Find the Super Key for the Relation R ( A, B, C, D, E, F, G, H )

Consider a relation R with the schema R = (A, B, C, D, E, F, G, H) and a set of functional dependencies F as:

{AB→C, A→DE, B →F, F→GH}

 

Find the super key for this relation.

Given Relation:

R (A, B, C, D, E, F, G, H)

Given Functional Dependencies (F):

AB → C
A → DE
B → F
F→ GH

 

Attributes A and B are not present on the right side of any Functional Dependency(FD). Hence, AB must be present in the candidate key and super key.

Let's find the closure of AB.

{AB}+

= {AB} (axiom of reflexivity)

= {ABC} (AB → C, given FD)

= {ABCDE} (A → DE, given FD)

= {ABCDEF} (B → F, given FD)

= {ABCDEFGH} (F → GH, given FD)

 

The closure of AB is:

{AB}+ = {ABCDEFGH}

 

This means AB can determine all other attributes of the relation R. Hence, {AB} is a Super Key for the relation R.

 

Besides {AB}, all the possible combinations of {AB} with other attributes are also Super Keys.

For example,

{ABCD}

{ABF}

{ABE}

{ABFG}

etc. are also Super Keys.

 

 

 

 

Asked in Year
2021
Course
BBM
University
TU