1. primary key constraint foreign key constraint
miniworld
Requirements & collection analysis
refinement
Database Requirements
Conceptual Design
Conceptual Schema ( ER diagram )
DBMS independent
Data Model Mapping
DBMS specific
Conceptual Schema ( Relations )
primary key constraint
foreign key constraint

2. Schema Refinement and Normal Forms
Conceptual database design gives us a set of relation schemas and integrity constraints
Given a design, how do we know it is good or not?
A design can be evaluated from various perspectives, our focus is on data redundancy
Conceptual design
Schemas
ICs

3. The Evils of Redundancy
Redundancy is at the root of several problems associated with relational schemas:
redundant storage
Insertion/update/deletion anomalies

4. Example Schema Constraints:
Hourly_Emps (ssn, name, lot, rating, hrly_wages, hrs_worked)
Constraints:
ssn is the primary key
If two tuples have the same value on rating, they have the same value on hrly_wages

5. Solution: Decomposition
If we break Hourly_Emps into Hourly_Emps2 and Wages, then we don’t have updates, insertion, deletion anomalies.
Wages
Hourly_Emps2

6. Decomposition Concerns
Should a relation be decomposed?
If a relation is not in certain form, some problems (e.g., redundancy) will arise, are these problems tolerable?
Aforementioned anomalies
Potential performance loss: Queries over the original relation may required to join the decomposed relations
How to decompose a relation? Two properties must be preserved:
lossless-join: the data in the original relation can be recovered from the smaller relations
dependency-preservation: all constraints on the original relation must still hold by enforcing some constraints on each of the small relations

7. Functional Dependencies (FDs)
In a relation schema R, a set of attributes X functionally determines a set of attributes Y if and only if whenever two tuples of R agree on X value, they must necessarily agree on the Y value.
XY
where r(R) is an instance of R,
XY: Y is functionally dependent on X, or
X uniquely determines Y or
X functionally determines Y, or
X determines Y

8. X Y Z X1 Y2 Z1 Z2 X2 Z3
Does this data set violate
X->Y?
Does this data set violate Z->Y? X Y Z X1 Y1 Z1 Z2 Y2
Does this data set violate X->Y?
Does this data set violate
XY->Z?
Does this data set violate Z->X?
No, No, Yes, Yes, No

9. An FD is a statement about all allowable relations.
Must be identified based on semantics of application.
Given some allowable instance r1 of R, we can check if it violates some FD f, but we cannot tell if f holds over R!
A primary key constraint is a special case of an FD
The attributes in the key play the role of X, and the set of all attributes in the relation plays the role of Y
K is a candidate key for R means that KR.
However, KR does not require K to be minimal!

10. Example 1 Hourly_Emps (ssn, name, lot, rating, hrly_wages, hrs_worked)
Notation: We will denote this relation schema by listing the attributes: SNLRWH
This is really the set of attributes {S,N,L,R,W,H}.
Sometimes, we will refer to all attributes of a relation by using the relation name. (e.g., Hourly_Emps for SNLRWH)
Some FDs on Hourly_Emps:
ssn is the key: SSNLRWH (or {S}{S,N,L,R,W,H})
rating determines hrly_wages : RW

11. Example 2 FD: did->lot Works_for(ssn,name,did,since)
dname
lot
name
ssn
did
budget
Works_for
Employees
Departments
Additional Constraints: Employees are assigned parking lots based on their department. All employees in the same department is given the same lot.
FD: did->lot
Works_for(ssn,name,did,since)
Department (did,dname,budget,lot);

12. Dependency Reasoning
A set of dependencies may imply some additional dependencies.
EMP_DEPT(ENAME,SSN,BDATE,ADDRESS,DNUMBER,DNAME,DMGRSSN)
F={SSN->{ENAME,BDATE,ADDRESS,DNUMBER},
DNUMBER->{DNAME,DMGRSSN} }
F infers the following additional functional dependencies: F
{SSN}->{DNAME,DMGRSSN} F
{SSN}->{SSN} F
{DNUMBER}->{DNAME}

13. Some important questions
Dependency Reasoning
A set of dependencies may imply some additional dependencies.
Some important questions
Given a set of attributes X, what attributes can be determined by X
Given an FD set, what other dependencies are implied
Given an FD set F, what is the minimum set of dependencies that is equivalent to F

14. Armstrong’s Axioms
Armstrong’s Axioms where X, Y, Z are sets of attributes:
Reflexivity: If X Y, then XY.
Augmentation: If XY, then XZ YZ for any Z.
Transitivity: If X Y and YZ, then XZ.

15. PROOFS Reflexive rule: If X Y, then XY. Let {t1,t2}
r(R) such that t1[X]=t2[X]
Since Y X, t1[X]=t2[X] t1[Y]=t2[Y]
XY.

16. PROOFS (Cont’d) Transitive rule: If XY and YZ, then XZ.
Let XY and (1)
YZ (2)
such that t1[X]=t2[X], (3)
we have:
(1) t1[Y]=t2[Y] (4)
(2)&(4) t1[Z]=t2[Z] (5)
(3)&(5) XZ

17. PROOFS (Cont’d) Augmentation rule: If XY, then XZYZ.
Assume that the Augmentation rule is not true.
t1[X] = t2[X] (1)
t1[Y] = t2[Y] (2)
t1[XZ] = t2[XZ] (3)
t1[YZ] != t2[YZ] (4)
(1)&(3)
t1[Z]=t2[Z] (5)
(2)&(5)
t1[YZ]=t2[YZ] (6)
(6) Contradicts (4)

18. Additional Inference Rules for Functional Dependencies
Union:
If X Y and X  Z, then X  YZ.
Decomposition:
If XYZ, then XY and XZ.
Pseudotransitive Rule:
If XY and WYZ then WXZ.

19. PROOFS (Cont’d) Union rule: If XY and XZ, then XYZ.
Given XY and (1)
XZ (2)
Applying Augmentation rule on (1), we have
XXXY XXY (3)
Applying Augmentation rule on (2), we have
XYZY XYYZ (4)
Applying Transitive rule on (3) and (4), we have
XYZ.

20. PROOFS (Cont’d) Decomposition rule: If XYZ then XY and XZ.
Given XYZ (1)
Since Y YZ, reflexive rule gives
YZY (2)
Applying Transitive rule on (1) and (2), we have XY.
XZ is derived in a similar way.

21. PROOFS (Cont’d) Pseudotransitive rule: If XY and WYZ, then WXZ.
Given XY (1)
and WYZ (2)
Applying Augmentation rule on (1), we have
WXWY (3)
Applying Transitive rule on (3)&(2), we have
WXZ.

22. Exercise Prove or disprove the following inference rules
{WY,XZ} |= {WXY}
{XY,XW,WYZ} |= {XZ}
{XY} |= {XYZ}
{XY, Z Y} |= {XZY}
Prove using inference rules
Disprove by showing a counter example