RELATIONAL ALGEBRAusers.abo.fi/soini/DBEngTest.pdf · Who_does_what_scheme = (teacher, code)1 Course_registration_scheme = (studentnumber, code) course course_registration code coursename

RELATIONAL ALGEBRA FOR BEGINNERS

Annamari Soini, dept. of Information Technology Åbo Akademi

Course database:

Course_scheme = (code, coursename, cu, period)Student_scheme = (studentnumber, name, UP, sex) Staff_scheme = (name, title)Who_does_what_scheme = (teacher, code)1

Course_registration_scheme = (studentnumber, code)

course course_registration

code coursename cu period

studentnumber code

A001 History of Art 8 1 SA01 A001

A002 Renaissance Art 5 2 SA01 A002

A003 Modern Art 5 3 SA02 A001

E001 English Grammar 5 1 SA02 A002

E002 English Literature 5 2 SA02 A003

E003 Shakespeare 1 5 3 SA03 A003

L001 Latin 1 5 1 SE01 E001

L002 Latin 2 5 2 SE01 E002

L003 Caesar 5 3 SE01 A001

L004 Catullus 5 4 SE02 E003

CS001 Introduction to CS 5 1 SE03 A001

CS002 Programming 1 5 1 SE03 L003

CS003 Programming 2 5 2 SL01 L003

CS004 Databases 5 3 SL02 L004

CS005 Data Structures 5 4 SL03 L004

SL03 A001

SCS01 CS004

SCS02 CS004

SCS03 CS005

SCS04 CS005

SCS05 A003

SCS05 L004

1 Allows several teachers per course, and several courses per teacher.

student staff

studentnumber

name UP sex name title

SA01 Mirella Conti Arts F Caravaggio Dr.

SA02 Fleur D'Amour Arts F Picasso M. A.

SA03 Belle Visiteur Arts F Austen Dr.

SA04 Carlo Straniero Arts M Eliot Dr.

SE01 Will Smith English M Nero M. A.

SE02 Angela Bowie English F Turing Dr.

SE03 Sharon Manners English F Church M. SC.

SE04 Jack Harper English M

SL01 Lucius Valerius Latin M who_does_what

SL02 Antonius Primus Latin M teacher code

SL03 Fulvia Morbida Latin F Caravaggio A001

SL04 Septimus Romanus Latin M Caravaggio A002

SCS01 Betty Buffer CS F Picasso A003

SCS02 Bill Bitwise CS M Austen E001

SCS03 Lily Float CS F Austen E002

SCS04 Judy Python CS F Eliot E003

SCS05 Percy Pascal CS M Nero L001

Nero L002

Nero L003

Nero L004

Church CS001

Church CS002

Church CS003

Turing CS004

Turing CS005

chooses columns (attributes)

coursename(course):

coursename

History of Art

Renaissance Art

Modern Art

English Grammar

English Literature

Shakespeare 1

Latin 1

Latin 2

Caesar

Catullus

Introduction to CS

Programming 1

Programming 2

Databases

Data Structures

chooses rows (tuples)

period = 3(course):


A003 Modern Art 5 3

E003 Shakespeare 1 5 3

L003 Caesar 5 3

CS004 Databases 5 3

temp period = 4(course)

temp:


L004 Catullus 5 4

CS005 Data Structures 5 4

What are the courses given by Turing?

• First pick the table that has that information (who_does_what).

• Then select () the rows where teacher = 'Turing'.

