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Given the following two statements: 

S1: Every table with two single-valued attributes is in 1NF, 2NF, 3NF and BCNF. 

S2: $AB \to C$, $D \to E$, $E \to C$ is a minimal cover for the set of functional dependencies $AB \to C$, $D \to E$, $AB \to E$, $E \to C$.

Which one of the following is CORRECT?

  1. S1 is TRUE and S2 is FALSE.
  2. Both S1 and S2 are TRUE.
  3. S1 is FALSE and S2 is TRUE.
  4. Both S1 and S2 are FALSE.
in Databases by Veteran (105k points)
edited by | 3.2k views
0
Here in question they have mentioned that the table contains two single valued attributes. But the table to be in 1NF, the attributes should be atomic and single valued. So, can we conclude that S1 is TRUE ???
0
how about a Relation R(A,B) with FD's like

A->B

B->A

? which normal form the table is in?
0
it is in BCNF form because both A and B are candidate keys ,uniquely defines table.Also there is no partial dependency and transitive dependencey
0
A B
1 2
1 3
2 2

So in above schema neither

A$\rightarrow$ B hold nor 

B $\rightarrow$ A holds

Then how we can say this is in 1NF ?

0

@shivam kasat .....From above PDF

If a relation has only two attributes then it is in BCNF.

Proof: Let a, b be the two attributes in a relation R.

The possible functional dependencies are:

case 1: LHS contains both attributes

a b -> ---

clearly this is a trivial functional dependency because RHS attributes form subset of LHS attributes .

Case 2 : LHS contains only one attribute.

a->---

b->---

clearly in this case, the LHS attribute will be candidate key

how can you decide it is candidate key.......plz explain

 

4 Answers

+31 votes
Best answer

(A) S1 is TRUE and S2 is FALSE.

A relation with 2 attributes is always in BCNF

The two sets of functional dependencies are not the same. can not derive $AB \to E$ from the 1st set

by Active (3.3k points)
edited by
+1
Suppose a two attribute table contains tuples like - {2,3}, {1,5},{2,4}, {2,3}. then this is clearly not in 1NF ?? so S1 should be false
+1
Every RDBMS table atleast is in 1NF.
0
can not derive AD→E from the 1st set

just correct it : AB → E
0
It's AB$\rightarrow E$ not AD $\rightarrow E$
+16 votes

S1 is true bcoz if there are only 2 attributes then relation is always in bcnf.

let R(A B) possible cases are:

(1) {A->B } here A is cand key so BCNF

(2) {B->A}  here B is cand key so BCNF

(3) { A->B  B->A} here A and B both are cand key so BCNF

​​​​​​​(4) no non-trivial FD's here AB is cand key so BCNF

S2 is false bcoz minimal cover should be { D->E , AB->E , E->C }

so ans is A

by Active (4.5k points)
+15 votes
Table with 2 single valued attributes will be in 1NF,

2NF because there can not be a partial key dependencies. Key can be 1 attribute or 2 attributes. If one is key then other can be dependent, if 2 attributes make key then it is trivial.

3NF, there is no question of Transitive dependencies.

BCNF, because if at all a dependency exists, the determinant will be a Key.
by (261 points)
0 votes
s1 is true and s2 is false.In s2 AB->E is not covered.
by (27 points)

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