1. Handout 4 Functional Dependencies
CIS 550
Handout 4
Functional Dependencies
CIS550 Handout 4

2. Why we need relational design theory
We don’t need it to design databases
ER diagrams and related tools are much more understandable and effective.
The theory is useful
as a check on our designs
to understand certain things that ER diagrams cannot do
to help us understand the consequences of redundancy (which we may use for efficiency)
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3. Not all designs are equally good
Why is this design bad?
And why is this one preferable?
Data(Id#, Name, Address, C#, Description, Grade)
Student(Id#, Name, Address)
Course(C#, Description)
Enrolled(Id#, C#, Grade)
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4. An example of “bad” design
Name
Jones
Smith
Brown
Address
Phila
NYC
Boston C#
Phil7
Math8
Eng12
Description
Plato
Topology
Chaucer
Grade A B C
Id#
124
456
789
Information is redundantly given. E.g. Name and Address
Some information, e.g., course information depends on the existence of some student.
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5. Functional Dependencies
Recall that a key is a set of attribute names. If two tuples agree on a key they agree everywhere -- they are the same.
In our “bad” design, if two tuples agree on Id#, they agree on Address, even though they are not the same.
We can say “Id# determines Address” -- written Id#  Address
This is a functional dependency
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6. Here are some functional dependencies that we expect to hold in our student-course database
Id#  Name, Address
C#  Description
Id#,C#  Grade
Note that any relation (good or bad design) should be constrained by these dependencies
A functional dependency X  Y is simply a pair of sets. Notice the “sloppy” notation A,B  C,D or AB  CD rather than {A, B}  {C,D }
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7. The Meaning of fd’s
Defn. Given a relation scheme R (a set of attributes) and subsets X,Y of R, an instance r of R satisfies X  Y if, for any two tuples t1, t2 in R, t1[X ]=t2[X ] implies t1[Y ] = t2[Y ]
N.B. We cannot look at a relation to determine which fd’s hold (we can tell if an it doesn’t satisfy an fd.
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8. Basic Intuition in Relational Design
A database scheme is “good” if all fd’s are of the form K  R where K is a key for R
Example: Our “bad” design is bad because, for example Id#  Address is not a key for the relation scheme in which these attributes occur.
However, it isn’t as simple as as this. A  A is a functional dependency for any attribute A. Are all attributes keys??
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9. Armstrong’s Axioms
Some fd’s occur as consequences of others These can be deduced by Armstrong’s axioms:
Reflexivity. If Y  X then X Y
(These are called trivial dependencies).
Example: Name, Address -> Address
Augmentation. If X  Y then XW  YW
Example: From C#  Description we deduce
C#,Id#  Description, Id#
Transitivity. If X  Y and Y  Z then X  Z
Example: From Id#,C#  C# and C#  Description, we deduce Id#,C#  Description
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10. Consequences of Armstrong’s Axioms
Union. If X Y and X  Z then X  YZ
Pseudotransitivity. If X  Y and WY  Z then XW  Z
Decomposition. If X  Y and Z  Y then X  Z
Prove these from Armstrong’s Axioms.
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11. {X  Y | X  Y can be deduced from F by Armstrong’s Axioms}
Closure of a set of fd’s
Defn. Let F be a set of fd’s. The closure of F, F + is the set of fd’s
{X  Y | X  Y can be deduced from F by Armstrong’s Axioms}
Which of the following are in the the closure of our Student-Course fd’s?
Address  Address
C#  Description
C#  Description, Name
C#, Id#  Description, Name
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12. Equivalence of fd sets
Defn. Two sets of fd’s, F and G, are equivalent if F + = G +
Example:
{AB  C, A  B } and
{A  C, A B }
are equivalent.
F + contains a huge number of fd’s (exponential in the size of the scheme). One naturally looks for small equivalent fd sets
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13. Minimal Cover Defn. A fd set F is minimal if
1. Every fd in F is of the form where A is a (single) attribute,
2. For no X  A 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.
Example (from previous slide)
{A  C, A B } is a minimal cover for
{AB  C, A  B }
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14. More on closures
Fact. If F is a set of fd’s and X  Y  F + then there exists an attribute A s.t. X A  F +.
Proof. Assume otherwise Let Y = {A1,..., An}. Then X  A1, ..., X  An are in F + . Therefore X  A1 ... An is in F +, i.e., X  Y is in F +
Notation: F (X ) for  {Y | X Y  F +}
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15. Why Armstrong’s Axioms?
Why are Armstrong’s axioms (or an equivalent rule set) appropriate for fd’s?
They are consistent and complete
“Consistent” means that any relation that satisfies the fd’s in F will satisfy the fd’s in F +
“Complete” means that if an fd X  Y cannot be derived by Armstrong’s axioms from F. Then there’s a relational instance satisfying F but not X  Y.
In other words, Armstrongs axioms derive all the fd’s that should hold.
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16. Proof of consistency
This comes directly from the definition. Consider augmentation, for example. This says that if XY then XW  YW.