• Finally project () the column(s) asked for (code).

who_does_what

teacher code

Caravaggio A001

Caravaggio A002

Picasso A003

Austen E001

Austen E002

Eliot E003

Nero L001

Nero L002

Nero L003

Nero L004

Church CS001

Church CS002

Church CS003

Turing CS004

Turing CS005

teacher = 'Turing' (who_doeswhat)

teacher code

Turing CS004

Turing CS005

code (teacher = 'Turing' (who_doeswhat))

code

CS004

CS005

What are these courses called?

temp code (teacher = 'Turing' (who_doeswhat))

tempcode

CS004

CS005

• Which table has the information required (the names of the courses)? • combine course and temp:

temp course (every row of temp paired with every row of course):

(2 x 15 rows, and most of these will be nonsense)

temp course

temp.code course.code coursename cu period

CS004 A001 History of Art 8 1


CS004 A002 Renaissance Art 5 2


CS004 A003 Modern Art 5 3


CS004 E001 English Grammar 5 1


CS004 E002 English Literature 5 2


... ...

CS004 CS004 Databases 5 3


CS004 CS005 Data Structures 5 4


Select only the sensible rows (where temp.code = course.code):

temp.code = course.code ( temp course)

Parenthesis () tell you which table(s) named or anonymous you go to for information.

temp.code = course.code ( temp course):

temp course












... ...





... giving you




Now you can write the expression:

coursename (temp.code = course.code ( temp course))

coursename (temp.code = course.code ( temp course)):

coursename

Databases

Data Structures

You can replace the cartesian product (cross product) temp course with a natural join (1): this will automatically choose only the “sensible” rows

(combinations) from the cross product.

This is how the natural join (1) works:

• Use the common attribute (or attributes, if several), in this case code.• Only the rows with the same value for this attribute get chosen:

temp 1 course:




• Finally duplicate attributes are removed:


CS004 Databases 5 3


Now we can write our expression as follows:

coursename (temp 1 course):

coursename (temp 1 course):

coursename

Databases

Data Structures

A couple more exercises on and 1 :

• Find the names of the students who take A001

name (student.student_number = course_registration.studentnumber

∧ code = 'A001' (student course_registration))

The cross product student course_registration will have 17 x 22 rows;every row from student (17 rows) will be combined with every row from course_registration (22 rows), giving you a table with 374 rows!

studentnumber

name UP sex studentnumber

code

SA01 Mirella Conti Arts F SA01 A001






SA01 Mirella Conti Arts F SE01 E001


SA01 Mirella Conti Arts F SE01 A001


SA01 Mirella Conti Arts F SE03 A001

SA01 Mirella Conti Arts F SE03 L003

SA01 Mirella Conti Arts F SL01 L003



SA01 Mirella Conti Arts F SL03 A001

SA01 Mirella Conti Arts F SCS01 CS004




SA01 Mirella Conti Arts F SCS05 A003

SA01 Mirella Conti Arts F SCS05 L004

SA02 Fleur D'Amour Arts F SA01 A001


... ... ... ... ... ...

SCS05 Percy Pascal CS M SCS05 L004

The result will be:

name

Mirella Conti

Fleur D'Amour

Will Smith

Sharon Manners

Fulvia Morbida

... or you can use natural join:

name (code = 'A001' ( student 1 course_registration))

student 1 course_registration:

• I studentnumber is shown here twice, but the natural join eliminates one of these identical columns (see next page!).

studentnumber


code






SA03 Belle Visiteur Arts F SA03 A003

SE01 Will Smith English M SE01 E001

SE01 Will Smith English M SE01 E002

SE01 Will Smith English M SE01 A001

SE02 Angela Bowie English F SE02 E003

SE03 Sharon Manners English F SE03 A001

SE03 Sharon Manners English F SE03 L003

SL01 Lucius Valerius Latin M SL01 L003

SL02 Antonius Primus Latin M SL02 L004

SL03 Fulvia Morbida Latin F SL03 L004

SL03 Fulvia Morbida Latin F SL03 A001

SCS01 Betty Buffer CS F SCS01 CS004

SCS02 Bill Bitwise CS M SCS02 CS004

SCS03 Lily Float CS F SCS03 CS005

SCS04 Judy Python CS F SCS04 CS005

SCS05 Percy Pascal CS M SCS05 A003

SCS05 Percy Pascal CS M SCS05 L004

student 1 course_registration:

