Relational Model and Algebra

Overview

Relational algebra is a mathematical query language for relations.
It is a set based query language:
- The input to each operator is one or more relations, sets of tuples.
- The output of each operator is a relation: a set of tuples.
Relational algebra is based on a minimal set of operators that can be combined to write complex queries.
The meaning (semantics) of other query languages, i.e. SQL, are defined in terms of relational algebra.
SQL queries are translated to relational algebra.
Databases implement relational algebra operators to execute SQL queries.
- Caveat: we will see a bag version of these operators for SQL later in the semester.
Mathematical definition of database design (normalization) uses relational algebra.

Relational Algebra Overview

Crucial operators (complete set):
- Union
- Set difference
- Selection
- Projection
- Cartesian Product
- Renaming
Derived operators (added for convenience):
- Theta join
- Natural join
- Set intersection
Each operator has a Greek symbol that starts with the same letter as the operator: Select-sigma, Project-Pi, etc.

Set operators

We use the term union compatible to mean that two relations have the same schema: exactly same attributes drawn from the same domain.
Given two relations R and S that are union compatible:
- Union:
  
  $R \cup S = \{ \mbox{the set of all tuples that are either in R or S} \}$
- Intersection:
  
  $R \cap S = \{ \mbox{the set of all tuples are both in R and S} \}$
- Set difference:
  
  $R - S = \{ \mbox{the set of all tuples that are in R but not in S} \}$
The result is a relation with same schema as R (and S).
If two relations are not union compatible, the set operation is not defined.
Note that set intersection is not technically necessary as it can driven from other operations:

$R \,\cap\, S = (R\,\cup\,S)- ((R-S)\,\cup\,(S-R))$

Set Operation Examples

The relations MarvelHeroes, DCHeroes, TVHeros are compatible, they have the same attributes.
The relations Movies and TVShows are not compatible, they have different attributes. Even if they had the same number of attributes, the name of attributes should match.
Find all heroes in the database

$MarvelHeroes \,\cup\, DCHeroes$
Find all heroes that are in both universes (which we know will be the empty set)

$MarvelHeroes \,\cap\, DCHeroes$

Output:

hid

hero

realname
Find all DC comics heroes that are in a TV show

$DCHeroes \,\cap\, TVHeroes$

Output:

hid

hero

realname

h8

Superman

Clark Kent

h9

Batman

Bruce Wayne

h10

Supergirl

Kara Danvers

h11

Flash

Barry Allen

h12

Arrow

Oliver Queen

h14

Wonder Woman

Diana Prince
Find all Marvel Comics heroes that are not in a TV show

$MarvelHeroes \,-\, TVHeroes$
Find all heroes that are not in a TV show (both solutions are equivalent)

$(MarvelHeroes \,\cup\, DCHeroes) \,-\, TVHeroes (MarvelHeroes\,-\, TVHeroes) \,\cup\, (DCHeroes\,-\, TVHeroes)$
Note that set subtraction is special, it is the only way to write negation (something that is NOT true for the whole relation)

Rename

Rename operation renames the relation and the attributes in it. It does not change the contents of the relation.
We can use any of two following notations for rename.
Option 1.

$\rho_{S(A_1,A_2,\ldots,A_n)} \, R$
Option 2.

$S(A_1,A_2,\ldots,A_n) := R$

If you did not list the attributes, you are simply changing the name of the relation:

$S := R$

This is used to give names to intermediate results of the relational algebra operations.

Example:

$X(id1, hero1, r1) := DCHeroes$

Output:

id1	hero1	r1
h8	Superman	Clark Kent
h9	Batman	Bruce Wayne
h10	Supergirl	Kara Danvers
h11	Flash	Barry Allen
h12	Arrow	Oliver Queen
h13	Green Lantern	Hal Jordan
h14	Wonder Woman	Diana Prince

Selection

Given a relation R and a boolean condition C over the attributes of R, the selection is given by:

$\sigma_C\, R = \{ \mbox{the set of all tuples in R that satisfy the condition C}\}$
- Selection returns a new relation with the same schema as R, but containing a subset of the tuples in R based on the condition C.
All movies made before 2010

$\sigma_{year < 2010}\, Movies$

Output:

mid

moviename

year

m2

Iron Man

2008

m4

The Incredible Hulk

2008

m13

Superman Returns

2006

m14

The Dark Knight

2008
All heroes who do not have an alias

$\sigma_{hero=realname} (MarvelHeroes \,\cup\, DCHeroes)$

Output:

hid

hero

realname

h15

Jessica Jones

Jessica Jones
All TV shows that are still running and have started after 2012.

$\sigma_{firstyear\geq 2013 \mbox{ and } ended=No}\, TVShows$

Output:

sid

showname

hid

Channel

FirstYear

LastYear

Ended

s3

Supergirl

h10

CBS

2015

2016

No

s4

Gotham

h9

FOX

2014

2016

No

s5

Jessica Jones

h15

Netflix

2015

2016

No

Projection

Projection does not change the tuples in the relation, but the schema of the relation.

$\Pi_{A_1,A_2,\ldots,A_n}\, R = \{\mbox{all tuples in R but only containing attributes} A_1,A_2,\ldots,A_n\}$
As each operator returns a set of tuples, any duplicate values will be removed. As a result, the result of a projection may contain fewer tuples than the input relation.
Find the real name of all heroes

$\Pi_{realname} (MarvelHeroes \,\cup\, DCHeroes)$
Find the id of all heroes in a TV show

$\Pi_{hid} TVHeroes$

Output:

hid

h3

h8

h9

h10

h11

h12

h14

h15
Find all years in which a hero movie was made

$\Pi_{year} Movies$
Find the id of all heroes that were both in a movie and a tv show.

