Relational Model and Algebra - Imperial College …Relational Algebra Five Primitive Operators...

Relational Model and Algebra

P.J. McBrien

Imperial College London

P.J. McBrien (Imperial College London) Relational Model and Algebra 1 / 45

The Relational Data Model Relations

Relations are sets of typed tuples

Relations

Relations take the form R(A,B, . . .) where

R is the name of the relation

A,B, ... is the set of attributes of the relation

Often write the set without commas: A,B, . . . ≡ AB . . ., and can refer to a set of

attributes as ~AThe number of attributes n is the arity of the relationCan call R(A1, . . . , An) an n-ary relationDomain(A) is the set of values (type) that the attribute can haveWill use Attrs(R) to find A,B, . . .

The extent of R(A,B, . . .) is the set of tuples{〈vA1 , vB1 , . . . 〉, 〈vA2 , vB2 , . . . 〉, 〈vA3 , vB3 , . . . 〉, . . . }

∀x.vAx ∈ Domain(A)No duplicate tuplesNot orderedAll tuples have the same arity

Relation=Table

A B . . .vA1 vB1 . . .vA2 vB2 . . .vA3 vB3 . . .

Set Semantics

Order of columnsnot significant

Order of rows notsignificant

No duplicate rows

Attribute=Column

Tuple=Row

Quiz 1: Equivalent Relations

Which is the odd one out?

branchsortcode bname cash

56 ’Wimbledon’ 94340.4534 ’Goodge St’ 8900.6767 ’Strand’ 34005.00

branchbname sortcode cash’Wimbledon’ 56 94340.45’Goodge St’ 34 8900.67’Strand’ 67 34005.00

34 ’Goodge St’ 8900.6756 ’Wimbledon’ 94340.4567 ’Strand’ 34005.00

56 ’Wimbledon’ 94340.4556 ’Wimbledon’ 94340.4534 ’Goodge St’ 8900.6767 ’Strand’ 34005.00

Handling ‘missing’ attribute values

Suppose we want to have a relation account(no,type,cname,rate,sortcode), but not allaccounts have a rate.

Solution 1: Separate relations

accountno type cname sortcode

100 ’current’ ’McBrien, P.’ 67101 ’deposit’ ’McBrien, P.’ 67103 ’current’ ’Boyd, M.’ 34107 ’current’ ’Poulovassilis, A.’ 56119 ’deposit’ ’Poulovassilis, A.’ 56125 ’current’ ’Bailey, J.’ 56

account rateno rate

101 5.25119 5.50

Solution 2: NULL values

accountno type cname rate sortcode

100 ’current’ ’McBrien, P.’ NULL 67101 ’deposit’ ’McBrien, P.’ 5.25 67103 ’current’ ’Boyd, M.’ NULL 34107 ’current’ ’Poulovassilis, A.’ NULL 56119 ’deposit’ ’Poulovassilis, A.’ 5.50 56125 ’current’ ’Bailey, J.’ NULL 56

Relational Keys

A key of a relation R(AB . . .) is a subset of the attributes for which the values inany extent are unique across all tuples

Every relation has at least one key, which is the entire set of attributes

A key is violated by there being two tuples in the extent which have the samevalues for the attributes of the key

If A is a key, then so must AB be a key

A minimal key is a set of attributes AB . . . for which no subset of theattributes is also a key

The primary key is one of the keys of the relation: serves as the default keywhen no key explicitly stated

Quiz 2: Violation of Relational Keys

movementmid no amount tdate1000 100 2300.00 5/1/19991001 101 4000.00 5/1/19991002 100 -223.45 5/1/19991004 107 -100.00 11/1/19991005 103 145.50 12/1/19991006 100 10.23 15/1/19991007 107 345.56 15/1/19991008 101 1230.00 15/1/19991009 119 5600.00 18/1/1999

Which key is violated?

movement(mid)

movement(no,amount)

movement(no,tdate)

movement(amount,tdate)

Quiz 3: Correct Keys for Relations

Which key makes most sense in a bank UoD?

movement(amount)

movement(no,amount)

movement(no,tdate)

movement(amount,tdate)

Relational Foreign Keys

Foreign Key

A foreign key R( ~X)fk⇒ S(~Y ) of a relation R(AB . . .) is a subset ~X ⊆ AB . . . of the

attributes for which the values in the extent of R also appear as values of attributes~Y in the extent of S, and ~Y is a key of S.

