relational algebra (continued) lecture 5. relational algebra (summary) basic operations: –...

Relational AlgebraRelational Algebra(continued)(continued)

Lecture 5

Relational Algebra (Summary)

• Basic operations:– Selection ( ) Selects a subset of rows from relation.– Projection ( ) Deletes unwanted columns from relation.– Cross-product ( ) Allows us to combine two relations.– Set-difference ( - ) Tuples in reln. 1, but not in reln. 2.– Union ( ) Tuples either in reln. 1 or in reln. 2 or in both.– Rename ( ) Changes names of the attributes

• Since each operation returns a relation, operations can be composed ! (Algebra is “closed”.)

• Use of temporary relations recommended.

3

Additional operators We define additional operations that do not add

any power to the relational algebra, but that simplify common queries.

– Natural join– Conditional Join– Equi-Join– Division

Also, we’ve already seen “Set intersection”:r s = r - (r - s)

All joins are really special cases of conditional join

4

Quick note on notation

customer-name loan-number

Patty 1234

Apu 3421

Selma 2342

Ned 4531


Seymour 3432

Marge 3467

Selma 7625

Abraham 3597

good_customers bad_customers

If we have two or more relations which feature the same attribute names, we could confuse them. To prevent this we can use dot notation.For example

good_customers.loan-number

5

Natural-Join Operation: MotivationNatural-Join Operation: Motivationcust-name l-number

Patty 1234

Apu 3421

l-number branch

1234 Dublin

3421 Irvine

borrower loan

cust-name borrower.l-number loan.l-number branch

Patty 1234 1234 Dublin

Patty 1234 3421 Irvine

Apu 3421 1234 Dublin

Apu 3421 3421 Irvine

cust-name borrower.l-number loan.l-number branch

Patty 1234 1234 Dublin

Apu 3421 3421 Irvine

borrower.l-number = loan.l-number(borrower x loan)))

Very often, we have a query and the answer is not contained in a single relation. For example, I might wish to know where Apu banks.The classic relational algebra way to do such queries is a cross product, followed by a selection which tests for equality on some pair of fields.

While this works…• it is unintuitive• it requires a lot of memory• the notation is cumbersome

Note that in this example the two relations are the same size (2 by 2), this does not have to be the case.

So, we have a more intuitive way of achieving the same effect, the natural join, denoted by the symbol

6

Natural-Join Operation: IntuitionNatural-Join Operation: IntuitionNatural join combines a cross product and a selection into one operation. It performs a selection forcing equality on those attributes that appear in both relation schemes. Duplicates are removed as in all relation operations.For the previous example (one attribute in common: “l-number”), we have…

borrower loan

• If the two relations have no attributes in common, then their natural join is simply their cross product.• If the two relations have more than one attribute in common, then the natural join selects only the rows where all pairs of matching attributes match. (let’s see an example on the next slide).

= borrower.l-number = loan.l-number(borrower x loan)))

7

l-name f-name age

Bouvier Selma 40

Bouvier Patty 40

Smith Maggie 2

Al-name f-name ID

Bouvier Selma 1232

Smith Selma 4423

B

l-name f-name age l-name f-name ID

Bouvier Selma 40 Bouvier Selma 1232

Bouvier Selma 40 Smith Selma 4423

Bouvier Patty 2 Bouvier Selma 1232

Bouvier Patty 40 Smith Selma 4423

Smith Maggie 2 Bouvier Selma 1232

Smith Maggie 2 Smith Selma 4423

l-name f-name age l-name f-name ID

Bouvier Selma 40 Bouvier Selma 1232

l-name f-name age ID

Bouvier Selma 40 1232A B =

Both the l-name and the f-name match, so select.

Only the f-names match, so don’t select.

Only the l-names match, so don’t select.

