ISI2011-PCB-CS-5b

Question

Suppose we have a relation $R(A, B, C, D, E)$ with the functional dependencies:
$A \rightarrow D, B \rightarrow C, D \rightarrow E, CE \rightarrow B$.

If we project $R$ and therefore its functional dependencies onto the schema $ABC$, what will the key(s) for $ABC$ be?

Comments

First, note that $AB^{+}$ → $ABCDE$ & $AC^{+}$ → $ABCDE$. Thus, when we project the FD’s onto ABC, AB & AC is certainly a key. However, there cannot be any other keys, because $B^{+}$→$BC$ & $C^{+}$→$C$.

Thus, AB & AC are the only key.

Answer 1

$A+   = ADE$      $B+ =BC     C+ =C $

$AB+ = ABCDE$         First Key : $AB+ =ABC$

$AC+ = ABCDE$         Second Key :  $AC+ =ACB$

$BC+ = BC$

So, on schema  $ABC$ we Got $2$ keys. $ABC$

Comments

How to interpret this question suppose if I project R(a b c) then only fd obtained is B-> C ?? Can someone explain me in more detail?

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here if we project R(ABC) first, then from the FD set only B->C is applicable.

i.e in that relation R(ABC), only AB should be the key.

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We can derive FD's from the original set using Inference Rules and Closure ...

Answer 2

A is not in RHS of any FD.
So A will be there for must.

Only focus on A, B, C.

Now (A)+ = ADE which does not contain B, C.

So check  (AC)+ and (AB)+.
Closure of both AB and AC contains ABC.  

Answer 3

They are asking FD's for sub relation R(ABC)

A+   = ADE      B+ =BC     C+ =C  

AB+ = ABC         First Key

AC+ = ABC      Second Key

BC+ = BC

Hence we got 2 Keys

Answer 4

Keys for (ABC) are

(AB)+ ={ABCDE}

(ACE)+ ={ABCDE}

These two keys for getting schema ABC .No other key gives schema (ABC)

So, keys are (AB) , (ACE)

Comments

yaa ACE is not CK

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@Tauhin check out my answer

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oh yes @Tuhin