• Duplicate column studentnumber eliminated:

studentnumber

name UP sex code

SA01 Mirella Conti Arts F A001


SA02 Fleur D'Amour Arts F A001



SA03 Belle Visiteur Arts F A003

SE01 Will Smith English M E001


SE01 Will Smith English M A001

SE02 Angela Bowie English F E003

SE03 Sharon Manners English F A001

SE03 Sharon Manners English F L003

SL01 Lucius Valerius Latin M L003

SL02 Antonius Primus Latin M L004

SL03 Fulvia Morbida Latin F L004

SL03 Fulvia Morbida Latin F A001

SCS01 Betty Buffer CS F CS004

SCS02 Bill Bitwise CS M CS004

SCS03 Lily Float CS F CS005

SCS04 Judy Python CS F CS005

SCS05 Percy Pascal CS M A003

SCS05 Percy Pascal CS M L004

Now choose only those rows where code equals A001:

code = 'A001' ( student 1 course_registration):

studentnumber

name UP sex code























... gives us the following table:

code = 'A001' ( student 1 course_registration):

studentnumber


code



SE01 Will Smith English M SE01 A001

SE03 Sharon Manners English F SE03 A001

SL03 Fulvia Morbida Latin F SL03 A001

name (code = 'A001' ( student 1 course_registration))

gives you now the same result:

name

Mirella Conti

Fleur D'Amour

Will Smith

Sharon Manners

Fulvia Morbida

Three more joins (ways to combine two tables):

• left outer join • right outer join • full outer join

• left outer join : student courseregistration:• First you perform natural join, then you fill in information from the left

side table (those rows that did not get paired with the natural join), in this case those students who are not registered on any course.

studentnumber

name UP sex code























SA04 Carlo Straniero Arts M NULL

SE04 Jack Harper English M NULL

SL04 Septimus Romanus Latin M NUL

• right outer join : courseregistration course:• First natural join, then fill in information from the right side table! (In this

case, courses that have no registered students.)studentnumber


SA01 A001 History of Art 8 1

SA01 A002 Renaissance Art 5 2



SA02 A003 Modern Art 5 3


SE01 E001 English Grammar 5 1

SE01 E002 English Literature 5 2

SE01 A001 History of Art 8 1

SE02 E003 Shakespeare 1 5 3


SE03 L003 Caesar 5 3

SL01 L003 Caesar 5 3

SL02 L004 Catullus 5 4


SL03 A001 History of Art 8 1

SCS01 CS004 Databases 5 3


SCS03 CS005 Data Structures 5 4


SCS05 A003 Modern Art 5 3

SCS05 L004 Catullus 5 4

NULL L001 Latin 1 5 1

NULL L002 Latin 2 5 2

NULL CS001 Introduction to CS 5 1

NULL CS002 Programming 1 5 1

NULL CS003 Programming 2 5 2

full outer join : scholarship student

To exemplify this join, we define a table scholarship, with information of which scholarships there are, and which students have received them (we assume that there may only be one student who gets the scholarship, and some scholarships have not yet been awarded to anyone):

Scholarship_scheme = (scholarshipname, studentnumber)

scholarship

scholarshipname studentnumber

Top Latin SL03

Logic Excellence SCS04

Arts Achievement NULL

Creative Writer SE01

Top Athlete NULL

First make the natural join: scholarship 1 student:


name UP sex

Top Latin SL03 Fulvia Morbida Latin F

Logic Excellence SCS04 Judy Python CS F

Creative Writer SE01 Will Smith English M

Then fill in information from the left side table (scholarship), namely those rows which did not find a pair in the natural join (that is, scholarships that are not awarded to any student as yet):


name UP sex


Logix Excellence SCS04 Judy Python CS F


Arts Achievement NULL NULL NULL NULL

Top Athlete NULL NULL NULL NULL

Finally, fill in with information from the right side table (those students who did not get a scholarship):


