I have a R= {A,B,C,D,E,F,G,H,I,J,K} F={ABGH->IJKF,JIGH->ABF, A->CDE} I need to find all minimal candidate keys of R, and How to normalize R to BCNF
I got the following answers so far:
ABGH, GHJIK and AGHIJK.
But I was check my answer from this site :checked site
I don't know why 'K' is not part of the answer and I am not sure if my answers were correct. Thanks!
There are two candidate keys of R: {ABGH} and {GHIJ}.
{GHJIK} is not a candidate key, but if it were, then {AGHIJK} would not be a minimal key.
The attribute K isn't part of the two candidate keys, because the closure of {ABGH} contains K, and the closure of {GHIJ} contains K. For example, for {ABGH} . . .
ABGH->ABGH (trivial)
ABGH->IJKF (given), therefore
ABGH->ABGHIJKF
A->CDE (given), therefore
ABGH->ABCDEGHIJKF, or in alpha order
ABGH->ABCDEFGHIJK
BCNF decomposition will be
R1 (ACDE) nd R2(ABFGHIJK)
because in R (A--> CDE) is partial dependency. So we decompose it in R1 where A is ck and R2 where ABGH and GHIJ are candidate keys.
Any no. of attributes added to candidate key forms a super key. We can also thus say that candidate key is the minimal super key. Here ABGH and GHIJ can determine all other attributes of the relation so they become candidate keys hence GHIJK becomes a super key (due to addition of K) and not a candidate key and on similar principles AGHIJK (due to addition of A and K) also becomes a super key as it has GHIJ which is a candidate key.
Related
I need help finding the candidate keys from the given relation:
R (A,B,C,D,E) with FD:
A -> BE
B -> BE
B -> D
STEPS:
I know all the attributes can identify themselves so: A,B,C,D,E -> {ABCDE}
Now I know A -> BE, so i can cross out BE and get this: ACD -> {ABCDE}
The prime attributes are (A,C,D)
D would now be replaced by B giving me my other candidate key of (ABC)
I have the answer has ACD and ABC but apparently, it's AC. What am I doing wrong?
A simple reasoning is the follow.
A and C must be present in any candidate key, since they never appear in the right part of some FD.
So, let’s see if they are already a candidate key by computing their closure, AC*.
By applying the rules for computing the closure of a set of attributes, we can see easily that AC* is equal to ABCDE, so AC is a candidate key. The only other attribute that appears in the left side of a FD is B, so we should check that no other candidate key contains it. But noting that the attributes AC must be present in any key and already they form a candidate key, so adding any attribute to them produces only superkeys, we can conclude that this is the only candidate key.
I was doing homework for DB class.
One of the questions bugs me out even though I got the answer(I think)..
Question was simple.
FOR GIVEN RELATION R(A,B,C,D,E) and Functional dependencies F(AB -> C, D->E, DE ->B)
1. IS R IN 2NF?
2. IS R IN 3NF?
3. IS R IN BCNF?
I thought since there's no A and D on right-hand side of all FDs in F, A and D must be part of Candidate keys.
So I checked the Closure of AD, and I got AD+ : {A,B,C,D,E}.
That means that AD is super key.
Also, since both A and D must be part of Candidate key and AD cannot be reduced(no closure of subset of AD is {A,B,C,D,E}), AD is a candidate key and only possible candidate key. (Am I doing this right?)
With candidate key AD, D->E is partial dependence on candidate key AD.
So it violates the condition of 2NF.
On DE -> B, is this FD is violating 2NF?
If that's true then..
Is it violating because we can get D->DE from D->E . so DE -> B is equivalent to D -> B. Is this D->B is violating 2NF ??
OR
DE->B itself violates the 2NF without any conversion because of D on left-hand side?
It really confuses me when XY -> Z X is part of Candidate key and Y,Z is non-prime key.
Because I can't say it is violating 2NF or not. I think it is violating 2NF but I can't say why clearly.
I've been looking for examples and explanations and clips for hours but I haven't got any satisfying answer.
It's okay if I don't care specific reason and just want credit . But I can't bare myself with that kind of attitude.
Also, since both A and D must be part of Candidate key and AD cannot be reduced(no closure of subset of AD is {A,B,C,D,E}), AD is a candidate key and only possible candidate key. (Am I doing this right?)
Yes
Is this D->B is violating 2NF ?
Yes, since B is a non prime attribute and D is part of a candidate key and the dependency holds, since it is implied by D -> E and DE -> B (in a relation in 2NF dependencies where the determinant is a proper subset of a candidate key and the determinant is a non-prime attribute cannot hold).
I have a set of functional dependencies:
V = {ABCDEF} F = {AB → CD,ABDE → F,BC → A,C → DF}
Candidate keys are: {ABE, BCE}
Canonical cover is: {AB→ C, BC→ A, C→ DF} [This is what I think, might be wrong]
However, as you can see an attribute of candidate key, E, is not in my canonical cover and as far as I know candidate keys should be same in the canonical cover.
