This content originally appeared on DEV Community and was authored by hebaShakeel
How to find a Super Key in a Relation
Consider a relation R(a1, a2, a3,.....an), where n = number of attributes in a relation.
Suppose Candidate Key = (a1)
Find total SKs.
No. of Super Keys = 2^(n-1)
Example 1
R(ABCDE), CK={B}
Find total SKs
Here n = 5
n-1 = 4
Therefore total number of SKs = 2^4 = 16
Example 2
R(a1, a2, a3,....an), CK={a1a2}
Find total SKs
No. of Super Keys = 2^(n-2), since here there are 2 attributes in the candidate Key.
Example 3
R(XYZMN), CK={ZM}
Find total SKs
Here n = 5
No. of attributes in Candidate Key = 2
Therefore total number of SKs = 2^(5-2) = 2^3 = 8
Example 4
R(a1, a2, a3,....an), CK={a1,a2}
Find total SKs
All SKs which have either a1 or a2 or a1a2
n(a1 U a2) = n(a1) + n(a2) - n(a1a2)
Therfore No. of Super Keys = 2^(n-1) + 2^(n-1) - 2^(n-2)
Example 4
R(A,B,C,D,E,F), CK={A,DF}
Find total SKs
Total number of SKs = 2^(6-1) + 2^(6-2) - 2^(6-3) = 32+16-8 = 40
This content originally appeared on DEV Community and was authored by hebaShakeel
hebaShakeel | Sciencx (2021-06-13T09:51:57+00:00) Finding Super Keys. Retrieved from https://www.scien.cx/2021/06/13/finding-super-keys/
Please log in to upload a file.
There are no updates yet.
Click the Upload button above to add an update.