Identifying Functional Dependencies II - database

Question

Here is an example which should clear things up for the last post.
hireDate & carReg are the primary keys. Are there extra functional dependencies (FDs) other than the ones I have identified
below? Modifications also welcome:
fd1 carReg -> make, model, outletNo, outletLoc
fd2 custNo -> custName
fd3 outletNo -> outletLoc
fd4 model -> make (only if we assume a model name is unique to a make)
fd5 carReg, hireDate -> make, model, custNo, custName, outletNo, outletLoc
I'm not sure if the above are correct and I am sure there are more.
Based on Mike Sherrill Cat Recall's answer... My question is this: How is custName -> custNo a valid FD? For the above relation, sure, a customer name maps onto exactly one customer number, but by intuition, we know more than one J SMith could be added to the table. If this is the case, this FD is void as it forms a 1..* relationship. Can we really say that custName -> custNo knowing this fact? Do we merely base FDs on the sample data? Or do we take into account the possible values that can be added?

Answer 1

At a glance . . .
custName -> custNo
model -> make
outletLoc -> outletNo
carReg, custNo -> hireDate
carReg, custName -> hireDate
And I'm sure there are others. The sample data isn't representative, and that's a problem when you try to determine functional dependencies from data. Let's say your sample data had only one row.
carReg hireDate make model custNo custName outletNo outletLoc
--
MS34 0GD 14/5/03 Ford Focus C100 Smith, J 01 Bearsden
FDs answer the question, "Given one value for 'x', do I know one and only one value for 'y'?" Based on that one-row set of sample data, every attribute determines every other attribute. custNo determines hireDate. hireDate determines outletLoc. custName determines model.
When sample data isn't representative, it's easy to turn up FDs that aren't valid. You need more representative sample data to weed out some invalid functional dependencies.
custName -> custNo isn't valid ('C101', 'Hen, P')
carReg, custNo -> hireDate isn't valid ('MS34 0GD', 'C100', '15/7/04')
carReg, custName -> hireDate isn't valid ('MS34 0GD', 'Hen, P', '15/8/03')
You can investigate functional dependencies in sample data by using SQL.
create table reg (
CarReg char(8) not null,
hireDate date not null,
Make varchar(10) not null,
model varchar(10) not null,
custNo char(4) not null,
custName varchar(10) not null,
outletNo char(2) not null,
outletLoc varchar(15) not null
);
insert into reg values
('MS34 OGD', '2003-05-14', 'Ford', 'Focus', 'C100', 'Smith, J', '01', 'Bearsden'),
('MS34 OGD', '2003-05-15', 'Ford', 'Focus', 'C201', 'Hen, P', '01', 'Bearsden'),
('NS34 TPR', '2003-05-16', 'Nissan', 'Sunny', 'C100', 'Smith, J', '01', 'Bearsden'),
('MH34 BRP', '2003-05-14', 'Ford', 'Ka', 'C313', 'Blatt, O', '02', 'Kelvinbridge'),
('MH34 BRP', '2003-05-20', 'Ford', 'Ka', 'C100', 'Smith, J', '02', 'Kelvinbridge'),
('MD51 OPQ', '2003-05-20', 'Nissan', 'Sunny', 'C295', 'Pen, T', '02', 'Kelvinbridge');
Does model determine make?
select distinct model
from reg
order by model;
model
--
Focus
Ka
Sunny
Three distinct models . . .
select model, make
from reg
group by model, make
order by model;
model make
--
Focus Ford
Ka Ford
Sunny Nissan
Yup. One make for each model. Based on the sample data, model -> make.
Does carReg, custName -> hireDate?
select distinct carReg, custName
from reg
order by custName;
carReg
--
MH34 BRP Blatt, O
MS34 OGD Hen, P
MD51 OPQ Pen, T
MS34 OGD Smith, J
NS34 TPR Smith, J
MH34 BRP Smith, J
Six distinct combinations of carReg and custName.
select carReg, custName, hireDate
from reg
group by carReg, custName, hireDate
order by custName;
carReg custName hireDate
--
MH34 BRP Blatt, O 2003-05-14
MS34 OGD Hen, P 2003-05-15
MD51 OPQ Pen, T 2003-05-20
MH34 BRP Smith, J 2003-05-20
NS34 TPR Smith, J 2003-05-16
MS34 OGD Smith, J 2003-05-14
Yup. One hireDate for each combination of carReg and custName. So based on the sample data, {carReg, custName} -> hireDate.

