Lecture 6:
Schema refinement: Functional
dependencies
www.cl.cam.ac.uk/Teaching/current/Databases/
1
Recall: Database design
lifecycle
• Requirements analysis
– User needs; what must database do?
• Conceptual design
– High-level description; often using E/R model
• Logical design
Next
two
• Schema refinement
lectures
– Check schema for redundancies and anomalies
– Translate E/R model into relational schema
• Physical design/tuning
– Consider typical workloads, and further optimise
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Today’s lecture
• Why are some designs bad?
• What’s a functional dependency?
• What’s the theory of functional
dependencies?
• (Next lecture: How can we use this theory
to classify redundancy in relation design?)
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Not all designs are
equally good
• Why is this design bad?
Data(sid,sname,address,cid,cname,grade)
• Why is this one preferable?
Student(sid,sname,address)
Course(cid,cname)
Enrolled(sid,cid,grade)
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An instance of our bad
design
sid sname
addres ci
s
d
124 Britney
USA
206 Database
A++
204
124
206
124
Essex
USA
London
USA
202
201
206
202
C
A+
BB+
Victoria
Britney
Emma
Britney
cname
Semantics
S/Eng I
Database
Semantics
grade
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Evils of redundancy
• Redundancy is the root of many problems
associated with relational schemas
– Redundant storage
– Update anomalies
– Insertion anomalies
– Deletion anomalies
– LOW TRANSACTION THROUGHPUT
• In general, with higher redundancy, if
transactions are correct (no anomalies),
then they have to lock more objects thus
causing greater contention and lower
throughput
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Decomposition
•
We remove anomalies by replacing the schema
Data(sid,sname,address,cid,cname,grade)
with
Student(sid,sname,address)
Course(cid,cname)
Enrolled(sid,cid,grade)
•
•
Note the implicit extra cost here
Two immediate questions:
1. Do we need to decompose a relation?
2. What problems might result from a decomposition?
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Functional dependencies
• Recall:
– A key is a set of fields where if a pair of tuples
agree on a key, they agree everywhere
• In our bad design, if two tuples agree on
sid, then they also agree on address,
even though the rest of the tuples may not
agree
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Functional dependencies
cont.
• We can say that sid determines address
– We’ll write this
sid  address
• This is called a functional dependency
(FD)
• (Note: An FD is just another integrity
constraint)
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Functional dependencies
cont.
• We’d expect the following functional
dependencies to hold in our Student
database
– sid  sname,address
– cid  cname
– sid,cid  grade
• A functional dependency X  Y is simply a
pair of sets (of field names)
– Note: the sloppy notation A,B  C,D rather
than {A,B}  {C,D}
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Formalities
• Given a relation R=R(A1:1, …, An:n), and
X, Y ({A1, …, An}), an instance r of R
satisfies XY, if
– For any two tuples t1, t2 in R, if t1.X=t2.X then
t1.Y=t2.Y
• Note: This is a semantic assertion. We
can not look at an instance to determine
which FDs hold (although we can tell if the
instance does not satisfy an FD!)
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Properties of FDs
• Assume that X  Y and Y  Z are known to
hold in R. It’s clear that X  Z holds too.
• We shall say that an FD set F logically implies
X  Y, and write F [X  Y
– e.g. {X  Y, Y  Z} [ X  Z
• The closure of F is the set of all FDs logically
implied by F, i.e.
F+ @ {XY | F [ XY}
• The set F+ can be big, even if F is small 
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Closure of a set of FDs
• Which of the following are in the closure of
our Student FDs?
