Functional Dependencies

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 Farkas CSCE 520

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 Farkas CSCE 520

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? Farkas CSCE 520

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

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 Farkas CSCE 520

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! Farkas CSCE 520

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. Farkas CSCE 520

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. Farkas CSCE 520

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 Farkas CSCE 520

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} Farkas CSCE 520

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 Farkas CSCE 520

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 Farkas CSCE 520

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}+ Farkas CSCE 520

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}+ Farkas CSCE 520

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. Farkas CSCE 520

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

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 Farkas CSCE 520

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. Farkas CSCE 520

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 Farkas CSCE 520

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+ . Farkas CSCE 520

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 Farkas CSCE 520

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 Farkas CSCE 520

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 Farkas CSCE 520

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 Farkas CSCE 520

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