From the Database Management Systems book: given the relation SNLRWH (each letter denotes an attribute) and the following functional dependencies:
S->SNLRWH (S is the PK)
R->W
My attempt:
First, it is not 3NF: for the second FD, neither R contains W, nor R contains a key, nor W is part of a key.
Second, it is/not 2NF. If we examine the second FD, W is dependent on R, which in turn is not part of a key. STUCK.
2NF is violated if some proper subset of a candidate key appears as a determinant on the left hand side of one of your (non-trivial) dependencies. Ask yourself whether any of your determinants is a subset of a candidate key.
Usually 2NF is violated only when a relation has a composite key - a key with more than one attribute. It is technically possible for a relation with only simple keys (single attribute keys) to violate 2NF if the empty set (∅) happens to be a determinant. Such cases are fairly unusual and rarely thought worthy of consideration because they are so obviously "wrong". For completeness, here's a fun example of that special case. In the following relation Circumference and Diameter are both candidate keys. The dependency in violation of 2NF is ∅ -> Pi, the ratio of the circumference to the diameter.
2NF has to do with partial key dependencies. In order for a relation to fail the test for 2NF, the relation has to have at least one candidate key that has at least two columns.
Since your relation has only one candidate key, and that candidate key has only one column, you can't possibly have a partial key dependency. It passes the test for 2NF.
Related
What's the main point of Normalization?
I mean if a normal form is not in 2NF, it is because of partial dependency i.e. a non key attribute is dependent on a part of a candidate key.
So, let's say, for a relation R(A,B,C) with FDs:
AB->C, B->C
Clearly, AB is the candidate key and B->C is the partial dependency.
Solution: Decompose the relation such that (B,C) forms a new relation with B as the key.
Now, if a relation is not in 3NF, it is because a non key attribute is dependent on another non key attribute i.e. to say
if FDs for a relation R(A,B,C) are:
A->B,B->C
Clearly, A is the key and B->C shows transitive dependency, so not in 3NF.
Solution: Decompose the relation such that (B,C) forms a new relation with B as the key.
So, what's the exact difference?
I mean, why such a marked distinction? Essentially in both of the cases the action is same.
Decompose the relation using the dependency where the determinant (B here) is either PART of a key or not.
Why have separate terms like partial dependency or transitive dependency?
Why not just see, if there exists a dependency wherein a non prime attribute is determined by a something which is NOT a candidate key( no matter whether it is a partial key or another non prime attribute )
Why can't we implement a method like this:
1 NF -- having all elements in the atomic form
X NF -- if there's any
dependency of the form non_key -> non_prime_attribute(s) ,
decompose the relation with one of the new relation having this
particular "non_key" as the key with those non_prime_attributes.
BCNF
: Where for all the dependencies of the form X->Y, X is a superkey?
Can we have such NF condition format? Does it combine all the conditions?
So, what's the exact difference?
2NF is not 3NF & definitions of 2NF are not definitions of 3NF. There isn't any particular semantic or syntactic structural similarity that would leave some kind of "difference" other than that a 2NF relation can have the sort of problem FD (functional dependency) that violates 3NF that a 3NF relation doesn't have. You can find definitions all over the place. You almost give them correctly here yourself. But a NF (normal form) is a condition, not a process. What do you mean "actions are the same"? Being in 3NF implies being in 2NF, so naturally decomposing to 3NF also gives 2NF. But there are relations that are in 2NF but not in 3NF, and there may be decompositions for a relation to 2NF that don't get to 3NF. Those decompositions will involve in a removal of all problem partial FDs that does not result in the removal of all problem transitive FDs.
(Because 3NF is always achievable and there are no other disadvantages compared to 2NF, 2NF isn't even useful. It's just a condition that was discovered first that is not as strong as 3NF.)
(3NF is frequently defined in terms of 2NF plus no transitive dependencies of non-prime attributes on CKs, but actually no such FDs implies no partial FDs of non-prime attributes on CKs, hence 2NF, so the first condition is redundant.)
Why not just see, if there exists a dependency wherein a non prime attribute is determined by a something which is NOT a candidate key
Why should that condition be helpful? It is not a description of just getting rid of the problem FDs of 2NF & 3NF--that's what putting into 3NF does.
Getting rid of non-trivial FDs that are not determined by superkeys happens to give BCNF. It implies 2NF & 3NF. But it is different from both of them. A BCNF relation exhibits no FD-based update anomalies. It is always achievable. However 3NF is alway achievable while "preserving FDs", whereas BCNF is not. There are cases where in order for a FD that held in the original to be enforced in a view/query that gives it via constraints on its components we need an EQD (equality dependency) constraint. That says two column sets have the same set of subrow values, which is more expensive to enforce than a FD. Either you have BCNF & an EQD & fewer update anomalies or you have 3NF/EKNF & a FD & certain update anomalies.
