Given these functional dependencies for
R: {A,B,C,D,E,F}
AC->EF
E->CD
C->ADEF
BDF->ACD
I got this as the canonical cover:
E->C
C->ADEF
BF->C
And then broke it down to Boyce Codd Normal Form:
Relation 1: {C,A,D,E,F}
Relation 2: {B,F,C}
I figured that this is lossless and dependency preserving? But is this true, since from the original functional dependencies BDF->ACD is no longer in any of my relations. But if I go from my calculated canonical cover then all my functional dependencies are preserved.
So that question is: Is this decomposition to BCNF dependency preserving?
A decomposition preserves the dependencies if and only if the union of the projection of the dependencies on the decomposed relations is a cover of the dependencies of the relation.
So, to know if a decomposition preserves or not the dependencies it is not sufficient to check if the dependencies of a particular cover have been preserved or not (for instance by looking if some decomposed relation has all the attributes of the dependency). For instance, in a relation R(ABC) with a cover F = {A→B, B→C, C→A} one could think that in the decomposition R1(AB) and R2(BC) the dependency C→A is not preserved. But if you project F on AB you obtain A→B, B→A, projecting it on BC you obtain B→C, C→B, so from their union you can derive also C→A.
The check is not simple, even if there exists polynomial algorithms that perform this task (for instance, one is described in J. Ullman, Principles of Database Systems, Computer Science Press, 1983).
Assuming the dependencies that you have given form a cover of the dependencies of the relation, the canonical cover that you have found is incorrect. In fact BF -> C cannot be derived from the original dependencies.
For this reason, your decomposition is not correct, since R2(BCF) is not in BCNF (actually, it is not in 2NF).
One possible canonical cover of R is the following:
BDF → C
C → A
C → E
C → F
E → C
E → D
Following the analysis algorithm, there are two possible decompositions in BCNF (according to the dependencies chosen for elimination). One is:
R1 = (ACDEF)
R2 = (BC)
while the other is:
R1 = (ACDEF)
R3 = (BE)
(note that BC and BE are candidate keys of the original relation, together with BDF).
A cover of the dependencies in R1 is:
C → A
C → E
C → F
E → C
E → D
while both in R2 and R3 no non-trivial dependencies hold.
From this, we can conclude that both decompositions do not preserve the dependencies; for instance the following dependency (and all those derived from it) cannot be obtained:
BDF → C
Related
In R(A,B,C, D),
let the dependencies be
A->B
B->C
C->D
D-> B
the decomposition of R into (A,B), (B,C) and (B,D) is lossless or dependency preserving?
My attempt : (A,B) and (B,C) can be combined lossless-ly because of B->C. However, for (A,B,C) and (B,D), B does not form a key for either. Hence the decomposition in lossy.
Also for dependency preserving, the relation (C-D) can't be gotten from any of the decomposed relations and hence the decomposition is not dependency preserving.
However the answer given is that the decomposition is both lossless and dependency preserving. So where am I wrong?
Also the key for relation R is only {A} isn't it?
You say:
However, for (A,B,C) and (B,D), B does not form a key for either. Hence the decomposition in lossy.
This is wrong, since B is a key for (B, D). We can see this by computing B+ from the original dependencies, assuming that they form a cover of the relation.
B+ = B
B+ = BC (for the dependency B->C)
B+ = BCD (for the dependency C->D)
so, since D is included in B+, we have that B->D can be derived from the original set of dependencies, and in the decomposition (B, D) B is a key (as it is D).
To be dependency preserving we must check that the union of all the projected dependencies of the decomposition is a cover of the original set of dependencies. Since three covers of the decomposed relations are, respectively, {A->B}, {C->B, B->C} and {D->B, B->D}, by uniting those three sets you can easily derive also D->C as well as C->D, so the dependencies are preserved.
Finally, yes {A} is the unique candidate key of the original relation.
I have the following relation and I need to normalize it to 4NF.
Relation
First I've tried to find all the FD's and MVD's that hold.
AB ->> C (MVD)
C -> D (FD)
D -> E (FD)
ABC -> F (FD)
Next, using these dependencies I've managed to find the candidate key: ABC.
Let me know if what I've done so far is right. Also, is it ok to have a multivalued dependency in 4NF? Like AB ->> C and ABC -> F?
Thanks.
In general dependencies describe important constraints on the data, for instance a functional dependency X → A means that a certain value of X determines uniquely a certain value of A (that is, each time we find in a tuple a certain value of X, we always find the same value of A). Such kinds of constraints cannot be inferred by (few) rows of a table, in which is unknown the meaning of the data.
At the best, we can infer a set of possible functional dependencies holding in that particular instance of the table, hoping (but without any particular reason) that those functional dependencies will hold on every instance of the table, which is the only condition for which we can “normalize” the relation (and not simply find a non-redundant way of storing a particular instance of that table).