name UP sex


Logix Excellence SCS04 Judy Python CS F


Arts Achievement NULL NULL NULL NULL

Top Athlete NULL NULL NULL NULL

NULL SA01 Mirella Conti Arts F

NULL SA02 Fleur D'Amour Arts F

NULL SA03 Belle Visiteur Arts F

NULL SA04 Carlo Straniero Arts M

NULL SE02 Angela Bowie English F

NULL SE03 Sharon Manners English F

NULL SE04 Jack Harper English M

NULL SL01 Lucius Valerius Latin M

NULL SL02 Antonius Primus Latin M

NULL SL04 Septimus Romanus Latin M

NULL SCS01 Betty Buffer CS F

NULL SCS02 Bill Bitwise CS M

NULL SCS03 Lily Float CS F

NULL SCS05 Percy Pascal CS M

Aggregate functions:

• Work on a collection of rows (“aggregate”) to produce one result(a set with just one value as a member).

• How many courses are given in period 1?

Gcount() ( period = 1 ( course)):

course


A001 History of Art 8 1

A002 Renaissance Art 5 2

A003 Modern Art 5 3

E001 English Grammar 5 1

E002 English Literature 5 2


L001 Latin 1 5 1

L002 Latin 2 5 2

L003 Caesar 5 3

L004 Catullus 5 4

CS001 Introduction to CS 5 1

CS002 Programming 1 5 1


CS004 Databases 5 3


• How many cu:s does one get if one takes all the courses that are given in period 1?

Gsum(cu) (period = 1 ( course))

5

Gsum(cu) (period = 1 ( course)):

course




A003 Modern Art 5 3




L001 Latin 1 5 1

L002 Latin 2 5 2

L003 Caesar 5 3

L004 Catullus 5 4




CS004 Databases 5 3


With aggregate functions one sometimes uses grouping:

• first you form the groups• then the operation requested is performed in each group separately:

28

• “How many courses are given in each period?”

periodGcount()(course)

• First, form the groups, using period as the criterion:




L001 Latin 1 5 1





L002 Latin 2 5 2


A003 Modern Art 5 3


L003 Caesar 5 3

CS004 Databases 5 3

L004 Catullus 5 4


• Then, count the rows in each group:

period count()

1 5

2 4

3 4

4 2

• “How many points will each student get if they pass all the courses they are registered on?”

◦ First form the groups: one group for each student, containing the courses this student is registered on, then add the cucolumns:

studentnumberGsum(cu)(course_registration 1 course) ... first:

studentnumber








SE01 E001 English Grammar 5 1

SE01 E002 English Literature 5 2


SE02 E003 Shakespeare 1 5 3


SE03 L003 Caesar 5 3

SL01 L003 Caesar 5 3



SL03 A001 History of Art 8 1





SCS05 A003 Modern Art 5 3

SCS05 L004 Catullus 5 4

Then: studentnumberGsum(cu)(course_registration 1 course)

studentnumber

sum(cu)

SA01 13

SA02 18

SA03 5

SE01 18

SE02 5

SE03 13

SL01 5

SL02 5

SL03 13

SCS01 5

SCS02 5

SCS03 5

SCS04 5

SCS05 10

As you can see, the attribute we use for grouping (studentnumber) and the result of the aggregate function (sum(cu)) are projected in the result.

Set operations

• union ∪• intersection ∩• difference • division (you will also get a separate pdf on this!)• multiplication you are already familiar with!

UNION: combines the elements of two sets

• the elements of each set must have the same type• duplicate elements are eliminated from the result

Example: “Which students take either A001 or L004?”

studentnumber(code = 'A001' ( course_registration))

studentnumber(code = 'L004' ( course_registration))

Those who take A001:

studentnumber

SA01

SA02

SE01

SE03

SL03

Those who take L004:

studentnumber

SL02

SL03

SCS05



studentnumber

SA01

SA02

SE01

SE03

SL03

SL02

SCS05

Note that the student SL03 who takes both courses shows in the result only once!