The NF that really matters is 5NF, which implies BCNF, with no update anomalies & with other benefits. (We might then decide to denormalize for performance reasons.)
PS Normalization to a given NF does not necessarily involve normalization to lower NFs.
It almost sounds as though you want to know why they called these two normal forms by different names instead of inventing just one form that covers both cases. If that's not the case, please ignore this answer.
Part of the answer is that the forms weren't discovered at the same time. And part of the answer is that the problem with 1NF that gave rise to 2NF is not the same as the problem with 2NF that gave rise to 3NF, even though they both exhibit harmful redundancy.
What might satisfy you a little more is BCNF. BCNF was actually discovered later than 4NF, so that name was already in use. But BCNF has to be placed between 3NF and 4NF, because it is more restrictive than 3NF but less restrictive than 4NF. So it was discovered "out of sequence", so to speak.
In BCNF, every (non trivial) determinant is a candidate key. That seems to be what you are looking for. I conjecture that any relation that is in 1NF and where every determinant is a candidate key, could be shown to be in 2NF and 3NF. But the proof is beyond me.
2NF and 3NF are essentially historical concepts and your question is a reasonable one. There is no real reason to apply them in practical database design because better tools exist today.
When it comes to teaching there is possibly some justification for mentioning 2NF and 3NF. Doing so allows students to explore the concepts involved (as you have done) while also teaching them a bit about the origins and rationale of design theory. In school maths lessons I was taught long division and differentiation from first principles. No one uses those techniques in practice, they are just teaching aids.
Before checking for 2NF the relation should be in 1NF. In simple words 2NF have only full dependencies only, no partial dependencies in relation. Full dependency means if x gives y, then by removal of any element in x, then y is not having any relation. If by removal of x, you are having relation with y then it is partial dependency. For 3NF we have to check for the 2NF, in 3NF we should not have any transitive relations like if x gives z, then there is no relation like x gives y and y gives z.
Solution for 2NF create a table for the partial dependcies and add foreign key in new relation which is primary key on the previous relation.
Solution for 3NF create a relation for both x gives y and y gives z. Add keys to relations.
1.
A table is automatically in 3NF if one of the following holds:
(i) If a relation consists of two attributes.
(ii) If 2NF table consists of only one non key attribute.
2.
If X → A is a dependency, then the table is in 3NF, if one of the following conditions exists:
(i) If X is a superkey
(ii) If A is a part of superkey
I got the above claims from this site.
I think that in both the claims, 2nd subpoint is wrong.
The first one says that a table in 2NF will be in 3NF if we have all non-key attributes and the table is in 2NF.
Consider the example R(A,B,C) with dependency A->B.
Here we have no candidate key, so all attributes are non-prime attributes and the relation is not in 3NF but in 2NF.
The second one says that for a dependency of the form X->A if A is part of a super key then it's in 3NF.
Consider the example R(A,B,C) with dependencies A->B, B->C . Here a CK is {A}. Now one of the super keys can be AC and the RHS of FD B->C contains part of AC but still the above relation R is not in 3NF.
I think it should be A should be part of a candidate key and not super key.
Am I correct?
Also can a particular relation be in 1NF, 3NF or 2NF if there are no functional dependencies present?
A CK (candidate key) is a superkey that contains no smaller superkey. A superkey is a unique set of attributes. A relation is a set of tuples. So every relation has a superkey, the set of all attributes. So it has at least one CK.
A FD (functional dependency) holds by definition when each value of a determining set of attributes appears always with the same value for its determined set. Every relation value or variable satisfies "trivial" FDs, the ones where the determined set is a subset of the determining set. Every set of attributes determines {}. So every relation satisfies at least one FD. However, the correct forms of definitions typically specifically talk about non-trivial FDs. Don't use the web, use textbooks, of which dozens are free online, although not all are well-written. Many textbooks also forget about FDs where the determinant and/or determined set is {}.
Your first point is not a correct definition of 3NF. Since its phrased "if..." instead of "if and only if", maybe it's not trying to be a definition. However, it is still wrong. (i) is wrong because a relation with two attributes is not in 3NF if one is a CK and the other has the same value in every tuple, ie it is determined by {}.
Similarly the second point is not a proper definition and also even if you treat it as only a consequence of 3NF (if...) it's false. It would be a definition if it used if and only if and talked about an FD that holds and it said it was a non-trivial FD and some other things were fixed.
Since those are neither correct definitions nor correct implications, there's a unlimited number of ways to disprove them. Read a book (or my posts) and get correct definitions.
Some comments re your reasoning:
First one says that, a table in 2NF will be in 3NF if we have all non key attributes and table is in 2NF.
