Discussion #23 Relational Algebra

Discussion #23 1/32

Discussion #23

Relational Algebra

Discussion #23 2/32

Topics• Algebras • Relational Algebra

– use of standard notation– set operators , , – renaming – selection – projection – cross product – join ||

• Queries (from English)• Query optimization• SQL

Discussion #23 3/32

Relational Algebra• What is an algebra?

– a pair: (set of values, set of operations) ADT type Class Object

e.g. stack: (set of all stacks, {pop, push, top, …})

integer: (set of all integers, {+, -, *, })

• What is relational algebra?– (set of relations, set of relational operators)

– {, , , , , , , ||}

Discussion #23 4/32

Relational Algebra is Closed

• Closed: all operations produce values in the value set– (reals, {+, *, }) closed

– (reals, {+, *, , }) not closed (divide by 0)

– (reals, {+, *, >}) not closed (T/F not in value set)

– (computer reals, {+, *, }) not closed (overflow, roundoff)

– (relations, relational operators) closed

• Implication: we can always nest relational operators; can’t for algebras that are not closed.– e.g. after overflow, can do nothing

– e.g. can’t always nest: (2 < 3) + 5 = ?

Discussion #23 5/32

Set Operations: , , and • Relations are sets; thus set operations should work.• Examples:

R = A B 1 2 2 2 2 3

S = A B 2 2 2 3 4 2 5 5

RS = A B 1 2 2 2 2 3 4 2 5 5

RS = A B 2 2 2 3

RS = A B 1 2

SR = A B 4 2 5 5

Discussion #23 6/32

Set Operations (continued …)• Definition: schema(R) = {A, B} = AB, i.e. the

set of attributes• We sometimes write R(AB) to mean the relation

R with schema AB.• Definition: union compatible

– schema(R) = schema(S)– required precondition for , ,

• Definitions: – R S = { t | t R t S}– R S = { t | t R t S}– R S = { t | t R t S}

Discussion #23 7/32

Tuple Restriction: [X]• Restriction is a tuple operator (not a relational

operator).• t[X] restricts tuple t to the attributes in X.

A B Ct = 1 2 3

t[A] = (1) t[AC] = (1,3)

t = (1,2,3)

t[A] = (1,2,3)[A]= {(A,1), (B, 2), (C,3)}[A]= {(A,1)}= (1)

Discussion #23 8/32

Renaming: ABR renames attribute A to be B.

– A must be in schema(R)– B must not be in schema(R)

• Example: let

CBQ = A B 2 2 3 2

RCBQ = A B 1 2 2 2 2 3 3 2

R = A B 1 2 2 2 2 3

Q = A C 2 2 3 2

• But with :

RQ = ?Not union compatible

Discussion #23 9/32

Renaming (continued…)

• Q = ABR renames attribute A to B; the result is Q.• Precondition:

– A schema(R)– B schema(R)

• Postcondition:– schema(Q) = (schema(R) {A}) {B}– Q = {t' | t (tR t' = (t – {(A, t[A])}) {(B, t[A])})}

R = {{(A,1), (C,2)} {(A,2), (C,2)}}

Q = ABR = {{(B,1), (C,2)}

{(B,2), (C,2)}}

Discussion #23 10/32

Selection: • The selection operation selects the tuples that

satisfy a condition.

A=1R = A B 1 2

B=2R = A B 1 2 2 2

A=2B2R = A B 2 2

2 3

A=3R = A B

Note: empty, but still retain the schema

PR = { t | t R P(t) }

• Precondition: each attribute mentioned in P must be in schema(R).

• Postcondition: PR = { t | t R P(t) }

schema(PR) = schema(R)

Meaning: apply predicate P to tuple t by substituting into P appropriate t values.

R = A B 1 2 2 2 2 3


Projection: The projection operation restricts tuples in a

relation to those designated in the operation.

R = A B 1 2 2 2 2 3Q = A B C 1 1 1 2 1 1 3 4 5

AR = A 1 2

BR = B 2 3

BCQ = B C 1 1 4 5

ABR = R = A,BR = {A,B}R

Precondition: X schema(R)

Postcondition: XR = { t' | t (t R t' = t[X]) }schema(XR) = X


Cross Product: Standard cartesian product adapted for

relational algebra

R = A B 1 2 2 2

S = C D 1 1 2 2 3 3

R S = A B C D 1 2 1 1 1 2 2 2 1 2 3 3 2 2 1 1 2 2 2 2 2 2 3 3


Cross Product (continued…)

R = A B 1 2 = t' 2 2

S = C D 1 1 2 2 3 3 = t''

Precondition: schema(R) schema(S) = Postcondition: R S = { t | t' t''(t' R t'' S t = t' t'')}

schema(R S) = schema(R) schema(S)

t' = { (A,1), (B,2) }

t'' = { (C,3), (D,3) }

t' t'' = { (A,1), (B,2), (C,3), (D,3) }


Cross Product (continued…)

R = A B 1 2 = t' = { (A,1), (B,2) } 2 2

S = C A 1 1 = t'' = { (C,1), (A,1) } 2 2 3 3 = t''' = { (C,3), (A,3) }

t' t'' = { (A,1), (B,2), (C,1), (A,1) }

What if R and S have the same attribute, e.g. A?

