relational algebra - university of virginia school of ...up3f/cs4750/slides/4750meet20-21-ra.pdf ·...

Fall 2020 – University of Virginia 1© Praphamontripong© Praphamontripong

Relational Algebra

CS 4750Database Systems

[H. Garcia-Molina, J.D. Ullman, J. Widom, Database Systems: The Complete Book, Ch.2]

[A. Silberschatz, H. F. Korth, S. Sudarshan, Database System Concepts, Ch.2.6][C.M. Ricardo, S.D. Urban, “Databases Illuminated, Ch.4.5]

Fall 2020 – University of Virginia 2© Praphamontripong

Relational Algebra (RA)• A data model is not just structure

• Needs a way to query the data • Needs a way to modify the data

• Relational algebra – “Procedural Query Language”

• Ways to build expressions by applying operators to atomic operands and/or other expressions of the algebra

• Atomic operands = variables that stand for relations or constants

• Mathematical foundation to define relationships and get data in and out of the system

• When a DBMS processes queries, a SQL query is translated into relational algebra

•Ways to construct •new relations from

given relations


Why? (Query Designers’ Perspective)

SELECT …FROM …[WHERE …][GROUP BY …][HAVING …][ORDER BY …]

SQL “What” to get from the data

RA“How” to get

the data


Why? (DBMS and Query Processors)

SELECT …FROM …[WHERE …][GROUP BY …][HAVING …][ORDER BY …]

SQL “What” to get from the data

1001010101100 0010111101010 001010100000 1010010100

RA“How” to get

the data

RDBMS

ComputersNeeds “how” to

get data

RA tree(logical plan) (physical plan)


RA Operator Categories1. Operations that remove parts of a relation

• Selection• Projection

2. Operation that changes the schema of a relation• Renaming

3. Operations that combine the tuples of two relations• Set operations: Union, Intersection, Difference• Cartesian product• Natural join

4. Extended operation• Division

Additional operators: Assignment, Aggregate functions


Selection ( σ )• Unary operation – take one operand

• Return all tuples that satisfy a condition (filter tuples)

• Conditions can be =, <, ≤, >, ≥, <> and combined with AND, OR, NOT

σC (R) where C is a set of conditions that involve the attribute(s) of R


Example: Selection ( σ )

Student_lectureS_id Address Course TA1234 57 Hockanum Blvd Database Systems Minnie2345 1400 E. Bellows Database Systems Humpty3456 900 S. Detroit Cloud Computing Dumpty1234 57 Hockanum Blvd Web Programming Lang. Mickey5678 2131 Forest Lake Ln. Software Analysis Minnie

S_id Address Course TA1234 57 Hockanum Blvd Database Systems Minnie

Find all tuples in Student_lecture relation that have a “Database Systems” course and “Minnie” as TA

SELECT * FROM Student_lectureWHERE TA=‘Minnie’ AND Course=‘Database Systems’;

σTA=‘Minnie’ AND Course=‘Database Systems’ (Student_lecture)

No “wild card” in RA


Example: Selection ( σ )

Student_lecture

σTA=‘Minnie’ ANDCourse =‘Database

Systems’

RA tree

Data source

Query output

Find all tuples in Student_lecture relation that have a “Database Systems” course and “Minnie” as TA

SELECT * FROM Student_lectureWHERE TA=‘Minnie’ AND Course=‘Database Systems’;

σTA=‘Minnie’ AND Course=‘Database Systems’ (Student_lecture)


σDept=“CS” AND Course=2910 (Schedule)

Consider a simplified version of Lou’s List. Write RA to find all tuples in the Schedule relation that have “CS” and 2910 for Department and Course. Assume course is stored as integer.

