GATE CSE 2013 | Question: 55

Question

Relation $R$ has eight attributes $\text{ABCDEFGH}$. Fields of $R$ contain only atomic values. $F = \{ CH$ $\rightarrow$ $G, A \rightarrow BC, B \rightarrow CFH, E \rightarrow A, F \rightarrow EG \}$ is a set of functional dependencies $(FDs)$ so that $F^+$ is exactly the set of $FDs$ that hold for $R$.

The relation $R$ is

• A. in $\text{1NF}$, but not in $\text{2NF}$.
• B. in $\text{2NF}$, but not in $\text{3NF}$.
• C. in $\text{3NF}$, but not in $\text{BCNF}$.
• D. in $\text{BCNF}$.

Comments

Of all the 4 candidate keys suppose one key only violates 2nf, is the entire relation still not a 2nf i.e. other 3 candidate key follow 2nf and 1 candidate key violates.

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Proper subsets of Candidate keys are : A,B,D,E,F

And non-prime attributes are : C,G,H

Now, can these attributes (any) be able to determine a non-prime attribute?

Yes A->BC, A->C therefore not in 2NF

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the give relation is in 1NF

option A is correct

Answer 1

Here, candidate keys are $AD, BD, ED$ and $FD$.

Partial dependency exists  $A$ $\rightarrow$ $BC$, $B$ $\rightarrow$ $CFH$ and $F$ $\rightarrow$ $EG$ etc. In the following $FDs.$

For example partial dependency $\color{blue}{A \rightarrow C}$  exists in $A$ $\rightarrow$ $BC$ and $\color{blue}{B  \rightarrow C}$ and $\color{blue}{B\rightarrow H}$ in $B$ $\rightarrow$ $CFH$. etc.

So, given relation is in $\text{1NF}$ ,but not in $\text{2NF}$.

Correct Answer: $A$

Comments

@MSG1999 i hope u watched go classes lectures , in those sir told that some ck partially determine some non prime attribute then its voilation of 2 NF

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@MSG1999 implication needs to be modify, proper subset of C.K ----> Non prime attribute

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for CKey  = BD ;  B+ = {BDCFHGA}   of which C, G,H  etc are non-prime attributes so violates for 2NF ,  similarity other ck can be taken as example

Answer 2

Correct Answer: $A$

as in $F \rightarrow G$, $G$ is non-prime attribute but $F$ is a proper subset of a candidate key ( this problem is known as partial dependency , which is not allowed in 2 NF ).so violates 2 NF condition,
similarly , $B\rightarrow CH$ and  $A\rightarrow C$ also violates 2 NF condition,
hence $R$ is not in 2NF  but in 1NF. ( since attributes of relation $R$ has only atomic values - it is property of 1NF )

Answer 3

The given relation has 4 candidate keys $AD, BD, ED, FD$

Prime attributes : $A, B, D, E, F$

Non-Prime attributes : $C, G, H$

 

Lets check for $BCNF$ first

The very first FD i.e. $CH \to G$ is a violation to BCNF. 

$R$ is not in $BCNF$

 

Lets check for $3NF$

Again the very first FD $CH \to G$ is a violation to $3NF$, since $CH$ is not a super key and it determines a non prime attribute $G$

Transitive functional dependency : If $\alpha \to \beta$ and $\beta \to \gamma$ and  $\beta \not\to \alpha$ and $\gamma \not\in \beta$ and $\gamma \not\in \alpha$, then we can say $\alpha \to \gamma$ transitively.

$AD \to CH$ and $CH \to G$, so one of the candidate keys $AD$ transitively determines a non-prime attribute $G$. 

 

Lets check for $2NF$

Since the candidate key $BD$ determines all of the attributes of $R$ i.e. ${(BD)}^+ = R$, lets consider the third FD which is $B \to CFH$, let's take by RHS splitting $B \to CH$.

$BD \to CH$ is a violating partial dependency since $B$ fully determines $CH$. Hence 2NF is violated. 

 

OMG, it's not even in 2NF. 

So the given relation $R$ is in 1NF only.

Correct answer : $A$

 

Comments

Underrated Answer

Answer 4

according to definition given in Korth 

 A Functional Dependency $A → B$  Is Called A Partial Dependency if there is
a proper subset $C$ of $A$ such that $C → B$.We say that $B$ is partially dependent
on $A.$

Here, candidate keys are $AD,BD,ED$ and $FD.$

So here Partial dependency are  $DA→BC, DB→CFH$

As we Can see Proper subset $A$ OF $DA$   And  Proper subset $B$ of $DB$ is there such that $A→ BC, B→CFH $

so we can say $BC$ is partially dependent on CK $DA$ and $CFH$  is partially dependent on CK $DB$

So, given relation is in 1NF,but not in 2NF.

Answer 5

COMMENT BELOW IF YOU HAVE ANY DOUBT

Answer 6

Correct Answer is option A.