Relational Algebra

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Relational Algebra

• Chapter 4 - part I

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Relational Query Languages

• Query languages Allow manipulation and
  retrieval of data from a database.
• Relational model supports simple powerful QLs
• Strong formal foundation based on logic.
• Allows for much optimization.
• Query Languages ! Programming languages!
• QLs not expected to be Turing complete.
• QLs not intended to be used for complex
  calculations.
• QLs support easy, efficient access to large data
  sets.

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Formal Relational Query Languages

• Two mathematical Query Languages form the basis
  for real languages (e.g. SQL) and for
  implementation
• Relational Algebra
• More operational, very useful for representing
  execution plans.
• Relational Calculus
• Lets users describe what they want, rather than
  how to compute it. (Non-operational, rather
  declarative.)

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Preliminaries

• A query is applied to relation instances.
• The result of a query is also a relation
  instance.
• Schemas of input relations for a query are fixed
  (but query will run regardless of instance!)
• The schema for result of given query is also
  fixed! Determined by definition of query language
  .

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Preliminaries

• Positional vs. named-field notation
• Positional field notation e.g., S.1
• Named field notation e.g., S.sid
• Pros/Cons
• Positional notation easier for formal
  definitions, named-field notation more readable.
  
• Both used in SQL

• Assume that names of fields in query results are
  inherited from names of fields in query input
  relations.

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Example Instances
Sailors
Reserves
R1
S1
Sailors
S2
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Relational Algebra

• Basic operations
• Selection ( ) Selects a subset of rows
  from relation.
• Projection ( ) Deletes unwanted columns
  from relation.
• Cartesian-product ( ) Allows us to combine
  two relations.
• Set-difference ( ) Tuples in reln. 1, but
  not in reln. 2.
• Union ( ) Tuples in reln. 1 and in reln. 2.
• Additional operations
• Intersection, join, division, renaming
  Not essential, but
  (very!) useful.
• Since each operation returns a relation,
  operations can be composed! (Algebra is
  closed.)

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Selection
Sailors
S2
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Selection

• ?condition (R)
• Selects rows that satisfy selection condition.
• attribute op constant
• attribute op attribute
• Op is lt,gt,lt,gt, ,
• No duplicates in result!
• Schema of result identical to schema of (only)
  input relation.

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Selection

• Result relation can be input for another
  relational algebra operation! (Operator
  composition.)

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Projection
Sailors
S2
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Projection

• ? projectlist (R)
• Deletes attributes that are not in projection
  list.
• Schema of result contains fields in projection
  list
• Projection operator has to eliminate duplicates!
  (Why??)
• Note real systems typically dont do duplicate
  elimination unless the user explicitly asks for
  it. (Why not?)

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Union, Intersection, Set-Difference

• All of these operations take two input relations,
  which must be union-compatible
• Same number of fields.
• Corresponding fields have same type.
• What is the schema of result?

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Example Instances Union
S1
S2
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Union, Intersection, Set-Difference

• All of these operations take two input relations,
  which must be union-compatible
• Same number of fields.
• Corresponding fields have the same type.
• What is the schema of result?

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Difference Operation
S1
S2
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Union, Intersection, Set-Difference

• All of these operations take two input relations,
  which must be union-compatible
• Same number of fields.
• Corresponding fields have the same type.
• What is the schema of result?

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Intersection Operation
S1
S2
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Union, Intersection, Set-Difference

• All of these operations take two input relations,
  which must be union-compatible
• Same number of fields.
• Corresponding fields have the same type.
• What is the schema of result?

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Cross-Product (Cartesian Product)

• S1 R1 Each row of S1 is paired with each row
  of R1.

Reserves
R1
Sailors
S1
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Cross-Product (Cartesian Product)

• S1 R1 Result schema has one field per field
  of S1 and R1, with field names inherited if
  possible.
• Conflict Both S1 and R1 have a field called sid.

• Renaming operator

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Why we need a Join Operator ?

