1. 376a. Database Design Dept. of Computer Science Vassar College
Class 3 – Relational Data Model and Relational Algebra
Prof. Billibon Yoshimi
9/18/2018

2. So far Covered chapters 1, 2, and Pages 409-424
Now covering chapter 6.
Prof. Billibon Yoshimi
9/18/2018

3. MySQL Free relational database system
Available for download from
Versions available for win32, unix, and Mac OS X.
You may wish to download a copy for your dorm machine to experiment with administering the database. You'll use the Solaris version in the lab for development.
Should start using it.
Prof. Billibon Yoshimi
9/18/2018

4. The Final Project
Prof. Billibon Yoshimi
9/18/2018

5. The Relational Data Model
Need to know Domains, Tuple, Attributes and Relations.
A relation is represented by a table.
Each row is a collection of related data values. A row is also called a tuple.
Each column holds the values for a single attribute of all the entities. Each column has the same type from the domain D (of all valid values for that type).
Prof. Billibon Yoshimi
9/18/2018

6. Domain A domain is a set of atomic values (indivisible)
Each domain has a name, data type, and format.
e.g USA_phone_number, character string, (ddd)-ddd-dddd
Social_security_numbers
Grade_point_averages
(where d are decimal digits and first 3 digits are valid area code)
The domain of attribute, Ai, is written as dom(Ai)
e.g. dom(HOME_PHONE) = USA_phone_numbers
Prof. Billibon Yoshimi
9/18/2018

7. Relational Schema
Used to describe a relation, R is the name of the relation.
Degree of relation is n, the number of attributes in the schema.
R(A1, A2, … An)
Each attribute Ai is a role played by domain D in schema R.
e.g. STUDENT(Name, SSN, HomePhone, Address, Age, GPA) has degree 6.
Prof. Billibon Yoshimi
9/18/2018

8. Relational Instances Denoted by r or r( R )
r has n-tuples r = {t1, t2,… tm}
Each n-tuple t is a list of values <v1, v2, … vn> where each vi is from dom(Ai).
vi can be null.
Remember:
intension – R and extension – r( R)
r( R)  (dom(A1) X dom(A2) X … dom(An))
Prof. Billibon Yoshimi
9/18/2018

9. Example of STUDENT relation
STUDENT(Name, SSN, HomePhone, Address, Age, GPA)
STUDENT schema
STUDENT
Name
SSN
HomePhone
Address
Age
GPA
Bill Y
19 J ST 35
3.5
Felica Z
Null
23 A ST 21
3.2
Mark M
8 B ST 19
3.8
STUDENT relation
Prof. Billibon Yoshimi
9/18/2018

10. Characteristics of Relations
Tuples have no implicit order (can be ordered by other means, but implicitly are not ordered).
Values within a tuple are ordered.
Attributes are listed in a specific order.
Alternative definition of relations use <attribute,value> pairs to remove this constraint.
Prof. Billibon Yoshimi
9/18/2018

11. Values in tuples All values are atomic.
Multivalued and composite values (from earlier) must be resolved
Multivalued attributes are represented by separate relations.
Composite attributes are reduced to simple component attributes.
Prof. Billibon Yoshimi
9/18/2018

12. Relational Model Notation
Q, R, S – relations
Q, r, s – relation instances.
T, u, v – tuples
STUDENT – represents the current set of tuples in the current relation instance.
STUDENT (Name, SSN …) – represent relation schema
STUDENT.Age – attribute names qualified with relation name.
Prof. Billibon Yoshimi
9/18/2018

13. Key attributes A relation is a set of tuples. No duplication in a set.
Subsets of attributes in R can be used to maintain this constraint.
SK – is a subset of attributes for relation schema R.
t1[SK] != t2[SK] where t1 and t2 are distinct tuples.
Prof. Billibon Yoshimi
9/18/2018

14. Super key
Every set has at least one superkey (the set of all attributes)
A key of relation schema R is a minimal superkey of R.
Minimal superkey – remove any attribute and its no longer a superkey.
Key are derived from data semantic.
Key should be time invariant.
Every potential key is a candidate key.
One candiate key designated primary key (used to identify tuples in a relation.)
Prof. Billibon Yoshimi
9/18/2018

15. Relational Database Schema
A set of relation schemas.
S = {R1, R2, … R3}
Relational data base instance of a relational database schema is a set of relation instances (such that earch ri is taken from Ri in S)
DB = {r1, r2, … r3}
Prof. Billibon Yoshimi
9/18/2018

16. Two key integrity constraints
Key constraint
Entity integrity constraint
Referential integrity
Prof. Billibon Yoshimi
9/18/2018

