore, the result = {B,D}. The condition “D” is a subset of {B,D} is satisfied for the functional dependency D → A Hence “A” is added to the result

Question

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Answer 1

Expert Solution & Answer

Chapter 7, Problem 30E

a.

Explanation of Solution

To compute B⁺:

To compute B^+, the closure of an attribute “B” over a set of functional dependencies “F”, refer the algorithm in Figure 7.19 of the textbook.

The result is initialized to {B}.

For the first repetition,

The condition is not satisfied in first and second functional dependency.

The condition B is a subset of {B} is satisfied for the functional dependency B→D Hence “D” is added to the result.

Therefore, the result = {B,D}.

The condition “D” is a subset of {B,D} is satisfied for the functional dependency D→A Hence “A” is added to the result

b.

Explanation of Solution

To prove:

“AG” is super key.

Proof:

This can be proved by proving the existence of functional dependency so that AG→ABCDEG It is given that A→BCD.

The union of the similar sets is the set with itself (AA=A).

To get A→ABCD, use Armstrong’s augmentation axiom with “A”.

BC→DE is given and to get ABCD→ABCDE, use Armstrong’s augmentation axiom with ABCD

c.

Explanation of Solution

Canonical cover:

“D” is irrelevant in first and second dependencies because of third dependency.

Hence remove the attribute “D” from the first and second functional dependencies to achieve A→BC and BC→E.

According to the sub-part “a”, B⁺ is computed as {B,D,A,C,E}.

The functional dependency B→E can be determined from the above set and from this the functional dependency BC→E becomes irrelevant

d.

Explanation of Solution

Third Normal Form (3NF) decomposition:

In the canonical cover, functional dependency is exists such that the attribute set id is a subset of other dependency.

Hence, each functional dependency will generate its own relation.

The relations according to the functional dependencies in the canonical cover are,

r1(A,B,C)

r2(B,D,E)

r3(D,A)

“G” has to part of all super keys as it is not dependent on any attribute...

e.

Explanation of Solution

Boyce-Codd Normal Form (BCNF) decomposition:

To compute the BCNF decomposition of the schema relation “r” using the original set “F” of functional dependency, follow the steps below,

For the relation not being in BCNF, the first original functional dependency is responsible.

So, according to first functional dependency , the original relation is decomposed as follows,
...

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Answer 2

Expert Solution & Answer

Chapter 7, Problem 30E

a.

Explanation of Solution

To compute B⁺:

To compute B^+, the closure of an attribute “B” over a set of functional dependencies “F”, refer the algorithm in Figure 7.19 of the textbook.

The result is initialized to {B}.

For the first repetition,

The condition is not satisfied in first and second functional dependency.

The condition B is a subset of {B} is satisfied for the functional dependency B→D Hence “D” is added to the result.

Therefore, the result = {B,D}.

The condition “D” is a subset of {B,D} is satisfied for the functional dependency D→A Hence “A” is added to the result

b.

Explanation of Solution

To prove:

“AG” is super key.

Proof:

This can be proved by proving the existence of functional dependency so that AG→ABCDEG It is given that A→BCD.

The union of the similar sets is the set with itself (AA=A).

To get A→ABCD, use Armstrong’s augmentation axiom with “A”.

BC→DE is given and to get ABCD→ABCDE, use Armstrong’s augmentation axiom with ABCD

c.

Explanation of Solution

Canonical cover:

“D” is irrelevant in first and second dependencies because of third dependency.

Hence remove the attribute “D” from the first and second functional dependencies to achieve A→BC and BC→E.

According to the sub-part “a”, B⁺ is computed as {B,D,A,C,E}.

The functional dependency B→E can be determined from the above set and from this the functional dependency BC→E becomes irrelevant

d.

Explanation of Solution

Third Normal Form (3NF) decomposition:

In the canonical cover, functional dependency is exists such that the attribute set id is a subset of other dependency.

Hence, each functional dependency will generate its own relation.

