Minimum closure of F (functional dependencies) help

[KataKoniK]

Homework Statement

F = {S -> NOPQZVWX, NX -> UV, NXT -> RY, YQ -> NXR, QZ -> S, PQ -> S}

Find the minimum closure of F.

Homework Equations

none

The Attempt at a Solution

So far I have made RHS of each FD into a single attribute and then I eliminated redundant attributes from LHS.

This is what I got. Is it correct so far? Do I still have to delete more stuff? I don't know what to do with QZ -> S and PQ -> S in terms of if they should be removed or not.

{NX -> U, NX -> V, S -> N, S -> O, S -> P, S -> Q, S -> Z, S -> V, S -> W, S -> X, NXT -> Y, QZ -> S, PQ -> S, YQ -> N, YQ -> X}

 

[KataKoniK]

Your approach is a good start, but there are a few more steps you can take to minimize the closure of F. Here are some suggestions:

1. Eliminate redundant attributes from the right-hand side of the functional dependencies. For example, in the FD NX -> UV, the attribute V is redundant since it is already implied by U. So the FD can be simplified to NX -> U.

2. Combine FDs with the same left-hand side. In this case, the FDs S -> NOPQZVWX and S -> V can be combined into one FD, S -> NOPQZVWXV.

3. Remove transitive dependencies. For example, in the FD NXT -> RY, the attribute Y is transitively dependent on NXT since it can be determined from the FD YQ -> NXR. So the FD can be simplified to NXT -> R.

After applying these steps, the minimum closure of F would be:

{S -> NOPQZVWX, NX -> U, NXT -> R, YQ -> N, YQ -> X}

Note that the FDs QZ -> S and PQ -> S are not redundant and should not be removed, as they cannot be implied by any other FD in the closure.

 

[piercas]

I would suggest using the Armstrong's axioms and closure algorithm to find the minimum closure of F. This method systematically eliminates redundant functional dependencies and ensures that the resulting set is a closure of F. In this case, the minimum closure of F would be:

{S -> NOPQZVWX, NX -> UV, NXT -> RY, YQ -> NXR}.

This is because all the other dependencies can be derived from these four using the Armstrong's axioms. Additionally, QZ -> S and PQ -> S cannot be removed as they are not redundant and are necessary for the closure of F.