1. Chapter 2: Introduction to Relational Model
Structure of Relational Databases
Fundamental Relational-Algebra-Operations

2. Example of the instructor Relation
attributes
(or columns)
tuples
(or rows)

3. Attribute Types
A relation is represented as a table. The term attribute (屬性) refers to a column of a table.
屬性一般指稱物件的特性, 或欲處理的資料.
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 (分割後沒有意義)
multivalued attribute values are not atomic (see page 1.17)
For example: author = {{Smith, Jones}, {Jones, Frick}}
composite attribute values are not atomic
For example: publisher = {(McGraw-Hill, New York), (Oxford, London)}
The special value null is a member of every domain, which signifies that the value is unknown or does not exist.
The null value causes complications in the definition of many operations, and will be discussed later.

4. 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  Di
D1 = {a, b, c}, D2 = {1, 2}, D1XD2 = {(a, 1), (a, 2), (b, 1), (b, 2), (c, 1), (c, 2)}
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.

5. Relations are Unordered
Order of tuples is irrelevant (tuples may be stored in an arbitrary order)
資料的排序方式屬physical level, 非logical level
寫query時不知資料如何排序;
Example: instructor relation with unordered tuples

6. Database A database consists of multiple relations
Information about an enterprise is broken up into parts, where each relation storing one part of the information
The university database example:
instructor (ID, name, dept_name, salary)
department (dept_name, building, budget)
student (ID, name, dept_name, tot_cred)
course (course_id, title, dept_name, credits)
prereq (course_id, prereq_id)
Bad design: (c.f. page 1.15) univ (instructor_ID, name, dept_name, salary, student_Id, ..) Normalization theory (Chapter 8) deals with how to design “good” relational schemas.

7. Keys
Let K  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}, {name}, and {ID,dept_name} are all superkeys of instructor. (see page 2.5)
Superkey K is a candidate key if K is minimal Example: {ID} , {name} are both candidate keys for Instructor
One of the candidate keys is selected to be the primary key.
which one? 通常依一般使用習慣,找最有代表性的.
Another example: {系別, 年級,班級,座號}, {學號}, {身分證字號)都是同學的candidate key, 選 {學號}做為primary key.
注意:
Key通常做為物件的代表性屬性
屬性值的唯一性會隨表格表示的資料不同而改變.,
如下一頁的dept_name SK CK PK

8. Foreign Keys
Foreign key constraint: Value in one relation must appear in another
Referencing relation: e.g., instructor
Referenced relation: e.g., department
Will discuss this again in Chapter 3 and Chapter 4.
department
instructor

9. Schema Diagram for University Database

10. Relational Query Languages
Language in which user requests information from the database.
Categories of languages
Procedural
non-procedural, or declarative
“Pure” languages:
Relational algebra: procedural
Tuple (Domain) relational calculus: declarative
“Algebra” is based on operators.
Example of arithmetic algebra: 1 + 5*3
How to write a query
Determine which relations to use
Determine which operators to use

11. Relational Algebra Relational operators select:  project: 
Natural join:
Cartesian product: x
union: 
Intersection: 
set difference: –
The operators take one or two relations as inputs and produce a new relation as a result.

12. Selection of tuples σ 限制式(r) Relation r Selection Select tuples with
A=B and D > 5
σ A=B ^ D > 5 (r)

13. Selection of Columns (Attributes)
Relation r:
Projection
 屬性名稱 (r)
Select columns A and C
 A, C (r)
-> duplicates are removed

14. More Examples
Return those instructors whose salaries are more than (see page 2.5, page2.6)
σ salary>=85000 (instructor)
Output the attributes ID and Salary of instructors.
Π ID, salary (instructor)
Find the ID and salary for those instructors
who have salary greater than $85000.
Π ID, salary (σ salary>=85000 (instructor))
<- composition

15. Joining two relations – Cartesian Product
Relations r, s:
r x s:
Example: instructor X department => 12*7 = 84 tuples
(see page 2.8) <-會希望同一列的dept_name要一樣!

16. Joining two relations – 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  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  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
Example:
R = (A, B, C, D)
S = (E, B, D)
Result schema = (A, B, C, D, E)
r s is defined as: r.A, r.B, r.C, r.D, s.E (r.B = s.B  r.D = s.D (r x s))

17. Natural Join Example
Relations r, s:
Natural Join
r s

18. Example Instructor department (c.f., page 2.8)
Example: Output the attributes ID and building of instructors.
 ID, building (Instructor department)

19. Union of two relations Relations r, s: r  s:
Find the names of instructors and students.
 name (instructor)   name (student)

20. Set difference of two relations
Relations r, s:
r – s:
Find the departments which do not hire instructors.
 dept _ name (department) –  dept _ name (instructor)

21. Set Intersection of two relations
Relation r, s:
r  s
Find the instructors who are also students.
 name (instructor)   name (student)