functional dependencies. farkascsce 5202 reading and exercises database systems- the complete book:...
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Functional Dependencies
Farkas CSCE 520 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 2002 -- Stanford
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Database Design Goal:
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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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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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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Example - FD
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
Pepper G.S. 1 70 04/17/02 Little Creek
Panka Vizsla 12 40 02/14/02 White Oak
Functional Dependencies: Name,Breed AgeName,Breed Weight
Dog-Kennels(Name,Breed,Age,Weight,Date,Kennel)
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FD with Multiple Attributes
Right side: can be more than 1 attribute – splitting/combining rule E.g., FD: Name, Breed Age
FD: Name, Breed Weightcombine into:
FD: Name, Breed Age,Weight
Left side cannot be decomposed!
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FD EquivalenceLet 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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Trivial FD
Given FD of the form A1,A2,…,AnB1,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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Keys and FD
K is a (primary) key for a relation R if1. K functionally determines all attributes in R2. 1 does not hold for any proper subset of K
Superkey: 1 holds, 2 does not hold
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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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Where do Keys Come From?
Assert a key K, then only FDs areK A for all attributes A (K is the only key from FDs)
Assert FDs and deduce the keysE/R gives FDs from entity set keys and many-one relationships
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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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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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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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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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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?
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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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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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Armstrong Axioms Reflexivity
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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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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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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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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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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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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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: AB, AC, BC Projection AC: A C