Relational Algebra

Lesson Outcomes2

By the end of the lesson students should be able to:

Relate the use of relational algebra in database manipulation

Identify the basic symbols used in relational algebra

Express database queries in the form of relational algebra expressions

What is Relational Algebra (RA) ?3

A query language (QL) is a language that allows users to manipulate and retrieve data from a database.

The relational model supports simple, powerful QLs (having strong formal foundation based on logics, allow for much optimization)

Two (mathematical) query languages form the basis for “real" languages (e.g., SQL) and for implementation. One of them is relational algebra

Relational Algebra is useful for representing query execution plans, and query optimization techniques.

Relational Algebra comprises a set of basic operations. An operation is the application of an operator to one or more

source (or input) relations to produce a new relation as a result. Relational algebra has the property of closure, that is the

output from each of the operations is another relation.

5 Basic RA Operations• Projection ( ) Retains only wanted columns from

relation (vertical).• Selection ( ) Selects a subset of rows from

relation (horizontal).• Cross-product (x) Allows us to combine two

relations.• Set-difference (–) Tuples in R1, but not in R2.• Union ( ) Tuples in R1 and/or in R2.

NOTE: R1 = relation/table 1 R2 = relation/table 2

s

p

Example InstancesS2Boats

bid bname color 101 Interlake blue 102 Interlake red 103 Clipper green 104 Marine red

sid bid day 22 101 10/10/96 58 103 11/12/96

sid sname rating age 28 yuppy 9 35.0 31 lubber 8 55.5 44 guppy 5 35.0 58 rusty 10 35.0

Rents

Projection

page S( )2

Retains only attributes that are in the “projection list”.

Examples:

The 1st query will produce a relation with only one column which is age. While the 2nd query produces a relation with two columns: sname and rating. The number of rows depends on the number of unique content. e.g. 1st query, 2 rows, 2nd query, 4 rows

Projection operator eliminates duplicates rows

p sname rating S, ( )2

Projection

)2(, Sratingsnamep

page S( )2

sname rating yuppy 9 lubber 8 guppy 5 rusty 10

age 35.0 55.5

Eliminates redundant

35.0

Selection (s)

s rating S>8 2( )

Selects rows that satisfy selection condition. Can also have multiple conditions

Result is a relation with same number of columns BUT probably lesser number of rows.

Example:

sid sname rating age 28 yuppy 9 35.0 31 lubber 8 55.5 44 guppy 5 35.0 58 rusty 10 35.0

Union and 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.R1 U R2 R1 - R2 R2 - R1

Union

S1

S2

sid sname rating age 22 dustin 7 45.0 31 lubber 8 55.5 58 rusty 10 35.0

sid sname rating age 28 yuppy 9 35.0 31 lubber 8 55.5 44 guppy 5 35.0 58 rusty 10 35.0

sid sname rating age 22 dustin 7 45.0 31 lubber 8 55.5 58 rusty 10 35.0 44 guppy 5 35.0 28 yuppy 9 35.0

S1 U S2

Eliminates duplicates

Set Difference

S1

S2

S S1 2-

S2 – S1

sid sname rating age 22 dustin 7 45.0 31 lubber 8 55.5 58 rusty 10 35.0

sid sname rating age 28 yuppy 9 35.0 31 lubber 8 55.5 44 guppy 5 35.0 58 rusty 10 35.0

sid sname rating age 22 dustin 7 45.0

sid sname rating age 28 yuppy 9 35.0 44 guppy 5 35.0

Cross-Product R1 x R2: Each row of R1 (m rows) paired with each row of

R2 (n rows). Q: How many rows in the result? m x n• Result schema has one field per field of R1 and R2, with

field names `inherited’ if possible. May have a naming conflict: Both R1 and R2 have a field

with the same name. In this case, can use the renaming operator (OPTIONAL):

X (E) - rename expression E as x (A1, A2, …, An) (E) = rename attributes in E as A1, A2,..,An

Cross Product Example

R1R2

R1 X R2 =

sid sname rating age 22 dustin 7 45.0 31 lubber 8 55.5 58 rusty 10 35.0

sid bid day 22 101 10/10/96 58 103 11/12/96

(sid) sname rating age (sid) bid day 22 dustin 7 45.0 22 101 10/ 10/96 22 dustin 7 45.0 58 103 11/ 12/96 31 lubber 8 55.5 22 101 10/ 10/96 31 lubber 8 55.5 58 103 11/ 12/96 58 rusty 10 35.0 22 101 10/ 10/96 58 rusty 10 35.0 58 103 11/ 12/96

Compound Operator: Intersection

In addition to the 5 basic operators, there are several additional “Compound Operators” These add no computational power to the

language, but are useful shorthands. Can be expressed solely with the basic

operations.

Intersection takes two input relations, which must be union-compatible.

Q: How to express it using basic operators?R S = R - (R - S)

Intersection

S1

S2

sid sname rating age 31 lubber 8 55.5 58 rusty 10 35.0

sid sname rating age 22 dustin 7 45.0 31 lubber 8 55.5 58 rusty 10 35.0

sid sname rating age 28 yuppy 9 35.0 31 lubber 8 55.5 44 guppy 5 35.0 58 rusty 10 35.0

S1 n S2

Compound Operator: Join Joins are compound operators involving cross

product, selection, and (sometimes) projection. Most common type of join is a “natural join” (often

just called “join”). R S conceptually is: Compute R X S Select rows where attributes that appear in both relations have equal

values and domains Project all unique atttributes and one copy of each of the common

ones. Note: Usually done much more efficiently than this. Useful for putting “normalized” relations back

together.

