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.
Related
I have a relation R(a,b,c,d) where (a,b) is a primary key so I have the determinant a,b -> c,d.
In addition to that I have the following determinants: a,c -> b,d and a,d -> b.
Let's be ok that this relation is in 3NF.
I'm wondering if it's in BCNF or not. I was using a definition for the BCNF that recognize:
a relation is in BCNF if it's in 3NF and there's no determinant X -> Y such as X is non-key attributes and Y is a part (or the totality) of the key
which is not applicable in my case for the determinant a,d -> b for example. Another definition is that
A relation, R, is in BCNF iff for every nontrivial FD (X->A) satisfied
by R the following condition is true:
(a) X is a superkey for R
which left me undecided insofar as in a,d->b it's clear that (a,d) is not a superkey (nor a key), but we have (a,d) clearly a key to the relation R!
So, my question is:
Is the relation R in BCNF or not, and why?
And what's the right process to determine whether a relation is or isn't in BCNF?
About terminology
You say:
I have the determinant a,b -> c,d
This is wrong therminology. a,b -> c,d is a functional dependency (sometimes abbreviated with FD), which has a determinant a,b (sometimes called left hand side (LHS) of the FD) and a determinate c,d (sometimes called right hand side, RHS, of the FD). This terminology is used since the values of the attributes a,b uniquely determinate the values of the attributes c,d.
About the key
The information that:
(a,b) is a primary key
can be irrelevant when normalizing a relation when you have enough information about the functional dependencies. From those dependencies you can calcolate which are the candidate keys: set of attributes that uniquely determines all the attributes of the relation and such that you cannot remove any attribute from them maintaining this property (in other words, minimal sets of attributes that uniquely determines all the attributes of the relation). The information about the primary key can be relevant when you have only partial information about the functional dependencies that hold in a relation, but in your case the information about all the (candidate) keys can be derived from the functional dependencies.
In your example, for instance, there are three candidate keys:
1. a, b
2. a, c
3. a, d
You can verify this fact by computing the closure of the attributes of a candidate key to see if it contains all the attributes. For instance, let's try to calculate the closure of a,d (called a,d *):
1. a,d * = a,d
2. a,d * = a,d,b (since a, d -> b)
3. a,d * = a,d,b,c (since a, b -> c, d)
So a,d is a candidate key (which is also a superkey, i.e. a set of attributes that determines all the attributes of the relation).
About the BCNF
There are different definitions of BCNF. Using for instance the second one that you cited, all the three dependencies have a determinant which is a candidate key (and so a superkey), and so the relation is in BCNF.
My doubt is for a given set of funtional dependencies F = { AE -> BCD, B -> E
}. Is this in BCNF or 3NF? It's a question from a test I have recently done and I would say that it is 3NF, but my teacher said it's neither 3NF nor BCNF. (I believe it is an error).
I have obtained as candidate keys AE and AB, and as in the first functional dependency the left side is a candidate key and in B -> E, E is contained in a candidate key, so it is in 3NF.
Is this in BCNF, 3NF or neither?
Assuming that all the attributes of the relations are A B C D and E, and that the only dependencies given are the two described (F), you are correct. Since the (only) candidate keys are correctly A E and A B, and since the functional dependency B → E has a determinant which is not a superkey, the relation is not in BCNF. Given one of the definitions of BNCF: “for all the non-trivial dependencies X → Y of F+, X is a superkey”, there is a theorem that shows that a necessary and sufficient condition for this is that the property of being a superkey holds for all the dependencies in F.
On the other hand, since E is a prime attribute, i.e. an attribute of a candidate key, the dependency B → E does not violate the 3NF, so that the relation is in 3NF. This, again, given one of the definitions of 3NF: “for all the non-trivial dependencies X → A in F+, then X is a superkey or A is a prime attribute”, is due to a theorem that says that this condition is equivalent to check, “for each functional dependency X → A1,...,An in F, and for each i in {1..n}, either Ai belongs to X, or X is a superkey or Ai is prime”. And this is satified by the two dependencies of F.
