2red reviewrelation

Review of the Relational Model Acknowledgements
• These slides were written by Richard T. Snodgrass
(University of Arizona, SIGMOD chair), Christian S.
Jensen (Aalborg University), and Curtis Dyreson (Utah
State University).

• Kristian Torp (Aalborg University) converted the slides
from island presents to Powerpoint.
CS 6800
Utah State University

The Relational Model R-2

Overview Sets
• The Relational Model (RM) • A set is an unordered collection of distinct objects.
Relations Examples
Properties of relations {3, 4, a} is a set
{3, 4, a} is the same set as {a, 4, 3}
• Relational Algebra (RA) {4, 4} is not a set
Operators
x Complete set – projection, selection, Cartesian product, difference,
• Set operations include
union Intersection, e.g., {3, 4, a} I {b, 4} = {4}
x Convenient – intersection, theta join, equijoin, natural join, semijoin, Union, e.g., {3, 4, a} U {b, 4} = {3, 4, a, b}
division
Difference, e.g., {3, 4, a} - {b, 4} = {3, a}
Example queries
Limitations
• DRC and TRC
The Relational Model R-3 The Relational Model R-4

Domains and Attributes Relations
Definition: A domain is a set of values of some "type" Definition: Given sets A1, A2, ... , An a relation r is a
Positive integers = {1, 2, 3, 4, …} subset of their Cartesian product.
Alphanumeric characters = {‘a’, ‘b’, …, ‘Z’, ‘0’, …, ‘9’} r ⊆ A1 × A2 × ... × An
• A common database restriction: A domain is atomic. • r is a set of n-tuples (a1, a2, ... , an) where ai ∈ Ai
The following programming types are atomic domains.
integer, char, float, varchar Definition: R(A1, A2, ..., An) is the schema of relation r.
Structs/records are a composite domain. A1, A2, ... , An are domains; their names are attributes
struct { a: int, b: char } record_definition
R is the name of the relation
Definition: An attribute is the name of a domain.
• Notation: r(R) is a relation on the relation schema R.


Example, The < Relation Relations as Tables
• Let O be the domain {1, 3} and E be {0, 2}. • Relations can be depicted as tables (approx.)
O × E = {(1,0), (1,2), (3,0), (3,2)} • ×(O,E) where O = {1,3} and E = {0,2}
× O E
1 0 1 0

3 2 1 0 1 2
3 0
3 2
3 2
• <(O,E) is the schema, O, E are attributes, {1,3} and • <(O,E)
{0,2} are domains
O < E = {(1,2)} ⊆ O × E 1 0 < O E
1 2
1 0 3 2

3 2

Tuples Characteristics of Relations
Definition: An element t of a relation r is called a tuple. • Tuples in a relation are unorderd.
• Example: The following two relations have the same
• We refer to component values of a tuple t by t[Ai] = vi information content.
(the value of attribute Ai for tuple t).
Alternatively, use ‘dot’ notation, e.g., t.Ai is the ith attribute
of tuple t
Name Age Name Age
• t[Ai, ... , Ak] refers to the subtuple of t containing the
Pat 1 Sue 3
values of attributes Ai, ... , Ak respectively. Fred 2 Pam 4
• Table metaphor Sue 3 Fred 2
Tuple is a row Pam 4 Pat 1
Attribute is a column


Characteristics of Relations (cont.) Characteristics of Relations (cont.)
• Attributes in a tuple/relation are ordered. • Values in a tuple are atomic (indivisible).
But, we can change the order at will (to be shown in future). • A value cannot be a structure, record, or relation
Example: The following relations do not have the same Example: All atomic values (string or integer).
information. Name Age
Pat 1
Name Age Age Name Fred 2
Pat 1 1 Pat
Fred 2 2 Fred Example: The following is not a relation, Name is not atomic
Sue 3 3 Sue Name Age
Pam 4 4 Pam First Last
1
Joe Doe