Schedule

Let’s Try: Selection ( σ )CNumber Dept Course computingID DateTime Location15783 CS 1111 up3f MoWe 2:00pm - 3:15pm Thornton Hall E303… … … …16633 CS 2110 nb3f MoWeFr 1:00pm - 1:50pm Mechanical Engr 20516636 CS 2910 up3f TuTh 12:30pm - 1:45pm Rice 340

20294 CS 2910 up3f Th 11:15am – 12:00pm, Fr 3:00pm-3:45pm Rice 536, Olsson 001

… … …16639 CS 4640 up3f TuTh 2:00pm - 3:15pm Clark Hall 10816455 CS 4750 nb3f MoWe 3:30pm– 4:45pm Mechanical Engr 205… … …


Schedule

Let’s Try: Selection ( σ ) cont.

σDept=“CS” AND Course=2910 (Schedule)

CNumber Dept Course computingID DateTime Location15783 CS 1111 up3f MoWe 2:00pm - 3:15pm Thornton Hall E303… … … …16633 CS 2110 nb3f MoWeFr 1:00pm - 1:50pm Mechanical Engr 20516636 CS 2910 up3f TuTh 12:30pm - 1:45pm Rice 340



CNumber Dept Course computingID DateTime Location16636 CS 2910 up3f TuTh 12:30pm - 1:45pm Rice 340



• Unary operation – take one operand

• Return specified attributes of a relation

πA1, A2, …, Am (R) where Ai is attribute of a relation R

Projection ( π )


Student_lecture

S_id Course1234 Database Systems2345 Database Systems3456 Cloud Computing1234 Web Programming Lang.5678 Software Analysis

S_id Address Course TA1234 57 Hockanum Blvd Database Systems Minnie2345 1400 E. Bellows Database Systems Humpty3456 900 S. Detroit Cloud Computing Dumpty1234 57 Hockanum Blvd Web Programming Lang. Mickey5678 2131 Forest Lake Ln. Software Analysis Minnie

Example: Projection ( π )Find all S_id and Course in Student_lecture relation

π S_ID, Course (Student_lecture)

SELECT S_id, Course FROM Student_lecture;


Example: Projection ( π )Find all S_id and Course in Student_lecture relation

π S_ID, Course (Student_lecture)

SELECT S_id, Course FROM Student_lecture;

Student_lecture

π S_ID, Course

RA tree

Data source

Query output


Student_lecture

TAMinnieHumptyDumptyMickey

S_id Address Course TA1234 57 Hockanum Blvd Database Systems Minnie2345 1400 E. Bellows Database Systems Humpty3456 900 S. Detroit Cloud Computing Dumpty1234 57 Hockanum Blvd Web Programming Lang. Mickey5678 2131 Forest Lake Ln. Software Analysis Minnie

π TA (Student_lecture)

More Example: Projection ( π )

Student_lecture

π TA

• Identical tuples collapse into a single tuple – no duplicates

Find all TAs in Student_lecture

relation

RA follows “Set” properties


Consider a simplified version of Lou’s List (Spring 2019). Write RA to find all Departments and Courses in the Schedule relation that is taught by “up3f”

Schedule

Let’s Try: Projection ( π )

πDept, Course( σcomputingID=“up3f”(schedule) )








Let’s Try: Projection ( π ) cont.

Dept CourseCS 1111CS 2910CS 4640

Schedule


Schedule

σcomputingID=‘up3f’

π Dept, Course

Notice: noduplicates


ρfriend_contact(ID, primary_email, alternative_email)(contact)

Renaming ( ρ )

ID email1mi1y [email protected]

mi1y [email protected]

[email protected]

[email protected]

contact

=ID primary_email



[email protected]

[email protected]

friend_contact

• Unary operation – take one operand• Change the schema, not the instance• Rename the result of expression E to R’• Rename the resultant attributes to B1, B2, …, Bn

ρR’(B1, B2, …, Bn)(R) where Bi is new attribute name of a relation R’


Union, Intersection, Difference

Binary operations – take left and right operands

Requirements: • Schemas of R and S must have same degree (number of attributes)

• Corresponding attributes of R and S must be based on the same domain / compatible data types

• Attributes are in the same order


Union (R U S)

Each tuple of R and each tuple of S is put into the resultant relation


Consider a simplified version of Lou’s List (Spring 2019). Find all Departments and Courses in the Schedule relation that is taught by “up3f”, “nb3f”, or both.