We remove duplicate attributes…

The natural join of A and B

Note that this is just a way to visualize the natural join, we don’t really have to do the cross product as in this example

8

Natural-Join OperationNatural-Join Operation• Notation: r s• Let r and s be relation instances on schemas R and S

respectively.The result is a relation on schema R S which is

obtained by considering each pair of tuples tr from r and ts from s.

• If tr and ts have the same value on each of the attributes in R S, a tuple t is added to the result, where

– t has the same value as tr on r

– t has the same value as ts on s

• Example:R = (A, B, C, D)S = (E, B, D)

• Result schema = (A, B, C, D, E)• r s is defined as:

r.A, r.B, r.C, r.D, s.E (r.B = s.B r.D = s.D (r x s))

9

Natural Join Operation – ExampleNatural Join Operation – Example• Relation instances r, s:

A B

12412

C D

aabab

B

13123

D

aaabb

E

r

A B

11112

C D

aaaab

E

s

r sHow did we get here?

Lets do a trace over the next few slides…

10

A B

12412

C D

aabab

B

13123

D

aaabb

E

r s

First we note which attributes the two relations have in common…

11

A B

12412

C D

aabab

B

13123

D

aaabb

E

r

A B

11

C D

aa

E

s

There are two rows in s that match our first row in r, (in the relevant attributes) so both are joined to our first row…

12

A B

12412

C D

aabab

B

13123

D

aaabb

E

r s

A B

11

C D

aa

E

…there are no rows in s that match our second row in r, so do nothing…

13

A B

12412

C D

aabab

B

13123

D

aaabb

E

r s

A B

11

C D

aa

E

…there are no rows in s that match our third row in r, so do nothing…

14

A B

12412

C D

aabab

B

13123

D

aaabb

E

r s

A B

1111

C D

aaaa

E

There are two rows in s that match our fourth row in r, so both are joined to our fourth row…

15

A B

12412

C D

aabab

B

13123

D

aaabb

E

r s

There is one row that matches our fifth row in r,.. so it is joined to our fifth row and we are done!

A B

11112

C D

aaaab

E

16

Natural Join on Sailors Example

sid sname rating age bid day

22 dustin 7 45.0 101 10/ 10/ 9658 rusty 10 35.0 103 11/ 12/ 96

11 RS

sid sname rating age

22 dustin 7 45.0

31 lubber 8 55.558 rusty 10 35.0

sid bid day

22 101 10/10/9658 103 11/12/96

S1 R1

17

Last week we saw…

Query: Find the name of the sailor who reserved boat 101.

€

Temp=ρ (sid→sid1, S1) × ρ(sid→sid2, R1)

Result=πSname

(σsid1=sid2 ∧bid=101

(Temp))

* Note my use of “temporary” relation Temp.

18

Query revisited using natural join

Query: Find the name of the sailor who reserved boat 101.

€

Result=πSname

(σbid=101

(S1 >< R1))

Or

Result=πSname

(S1 >< σbid=101

(R1))

19

Conditional-Join Operation:Conditional-Join Operation:The conditional join is actually the most general type of join. We introduced the natural join first only because it is more intuitive and... natural!

Just like natural join, conditional join combines a cross product and a selection into one operation. However instead of only selecting rows that have equality on those attributes that appear in both relation schemes, we allow selection based on any predicate.

r c s = c(r x s) Where c is any predicate on the attributes of r and/or s

Duplicate rows are removed as always, but duplicate columns are not removed!

20

l-name f-name marr-Lic age

Simpson Marge 777 35

Lovejoy Helen 234 38

Flanders Maude 555 24

Krabappel Edna 978 40


Simpson Homer 777 36

Lovejoy Timothy 234 36

Simpson Bart null 9

r r.age < s.age AND r.Marr-Lic = s.Marr-Lic s

r.l-name r.f-name r.Marr-Lic r.age s.l-name s.f-name s.marr-Lic s.age

Simpson Marge 777 35 Simpson Homer 777 36

We want to find all women that are younger than their husbands…

Conditional-Join Example:Conditional-Join Example:

r s

Note we have removed ambiguity of attribute names by using “dot” notationAlso note the redundant information in the marr-lic attributes

21

Equi-Join

• Equi-Join: Special case of conditional join where the conditions consist only of equalities.

