1. Chapter 16: Relational Database Design and Further Dependencies
TA: Zhe Jiang
Ref: Elmasri, Navathe, Fundamentals of Database Systems, 6th, Addison Wesley, ISBN-10:

2. Outline Big picture & motivation
Simple case algorithm (part of )
Formal algorithm
Basic concepts (16.1):
General case algorithm (16.3.3)

3. Database Design Phases
Big Picture:
Database Design Phases
ER-Diagram
Relational Tables
Which choice is good?
How to guarantee it?
Formal Norm Theory

4. 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?

5. 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

6. Simple Case Decomposition Algorithm
Suppose the FD set given is already “good” minimal cover (defined later)
Approach:
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.
If none of the relation schemas in D contains a key of R, create one relation with key. (How? Introduce later)
Eliminate redundant relations.

7. 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:
R1(P,L,C); R2(L,C,A,P); R3(A,C)
Already contains key.
Remove redundant relations R1 and R3, final answer is R2(L,C,A,P).

8. 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

9. 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;

10. Basic Concepts Closure of left-hand-side under dependency set
Algorithm 16.1
Start: X+={X}
Grow X+ with new attributes determined by elements in X+
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+ ?

11. 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

12. Basic Concept Minimal Cover of dependency set F Example:
break down right-hand-side, X{A1,A2,…An} to XA1, XA2, …XAn
Try reduce size of LHS X in F, e.g. changing X into {X-B} still equivalent to F?
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?

13. Basic Concepts Algorithm to find key of R&F Example: 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?

14. Decomposition Algorithm: Exercise
FD3
FD1: Property_id Lot#, County, Area
FD2: Lot#, County Area, Property_id
FD3: AreaCounty
Simpler Version:
F={PLCA, LCAP, AC}
What is the minimal cover G?
Decompose G

15. Decomposition Algorithm Example
Simpler Version:
F={PLCA, LCAP, AC}
First Case:
Minimal cover GX:
F: {PL, PC, PA, LCA, LCP,AC}
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)

16. Decomposition Algorithm Example
Simpler Version:
F={PLCA, LCAP, AC}
Second Case:
Minimal cover GX:
F: {PL, PC, PA, LCA, LCP,AC}
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.

17. 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?