Question

10 points QUESTION 3 Given R=(A, B, C) is a schema and F = {2C-A AB) is a set of FDs that hold on R. Which of the following statements is not true? d=(AB. AC) is a decomposition of that is in BCNF. O Ris in 3NF O BC is a candidate key for R Ris in BCN 10 points QUESTION 4 Given R= (A. 3. CD. E) is a schema and F= (A -0.8-E DE-C) is a set of FDs that hold on R. Which of the following statements is not true? Oo= {AD. BE ABC) is a lossless-join BCNF decomposition of R Od = {DEC, AD, AB. BE) is a lossless-join and dependency preserving BCNF decomposition of Every candidate key of R must contain the attributes A and B A3 is the only candidate key for R Od = {AD. BE CDE is a 3NF lossless-join and dependency-preserving decomposition of R

my choices for these are wrong.

Answer 1

3.

Here, the problem requires the option which is not true.

For the first option, on decomposition the relaiton into AB and AC, the following functional dependencies are noted.

The set of functional dependencies with respect to decomposed relation AB is :

A -> B

Here, attribute A is the superkey in decomposed relation AB as attribute closure of A will give the attributes A and B and this follows BCNF definition.

The decomposed relation AC does not have any functional dependencies.

So, the decomposition follows BCNF definition.

For the second option, the set of functional dependencies of R given is :

BC -> A

A -> B

The candidate key of R are BC and AC as the closure of both the attributes will give all the attributes. So, the prime attributes are A, B and C.

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.

Since all the attributes are prime, the relation R is in 3NF.

For the third option, it is correct that BC is a candidate key of R as attribute closure of BC gives { A, B, C }.

For the fourth option, it is wrong because in BCNF, the left side attribute of every functional dependency must be a superkey. But in the given relation R, the functional dependency A -> B, the attribute A is not a superkey.

So, the answer is option 4.

4.

Here, the problem requires the option which is not true.

For the first option, the BCNF decomposition is given as :

In Relation AD,

The set of F.D.s is :

A -> D

In Relation BE,

The set of F.D.s is :

B -> E

In Relation ABC,

There are no FDs.

BCNF property is maintained.

In order to check the lossless property, the common attribute between relation BE and ABC is B and B is the candidate key in relation BE. The common attribute between ABCE and AD is A and A is the candidate key in relation AD. So, the BCNF decomposition is lossless.

For the second option, the BCNF decomposition is given as :

In relation DEC,

The set of F.D.s is :

DE -> C

DE is a superkey in DEC and so it follows BCNF.

In relation AD,

The set of F.D.s is :

A -> D

A is a superkey in AD and so it follows BCNF.

In relation AB,

There are no functional dependencies.

In relation BE,

The set of F.D.s is :

B -> E

B is a superkey in BE and so it follows BCNF.

The common attribute between relation AB and BE is B and B is the candidate key in relation BE. The common attribute between relations ABE and AD is A and A is the candidate key in AD. The common attribute between relations ABED and DEC is DE and DE is the candidate key in DEC. So, the given BCNF decomposition is a lossless decomposition.

The closure of the functional dependencies of the decomposed relations is same as the closure of the original functional dependency.

For the third option, the set of functional dependencies of R is :

A -> D

B -> E

DE -> C

The candidate key is AB as the attribute closure of AB is { A, B, D, E, C }.

So, it is correct that every candidate key of R must contain the attributes A and B.

For the fourth option, the candidate key of R is AB.

For the fifth option, the decomposition is :

In relation AD,

The set of F.D.s is :

A -> D

In relation BE,

The set of F.D.s is :

B -> E

In relation CDE,

The set of F.D.s is :

DE -> C

The relations are in 3NF.

The common attribute between relations BE and CDE is E but E is not the candidate key in either BE or CDE. The common attribute between relations AD and CDE is D but D is not the candidate key in either AD and CDE.

The 3NF decomposition is not a lossless decomposition.

So, the answer is option 5.