First Last
2
Sue Doe


Degree and Cardinality Relational Algebra
Ord CID Date Salesperson
• Five primary operators (complete)
102 42 860207 Johnson Selection: σ
103 39 860211 Strong
Projection: π
104 24 860228 Boggs
105 42 860303 Strong Union: ∪
107 34 860308 West Tuples (or records, rows)
108 46 860308 Wset
[Cardinality is # of them] Difference: –
110 24 860312 Boggs Cartesian Product: ×
111 21 860318 Strong
112 29 860320 Clark
• Convenient derived operators (closed)
degree/arity Intersection: ∩
Theta join nθ, equijoin n= , and natural join n
Semijoins: left l and right r
Relational division: ÷

RA Operations Produce Relations Selection, Formal Description
σP(r) =
• P is a formula in propositional calculus, dealing with terms of
Name Age Name Age Name the form:
joe 4 sue 10 sue
attribute (or constant) = attribute (or constant)
sue 10 sam 9 sam
sam 9 operation1 operation2 attribute ≠ attribute
attribute < attribute
attribute ≤ attribute
attribute ≥ attribute
attribute > attribute
Notation: operation2( operation1(R) ) term ∧ term (Note: ∧ is AND)
term ∨ term (Note: ∨ is OR)
¬(term)


Projection, Formal Description Switching Columns
πX(r) = • Assume r is the following
• Let X = {Ai1, Ai2 , ... , Ain} Name Age

• The result is a relation of n columns obtained by Pat 1
Fred 2
keeping the columns that are specified.
Sue 3
• Example
A B C Pam 4
r: 1 y ω
3 x γ • To switch columns use projection
2 y β
Age Name
1 Pat
πA,C (r): A C πB (r): B πAge,Name (r) = 2 Fred
1 ω y 3 Sue
3 γ x
4 Pam
2 β

Union, Formal Description Difference, Formal Description
r∪s= r–s=
• Assume r and s are union-compatible
r and s have the same arity.
The attributes of r and s are same over domains.
• Assume r and s are union-compatible.
• Example • Example
A B C A B C A B C A B C
r: 1 y ω s: 4 w ζ r: 1 y ω s: 4 w ζ
3 x γ 3 x γ 3 x γ 3 x γ
2 y β 2 y β

r∪s:
A B C A B C
1 y ω r–s: 1 y ω
3 x γ 2 y β
2 y β
4 w ζ

Cartesian Product Intersection, Formal Description
r×s= r∩s=
• Assume r and s are union-compatible
• The schema of the resulting relation is R S • Example
A B C D A
r: 1 y ω s: “Tom” 1
3 x γ “Eric” 2
A B C A B C
2 y β r: 1 y ω s: 4 w ζ
3 x γ 3 x γ
r×s: A
1
B
y
C
ω
D
“Tom”
A
1
2 y β

3 x γ “Tom” 1
2 y β “Tom” 1 A B C
1 y ω “Eric” 2 r∩s: 3 x γ
3 x γ “Eric” 2
2 y β “Eric” 2

Joins Theta join, Formal Description
• Joins are Cartesian products coupled with selections r nθ s =
and projections.
• Theta join
• Example
r nθ s = σθ (r × s) r: A
1
B
y
C
ω
s: D
“Tom”
E
3
Example: r nA < E s = σA < E (r × s) 0 x γ “Eric” 1
• Equijoin 2 y β

A theta join in which θ is an equality predicate A B C D E
Example: r nA = E s = σA = E (r × s) r n A<E s : 1 y ω “Tom” 3
• Natural join n 0 x γ “Tom” 3
0 x γ “Eric” 1
• Semijoins 2 y β “Tom” 3
Left semijoin l • Equijoin
Right semijoin r A theta join in which θ is an equality predicate
Example: r nA = E s

