I'm studying normalization and was wondering if this table could be considered to be normalized to 2NF?
Yes, it is:
☑ 1NF: A relation is in first normal form if and only if the domain of each attribute contains only atomic (indivisible) values, and the value of each attribute contains only a single value from that domain.
Since the names of the lakes/creeks, though containing the fishes' names, are not dividable, they are atomic in themselves.
In other words: the first words of the lake/creek names alone are not sufficient to identify the lakes/creeks properly and so are the second words.
☑ 2NF: [...] a relation is in 2NF if it is in 1NF and no non-prime attribute is dependent on any proper subset of any candidate key of the relation. A non-prime attribute of a relation is an attribute that is not a part of any candidate key of the relation.
There is no proper subset of the PK attribute Fish , since it is just one (apart from the empty set {}, see comments). Best Lake is non-prime since it doesn't belong to the PK and it doesn't depend on a subset (because there is no proper one apart from the empty one) but on the whole PK.
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.
Consider the following table:
The primary key is a composite key consisting of PatID and PhysName. My professor says this table is in 3rd normal form. I thought it's not even in second normal form because the non-key attribute, Name, is not dependent on the entire primary key. You can identify the Name simply by looking at PatID. It is not dependent on PhysName.
In order to really know whether the table is in 2NF or not, you would have to have the functional dependencies explicitly laid out for you.
Inferring the FDs from a small sample of data is a risky business. The smaller the sample, the greater the risk.
We would have to see a patient with two physicians here to see whether the name is the same. I expect it would be, but that's only common sense.
When you move on from classroom exercises to million dollar projects, you'll find that common sense is an unreliable substitute for data analysis.
Given a table value we can see what FDs (functional dependencies) hold in it, hence what its CKs (candidate keys) are and what NFs (normal forms) it satisfies (up to BCNF). (We can't know the CKs & NFs without knowing the FDs.)
A FD (or any constraint) holds in a variable when it holds in every value that can arise. Then its CKs and satisfied NFs are based on those FDs. So for a variable, example data tells us that certain FDs don't hold, and the "trivial" FDs must hold, but for the other FDs example data just doesn't tell us whether they hold.
Since the table value doesn't have {PatId, PhysName} as CK, your instructor must mean that that some variable with that value has that CK. (Of course, you should have got value vs variable straight anyway.) In order to consider that that variable has that CK, they must have decided something like:
the table holds rows that make a true statement from "a physician named PhysName tends a patient they identify as PatId and know by name PatName"
the physicians with a given name each only knows their patients with a given id by one name
(we don't know that it's false that) two different physicians could identify two different patients by the same id
likely that each physician has a unique name
likely that each physician identifies every one of their patients by an id
likely that a physician identifies just one patient via a given id
likely that a physician identifies a patient via only one id
likely that "identifies" always means a 1:1 correspondence of entities & ids
likely that each patient has only one name
etc
You need to know whether it's value vs variable, and it's pointless to argue about a variable and constraints (including FDs) until you agree on the predicate and the BRs (Business rules).
PS Re BRs, predicates & constraints:
A proposition is a statement about a situation: "a physician named 'Scholl, F.' tends a patient they identify as 99999 and know by name 'Gore, Z.'". A predicate is a statement template mapping from a row of column names & values to a proposition: "a physician named PhysName tends a patient they identify as PatId and know by name PatName". A table variable holds the rows that form true propositions in a situation.
BRs (business rules) give variable predicates and characterize what situations can arise. Hence what table variable values can arise, hence what FDs hold, hence the CKs, etc.
I have been learning Normalization from "Fundamentals of Database Systems by Elmasri and Navathe (6th edition)" and I am having trouble understanding the following part about 2NF.
The following image is an example given under 2NF in the textbook
The candidate key is {SSN,Pnumber}
The dependencies are
SSN,Pnumber -> hours, SSN -> ename, pnumber->pname, pnumber -> plocation
The formal Definition:
A relation schema R is in 2NF if every nonprime attribute A in R is
fully functionally dependent on the primary key of R.
for example in the above picture:
if suppose, I define an additional functional dependency SSN -> hours, then taking the two functional dependencies,
{SSN,Pnumber} -> hours and SSN -> hours
the relation wont be in 2NF, because now SSN ->hours is now a partial functional dependency as SSN is a proper subset for the given candidate key {SSN,Pnumber}.
