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Attribute Closure in DBMS - GeeksforGeeks

Attribute Closure in DBMS

Last Updated : 23 Jul, 2025

Functional dependency and attribute closure are essential for maintaining data integrity and building effective, organized and normalized databases. Attribute closure of an attribute set can be defined as set of attributes which can be functionally determined from it.

How to find attribute closure of an attribute set?

To find attribute closure of an attribute set:

Add elements of attribute set to the result set.
Recursively add elements to the result set which can be functionally determined from the elements of the result set.

Using FD set of table 1, attribute closure can be determined as:

(STUD_NO)⁺+= {STUD_NO, STUD_NAME, STUD_PHONE, STUD_STATE, STUD_COUNTRY, STUD_AGE}
(STUD_STATE)⁺ = {STUD_STATE, STUD_COUNTRY}

Important Points About Attribute Closure

Helps to identify all possible attributes that can be derived from a set of given attributes.
Helps in database design by showing how attributes and tables are related, which can improve query performance.
Can be computationally expensive, especially for large datasets.
Become complex to manage as the number of attributes and tables increases.

How to Find Candidate Keys and Super Keys Using Attribute Closure?

If attribute closure of an attribute set contains all attributes of relation, the attribute set will be super key of the relation.
If no subset of this attribute set can functionally determine all attributes of the relation, the set will be candidate key as well. For Example, using FD set of table 1

(STUD_NO, STUD_NAME)⁺ = {STUD_NO, STUD_NAME, STUD_PHONE, STUD_STATE, STUD_COUNTRY, STUD_AGE}

(STUD_NO)⁺= {STUD_NO, STUD_NAME, STUD_PHONE, STUD_STATE, STUD_COUNTRY, STUD_AGE}

(STUD_NO, STUD_NAME) will be super key but not candidate key because its subset (STUD_NO)⁺ is equal to all attributes of the relation. So, STUD_NO will be a candidate key.

Prime and Non-Prime Attributes

Attributes which are parts of any candidate key of relation are called as prime attribute, others are non-prime attributes. For Example, STUD_NO in STUDENT relation is prime attribute, others are non-prime attribute.

GATE Questions Q.1: Consider the relation scheme R = {E, F, G, H, I, J, K, L, M, N} and the set of functional dependencies {{E, F} -> {G}, {F} -> {I, J}, {E, H} -> {K, L}, K -> {M}, L -> {N} on R. What is the key for R? (GATE-CS-2014)

A. {E, F}
B. {E, F, H}
C. {E, F, H, K, L}
D. {E}

Solution:

Finding attribute closure of all given options, we get:
{E,F}⁺ = {EFGIJ}
{E,F,H}⁺= {EFHGIJKLMN}
{E,F,H,K,L}⁺ = {{EFHGIJKLMN}
{E}⁺= {E}
{EFH}⁺ and {EFHKL}⁺ results in set of all attributes, but EFH is minimal. So it will be candidate key. So correct option is (B).

Q.2: How to check whether an FD can be derived from a given FD set? Solution:

To check whether an FD A->B can be derived from an FD set F,

Find (A)⁺ using FD set F.

If B is subset of (A)⁺, then A->B is true else not true.

Q.3: In a schema with attributes A, B, C, D and E following set of functional dependencies are given
{A -> B, A -> C, CD -> E, B -> D, E -> A}
Which of the following functional dependencies is NOT implied by the above set? (GATE IT 2005)

A. CD -> AC
B. BD -> CD
C. BC -> CD
D. AC -> BC

Solution:

Using FD set given in question,
(CD)+ = {CDEAB} which means CD -> AC also holds true.
(BD)+ = {BD} which means BD -> CD can't hold true. So this FD is no implied in FD set. So (B) is the required option.
Others can be checked in the same way.

Q.4: Consider a relation scheme R = (A, B, C, D, E, H) on which the following functional dependencies hold: {A–>B, BC–> D, E–>C, D–>A}. What are the candidate keys of R? [GATE 2005]

(a) AE, BE
(b) AE, BE, DE
(c) AEH, BEH, BCH
(d) AEH, BEH, DEH

Solution:

(AE)⁺ = {ABECD} which is not set of all attributes. So AE is not a candidate key. Hence option A and B are wrong.
(AEH)⁺= {ABCDEH}
(BEH)⁺ = {BEHCDA}
(BCH)⁺= {BCHDA} which is not set of all attributes. So BCH is not a candidate key. Hence option C is wrong.
So correct answer is D.

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