can superkeys be considered?

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consider R(A,B,C,D,E) with FDs

AB->C,C->D,D->B,D->E

if number of keys in relation R is ‘a’ and the number of relation in 3NF decomposition is ‘b’ what is the value of a-b?

i am getting a=3,b=4

a-b=-1

am i correct?

AB->C,C->D,D->B,D->E

if number of keys in relation R is ‘a’ and the number of relation in 3NF decomposition is ‘b’ what is the value of a-b?

i am getting a=3,b=4

a-b=-1

am i correct?

0 votes

**Candidate keys**: AB,AD,AC

AB->C, (Full)

C->D, (key)

D->B, (key)

D->E (Partial)

Decompose into ABCD, DE

**ABCD (keys: A,B,C,D)**

AB->C, (Full)

C->D, (key)

D->B, (key)

when in a relation all the attributes are prime then the minimum normal form the relation satisfies is 3NF

**DE**

D->E (Full)

So the no. of Relations needed will be 2.

a=3, b=2, So the answer will be 1.

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