Looking at the relation and its general definition on 2NF, i presume that the above relation is in 2NF
As far as my understanding goes and how i understand what 2NF is,
A relation is in 2NF if one cannot find a proper subset (prime attributes)
of the on the left hand side (candidate key) of a functional dependency
which defines the NPA(non prime attribute).
My first question is, Why is the above relation not in 2NF? (The textbook has considered the above relation as not in 2NF)
There is, however, a informal ways(steps as per the textbook where a normal person not knowing normalization can take to reduce redundancy) being defined at the beginning of this chapter which are:
■ Making sure that the semantics of the attributes is clear in the schema
■ Reducing the redundant information in tuples
■ Reducing the NULL values in tuples
■ Disallowing the possibility of generating spurious tuples
The guideline mentioned is as follows:
My second question is, If the above steps described are taken into account, and consider why the following relation is not in 2NF, do you assume the following functional dependencies, which are,
{SSN,Pnumber} -> Pname
{SSN,Pnumber} -> Plocation
{SSN,Pnumber} -> Ename
making the decomposition of the relation correct? If the functional dependencies assumed are incorrect, then what are the factors leading for the relation to not satisfy 2NF condition?
When looked at a general point of view ... because the table contains more than one primary attributes and the information stored is concerned with both employee and project information, one can point out that those need to be separated, as Pnumber is a primary attribute of the composite key, the redundancy can somehow be intuitively guessed. This is because the semantics of the attributes are known to us.
what if the attributes were replaced with A,B,C,D,E,F
My Third question is, Are functional dependencies pre-determined based on "functionalities of database and a database designer having domain knowledge of the attributes" ?
Because based on the data and relation state at a given point the functional dependencies can change which was valid in one state can go invalid at a certain state.In general this can be said for any non primary attribute determining non primary attribute.
The formal definition :
A functional dependency, denoted by X → Y, between two sets of
attributes X and Y that are subsets of R specifies a constraint on the
possible tuples that can form a relation state r of R. The constraint is
that, for any two tuples t1 and t2 in r that have t1[X] = t2[X], they must
also have t1[Y] = t2[Y].
So won't predefining a functional dependency be wrong as on cannot generalize relation state at any given point?
Pardon me if my basic understanding of things is flawed to begin with.
Why is the above relation not in 2NF?
Your original/first/informal "definition" of 2NF is garbled and not helpful. Even the quote from the textbook is wrong since 2NF is not defined in terms of "the PK (primary key)" but rather in terms of all the CKs (candidate keys). (Their definition makes sense if there is only one CK.)
A table is in 2NF when there are no partial dependencies of non-prime attributes on CKs. Ie when no determinant of a non-prime attribute is a proper/smaller subset of a CK. Ie when every non-prime attribute is fully functionally dependent on every CK.
Here the only CK is {Ssn, Pnumber}. But there are FDs (functional dependencies) out of {Ssn} and {Pnumber}, both of which are smaller subsets of the CK. So the original table is not in 2NF.
If the above statement is taken into account, do you assume the following functional dependencies
so won't the same process of the decomposition shown based on the informal way alone be difficult each time such a case arrives?
A table holds the rows that make some predicate (statement template parameterized by column names) into a true proposition (statement). Given the business rules, only certain business situations can arise. Then given the table predicates, which give table values from a business situation, only certain database values can arise. That leads to certain tables having certain FDs.
However, given some FDs that hold, we can formally use Armstrong's axioms to get all other FDs that must also hold. So we can use both informal and formal ways to find which FDs hold and don't hold.
There are also shorthand rules that follow from the axioms. Eg if a set of attributes has a different subrow value in each tuple then so does every superset of it. Eg if a FD holds then every superset of its determinant determines every subset of its determined set. Eg every superset of a superkey is a superkey & no proper subset of a CK is a CK. There are also algorithms.