In your case, for instance, since the table has very few rows, many functional dependencies could be seen as holding in it, for instance at least the following:
F → AB
E → AD
D → AE
C → ADE
B → A
EF → ABCD
DF → ABCE
CF → ABDE
CB → ADEF
(while ABC → F can be derived from CB → ADEF, and AB →→ C does not hold).
And if we should apply a normalization algorithm to that instance (for instance the synthesis algorithm for 3NF), we will decompose the relation in an exaggerate number of subschemas:
R1(AB), R2(BCF), R3(CD), R4(ADE), R5(CEF),
five relations for a table with six attributes!
Given the set of dependencies AB->C, BD->EF, AD->GH, A->I, H->J.
How would you find a minimal cover? By applying a process described in a book I get: AB->C, A->I, BD->EF, AD->GH, H->J instead of AB->CI, BD->EF, AD->GHIJ. Is it possible to combine AB->C and A->I into AB->CI and get rid of A->I?
The functional dependencies of a minimal cover of a set of dependencies F must satisfy four conditions:
They must be a cover of F (of course!)
Each right part has only one attribute
Each left part must not have extraneous attributes (that is attributes such that original dependency can be derived even if we remove them)
No dependency of the cover is redundant (i.e. can be derived from the remaining dependencies).
So this means that a minimal cover of the example is (note that there can be more then one minimal cover, that is set satisfying the above conditions):
{ AB → C
AD → G
AD → H
A → I
BD → E
BD → F
H → J }
Of course to this set you can apply the Armstrong’s axioms to derive many other dependencies (for instance AD → GH, AB → CI, AD → GHIJ, ABD → EJ, etc.) but these are not part of any minimal cover of F (that is, they do not satisfy the above definition).
I am working on a textbook question and it asks the following:
Let R(A,B,C,D,E) be decomposed into relations with the following three sets of attributes:
{A,B,C} , {B,C,D}, {A,C,E}
For each of the following sets of functional dependencies, determine if the dependencies are preserved by the decomposition.
AC -> E and BC -> D
How do I solve this?
The textbook doesn't provide a clear enough explanation on dependency preserving.
R = {A,B,C,D,E} decomposed into R1 ={A,B,C} , R2 ={B,C,D} and R3 ={A,C,E}.
"determine if the dependencies are preserved by the decomposition."
Yes they are as BC->D is preserved in R2 and AC->E is preserved in R3 as is very apparent!Note - Although a decomposition may be dependency-preserving it is not necessary that it is in a higher normal form.
There is an easy method to check whether a decomposition is dependency-preserving. Check this video.
I am reading this topic Functional dependency and Normalization in Database Management Subject. I came across this example.
Relation R(A,B,C,D) Which one is Lossy join but Dependency Preserving BCNF Decomposition?
a. A ->B, B -> CD
b. A -> B, B -> C, C->D
c. AB -> C, C -> AD
d. A -> BCD
Now answer given is option C.
How can option C. be a lossy decomposition. if you do ABC union CAD = ABCD This satisfies first condition.
if we do ABC intersection CAD = AC which is perfectly fine, since in AC, C is key for (CAD) C -> AD decomposition. which also satisfies the second condition. Am i making any mistake in understanding this concept.
Usually for a Normalisation/decomposition exercise, you are given:
The full relation and its attributes. [yes: R(A, B, C, D)]
The Functional dependencies. [yes? it looks like a., b., c., d. are possible sets of Fun Deps.]
The proposed decomposition. [Often named R1, R2, etc. I don't see those. I can't interpret option d. to be proposing a decomposition.]
Perhaps your post has missed out part of the exercise? Perhaps the exercise wants you to decide which decomp preserves the dependencies in BCNF? (But results in a lossy join.)
[editted in response to Nikhil's comment] Note that the list of FD's alone doesn't amount to a decomposition: the FD C -> AD is short-hand for C -> A, C -> D. Does that mean two decomposing relations? No, because A and C are already in the FD AB -> C. So we have R1= (A, B, C), R2 = (C, D). But I don't know if that is what the exercise is asking. Think about it. What does option d. mean in terms of decompositions?
Perhaps the exercise is asking (for example): given a proposed decomposition into R1 = (A, B) and R2 = (B, C, D), which of the sets of FD's would give a lossy decomposition?
There's a worked example here: http://en.wikipedia.org/wiki/Lossless-Join_Decomposition.
It points to a previous q Lossless Join Property.
And there's further references.
By the way, options a., b., include the same Fun Deps as option d., by the transitivity of dependencies (Armstrong's Axioms http://en.wikipedia.org/wiki/Armstrong%27s_axioms see also http://en.wikipedia.org/wiki/Heath%27s_theorem). This is a clue.