Natural Join Semijoins
r n s= r ls =
• Schema of result is R∪S • The result has the same schema as the left-hand
• Let t be a tuple in the result. argument, r.
t[R] has the same value as a tuple tR ∈ r. • Example
t[S] has the same value as a tuple ts ∈ s.
A B C
• Example r: 1 y ω
s: D
“Tom”
B
y
A B C
r: 1 y ω
s: D
“Tom”
B
y
3 x γ
2 y β
3 x γ “Eric” x
2 y β
rls: A B C
A B C D 1 y ω
rns: 1 y ω “Tom” 2 y β
3 x γ “Eric”
2 y β “Tom” • Right semijoin: r r s =

Relational Division Relational Division, cont.
r ÷ s = { t | ∀u ∈s (t u ∈ r)} • Let r and s be relations on schemes R and S
respectively, where
• Example
R = (A1, A2, ..., An, B1, B2, ..., Bn,)
A B C
S = (B1, B2, ..., Bn,)
B
r: 1 y ω
s: y
3 x γ z
2 y β • The result of r divided by s is a relation on scheme
1 z ω R ÷ S = (A1, A2, ..., An)

r÷s: A
1
C
ω
• R ÷ S = πR – S (r) – πR – S ((πR – S (r) × s) – r)


Renaming Operator Limitations of the Algebra
• Find the film(s) with the highest rental price. • Can't do arithmetic.
Find the rental price assuming a 10% increase.
• We need a renaming operator: ρName • Can't do aggregates.
How many films has each customer reserved?
• Alternative formulation: Find the film(s) with a rental • Can't handle “missing” data.
price for which no other rental price is higher. Make a list of the films, along with who reserved it, if
applicable.
• Can't perform transitive closure.
For a partof(Part, ConstituentPart) relation, find all parts in
the car door.
πTitle(Film) –
πF2.Title (σFilm.RentalPrice > F2.RentalPrice (Film×ρF2(Film))) • Can't sort, or print in various formats.
Print a reserved summary, sorted by customer name.
• Can't modify the database.
Increase all $3.25 rentals to $3.50.

Extending the Projection Operator Aggregates
• The generalized projection operator allows expressions AGagg1, ..., aggk (r) =
in addition to column names in the subscript. t a1 ... ak | t ∈ r ∧
a1 = agg1 ({u | u ∈ r ∧ u[G] = t[G]})
• Find the rental price, assuming a 10% increase. ....
ak = aggk ({u | u ∈ r ∧ u[G] = t[G]})

πTitle, RentalPrice*1.1(Film) • Appends new attributes.
• Each aggregate function aggi decides which attribute
to aggregate over.
• The grouping attributes A are optional.


Outer Joins Outer Join Example
• In a regular equijoin or natural join, tuples in r or s that do not
having matching tuples in the other relation do not appear in the
result. A B C
r: 1 y ω
s: D
“Tom”
B
y
• The outer joins retain these tuples, and place nulls in the missing
3 x γ “Eric” x
attributes.
2 z β “Melanie” w
• Left outer join:
r L s = r n s ∪ ((r – (r l s)) × (null, ..., null))
• Right outer join: r N s: A B C D
r R s = r n s ∪ ((null, ..., null) × (s – (s l r))) 1 y ω “Tom”
3 x γ “Eric”
• Full outer join: 2 z β Null
r N s = r n s ∪ ((r – (r l s)) × (null, ..., null)) Null w Null “Melanie”

∪ ((null, ..., null) × (s - (s l r)))

Relational Completeness Overview
• All the operators can be expressed in terms of the five • Relational Calculus
basic operators: σ, π, −, ×, ∪ Tuple relational calculus
x Safety
Domain relational calculus
• This set is called a complete set of relational algebraic x Equivalence
operators. Completeness and expressiveness

• Any query language that is at least as powerful as these
operators is termed (query) relationally complete.


Relational Calculus Relational Calculus, cont.
• The relational algebra is procedural, specifying a • Two forms of calculi
sequence of operations to derive the desired results.
• Relational calculus is based on first-order predicate
calculus.
• Relational calculus is more declarative, specifying Tuple Relational Calculus (TRC)
what is desired.
• The expressive power of the two languages is identical.
This implies that relational calculus is relationally complete. Domain Relational Calculus (DRC)
• Many commercial relational query languages are based
on the relational calculus.
• The implementations are based on the relational
algebra.


