RELATIONAL ALGEBRA III

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RELATIONAL ALGEBRA (III)
CS157A

• Prof. Sin-Min LEE
• Department of Computer Science

Lecture 10
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Unary Relational Operations SELECT and PROJECT

• The PROJECT Operation
• Sequences of Operations and the RENAME Operation
• The SELECT Operation

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Relational Algebra Operations from Set Theory

• The UNION, INTERSECTION, and MINUS Operations
• The CARTESIAN PRODUCT (or CROSS PRODUCT) Operation

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Binary Relational Operations JOIN and DIVISION

• The JOIN Operation
• The EQUIJOIN and NATURAL JOIN Variations of JOIN
• A Complete Set of Relational Algebra Operations
• The DIVISION Operation

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Additional Relational Operations

• Aggregate Functions and Grouping
• Recursive Closure Operations
• OUTER JOIN Operations
• The OUTER JOIN Operation

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SPECIAL RELATIONAL OPERATORS

• The following operators are peculiar to
  relations
• - Join operators
• There are several kind of join operators. We only
  consider three of these here (others will be
  considered when we discuss null values)
• - (1) Condition Joins
• - (2) Equijoins
• - (3) Natural Joins
• - Division

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JOIN OPERATORS

• Condition Joins
• - Defined as a cross-product followed by a
  selection
• R ?c S sc(R ? S)
  (? is called the bow-tie)
• where c is the condition.
• - Example
• Given the sample relational instances S1 and R1

The condition join S ?S1.sidltR1.sid R1 yields
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JOIN OPERATORS

• Condition Joins
• - Defined as a cross-product followed by a
  selection
• R ?c S sc(R ? S)
  (? is called the bow-tie)
• where c is the condition.
• - Example
• Given the sample relational instances S1 and R1

The condition join S ?S1.sidltR1.sid R1 yields
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• Equijoin
• Special case of the condition join where the join
  condition consists solely of equalities between
  two fields in R and S connected by the logical
  AND operator (?).
• Example Given the two sample relational
  instances S1 and R1

The operator S1 R.sidSsid R1 yields
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• Natural Join
• - Special case of equijoin where equalities are
  implicitly specified on all fields having the
  same name in R and S.
• - The condition c is now left out, so that the
  bow tie operator by itself signifies a natural
  join.
• - N. B. If the two relations have no attributes
  in common, the natural join is simply the
  cross-product.

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DIVISION

• - The division operator is used for queries which
  involve the all
• qualifier such as Find the names of sailors who
  have reserved all boats.
• - The division operator is a bit tricky to
  explain, and perhaps best approached through
  examples as will be done here.

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EXAMPLES OF DIVISION

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DIVISION

• Interpretation of the division operation A/B
• - Divide the attributes of A into 2 sets A1 and
  A2.
• - Divide the attributes of B into 2 sets B2 and
  B3.
• - Where the sets A2 and B2 have the same
  attributes.
• - For each set of values in B2
• - Search in A2 for the sets of rows (having the
  same A1 values) whose A2 values (taken together)
  form a set which is the same as the set of B2s.
• - For all the set of rows in A which satisfy the
  above search, pick out their A1 values and put
  them in the answer.

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DIVISION

• Example Find the names of sailors who have
  reserved all boats
• (1) A ?sid,bid(Reserves). A1 ?sid(Reserves)
  A2 ?bid(Reserves)
• (2) B2 ?bid(Boats) B3 is the rest of B.
• Thus, B2 101, 102, 103, 104
• (3) Find the rows of A such that their A.sid is
  the same and their combined A.bid is the set B2.
• Thus we find A1 22
• (4) Get the set of A2 corresponding to A1 A2
  Dustin

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FORMAL DEFINITION OF DIVISION

• The formal definition of division is as follows
• A/B ?x(A) - ?x((?x(A) ? B) A)

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EXAMPLES OF ALGEBRA QUERIES

• In the rest of this chapter we shall illustrate
  queries using the following new instances S3 of
  sailors, R2 of Reserves and B1 of boats.

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QUERY Q1

• Given the relational instances

(Q1) Find the names of sailors who have reserved
boat 103 ?sname((sbid103 Reserves) ? Sailors)
The answer is thus the following relational
instance ltDustingt, ltLubbergt, ltHoratiogt
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QUERY Q1 (contd)

• There are of course several ways to express Q1 in
  relational algebra.
• Here is another
• ?sname(sbid103(Reserves? Sailors))

Which of these expressions should we use? That is
a question of optimization. Indeed, when we
describe how to state queries in SQL, we can
leave it to the optimizer in the DBMS to select
the nest approach.
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QUERY Q2

• (Q2) Find the names of sailors who have reserved
  a red boat.

?sname((scolorredBoats) ? Reserves ? Sailors)

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QUERY Q3

• (Q3) Find the colors of boats reserved by Lubber.

?color((ssnameLubberSailors)Sailors ? Reserves
? Boats)
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QUERY Q4

• (Q4) Find the names of Sailors who have reserved
  at least one boat

?sname(Sailors ? Reserves)
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QUERY Q5

• (Q5) Find the names of sailors who have reserved
  a red or a green boat.

?(Tempboats, (scolorredBoats) ?
(scolorgreenBoats))
?sname(Tempboats ? Reserves ? Sailors)
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QUERY Q6

• (Q6) Find the names of Sailors who have reserved
  a red and a green boat.
• It seems tempting to use the expression used in
  Q5, replacing simply ? by n. However, this wont
  work, for such an expression is requesting the
  names of sailors who have requested a boat that
  is both red and green! The correct expression is
  as follows
• ?(Tempred, ?sid((scolorredBoats) ?
  Reserves))
• ?(Tempgreen, ?sid((scolorgreenBoats
  ) ? Reserves))
• ?sname ((Tempred n Tempgreen) ?
  Sailors)

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QUERY Q7

• (Q7) Find the names of sailors who have reserved
  at least two boats.

?(Reservations, ?sid,sname,bid(Sailors ?
Reserves)) ?(Reservationpairs(1?sid1, 2?sname,
3?bid1, 4?sid2, 5?sname, 6?bid2),
Reservations?Reservations) ?sname1s(sid1sid2)?(bi
d1?bid2)Reservationpairs)
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QUERY 8

• (Q8) Find the sids of sailors with age over 20
  who have not reserved a red boat.

?sid(sagegt20Sailors) - ?sid((scolorredBoats) ?
Reserves ? Sailors)
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QUERY 9

• (Q) Find the names of sailors who have reserved
  all boats.

?(Tempsids, (?sid,bidReserves) /
(?bidBoats)) ?sname(Tempsids ? Sailors
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QUERY Q10

• (Q10) Find the names of sailors who have reserved
  all boats called Interlake.

?(Tempsids, (?sid,bidReserves)/(?bid(sbnameInter
lakeBoats))) ?sname(Tempsids ? Sailors)
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• Natural Join
• - combines s, p, ?
• - very commonly used
• Natural Join forms the cross product of its two
  arguments, does a selection to enforce equality
  of columns with the same name and removes
  duplicate columns.
• Eg show all transactions done by account owner
  Bob
• s ownerBob (account JOIN transaction)

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Rename operation

• What if you need to access the same relation
  twice in a query?
• eg. person(ss, name, mother_ss, father_ss)
• Find the name of Bobs mother needs the
  person table to be accessed twice.
• The operation ? x (r) evaluates to a second
  logical copy of relation r renamed to x.

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Rename operation (contd)

• eg
• p mother.name (
• (? mother (person))
• JOIN mother.ss person.mother_ss
• (s nameBob (person)))

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