DBMS Counting superkeys

Question

R(ABCDEF) with candidate keys set { A,BC,CD}. Find number of superkeys?

Comments

Is 44 answer?

Answer 1

We have CK = {A, BC, CD}

Then SK = SK(A) + SK(BC) + SK(CD) - SK(ABC) - SK(BCD) - SK(ACD) + SK(ABCD)

n = number of attributes = 5.

SK = 2^n-1 + 2^n-2 + 2^n-2 - 2^n-3 - 2^n-3 - 2^n-3 + 2^n-4

 = 32 + 16 + 16  - 8 - 8 - 8 + 4 

= 68 - 24

= 44.

Please correct me if i am wrong

Answer 2

1)When A is candidate Key

then any combination from BCDEF along with A form  Super  Key.

Therefore ,total super key = $2^{5}$ 

2)When BC is candidate Key

In this case any combination from DEF along with BC form  Super  Key.

A is not considered in the case because all the combination of BC with A is counted in first case.

Ex={ABC,ABCD,....ABCDEF} all are counted in case 1.

Therefore ,total super key = $2^{3}$ (All combination from DEF)

3)When CD is candidate Key

In this case any combination from BEF along with CD form  Super  Key.

A is not considered (Similar logic as case B).

Along with we have also find the keys which are counted in case B.

{CDB,CDBE,CDBF,CDBEF} all these values are counted in case B,because all have BC(candidate key of case B).

So,these keys are counted twice because all have CD(candidate key )

Therefore ,total super key = $2^{3}$ (All combination from BEF)-$2^{2}$(ABCD as KEY)

total key = $2^{5}+2^{3}+2^{3}-2^{2}$