The relations according to the functional dependencies in the canonical cover are,

r1(A,B,C)

r2(B,D,E)

r3(D,A)

“G” has to part of all super keys as it is not dependent on any attribute...

e.

Explanation of Solution

Boyce-Codd Normal Form (BCNF) decomposition:

To compute the BCNF decomposition of the schema relation “r” using the original set “F” of functional dependency, follow the steps below,

For the relation not being in BCNF, the first original functional dependency is responsible.

So, according to first functional dependency , the original relation is decomposed as follows,
...

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 3

Expert Solution & Answer

Chapter 7, Problem 30E

a.

Explanation of Solution

To compute B⁺:

To compute B^+, the closure of an attribute “B” over a set of functional dependencies “F”, refer the algorithm in Figure 7.19 of the textbook.

The result is initialized to {B}.

For the first repetition,

The condition is not satisfied in first and second functional dependency.

The condition B is a subset of {B} is satisfied for the functional dependency B→D Hence “D” is added to the result.

Therefore, the result = {B,D}.

The condition “D” is a subset of {B,D} is satisfied for the functional dependency D→A Hence “A” is added to the result

b.

Explanation of Solution

To prove:

“AG” is super key.

Proof:

This can be proved by proving the existence of functional dependency so that AG→ABCDEG It is given that A→BCD.

The union of the similar sets is the set with itself (AA=A).

To get A→ABCD, use Armstrong’s augmentation axiom with “A”.

BC→DE is given and to get ABCD→ABCDE, use Armstrong’s augmentation axiom with ABCD

c.

Explanation of Solution

Canonical cover:

“D” is irrelevant in first and second dependencies because of third dependency.

Hence remove the attribute “D” from the first and second functional dependencies to achieve A→BC and BC→E.

According to the sub-part “a”, B⁺ is computed as {B,D,A,C,E}.

The functional dependency B→E can be determined from the above set and from this the functional dependency BC→E becomes irrelevant

d.

Explanation of Solution

Third Normal Form (3NF) decomposition:

In the canonical cover, functional dependency is exists such that the attribute set id is a subset of other dependency.

Hence, each functional dependency will generate its own relation.

The relations according to the functional dependencies in the canonical cover are,

r1(A,B,C)

r2(B,D,E)

r3(D,A)

“G” has to part of all super keys as it is not dependent on any attribute...

e.

Explanation of Solution

Boyce-Codd Normal Form (BCNF) decomposition:

To compute the BCNF decomposition of the schema relation “r” using the original set “F” of functional dependency, follow the steps below,

For the relation not being in BCNF, the first original functional dependency is responsible.

So, according to first functional dependency , the original relation is decomposed as follows,
...

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 4

Expert Solution & Answer

Chapter 7, Problem 30E

a.

Explanation of Solution

To compute B⁺:

To compute B^+, the closure of an attribute “B” over a set of functional dependencies “F”, refer the algorithm in Figure 7.19 of the textbook.

The result is initialized to {B}.

For the first repetition,

The condition is not satisfied in first and second functional dependency.

The condition B is a subset of {B} is satisfied for the functional dependency B→D Hence “D” is added to the result.

Therefore, the result = {B,D}.

The condition “D” is a subset of {B,D} is satisfied for the functional dependency D→A Hence “A” is added to the result

b.

Explanation of Solution

To prove:

“AG” is super key.

Proof:

This can be proved by proving the existence of functional dependency so that AG→ABCDEG It is given that A→BCD.

The union of the similar sets is the set with itself (AA=A).

To get A→ABCD, use Armstrong’s augmentation axiom with “A”.

BC→DE is given and to get ABCD→ABCDE, use Armstrong’s augmentation axiom with ABCD

c.

Explanation of Solution

Canonical cover:

“D” is irrelevant in first and second dependencies because of third dependency.

Hence remove the attribute “D” from the first and second functional dependencies to achieve A→BC and BC→E.

According to the sub-part “a”, B⁺ is computed as {B,D,A,C,E}.

