Question

Q2: Explain with example the difference between super keys and functional dependencies in relational databases. Show how to find a key (super/candidate) for the following functional dependencies [10 points]: R = {A, B, C, D, E, F, G, H, I, J, K,L,M} F {A B C D E, E F G H I J,AI →K,AL →M} =

Answer 1

Super key is a set of one or more attribute which can uniquely identify a row in the relational schema while function dependecies can not identify the whole row but the can identify the attributes that are dependent.

For a relation schema R = {A, B, C, D, E} and functional dependencies F = {CE $\rightarrow$ D, D $\rightarrow$ B, C $\rightarrow$ A} super key for this relational schema can be ACE by using ACE we can uniquely identify a row but by using a functional dependecies we can not uniquely identify a row.

Candidate key in the minimal super key or minimal set of attributes that can uniquely identify a row in the schema. Every candidate key is a super key but vice versa is not true.

Given relational schema and its functional dependencies:

R = {A, B, C, D, E, F, G, H, I, J, K, L, M}

F = {A $\rightarrow$ BCDE, E $\rightarrow$ FGH, I $\rightarrow$ J, AI $\rightarrow$ K, AL $\rightarrow$ M}

Check all the attributes that are not available on the right side of the functional dependencies. A, I, L are not present on the right side of the functional dependencies. Now find the Attribute Closure of A, I, L.

(A, I, L)⁺ = {A, I, L}

= {A, B, C, D, E, I, J, K, L, M} ( Using A $\rightarrow$ BCDE, I $\rightarrow$ J, AI $\rightarrow$ K, AL $\rightarrow$ M)

= {A, B, C, D, E, F, G, H, I, J, K, L, M} ( Using E $\rightarrow$ FGH)

As (A, I, L)⁺ give the set of all attribute of relation R. So it is an candidate/super key.

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