Natural Join Example

R1S1

S1 R1 =

sid sname rating age bid day22 dustin 7 45.0 101 10/10/9658 rusty 10 35.0 103 11/12/96

sid sname rating age 22 dustin 7 45.0 31 lubber 8 55.5 58 rusty 10 35.0

sid bid day 22 101 10/10/96 58 103 11/12/96

“Theta” Join Examplesid sname rating age22 dustin 7 45.031 lubber 8 55.558 rusty 10 35.0

sid bid day22 101 10/10/9658 103 11/12/96

R1 S1

 

S1>< S1.sid < R1.sid R1 =

(sid) sname rating age (sid) bid day22 dustin 7 45.0 58 103 11/12/9631 lubber 8 55.5 58 103 11/12/96

Outer Join

• An extension of the join operation that avoids loss of information.

• Computes the join and then adds tuples form one relation that does not match tuples in the other relation to the result of the join.

• Uses null values:null signifies that the value is unknown or does

not exist

Outer Join – Example

Relation loan

customer-name loan-number

JonesSmithHayes

L-170L-230L-155

300040001700

loan-number amount

L-170L-230L-260

branch-name

DowntownRedwoodPerryridge

Relation borrower

Left Outer JoinJoin

loan Borrower

loan-number amount

L-170L-230

30004000

customer-name

JonesSmith

branch-name

DowntownRedwood

JonesSmithnull

loan-number amount

L-170L-230L-260

300040001700

customer-namebranch-name

DowntownRedwoodPerryridge

Left Outer Join

loan Borrower

Right Outer Join, Full Outer JoinRight Outer Join loan borrower

loan borrowerFull Outer Join

loan-number amount

L-170L-230L-155

30004000null

customer-name

JonesSmithHayes

branch-name

DowntownRedwood

null

loan-number amount

L-170L-230L-260L-155

300040001700null

customer-name

JonesSmithnull

Hayes

branch-name

DowntownRedwoodPerryridge

null

Compound Operator: Division

Division is useful for expressing “for all” queries like: Find sids of sailors who have reserved all boats.

For A/B attributes of B are subset of attributes of A. May need to “project” to make this happen.

E.g., let A have 2 fields, x and y; B have only field y:

A/B contains all x tuples such that for every y tuple in B, there is an xy tuple in A.

Examples of Division A/B (AB)sno pno s1 p1 s1 p2 s1 p3 s1 p4 s2 p1 s2 p2 s3 p2 s4 p2 s4 p4

pnop2

pnop2p4

pnop1p2p4

snos1s2s3s4

snos1s4 sno

s1A

B1B2

B3

A/B1 A/B2 A/B3

Another Division Example

A B

aaaaaaaa

C D

aabababb

E11113111

Relations r, s:

r s:

Dab

E11

A B

aa

C

s

Aggregate Functions and Operations

• Aggregation function takes a collection of values and returns a single value as a result.avg: average valuemin: minimum valuemax: maximum valuesum: sum of valuescount: number of values

• Aggregate operation in relational algebra G1, G2, …, Gn ℱ F1( A1), F2( A2),…, Fn( An) (E)– E is any relational-algebra expression– G1, G2 …, Gn is a list of attributes on which to group (can be

empty)– Each Fi is an aggregate function– Each Ai is an attribute name

Aggregate Operation – Example

Relation r:A B

C

77310

ℱ sum(c) (r)sum-C

27ℱ

Aggregate Operation – Example

Relation account grouped by branch-name:

branch-name ℱ sum(balance) (account)

branch-name account-number balance

PerryridgePerryridgeBrightonBrightonRedwood

A-102A-201A-217A-215A-222

400900750750700

branch-name balance

PerryridgeBrightonRedwood

13001500700

ℱ

Assignment Operation

• The assignment operation () provides a convenient way to express complex queries. – Write query as a sequential program consisting

of a series of assignments followed by an expression whose value is displayed as a result of the query.

– Assignment must always be made to a temporary relation variable.

• Example: Write r s as temp1 R-S (r) temp2 R-S ((temp1 x s) – R-S,S (r))result = temp1 – temp2

References32

www.cs.kent.edu/~yuri/cs43005.s2008/lectures/lecture2.ppt

inst.eecs.berkeley.edu/~cs186/sp06/lecs/lecture8Alg.ppt

and others

Exercises33

Given the following relational schema: STUDENT (sid, sname) COURSE (ccode,cname, ccredithr, lno) LECTURER (lno, lname) ENROLMENT(sid,ccode) Write the relational algebra expression to extract

the following: The name of all courses offered The name of 3 credit hours’ courses

Exercises34

The name of courses along with the name of the lecturers who teach them

The name of courses and the name of lecturers

The name of lecturers who teach 3 cr hr’s courses as well as 4 cr hr’s courses

The name of lecturers who teach 3 cr hr’s courses but not 4 cr hr’s courses

Exercises35

The name of students who take SCB1033 and SCB1043.

The name of students who take all courses taught by Dr Samuel.

The maximum number of credit hours offered The number of students enrolled in each

course The number of credit hrs taken by a student

with id 1121.