I'm struggling with finding a minimal basis for the given relation and FD's. Can you please confirm I'm doing it right?
I've got R(A, B, C, D, E, F, G)
A -> G
E -> F,G
C,D -> B
B -> E
C,D -> A
After checking each FD there is nothing to remove, so my minimal basis is
A -> G
E -> F
E -> G
C,D -> B
B -> E
C,D -> A
and the key is (C,D) as (C,D)+ = (C,D,A,B,E,F,G)
Many thanks
In American English, I think the term is minimal cover.
CD is the only candidate key in R.
I'm not sure what you mean by "checking each FD", but your result is right.
Related
I can't solve it.please help
Q In a relation schema R = (A, B, C, D, E), the following functional dependencies are held:
A → BC
CD → E
B → D
E → A
Which of the following is a combination of two candidate keys for R?
Answer Choice:
a) A and E
b) B and C
c) B and D
d) C and D
The Correct Answer is a) A and E
But if E depend on A ( E → A), then why option a) A and E is correct? please help
We can compute the closure of the functional dependencies for each set of attributes on the left-hand side of an FD:
A -> A (Trivial)
A -> ABC (since A -> BC)
A -> ABCD (since B -> D)
A -> ABCDE (since CD -> E)
So we can see that A determines all attributes in the relation. A is a candidate key. For CD, we get:
CD -> CD (trivial)
CD -> CDE (since CD -> E)
CD -> CDEA (since E -> A)
CD -> CDEAB (since A -> BC)
So CD is a candidate key. For B, the most we can derive is:
B -> B (trivial)
B -> BD (since B -> D)
For E:
E -> E (trivial)
E -> EA (since E -> A)
E -> EABC (since A -> BC)
E -> EABCD (since B -> D)
Again, we can determine all attributes in R so E is a candidate key.
So we've got candidate keys A, E and CD. Now, read the question carefully. Which of the possible answers are a combination of two candidate keys? Option d has two attributes of a single candidate key. Options b and c contain B which isn't a candidate key. The only correct answer is a) A and E, which lists two separate candidate keys.
I have 2 sets
R = {A B C D}
H = {AB-> C , AB-> D, D-> B}
Want to find all minimal keys in R set
My answer for minimal keys is : { A D }
this is because
AB -> C and AB -> D then AB -> CD
since D -> B then AD is the minimal keys
when i check my answer with this site. the site giving wrong answer.
can explain?
The site says, "Set of found candidate-keys: {{A, C, F}, {B, C, F}}." That's clearly wrong; F isn't even in R.
In any case, your answer is incomplete. AD is one of two candidate keys.
Consider the following relation and the set of functional dependencies (FDs).
R = (A B C D E F G H I)
D -> H
DH -> A
EFH -> C
AF -> IG
CD -> ABG
IB -> C
G -> I
IBC->E
F -> H
C -> IE
Find the candidate key(s) of R.
Find the minimal cover of FD set.
I think "DF" is one of candicate keys.
The ONLY candidate key is DF.
And the Minimal cover is as follows:
D->AH;
EF->C;
AF->G;
CD->ABG;
IB->C;
G->I;
F->H;
C->IE
Given the following functional dependencies
F = { A -> BC, CD -> E, B -> D, E -> A }
I cannot wrap my head that F |= E -> BC is false. If you have E, you can go to A, if you have A, you can go to BC. What is incorrect with this logic?
There is nothing wrong with your logic. F |= E->BC.
I have this problem:
R = (A, B, C, D, E)
and
F = (A -> BC, B -> CD, E -> AD)
the book says that this in BCNF because all FDs are trivial.
My question is: how this FDs is trivial?
by what you've posted :
R = ABCDE
A -> BC
B -> CD
E -> AD
the candidate key is E
All the left hand side FDs do not show E so this is not a BCNF.
and hence the non trivial FD.