account(sortcode)fk⇒ branch(sortcode)

accountno type cname rate sortcode100 ’current’ ’McBrien, P.’ NULL 67101 ’deposit’ ’McBrien, P.’ 5.25 67103 ’current’ ’Boyd, M.’ NULL 34107 ’current’ ’Poulovassilis, A.’ NULL 56119 ’deposit’ ’Poulovassilis, A.’ 5.50 56125 ’current’ ’Bailey, J.’ NULL 56

key branch(sortcode)

Quiz 4: Foreign Key Violation

accountno type cname rate sortcode100 ’current’ ’McBrien, P.’ NULL 67101 ’deposit’ ’McBrien, P.’ 5.25 67103 ’current’ ’Boyd, M.’ NULL 34107 ’current’ ’Poulovassilis, A.’ NULL 56119 ’deposit’ ’Poulovassilis, A.’ 5.50 56125 ’current’ ’Bailey, J.’ NULL 56

key branch(sortcode)

Which update violates the foreign key?

insert into account(126,’business’,’McBrien, P.’,1.00,67)

insert into branch(78,’Ealing’,1000.00)

delete from branch(67,’Strand’,34005.00)

delete from account(103,’current’,’Boyd, M.’,NULL,34)

Example Relational Schema

accountno type cname rate? sortcode

key branch(sortcode)key branch(bname)key movement(mid)key account(no)

movement(no)fk⇒ account(no)

Relational Algebra Five Primitive Operators

Relational Algebra: A Query Language for the Relational Model

Primitive operators of the Relational Algebra

Symbol Name Typeπ Project Unaryσ Select Unary× Cartesian Product Binary∪ Union Binary− Difference Binary

All operators take relations as input

All operators produce one relation as their output

Other (useful) operators may be defined in terms of the five primitive operators

Relational Algebra: Project π

Project Operator

πno,typeaccountno type100 ’current’101 ’deposit’103 ’current’107 ’current’119 ’deposit’125 ’current’

πsortcodeaccountsortcode

673456

Relational Algebra: Select σ

Select Operator

σrate>0accountno type cname rate sortcode101 ’deposit’ ’McBrien, P.’ 5.25 67119 ’deposit’ ’Poulovassilis, A.’ 5.50 56

Relational Algebra: Product ×

σrate>0accountno type cname rate sortcode101 ’deposit’ ’McBrien, P.’ 5.25 67119 ’deposit’ ’Poulovassilis, A.’ 5.50 56

Product Operator

branch × σrate>0accountsortcode bname cash no type cname rate sortcode

56 ’Wimbledon’ 94340.45 101 ’deposit’ ’McBrien, P.’ 5.25 6756 ’Wimbledon’ 94340.45 119 ’deposit’ ’Poulovassilis, A.’ 5.50 5634 ’Goodge St’ 8900.67 101 ’deposit’ ’McBrien, P.’ 5.25 6734 ’Goodge St’ 8900.67 119 ’deposit’ ’Poulovassilis, A.’ 5.50 5667 ’Strand’ 34005.00 101 ’deposit’ ’McBrien, P.’ 5.25 6767 ’Strand’ 34005.00 119 ’deposit’ ’Poulovassilis, A.’ 5.50 56

Quiz 5: RA Queries

Which RA query lists the name of branches that have deposit accounts?

πsortcode σtype=‘deposit’ account

Bπbname

σaccount.sortcode=branch.sortcode∧type=‘deposit’

(account× branch)

πbname(branch× σtype=‘deposit’ account)

πbname σtype=‘deposit’(account× branch)

SPJ Queries

Select Project Join (SPJ) queries

If a product of tables is formed, where a selection is then done that compares theattributes of those tables, we say that a join has been performed.Normally not all columns of the product are returned, and therefore a project is alsorequired.

Branches with current accounts

πbname,no σbranch.sortcode=account.sortcode∧account.type=‘current’(branch× account)bname no‘Goodge St’ 103‘Wimbledon’ 107‘Wimbledon’ 125‘Strand’ 100

Relational Algebra: Union ∪

πsortcode as idaccountid673456

πno as idaccountid

100101103107119125

Union Operator

πsortcode as idaccount ∪ πno as idaccountid673456

100101103107119125

relations must be union compatible

Relational Algebra: Difference −

πnoaccountno

100101103107119125

πnomovementno

100101103107119

Difference Operator

πnoaccount− πnomovementno125

relations must be union compatible

Rules for Combining Operators

Since all operators produce a relation as output, any operator may produce one ofthe inputs to any other operator.

well formed RA query

the output of the nested operator must contain the attributes required by anouter π or σ

the two inputs to a ∪ or − must contain the same number of attributes

Quiz 6: Well formed queries

Which RA query is well formed?

