Chapter 16: Relational Database
Design and Further Dependencies
TA: Zhe Jiang
zhe@cs.umn.edu
Ref: Elmasri, Navathe, Fundamentals of Database Systems, 6th, Addison Wesley,
ISBN-10: 0-13-608620-9.
Outline
• Big picture & motivation
• Simple case algorithm (part of 16.3.3)
• Formal algorithm
– Basic concepts (16.1):
– General case algorithm (16.3.3)
Big Picture:
Database Design Phases
ER-Diagram
Relational Tables
Which choice is good?
How to guarantee it?
Formal Norm Theory
Motivation
We have:
•Universal relational schema U(A1,A2, … An).
•A set of functional dependencies (FDs) from
domain knowledge.
Question:
How do we decompose U into sub-relations, so
as to satisfy 3NF?
Simple Case Decomposition Algorithm
• Motivation:
– Decompose universal relational schema into sub
relations which satisfy 3NF
• Properties:
– Preserve dependencies (nonlossy design)
– Non-additive join property (no spurious tuples)
– Resulting relational schemas are in 3NF
• Problem Definition:
– Input: Universal Relation R and a set of functional
dependencies F on the attributes of R
– Output: Sub-relations, FDs.
– Constraint: the three properties above
Simple Case Decomposition Algorithm
• Suppose the FD set given is already “good”
minimal cover (defined later)
• Approach:
1. For each LHS X in F, create a relation schema
in D {X U {A1} U {A2} … U {Ak} }.where
XAi only dependency with X as LHS.
2. If none of the relation schemas in D contains a
key of R, create one relation with key. (How?
Introduce later)
3. Eliminate redundant relations.
Simple Case Decomposition Algorithm
• Exercise:
– Universal relation
– FD: {PLC, LCAP, AC}
– Q: Does it satisfy 1NF, 2NF, 3NF?
– Q: How to decompose the relation to satisfy 3NF?
• Solution:
1. R1(P,L,C); R2(L,C,A,P); R3(A,C)
2. Already contains key.
3. Remove redundant relations R1 and R3, final
answer is R2(L,C,A,P).
General Case Decomposition
Algorithm
• New info: Transform the given FD set into
minimal cover
• New info: If no key exists, find key of U, then
create a relation contain key
• We will introduce some basic concepts, then
formal algorithm
Basic Concept
• Inference rules: One FD could infer another
– trivial: IR1: IR1 (reflexive rule)
• If X Y, then X Y.
– non-trivial: IR2-IR4
• {XY} |= XZYZ
• {XY, YZ} |= XZ
• {XYZ} |=XY
• Closure of set of dependencies
– Closure of F: F+, set of all FDs could be inferred.
– Use IR1 to IR3;
Basic Concepts
• Closure of left-hand-side under dependency set
Algorithm 16.1
1. Start: X+={X}
2. Grow X+ with new attributes determined by elements in X+
3. Repeat 2 until can’t grow any more.
Exercise:
Given: F={XYZ, XW, WU, YV}, U(X,Y,Z,W,U,V)
Find: X+ ?
Basic Concepts
• Equivalence of functional dependencies sets
– Definition
• Cover: F covers E if F+ contains E.
• Equivalent FD sets:
– Algorithm
• Check if all left-hand-sides’ closures are same
• Minimal Cover of dependency set F
– definition: Can’t find subset that is equivalent to F
Basic Concept
• Minimal Cover of dependency set F
1. break down right-hand-side, X{A1,A2,…An}
to XA1, XA2, …XAn
2. Try reduce size of LHS X in F, e.g. changing X
into {X-B} still equivalent to F?
3. Try reduce unnecessary FD in F, e.g. remove
XA in F, if result still equivalent to F.
• Example:
– F={PLCA, LCAP, AC}
– What is “minimal cover” of F?
Basic Concepts
• Algorithm to find key of R&F
– Start with K=R.
– Find A in R such that (K-A)+ contain all attributes.
– Repeat until size of K is as small as possible
• Example:
– U(Emp_ssn, Pno, Esal, Ephone, Dno, Pname,
Plocation)
– F={Emp_ssnEsal, Ephone, Dno; PnoPname,
plocation};
– What is the key?
Decomposition Algorithm: Exercise
FD3
FD1: Property_id Lot#, County, Area
FD2: Lot#, County Area, Property_id
FD3: AreaCounty
1. What is the minimal cover
G?
Simpler Version:
2. Decompose G
F={PLCA, LCAP, AC}
Decomposition Algorithm Example
Simpler Version:
F={PLCA, LCAP, AC}
First Case:
Minimal cover GX:
1. F: {PL, PC, PA, LCA, LCP,AC}
2. Minimal cover GX: {PLC, LCAP, AC}
Design X:
3. R1(P,L,C), R2(L,C,A,P), and R3(A,C)
4. R2(L,C,A,P)
Decomposition Algorithm Example
Simpler Version:
F={PLCA, LCAP, AC}
Second Case:
Minimal cover GX:
1. F: {PL, PC, PA, LCA, LCP,AC}
2. Minimal cover GX: {PLA, LCP, AC}
Design Y:
3. S1(P,A,L), S2(L,C,P), and S3(A,C)
4. No redundant relations.
Exercise
• Given:
– Universal relation U(A,B,C,D,E,F,G,H,I,J)
– Functional dependencies F={ {A,B}{C},
{B,D}{E,F}, {A,D}{G,H}, {A}{I},
{H}{J} }.
– Decompose it into 3NF?