Are functional dependencies pre-determined based on "functionalities of database and a database designer having domain knowledge of the attributes" ?
When normalizing we are concerned with the FDs that hold no matter what the business situation is, ie what the database state is. Each table for each business can have its own particular FDs per the table predicate & the possible business situations.
PS Do "make sense" of formal things in terms of the real world when their definitions are in terms of the real world. Eg applying a predicate to all possible situations to get all possible table values. But once you have the necessary formal information, only use formal definitions and procedures. Eg determining that a FD holds for a table because it holds in every possible table value.
so would any general table be in 2NF based on a solo condition of a table having a composite primary key?
There are tables in 5NF (hence too all lower NFs) with all sorts of mixes of composite & non-composite CKs. PKs don't matter.
It is frequently wrongly said that having no composite CKs guarantees 2NF. A table without composite keys and where {} does not determine any attribute is in 2NF. But if {} determines an attribute then it's a proper/smaller subset of any/every CK with any attributes. {} determines an attribute when every row has to have the same value for that attribute.
Why is the above relation in 2NF?
EP1, EP2, and EP3 are in 2NF because, for each one, the key identifies the non-key. No part of any key identifies any part of any non-key. That is what is meant by for any two tuples t1 and t2 in r that have t1[X] = t2[X], they must also have t1[Y] = t2[Y].
By contrast, you might say EMP_PROJ is over-specified. If ssn identifies, ename (as the text says it does), then the combination of {ssn, pnumber} is too much. There exists a subset of the key {ssn,pnumber} that identifies a part of the non-key, {ename}. That situation does not occur in a table conforming to 2NF, as EP1, EP2, and EP3 illustrate.
Are functional dependencies ... based on ... domain knowledge of the attributes?
Emphatically, yes! That's all they're based on. The DBMS is just a logic machine. The ideas of "employee" and "hours" don't exist for it. The database designer chooses to define tables that model some real-world universe of discourse, and imposes meaning on the columns. He gives names to the attributes (above) in X and Y. He decides which columns serve to identify a row based on what is true about the universe being modeled.
if a table has a composite primary key, regardless of the functional dependencies is not in 2NF?
No. Remember, 2NF is defined in terms of FDs. What could it mean to speak of conforming to 2NF "regardless" of them?
The number of columns in the key is immaterial. It's some set, X, identifying the complement, Y.
I'm not sure if I thoroughly understand your questions, but I'll give a try to explain.
Your first statement about 2NF:
a relation is in 2NF if one cannot find a proper subset on the left hand side of a functional dependency which defines the NPA
is correct, as well as your supposition
if {SSN,Pnumber} -> hours and SSN -> hours then this relation wont be in 2NF
because what that means that you could determine 'hours' from 'SSN' alone, so using the composite key {SSN,Pnumber} to determine 'hours' will be redundant, and thus violates the 2NF requirements.
What you call the left hand side of an FD is usually called a key. You use the key to find the related data. In order to save space (and reduce complexity), you should always try to find a minimal key, and break up larger tables into smaller ones if possible, so you do not have to save information in more places than necessary. This is what normalization to the normal forms is all about, and being studied for about half a century now, substantial theory on the matter has been developed, and some rules chrystalized from it, like 1NF, 2NF, 3NF etc.
Your second question confuses me a lot, because from what you are saying, it seems you already understands this.
Could there be some confusion about the FD's? From the figure, it seems to me as they are defined like this:
{SSN,Pnumber} -> hours
{SSN} -> ename
{Pnumber} -> Pname,Plocation
Just like the three lower tables are modeled, together they add up to the relation (table) modeled above.
So, in the first table, you would need the composite key {SSN,Pnumber} to access any data in the relation (search in the table), while that clearly is not necessary for most of the fields.
Now, I'm not sure about what purpose that table would fulfill in real life. While that is not formally necessary, as long as the FD's are given, it might be easier to imagine why the design will benefit from normalization.
