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GATE CSE 2014 Set 2 | Question: 21

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34 votes
34 votes
The maximum number of superkeys for the relation schema $R(E,F,G,H)$ with $E$ as the key is _____.
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8 Answers

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3 votes

THE BEST METHOD FOR CALCULATING NO OF SK's:

total no of sk's= (# sk's over prime attributes) *  (2^#no of non-prime)

total no of sk's= 1*(2^3)

total no of sk's=8

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3 votes
3 votes

$No. of superkeys = 2^{M - N}$

where, M = Total number of attributes, N= number of attributes in given candidate key

In this question, M = 4 , N = 1, Therefore, no. of superkeys = $2^{4-1} = 2^{3} = 8$

Note  that if multiple candidate keys are given we need to consider set theory to find the superkeys possible.

For example, R(A1,A2,A3,......,An) and Candidate keys = {A1,A2}

then superkeys, $SKs = SK(A1) + SK(A2) - SK(A1A2)= 2^{n-1}+2^{n-1}-2^{n-2} $

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1 votes
1 votes
Given E is the key,So if we add any element  with E it will become super key. Now we have 3 element F,G,H. for each element we have 2 option either to include it or not to include it . So resulting no. of superkey is 2 * 2 * 2= 8
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–6 votes
–6 votes
15 super keys are possible with E as compulsory attrubite we can add any number of attrubuts from 3 of them.

If we add one of them = 3

If we add two of them = 3*2

If we add 3 of them = 3*2
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