Schema Refinement and Normalization Zachary G. Ives University of Pennsylvania CIS 550 – Database & Information Systems October 6, 2004 Some slide content courtesy of Susan Davidson & Raghu Ramakrishnan

2 ISA Relationships: Subclassing (Structurally)  Inheritance states that one entity is a “special kind” of another entity: “subclass” should be member of “base class” name ISA People id Employees salary

3 But How Does this Translate into the Relational Model? Compare these options:  Two tables, disjoint tuples  Two tables, disjoint attributes  One table with NULLs  Object-relational databases

4 Weak Entities A weak entity can only be identified uniquely using the primary key of another (owner) entity.  Owner and weak entity sets in a one-to-many relationship set, 1 owner : many weak entities  Weak entity set must have total participation People Feeds Pets ssn name weeklyCost name species

5 Translating Weak Entity Sets Weak entity set and identifying relationship set are translated into a single table; when the owner entity is deleted, all owned weak entities must also be deleted CREATE TABLE Feed_Pets ( name VARCHAR(20), species INTEGER, weeklyCost REAL, ssn CHAR(11) NOT NULL, PRIMARY KEY (pname, ssn), FOREIGN KEY (ssn) REFERENCES Employees, ON DELETE CASCADE)

6 N-ary Relationships  Relationship sets can relate an arbitrary number of entity sets: StudentProject Advisor Indep Study

7 Summary of ER Diagrams  One of the primary ways of designing logical schemas  CASE tools exist built around ER (e.g. ERWin, PowerBuilder, etc.)  Translate the design automatically into DDL, XML, UML, etc.  Use a slightly different notation that is better suited to graphical displays  Some tools support constraints beyond what ER diagrams can capture  Can you get different ER diagrams from the same data?

8 Schema Refinement & Design Theory  ER Diagrams give us a start in logical schema design  Sometimes need to refine our designs further  There’s a system and theory for this  Focus is on redundancy of data  Causes update, insertion, deletion anomalies

9 Not All Designs are Equally Good Why is this a poor schema design? And why is this one better? Stuff(sid, name, serno, subj, cid, exp-grade) Student(sid, name) Course(serno, cid) Subject(cid, subj) Takes(sid, serno, exp-grade)

10 Focus on the Bad Design  Certain items (e.g., name) get repeated  Some information requires that a student be enrolled (e.g., courses) due to the key sidnamesernosubjcidexp-grade 1Sam570103AI570B 23Nitin550103DB550A 45Jill505103OS505A 1Sam505103OS505C

11 Functional Dependencies Describe “Key-Like” Relationships A key is a set of attributes where: If keys match, then the tuples match A functional dependency (FD) is a generalization: If an attribute set determines another, written A ! B then if two tuples agree on attribute set A, they must agree on B: sid ! name What other FDs are there in this data?  FDs are independent of our schema design choice

12 Formal Definition of FD’s Def. Given a relation schema R and subsets X, Y of R: An instance r of R satisfies FD X  Y if, for any two tuples t1, t2 2 r, t1[X ] = t2[X] implies t1[Y] = t2[Y]  For an FD to hold for schema R, it must hold for every possible instance of r (Can a DBMS verify this? Can we determine this by looking at an instance?)

13 General Thoughts on Good Schemas We want all attributes in every tuple to be determined by the tuple’s key attributes, i.e. part of a superkey (for key X  Y, a superkey is a “non-minimal” X) What does this say about redundancy? But:  What about tuples that don’t have keys (other than the entire value)?  What about the fact that every attribute determines itself?

14 Armstrong’s Axioms: Inferring FDs Some FDs exist due to others; can compute using Armstrong’s axioms:  Reflexivity: If Y  X then X  Y (trivial dependencies) name, sid  name  Augmentation: If X  Y then XW  YW serno  subj so serno, exp-grade  subj, exp-grade  Transitivity: If X  Y and Y  Z then X  Z serno  cid and cid  subj so serno  subj

15 Armstrong’s Axioms Lead to…  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 Let’s prove these from Armstrong’s Axioms

16 Closure of a Set of FD’s Defn. Let F be a set of FD’s. Its closure, F +, is the set of all FD’s: {X  Y | X  Y is derivable from F by Armstrong’s Axioms} Which of the following are in the closure of our Student-Course FD’s? name  name cid  subj serno  subj cid, sid  subj cid  sid

17 Attribute Closures: Is Something Dependent on X? Defn. The closure of an attribute set X, X +, is: X + =  {Y | X  Y  F + }  This answers the question “is Y determined (transitively) by X?”; compute X + by:  Does sid, serno  subj, exp-grade ? closure := X; repeat until no change { if there is an FD U  V in F such that U is in closure then add V to closure}

18 Equivalence of FD sets Defn. Two sets of FD’s, F and G, are equivalent if their closures are equivalent, F + = G + e.g., these two sets are equivalent: { XY  Z, X  Y } and { X  Z, X  Y }  F + contains a huge number of FD’s (exponential in the size of the schema)  Would like to have smallest “representative” FD set

19 Minimal Cover Defn. A FD set F is 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 Defn. F is a minimum cover for G if F is minimal and is equivalent to G. e.g., {X  Z, X  Y} is a minimal cover for {XY  Z, X  Y} in a sense, each FD is “essential” to the cover we express each FD in simplest form

