Test by Bikram | Mock GATE | Test 3 | Question: 25

Question

Consider these relations :

$R = ABCDE$

$A\rightarrow BC$
$B\rightarrow E$
$D\rightarrow E$

What is the candidate key for the above relations?

• A. CA
• B. AB
• C. BD
• D. AD

Comments

Useful analysis (Thanks to @Deepak Poonia and @GO Classes)

There is no FD that implies A or D. Hence A and D both should be part of the Candidate Keys because of the trivial FD:

$AD \rightarrow AD$

Now,

$(AD)^+ \rightarrow ABCDE$

i.e. AD uniquely identifies all the attributes and is the minimal SK (removing any of A or D is not possible)

We can also conclude that AD is the only CK possible from the fact that there is no non-trivial FD implying A or D. Hence making them necessary attrbutes in the CK. Adding any new attribute will not keep it minimal.

Basically, we can't add anything nor can we remove anything. Hence AD is the only CK possible.

Answer 1

we can solve it by verifying options.

Option A) $(CA)^+\rightarrow CABE$

Option B) $(AB)^+\rightarrow AB\rightarrow ABC\rightarrow ABCE$

Option C) $(BD)^+\rightarrow BDE$

Option D) $(AD)^+\rightarrow ABD\rightarrow ABCD\rightarrow ABCDE $

So this option covers all the attributes, unlike others.

Option (D) is correct.