$(\Pi_{hid} TVShows) \,\cap\, (\Pi_{hid} HeroInMovie)$

Output:

hid

h12

h11

h10

h9

h15

Find the id of all movies with no heroes in them (according to our database instance)

$(\Pi_{mid} Movies) - (\Pi_{mid} HeroInMovie)$

Output:

mid

m10

Find start and end year of all TV shows with name The Flash.

$\Pi_{firstyear, lastyear} (\sigma_{showname=The Flash}\, TVShows)$

Output:

FirstYear

LastYear

2012

2016

1990

1991
Projection is crucial for changing the schema of relations, especially before a set operation!

Cartesian Product

Given two relations R and S that have no attributes in common,

$R \times S = \{\mbox{tuples (r,s) for all tuples r in R and s in S}\} Note that (r,s) contains a value for attributes in R and all attributes in S.$
The schema of RxS is the union of the attributes in R and S. As a result, the attributes in R and S must have unique names to distinguish them from each other.
Cartesian product is a multiplication. If R has n tuples and S has m tuples, the Cartesian product will produce n*s tuples.
Cartesian product is the only way to put tuples from two relations side by side.

Cartesian Product Example

Given the following smaller relation instances:

HeroInMovie

hid	mid
h1	m1
h2	m2

R1(h1, hero, realname) := MarvelHeroes

h1	hero	realname
h1	Captain America	Steve Rogers
h2	Iron Man	Tony Stark
h3	Hulk	Bruce Banner

HeroInMovie x R1:

hid	mid	h1	hero	realname
h1	m1	h1	Captain America	Steve Rogers
h1	m1	h2	Iron Man	Tony Stark
h1	m1	h3	Hulk	Bruce Banner
h2	m2	h1	Captain America	Steve Rogers
h2	m2	h2	Iron Man	Tony Stark
h2	m2	h3	Hulk	Bruce Banner

Notice that we renamed the id attribute in MarvelHeroes to make sure the schema of the two relations had no attributes in common.
If we wanted to return only the tuples with matching hero ids, we need to do a selection:

$R1(h1, hero, realname) := MarvelHeroes \Pi_{hero} (\sigma_{hid=h1}\, (HeroInMovie \times R1))$

This query returns the name of all Marvel heroes who have a movie in the database.

Theta-Join (or simply Join)

Cartesian product is often (but not always) followed by a selection. We can define a short cut for this combination for simplicity:

$R \,\bowtie_C\, S$

which is read as R theta join S on join condition C.
Join is the same as a Cartesian product, followed by a selection:

$R \,\bowtie_C\, S = \sigma_C (R\times S)$
A join condition must involve expressions that compare attributes from R to S.
- Given R(A,B) and S(C,D):
  - (A=C or B>D) is a join condition.
  - A=5 or B=4 is not a join condition.
We can rewrite the above query as:

$R1(h1, hero, realname) := MarvelHeroes \Pi_{hero} (HeroInMovie \bowtie_{hid=h1} R1))$
Note that you equally have to rename attributes before using theta-join to make sure join conditions can be written unambiguously.

Natural Join

Natural join of two relations R and S (R*S) is given by a join on the equality of all attributes in common. The common attributes are not repeated.
For example:

$R := DCHeroes * HeroInMovie$

or

$R := DCHeroes \bowtie HeroInMovie$

will return a relation R with schema:

R(hid, hero, realname, mid)

such that

R matches hero ids from DCHeroes and HeroInMovie

R only contains heroes with a movie and movies with a matching hero

hid is not repeated

Output:

hid

hero

realname

mid

h8

Superman

Clark Kent

m13

h9

Batman

Bruce Wayne

m14

h13

Green Lantern

Hal Jordan

m15

h14

Wonder Woman

Diana Prince

m16

Find the name of all movies with the hero whose real name is Tony Stark or Bruce Wayne.

$R1 := (MarvelComics \,\cup\, DCComics) * HeroInMovie * Movies R2 := \sigma_{\mbox{realname = Tony Stark or realname=Bruce Wayne}} R R3 := \Pi_{moviename} R2$

Exercises:

Write the following queries using relational algebra. Try the simplest possible expression possible. There may be multiple solutions to the same query.

Find the name of all movies that are released after 2014 and has a Marvel comic hero in them.
Find pairs of heroes who have starred in the same movie. Return their aliases.

Try to write this so that you only return each pair only once (do not return both a,b and b,a)
Find the name of all movies that has more than one comic hero in them.
Find the name of all movies that has only one comic hero in them.
Find the TV shows that star heroes who have also been in a movie.

Summary

Observe that there are many ways to write the same query using different relational algebra operators or different ordering of the same operators.
Logically equivalent operations may have different time complexity, that is what query optimization is about.

mid	moviename	year
m2	Iron Man	2008
m4	The Incredible Hulk	2008
m13	Superman Returns	2006
m14	The Dark Knight	2008

sid	showname	hid	Channel	FirstYear	LastYear	Ended
s3	Supergirl	h10	CBS	2015	2016	No
s4	Gotham	h9	FOX	2014	2016	No
s5	Jessica Jones	h15	Netflix	2015	2016	No

FirstYear	LastYear
2012	2016
1990	1991