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2 votes
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Consider a universal relation R={ABCDEFGHIJ} and set of functional dependencies  G = { {A,B} -> {C} , {B,D}->{E,F} ,{A,D} ->{G,H} , {A} -> {I} ,{H} -> {J} } .

a) What is  the key for the above relation ?

b) Decompose it into 2NF relation .

c) Then decompose it into 3NF relations .

4 Answers

3 votes
3 votes
a)))    (ABD) is key\

b))) in 2NF

R1(ABC)  ,  R2(BDEF) , R3(ADGHJ) , R4(AI) , R5(ABD)

 

C))) in 3NF

 R1(ABC)  ,  R2(BDEF) , R3(ADGH) , R4(HJ) , R5(AI) , R6(ABD)
2 votes
2 votes
<a> As A,B,D are not present ate the Right hand side of the given functional dependencies So this three attributes should include in the candidate key.Now find, closure(ABD)=ABDEFGHIJ.We see  that closure(ABD) include all the attributes of the given relation.So ABD is the only key.(No other key possible you can check it.)

<b>As key attributes are A,B,D.So non-key atrributes are C,E,F,G,H,I,J.From the given set of functional dependencies the non-key attributes C,E,F,G,H,I are not fully functionally dependent on ABD.So we split the relation So that it can transform to 2NF.After spliting we get the following relation which are in 2NF---

R1(ABC),R2(BDEF),R3(ADGHJ),R4(AI).

<c>From the 2NF relation only R3 are not in 3NF others are in 3NF.Sowe split only relation R3.After spliting we get---

R3(ADGH)    R5(HJ).

So all the relation are in 3NF {  R1(ABC)  R2(BDEF) R3(ADGH) R4(AI) R5(HJ) }
0 votes
0 votes
I think it should be

a) Candidate key = ABD

b) 2NF = R1(ABD) R2(ABCGHJ) R3(BDEF) R4(AI)

C) 3NF = R1(ABD) R2(ABCGH) R3(BDEF) R4(AI) R5(HJ)
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In Short::

Ck;ABD

find closure of ABD and varify.

2..

A->I,    BD->EF,     AD->GH    AB->C all are PD

After decomposition 

R1(AI)   R2(BDEF)     R3(ADGHIJ)    R4(ABC)    R5(ABD) these are in 2NF.

For 3NF find out transitive dependency and do it yourself.

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