If you consider Augmentation rule from Armstrong calculus we can say it is correct but I am confused. Does attribute E have to be represented in the canonical cover?
You say:
as far as I know candidate keys should be same in the canonical cover
This is not true. On the contrary, if an attribute does not belong to any right part of the functional dependencies of a canonical cover, it must be present in any candidate key (this is because it cannot be derived from any other subset of attributes, so, since a candidate key must determines all the attributes, it should be present in any key). Your canonical cover and candidate keys are correct.
Note that if an attribute does not belong to any functional dependency (both in the left and right part), as E in your example, this is a special case of above (it does not belong the a right part side), and must be present in any candidate key.
Finally, note that this can be considered a “symptom” of something wrong in the relation and in fact the schema is not in 3NF or BCNF.
Well, when I try to do Bernnstein synthesis from this relation (ABCDEF) I have to use basis: {AB→C,BC→A,C→DF} I need to add candidate keys since no candidate key exist when we form a relation from functional dependencies : R1(ABC) and R2(CDF) and I was wondering if we need to add E here since our basis doesn't contain E and we consider basis when we do synthesis. That's why I was little confused. But, I think we need to add E since we are doing a synthesis from original R(ABCDEF) so it should be R1(ABC), R(CDF) and R3( ABCE). R3 contains all candidate keys.
Given a relation schema R with n attributes R(A1, A2, ..., An). What’s the maximum number of possible super-keys for R? Please justify your answer.
Given a relation schema R with n attributes R(A1, A2, ..., An). What’s the maximum number of possible candidate keys for R? Please justify your answer.
I am still wondering on how to answer both of these questions. What I have thought as answer for the first question would be (2^n) - 1 because empty set is not included.
As for the second question. My answer would be n atrributes.
What do you guys think?
Maximum number of superkeys on relation with n attributes would be number of all possible combinations of attributes. This turns out to be (2^n)-1.
This is nothing but taking
1 attribute from n (nC1) + 2 attributes from n (nC2) + ... + nCn = (2^n)-1
Or we can simply think it as follows: we have each of n attributes represented as a bit. We can put 1 when an attribute has to be a part of superkey or 0 otherwise. So this will be (2^n), because we have two choices (1 or 0) for each of the n bits/attributes. We subtract 1 to avoid all 0's, that is considering 'no-attribute' as a superkey. So (2^n)-1.
This situation can occur when all attributes can functionally determine all other attributes. This occurs when there is a cycle of functional dependencies among attributes. For example if there is a relation R(A,B,C,D), then the FD cycle would be:
A->B
B->C
C->D
D->A
The superkeys would are A,B,C,D,(AB),(AC),(AD),(BC),(BD),(CD),(ABC),(ACD),(ABD),(BCD),(ABCD), total (2^4)-1=15
The maximum possible number of candidate keys will occur for size-r keys where nCr is biggest. Or in other words, when all size-r combinations of attributes are candidate keys, maximum number of candidate keys occur.
This can be seen from above example. Above A,B,C,D are all candidate keys, so none of their superkeys (say (AB), or (BCD) or (ABCD)) are candidate keys. Similarly if, in any relation (AB) is a candidate key, then none of its superkey (say ABC or ABD) can be a candidate key.
In general, nCfloor(n/2) is the maximum number of possible candidate key for relation on n attributes.
PS: this considers the definition that candidate key is a minimal superkey (one from which no attribute can be removed while still leaving it capable to uniquely identify / functionally determine all other attributes)
The maximum number of superkeys for R with n attributes is 2^n, which is the size of the power set of R's attributes. This is obvious when you realise that ∅ (the empty set) may be a candidate key and that ∅ is a subset of every set of attributes.
The maximum number of candidate keys is given by nC(n/2) (binomial coefficient).
I have a problem about the 2nd normal form. The rule says : “A relation is in second formal form when it is in 1NF and there is no such non-key
attribute that depends on part of the candidate key, but on the entire candidate key.” (Neeraj Sharma, 2010) My problem is about the candidate key. It is only the primary key of a relation or all possible candidate keys.
Thank you for your help
It counts for any candidate key. If it counted only for the primary key, simply adding a surrogate id would be enough to put any table into 3NF. However, that wouldn't help to ensure that each fact is recorded once only and independent of other facts.
Trying to clear your doubt by an example:
According to 2NF "Partial Dependencies are not allowed in a relation."
Assume this relation: R(A,B,C,D)
lets suppose there are 3 CK's related to this relation (Assume CK's: AB,AC,B).
Then first write all the attributes that are present in any of CK's,these are called Prime attributes.Other than that are called non prime attributes.
Here:
Prime Attributes (3)= {A,B,C}
Non Prime Attributes (1)={D}
Now According to 2 NF, any FD should not be in this form:
This kind of FD's aren Not allowed in 2NF:
"Part of any candidate key(Partial Dependency) ---> Non Prime attribute"
Means:
Here : C---> D(Not allowed in 2 NF because C is a part of CK "AC" and D is non prime attribute)
Hope this helps. For more detail, you can also refer : Detailed explanation of Normal forms