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52 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

  1. Zero
  2. More than zero but less than that of an equivalent $3NF$ decomposition
  3. Proportional to the size of F
  4. Indeterminate
in Databases edited by


Question is 19?
Yes, thanks for catching it.  It has been corrected. You can solve it now. :-)

typo in F*

corrected :)
which book?
this is a error i have confirmed from vani institute's teacher .this was there in there study material.
Any good references to study MVDs?
based on  functional dependency BCNF has 0 percent redundancy ,but is there any chance for other type redundancy m not getting other type ?explain pls
edited by


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.


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7 Answers

61 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:

edited by


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 ??
Yes .ans is A

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..

if BCNF has multivalued dependencies..then redundancy will happen? right..?

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.


Redundancy allowed by 3NF is disallowed by BCNF

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

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 option b correct.

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  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.


@Srestha mam

Can you please update the NPTEL link?



Multi valued dependency is also a type of FD right?
edited by

(C) Proportional to the size of F+

I really didn't understand this option. What does it mean by the size of the F closure? 

Yes, it's the size of closure.
37 votes
should be zero..
BCNF can have Multi valued dependency but no redundancy due to FDs..
edited by


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

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


    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 Zero coz in BCNF we cannot have redundancies due to Functional dependencies but can have other redundancies like Multi-valued Dependency. Correct me, if i am wrong.

One more thing, basic definition of a Relation says that a Table should have atomic attributes i.e. neither Multi-valued nor composite. Then, how can we relate in BCNF as a Relation in BCNF qualifies to be in 1NF. 


BCNF has "Multi-valued Dependency.".

No , it is not correct. MVD is for4NF


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 Multi-valued Attributes. Please, Confirm. 

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


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.

edited by

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
Yes , the answer should be zero. There should not be any redundancy due to FD in BCNF.
edited by


There should not be any //dependency due to FD in BCNF.
ohh yaa.. typos there.. correcting it..thanks :)
0 votes

Explain please

BCNF can have Multi valued dependency but no redundancy due to FDs

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 ?

1 comment

there can be other forms of dependency too, apart from FD. In BCNF, FDs are preserved.
–2 votes
Increasing order of redundancy:

4NF < BCNF < 3NF < 2NF < 1NF < un-normalized

Answer is B.


This should be 0. Ans -> A
Yes Answer is b because there can be redundancy beacuse of MVDS
@Akash please share us link if you have any, how 3NF redundancy became zero.

@ Akash Kanase b is right


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