17. Key constraints Relate to candidate keys of each relation.
Must be unique for each tuple in each relation instance
Prof. Billibon Yoshimi
9/18/2018

18. Entity integrity constraint
No primary key may be null
Can’t identify a tuple with a null key.
Prof. Billibon Yoshimi
9/18/2018

19. Referential integrity constraint
Maintains consistency between relations.
Tuples can only refer to existing tuples.
Foreign key (FK) in R1
A set of attributes (FK) in relation schema R1 having the same domain as primary key PK on relation R2. The attributes FK “refers to” R1.
FK’s value for t1 in R1 is either a PK for t2 in R2 or is null. In first case, t1 “refers to” t2.
Prof. Billibon Yoshimi
9/18/2018

20. Foreign keys Can refer to their own relation.
Diagrammatically display referential integrity using a directed graph. Draw an arrow from the foreign key to the primary key.
Prof. Billibon Yoshimi
9/18/2018

21. Foreign key example EMPLOYEE FNAME MINIT LNAME SSN BDATE ADDR SUPERSSN
DEPT_LOCATIONS
DEPT
LOCATION
DEPARTMENT
DNAME
DNUMBER
MGRSSN
MGRSTARTDATE
PROJECT
PNAME
PNUM
PLOC
DNUM
WORKS_ON
ESSN
PNO
HOURS
DEPENDENT
ESSN
DEPENDENT_NAME
SEX
BDATE
RELATIONSHIP
Prof. Billibon Yoshimi
9/18/2018

22. Foreign key example EMPLOYEE FNAME MINIT LNAME SSN BDATE ADDR SUPERSSN
DEPT
DEPT_LOCATIONS
DEPT
LOCATION
DEPARTMENT
DNAME
DNUMBER
MGRSSN
MGRSTARTDATE
PROJECT
PNAME
PNUM
PLOC
DNUM
WORKS_ON
ESSN
PNO
HOURS
DEPENDENT
ESSN
DEPENDENT_NAME
SEX
BDATE
RELATIONSHIP
Prof. Billibon Yoshimi
9/18/2018

23. Update operations on relations
Look at three basic operations, insert, delete and modify.
Verify that constraints are satisfied for these operations.
Prof. Billibon Yoshimi
9/18/2018

24. Try insert Given this relations schema, do
EMPLOYEE
FNAME
MINIT
LNAME
SSN
BDATE
ADDR
SUPERSSN
DEPT
Given this relations schema, do
Insert <‘Bill’, ‘H’, ‘Yoshimi’, ‘ ’, ’ ’, ‘Ave. B, Morgan, TX’, null, 4> into EMPLOYEE
What is the required to succeed?
Prof. Billibon Yoshimi
9/18/2018

25. On insert failure Many options Abort the transaction.
Ask user for more information (like null primary key).
If reference made to non-existant relation, ask user if they’d like to add the relation. May cascade.
Prof. Billibon Yoshimi
9/18/2018

26. Try delete Delete WORKS_ON(‘999887777’ 10)
Delete EMPLOYEE(SSN=` ’)
Delete EMPLOYEE(SSN=‘ ’)
Prof. Billibon Yoshimi
9/18/2018

27. On delete failure Reject the delete. Try cascading the delete
Change references to null or set to valid tuple.
Prof. Billibon Yoshimi
9/18/2018

28. Relational algebra Manipulating operations
Operations on sets of tuples including union, intersection, difference and Cartesian product
Operations like SELECT, PROJECT and JOIN
Prof. Billibon Yoshimi
9/18/2018

29. SELECT operation (s) s<condition>(RELATION)
-select a subset of tuples from RELATION where <condition> is satisfied.
e.g. sSSN=‘ ’(EMPLOYEE), sDEPT=5(EMPLOYEE)
Degree of result is same as R.
# of relations returned <= # of relations in R.
All relations in R are evaluated.
SELECT is commutative.
Prof. Billibon Yoshimi
9/18/2018

30. Conditions
<condition> can be a composite predicate like ((DNO=4 AND SALARY>12000) or (DNO=5 AND FNAME=‘Bill’)) using boolean operators AND, OR and NOT.
Operators include =, >, <, , , and  if ordered.
If unordered can only do comparison (= or ).
Strings can use SUBSTRING-OF
Prof. Billibon Yoshimi
9/18/2018

31. PROJECT operation (s) p<list of attributes>(RELATION)
-selects <list of attributes> columns from RELATION
e.g. pSEX, SALARY(EMPLOYEE)
PROJECT removes duplicate tuples from result set.
Degree of result is = length of <list of attributes>.
If one attribute is a key of RELATION, result will have same number of relations as RELATION.
Prof. Billibon Yoshimi
9/18/2018

32. Prof. Billibon Yoshimi
9/18/2018