No of Super key possible

Question

$R=(A,B,C,D)$
Candidate key $(AB,DB)$
How many super key possible?

Comments

correct!!1

Answer 1

Q1)R=(A,B,C,D)
Candidate key(AB,DB)- 2CK

Superkeys:-We can find as:-

CASE:1

X=1st Candidate key AB = AB_ _ =2²  

Y=2nd Candidate key DB = _ B_D =2²

CASE:2

X∩ Y=   Due to both AB and DB candidate key AB_D=2

According to Inclusion-Exclusion

n(X U Y) = n(X) + n(Y) - n(X∩ Y)

So,no of superkeys = 4+4 -2 = 6

AB,ABC,ABD,DB,ADB,CDB 

Comments

if no. of possibilites is 2 then why 2 square is taken?

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@Krishna R=(A,B,C,D)  so if we take AB_ _ then we have a choice in two places and in each place we have two choice either we take or not thats y 2²

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Good job...nice explanation

Answer 2

6 sk

total # sk's = # sk's over prime attributes *(2^# non-primes attributes)

here 3 sk's possible over prime attributes {AB ,DB, ABD}

here only one non-prime remains C

so total # sk's = 3*(2^1)

=6

Comments

anyone explain plz how to find it??

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check now@ https://gateoverflow.in/user/Hira+Thakur