Ace Test Series: Databases - Functional Dependencies

Question

Comments

Ans:(B) NO

Explanation:

Since AB  determine all the attributes of the Relation R(A,B,C,D) , AB is called Key.

As  AB is the minimal Key it can be called as Candidate key (or primarykey)

Intailly the decomposed relations be R1(A,C,D), R2(B,C)

Now from R1(ACD) the Functional Dependencies determined are C - >A, C - >D 

and From R2(BC) no functional dependencies are possible

now to test whether AB ->C is in the  functional dependencies of (R1 U R2)  compute {AB}⁺

(AB)⁺={A,B} 

As (AB)⁺ doesn't contains C, the decomposition is not dependendency preserving.

Hope this makes you clear, Thanks for your question :)

---

but (AB)+ contains C.

Also  we can reduce AB -> C as B->C since C->A so dependacy is preserved.Correct me if wrong

---

(AB)⁺={A,B} (it is with respect to functional dependecies of R1 U R2 , that is the newly decomposed tables but not with respect to R(ABCD) )

secondly you are confused with Decomposition rule

According to Decomposition rule  we can not reduce AB -> C to B ->C

for example if functional dependecy is X -> YZ then according to Decomposition rule we can reduce it to X ->Y and X ->Z.

Hope this makes you clear :)

Answer 1

Ans:(B) NO

Explanation:

Since AB  determine all the attributes of the Relation R(A,B,C,D) , AB is called Key.

As  AB is the minimal Key it can be called as Candidate key (or primarykey)

Intailly the decomposed relations be R1(A,C,D), R2(B,C)

Now from R1(ACD) the Functional Dependencies determined are C - >A, C - >D 

and From R2(BC) no functional dependencies are possible

now to test whether AB ->C is in the  functional dependencies of (R1 U R2)  compute {AB}⁺

(AB)⁺={A,B} 

As (AB)⁺ doesn't contains C, the decomposition is not dependendency preserving.

Hope this makes you clear, Thanks for your question :)

Comments

yes , I read it wrong

---

I see, Anyway thank you !! :)