Tuple Relational Calculus Syntax of Tuple Relational Calculus
• A tuple variable ranges over tuples of a particular {t1 .A1, t2.A2, ... , tj.Aj | P (t1, t2 , ... , tj)}
relation.
• Example: List the information about expensive films.
• t1, t2 , ... , tj are tuple variables.
{t | t ∈ Film ∧ t[RentalPrice] > 4}
• t ∈ Film specifies the range relation Film for the tuple • Each tk is an tuple of the relation over which it ranges.
variable t.
• Each tuple satisfying t[RentalPrice] > 4 is retrieved.
• The entire tuple is retrieved.
• P is a predicate, which is made up of atoms, of the
following types.
• List the titles of expensive films.
tk ∈ R or R (tk) identifies R as the range of tk
{t | x ∈ Film (x[RentalPrice] = t[RentalPrice]
Atoms from predicate calculus: ∧, ∨, ¬,∀, ∃
∀
∧ x[RentalPrice] > 4)}

Safety Domain Relational Calculus
• It is possible to write tuple calculus expressions that • Each query is an expression of the form
generate infinite relations.
{ 〈x1, x2, ... , xn〉 |P (x1, x2, ... , xn)}
• {t | ¬ t ∈ r} results in an infinite relation if the domain
of any attribute of relation r is infinite. • P is a formula similar to the formula in the predicate
• We wish to ensure that an expression in relational calculus.
calculus yields only a finite number of tuples.
• The domain of a tuple relational calculus expression is
the set of all values that either appear as constant
values in the expression or that exist in any tuple of the
relations referenced in the expression.
• An expression is safe if all values in its result are from
the domain of the expression.


Summary Expressive Power
• Relational algebra • Theorem: The following four languages define the same class of
Objects are relations (sets of n-tuples). functions.
Relational algebra expressions
• Tuple relational calculus Safe relational tuple calculus formulas
Variables range over relations (sets of tuples). Safe relational domain calculus formulas
Each variable is associated with an individual tuple.
• Domain relational calculus • Proof:
relational algebra ≤ (safe) tuple relational calculus
Variables range over domains (sets of values). x Show via induction that every relational algebra expression has a counterpart
Each variable is associated with an individual value. in the tuple relational calculus.
tuple relational calculus ≤ domain relational calculus
• Misnamed domain relational calculus ≤ relational algebra
Tuple relational calculus and Value relational calculus
Relational relational calculus and Domain relational calculus • Corollary: All four languages are relationally complete.

Completeness Completeness, Example
• Theorem: The tuple relational calculus is complete. • List the titles of all reserved films.
πTitle (σFilm,FilmID = Reserved.FilmID(Film × Reserved))
πTitle (Film n Reserved)
• Proof:
Selection σP(r) ⇒ { t | t ∈ r ∧ P(t) }
Projection πX(r) ⇒ {t | s ∈ r (s[X] = t[X] )} • Replace projection:
Union r∪s⇒{t|t∈r∨t∈s}
Difference r – s ⇒ { t | t ∈ r ∧ ¬t ∈ s }
Cartesian product r×s⇒ {t q|t∈r∧ q∈s} {t.Title | t ∈ σFilm,FilmID = Reserved.FilmID
Induct on length of the algebraic expression. (Film × Reserved)}


Completeness, Example, Cont. Equivalence of Expressive Power
• Replace selection: • Theorem: The relational algebra is as expressive as the
(safe) tuple relational calculus.
{t.Title | t ∈ (Film × Reserved)
∧ t[Film.FilmID] = t[Reserved.FilmID] }
• (Informal) Proof: by induction on the number of
operators in the calculus predicate
• Replace the Cartesian product:

{t.Title. | q ∈ Film ∧ r ∈ Reserved
∧ q.FilmID = r.FilmID ∧ t = q r }


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