GATE IT 2005 | Question: 22

Question

A table has fields $F_1, F_2, F_3, F_4, F_5$ with the following functional dependencies  

• $F_1 \to F_3,  F_2\to F_4,   (F_1 . F_2) \to F_5$

In terms of Normalization, this table is in

• A. $1\;\text{NF}$
• B. $2\;\text{NF}$
• C. $3\;\text{NF}$
• D. None of these

Comments

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Detailed Video Solution

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Answer 1

Answer is A 1NF

Key is $\{F_1,F_2\}$

$F_1\rightarrow F_3, F_2 \rightarrow F_4$ are partial dependencies (a proper subset of candidate key determining a non-key attribute) thus violating $2\;\text{NF}$ requirement.

Comments

Can't it be 3NF as there are no non prime attributes depending on non prime attributes?

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Do we assume that all the fields are atomic ( single valued) for it to be 1NF . The question doesnt state nature of the fields. The ans can also be none of these!

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@Harsh Chaudhari
By the definition of relation, all tables are in 1NF

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@craft, If it is not in 2NF, then it will not be in higher normal form(3NF, BCNF….)

Also For 3NF:  In FD, X → Y: Either X has to be super key, or Y has to be primary attribute.

• F1→ F3, and F2→ F4 is violating this.

Answer 2

Answer is 1 NF.

(F1F2)⁺ = F1F2F3F4F5

F1 -> F3 : Partial Dependency : it is in 1NF

Answer 3

Detailed Video Solution: https://youtu.be/1UnTwHvQ3y8 

For the given relation $R$, the only candidate key is $F_1F_2.$ So, the non-prime attributes are $F_3,F_4,F_5.$

Now, we can check whether $R$ is in 2NF or Not, in one of the two way, given below:

Method 1:

The given relation is Not in 2NF because there exists a non-trivial functional dependency where a proper subset of some candidate key determines some non-prime attribute, & that is: $F_1 \rightarrow F_3.$

Method 2:

The given relation is Not in 2NF because some non-prime attribute (i.e. $F_3$) is partially dependent on some candidate key (i.e. $F_1F_2$).

NOTE: For the given relation, $F_1 \rightarrow F_3$ is NOT partial dependency. It is actually a Full functional dependency. The partial dependency is $F_1.F_2 \rightarrow F_3$ because we can remove $F_2$ from LHS & still derive $F_3.$ 

$\color{red}{\text{Misconception:}}$ A very common misconception is that $F_1 \rightarrow F_3$ is a partial dependency. This is wrong. Refer this lecture & the definitions below.

Partial Dependency Definition (Source: Silberschatz, Korth):

Partial Dependency Definition (Source: Elmasri, Navathe)

Partial Dependency Definition (Source: Codd’s Original Research Paper)

https://forum.thethirdmanifesto.com/wp-content/uploads/asgarosforum/987737/00-efc-further-normalization.pdf 

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Detailed Video Solution of this question: https://youtu.be/1UnTwHvQ3y8 

Answer 4

1NF is the answer.