1. Lecture 2 Relational Database
Reason of RDB's popularity: simplicity, easy to use, properties of rob, tables for entity and their relation
Concepts, relational database

2. Outline Relational Data Model Schema and Instance
Mathematica Fundament
select: s
project: Õ
union: È
set difference: –
Cartesian product: x
rename: r

3. Relational Data Model R

4. Example of a Relation attributes (or columns) tuples (or rows)
Relations, attributes, tuples, ids, the semantics of a row as well as a column, take the table as an example
A tuple is a sequence of values
Similar to instructors, departments….

5. Relationship between two entities
Related tuples(tables)

6. Attribute Types
The set of allowed values for each attribute is called the domain of the attribute
Attribute values are (normally) required to be atomic; that is, indivisible
The special value null is a member of every domain. Indicated that the value is “unknown” or does not exist
The null value causes complications in the definition of many operations
domain: Permitted values for a column
Atomic: depends the real application, for example, Address
Null: causes a lot of difficulties from operations tables, should be eliminated

7. Relation Schema and Instance
A1, A2, …, An are attributes
R = (A1, A2, …, An ) is a relation schema
Example:
instructor = (ID, name, dept_name, salary)
Formally, given sets D1, D2, …. Dn a relation r is a subset of D1 x D2 x … x Dn
Thus, a relation is a set of n-tuples (a1, a2, …, an) where each ai Î Di
Database Schema: the logical design of the database and consists of a list of attributes and their corresponding domains ,does not generally change
database instance, which is a snapshot of the data in the database at a given instant in time, change with time
Relation between tuples
The current values (relation instance) of a relation are specified by a table
An element t of r is a tuple, represented by a row in a table

8. Relations are Unordered
Order of tuples is irrelevant (tuples may be stored in an arbitrary order)
Example: instructor relation with unordered tuples
Tuples is in set, the properties of a set
the order in which the tuples appear in a relation is irrelevant, since a relation is a set of tuples. Any two relations, if they have the same schema and same tuples, they are the same as long as they have the same set of tuples whatever their order is.

9. Keys
Let K Í R
K is a superkey of R if values for K are sufficient to identify a unique tuple of each possible relation r(R)
Example: {ID} and {ID,name} are both superkeys of instructor.
Superkey K is a candidate key if K is minimal Example: {ID} is a candidate key for Instructor
One of the candidate keys is selected to be the primary key
which one?
Foreign key constraint: Value in one relation must appear in another
referencing relation
referenced relation
example – dept_name in instructor is a foreign key from instructor referencing department
referential integrity constraint;
Tuples in the same tables should be distinguishable.
A superkey is a set of one or more attributes that, taken collectively, allow us to identify uniquely a tuple in the relation, maybe have extraneous attributes
(ID, name) vs. Id
candidate keys: minimal superkeys
primary key to denote a candidate key that is chosen by the database designer as the principal means of identifying tuples within a relation
attribute values are never, or very rarely, changed
Primary key attributes are also underlined ， no duplication, unique

10. Schema Diagram for University Database
schema diagrams depicted, A database schema, along with primary key and foreign key dependencies, can be depicted by schema diagrams
The parts of a table
The underlined attributes are the PK
The arrow from… to…
E-R model

11. Expression of relation schema

12. Mathematica Fundament:
Relational Algebra

13. Relational Query Languages
Procedural vs .non-procedural, or declarative
“Pure” languages:
Relational algebra
Tuple relational calculus
Domain relational calculus
The above 3 pure languages are equivalent in computing power
We will concentrate in this chapter on relational algebra
Not turning-machine equivalent
consists of 6 basic operations
In a procedural language, the user instructs the system to perform a sequence of operations on the database to compute the desired result. In a nonprocedural language, the user describes the desired information without giving a specific procedure for obtaining that information
There are a number of “pure” query languages:
The relational algebra is procedural(fundamental techniques for extracting data from the database
),whereas the tuple relational calculus and domain relational calculus are nonprocedural.

14. Relational Algebra Procedural language Six basic operators
select: s
project: Õ
union: È
set difference: –
Cartesian product: x
rename: r
The operators take one or two relations as inputs and produce a new relation as a result.

15. Select Operation Notation: s p(r) p is called the selection predicate
Defined as: sp(r) = {t | t Î r and p(t)}
Where p is a formula in propositional calculus consisting of terms connected by : Ù (and), Ú (or), Ø (not) Each term is one of:
<attribute> op <attribute> or <constant>
where op is one of: =, ¹, >, ³. <. £

16. Select Operation – selection of rows (tuples)
Relation r
sA=B ^ D > 5 (r)

17. Example of Selection
s dept_name=“Physics”(instructor)

18. Project Operation Notation:
where A1, A2 are attribute names and r is a relation name.
The result is defined as the relation of k columns obtained by erasing the columns that are not listed
Duplicate rows removed from result, since relations are sets

19. Example of Selecting columns (Attributes)
Relation r:
ÕA,C (r)
To eliminate the dept_name attribute of instructor ÕID, name, salary (instructor)

20. Union Operation Notation: r È s Defined as:
r È s = {t | t Î r or t Î s}
For r È s to be valid.
r, s must have the same arity (same number of attributes)
The attribute domains must be compatible (example: 2nd column of r deals with the same type of values as does the 2nd column of s)