You need to use a definition of a NF when you claim/show that a relation is in it.
You don't actually say what all the attributes are. I'll assume the attributes are A through E. Otherwise, the CKs (candidate keys) are not what you say.
You are right in your argument against BCNF. You are using the definition that all determinants of FDs (functional dependencies) are out of superkeys. You found a counterexample FD B → E.
If it were an either-or question re BCNF vs 3NF you could stop there.
in the first functional dependency the left side is a candidate key and in B -> E, E is contained in a candidate key
You don't show that the table meets the conditions of either of the following definitions (from Wikipedia that happen to be correct) that a table is in 3NF if and only if:
both of the following conditions hold:
The relation is in 2NF
Every non-prime attribute is non-transitively dependent on every [candidate] key
for each of its functional dependencies X → A, at least one of the following conditions holds:
X contains A
X is a superkey
each attribute in A-X is prime
You seem to using definition 2 (but not saying so). You show bullet 2 holds for AE → BCD. Pointing out that E is prime in B → E seems to be part of showing that E-B is all prime. But you need to show every FD satisfies a bullet. Note that more FDs hold than the given ones. Armstrong's axioms tell you what all the FDs are.
In practice it can be easier to show a schema is in 3NF by applying a 3NF algorithm.
Hey all I have an assignment that says:
Let R(ABCD) be a relation with functional dependencies
A → B, C → D, AD → C, BC → A
Which of the following is a lossless-join decomposition of R into Boyce-Codd Normal Form (BCNF)?
I have been researching and watching videos on youtube and I cannot seem to find how to start this. I think I'm supposed to break it down to subschemas and then fill out a table to find which one is lossless, but I'm having trouble getting started with that. Any help would be appreciated!
Your question
Which of the following is a lossless-join decomposition of R into
Boyce-Codd Normal Form (BCNF)?
suggests that you have a set of options and you have to choose which one of those is a lossless decomposition but since you have not mentioned the options I would first (PART A) decompose the relation into BCNF ( first to 3NF then BCNF ) and then (PART B) illustrate how to check whether this given decomposition is a lossless-join decomposition or not. If you are just interested in knowing how to check whether a given BCNF decomposition is lossless or not jump directly to PART B of my answer.
PART A
To convert a relation R and a set of functional dependencies(FD's) into 3NF you can use Bernstein's Synthesis. To apply Bernstein's Synthesis -
First we make sure the given set of FD's is a minimal cover
Second we take each FD and make it its own sub-schema.
Third we try to combine those sub-schemas
For example in your case:
R = {A,B,C,D}
FD's = {A->B,C->D,AD->C,BC->A}
First we check whether the FD's is a minimal cover (singleton right-hand side , no extraneous left-hand side attribute, no redundant FD)
Singleton RHS: All the given FD's already have singleton RHS.
No extraneous LHS attribute: None of the FD's have extraneous LHS attribute that needs to e removed.
No redundant FD's: There is no redundant FD.
Hence the given set of FD's is already a minimal cover.
Second we make each FD its own sub-schema. So now we have - (the keys for each relation are in bold)
R1={A,D,C}
R2={B,C,A}
R3={C,D}
R4={A,B}
Third we see if any of the sub-schemas can be combined. We see that R1 and R2 already have all the attributes of R and hence R3 and R4 can be omitted. So now we have -
S1 = {A,D,C}
S2 = {B,C,A}
This is in 3NF. Now to check for BCNF we check if any of these relations (S1,S2) violate the conditions of BCNF (i.e. for every functional dependency X->Y the left hand side (X) has to be a superkey) . In this case none of these violate BCNF and hence it is also decomposed to BCNF.
PART B
When you apply Bernstein Synthesis as above to decompose R the decomposition is always dependency preserving. Now the question is, is the decomposition lossless? To check that we can follow the following method :
Create a table as shown in figure 1, with number of rows equal to the number of decomposed relations and number of column equal to the number of attributes in our original given R.
We put a in all the attributes that our present in the respective decomposed relation as in figure 1. Now we go through all the FD's {C->D,A->B,AD->C,BC->A} one by one and add a whenever possible. For example, first FD is C->D. Since both the rows in column C has a and there is an empty slot in second row of column D we put a a there as shown in the right part of the image. We stop as soon as one of the rows is completely filled with a which indicates that it is a lossless decomposition. If we go through all the FD's and none of the rows of our table get completely filled with a then it is a lossy decomposition.
Also, note if it is a lossy decomposition we can always make it lossless by adding one more relation to our set of decomposed relations consisting of all attributes of the primary key.
I suggest you see this video for more examples of this method. Also other way to check for lossless join decomposition which involves relational algebra.
I am struggling with the concepts of 4NF and Multivalued Dependencies (MVDs).
I am looking at a supplementary book for the course I am currently taking and one of the examples is below.
The book states the asterisks refer to a unique key or a composite attribute key.
Given: R(A*,B,C*) and the set {(A,B):R,(B,C):R} satisfies the
lossless decomposition property.
Does the multivalued dependency B->>C hold?
Is B definitely a unique key?
Is R in 4NF?
I understand lossless decomposition - if you take the natural join of the two sets above - you are given the original dataset i.e in this case A,B,C.
But I just cannot grasp how to take the given information and prove/confirm that B->>C holds or if it does not.
I emailed my professor about my confusion and he just told me to look over his notes (which I've obviously done numerous times) and it's gotten me nowhere.
Does the multivalued dependency B->>C hold?
You have been told some things about MVDs. One of them is probably:
A decomposition of R into (X, Y) and (X, R − Y) is a lossless-join
decomposition if and only if X ->> Y holds in R.
In your case R is {A,B,C}, X is {B}, Y is {A} and R - Y = {A,B,C} - {A} = {B,C}. So a decomposition of R into (B,A) and (B,C) is a lossless-join decomposition if and only if {B,A} ->> {B,C} holds in R. But we are given that decomposition of R into (B,A) and (B,C) is a lossless-join decomposition. So {B,A} ->> {B,C} holds in R.
Is B definitely a unique key?
I can't make sense of that.
Maybe you are trying to say that we are given that {A,C} is a CK (candidate key) of R but that there might be other CKs and you are trying to ask whether the decompositionality means that {B} must also be a CK of R. Let's look for a counterexample. Pick a simplest example. Suppose R is {(a,b,c1),(a,b,c2)} = {(a,b)} JOIN {(b,c1),(b,c2)}. This agrees with R CK {A,C} & R MVD {B,A} ->> {B,C}. But b appears with c1 and c2 so {B} does not functionally determine all other attributes so {B} is not a CK of R. So that CK and that MVD do not force that {B} is a CK.
Is R in 4NF?
You have been told some things about 4NF. One is probably that:
A Table is in 4NF if and only if, for every one of its non-trivial
multivalued dependencies X ->> Y, X is a superkey
The MVD {B,A} ->> {B,C} is non-trivial. But to show whether R must be in 4NF or must not be in 4NF or that we can't tell, you are going to have to address possible sets of non-trivial MVDs that could hold in R and CKs that R could have.
Let R be a relation with Schema R(X,Y,Z)
and it's FDs are
{XY -> Z, Z -> Y}
I am not able to decompose it into BCNF .
Because r1(Z,Y), r2(Z,X) will lose FD XY -> Z and
R(X,Y,Z) itself is not the solution as Z->Y shows that Z should be a key ..
How to do this ???
Every conversion into BCNF may not be dependency preserving
We only need to give a counter example: Consider the following schema;
a b c and c->b
Clearly the above schema is in 3NF, because ab->c is a superkey dependency and ,from c->b we
can see that b-c=b, which is a subset of the primary key (such dependency is also allowed in 3NF).
But, the above schema is not in BCNF because c->b is neither super-key nor trivial dependency.
So we decompose above schema , keeping it lossless.
Only possible lossless decomposition is: ac and cb. (because,their intersection c is primary key for the 2nd table).
But clearly the dependency ab->c is lost.
Hence, proved.