A001

L004

SA01

SA02

SE01

SE03

SL03

SL02

SCS05

INTERSECTION: chooses the elements that are in both sets:

Example: “Which students take both A001 and L004?”



Those who take A001: Those who take L004:

studentnumber studentnumber

SA01 SL02

SA02 SL03

SE01 SCS05

SE03

SL03

The result:

studentnumber

SL03

A001

L004

SA01

SA02

SE01

SE03

SL03

SL02

SCS05

SET DIFFERENCE: Which elements are in the first set but NOT in the other?

Example: “Which students take A001 but not L004?”

studentnumber(code = 'A001' ( course_registration)) -


Those who take A001: Those who take L004:

studentnumber studentnumber

SA01 SL02

SA02 SL03

SE01 SCS05

SE03

SL03

Result:

studentnumber

SA01

SA02

SE01

SE03

This part not this part

A001

L004

SA01

SA02

SE01

SE03

SL03

SL02

SCS05

DIVISION: Which elements share ALL of another relation? I. e., which elements have ALL of this other relation in common.

• Used for queries with “all”:• “Which students take ALL the courses given in period 1?”

◦ All the courses given in period 1 is the set (A001, E001, L001, CS001, CS002)and we ask which students, if any, are registered in ALL of these courses. They have all of these courses in common.

• “Which female students take ALL the courses that student nr. SE01 is registered on?” (CSI might be interested in finding out this, if SE01, Will Smith, gets murdered by a mystery woman ...)

◦ All the courses SE01 is registered on is the set(E001, E002, A001)and we ask which female students, if any, are registered on ALL of these courses. They have all of these courses in common.

• “Which students take ALL the courses taught by Caravaggio?” (We might give this query if we want to find out who are Caravaggio's greatest fans.)

◦ All the courses taught by Caravaggio is the set(A001, A002)and we ask which students, if any, are registered on ALL of these courses. They have all of these courses in common.

• BUT NOT: “Which students have passed (if we had this information) ALL the courses they are registered on?”

◦ In this case there is no COMMON set of courses; the set to be tested varies according to the student in question. There is (most probably) no set of courses they would all have in common.

Example: “Which students take ALL the courses taught by Caravaggio?”

• First, find out which courses are taught by Caravaggio:

Caravaggios_courses code (teacher = 'Caravaggio' (who_doeswhat))

who_does_what

teacher code

Caravaggio A001

Caravaggio A002

Picasso A003

Austen E001

Austen E002

Eliot E003

Nero L001

Nero L002

Nero L003

Nero L004

Church CS001

Church CS002

Church CS003

Turing CS004

Turing CS005

Caravaggios_courses:

code

A001

A002

Now divide the table course_registration by Caravaggios_courses:

course_registration Caravaggios_courses:

code

A001

A002

course_registration:

studentnumber code

SA01 A001

SA01 A002

SA02 A001

SA02 A002

SA02 A003

SA03 A003

SE01 E001

SE01 E002

SE01 A001

SE02 E003

SE03 A001

SE03 L003

SL01 L003

SL02 L004

SL03 L004

SL03 A001

SCS01 CS004

SCS02 CS004

SCS03 CS005

SCS04 CS005

SCS05 A003

SCS05 L004

The only students who have ALL (both) of these courses are numbers SA01 and SA02. They are the result of our query.

course_registration Caravaggios_courses:

studentnumber

SA01

SA02

That SA02 also takes a third course (not by Caravaggio) does not affect the result.

• Observe that the column used for division, code, is eliminated from the result!

Documents

RELATIONAL ALGEBRAusers.abo.fi/soini/DBEngTest.pdf · Who_does_what_scheme = (teacher, code)1 Course_registration_scheme = (studentnumber, code) course course_registration code coursename