20 More on Closures If F is a set of FD’s and X  Y  F + then for some attribute A  Y, X  A  F + Proof by counterexample. Assume otherwise and let Y = {A 1,..., A n } Since we assume X  A 1,..., X  A n are in F + then X  A 1... A n is in F + by union rule, hence, X  Y is in F + which is a contradiction

21 Why Armstrong’s Axioms? Why are Armstrong’s axioms (or an equivalent rule set) appropriate for FD’s? They are:  Consistent: any relation satisfying FD’s in F will satisfy those in F +  Complete: if an FD X  Y cannot be derived by Armstrong’s axioms from F, then there exists some relational instance satisfying F but not X  Y  In other words, Armstrong’s axioms derive all the FD’s that should hold

22 Proving Consistency We prove that the axioms’ definitions must be true for any instance, e.g.:  For augmentation (if X  Y then XW  YW ): If an instance satisfies X  Y, then:  For any tuples t 1, t 2  r, if t 1 [X] = t 2 [X] then t 1 [Y] = t 2 [Y] by defn.  If, additionally, it is given that t 1 [W] = t 2 [W], then t 1 [YW] = t 2 [YW]

23 Proving Completeness Suppose X  Y  F + and define a relational instance r that satisfies F + but not X  Y:  Then for some attribute A  Y, X  A  F +  Let some pair of tuples in r agree on X + but disagree everywhere else: x 1 x 2... x n a 1,1 v 1 v 2... v m w 1,1 w 2,1... x 1 x 2... x n a 1,2 v 1 v 2... v m w 1,2 w 2,2... XAX+ – XX+ – X R – X + – {A}

24 Proof of Completeness cont’d  Clearly this relation fails to satisfy X  A and X  Y. We also have to check that it satisfies any FD in F +.  The tuples agree on only X +. Thus the only FD’s that might be violated are of the form X’  Y’ where X’  X + and Y’ contains attributes in R – X + – {A}.  But if X’  Y’  F + and X’  X + then Y’  X + (reflexivity and augmentation). Therefore X’  Y’ is satisfied.

25 Decomposition  Consider our original “bad” attribute set  We could decompose it into  But this decomposition loses information about the relationship between students and courses. Why? Stuff(sid, name, serno, subj, cid, exp-grade) Student(sid, name) Course(serno, cid) Subject(cid, subj)

26 Lossless Join Decomposition R 1, … R k is a lossless join decomposition of R w.r.t. an FD set F if for every instance r of R that satisfies F,  R 1 (r) ⋈... ⋈  R k (r) = r Consider: What if we decompose on (sid, name) and (serno, subj, cid, exp-grade)? sidnamesernosubjcidexp-grade 1Sam570103AI570B 23Nitin550103DB550A

27 Testing for Lossless Join R 1, R 2 is a lossless join decomposition of R with respect to F iff at least one of the following dependencies is in F+ (R 1  R 2 )  R 1 – R 2 (R 1  R 2 )  R 2 – R 1 So for the FD set: sid  name serno  cid, exp-grade cid  subj Is (sid, name) and (serno, subj, cid, exp-grade) a lossless decomposition?

28 Dependency Preservation Ensures we can “easily” check whether a FD X  Y is violated during an update to a database:  The projection of an FD set F onto a set of attributes Z, F Z is {X  Y | X  Y  F +, X  Y  Z} i.e., it is those FDs local to Z’s attributes  A decomposition R 1, …, R k is dependency preserving if F + = (F R 1 ...  F R k ) + The decomposition hasn’t “lost” any essential FD’s, so we can check without doing a join

29 Example of Lossless and Dependency-Preserving Decompositions Given relation scheme R(name, street, city, st, zip, item, price) And FD setname  street, city street, city  st street, city  zip name, item  price Consider the decomposition R 1 (name, street, city, st, zip) and R 2 (name, item, price)  Is it lossless?  Is it dependency preserving? What if we replaced the first FD by name, street  city?

30 Another Example Given scheme: R(sid, fid, subj) and FD set: fid  subj sid, subj  fid Consider the decomposition R 1 (sid, fid) and R 2 (fid, subj)  Is it lossless?  Is it dependency preserving?

31 FD’s and Keys  Ideally, we want a design s.t. for each nontrivial dependency X  Y, X is a superkey for some relation schema in R  We just saw that this isn’t always possible  Hence we have two kinds of normal forms

32 Two Important 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

33 Normal Forms Compared  BCNF is preferable, but sometimes in conflict with the goal of dependency preservation  It’s strictly stronger than 3NF  Let’s see algorithms to obtain:  A BCNF lossless join decomposition  A 3NF lossless join, dependency preserving decomposition

34 BCNF Decomposition Algorithm (from Korth et al.; our book gives recursive version) result := {R} compute F+ while there is a schema R i in result that is not in BCNF { let A  B be a nontrivial FD on R i s.t. A  R i is not in F+ and A and B are disjoint result:= (result – R i )  {(R i - B), (A,B)} }

35 3NF Decomposition Algorithm Let F be a minimal cover i:=0 for each FD A  B in F { if none of the schemas R j, 1  j  i, contains AB { increment i R i := (A, B) } if no schema R j, 1  j  i contains a candidate key for R { increment i R i := any candidate key for R } return (R 1, …, R i ) Build dep.- preserving decomp. Ensure lossless decomp.

36 Summary  We can always decompose into 3NF and get:  Lossless join  Dependency preservation  But with BCNF we are only guaranteed lossless joins  BCNF is stronger than 3NF: every BCNF schema is also in 3NF  The BCNF algorithm is nondeterministic, so there is not a unique decomposition for a given schema R