miniworld
Requirements & collection analysis
Database Requirements
Conceptual Design
Conceptual Schema ( ER diagram )
DBMS independent
DBMS specific
Data Model Mapping
Conceptual Schema ( Relations )
• primary key constraint
• foreign key constraint
refinement
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
The Evils of Redundancy
• Redundancy is at the root of several
problems associated with relational
schemas:
– redundant storage
– Insertion/update/deletion anomalies
Example
•
Schema
–
•
Hourly_Emps (ssn, name, lot, rating, hrly_wages, hrs_worked)
Constraints:
1.
2.
ssn is the primary key
If two tuples have the same value on rating, they have the
same value on hrly_wages
SSN
Name
H
Attishoo
Lot Rd W
d
48 8 10
123-22-3666
231-31-5368
Smiley
22
8
10
30
131-24-3650
Smethurst
35
5
7
30
434-26-3751
Guldu
35
5
7
32
612-67-4134
Madayan
35
8
10
40
40
•Solution: Decomposition
If we break Hourly_Emps into Hourly_Emps2 and Wages,
then we don’t have updates, insertion, deletion anomalies.
Wages
Hourly_Emps2
S
N
L
R H
123-22-3666
Attishoo
48 8
40
231-31-5368
Smiley
22 8
30
131-24-3650
Smethurst 35 5
30
434-26-3751
Guldu
35 5
32
612-67-4134
Madayan
35 8
40
R W
8
10
5
7
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
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
 t1, t 2  r( R),
where r(R) is an instance of R,
t1[ X ]  t 2[ X ]  t1[Y ]  t 2[Y ]
X  R and Y  R
XY: Y is functionally dependent on X, or
X uniquely determines Y or
X functionally determines Y, or
X determines Y
X
X1
X1
X2
Y
Y2
Y2
Y2
Z
Z1
Z2
Z3
X
X1
X1
X1
Y
Y1
Y1
Y2
Z
Z1
Z2
Z1
Does this data set violate
X->Y?
Does this data set violate
Z->Y?
Does this data set violate
X->Y?
Does this data set violate
XY->Z?
Does this data set violate
Z->X?
• 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
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
Example 2
since
dname
name
ssn
did
lot
Employees
Works_for
budget
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);
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}
Dependency Reasoning
A set of dependencies may imply some
additional dependencies.
Some important questions
1. Given a set of attributes X, what attributes can
be determined by X
2. Given an FD set, what other dependencies are
implied
3. Given an FD set F, what is the minimum set of
dependencies that is equivalent to F
Armstrong’s Axioms
• Armstrong’s
attributes:
Axioms where X, Y, Z are sets of
–Reflexivity:
If X  Y, then XY.
–Augmentation:
–Transitivity:
If XY, then XZ YZ for any Z.
If X Y and YZ, then XZ.
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.
PROOFS (Cont’d)
•Transitive rule: If XY and YZ, then XZ.
Let XY and
YZ
t1, t 2  r ( R) such that t1[X]=t2[X],
we have:
(1)
t1[Y]=t2[Y]
(2)&(4) t1[Z]=t2[Z]
(3)&(5) XZ
(1)
(2)
(3)
(4)
(5)
PROOFS (Cont’d)
•Augmentation rule: If XY, then XZYZ.
Assume that the Augmentation rule is not true.
 t1, t 2  r ( R)
t1[X] = t2[X]
t1[Y] = t2[Y]
t1[XZ] = t2[XZ]
t1[YZ] != t2[YZ]
(1)&(3)
(2)&(5)


(1)
(2)
(3)
(4)
t1[Z]=t2[Z]
(5)
t1[YZ]=t2[YZ]
(6)
(6) Contradicts (4)
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.
PROOFS (Cont’d)
•Union rule: If XY and XZ, then XYZ.
Given XY and
XZ.
(1)
(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.
PROOFS (Cont’d)
•Decomposition rule: If XYZ then XY and XZ.
Given XYZ.
Since Y  YZ, reflexive rule gives
YZY.
(1)
(2)
Applying Transitive rule on (1) and (2), we have XY.
XZ is derived in a similar way.
PROOFS (Cont’d)
•Pseudotransitive rule: If XY and WYZ, then WXZ.
Given XY
and WYZ.
(1)
(2)
Applying Augmentation rule on (1), we have
WXWY.
(3)
Applying Transitive rule on (3)&(2), we have
WXZ.
Exercise
•
Prove or disprove the following inference rules
1.
2.
3.
4.
•
•
{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