If a relation instance satisfies X  Y then for any tuples t1, t2 r. If t1[X]=t2[X] then t1[Y] = t2[Y]. If, in addition, t1[W]=t2[W] then t1[YW]=t2[YW]
(remember that we are using “sloppy” notation -- YW for YW)
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17. Proof of Completeness
To prove completeness we suppose X  Y  F + and construct a relation instance that satisfies F + but not X  Y. By our previous result, we know there is an attribute A  X such that X  A  F +. Our relation has 2 tuples. They agree on F (X ) but disagree everywhere else.
x1 x xn a1,1 v1 v vm w1,1 w2,1...
x1 x xn a1,2 v1 v vm w1,2 w2,2... X A
F(X) \ X
rest of R
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18. Proof of Completeness cont’d
It is immediate that this relation fails to satisfy XA and hence X  Y. We also have to check that it does satisfy any fd in F + .
The tuples agree on only F (X ) . Thus the only fd’s that might be violated are of the form X’  Y’ where X’  F (X ).
But if X’  Y’ F + and X’  F (X ) then Y’  F (X ) (reflexivity and augmentation). Therefore X’  Y’ is satisfied.
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19. Data(Id#, Name, Address, C#, Description, Grade)
Decomposition
Consider our attribute set
We could decompose it into
But this decomposition loses information about the relationship between students and courses. Why?
Data(Id#, Name, Address, C#, Description, Grade)
R1 (Id#, Name, Address,)
R2(C#, Description, Grade)
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20. Lossless Join Decomposition
R1, … Rk is a lossless join of R with respect to a fd set F if for every instance r of R that satisfies F,
R1 r R1 r= r
Consider
What happens if we decompose on (Id#, Name,Address) and (C#,Description, Grade)?
Name
Jones
Brown
Address
Phila
Boston C#
Phil7
Math8
Description
Plato
Topology
Grade A C
Id#
124
789
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21. Testing for lossless join
Fact. R1, R2 is a lossless join decomposition of R with respect to F iff at least one of the following dependencies is in F
(R1  R2)  R1 \ R2
Example: WRT the fd set
Id#  Name, Address
C#  Description
Id#,C#  Grade
Is (Student,Name,Address) and (Student, C#, Description, Grade) a lossless decomposition?
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22. Dependency preservation
Suppose we update a relation in a database. Can we easily check whether a fd XY is violated. We can if X Y is contained within set of attributes
The projection of an fd set F onto a set of attributes Z, FZ is
{XY | XYF + and X Y Z }
A decomposition R1, …, Rk is dependency preserving if F + = (FR1...FRk)+
This means that the decomposition hasn’t “lost” any essential fd’s
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23. {Sname, Sadd, City, Zip, Item, Price}
An example
A relation scheme
{Sname, Sadd, City, Zip, Item, Price}
A fd set Sname  Sadd, City
Sadd,City  Zip
Sname,Item  Price
Consider the decomposition
{Sname,Sadd, City,Zip} and{Sname,Item,Price}
Is it lossless?
Is it dependency preserving?
What if we replaced the first fd by
Sname, Sadd  City ?
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24. Another example The scheme: {Student, Teacher, Subject}
The fd set: Teacher  Subject
Student, Subject  Teacher
The decomposition:
{Student, Teacher} and {Teacher, Subject}
Is it lossless?
Is it dependency preserving?
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25. Fd’s and keys
Earlier we stated that the idea in relational database design (from fd’s) is to obtain a design such that for each nontrivial dependency XY , X is a super-key for some relation scheme in R
The last example shows that this cannot always be achieved in a way that preserves dependencies.
This leads to two notions of normal forms
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26. Normal forms
Boyce-Codd Normal Form (BCNF). For every relation scheme R and for every X  A that holds over R,
either A  X (it is trivial) ,or
or X is a superkey for R
Third Normal Form (3NF) For every relation scheme R and for every X  A that holds over R,
either A  X (it is trivial), or
X is a superkey for R, or
A is a member of some key for R.
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27. Normal Forms contd.
BCNF is clearly desirable, but the teacher/student/subject example shows that it is not always obtainable.
BCNF is stronger than 3NF
There are algorithms to obtain
A BCNF lossless join decomposition
A 3NF lossless join, dependency preserving decomposition
The 3NF algorithm uses a minimal cover.
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28. BCNF Decomposition Algorithm
RES:= {R} //R = set of all attributes
while there is a scheme S in RES that is not in BCNF do
begin
let A  B be a nontrivial functional dependency
that holds on S such that A  S is not in F+
and A and B are disjoint
RES:= (RES-{S})  {S-B}  {AB}
end
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29. 3NF Decomposition Algorithm
let F be a minimal cover.
RES = {}
for each A  B in F do
if none of the schemes in RES contains AB
then RES:= RES {AB}
if none of the schemes in RES contains a candidate key for R
then RES:= RES  {any candidate key for R}
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