I have no idea why you think that.
Here we have no candidate key
There's always one or more CKs. You need to read a definition of CK. There are also non-brute-force algorithms for finding them all.
Second one says that, for the dependency of form X->A if A is part of super key then it's in 3NF.
I have no idea why you think that.
A should be part of candidate key and not super key.
A correct defintion like the second point does normally say "... or (ii) A-X is part of a CK". But I can't follow your reasoning.
Sound reasoning involves starting from assumptions and writing new statements that we know are true because we applied a definition, a previously proved statement (theorem) or a sound rule of reasoning, eg from 'A implies B' and 'A' we can derive 'B'. You seem to need to read about how to do that.
Somewhere the definition of 2NF is given as -
A relation schema R is in 2NF if every nonprime attribute A in R is fully functionally dependent on the primary key of R.
And somewhere it is given as -
A relation schema R is in 2NF if every nonprime attribute A in R is fully functionally dependent on any key of R.
Which is correct ?
Only primary key is to be considered or all the keys are considered when checking partial dependency.
A relation R is in second normal form if every non-prime attribute of R is fully dependent on each candidate key
of R.
E.F.Codd, 1971, Further Normalization of the Data Base Relational Model
2NF is not particularly significant. With respect to functional dependencies the normal forms that matter are Elementary Key Normal Form and Boyce-Codd Normal Form.
A movie has one and only one genre.
A genre can be assigned to many movies.
Movies(MovieName, Genre)
MovieName is the only key
MovieName->Genre is the only determinant
If MovieName is the key, and the only dependence (not the only determinant!) is:
MovieName → Genre
then the relation is in First, Second, and Third Normal Form, as well as in Boyce-Codd Normal Form, and also in higher level forms (for instance Fourth Normal Form).
This is because in the (unique) dependency the determinant (MovieName) is the key, so all the definition of normal forms are respected.
You don't really mean "What level of normalization is this relation [in]". A relation variable or value can be in lots of normal forms at once. When it's in one, it's in all the lower ones, and it could be in higher ones. So presumably what you mean is "What is the highest level of normalization this relation must be in".
MovieName->Genre is not a determinant. It is a FD (functional dependency). Its determinant is {MovieName}. That is the only determinant of a CK (candidate key). The non-trivial FDs are
{MovieName}->{Genre}
{MovieName}->{MovieName,Genre}
Since every determinant of a non-trivial FD is a superkey, this is in BCNF. Since it's in BCNF, it's in every lower normal form.
There are two theorems by Date & Fagin relevant here:
if a relation is in 3NF (or BCNF) and every CK is simple, then it is in 5NF
if a relation is in BCNF and some CK is simple, then it is in 4NF
Since your relation is in BCNF it is 3NF and since every CK is simple it is in 5NF.
PS What if we didn't know these theorems?
If the only constraints on this value/variable were the ones implied by what you gave then it is in 5NF. Because by definition it's in 5NF when it doesn't satisfy any JDs (join dependencies) other than those that are implied by its having that set of candidate keys.
But if all we knew about the value/variable is what you gave then how could we show that it still must be in 5NF?
If the JD *{{MovieName},{Genre}} also held then it wouldn't be in 4NF (or higher). Ie if it were also equal to the join of its projections on {MovieName} & {Genre}. In the join/original every input MovieName value would be in a tuple with every Genre value. The CK says there is only one such pair per MovieName. So there would only be one Genre value in the input. Then FK {} -> {Genre} would hold. BCNF implies 2NF, which says there are no partial dependencies on CKs, so FK {} -> {Genre} doesn't hold. So we have a contradiction re it holding. So the JD doesn't hold. So the relation is in 4NF because there are no other non-trivial binary JDs that could hold to violate 4NF. It is also in 5NF because there are no JDs with more than two elements that could hold.
I have a question:
Considering a relation R{A,B,C,D,E,F} with the next set of functional dependencies {ABC->DEF,D->E,ABC->A}. A, B and C are Prymary Keys.
Can you explain me why this is on 2nd NF? Thanks.
Can you explain me why this is on 2nd NF?
I'm not quite sure what "why this is on 2nd NF" means. (Typo?) But the relation R is not in 3NF, because there's a transitive dependency: ABC->D, and D->E. So relation R must be in either 1NF or 2NF.
Relation R is in 2NF if and only if
it's in 1NF, and
there are no partial key dependencies.
ABC->A might look like a partial key dependency, but it's not, because "A" is a prime attribute. (ABC->A is a trivial dependency, because A->A.) The non-prime attributes are {DEF}. None of those attributes are functionally dependent on only part of any candidate key (a more general way of saying they're not functionally dependent on part of this relation's primary key).
So relation R is in 2NF.