The functional dependency B→E can be determined from the above set and from this the functional dependency BC→E becomes irrelevant

d.

Explanation of Solution

Third Normal Form (3NF) decomposition:

In the canonical cover, functional dependency is exists such that the attribute set id is a subset of other dependency.

Hence, each functional dependency will generate its own relation.

The relations according to the functional dependencies in the canonical cover are,

r1(A,B,C)

r2(B,D,E)

r3(D,A)

“G” has to part of all super keys as it is not dependent on any attribute...

e.

Explanation of Solution

Boyce-Codd Normal Form (BCNF) decomposition:

To compute the BCNF decomposition of the schema relation “r” using the original set “F” of functional dependency, follow the steps below,

For the relation not being in BCNF, the first original functional dependency is responsible.

So, according to first functional dependency , the original relation is decomposed as follows,
...

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 5

Expert Solution & Answer

Answer 6

Expert Solution & Answer

Answer 7

Chapter 7, Problem 30E

a.

Explanation of Solution

To compute B⁺:

To compute B^+, the closure of an attribute “B” over a set of functional dependencies “F”, refer the algorithm in Figure 7.19 of the textbook.

The result is initialized to {B}.

For the first repetition,

The condition is not satisfied in first and second functional dependency.

The condition B is a subset of {B} is satisfied for the functional dependency B→D Hence “D” is added to the result.

Therefore, the result = {B,D}.

The condition “D” is a subset of {B,D} is satisfied for the functional dependency D→A Hence “A” is added to the result

b.

Explanation of Solution

To prove:

“AG” is super key.

Proof:

This can be proved by proving the existence of functional dependency so that AG→ABCDEG It is given that A→BCD.

The union of the similar sets is the set with itself (AA=A).

To get A→ABCD, use Armstrong’s augmentation axiom with “A”.

BC→DE is given and to get ABCD→ABCDE, use Armstrong’s augmentation axiom with ABCD

c.

Explanation of Solution

Canonical cover:

“D” is irrelevant in first and second dependencies because of third dependency.

Hence remove the attribute “D” from the first and second functional dependencies to achieve A→BC and BC→E.

According to the sub-part “a”, B⁺ is computed as {B,D,A,C,E}.

The functional dependency B→E can be determined from the above set and from this the functional dependency BC→E becomes irrelevant

d.

Explanation of Solution

Third Normal Form (3NF) decomposition:

In the canonical cover, functional dependency is exists such that the attribute set id is a subset of other dependency.

Hence, each functional dependency will generate its own relation.

The relations according to the functional dependencies in the canonical cover are,

r1(A,B,C)

r2(B,D,E)

r3(D,A)

“G” has to part of all super keys as it is not dependent on any attribute...

e.

Explanation of Solution

Boyce-Codd Normal Form (BCNF) decomposition:

To compute the BCNF decomposition of the schema relation “r” using the original set “F” of functional dependency, follow the steps below,

For the relation not being in BCNF, the first original functional dependency is responsible.

So, according to first functional dependency , the original relation is decomposed as follows,
...

Answer 8

Chapter 7, Problem 30E

a.

Explanation of Solution

To compute B⁺:

To compute B^+, the closure of an attribute “B” over a set of functional dependencies “F”, refer the algorithm in Figure 7.19 of the textbook.

The result is initialized to {B}.

For the first repetition,

The condition is not satisfied in first and second functional dependency.

The condition B is a subset of {B} is satisfied for the functional dependency B→D Hence “D” is added to the result.

Therefore, the result = {B,D}.

The condition “D” is a subset of {B,D} is satisfied for the functional dependency D→A Hence “A” is added to the result

Answer 9

Chapter 7, Problem 30E

a.

Explanation of Solution

To compute B⁺:

To compute B^+, the closure of an attribute “B” over a set of functional dependencies “F”, refer the algorithm in Figure 7.19 of the textbook.

The result is initialized to {B}.

For the first repetition,

The condition is not satisfied in first and second functional dependency.

