Normal Form

Question

Consider the following relations R1 and R2 given below

R1(ABCD) And functional dependency are {AB→C  ,C→A ,AC→D} and

R2(ABCD) functional dependency are {AB→C ,BC→D}.

(a)Both R1 and R2 in 3NF

(b)R1 is not in 2NF and R2 in 2NF

(c)R1 in 2NF and R2 in 3NF

(d)None of the above

Comments

I think its option d none of the above

Answer 1

R1 : AB+ = {ABCD}  --------  CK   ,     BC IS CK

       AC+={ACD} ------------ NOT CK

FOR 2NF X->Y either subset of X is not  CK or Y is prime attribute

AB→C  ,C→A ,AC→D

 since AB is CK not subset of CK  ,C,AC is not subset of CK but C+={ACD} 

here C can alone derive A which implies AC->D as PD

so R1 is not 2NF 

FOR 3NF X->Y  either X is superkey or Y is prime attribute

 BUT AC->D here AC is not superkey and D is not prime attribute

Therefore R1 is not 3NF

R2: AB+ ={ABCD}  ------- CK

      BC+={BCD}  ---------- NOT CK

AB->C ,BC->D here AB is CK  A+={A},B+={B} AB+={ABCD} ,BC is not CK B+={B} ,C+={C},BC+={BCD} 

No PD's in R2

Therefore R2 is in 2NF

BC->D here BC is not superkey and D is not prime attribute

R2 is not 3NF

So R1 is not 2NF ,R2  is 2NF 

Therefore Option B 

Let me know if anything wrong

Comments

option B is correct .Tell me how ?

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yeap may be i checked only 3NF

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I have edited please check it let me know if anything wrong

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Sorry for options ?

when we say exactly a relation is not in 2NF ?

According to you AC→D is not subset of any key .But here C→A. can we write C→D.So it become PD.

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Partial dependency

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yes yes correct C->D is PD

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In second relation AB is key

BC->D//what is problem with this dependency ? why its not in 3NF ?

Is BC should be always SK and D should be prime attribute ?

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The condition for 3Nf is if X->Y is 3NF means that either X has to be SK or CK  or Y has to be Prime Attribute

Answer 2

Answer  (B)