**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

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