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Relation $R$ has eight attributes $\text{ABCDEFGH}$. Fields of $R$ contain only atomic values. $F = \text{{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

  1. in $\text{1NF}$, but not in $\text{2NF}$.
  2. in $\text{2NF}$, but not in $\text{3NF}$.
  3. in $\text{3NF}$, but not in $\text{BCNF}$.
  4. in $\text{BCNF}$.
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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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5 Answers

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56 votes
Best answer
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: $A$
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4 Comments

Answer is correct but the explanation seems weird. Maybe re-word it a little? Infact just remove the line: "For example partial dependency A→C  exists in A → BC and B→C and B→H in B → CFH. etc."  

 

Partial dependency is from LHS side. Currently it seems to imply it is from RHS.
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@Lakshman Patel RJIT Sir, This is misconception that A→ C is partial dependency, AD is key & AD→ C is partial dependency due to A→C. So please edit it. Reference: Navathe

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@Deepak Poonia Sir , @Lakshman Bhaiya bro , Please make the necessary changes in the answer as pointed out by @Pratik2404 and @mkagenius . 

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@Deepak Poonia

Explanation is not proper here

for 2NF i am searching for violation condition but not getting:
 

some ck -----→ some non prime attribute   ?

 

 

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13 votes
13 votes

A.

as in F-->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 2NF ).so violates 2NF condition,
similarly , B-->CH and  A-->C also violates 2NF condition,
hence R is not in 2NF but in 1NF. ( since attributes of relation R has only atomic values - it is property of 1NF )

0 votes
0 votes

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.

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–1 vote
–1 vote
telation is 1nf but not in 2nf
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