Can’t do cross productSolution: RenameAAS

R AAS = A B C A 1 2 1 1 1 2 2 2 1 2 3 3 2 2 1 1 2 2 2 2 2 2 3 3


Natural Join: ||

R = A B 1 2 2 2

S = B C 1 2 2 1 3 2

R || S = A B C 1 2 1

2 2 1

(R )

Cross Product A B

1 2 1 2 1 2 2 2 1 2 2 2 2 1 2 2 3 2

R || S = ABC

Projection

B=B'

Selection Renaming

BB'S

B' C1 22 13 2

1 2

2 2 1

1


Join (continued …)

• In general, we can equate 0, 1, 2, or more attributes using || .

• A join is defined as:

schema (R || S) = schema(R) schema(S)

R || S = {t | t[schema(R)] R

t[schema(S)] S}

• There are no preconditions join always works.


Join (continued…)

R = A B 1 1 2 3 4 1

S = C D 1 1 1 5

R || S = A B C D 1 1 1 1 1 1 1 5 2 3 1 1 2 3 1 5 4 1 1 1 4 1 1 5

R = A B 1 2 2 2 2 3

S = B C 1 1 2 2 3 3

R || S = A B C 1 2 2 2 2 2 2 3 3

R = A B C 1 2 3 2 2 4 2 3 5

S = A B D 1 1 1 2 2 2 2 2 1

R || S = A B C D

2 2 4 2 2 2 4 1

0 attributes in common (full cross product)

1 attribute in common

2 attributes in common


Join (continued…)

• We can use renaming to control the ||

R = A B 1 2 2 2

S = B C 1 2 2 1 3 2

R || CAS = A B 1 2

S' = B A 1 2 2 1 3 2

= A B 2 1 1 2 2 3

R || S' = A B 1 2

• BTW, observe equivalence with intersection


Relational Algebra Expressions• Relational operators are closed. Thus we can nest

expressions:

R = A B 1 2 3 4

S = B C D 2 5 1 2 7 2 3 2 3 4 5 4

DC=5(R || S) = A B C D 1 2 5 1 1 2 7 2 3 4 5 4

• Unary operators have precedence over binary operators; binary operators are left associative.

• We can now do something very useful: ask and answer with relational algebra (almost) any query we can dream up.

= D 1 4


Relational Algebra Queries• List the prerequisites for EE200.

PrerequisiteCourse='EE200'cp = PrerequisiteEE005CS100

• When does CS101 meet?Day,HourCourse='CS101'cdh = Day Hour

M 9AM W 9AM F 9AM

• When and where does EE200 meet?Day,Hour,RoomCourse='EE200'(cdh || cr) = Day Hour Room

Tu 10AM 25 Ohm Hall W 1PM 25 Ohm Hall Th 10AM 25 Ohm Hall

Our answers are in (cdh || cr).We select Course to be EE200.Then, project on Day, Hour, Room.


Queries (continued…)

• Can we rewrite the query more optimally?• What rules should we use?

– Associativity and commutivity of join– Distributive laws for select and project

• What strategy should we use?– Eliminate unnecessary operations– Make joins as small as possible before execution

RoomName='Snoopy' Day='M' Hour='9AM' (snap || csg || cr || cdh)

StudentID Name'Snoopy' Address PhoneCourse StudentID GradeCourse Room*Course Day'M' Hour'9AM'

= Room Turing Aud.

• Where can I find Snoopy at 9 am on Monday?


Query Optimization• “Intuitively” we can write

Room(Name='Snoopy'snap || csg || cr || Day='M' Hour='9AM'cdh)

• Why does this execute faster?

• What laws hold that will let us do this?

R || S = S || R

P1P2E = P1P2E

P(R |×| S) = R || PS (if all the attributes of P are in S)

• How do we know they hold?

RoomName='Snoopy' Day='M' Hour='9AM' (snap || csg || cr || cdh)

as


Proofs for Laws• To prove P1P2E = P1P2E, we need to prove that two

sets are equal. We prove A = B by showing AB BA. We show that AB by showing that xA xB.