The condition B is a subset of {B} is satisfied for the functional dependency B→D Hence “D” is added to the result.

Therefore, the result = {B,D}.

The condition “D” is a subset of {B,D} is satisfied for the functional dependency D→A Hence “A” is added to the result

Answer 10

b.

Explanation of Solution

To prove:

“AG” is super key.

Proof:

This can be proved by proving the existence of functional dependency so that AG→ABCDEG It is given that A→BCD.

The union of the similar sets is the set with itself (AA=A).

To get A→ABCD, use Armstrong’s augmentation axiom with “A”.

BC→DE is given and to get ABCD→ABCDE, use Armstrong’s augmentation axiom with ABCD

Answer 11

b.

Explanation of Solution

To prove:

“AG” is super key.

Proof:

This can be proved by proving the existence of functional dependency so that AG→ABCDEG It is given that A→BCD.

The union of the similar sets is the set with itself (AA=A).

To get A→ABCD, use Armstrong’s augmentation axiom with “A”.

BC→DE is given and to get ABCD→ABCDE, use Armstrong’s augmentation axiom with ABCD

Answer 12

c.

Explanation of Solution

Canonical cover:

“D” is irrelevant in first and second dependencies because of third dependency.

Hence remove the attribute “D” from the first and second functional dependencies to achieve A→BC and BC→E.

According to the sub-part “a”, B⁺ is computed as {B,D,A,C,E}.

The functional dependency B→E can be determined from the above set and from this the functional dependency BC→E becomes irrelevant

Answer 13

c.

Explanation of Solution

Canonical cover:

“D” is irrelevant in first and second dependencies because of third dependency.

Hence remove the attribute “D” from the first and second functional dependencies to achieve A→BC and BC→E.

According to the sub-part “a”, B⁺ is computed as {B,D,A,C,E}.

The functional dependency B→E can be determined from the above set and from this the functional dependency BC→E becomes irrelevant

Answer 14

d.

Explanation of Solution

Third Normal Form (3NF) decomposition:

In the canonical cover, functional dependency is exists such that the attribute set id is a subset of other dependency.

Hence, each functional dependency will generate its own relation.

The relations according to the functional dependencies in the canonical cover are,

r1(A,B,C)

r2(B,D,E)

r3(D,A)

“G” has to part of all super keys as it is not dependent on any attribute...

Answer 15

d.

Explanation of Solution

Third Normal Form (3NF) decomposition:

In the canonical cover, functional dependency is exists such that the attribute set id is a subset of other dependency.

Hence, each functional dependency will generate its own relation.

The relations according to the functional dependencies in the canonical cover are,

r1(A,B,C)

r2(B,D,E)

r3(D,A)

“G” has to part of all super keys as it is not dependent on any attribute...

Answer 16

e.

Explanation of Solution

Boyce-Codd Normal Form (BCNF) decomposition:

To compute the BCNF decomposition of the schema relation “r” using the original set “F” of functional dependency, follow the steps below,

For the relation not being in BCNF, the first original functional dependency is responsible.

So, according to first functional dependency , the original relation is decomposed as follows,
...

Answer 17

e.

Explanation of Solution

Boyce-Codd Normal Form (BCNF) decomposition:

To compute the BCNF decomposition of the schema relation “r” using the original set “F” of functional dependency, follow the steps below,

For the relation not being in BCNF, the first original functional dependency is responsible.

So, according to first functional dependency , the original relation is decomposed as follows,
...

Answer 18

Expert Solution & Answer

Answer 19

Expert Solution & Answer

Answer 20

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Answer 21

Ch. 7 - Prob. 1PE Ch. 7 - Prob. 2PE Ch. 7 - Explain how functional dependencies can be used to...Ch. 7 - Prob. 4PE Ch. 7 - Prob. 5PE Ch. 7 - Prob. 6PE Ch. 7 - Prob. 7PE Ch. 7 - Prob. 8PE Ch. 7 - Prob. 9PE Ch. 7 - Prob. 10PE

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Chapter 7 Solutions