Decomposition

Question

Consider following set of FDs on $R(A,B,C,D,E,F)$

$A \to BCD$

$BC \to DE$

$B \to D$

$D \to A$

1. Compute the canonical cover.
2. Give 3NF decomposition of $R$ based on canonical cover.
3. Give BCNF decomposition of $R$ based on original set of FD.
4. Can you get same decomposition of $R$ as above using canonical cover?

Answer 1

1.A−>BC
 B−>DE
 D−>A
CK{AF , BF, DF}

2.R1(ABCDE)R2(AF)

3..R1(ABCDE)R2(AF)

4..R1(ABCDE)R2(AF)
Verify once!
 

 

Comments

how ?
In the table ABCDE {A,B,D} are candidate key .

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for 2 part B->DE  violates 3NF property

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@ Umang A -> C violates 2NF (proper subset of candidate key (AF) gives non prime)

Answer 2

We always decompose after getting canonical cover.

Canonical Cover : $$\begin{align*} A &\rightarrow B \\ A &\rightarrow C \\ B &\rightarrow E \\ B &\rightarrow D \\ D &\rightarrow A \\ \end{align*}$$

after decomposition into  3NF:

because of attribute $F$ no LHS of any FD available in table like $R_1$ which contains $F$ could be a superKey, so it cannot be in BCNF, hence we can say that BCNF decomposition is not possible but 3NF decomposition is.

Comments

how to find canonical for this

Answer 3

1. Canonical Cover : { $A\rightarrow BC$, $D\rightarrow A$, $B\rightarrow DE$ }
2. 3NF : {A,B,C} {B,E,D} {D,A} {F,A}
3. BCNF : {A,B,C} {B,D} {A,E,F}
4. If you are generating the relations usinf some algorithm then no you can't gernerate same sets using different set of functional dependencies.

Comments

i think these are the correct 3nf and bcnf decomposition. i don't know how amar vashisht got 3nf decomposition.