• Natural Join: Special case of equi-join in which equalities are specified on ALL fields having the same names in both relations.

22


Simpson Marge 777 35

Lovejoy Helen 234 38

Flanders Maude 555 24

Krabappel Edna 978 40


Simpson Homer 777 36

Lovejoy Timothy 234 36

Simpson Bart null 9

r r.Marr-Lic = s.Marr-Lic s

r.l-name r.f-name Marr-Lic r.age s.l-name s.f-name s.age

Simpson Marge 777 35 Simpson Homer 36

Lovejoy Helen 234 38 Lovejoy Timothy 36

Equi-JoinEqui-Join

r s

23

Review on Joins• All joins combine a cross product and a selection into

one operation.• Conditional Join

– the selection condition can be of any predicate (e.g. rating1 > rating2)

• Equi-Join: – Special case of conditional join where the conditions consist

only of equalities.

• Natural Join– Special case of equi-join in which equalities are specified on

ALL fields having the same names in both relations.

24

Banking ExamplesBanking Examples

branch (branch-name, branch-city, assets)

customer (customer-name, customer-street, customer-only)

account (account-number, branch-name, balance)

loan (loan-number, branch-name, amount)

depositor (customer-name, account-number)

borrower (customer-name, loan-number)

Note that I have not indicated primary keys here for simplicity.

25

Example QueriesExample Queries• Find all loans of over $1200

amount > 1200 (loan)

loan

loan-number branch-name amount

1234 Riverside 1,923.03

3421 Irvine 123.00

2342 Dublin 56.25

4531 Prague 120.03

1234 Riverside 1,923.03amount > 1200 (loan)

“select from the relation loan, only the rows which have an amount greater than 1200”

26

Example QueriesExample Queries• Find the loan number for each loan of an amount greater than $1200

loan-number (amount > 1200 (loan))

loan


1234 Riverside 1,923.03

3421 Irvine 123.00

2342 Dublin 56.25

4531 Prague 120.03

amount > 1200 (loan) 1234 Riverside 1,923.03

loan-number (amount > 1200 (loan)) 1234

“select from the relation loan, only the rows which have an amount greater than 1200, then project out just the loan_number”

27

Example QueriesExample Queries• Find all loans greater than $1200 or less than $75

amount > 1200 amount < 75(loan)

loan


1234 Riverside 1,923.03

3421 Irvine 123.00

2342 Dublin 56.25

4531 Prague 120.03

amount > 1200 amount < 75(loan)

“select from the relation loan, only the rows which have an amount greater than 1200 or an amount less than 75

1234 Riverside 1,923.03

2342 Dublin 56.25

28

Example QueriesExample Queries• Find the names of all customers who have a loan, an account, or both, from

the bank

customer-name (borrower) customer-name (depositor)


Patty 1234

Apu 3421

Selma 2342

Ned 4531

depositorborrower

customer-name account-number

Moe 3467

Apu 2312

Patty 9999

Krusty 3423

Patty

Apu

Selma

Ned

Moe

Apu

Patty

Krusty

Moe

Apu

Patty

Krusty

Selma

Ned

customer-name (borrower) customer-name (depositor)

29

Example QueriesExample QueriesFind the names of all customers who have a loan at the Riverside branch.

customer-name (branch-name=“Riverside ” (borrower.loan-number = loan.loan-number(borrower x loan)))


Patty 1234

Apu 3421


1234 Riverside 1,923.03

3421 Irvine 123.00

borrower loan

Note this example is split over two slides!

