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

### 9 Comments

A table is in 4NF if it is in BCNF and it should not have any multivalued dependencies. Databases with multivalued dependencies exhibit redundancy.

So redundancy **arising out of functional dependencies** is 0 in BCNF but if the table has any multivalued dependencies then it has redundancy due to the multivalued dependencies.

See Ref: 1. https://en.wikipedia.org/wiki/Multivalued_dependency#Example

2. https://www.studytonight.com/dbms/fourth-normal-form.php

## 7 Answers

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

### 12 Comments

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

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?

@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.

### 9 Comments

Hello hemant

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

**x**is

**super key**and in that way when attributes are dependent on super key , redundancy can't exist.

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 )

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

*coz in BCNF we cannot have*

**Zero***but can have*

**redundancies due to Functional dependencies***. Correct me, if i am wrong.*

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

*. Then, how can we relate in BCNF as a Relation in BCNF qualifies to be in 1NF.*

**composite**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

*. Please, Confirm.*

**Multi-valued Attributes**@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.

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*