Answer 2

Well, since you asked for a second opinion, I'll give you one.
The second opinion is that the first (CatCall's) is entirely correct.
Sample data do not suffice to identify/determine functional dependencies in the data. What is needed to identify/determine functional dependencies in the data, are user requirements, descriptions/definitions of the business environment the database is intended to support, ...
Only your users can tell you, one way or another, what functional dependencies apply. (Don't interpret this as meaning that you should be telling your users that they should be telling you "what the applicable FDs are", because your users will typically not know what the term means. However, what the applicable FDs are, can still be derived from nothing else than the business specs the user provides you with.)
(PS sample data may on the contrary indeed suffice to demonstrate that a certain given FD certainly will NOT apply. But that's not your question.)

Answer 3

A FD (functional dependency) expresses a certain property of a relation value or variable. We can say that it holds for or doesn't hold for (is satisfied by or isn't satisfied by) (is true of or is not true of) a given relation value. When we say it holds or doesn't hold for a relation variable we mean it holds or doesn't hold for every possible value for the variable that can arise in an application.
Also if we are given a value and we are told that the FDs it satisfies are the FDs that a variable that could hold it satisfies then by that assumption the variable's FDs are the value's FDs. (This is sometimes called "representative data" for the variable.) But if we are just given a value that might arise for a variable then we only know that
the FDs that don't hold in the value also don't in the variable
the trivial FDs of both hold
(the ones of the form S -> subset of S)
(the ones that must hold regardless of the value, based only on the attributes)
(which must be the same for the value & the variable)
From my answer to What did I do wrong? (Find FD from table):
We say that a FD (functional dependency) expression S -> T has a
"determinant" set of attributes S and a "determined" set of
attributes T. It says that a given subtuple value for S appears in a
given relation value or variable/schema always with the same subtuple
value for T. For S -> {A} we can say S -> A. For {A} -> T we can say A
-> T.
Given a relation, we say that a FD "holds in" it or "is satisfied by"
it or "is true" in it or (sloppily) "is in" it or (sloppily) it "has"
a FD when what the FD says is true about it. Every FD that can be
expressed using attributes of a relation value/variable/schema will
either hold or not hold.
We can find all the FDs S -> T that hold in a relation by checking
every subset of the set of attributes as S with every subset of
attributes as T. There are also algorithms. FDs where S is a superset
of T must hold and are called "trivial".
We can find all the FDs S -> A that hold in a relation by checking
every subset of the set of attributes as S with every attribute as A.
There are also algorithms. (Then to find all FDs that hold: FDs S ->
{} hold trivially & whether S -> T for T with multiple elements can be
found from the FDs S -> A.)
Here are some shortcuts: A set determines itself. If S -> T then every
superset of S determines every subset of T. If S doesn't determine T
then no subset of S determines any superset of T. If a set has a
different subtuple of values in every tuple (ie it is "unique", ie it
is a superkey) (including if it is a candidate key) then it determines
every set. {} -> T when/iff every tuple has the same T subtuple value.
Given some FDs that hold, Armstrong's axioms generate all FDs that
must also hold. The latter is called the "closure" of the former. A
set of FDs that generates a certain closure is called a "cover". A
cover is "minimal" or "irreducible" when removing any FD from it gives
a set that is not a cover. A minimal/irreducible cover with every
determinant unique is "canonical".
Usually we are not asked to give a closure for all FDs that hold in a
schema, we are asked to give a canonical cover for them. In general if
we only know some FDs that hold in a schema then we don't know that
its closure is all the FDs that hold.
Assuming not every possible table value for a table variable is given, determining FDs for a table variable requires its meaning/predicate & the business rules to be given.
See my answer to Identifying functional dependencies (FDs).

Answer 4

Here's my attempt at relationships: