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Consider the relation R(A,B,C,D,E) with the functional dependencies :

A → B, B → C, C → A, D → E, and E → D

The Maximum possible super-keys of R is ______________?
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ANSWER IS 21

Candidate Keys are AD AE BD BE CD CE.

Now, we have to find out Super keys. Let's consider the first candidate key(CK);

1st CK= AD {_ _ _} we can fill 3 space with B, C, E so combinations will be 2^3 = 8 

8 Super keys from first CK.

2nd CK= AE{_ _ _} B C D 

Now, the important point here is, in the first candidate key attribute A with attribute D

and in Second candidate key, we have attribute A (So, the combination of AE with attribute D we can eliminate directly because those combinations we already considered in first CK).

so combinations will be 2^2 = 4

4 Super keys from 2nd CK.

3rd CK= BD{_ _ _} A C E 

Similarly here, in the first candidate key attribute A with attribute D

So, the combination of BD with the attribute A we can eliminate directly because those combinations we already considered in first CK).

so combinations will be 2^2 = 4

4 Super keys from 3rd CK.

4th CK= BE{_ _ _} A C D 

Similarly here, in the 2nd and 3rd candidate key's attributes A  and D with attribute B

So, the combination of BE with the attributes A and D we can eliminate directly because those combinations we already considered in 2nd and 3rd CKs).

so combinations will be 2^1 = 2

2 Super keys from 4th CK.

Similarly for CD, 2 Super keys from 5th CK

and 1 Super key from 6th CK.

 

Total Super keys= 8+4+4+2+2+1= 21

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