• In many cases,
• Join Cross-Product Select Project
• However
• Cross-product is too large to materialize
• Apply Select and Project "On-the-fly"

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Condition Join / Theta Join

• Condition Join
• Result schema same as that of cross-product.
• Fewer tuples than cross-product, more efficient.

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EquiJoin

• Equi-Join A special case of condition join
  where the condition c contains only equalities.
• Result schema similar to cross-product, but only
  one copy of fields for which equality is
  specified.
• An extra project PROJECT ( THETA-JOIN)

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Natural Join

• Natural Join Equijoin on all common fields.

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Division

• Not supported as a primitive operator, but
  useful for expressing queries like
• 
  
  Find sailors who have reserved all boats.

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Division

• Let A have 2 fields x and y B have only field
  y
• A/B
• A/B contains all x tuples (sailors) such that for
  every y tuple (boat) in B, there is an xy tuple
  in A.
• If set of y values (boats) associated with an x
  value (sailor) in A contains all y values in B,
  then x value is in A/B.
• A/B is the largest relation instance Q such that
  Q?B ?A.

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Division Example

• e.g., A all parts supplied by
  suppliers,
• B relation parts
• A/B suppliers who supply all
  parts listed in B

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Examples of Division A/B
B1
B2
B3
A/B1
A/B2
A/B3
A
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Expressing A/B Using Basic Operators

• Idea For A/B, compute all x values that are not
  disqualified by some y value in B.
• x value is disqualified if by attaching y value
  from B,
• we obtain an xy tuple that is not in A.

Disqualified x values
A/B
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A few example queries
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Find names of sailors whove reserved boat 103
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Find names of sailors whove reserved a red boat
Reserves
R1
Sailors
S1
Boats
bid bname color
101 Interlake blue
103 Clipper red
B1
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Find names of sailors whove reserved a red boat

• Information about boat color only available in
  Boats so need an extra join

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Find sailors whove reserved a red or a green boat

• Can identify all red or green boats, then find
  sailors whove reserved one of these boats

Can also define Tempboats using union! (How?)
What happens if is replaced by in this
query?
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Find sailors whove reserved a red and a green
boat

• Previous approach wont work!
• Must identify sailors who reserved red boats,
  sailors whove reserved green boats,
  then find their intersection

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Find the names of sailors whove reserved all
boats

• Uses division schemas of the input relations to
  / must be carefully chosen

• To find sailors whove reserved all Interlake
  boats

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Relational Algebra Some More Operators

• Beyond Chapter 4

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Generalized Projection

• ? F1, F2, (R)

? sname, (rating2 as myrating) (Sailors)
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Aggregation operators

• MIN, MAX, COUNT, SUM, AVG
• AGGB (R) considers only non-null values of R.

SUMB (R)
COUNTB (R)
MINB (R)
R
SUMB (R)
9
MINB (R)
2
COUNTB (R)
3
A B
1 2
3 4
1 null
1 3
AVGB (R)
COUNT (R)
MAXB (R)
AVGB (R)
3
COUNT (R)
4
MAXB (R)
4
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Aggregation Operators

• MIN, MAX, SUM, AVG must be on any 1 attribute.
  COUNT can be on any 1 attribute or COUNT (R)
• An aggregation operator returns a bag, not a
  single value ! But SQL allows treatment as a
  single value.

sBMAXB (R) (R)
A B
3 4
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Grouping Operator ?GL, AL (R)

• ?GL, AL (R) groups all attributes in GL, and
  performs the aggregation specified in AL.

? starName, MIN (year)?year, COUNT(title) ?num
(StarsIn)
starName year num
HF 77 3
KR 94 2
StarsIn
title year starName
SW1 77 HF
Matrix 99 KR
6D7N 93 HF
SW2 79 HF
Speed 94 KR
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Summary

• The relational model has rigorously defined query
  languages that are simple and powerful.
• Relational algebra is operational useful as
  internal representation for query evaluation
  plans.
• Several ways of expressing a given query a query
  optimizer should choose most efficient version.