![15
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?)](https://image.slidesharecdn.com/9-normalization-191211045750/85/9-normalization-15-320.jpg)
![25
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 t1, t2 r,
if t1[X] = t2[X] then t1[Y] = t2[Y] by defn.
If, additionally, it is given that t1[W] = t2[W],
then t1[YW] = t2[YW]](https://image.slidesharecdn.com/9-normalization-191211045750/85/9-normalization-25-320.jpg)
9 normalization
- 1. ER Diagrams (Concluded), Schema Refinement, and Normalization Zachary G. Ives University of Pennsylvania CIS 550 – Database & Information Systems October 6, 2005 Some slide content courtesy of Susan Davidson & Raghu Ramakrishnan
- 2. 2 Examples of ER Diagrams Please interpret these ER diagrams: COURSESSTUDENTS Takes COURSESSTUDENTS Takes STUDENTS COURSESTakes
- 3. 3 Converting ER Relationship Sets to Tables: 1:n Relationships CREATE TABLE Teaches( fid INTEGER, serno CHAR(15), semester CHAR(4), PRIMARY KEY (serno), FOREIGN KEY (fid) REFERENCES PROFESSORS, FOREIGN KEY (serno) REFERENCES Teaches) CREATE TABLE Teaches_Course( serno INTEGER, subj VARCHAR(30), cid CHAR(15), fid CHAR(15), when CHAR(4), PRIMARY KEY (serno), FOREIGN KEY (fid) REFERENCES PROFESSORS) • “1” entity = key of relationship set: • Or embed relationship in “many” entity set: COURSES PROFESSORS Teaches
- 4. 4 1:1 Relationships If you borrow money or have credit, you might get: What are the table options? CreditReport Borrower delinquent? ssn namedebt Describesrid
- 5. 5 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
- 6. 6 But How Does thisTranslate into the Relational Model? Compare these options: Two tables, disjoint tuples Two tables, disjoint attributes One table with NULLs Object-relational databases (allow subclassing of tables)
- 7. 7 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
- 8. 8 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)
- 9. 9 N-ary Relationships Relationship sets can relate an arbitrary number of entity sets: Student Project Advisor Indep Study
- 10. 10 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?
- 11. 11 Schema Refinement & DesignTheory 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
- 12. 12 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)
- 13. 13 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 sid name serno subj cid exp-grade 1 Sam 570103 AI 520 B 23 Nitin 550103 DB 550 A 45 Jill 505103 OS 505 A 1 Sam 505103 OS 505 C
- 14. 14 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 X !Y then if two tuples agree on attribute set X, they must agree on X: sid ! name What other FDs are there in this data? FDs are independent of our schema design choice
- 15. 15 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?)
- 16. 16 GeneralThoughts 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?
- 17. 17 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 andY Z then X Z serno cid and cid subj so serno subj
- 18. 18 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
- 19. 19 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
- 20. 20 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 “isY 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}
- 21. 21 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
- 22. 22 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 Z, X Y} in a sense, each FD is “essential” to the cover we express each FD in simplest form
- 23. 23 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 = {A1,..., An} Since we assume X A1, ..., X An are in F + then X A1 ...An is in F + by union rule, hence, X Y is in F + which is a contradiction
- 24. 24 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
- 25. 25 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 t1, t2 r, if t1[X] = t2[X] then t1[Y] = t2[Y] by defn. If, additionally, it is given that t1[W] = t2[W], then t1[YW] = t2[YW]
- 26. 26 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: x1 x2 ... xn a1,1 v1 v2 ... vm w1,1 w2,1... x1 x2 ... xn a1,2 v1 v2 ... vm w1,2 w2,2... X A X+ – X R – X+ – {A}
- 27. 27 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.
- 28. 28 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)
- 29. 29 Lossless Join Decomposition R1, … Rk is a lossless join decomposition of R w.r.t. an FD set F if for every instance r of R that satisfies F, R1 (r) ⋈ ... ⋈ Rk (r) = r Consider: What if we decompose on (sid, name) and (serno, subj, cid, exp-grade)? sid name serno subj cid exp-grade 1 Sam 570103 AI 570 B 23 Nitin 550103 DB 550 A
- 30. 30 Testing for Lossless Join 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 (R1 R2) R2 – R1 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?
- 31. 31 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, FZ is {X Y | X Y F +, X Y Z} i.e., it is those FDs local to Z’s attributes A decomposition R1, …, Rk is dependency preserving if F + = (FR1 ... FRk)+ The decomposition hasn’t “lost” any essential FD’s, so we can check without doing a join
- 32. 32 Example of Lossless and Dependency-Preserving Decompositions Given relation scheme R(name, street, city, st, zip, item, price) And FD set name street, city street, city st street, city zip name, item price Consider the decomposition R1(name, street, city, st, zip) and R2(name, item, price) Is it lossless? Is it dependency preserving? What if we replaced the first FD by name, street city?
- 33. 33 Another Example Given scheme: R(sid, fid, subj) and FD set: fid subj sid, subj fid Consider the decomposition R1(sid, fid) and R2(fid, subj) Is it lossless? Is it dependency preserving?
- 34. 34 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
- 35. 35 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
- 36. 36 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
- 37. 37 BCNF Decomposition Algorithm (from Korth et al.; our book gives recursive version) result := {R} compute F+ while there is a schema Ri in result that is not in BCNF { let A B be a nontrivial FD on Ri s.t. A Ri is not in F+ and A and B are disjoint result:= (result – Ri) {(Ri - B), (A,B)} }
- 38. 38 3NF Decomposition Algorithm by Phil Bernstein, now @ MS Research Let F be a minimal cover i:=0 for each FD A B in F { if none of the schemas Rj, 1 j i, contains AB { increment i Ri := (A, B) } } if no schema Rj, 1 j i contains a candidate key for R { increment i Ri := any candidate key for R } return (R1, …, Ri) Build dep.- preserving decomp. Ensure lossless decomp.
- 39. 39 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