ch2.ppt ch2.pptch2.pptch2.pptch2.pptch2.pptch2.ppt

Database System Concepts, 6th
Ed.
©Silberschatz, Korth and Sudarshan
See www.db-book.com for conditions on re-use
Chapter 2: Intro to Relational Model
Chapter 2: Intro to Relational Model

2.2
Database System Concepts - 6th
Edition
Example of a Relation
Example of a Relation
attributes
(or columns)
tuples
(or rows)

2.3
Edition
Attribute Types
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”
 The null value causes complications in the definition of many
operations

2.4
Edition
Relation Schema and Instance
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
 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

2.5
Edition
Relations are Unordered
Relations are Unordered
 Order of tuples is irrelevant (tuples may be stored in an arbitrary order)
 Example: instructor relation with unordered tuples

2.6
Edition
Keys
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} 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

2.7
Edition
Schema Diagram for University Database
Schema Diagram for University Database

2.8
Edition
Relational Query Languages
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

2.9
Edition
Select Operation – selection of rows (tuples)
Select Operation – selection of rows (tuples)
 Relation r
 A=B ^ D > 5 (r)

2.10
Edition
Project Operation – selection of columns (Attributes)
Project Operation – selection of columns (Attributes)
 Relation r:
 A,C (r)

2.11
Edition
Union of two relations
Union of two relations
 Relations r, s:
 r  s:

2.12
Edition
Set difference of two relations
Set difference of two relations
 Relations r, s:
 r – s:

2.13
Edition
Set intersection of two relations
Set intersection of two relations
 Relation r, s:
 r  s
Note: r  s = r – (r – s)

2.14
Edition
joining two relations -- Cartesian-product
joining two relations -- Cartesian-product
 Relations r, s:
 r x s:

2.15
Edition
Cartesian-product – naming issue
Cartesian-product – naming issue
 Relations r, s:
 r x s: s.B
B
r.B

2.16
Edition
Renaming a Table
Renaming a Table
 Allows us to refer to a relation, (say E) by more than one name.
 x (E)
returns the expression E under the name X
 Relations r
 r x  s (r)
α
α
β
β
1
1
2
2
α
β
α
β
1
2
1
2
r.A r.B s.A s.B

2.17
Edition
Composition of Operations
Composition of Operations
 Can build expressions using multiple operations
 Example: A=C (r x s)
 r x s
 A=C (r x s)

2.18
Edition
Joining two relations – Natural Join
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

2.19
Edition
Natural Join Example
Natural Join Example
 Relations r, s:
 Natural Join
 r s
 A, r.B, C, r.D, E ( r.B = s.B r.D = s.D
˄ (r x s)))

2.20
Edition
Notes about Relational Languages
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

2.21
Edition
Summary of Relational Algebra Operators
Summary of Relational Algebra Operators
Symbol (Name) Example of Use
(Selection) σ
salary > = 85000 (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.
(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.
⋈
-
(Set Difference) Π
name (instructor) -- Π
name (student)
Output the set difference of tuples from the two input relations.

Database System Concepts, 6th
Ed.
See www.db-book.com for conditions on re-use
End of Chapter 2
End of Chapter 2

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