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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?
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can superkeys be considered?

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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i also got a=3 but how u got b??

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

ABCD  (keys: A,B,C,D)

AB->C

C->D

D->B

In this relation all the attributes are prime/key, so the Relation will be in 3-NF.

+1 vote