BCNF Decompositions and Lossless joins for Databases - database

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.

Related

Simple relation decomposition example - insufficient information?

The entire question is:
The relation R(A, B, C, D, E) with F = {A→B ; A→C ; B,D→A} is decomposed considering the functional dependency A→B from the beginning.
What is the functional dependency that cannot be preserved by such a decomposition?
It seems to me that I have insufficient data to answer this question. Firstly, doesn't knowing which FD cannot be preserved depend on the normal form we are using? Secondly, the decomposition is unfinished; doesn't the preservation depend on which decomposition we use?
The question is certainly poorly phrased. Presumably F is supposed to be a cover for the FDs that hold in R--ie the FDs that hold in R are the ones implied by the ones in F. It's not clear what is meant by "decomposed considering the functional dependency A → B from the beginning". We would reasonably say that "decomposing considering a FD" means decomposing introducing a component with the FD's attributes. I suspect that they are trying to say, if you binary decompose to component AB plus some other component(s) then what FD "cannot be preserved by such a decomposition"?
When we losslessly decompose we really mean we are only interested in decompositions where all the components are smaller than the original & no component is a subset of another. So in a decomposition having component AB the other components would join to either ACDE or BCDE.
Look at the remaining given FDs: A → C & B, D → A. For each of those projections/components, if it were losslessly joinable with AB then the FDs that are implied by those per Armstrong's axioms & that have all their attributes in it will hold in it. (And no others will hold in it.) But there's one of those two FDs that can't hold in either of the two projections. So that's the one that "cannot be preserved by such a decomposition".
Firstly, doesn't knowing which FD cannot be preserved depend on the normal form we are using?
They seem to be trying to ask about a decomposition of a certain nature, so the resulting NFs don't matter.
Secondly, the decomposition is unfinished; doesn't the preservation depend on which decomposition we use?
They are telling you that a certain FD "cannot be preserved by such a decomposition". So we can expect that as long as we have "such a decomposition" the answer doesn't further "depend on which decomposition we use".
(Note that a decomposition to 2NF or 3NF can fail to preserve FDs. But there's always a decomposition that doesn't fail.)

Decomposition to BCNF and set of super key

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.

Specific scenario regarding BCNF decomposition

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.

boyce codd and finding candidate keys

needing desperate help with understanding boyce codd and finding the candidate keys.
i found a link here http://djitz.com/neu-mscs/how-to-find-candidate-keys/ which i have understood for most part but i get stuck
e.g
(A B C D E F)
A B → C D E
B C D → A
B C E → A D
B D → E
right as far as i understand from the link i know you find the common sets from the left which is only B, and common sets from the right which are none
now where do i go from here? i know all candidate sets will have B in them but i need guidance on finding candidate sets after that. someone explain in simple language
The linked article isn't written particularly well. (That's an observation, not a criticism. The author's first language isn't English.) I'll try to rewrite the algorithm. This isn't me telling you how to do this. It's my interpretation of how the original author is telling you to do this.
Identify the attributes that are on neither the left side nor right side of any FD.
Identify the attributes that are only on the right side of any FD.
Identify the attributes that are only on the left side of any FD.
Combine the attributes from steps 1 and 3.
Compute the closure of the attributes from step 4. If the closure comprises all the attributes, then the attributes from step 4 make up the only candidate key. (No matter how many candidate keys there are, every one of them must contain these attributes.)
Identify the attributes not included in step 4 and step 2.
Compute the closure of the attributes from step 4 plus every possible combination of attributes from step 6.
So for the FDs you posted, you'd end up with this.
{F}
{}
{B}
{BF}
The closure of {BF} is {BF}. That's not all the attributes. (But every candidate key must contain {BF}.)
{ACDE}
Compute the closure of these sets of attributes.
{ABF}
{CBF}
{DBF}
{EBF}
{ACBF}
{ADBF}
{AEBF}
{CDBF}
{CEBF}
{DEBF}
{ACDBF}
{ADEBF}
{CDEBF}
If I got those combinations right, every candidate key will be found among the possibilities in step 7. In your example, there are 3 candidate keys.
http://www.sroede.nl/projects/fdhelper.aspx
this would help'just put in ur relation and FD's
click generate at the bottom

how to produce a third normal form and BCNF decompositions

I'm trying to produce a 3NF and BCNF decomposition of a schema. I have been looking at the algorithms but I am very confused at how to do this.
If I have my minimal cover say: F' = {A->F, A->G, CF->A, BG->C) and I have identified one candidate key for the relation, say it is A. Then what exactly do I do?
I have been looking at examples, one which has the following:
F = {A → AB,A → AC,A → B,A → C,B → BC}
Minimal cover: F′ = {A → B,B → C}
And the final result was: (AB,A → B), (BC,B → C). How did they get to this?
If I have my minimal cover say: F' = {A->F, A->G, CF->A, BG->C) and I
have identified one candidate key for the relation, say it is A. Then
what exactly do I do?
F' is not a minimal cover: you have to combine A->F and A->G to A->FG
Even worth A cannot be a candidate key given F' since B does not belong yo the closure of A. A possible candidate key would be AB.
For 3NF you start with creating tables for each one of the dependencies in F', i.e.,
R1(A,F,G) R2(A,C,F) R3(B,C,G)
Next you check whether one of the tables contains a candidate key. Since B appears only on the left side of the dependencies, B should always be a part of a candidate key. The only table with B is R3 and it does not contain candidate keys (check it!). Hence, we add a new table R4 with a candidate key as attributes
R4(A,B)
Finally, we check whether the set of attributes of one of the tables is contained in the set of attributes of another table. This is not the case for our running example.
Hence, our 3NF decomposition is
R1(A,F,G) R2(A,C,F) R3(B,C,G) R4(A,B)
For BCNF you start with R(A,B,C,F,G) and look for BCNF violations.
For instance A->FG is a violation of BCNF because this dependency is not trivial and A is not a superkey. Hence we split R into
R1(A,F,G) and R2(A,B,C)
None of the relations obtained contains BCNF violations, so the process stops here.

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