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+43 votes

Relation $R$ with an associated set of functional dependencies, $F$, is decomposed into $\text{BCNF}$. The redundancy (arising out of functional dependencies) in the resulting set of relations is

- Zero
- More than zero but less than that of an equivalent $3NF$ decomposition
- Proportional to the size of F
^{+ } - Indeterminate

+52 votes

Best answer

Answer is **A**.

If a relation schema is in **BCNF** then all redundancy based on functional dependency has been removed, although other types of redundancy may still exist. A relational schema *R* is in Boyce–Codd normal form if and only if for every one of its dependencies* X → Y*, at least one of the following conditions hold:

*X → Y*is a trivial functional dependency (Y ⊆ X)*X*is a super key for schema*R*- http://en.wikipedia.org/wiki/Boyce%E2%80%93Codd_normal_form

0

question is redundancy (arising out of functional dependencies) in the resulting set of relations is ?

there is no redundancy due to FDs.

Should be A ??

there is no redundancy due to FDs.

Should be A ??

+9

please give an example where redundancy due to functional dependencies exist. ?

I am not getting what is diff between **redundancy due to functional dependency **and **due to other reasons..**

+2

Relation that have redundant data problem called anomalies. Anomalies can only removed by normalization.

**1st NF **talks about atomic values and non-repeating groups.

**2nd NF** enforces that a non-Key attribute should belong to entire Key attribute.

**3rd NF** makes sure that there should be no transitive dependency between a non-Key and a Key attribute.

and

Redundancy allowed by 3NF is disallowed by BCNF

BCNF is stricter than 3NF 3NF is stricter than 2NF

https://www.slideshare.net/jenspatel/data-redundancy-update-anomalies

0

Redundancy more than 0 Because

Example table R(A,B)

Tuples are (A1,b1),(A2,b1),(A3,b1),(A4,b1)(A5,b1).............(An,b1)

Here Attribute A is primary key so relation in bcnf but if you see attribute B there redundancy so we can't say bcnf has 0 redundancy.so option b correct.

Example table R(A,B)

Tuples are (A1,b1),(A2,b1),(A3,b1),(A4,b1)(A5,b1).............(An,b1)

Here Attribute A is primary key so relation in bcnf but if you see attribute B there redundancy so we can't say bcnf has 0 redundancy.so option b correct.

0

So why is the answer to this question is false? If the number of redundancies are zero, then the answer to this should have been true?

0

@srestha mam

If we consider the rajinder singh case in the previous comment, then we can form a FD like A->B, but then also the relation is in BCNF & it has duplicate tuples.

here A is a key & A->B satisfies the BCNF condition, but attribute B has redundant values.

+34 votes

+1

@digvijay pandey can you please explain what is multi valued dependency? And how it can be present in BCNF?

+14

Hello hemant

A relation would be BCNF when functional dependencies are like * $x->y$* ; where

But what will happen when we have such a relation that doesn't contain any non-trivial functional dependency?Hope you can guess that in such case super key will be the whole set of attributes.So eventually we can say such a relation will be BCNF relation. Can we say that this kind of relation will be free of redundancies ? There comes the term call 'Multivalued dependency', whose presence make the relation redundant. like

**R(course_id,professor_name,Books_name)**

Course_id |
Professor_name |
Books_name |

C1 | P1/P2 | B1/B2/B3 |

C2 | P2/P3 | B4/B5 |

C3 | P4 | B4 |

You can see , this relation is in BCNF as no functional dependencies exist there.

But here exists multivalued dependencies (basically multivalued dependency means multiple values of some attributes are dependent on some attribute value )

+1

@Rupendra Choudhary @srestha @Priya_das * Answer* should be

One more thing, basic definition of a Relation says that a **Table** should have **atomic** attributes i.e. neither * Multi-valued* nor

+1

Thanks @srestha for Replying.

Can you share the Resource where it is mentioned? Coz, What i have learnt is that, by definition, a Relation is always in 1NF and in 1NF, we cannot have either * Composite* or

0

1NF is atomic relation. I donot think attribute depends on it. Multivalued dependency is not in 1NF.

0

Thanks. I got it now. In 1NF we make attributes atomic, either by removing

**Composite attributes (by breaking them into simpler attributes), or,***Multi-valued attributes (by making a new tuple for each 'value' in Multi-valued attribute)*

Multi-valued Dependency is removed in 4NF.

0

@srestha

In 4NF we remove the multivalued dependency. That means in BCNF multi-valued dependencies might be present. Because in 4NF, the table must be present in the form of BCNF. Correct me if I am wrong.

+6 votes

0 votes

The answer is **A**.

If a relation schema is in **BCNF** then all redundancy based on functional dependency has been removed, although other types of redundancy may still exist. A relational schema *R* is in Boyce–Codd normal form iff for every one of its dependencies* X → Y*, at least one of the following conditions hold:

*X → Y*is a trivial functional dependency (Y ⊆ X)*X*is a super key for schema*R*

–1 vote

in above question, I don't understand the meaning of ( arising out of functional dependencies ) I am thinking like "arising out of" mean, not due to functional dependencies and you all explained that it due to functional dependencies how to get it please help ?

–2 votes

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