1. Functional Dependencies

2. Reading and Exercises
Database Systems- The Complete Book: Chapter 3.1, 3.2, 3.3., 3.4
Following lecture slides are modified from Jeff Ullman’s slides for Fall Stanford
Farkas
CSCE 520

3. Database Design Goal: Anomalies: Represent domain information
Avoid anomalies
Avoid redundancy
Anomalies:
Update: not all occurrences of a fact are changed
Deletion: valid fact is lost when tuple is deleted
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CSCE 520

4. Functional Dependencies
FD: X  A for relation R
X functional determines A, i.e., if any two tuples in R agree on attributes X, they must also agree on attribute A.
X: set of attributes
A: single attribute
If t1 and t2 are two tuples of r over R and t1[X]= t2[X] then t1[A]= t2[A]
What is the relation between functional dependencies and primary keys?
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CSCE 520

5. Functional Dependency Example
Owner(Name, Phone)
FD: Name  Phone
Dog(Name, Breed, Age, Weight)
FD: Name, Breed  Age
FD: Name, Breed  Weight
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CSCE 520

6. Example - FD Dog-Kennels(Name,Breed,Age,Weight,Date,Kennel)
Pepper
G.S. 1 70
01/01/02
White Oak
Buddy
Mix 4 50
03/04/01
Little Creek
04/17/02
Panka
Vizsla 12 40
02/14/02
Functional Dependencies: Name,Breed  Age
Name,Breed  Weight
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CSCE 520

7. FD with Multiple Attributes
Right side: can be more than 1 attribute – splitting/combining rule
E.g., FD: Name, Breed  Age
FD: Name, Breed  Weight
combine into:
FD: Name, Breed  Age,Weight
Left side cannot be decomposed!
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CSCE 520

8. FD Equivalence Let S and T denote two sets of FDs.
S and T are equivalent if the set of relation instances satisfying S is exactly the same as the set of instances satisfying T.
A set of FDs S follows from a set of FDs T if every relation instance that satisfies all FDs in T also satisfies all FDs in S.
Two sets of FDs S and T are equivalent if S follows from T and T follows from S.
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CSCE 520

9. Trivial FD Given FD of the form A1,A2,…,AnB1,B2,…,Bk FD is
Trivial: if the Bs are subset of As
Nontrivial: if at least one of the Bs is not among As
Completely nontrivial: if none of the Bs is in As.
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CSCE 520

10. Keys and FD K is a (primary) key for a relation R if
K functionally determines all attributes in R
1 does not hold for any proper subset of K
Superkey: 1 holds, 2 does not hold
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CSCE 520

11. Example Dog-Kennels(Name,Breed,Age,Weight,Date,Kennel)
Name,Breed,Date is a key:
K={Name,Breed,Date} functionally determines all other attributes
The above does not hold for any proper subset of K
What are?
{Name,Breed,Kennel}
{Name,Breed,Date,Kennel}
{Name,Breed,Age}
{Name,Breed,Age,Date}
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CSCE 520

12. Where do Keys Come From? Assert a key K, then only FDs are
K  A for all attributes A (K is the only key from FDs)
Assert FDs and deduce the keys
E/R gives FDs from entity set keys and many-one relationships
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CSCE 520

13. E/R and Relational Keys
E/R keys: properties of entities
Relation keys: properties of tuples
Usually: one tuple corresponds to one entity
Poor relational design: one entity becomes several tuples
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CSCE 520

14. Closure of Attributes
Let A1,A2,…,An be a set of attributes and S a set of FDs. The closure of A1,A2,…,An under S is the set of attributes B such that every relation that satisfies S also satisfies A1,A2,…,An  B.
Closure of attributes A1,A2,…,An is denoted as {A1,A2,…,An}+
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CSCE 520

15. Algorithm – Attribute Closure
Let X = A1,A2,…,An
Find B1,B2,…,Bk  C such that B1,B2,…,Bk all in X but C is not in X
Add C to X
Repeat until no more attribute can be added to X
X= {A1,A2,…,An}+
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CSCE 520

16. Closures and Keys
{A1,A2,…,An}+ is a set of all attributes of a relation if and only if A1,A2,…,An is a superkey for the relation.
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CSCE 520

17. Projecting FDs Some FD are physical laws
E.g., no two courses can meet in the same room at the same time
A professor cannot be at two places at the same time.
How to determine what FDs hold on a projection of a relation?
Farkas
CSCE 520

18. FD on Relation Relation schema design: which FDs hold on relation
Given: X1  A1, X2  A2, …, Xn  An whether Y  B must hold on relations satisfying X1  A1, X2  A2, …, Xn  An
Example: A  B and B  C, then A C must also hold
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CSCE 520

19. Inference Test Test whether Y  B
Assume two tuples agree on attributes Y
Use FDs to infer these tuples also agree on other attributes
If B is one of the “other” attributes, then Y  B holds.
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CSCE 520

20. Armstrong Axioms Reflexivity Augmentation Transitivity
If {A1,A2,…,Am} superset of {B1,B2,…,Bn} then A1,A2,…,Am  B1,B2,…,Bn
Augmentation
If A1,A2,…,Am  B1,B2,…,Bn then A1,A2,…,Am,C1,…,Ck  B1,B2,…,Bn,C1,…,Ck
Transitivity
If A1,A2,…,Am  B1,B2,…,Bn and B1,B2,…,Bn  C1,C2,…,Ck then A1,A2,…,Am  C1,C2,…,Ck
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CSCE 520

21. FD Closure Compute the closure of Y, denoted as Y+ Basis: Y+ = Y
Induction: look FD, where left side X is subset of Y+ . If FD is X  A then add A to Y+ .
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CSCE 520

22. Finding All FDs
Normalization: break a relation schema into two or more schemas
Example:
R(A,B,C,D)
FD: AB C, C D, D  A
Decompose into (A,B,C), (A,D)
FDs of (ABC): A,B C and C  A
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CSCE 520

23. Basic Idea What FDs hold on a projections:
Start with FDs
Find all FDs that follow from given ones
Restrict to FDs that involve only attributes from a schema
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CSCE 520

24. Algorithm For each X compute X+ Add X  A for all A in X+ - X
Drop XY  A if X  A
Use FD of projected attributes
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CSCE 520

25. Tricks
Do not compute closure of empty set of the set of all attributes
If X+ = all attributes, do not compute closure of the superset of X
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CSCE 520

26. Example ABC with A  B, B  C and projection on AC:
A+ = ABC, yields A  B and A  C
B+ = BC, yields B  C
C+ = C, yields nothing
BC+ = BC, yields nothing
Resulting FDs: AB, AC, BC
Projection AC: A  C
Farkas
CSCE 520