Closure of an Attribute Set-
- The set of all those attributes which can be functionally determined from an attribute set is called as a closure of that attribute set.
- Closure of attribute set {X} is denoted as {X}+.
Steps to Find Closure of an Attribute Set-
Following steps are followed to find the closure of an attribute set-
Step-01:
Add the attributes contained in the attribute set for which closure is being calculated to the result set.
Step-02:
Recursively add the attributes to the result set which can be functionally determined from the attributes already contained in the result set.
Example-
Consider a relation R ( A , B , C , D , E , F , G ) with the functional dependencies-
A → BC
BC → DE
D → F
CF → G
Now, let us find the closure of some attributes and attribute sets-
Closure of attribute A-
A+ = { A }
= { A , B , C } ( Using A → BC )
= { A , B , C , D , E } ( Using BC → DE )
= { A , B , C , D , E , F } ( Using D → F )
= { A , B , C , D , E , F , G } ( Using CF → G )
Thus,
A+ = { A , B , C , D , E , F , G }
Closure of attribute D-
D+ = { D }
= { D , F } ( Using D → F )
We can not determine any other attribute using attributes D and F contained in the result set.
Thus,
D+ = { D , F }
Closure of attribute set {B, C}-
{ B , C }+= { B , C }
= { B , C , D , E } ( Using BC → DE )
= { B , C , D , E , F } ( Using D → F )
= { B , C , D , E , F , G } ( Using CF → G )
Thus,
{ B , C }+ = { B , C , D , E , F , G }
Finding the Keys Using Closure-
Super Key-
- If the closure result of an attribute set contains all the attributes of the relation, then that attribute set is called as a super key of that relation.
- Thus, we can say-
“The closure of a super key is the entire relation schema.”
Example-
In the above example,
- The closure of attribute A is the entire relation schema.
- Thus, attribute A is a super key for that relation.
Candidate Key-
- If there exists no subset of an attribute set whose closure contains all the attributes of the relation, then that attribute set is called as a candidate key of that relation.
Example-
In the above example,
- No subset of attribute A contains all the attributes of the relation.
- Thus, attribute A is also a candidate key for that relation.
Also Read- How To Find Candidate Keys?
PRACTICE PROBLEM BASED ON FINDING CLOSURE OF AN ATTRIBUTE SET-
Problem-
Consider the given functional dependencies-
AB → CD
AF → D
DE → F
C → G
F → E
G → A
Which of the following options is false?
(A) { CF }+ = { A , C , D , E , F , G }
(B) { BG }+ = { A , B , C , D , G }
(C) { AF }+ = { A , C , D , E , F , G }
(D) { AB }+ = { A , C , D , F ,G }
Solution-
Let us check each option one by one-
Option-(A):
{ CF }+ = { C , F }
= { C , F , G } ( Using C → G )
= { C , E , F , G } ( Using F → E )
= { A , C , E , E , F } ( Using G → A )
= { A , C , D , E , F , G } ( Using AF → D )
Since, our obtained result set is same as the given result set, so, it means it is correctly given.
Option-(B):
{ BG }+ = { B , G }
= { A , B , G } ( Using G → A )
= { A , B , C , D , G } ( Using AB → CD )
Since, our obtained result set is same as the given result set, so, it means it is correctly given.
Option-(C):
{ AF }+ = { A , F }
= { A , D , F } ( Using AF → D )
= { A , D , E , F } ( Using F → E )
Since, our obtained result set is different from the given result set, so,it means it is not correctly given.
Option-(D):
{ AB }+ = { A , B }
= { A , B , C , D } ( Using AB → CD )
= { A , B , C , D , G } ( Using C → G )
Since, our obtained result set is different from the given result set, so,it means it is not correctly given.
Thus,
Option (C) and Option (D) are correct.
Next Article- 10 Different Kinds of Keys in DBMS
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