So let's day it's about recording workhours per emplyee per project in some organization. SSN identifies the employee, (whose name also is stored as ename because it is easier to remember, but could be duplicate), Pnumber identifies the project, which name and location is also stored much for the same reason.
Then if you as a manager need to register that an employee worked another few hours on some project, you would use your manager app on your device, which in turn will update the tables seamlessly (you cannot expect managers to understand the logics of normalization)
Behind the scenes, however, it would amount to some query, in SQL that would be an 'INSERT' statement which added another row to the relevant table(s).
Now you can see that in the above table, you would have to insert all the six attributes, while with the normalized tables below, you will only need to add a row to table EP1,consisting of three attributes. In a large organization with thousands of employees delivering their worksheets every week, that will quickly become huge differences in storage requirements. That has a number of benefits, perhaps the most significant beeing search speed.
Your third question I don't understand at all, I'm afraid. In a way you could say FD's are predetermined once you have decided what data you will save in your database. The FD's are not dupposed to change. When modeled in the DB, they will not change. If you later find you will alter the design, then that will be new relations with new FD's.
The text you seem to be quoting from somewhere simply says that if you have the FD X -> Y (X gives or determines Y) then if you have any two tuples (records) in that relation (table) that have the same value of X, they must also hve the same value of Y. Or in our example, if Pnumber somewhere is given the value of 888, Pname is 'Battleship' and Plocation is 'Kitchen Sink', then if somewhere else (some other record) the Pnumber 888 is used then also there Pname must be 'Battleship' and Plocation must be 'Kitchen Sink' because Pname and Plocation is functionally dependant on Pnumber.
Now that was almost another chapter in your textbook, or what? Hope it helps, because it took me some time to write :-)
A table can be said to be in 2NF, if the primary key is composed of multiple columns, and that if for each row these columns were concatenated together into a single string, then the resulting column would qualify as the primary key. Alternatively a single column primary key will also qualify as 2NF.
In this case the same employee could have multiple phone numbers (PNUMBER), so a you cannot have a compound primary key that includes the phone number.
Functional dependencies are the attributes that their values are determined in a unique way by another attribute.Given that, can a multivalued attribute be dependent upon a primary key?
"FDs are the attributes that their values are determined in a unique way by another attribute" is unintelligible. Find a way to say it correctly or how can you understand it?
An attribute (or set of attributes) is functionally determined by a set of attributes.
There is no such thing as a "multivalued attribute" in a relation. A tuple has an attribute value for each attribute name. (Maybe you mean, a set of attributes is being determined? Maybe you mean, a multi-valued dependency?) If you have an attribute that you consider to contain multiple parts, ie you want to generically query about the parts without using operators with parameters of their types, then it is usually good design to have a separate table with attributes for those parts. But that's not addressed by normalization. Any value can be considered to have multiple parts in multiple ways and it is your application/queries that determine when you stop making tables whose attributes are the values of parts of other values and just have an attribute for a value. Similarly, if you have a bunch of attributes that play a similar role (often with similar names) then it is usually good design to have a separate table with just one attribute for the role. But that's not addressed by normalization.
Candidate keys matter to FDs, MVDs, JDs and normalization. PKs don't. You can pick one CK as "the PK" but its primariness is irrelevant to the relational model. It might be relevant to some information modeling method or product.
Superkeys are sets of columns that determine every column. Since every set of attributes always determines the attributes in it, superkeys are sets of columns that determine every other column. CKs are superkeys that contain no smaller superkey. (So CKs are sets of columns that are unique but contain no smaller set of columns that are unique.)
You don't know all the CKs until you find all the FDs. But you might know that a particular set of attributes is unique and has no smaller unique set, so that you know that it is a CK and you can call it "the PK". (Eg an id attribute in a relation variable that can have more than one row.)
can a multivalued attribute be dependent upon a primary key?
Every attribute is dependent on every CK by definition of CK. So every attribute is dependent on every PK by definition of PK.(But you must clarify what you mean by "multivalued attribute" and "dependent".)