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Question

R(A,B,C,D,E) is a relation with candidate keys AB BC CD.Find the total number of super keys possible?

Comments

Is your question about no of superkeys or no of candidate keys since no of candidate keys is given already , makes no sense.I m writing the answer for superkeys plz check it

Answer 1

We use the inclusion exclusion principle.Before that,we calculate the following:

a) Superkeys due to 1 candidate key at a time:

Superkeys due to AB = 2³(since these 2 attributes will be there + each of C,D and E may or may not be included,so 2 ways for each of C,D and E) = 8

Similarly , keys due to BC = 2³(due to 2 ways each for A,D and E) = 8

Keys due to CD = 2³ = 8 

Now , taking 2 candidate keys intersection at a time:

Due to AB and BC i.e. ABC = 2²(choice is remaining for D and E only) = 4

Similarly,due to BC and CD i.e. BCD = 2² = 4(choice remains for A and E)

Due to AB and CD i.e. ABCD = 2(choice remains for E only)

Now,taking all 3 candidate keys intersection:

Due to AB,BC and CD i.e. ABCD = 2 ways(choice only for E attribute)

Now,we apply inclusion exclusion principle.

We know ,if n(A) , n(B) and n(C) are cardinalities of set A,B and C respectively, then

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A∩B) - n(B∩C) - n(A∩C) + n(A∩B∩C)

 So,in this case, set A,B and C are the superkeys due to candidate keys AB,BC and CD respectively.

So, total no. of superkeys = 8 + 8 + 8 - 4 - 4 - 2 + 2 = 16 superkeys

Hence,16 superkeys possible for the given candidate keys

Comments

Thanks bhai

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My question is regarding the number of super keys ,not regarding the candidate keys and thanks for the answer.

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nice explanation

Answer 2

I guess the question should be the number of super keys possible for given candidate keys AB,BC,CD so now it will be 2^3  + 2^2 + 2^2 = 16 super keys possible total=AB,ABC,ABD,ABE,ABCD,ABDE,ABCE,ABCDE,BC,BCD,BCE,BCDE,CD,CDE,CDA,CDAE.

Comments

Your answer has come right but the method is wrong . Plz have a look at my answer.