• Thus, we can do two proofs to prove P1P2E = P1P2E as follows:

1. t P1P2E premise2. t E (P1P2)(t) def.: PR = {t | tR P(t)}3. t E P1(t) P2(t) identical substitutions & operations4. t E P2(t) P1(t) commutative

5. t P2E P1(t) def. of 6. t P1P2E def. of

1. t P1P2E premise2. … just go backwards from 6 to 1 in

the proof above


Alternate Proof

Thus, we can prove P1P2E = P1P2E as follows:

P1P2E= {t | t E (P1P2)(t)} def.: PR = {t | tR P(t)}= {t | t E P1(t) P2(t)} identical substitutions & operations= {t | t E P2(t) P1(t)} commutative

= {t | t P2E P1(t)} def. of = {t | t P1P2E} def. of = P1P2E def. of a relation

(Derive the right-hand side from the left-hand side.)


Proofs for Laws (continued …)• To prove P(R || S) = R || PS, where all attributes of P are

in S, we again need to prove that two sets are equal.• As before, we can convert the lhs to the rhs.

P(R || S) = {t | t P(R || S)} def. of a relation

= {t | t R || S P(t)} def.: PR={t | tRP(t)}= {t | t[schema(R)] R t[schema(S)] S P(t)}

def.: R||S={t | t[schema(R)]Rt[schema(S)]S}= {t | t[schema(R)] R

t[schema(S)] S P(t[schema(S)])} all attributes of P are in S

= {t | t[schema(R)] R t[schema(S)] PS}def. of

= {t | t R || PS} def. of ||

= R || PS def. of a relation


SQL

Correspondence with Relational Algebra

select Afrom Rwhere B = 1

Assume we have relations R(AB) and S(BC).

select B from Rexceptselect B from S

select A, R.B, Cfrom R, Swhere R.B = S.B

A B = 1 R

B R B S

A, R.B, C R.B = S.B (R S) = R || S


SQL

Correspondence with Relational Algebra

select Afrom Rwhere B = 1

Assume we have relations R(AB) and S(BC).

select R.B from Rwhere R.B not in (select S.B from S)

select *from R natural join S

A B = 1 R

B R B S

R || S


SQL Queries• List the prerequisites for EE200.

select Prerequisite Prerequisitefrom cp EE005where Course='EE200' CS100

• When does CS101 meet?select Day, Hour Day Hourfrom cdh M 9AMwhere Course= 'CS101' W 9AM

F 9AM

• When and where does EE200 meet?select cdh.Course, Day, Hour, Room Course Day Hour Roomfrom cdh, cr EE200 Tu 10AM 25 Ohm Hall where cdh.Course= 'EE200' EE200 W 1PM 25 Ohm Hall and cdh.Course=cr.Course EE200 Th 10AM 25 Ohm Hall


SQL Queries• List the prerequisites for EE200.

select Prerequisite Prerequisitefrom cp EE005where Course='EE200' CS100

• When does CS101 meet?select Day, Hour Day Hourfrom cdh M 9AMwhere Course= 'CS101' W 9AM

F 9AM

• When and where does EE200 meet?select Course, Day, Hour, Room Course Day Hour Roomfrom cdh natural join cr EE200 Tu 10AM 25 Ohm Hall where cdh.Course= 'EE200' EE200 W 1PM 25 Ohm Hall

EE200 Th 10AM 25 Ohm Hall


SQL Queries• List all prerequisite courses.

select Prerequisite Prerequisitefrom cp CS100

EE005CS100CS101CS120CS101CS121CS205

select distinct Prerequisite Prerequisitefrom cp CS100

CS101CS120CS121CS205EE005


SQL Queries• Where can I find Snoopy at 9 am on Monday?

• List all prereqs of CS750 (including prereqs of prereqs.)• Not possible with standard SQL (unless nesting depth is

known)

• Is possible with Datalog

Rules: prereqOf(x, y) :- cp(y, x).

prereqOf(x, y) :- prereqOf(x, z), cp(y, z).

Query: prereqOf(x, 'CS750')?

• To gain more power and flexibility, we typically embed SQL in a high-level language.

select Room Roomfrom snap, csg, cr, cdh Turing Aud.where Name='Snoopy' and Day='M' and Hour='9AM' and snap.StudentID=csg.StudentID and csg.Course=cr.Course and cr.Course=cdh.Course


SQL Queries

• List all prereqs of CS750 (including prereqs of prereqs.)select cp.Prerequisitefrom cpwhere cp.Course = 'CS750'

union

select cp1.Prerequisitefrom cp cp1, cp cp2where cp1.Course = cp2.Prerequisite and cp2.Course = 'CS750'

union

…

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Discussion #23 Relational Algebra