σtype=‘current’ πno account

πno account− πno,mid movement

πno σtype=‘current’ account

πno πtype account

Worksheet: Primitive Relational Algebra Operators

Derived Operators

Derived Relational Algebra: Natural Join ⋊⋉

Natural Join

R ⋊⋉ S = σR.A1=S.A1∧...∧R.Am=S.AmR × S

Natural Join

branch ⋊⋉ account = σbranch.sortcode=account.sortcodebranch× account

branch ⋊⋉ accountsortcode bname cash no type cname rate

34 ’Goodge St’ 8900.67 103 ’current’ ’Boyd, M.’ NULL56 ’Wimbledon’ 94340.45 107 ’current’ ’Poulovassilis, A.’ NULL56 ’Wimbledon’ 94340.45 119 ’deposit’ ’Poulovassilis, A.’ 5.5056 ’Wimbledon’ 94340.45 125 ’current’ ’Bailey, J.’ NULL67 ’Strand’ 34005.00 100 ’current’ ’McBrien, P.’ NULL67 ’Strand’ 34005.00 101 ’deposit’ ’McBrien, P.’ 5.25

Derived Operators

Quiz 7: Natural Join

What is the result of πno(account ⋊⋉ movement)?

πno(account ⊲⊳ movement)no

100101103107119125

100101103107119

Derived Operators

Derived Relational Algebra: Semi Join ⋉

Semi Join

R⋉ S = R ⋊⋉ πAttr(R)∩Attr(S)(S)

Semi Join

account⋉movement = account ⋊⋉ πno(movement)

account⋉movementno type cname rate sortcode100 ’current’ ’McBrien, P.’ NULL 67101 ’deposit’ ’McBrien, P.’ 5.25 67103 ’current’ ’Boyd, M.’ NULL 34107 ’current’ ’Poulovassilis, A.’ NULL 56119 ’deposit’ ’Poulovassilis, A.’ 5.50 56

Derived Operators

Derived Relational Algebra: Joins

Natural Join

R ⋊⋉ S = σR.A1=S.A1∧...∧R.Am=S.AmR × S

Equi Join

⋊⋉ S = σR.A=S.BR × S

Semi Join

R⋉ S = R ⋊⋉ πAttr(R)∩Attr(S)(S)

Theta Join

⋊⋉ S = σθR× S

Derived Operators

Quiz 8: Understanding join operators

Which RA query produces the most tuples?

branchbranch.sortcode<account.sortcode

⋊⋉ account

branch ⋊⋉ account

branch⋉ account

branchbranch.sortcode=account.sortcode

⋊⋉ account

Derived Operators

Quiz 9: Foreign Keys and Natural Joins (1)

Suppose R and S only share attribute A, and there is a foreign key S(A)fk⇒ R(A).

If |R| = 100 and |S| = 1, 000, what is |R ⋊⋉ S|?

100,000

Note that |R| returns the number of tuples in the current extent of R

Derived Operators

Quiz 10: Foreign Keys and Natural Joins (2)

Suppose R and S only share attribute A, and there is a foreign key R(A)fk⇒ S(A).

If |R| = 100 and |S| = 1, 000, what is |R ⋊⋉ S|?

100,000

Derived Operators

Derived Relational Algebra: Intersection ∩

Intersection

R ∩ S = R− (R − S)

πnoaccount ∩ πnomovement

πnoaccountno

100101103107119125

πnoaccount − πnomovementno

πnoaccount ∩ πnomovementno

100101103107119

Derived Operators

Quiz 11: Intersection

emailname address’McBrien, P.’ p.mcbrien@imperial.ac.uk’Poulovassilis, A.’ ap@dcs.bbk.ac.uk’Pietzuch, P.’ prp@doc.ic.ac.uk

What is the result of πcname account ∩ πname email?

cname’McBrien, P.’’Boyd, M.’’Poulovassilis, A.’’Bailey, J.’’Pietzuch, P.’

cname’McBrien, P.’’Boyd, M.’’Poulovassilis, A.’’Bailey, J.’

cname’McBrien, P.’’Poulovassilis, A.’’Pietzuch, P.’

cname’McBrien, P.’’Poulovassilis, A.’

