retagged by
1,163 views
3 votes
3 votes
Let the relation R(X, Y, Z, A, B) with given functional dependencies
X -> YZ
Z -> A
A -> B
AZ -> X
The number of super keys possible__________ ?

Please answer with an explanation.
retagged by

3 Answers

0 votes
0 votes
24 bcz candidate key are X,Z so 2^4+2^4-2^3  {X2^(n-1)+Z2^(n-1)-2^(n-2)}

OR

Direct (Superkey Among Prime)*2^(n-2) #Of non prime key

3*2^(5-2)

24
0 votes
0 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)

here ck's are { X , Z }

# SK's over prime attributes are= { X , Z , XZ } =3 SK's over prime attributes

non-prime attributes are Y , A , B =3

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

=24 SK's

Related questions

4 votes
4 votes
1 answer
1
naveen bhatt asked Dec 7, 2018
1,364 views
Consider the relation R(A,B,C,D,E) with the functional dependencies :A → B, B → C, C → A, D → E, and E → DThe Maximum possible super-keys of R is ______________...
0 votes
0 votes
6 answers
2
kaustubh7 asked Sep 20, 2023
811 views
Consider a relation $R(ABCDEFGH)$.How many superkeys will be there in relation $R$ if the candidate keys for relation are ${A, BC, CDE, EF}$?
4 votes
4 votes
1 answer
3
Akash Kanase asked Jan 15, 2016
710 views
I think here even with Ph is removed from table we still get 12 Super keys. Because Salary & One of CK must be there.
6 votes
6 votes
5 answers
4