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Let R{A,B,C,D,E} be a relation with functional dependencies:

B->EA
EBC->D
BED->A

Question: What's the higher normal form that R satisfies?

Attempt:

  • 1NF: satisfied, since all the attributes are defined as single-valued.
  • 2NF: I'm not exactly sure. 2NF states that there must not be partial dependecies. The candidate key is BC, so is B->EA considered a partial dependency or does it have to be one attribute on the rhs?
  • 3NF: Assuming 2NF is satisfied, then for the dependency: B->EA, B is not a superkey and E,A are not prime attributes. Thus, 3NF is not satisfied.

Why (or why not) the 2NF is satisfied?

Andrew
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  • Your "I have these FDs" doesn't make sense. "These are all the FDs that hold"?--Not possible. "These are all the non-trivial FDs that hold"?--Not possible. "These are some FDs that hold"?--Question can't be answered. Find out what a *cover* is & what the exact conditions are to apply a particular definition/rule/algorithm. To determine CKs & NFs we must be given FDs that form a cover. Sometimes a minimal/irreducible cover. And the set of all attributes must be given. [See this answer.](https://stackoverflow.com/a/53386492/3404097) – philipxy May 20 '21 at 15:45
  • "2NF states that there must not be partial dependencies. " That is not correct. W;hy do you think that? What is your (authoritative) reference? Where are you stuck applying definitions of FD, partial FD & 2NF? – philipxy May 20 '21 at 15:49

1 Answers1

2
B -> EA

is equivalent to the two dependencies:

B -> E
B -> A

for the decomposition rule. Since neither E nor A are prime attribute and B is a proper subset of a candidate key, the relation is not in 2NF.

Renzo
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