21. Example of Union Operation
Relations r, s:
r È s:

22. Example of Union Operation
Example: to find all courses taught in the Fall 2009 semester, or in the Spring 2010 semester, or in both Õcourse_id (s semester=“Fall” Λ year=2009 (section)) È Õcourse_id (s semester=“Spring” Λ year=2010 (section))

23. Set Difference Operation
Notation r – s
Defined as:
r – s = {t | t Î r and t Ï s}
Set differences must be taken between compatible relations.
r and s must have the same arity
attribute domains of r and s must be compatible

24. Example of Set difference of two relations
Relations r, s:
r – s:

25. Example of Set difference of two relations
Example: to find all courses taught in the Fall 2009 semester, but not in the Spring 2010 semester Õcourse_id (s semester=“Fall” Λ year=2009 (section)) − Õcourse_id (s semester=“Spring” Λ year=2010 (section))

26. Set-Intersection Operation
Notation: r Ç s
Defined as:
r Ç s = { t | t Î r and t Î s }
Assume:
r, s have the same arity
attributes of r and s are compatible
Note: r Ç s = r – (r – s)

27. Example of Set intersection of two relations
Relation r, s:
r Ç s
Note: r Ç s = r – (r – s)

28. Cartesian-Product Operation
Notation r x s
Defined as:
r x s = {t q | t Î r and q Î s}
Assume that attributes of r(R) and s(S) are disjoint.
(That is, R Ç S = Æ).
If attributes of r(R) and s(S) are not disjoint, then renaming must be used.

29. Example of Cartesian-Product Operation
Relations r, s:
r x s:

30. Cartesian-product – naming issue
Relations r, s: B
r.B
s.B
r x s:

31. Rename Operation
Allows us to name, and therefore to refer to, the results of relational- algebra expressions.
Allows us to refer to a relation by more than one name.
Example:
r x (E)
returns the expression E under the name X
If a relational-algebra expression E has arity n, then
returns the result of expression E under the name X, and with the attributes renamed to A1 , A2 , …., An .

32. Renaming a Table
Allows us to refer to a relation, (say E) by more than one name
r x (E)
returns the expression E under the name X
Relations r
r.A r.B s.A s.B
r x r s (r) α β 1 2 α β 1 2

33. Composition of Operations
Can build expressions using multiple operations
Example: sA=C (r x s)
r x s
sA=C (r x s)

34. Natural Join
Let r and s be relations on schemas R and S respectively. Then, the “natural join” of relations R and S is a relation on schema R È S obtained as follows:
Consider each pair of tuples tr from r and ts from s.
If tr and ts have the same value on each of the attributes in R Ç S, add a tuple t to the result, where
t has the same value as tr on r
t has the same value as ts on s

35. Natural Join Example
Relations r, s:
Natural Join
r s
Õ A, r.B, C, r.D, E (s r.B = s.B ˄ r.D = s.D (r x s)))

36. Notes about Relational Languages
Each Query input is a table (or set of tables)
Each query output is a table.
All data in the output table appears in one of the input tables
Relational Algebra is not Turning complete
Can we compute:
SUM
AVG
MAX
MIN

37. Summary of Relational Algebra Operators
Symbol (Name) Example of Use
(Selection) σ salary > = (instructor) σ
Return rows of the input relation that satisfy the predicate. Π
(Projection) Π ID, salary (instructor)
Output specified attributes from all rows of the input relation. Remove duplicate tuples from the output. x
(Cartesian Product) instructor x department
Output pairs of rows from the two input relations that have the same value on all attributes that have the same name. ∪
(Union) Π name (instructor) ∪ Π name (student)
Output the union of tuples from the two input relations. -
(Set Difference) Π name (instructor) -- Π name (student)
Output the set difference of tuples from the two input relations. ⋈
(Natural Join) instructor ⋈ department
Output pairs of rows from the two input relations that have the same value on all attributes that have the same name.

38. Quiz Sno Sname Ssex Sage Sdept
Course
Student
Cno Cname Cpqo Ccredit
database
2　　　　math
information system　
4　　Operating System
5　　Data structure
6　　Data Processing
7　　 　Pascal 　　
Sno Sname Ssex Sage Sdept
Yong Li　　　　male　　20　　 CS　　　
Chen liu　　　　female IS
95003　　 Ming Wang　　 female MA
95004　 li Zhang　　　　 male IS
1.Give the names and departments of all students
Sno Cno Grade SC
2.Give the students whose department is ‘IS’
3.Give the student number who study the course ‘1’
4.Give the student names who study the course ‘1’ or ‘3’
5. Give the students’ name, name and the grades of the courses they took

39. • Table • Relation • Tuple
• Attribute • Domain • Atomicdomain
Nullvalue Database schema Database instance
Relation schema Relation instance
Keys
Superkey ◦ Candidate key
Primary key
Foreign key
Referencing relation
Referenced relation
Referential integrity constraint
Schema diagram
• Querylanguage
◦ Procedural language
Nonprocedural language
Operations on relations
Selection of tuples Selection of attributes ◦ Natural join Cartesian product Set operations
Relational algebra

40. What have you learned from this lecture?
Any Questions?