customer-name borrower.loan-number

loan.loan-number

branch-name amount

Patty 1234 1234 Riverside 1,923.03

Patty 1234 3421 Irvine 123.00

Apu 3421 1234 Riverside 1,923.03

Apu 3421 3421 Irvine 123.00

We retrieve borrower and loan…

…we calculate their cross product…

30


loan.loan-number

branch-name amount


Patty 1234 3421 Irvine 123.00

Apu 3421 1234 Riverside 1,923.03

Apu 3421 3421 Irvine 123.00

…we calculate their cross product…

…we select the rows where borrower.loan-number is equal to loan.loan-number…

…we select the rows where branch-name is equal to “Riverside”

…we project out the customer-name.


loan.loan-number

branch-name amount


Apu 3421 3421 Irvine 123.00


loan.loan-number

branch-name amount


Patty


31

Now Using Natural JoinNow Using Natural JoinFind the names of all customers who have a loan at the Riverside branch.


=> customer-name (branch-name=“Riverside ” borrower loan))


Patty 1234

Apu 3421


1234 Riverside 1,923.03

3421 Irvine 123.00

borrower loanWe retrieve borrower and loan…

1234 in borrower is matched with 1234 in loan…

3421 in borrower is matched with 3421 in loan…

The rest is the same.


loan.loan-number

branch-name amount


Apu 3421 3421 Irvine 123.00


loan.loan-number

branch-name amount


32

Example QueriesExample QueriesFind the largest account balance

...we will need to rename account relation as d...

balance(account) - account.balance(account.balance < d.balance (account x (d,account)))

account-number

balance

Apu 100.30

Patty 12.34

Lenny 45.34

Note this example is split over three slides!

accountaccount-number

balance

Apu 100.30

Patty 12.34

Lenny 45.34

d

We do a rename to get a “copy” of account which we call d…

… next we will do a cross product…

33

account.account-number

account.balance

d.account-number

d.balance

Apu 100.30 Apu 100.30

Apu 100.30 Patty 12.34

Apu 100.30 Lenny 45.34

Patty 12.34 Apu 100.30

Patty 12.34 Patty 12.34

Patty 12.34 Lenny 45.34

Lenny 45.34 Apu 100.30

Lenny 45.34 Patty 12.34

Lenny 45.34 Lenny 45.34



account.balance

d.account-number

d.balance

Patty 12.34 Apu 100.30


Lenny 45.34 Apu 100.30

… do a cross product…

…select out all rows where account.balance is less than d.balance…

.. next we project…

34



account.balance

d.account-number

d.balance

Patty 12.34 Apu 100.30


Lenny 45.34 Apu 100.30

.. next we project out account.balance…

…then we do a set difference between it and the original account.balance from the account relation…

… the set difference leaves us with one number, the largest value!

account.balance

12.34

12.34

45.34account-number

balance

Apu 100.30

Patty 12.34

Lenny 45.34

account

100.30

35

Now Using Conditional JoinNow Using Conditional Join

Find the largest account balance ...we will need to rename account relation as d...


(d,account)

balance(account) - account.balance(account account.balance < d.balance d)

36

More Examples on Sailors Relations

Sailors(sid, sname, rating, age)

Boats(bid, bname, color)

Reserves(sid, bid, day)

37

Find names of sailors who’ve reserved boat #103

• Solution 1:

• Solution 2:

• Solution 3:

)))Re(( (103

Sailorsservesbidsname

)Re1 (103

servesTempbid

SailorsTempTemp 12

)2(Re Tempsult sname

sname bidserves Sailors( (Re ))

103

38

Find names of sailors who’ve reserved a red boat

• Information about boat color only available in Boats; so need an extra join:

• A more efficient solution:

sname color redBoats serves Sailors((

' ') Re )

sname sid bid color redBoats s Sailors( ((

' ') Re ) )

A query optimizer can find this given the first solution!

Find sailors who’ve reserved a red or a green boat

• Can identify all red or green boats, then find sailors who’ve reserved one of these boats:

)(''''

Boatsgreencolorredcolor

Tempboats

)Re(Re SailorsservesTempboatssnamesult Can also define Tempboats using union! (How?)

What happens if is replaced by in this query?

Find sailors who’ve reserved a red and a green boat

• Previous approach won’t work! Must identify sailors who’ve reserved red boats, sailors who’ve reserved green boats, then find the intersection (note that sid is a key for Sailors):

)Re))(''

(( servesBoatsredcolorsid

Tempred

))((Re SailorsTempgreenTempredsnamesult

)Re))(''

(( servesBoatsgreencolorsid

Tempgreen

Consider yet another query

• Find the sailor(s) (sid) who reserved all the red boats.

sid bid day

22 101 10/10/96 22 103 10/11/96 58 102 11/12/96

R1

bid color

101 Red 102 Green 103 Red

B

Start an attempt

• who reserved what boat:

All the red boats:

sid bid

22 101 22 103 58 102

)1(,

1 Rbidsid

S

))((2 Bredcolorbid

S bid

101 103

43

Division OperationDivision Operation• Suited to queries that include the phrase “for all”, e.g. Find sailors who have

reserved all red boats.• Let S1 have 2 fields, x and y; S2 have only field y:

– S1/S2 =

– i.e., S1/S2 contains all x tuples (sailors) such that for every y tuple (redboat) in S2, there is an xy tuple in S1 (i.e, x reserved y).

• In general, x and y can be any lists of fields; y is the list of fields in S2, and xy is the list of fields of S1.

• Let r and s be relations on schemas R and S respectively where

– R = (A1, …, Am, B1, …, Bn)

– S = (B1, …, Bn)

The result of r / s is a relation on schema

R – S = (A1, …, Am)

r / s

€

x |∀ y in S2 (∃ x,y in S1) ⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪

44

Division Operation – ExampleDivision Operation – Example

Relations r, s:

r / s:

A

B

1

2

A B

12311134612

r

s

occurs in the presence of both 1 and 2, so it is returned. occurs in the presence of both 1 and 2, so it is returned. does not occur in the presence of both 1 and 2, so is ignored....

45

Another Division ExampleAnother Division Example

A B

aaaaaaaa

C D

aabababb

E

11113111

Relations r, s:

r /s:

D

ab

E

11

A B

aa

C

r

s

<, a , > occurs in the presence of both <a,1> and <b,1>, so it is returned.< , a , > occurs in the presence of both <a,1> and <b,1>, so it is returned.<, a , > does not occur in the presence of both <a,1> and <b,1>, so it is ignored.

Examples of Division A/Bsno pnos1 p1s1 p2s1 p3s1 p4s2 p1s2 p2s3 p2s4 p2s4 p4

pnop2

pnop2p4

pnop1p2p4

snos1s2s3s4

snos1s4

snos1

A

B1B2

B3

A/B1 A/B2 A/B3

47

Find the sailor(s) who reserved ALL red boats

• who reserved what boat:

• All the red boats:

sid bid

22 101 22 103 58 102

)1(,

1 Rbidsid

S

))((2 Bredcolorbid

S bid

101 103

=> S1/S2

Find the names of sailors who’ve reserved all boats

• Uses division; schemas of the input relations must be carefully chosen:

))((/))(Re,

(Tempsids Boatsbid

servesbidsid

)(Result SailorsTempsidssname

To find sailors who’ve reserved all ‘Interlake’ boats:

))(''

(/ BoatsInterlakebnamebid

.....

Some PropertiesSome Properties• Selection

• Projection

where li is a subset of attributes in R, li li+1

• A projection commutes with a selection that only uses attributes retained by the projection:

where c(a1,…,an) is a condition on attributes a1,…,an

c c c cR R1 2 2 1

(Commute)

l1 lnR R

l1

. . . (Cascade)

c cn c cnR R1 1 ... . . . (Cascade)

a1,..,an(c(a1,..,an) (R))= c(a1,..,an) (a1,..,an( R ))

Properties of JoinsProperties of Joins

Joins

where c(a1,…,an) is a condition on attributes only in R

(Commute)

(Associative)

(R cS) (S cR) R c1 (S c2T) (R c1S) c2T

c(R c S) c(R) c S

•

•

•

Additional Operator: Outer Additional Operator: Outer JoinJoin

• Joins: : match tuples satisfying a join condition..– Tuples without a matching are

discarded.– Tuples with Null value in at least one

join attribute are discarded.• Outer Joins: : useful when we want to keep

tuples in one or both of the involved relations in a join operation that do not satisfy the join condition. – Left outer join– Right outer join– Full outer join

• Schema of any Outer Join: Schema of Natural Join

52

Additional Operator: Outer Additional Operator: Outer JoinJoin

• An extension of the join operation that avoids loss of information.

• Computes the join and then adds tuples from one relation that does not match tuples in the other relation to the result of the join.

• Uses null values:– null signifies that the value is unknown or does not exist

– All comparisons involving null are (roughly speaking) false by definition.

• Will study precise meaning of comparisons with nulls later

53

Outer Join – ExampleOuter Join – Example

• Relation loan

Relation borrower customer-name loan-number

SimpsonWiggumFlanders

L-170L-230L-155

loan-number amount

L-170L-230L-260

300040001700

branch-name

SpringfieldShelbyville Dublin

54


• Inner Join

loan Borrower

loan borrower

• Left Outer Join

loan-number amount

L-170L-230

30004000

customer-name

SimpsonWiggum

branch-name

SpringfieldShelbyville

loan-number amount

L-170L-230L-260

300040001700

customer-name

SimpsonWiggumnull

branch-name

SpringfieldShelbyvilleDublin



L-170L-230L-155

loan-number amount

L-170L-230L-260

300040001700

branch-name


55


Right Outer Join loan borrower

loan-number amount

L-170L-230L-155

30004000null

customer-name


loan-number amount

L-170L-230L-260L-155

300040001700null

customer-name

SimpsonWiggumnullFlanders

loan borrower

Full Outer Join

branch-name

SpringfieldShelbyvillenull

branch-name

SpringfieldShelbyvilleDublinnull



L-170L-230L-155

loan-number amount

L-170L-230L-260

300040001700

branch-name


Full Outer JoinFull Outer Join

• Full Outer Join S R : keeps every tuple in both R and S, if not matching tuples are found in R (S), then the attributes in R (S) are set to Null

(S R) (S R)

Cost of Relational Algebra Cost of Relational Algebra OperationsOperations

• Given two relations R, S, such as :

CardinalityCardinality (R) = n, CardinalityCardinality(S) = m

• Cost O(n) c ( R )

a1,…,an ( R )

• Cost O(n x m)

R x S

When working with Cross-product (Joins), keep input relations as small as possible!

Cost of Relational Algebra Cost of Relational Algebra OperationsOperations

Let q= max(m,n),

Set Operations: R S, R S, R – S:Cost O(q log q)

Join Operations: R S, R cSOuter Joins Operations:

R S, R S, R S

Best Case : attributes in the join condition include the primary key in S: Cost O(q log q)

Otherwise : worst case : Cost O(m x n)

When working with these operations, keep input relations as small as possible

59

Summary• The relational model has rigorously defined query

languages that are simple and powerful.• Relational algebra is more operational; useful as

internal representation for query evaluation plans.• Several ways of expressing a given query; a query

optimizer should choose the most efficient version.

• Operations covered: 5 basic operations (selection, projection, union, set difference, cross product), rename, joins (natural join, equi-join, conditional join, outer joins), division

relational algebra (continued) lecture 5. relational algebra (summary) basic operations: –...

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