Schedule

Let’s Try: Union ( U )CNumber Dept Course computingID DateTime Location15783 CS 1111 up3f MoWe 2:00pm - 3:15pm Thornton Hall E303… … … …16633 CS 2110 nb3f MoWeFr 1:00pm - 1:50pm Mechanical Engr 20516636 CS 2910 up3f TuTh 12:30pm - 1:45pm Rice 340




πDept, Course( σcomputingID=“nb3f”(schedule) )

U


Schedule

Let’s Try: Union ( U ) cont.


πDept, Course( σcomputingID=“nb3f”(schedule) )

U

Dept courseCS 1111CS 2910CS 4640

Dept courseCS 2110CS 4750

Dept courseCS 1111CS 2910CS 4640CS 2110CS 4750

U =





Intersection(R∩S)

A tuple is in the resultant relation if and only if it is in both R and S


Consider a simplified version of Lou’s List (Spring 2019). Find all Departments in the Schedule relation that offer courses taught by “up3f” and “nb3f”.

Schedule

Let’s Try: Intersection (∩)CNumber Dept Course computingID DateTime Location15783 CS 1111 up3f MoWe 2:00pm - 3:15pm Thornton Hall E303… … … …16633 CS 2110 nb3f MoWeFr 1:00pm - 1:50pm Mechanical Engr 20516636 CS 2910 up3f TuTh 12:30pm - 1:45pm Rice 340



πDept ( σcomputingID=“up3f”(schedule) )

πDept ( σcomputingID=“nb3f”(schedule) )

∩


Schedule

Let’s Try: Intersection (∩) cont.

πDept ( σcomputingID=“up3f”(schedule) )




πDept ( σcomputingID=“nb3f”(schedule) )

∩

DeptCS

DeptCS∩ =

Dept

CS


Difference (R − S)

A tuple is in the resultant relation if and only if it is in R but not in S


ID name

go9y Goofy

mi2e Minnie

do3d Donald

pl4o Pluto

do5d Donald

pe6e Pete

Example: Difference (−)

CS1_students

ID name

mi1y Mickey

mi2e Minnie

do3d Donald

da4y Daisy

do5d Donald

Find tuples that are in one relation but are not in another

Same number of attributes

Attribute names match, Compatible type

CS1_students – CS2_students

ID name

mi1y Mickey

da4y Daisy

− =

CS2_students


Let’s Try: Difference (−)

πmaker( σtype=“laptop”(Product) )

πmaker( σtype=“PC”(Product) )

－

Consider the following schema statements. Write RA to find manufacturers (makers) that sell Laptops but not PC’s

Product (maker, model, type) -- type is stored as “laptop” or “PC”


Cartesian Product (R × S)

R x S where R(a1, a2, ..., an) and S(b1, b2, ..., bk)

ID Namehm1y Humpty

dm2y Dumpty

• Binary operations – take two operand• So-called “cross-product” or “product”• Combine two relations• Usually not meaningful when it is performed alone

R x S ≠ S x R

Employee

EID DeptNamehm1y CS

dm2y EE

Department

ID Name EID DeptNamehm1y Humpty hm1y CS

hm1y Humpty dm2y EE

dm2y Dumpty hm1y CS

dm2y Dumpty dm2y EE

Employee x Department


More Example: Product (×)

ID email


mi2e [email protected]

mi1y [email protected] ×

C1× C2

=ID email




Contact (Let’s call it C1) Contact (Let’s call it C2)

Usually not meaningfulwhen it is performedalone

ρC1(ID1, email1)(contact) ρC2(ID2, email2)(contact)


More Example: Product (×)

×

C1× C2

=

Find all students who have more than one email

(2 steps: cross product, then select tuples)

Meaningful when it is followed by other operations.