Derived Operators

Derived Relational Algebra: Division ÷

Division

R÷ S = πAttrs(R)−Attrs(S)R − πAttrs(R)−Attrs(S)((πAttrs(R)−Attrs(S)R × S) −R)

Division

πcname,type account÷ πtype account = πcname πcname,type account−πcname((πcname πcname,type account× πtype account)− πcname,type account)

πcname,type account÷ πtype account = πcname account−πcname((πcname account× πtype account)− πcname,type account)

πcname,type accountcname type’McBrien, P.’ ’current’’McBrien, P.’ ’deposit’’Boyd, M.’ ’current’’Poulovassilis, A.’ ’current’’Poulovassilis, A.’ ’deposit’’Bailey, J.’ ’current’

πtype accounttype’current’’deposit’

πcname,type account ÷ πtype accountcname’McBrien, P.’’Poulovassilis, A.’

Derived Operators

Evaluation of Division

πcname accountcname’McBrien, P.’’Boyd, M.’’Poulovassilis, A.’’Bailey, J.’

πtype accounttype

’current’’deposit’

πcname account × πtype accountcname type’McBrien, P.’ ’current’’McBrien, P.’ ’deposit’’Boyd, M.’ ’current’’Boyd, M.’ ’deposit’’Poulovassilis, A.’ ’current’’Poulovassilis, A.’ ’deposit’’Bailey, J.’ ’current’’Bailey, J.’ ’deposit’

Derived Operators

πcname account × πtype accountcname type’McBrien, P.’ ’current’’McBrien, P.’ ’deposit’’Boyd, M.’ ’current’’Boyd, M.’ ’deposit’’Poulovassilis, A.’ ’current’’Poulovassilis, A.’ ’deposit’’Bailey, J.’ ’current’’Bailey, J.’ ’deposit’

πcname,type accountcname type’McBrien, P.’ ’current’’McBrien, P.’ ’deposit’’Boyd, M.’ ’current’’Poulovassilis, A.’ ’current’’Poulovassilis, A.’ ’deposit’’Bailey, J.’ ’current’

(πcname account × πtype account)− πcname,type accountcname type’Boyd, M.’ ’deposit’’Bailey, J.’ ’deposit’

Derived Operators

(πcname account × πtype account)− πcname,type accountcname type’Boyd, M.’ ’deposit’’Bailey, J.’ ’deposit’

πcname((πcname account × πtype account)− πcname,type account)cname’Boyd, M.’’Bailey, J.’

Derived Operators

πcname accountcname’McBrien, P.’’Boyd, M.’’Poulovassilis, A.’’Bailey, J.’

πcname((πcname account × πtype account)− πcname,type account)cname’Boyd, M.’’Bailey, J.’

πcname account − πcname((πcname account × πtype account)− πcname,type account)cname’McBrien, P.’’Poulovassilis, A.

Derived Operators

Worksheet: Derived Relational Algebra Operators

Relational Algebra Equivalences

Equivalences Involving Project

Project and Project

π~X π~Y R ≡ π~X R

You can eliminate any inner project (note that to be well formed ~X ⊆ ~Y )

Project and Select

π~X σP(~Y) R ≡ σP(~Y) π~X R

You can move a project of attributes ~X inside a select, provided the select predicatecan be answered from those attributes, i.e. ~Y ⊆ ~X

Project and Product

π~X(R× S) ≡ π~X∩Attrs(R) R× π~X∩Attrs(S) S

Project and Union

π~X(R ∪ S) ≡ π~X R ∪ π~X S

Project and Difference

π~X(R− S) ⊇ π~X R− π~X S

Equivalences Involving Select

Select and Project

σP(~X) π~X R ≡ π~X σP(~X) R

Select and Select

σPx(~X)σPy(~Y)

R ≡ σPx(~X)∧Py(~Y)R

Select and Product

σP(~X)(R× S) ≡ σP(~X) R× S ⇐⇒ ~X ⊆ Attrs(R)

You can move a select predicate P(~X) onto one of the relations inside a product

provided ~X ⊆ Attrs(R).

Select and Union

σP(~X)(R ∪ S) ≡ σP(~X) R ∪ σP(~X) S

Select and Difference

σP(~X)(R− S) ≡ σP(~X) R− S

Quiz 12: Equivalent RA Expressions (Unary Operators)

Which RA expression is not equivalent to the other three?