– addressaddress
– cidcname
– cidcname,sname
– cid,sidcname,sname
13
Candidate keys and FDs
• If R=R(A1:1, …, An:n) with FDs F and
X{A1, …, An}, then X is a candidate key
for R if
– X  A1, …,An  F+
– For no proper subset YX is
Y  A1, …,An  F+
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Armstrong’s axioms
• Reflexivity: If YX then F \ XY
– (This is called a trivial dependency)
– Example: sname,addressaddress
• Augmentation: If F \ XY then
F \ X,WY,W
– Example: As cidcname then
cid,sidcname,sid
• Transitivity: If F \ XY and F \ YZ then
F \ XZ
– Example: As sid,cidcid and
cidcname, then sid,cidcname
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Consequences of
Armstrong’s axioms
• Union: If F \ XY and F \ XZ then
F \ XY,Z
• Pseudo-transitivity: If F \ XY and
F \ W,YZ then F \ X,WZ
• Decomposition: If F \ XY and ZY then
F \ XZ
Exercise: Prove that these are
consequences of Armstrong’s axioms
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Proof of Union Rule
Suppose that F \ XY and F \ XZ.
By augmentation we have
F \ XX,Y
since X U X = X. Also by augmentation
F \ X,YZ,Y
Therefore, by transitivity we have
F \ XZ,Y
QED
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Functional Dependencies Can be
useful in Algebraic Reasoning
Suppose R(A,B,C) is a relation schema
with dependency AB, then
R   A, B ( R)
A
 A,C ( R)
(This is called Heath’s rule.)
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Proof of Heath’s Rule
First show that
A
 A,C ( R)
Suppose
then
and
Since
we have
A
 A,C ( R)
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Proof of Heath’s Rule (cont.)
In the other direction, we must show that
Suppose
A
Then there must exist records
and
There must also exist
so that
But the functional dependency tells us that
Therefore, we have
QED
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Equivalence
• Two sets of FDs, F and G, are said to be
equivalent if F+=G+
• For example:
{(A,BC), (AB)} and
{(AC), (AB)}
are equivalent
• F+ can be huge – we’d prefer to look for
small equivalent FD sets
21
Minimal cover
• An FD set, F, is said to be minimal if
1. Every FD in F is of the form XA, where A is
a single attribute
2. For no XA in F is F-{XA} equivalent to F
3. For no XA in F and ZX is
(F-{XA}){ZA} equivalent to F
• For example, {(AC), (AB)} is a
minimal cover for {(A,BC), (AB)}
22
More on closures
• FACT: If F is an FD set, and XYF+ then
there exists an attribute AY such that
XAF+
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Why Armstrong’s axioms?
• Soundness
– If F \ XY is deduced using the rules, then
XY is true in any relation in which the
dependencies of F are true
• Completeness
– If XY is is true in any relation in which the
dependencies of F are true, then F \ XY can
be deduced using the rules
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Soundness
• Consider the Augmentation rule:
– We have XY, i.e. if t1.X=t2.X then t1.Y=t2.Y
– If in addition t1.W=t2.W then it is clear that
t1.(Y,W)=t2.(Y,W)
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Soundness cont.
Consider the Transitivity rule:
– We have XY, i.e. if t1.X=t2.X then t1.Y=t2.Y
(*)
– We have YZ, i.e. if t1.Y=t2.Y then t1.Z=t2.Z
(**)
– Take two tuples s1 and s2 such that s1.X=s2.X
then from (*) s1.Y=s2.Y and then from (**)
s1.Z=s2.Z
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Completeness
• Exercise
– (You may need the fact from slide 23)
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Attribute closure
• If we want to check whether XY is in a closure of the
set F, could compute F+ and check – but expensive 
• Cheaper: We can instead compute the attribute
closure, X+, using the following algorithm:
closure:= X;
repeat until no change{
if UVF, where Uclosure
then closure:=closureV
};
• Then F \ XY iff Y is a subset of X+
Try this with sid,snamecname,grade
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Preview of next lecture:
Goals of normalisation
• Decide whether a relation is in “good form”
• If it is not, then we will “decompose” it into
a set of relations such that
– Each relation is in “good form”
– The decomposition has not lost any
information that was present in the original
relation
• The theory of this process and the notion
of “good form” is based on FDs
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Summary
You should now understand:
• Redundancy and various forms of
anomalies
• Functional dependencies
• Armstrong’s axioms
Next lecture: Schema refinement:
Normalisation
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