I have to determine the functional dependencies for the below ER:
As shown It's a one-to-many relationship. Depending on the functional dependency definition:
"X->Y means, each possible value of X can correspond to exactly one
value of Y"
, I can write:
A -> B
P -> Q,R
But I'm not sure how should the FD should be written for Ent1 and Ent1 with the relation. If I write A -> X,Pis it correct?. Can someone explain?
Thank you.
In any relationship, the combination of entity sets in many-roles determine the other entity sets and attributes in the relationship.
In your example, Ent2 (represented by P) determines Ent1 (represented by A) as well as X.
So, you would write P -> A,X.
Related
So I have this set of relation
AB->CDEF
G->H,I
ABJ->K
C->L
How should I decompose this? I am so confused. Am I supposed to find the set of super key first?
We can first convert the relation R to 3NF and then to BCNF.
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,E,F,G,H,I,J,K,L}
FD's = {AB->CDEF,G->HI,ABJ->K,C->L}
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: We write the FD's with singleton RHS. So now we have FD's as {AB->C, AB->D, AB->E, AB->F, G->H, G->I, ABJ->K, C->L}
No extraneous LHS attribute: We remove the extraneous LHS attribute if any. There are no extraneous LHS attributes here.
No redundant FD's: We remove the redundant dependencies if any. Now FD's are {AB->C, AB->D, AB->E, AB->F, G->H, G->I, ABJ->K, C->L}
Second we make each FD its own sub-schema. So now we have - (the keys for each relation are in bold)
R1={A,B,C}
R2={A,B,D}
R3={A,B,E}
R4={A,B,F}
R5={G,H}
R6={G,I}
R7={A,B,J,K}
R8={C,L}
Third we combine all sub-schemas with the same LHS. So now we have -
S1 = {A,B,C,D,E,F}
S2 = {G,H,I}
S3 = {A,B,J,K}
S4 = {C,L}
Since none of the above decomposed relations contain contain a key of R, we need to create an additional relation schema that contains attributes that form of a key of R. This is to ensure lossless join decomposition that preserves dependencies. So we add -
S5 = {A,B,G,J}
ABGJ is the key of the original relation R
This is in 3NF. Now to check for BCNF we check if any of these relations (S1,S2,S3,S4,S5) 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.
Note - The importance of some of these steps may not be clear in this example. Have a look at other examples here and here.
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 having a hard time understanding the 3 Normal form.
3 NF: 2 NF + No transitions
So, for eg: If I have,
A -> B
B -> C
Then the above is sort of a transition relation and hence won't be in 3 NF?
Am I understanding it correctly?
But in this answer What exactly does database normalization do? , by paxdiablo, it says,
Third normal form (3NF) - 2NF and every non-key column in a table depends on nothing but the key.According to this, it will be in 3 NF. Where am I going wrong?
A relation is in 3NF if it is in 2NF and:
either each attribute depends on a key,
or, if an attribute depends on a non-key, then it is prime.
(being prime means that it belongs to a key).
See for instance Wikipedia.
A relation is in Boyce-Codd normal form if only the first condition hold, that is:
each attribute depends on a key
So, in your example, if the relation has only three attributes A, B and C and the two dependencies, it is not in 3NF, since C is not prime, and depends on B, which is a not a key. On the other hand, if there are other attributes, and C is a key or part of a key, then it could be in 3NF (but this depends on the other functional dependencies, that should satisfy the above conditions).
The 2NF says that each non-prime attribute depends on each whole candidate key, and not by part of it. For instance, if a relation has attributes A, B and C, the only key is AB, and B -> C, then this relation is not in 2NF.
The 2-part 3nf definition you are trying for is:
2NF holds and every non-prime attribute of R is non-transitively dependent on every superkey. (X transitively determines Z when there's a Y where X → Y and Y → Z and not Y → X.)
The other definition of 3NF is:
For every non-trivial FD X → Y, either X is a superkey or the attributes in Y but not in X are prime. (X → Y is trivial when X contains Y.)
Then BCNF is:
For every non-trivial FD X → Y, X is a superkey
See this answer.
If your example's only columns are A, B and C and your two FDs form a minimal cover then the only candidate key is A and C is dependent on a non-superkey so it is not in 3NF (or BCNF).
You are (mis)using terms so sloppily that your sentences don't mean anything. Learn the terms and how they are used in their definitions to refer to various things and use them that way in reference to appropriate things. And get your definitions from a (reputable) textbook.
Say I have a relation ABCD with FD's (A->D and AB -> ABCD)
Will a decomposed relation ABC be in BCNF? According to the second FD, AB form a key and is therefore in BCNF, but if you only look at the FD A -> D, is the relation no longer in BCNF then?
If you decompose a given relation schema (to which given dependencies apply), the next task is to determine, for each individual dependency in the original set :
(a) which (if any) of the new, decomposed, schemas does it apply to ?
(b) how has the decomposition affected the very definition of the FD ?
Question (a) applies to your original A->D dependency.
Question (b) applies, sort of, to your original AB->ABCD dependency. I say "sort of" because that version is quite "overstated". Given that A->D was already a given, it could just as well just say AB->C.
Let's consider, for instance, the following relation:
R (A,B,C,D,E,F)
where the bold denotes that it is a primary key attribute
with
F = {AB->DE, D->E}
Now, this looks to be in the first normal form. It can't be on the third normal form as I have a transitive dependency and it cannot be in the second form as not all non-key attributes depend on the whole primary key.
So my questions are:
I don't know what to make of F and C. I don't have any functional dependency info on them! F doesn't depend on anything? If that is the case, I can't think of any solution to get R into the 2nd normal form without taking it out!
What about C? C also suffers from the problem of not being referred on the functional dependencies list. What to do about it?
My attempt to get R into the 2nd normal form would be something like:
R(A,B,D)
R' (D,E)
but as stated earlier, I don't have a clue of what to do of C and F. Are they redundant so I simply take them out and the above attempt is all I have to do to get it into the 2nd form (and 3rd!)?
Thanks
Given the definition of R that { A, B, C } is the primary key, then there is inherently a functional dependency:
ABC → ABCDEF
That says that the values of A, B and C inherently determine or control the values of D, E and F as well as the trivial fact that they determine their own values.
You have a few additional dependencies, identified by the set F (which is distinct from the attribute F - the notation is not very felicitous, and could be causing confusion*):
AB → DE
D → E
As you rightly diagnose, the system is in 1NF (because 1NF really means "it is a table"). It is not in 2NF or 3NF or BCNF etc because of the transitive dependency and because some of the attributes only depend on part of the key.
You are right that you will end up with the following two relations as part of your decomposition:
R1(D, E)
R2(A, B, D)
You also need the third relation:
R3(A, B, C, F)
From these, you can recreate the original relation R using joins. The set of relations { R1, R2, R3 } is a non-loss decomposition of the original relation R.
* If the F identifying the set of subsidiary functional dependencies is intended to be the same as the attribute F, then there is something very weird about the definition of that attribute. I'd need to see sample data for the relation R to have a chance of knowing how to interpret it.
I think the primary key of R is set wrong. If F isn't functionally related to anything it has to be a part of the key
So you have R( ABCF DE) which is now in the first normal form (with F = {AB->DE, D->E}) Now you can change it to the second normal form. DE isn't dependant on the whole key (partial dependency) so you put it in another relation to get to second normal form:
R( ABCF ) F = {}
R1( #AB DE) F = {AB->DE}
Now this relation doesn't have any transitive dependencies so it is already in third normal form.
F doesn't depend on anything?
No, you just haven't been given any explicit information about it in the form
{something -> F}
And essentially the same can be said for C. You're expected to infer the other dependencies by applying Armstrong's axioms. (Probably.)
Think about how to finish this:
Given R (A,B,C,D,E,F)
{ABC -> ?}
[Later . . . I see that Jonathan Leffler has broken the suspense, so I'll just finish this.]
{ABC -> DEF} (By definition) therefore,
{ABC -> F} (By decomposition. Here's where F and C come in. And this is your third relation. ).