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Consider the universal relation R = {A, B, C, D, E, F, G, H, I, J}. What is the key for R? Decompose R into 2NF and then 3NF relations?

philipxy
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Ahsan Shabbir
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  • Right now you are just asking for us to rewrite your textbook with a bespoke tutorial & do your (home)work & you have shown no research or other effort. Please see [ask], hits googling 'stackexchange homework' & the voting arrow mouseover texts. Show the steps of your work following your textbook with justification & ask 1 specific question re the first place you are stuck. – philipxy Dec 30 '19 at 00:28
  • Possible duplicate of [Consider the universal relation R = {A, B, C, D, E, F, G, H, I, J}. What is the key for R? Decompose R into 2NF and then 3NF relations](https://stackoverflow.com/q/25653579/3404097) Before considering posting please always google any error message or many clear, concise & precise phrasings of your question/problem/goal, with & without your particular strings/names & site:stackoverflow.com & tags, & read many answers. If you post a question, use one phrasing as title. See [ask] & the voting arrow mouseover texts. – philipxy Dec 30 '19 at 04:55

1 Answers1

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Attributes appearing on the left of FDs are {A, B, D, H}. From these all but {H} seem plausibly part of keys. Calculating the closures of likely candidates gives:

{A, B}
+ = {A, B, C, I}
{B, D}
+ = {B, D, E, F}
{A, D}
+ = {A, D, G, H, I, J}
{A, B, D}
+ = {A, B, C, D, E, F, G, H, I, J}

So {A, B, D} is the only candidate key

Decomposing attributes based on relations partially dependent on the key gives:

R1 = {A, B, C}
R2 = {B, D, E, F}
R3 = {A, D, G, H, J}
R4 = {A, I}
R5 = {A, B, D}

Relation R5 is kept to preserve the original primary key

Further decomposing attributes base on transitive dependencies keeps R1, R2, R4, and R5 from above but splits R3 into:

R3a = {A, D, G, H}
R3b = {H, J}
Khaled Alajmi
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  • The question does not give a FD cover. It would be interesting to hear how you came up with this answer. – philipxy Dec 30 '19 at 04:46