View
216
Download
0
Embed Size (px)
Citation preview
Schema Refinement and Normalization
Zachary G. IvesUniversity 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
Peopleid
Employees salary
3
But How Does this Translateinto 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:
Student Project
Advisor
IndepStudy
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
sid
name
serno
subj
cid
exp-grade
1 Sam 570103
AI 570
B
23 Nitin 550103
DB 550
A
45 Jill 505103
OS 505
A
1 Sam 505103
OS 505
C
11
Functional DependenciesDescribe “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 YWserno subj so serno, exp-grade subj, exp-grade
Transitivity: If X Y and Y Z then X Zserno 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 SomethingDependent 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 attribute2. For no X A in F is:
F – {X A } equivalent to F3. For no X A in F and Z X is: F – {X A } {Z A } equivalent to FDefn. 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 expresseach FD insimplest 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 = {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
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 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]
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:
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}
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
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
27
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 nameserno cid, exp-gradecid 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, 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
29
Example of Lossless and Dependency-Preserving Decompositions
Given relation scheme R(name, street, city, st, zip, item, price)
And FD set name street, citystreet, city ststreet, city zipname, 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?
30
Another Example
Given scheme: R(sid, fid, subj)and FD set: fid subj
sid, subj fidConsider the decomposition
R1(sid, fid) and R2(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) ,oror 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 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)}}
35
3NF Decomposition Algorithm
Let F be a minimal coveri:=0for 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.-preservingdecomp.
Ensurelosslessdecomp.
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