ID email




ID email




Contact (Let’s call it C1) Contact (Let’s call it C2)ρC1(ID1, email1)(contact) ρC2(ID2, email2)(contact)


Let’s Try: Product (×)Find all students who have more than one email(2 steps: cross product, then select tuples)

SELECT * FROM C1, C2WHERE C1.ID1 = C2.ID2AND C1.email1 <> C2.email2

σID1=ID2 ^ email1≠ email2 (C1× C2)

ID email




ID email




Contact (Let’s call it C1) Contact (Let’s call it C2)ρC1(ID1, email1)(contact) ρC2(ID2, email2)(contact)Write RA to solve this

problem

(for more practice, also write SQL to solve this problem)


Natural Join (R ⋈ S)• Binary operations – take two operand• Merge relations on the specified condition

R ⋈ S where R and S has a set of attributes that are in common

R ⋈ S = πA ( σC (R x S) )

2. Check equality of all common attributes

(attributes with same names)

3. Eliminate duplicate common attributes

(attributes with same names)

1. Cross product


A B C1 2 3

6 7 8

9 7 8

B C D A2 3 4 1

2 3 5 1

7 8 10 6

7 8 10 9

RB C D2 3 4

2 3 5

7 8 10

7 9 9

S

A B C1 2 3

6 7 8

9 7 8

Example: Natural Join ( ⋈ )

B C D2 3 4

2 3 5

7 8 10

7 9 9

R SA B C D1 2 3 4

1 2 3 5

6 7 8 10

9 7 8 10

R ⋈ S

⋈ =

S ⋈ R

⋈ =


Let’s Try: Natural Join ( ⋈ )Consider the following schema statements. Write RA to find manufacturers (makers) that make laptops with a hard disk (hd) of at least 100GB

Product (maker, model, type) -- type is stored as “laptop” or “PC”

Laptop (model, speed, ram, hd, screen, price)-- hd stores size of hard disk

πmaker ( Product ⋈ ( σ hd ≥ 100 (Laptop) ) )


Division (R ÷ S)

A B C D E D E A B CR S R ÷ S

=÷

• Binary operations – take two operand• Use to find “for all” queries• Find “A” for all “B” where “A” and “B” are sets of attributes

AB ÷ B = A

R ÷ S = πA-B (R) - πA-B ( (πA-B (R) × S) – R )


Breakdown: Division (R ÷ S)R ÷ S = πA-B (R) - πA-B ( (πA-B (R) × S) – R )

1. Project attributes of R that are not in S, and then product with the divisor S

A B C Da 3 x ma 1 x oa 3 y ob 1 x mc 4 y ob 2 y nc 4 x m

C Dy ox m

R S πA-B (R) SA B C Da 3 y oa 3 x ma 1 y oa 1 x mb 1 y ob 1 x mc 4 y oc 4 x mb 2 y ob 2 x m

A Ba 3a 1b 1c 4b 2

C Dy ox m

where A and B are sets of attributes of R and S

× =

πA-B (R) x S



2. Extract tuples from the product that were not in R


C Dy ox m

R S


πA-B (R) x S (πA-B (R) x S) – RA B C Da 3 y oa 3 x ma 1 y oa 1 x mb 1 y ob 1 x mc 4 y oc 4 x mb 2 y ob 2 x m

A B C Da 1 y oa 1 x mb 1 y ob 2 y ob 2 x m



3. Project only attributes that are in the original R but not in S


C Dy ox m

R S


πA-B(πA-B (R) x S) – R)(πA-B (R) x S) – RA B C Da 1 y oa 1 x mb 1 y ob 2 y ob 2 x m

A Ba 1b 1b 2



4. Extract only the tuples that are in the original R


C Dy ox m

R S


πA-B(πA-B (R) x S) – R)πA-B (R)A Ba 3a 1b 1c 4b 2

– =A Ba 1b 1b 2

A Ba 3c 4


A B C D

a 3 x m

a 1 x o

a 3 y o

b 1 x m

c 4 y o

b 2 y n

c 4 x m

C D

y o

x m

A B

a 3

c 4

R S R ÷ S=÷

Short cut: Division (R ÷ S)


Let’s Try: Division ( ÷ )

πsname,bid ( Sailors ⋈ Reserves) ÷ (Boats)

sname bid

… …

bid bname color

… … …AB ÷ B = A

The divisor is B. Don’t care the

remaining columns.