πno σtype=‘current’ account

πno σtype=‘current’ πno,type,cname account

πno σtype<>‘deposit’ πno,type,cname account

πno σtype=‘current’ σtype<>‘deposit’ account

Quiz 13: Query Evaluation

Which RA means that the × operator handles fewer tuples?

Aσaccount.no=movement.no

(σsortcode=67 account× σamount<0 movement)

Bσaccount.no=movement.no∧sortcode=67

(account× σamount<0 movement)

Cσaccount.no=movement.no∧amount<0

(σsortcode=67 account×movement)

Dσaccount.no=movement.no∧sortcode=67∧amount<0

(account×movement)

Equivalences Involving Binary Operators

Product and Union

R× (S ∪ T) ≡ (R× S) ∪ (R× T)

Product and Difference

R× (S− T) ≡ (R× S)− (R× T)

Union and Product

R ∪ (S× T) unable to move ∪ inside ×

Union and Difference

R ∪ (S− T) unable to move ∪ inside −

Difference and Product

R− (S× T) unable to move − inside ×

Difference and Union

R− (S ∪ T) ≡ (R− S)− T

Quiz 14: Equivalent RA Expressions (Binary Operators)

Which equivalence does not hold?

(R× S)× T ≡ R × (S × T )

(R − S)− T ≡ R− (S − T )

(R ∪ S) ∪ T ≡ R ∪ (S ∪ T )

(R ∩ S) ∩ T ≡ R ∩ (S ∩ T )

Worksheet: Equivalences Between RA Expressions

1 πno,type σsortcode=56 πno,type,sortcode σtype=‘deposit’ account

2 σaccount.no=movement.no(πno,cname account× πmid,no σamount>1000 movement)

σaccount.no=movement.no(πno,cname,rate account×(σamount>1000 πmid,no movement ∪ σamount<100 πmid,no movement))

4 πno,cname,tdate σamount<0∧account.no=movement.no account×movement

Quiz 15: Monotonic and non-monotonic operators

A monotonic operator has the property that an additional tuple put into any inputrelation which only cause additional tuples to be generated in the output relation.

A non-monotonic operator has the property that an additional tuple put into aninput relation may remove tuples from the output relation

Which RA operator is non-monotonic?

R ∪ S

R− S

Incremental Query Evaluation

Suppose we add rows ∆R to extent of relation R so it becomes R′

If we represent ∆R as a relation (with the same attributes as R) then

R′ = R ∪∆R

π ~X R′ ≡ π ~X R ∪ π ~X ∆R

σP ( ~X) R′ ≡ σP ( ~X) R ∪ σP ( ~X) ∆R

R′ × S ≡ (R × S) ∪ (∆R × S)

R′ ∪ S ≡ (R ∪ S) ∪∆R

R′ − S ≡ (R − S) ∪ (∆R − S)

S −R′ ≡ (S −R)−∆R

Example: Query result after update to account (1)

1 Suppose that we had already evaluated query Qπbname,no σbranch.sortcode=account.sortcode∧account.type=‘current’(branch × account)bname no‘Goodge St’ 103‘Wimbledon’ 107‘Wimbledon’ 125‘Strand’ 100

2 If ∆account is added to account to get account′:πbname,no σbranch.sortcode=account.sortcode∧account.type=‘current’(branch × account′)πbname,no σbranch.sortcode=account.sortcode∧account.type=‘current’((branch × account) ∪ (branch ×

∆account))πbname,no σbranch.sortcode=account.sortcode∧account.type=‘current’(branch × account) ∪

πbname,no σbranch.sortcode=account.sortcode∧account.type=‘current’(branch ×∆account)

3 Thus if ∆account is added to account, we only need evaluateπbname,no σbranch.sortcode=account.sortcode∧account.type=‘current’(branch×∆account)

Example: Query result after update to account (2)

4 Suppose we have∆account

126 ’business’ ’McBrien, P.’ 1.00 67127 ’current’ ’Pietzuch, P.’ NULL 34

Thenπbname,no σbranch.sortcode=account.sortcode∧account.type=‘current’(branch×∆account)bname no’Goodge St’ 127

5 Thus since Q′ = Q ∪∆Q

πbname,no σbranch.sortcode=account.sortcode∧account.type=‘current’(branch × account′)bname no‘Goodge St’ 103‘Wimbledon’ 107‘Wimbledon’ 125‘Strand’ 100’Goodge St’ 127

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