Applying second and third Normal Form

Question

Let be a relation R(NameFile, Size, Directory, DateCreation,HourCreation,Login,AutAccess,DateAccess,TypeAcces

F={NF,D→S;
   NF, D →DC;
   NF, D→HC;
   L,NF,D→A;
   L,NF,D,DA,HA→TA;
}

Part of the instances are given in the following array:

1. What are the minimal keys of R? Why?

2. In which normal type is R? Why?

I only know that:

• a key is minimal when every attributes that appears in no functional dependency is in every minimal keys
• Every attributes that appears only on the left of the FD is in every keys.
• Every attributes that appears only on the right of the DF isn't in any key.

Therefore we should have K1={NF,R,L,DA,HA} as the only minimal key.

What type of Normal Form is relation R?

I know that

1. something is 1NF if forall attributes of R, there is an atomic value

Therefore, it is at least 1NF,

2. something is 2NF iif:

   • The scheme is at least 1NF
   • Forall An attribute, not in any key, A doesn't depends to part of the key. that is to say not in any functional dependency.

3. A scheme of relation R is 3NF iif

   • The scheme is 2NF
   • Not exists a transitive functional dependency

But I don't know what to do from there... I don't even understand the two last definitions

Answer 1

The only candidate key of your relation is {D DA HA L NF} (perhaps with R you mean D?)

You can verify this by calculating the closure of those attributes, {D DA HA L NF}⁺, and seeing that it contains all the attributes, while, if you remove any one of them, the closure of the remaining set does not contains all the attributes (this is the definition of a candidate key).

The relation is only in first normal form, since the second normal form requires the absence of partial dependencies, that is of dependencies in which non-prime attributes (i.e. attributes not belonging to any key) depends only on part of a key. In this case only the last functional dependency is not partial.