Consider the following schema statements. Write RA to find the names of sailors who have reserved all the boats

Boats (bid, bname, color) Sailors (sid, sname, rating, age)Reserves (sid, bid, day)

÷

= sname

…


Assignment ( ← )

Similar to assignment statement in programming

v ← E where v is a temporary variable representing a relation, E is an expression

Example: Find manufacturers (makers) that make laptops with a hard disk (hd) of at least 100GB

Product (maker, model, type) Laptop (model, speed, ram, hd, screen, price)

R1← σ hd ≥ 100 (Laptop) ) )

πmaker ( R2 )

R2← Product ⋈ ( R1)πmaker (Product ⋈ (σ hd ≥ 100 (Laptop)))


Aggregate Function ( G )• Not relational operators• Use Group by to help summarize a column in some way• Five standard operators: sum, avg, count, min, and max

where

A1, A2, …, An are attributes of a relation RG1, G2, .., Gm are attributes on which to group;F1, F2, …, Fn are aggregation functions on an attribute(Ai)

𝐆34,36,…,38, (R)<4 =4 ,<6 =6 ,…,<>(=>)


Example: Aggregate FunctionConsider the following schema statements. Write RA to find the number of each of the colors of the boats

Boats (bid, bname, color) Sailors (sid, sname, rating, age)Reserves (sid, bid, day)

color G count(bid) (Boats)

color count(bid)

… …Result in


Let’s Try: Aggregate FunctionConsider the following schema statements. Write RA to find the the average and max loan amount of each customer

Loan (loan_number, branch_name, amount)Borrower (customer_name, loan_number)

customer_name G avg(amount), max(amount) (Loan ⋈ Borrower)

customer_name avg(amount) max(amount)

… …Result in


Binaryoperations

Union rUs -- “or”

Intersection r∩ s-- “and”

Setdifference r− s

AdditionalExpressiveness

AssignmentAggregatefunctions

Takeonerelation, returnanewrelationUnaryoperations

Selection σp(r)

Project πA1, A2, …, Am (r)

Rename ρx(A1, A2, …, An)(E)

Findtuplesthatsatisfyagivencondition

Slicearelation,returnanewrelationwithcertainattributes

RenametheresultofexpressionEtox;renameresultantattributestoA1,A2,…

Combinetuplesfrom2relations[require:samenumberofattributes;compatibledomains][result:sameattributes]

Combinetuplesfrom2relations[require:samenumberofattributes;compatibledomains][result:sameattributes]

Findtuplesthatareinonerelationbutarenotinanother[require:samenumberofattributes;compatibledomains][result:sameattributes]

Cartesianproduct r × s

Naturaljoin r⋈ s

Division r÷ s

Taketworelation, returnanewrelation

Combine2relations,allcombination[result:combinedattributes]

Selecttuplesthatsatisfythematchingconditionsfromcombinedrelations[result:combinedattributes]

SimilartoAB÷ BFind“A”forall“B”[require:thereexistsB’sattributesinA][result:Aschema]

Summary RA Operators


Wrap-UpRelational operators

• Selection, projection • Renaming• Set operations, Cartesian product, Natural join• Division

Additional operators• Assignment, aggregate function

What’s next?• Translating between SQL and RA• RA tree

relational algebra - university of virginia school of ...up3f/cs4750/slides/4750meet20-21-ra.pdf ·...

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