Question

The number of superkeys possible for the relation R(A B C D E) with {A, BC, CDE} as three candidate keys are _________.

Please EXPLAIN the solution.

Answer 1

There are three Ck : A, BC, CDE

Take Each CK alone

No. of superkeys when CK is A alone = 2⁴ (Because each remaining 4 values may or may not be in the SK, hence 2 posibilities for each)

No. of superkeys when CK is BC alone = 2³ (Because each remaining 3 values may or may not be in the SK, hence 2 posibilities for each)

No. of superkeys when CK is CDE alone = 2² (Because each remaining 2 values may or may not be in the SK, hence 2 posibilities for each)

Now, take combinations from given CK

No. of superkeys when CK is ABC = 2² (Because each remaining 2 values may or may not be in the SK, hence 2 posibilities for each)

No. of superkeys when CK is BCDE = 2¹ (Because each remaining 2 values may or may not be in the SK, hence 2 posibilities for each)

No. of superkeys when CK is ACDE = 2¹ (Because each remaining 2 values may or may not be in the SK, hence 2 posibilities for each)

Now, take all three CK together

No. of superkeys when CK is ABCDE = 2⁰ (Because each remaining 2 values may or may not be in the SK, hence 2 posibilities for each)

Hence total SK = 2⁴ + 2³ + 2² - 2² - 2¹ - 2¹ + 2⁰ = 16 + 8 + 4 - (4+2+2) + 1 = 28 - 8 + 1 = 21

Hence total super keys are 21

Answer 2

using inclusion exclusion principle

no of superkeys :

2^4 +2^3 +2^2 - 2^2 -2^1 -2^1 +2^0

=21