Repeat exercise 15.24 for the following different set of functional dependencies G = { {A, B} -> {C}, {B, D} -> {E, F}, {A, D} -> {G, H}, {A} -> {I}, {H} -> {J} }.

Question

What will be an ideal response?

Answer 1

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To help in solving this problem systematically, we can first find the closures of all
single attributes to see if any is a key on its own as follows:
{A}+ -> {A, I}, {B}+ -> {B}, {C}+ -> {C}, {D}+ -> {D}, {E}+ -> {E}, {F}+ -> {F},
{G}+ -> {G}, {H}+ -> {H, J}, {I}+ -> {I}, {J}+ -> {J}
Since none of the single attributes is a key, we next calculate the closures of pairs of
attributes that are possible keys:
{A, B}+ -> {A, B, C, I}, {B, D}+ -> {B, D, E, F}, {A, D}+ -> {A, D, G, H, I, J}
None of these pairs are keys either since none of the closures includes all attributes. But
the union of the three closures includes all the attributes:
{A, B, D}+ -> {A, B, C, D, E, F, G, H, I}
Hence, {A, B, D} is a key. (Note: Algorithm 15.4a (see chapter 15 in the textbook) can
be used to determine a key).
Based on the above analysis, we decompose as follows, in a similar manner to problem
14.26, starting with the following relation R:
R = {A, B, D, C, E, F, G, H, I}
The first-level partial dependencies on the key (which violate 2NF) are:
{A, B} -> {C, I}, {B, D} -> {E, F}, {A, D}+ -> {G, H, I, J}
Hence, R is decomposed into R1, R2, R3, R4 (keys are underlined):
R1 = {A, B, C, I}, R2 = {B, D, E, F}, R3 = {A, D, G, H, I, J}, R4 = {A, B, D}
Additional partial dependencies exist in R1 and R3 because {A} -> {I}. Hence, we remove
{I} into R5, so the following relations are the result of 2NF decomposition:
R1 = {A, B, C}, R2 = {B, D, E, F}, R3 = {A, D, G, H, J}, R4 = {A, B, D}, R5 = {A, I}
Next, we check for transitive dependencies in each of the relations (which violate 3NF).
Only R3 has a transitive dependency {A, D} -> {H} -> {J}, so it is decomposed into R31
and R32 as follows:
R31 = {H, J}, R32 = {A, D, G, H}
The final set of 3NF relations is {R1, R2, R31, R32, R4, R5}
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