I am working on another question here and it asks for the candidate keys and the superkeys. I believe I have the correct candidate keys, but if someone could verify that would be great. I am also just having trouble finding superkeys. I don't know if there is a general formula to find them or whatnot.
Here is the question:
Relation Schema: U(A,B,C,D)
FD's:
A-->B
B-->C
C-->D
D-->A
I found the Candidate keys to be: {A},{B},{C},{D}. If someone could verify if this is correct, I would greatly appreciate it. Also if you could let me know how to find the superkeys of this, that would be great. I am not looking directly for the answers, I just want to know if my answer for the candidate keys is correct and how to find the superkeys for this and possibly other questions (if there is a generic way to find them).
Hope someone can help. Will rate best answer once someone helps me figure this out.
Cheers everyone.
Let me explain how to find candidate keys in a simple manner:
Form a three columns,left,right and middle
In left column,add the attributes which appear only on left hand side of FD
In right column,add attributes which appear only on right hand side of FD
In middle column,add attributes which appear both on right and left hand side of FD
Explanation:
Attributes on left column indicates,every possible candidate keys must include these attributes
and
Attributes on right column indicates candidate keys should not include it
and
Attributes on middle may or may not be included in super keys
In the given example,
All the attributes come under middle column
Then,apply Closure property,
A+=ABCD
since
A->A Reflexivity rule
A->B given
A->C Transitivity rule A->B and B->C
A->D Transitivity rule A->B and B->C and C->D
Similarly,we can show other keys also as Candidate keys.
Super keys are nothing but a superset of candidate keys and non-prime attributes
Thus,your answer is right
And super keys of this relation is a super set of candidate keys,therefore
Super keys are AB,AC,ABC and so on.
Hope this helps.
