Is this table in 1NF or 2NF?

Question

Supposed i have a relational table:

FinalYearProject(supervisor, researchTopic, consultationDay, student)

and the following functional dependencies:

    student -> researchTopic
    student, researchTopic -> supervisor
    researchTopic,supervisor -> consultationDay

From there i determined my minimal superkey is: student where

student -> researchTopic,supervisor, consultationDay

Is it right for me to say that there is partial dependency as supervisor do not depend solely on student based on the functional dependency:

student, researchTopic -> supervisor

Any help will be greatly appreciated.

What does "I have these FDs" mean? "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 given. [See this answer.](https://stackoverflow.com/a/53386492/3404097) — philipxy, Oct 25 '19 at 18:04
"not depend solely on student" is not clear, and neither is your reasoning using it. X -> A is partial if A depends on a proper subset of X. So {student, researchTopic} -> supervisor is partial if supervisor is functionally dependent on any of {student}, {researchTopic} or {}. Also you ask one question in the title but another one in the body. Also 2NF is not based just on partial FDs, but partial FDs of non-prime attributes on CKs. Also a schema can have more than one CK. So maybe you can see that you should give all the steps & justification of your reasoning when you ask "is this right". — philipxy, Oct 25 '19 at 18:12

Renzo · Accepted Answer · 2019-10-25T09:43:52.473

Since student is the candidate key, supervisor depends on it.

In fact, consider that, given student -> researchTopic, the research topic depends on the student: so in the dependency student, researchTopic -> supervisor the attribute researchTopic is superfluous (the depedency student -> supervisor holds). It is easy to show this by using the Armstrong’s Axioms.

And since a relation is not in 2NF when a non-prime attribute is functionally dependent on a proper subset of a candidate key, there is no such case in this example, and the relation is in 2NF.

Is this table in 1NF or 2NF?

1 Answers1