CHAPTER 2CHAPTER 2 RELATIONAL MODELRELATIONAL MODEL

CHAPTER 2CHAPTER 2RELATIONAL MODELRELATIONAL MODEL

Chapter 2: Relational Modelp

Structure of Relational DatabasesFundamental Relational Algebra OperationsAdditional Relational Algebra OperationsAdditional Relational Algebra OperationsExtended Relational-Algebra-OperationsNull ValuesModification of the Database

Original Slides:© Silberschatz, Korth and Sudarshan

Advanced DB (2008-1)Copyright © 2006 - 2008 by S.-g. Lee Silberschatz Chap 2 - 2

Basic Structure

Formally,given sets D1, D2, …. Dn a relation r is a subset of D1 x D2 x … x Dn

R ⊆ D1, X ….. X D (n: degree of R)R ⊆ D1, X ….. X Dn (n: degree of R)or

R = { <d1, ..., dn> | d1∈ D1, ..., dn∈ Dn } (set of tuples)

Example: customer-name = {Jones, Smith, Curry, Lindsay}

customer-street = {Main, North, Park}

customer-city = {Harrison, Rye, Pittsfield}y { , y , }

Then r = { (Jones, Main, Harrison), (Smith, North, Rye),

(Curry North Rye) (Lindsay Park Pittsfield)}(Curry, North, Rye), (Lindsay, Park, Pittsfield)}

is a relation over customer-name x customer-street x customer-city



Attribute Typesyp

Each attribute of a relation has a nameThe set of allowed values for each attribute is called the domainof the attributeof the attributeAttribute values are (normally) required to be atomic, that is, indivisibleindivisible

E.g. multivalued attribute values are not atomicE g composite attribute values are not atomicE.g. composite attribute values are not atomic

The special value null is a member of every domainTh ll l li i i h d fi i i fThe null value causes complications in the definition of many operations

dwe shall ignore the effect of null values in our main presentation and consider their effect later



Relation Schema

A1, A2, …, An are attributes1 2 n

R = (A1, A2, …, An ) is a relation schemaE g Customer schema = (customer name customer street customer city)E.g. Customer-schema = (customer-name, customer-street, customer-city)

r(R) is a relation (variable) on the relation schema RE.g. customer(Customer-schema)



Relational Algebrag

Algebra : operators and operandsRelational algebra

operands : relationsb i (+ ddi i l i )operators : basic operators (+ additional operations)

take two or more relations as inputs and give a new relation as a result.

Procedural languageProcedural language6 Fundamental Operators

selectselectprojectunionset differenceCartesian productrename



Example Queriesp Q

Find all loans of over $1200σamount >1200 (loan)

Find the loan number for each loan of an amount greater than $1200$1200

∏loan-number (σamount>1200 (loan))



Example Queriesp Q

Find the names of all customers who have a loan, an account, or both, from the bank

∏customer-name (borrower) ∪ ∏customer-name (depositor)

Find the names of all customers who have a loan and an account at bank.

∏customer-name (borrower) ∩ ∏customer-name (depositor)customer name customer name



Example Queriesp Q

Find the names of all customers who have a loan at the Perryridge branch.

∏customer-name (σbranch-name=“Perryridge”

(σb l b = l l b (borrower x loan)))(σborrower.loan-number = loan.loan-number(borrower x loan)))

Find the names of all customers who have a loan at the Perryridge branch but do notFind the names of all customers who have a loan at the Perryridge branch but do not have an account at any branch of the bank.

∏ t (σb h “P id ”∏customer-name (σbranch-name = “Perryridge”

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

– ∏customer-name(depositor)



Example Queriesp Q

Find the names of all customers who have a loan at the Perryridge branch.

Query 1∏ (∏customer-name(σbranch-name = “Perryridge”

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

Query 2∏ (∏customer-name(

(σloan.loan-number = borrower.loan-number(( σbranch name = “Perryridge”(loan) ) x borrower( branch-name = Perryridge ( ) ) x w

) ))



Example Queriesp Q

Find the largest account balancegRename account relation as dThe query is:

∏balance(account) –∏balance( )∏account.balance (

σ ( t x ρ ( t) )σaccount.balance < d.balance (account x ρd (account) ))



Formal Definition

A basic expression in the relational algebra consists of either one of the following:

A relation in the databaseA lA constant relation

Let E1 and E2 be relational-algebra expressions; the following are ll l ti l l b p iall relational-algebra expressions:

E1 ∪ E2E1 – E2E1 x E2σp (E1), P is a predicate on attributes in E1∏s(E1), S is a list consisting of some of the attributes in E1ρN (E1), N is the new name for the result of E1



Additional Operationsp

We define additional operations that do not add any power to thep y prelational algebra, but that simplify common queries.

Set intersection

Natural joinNatural joinDivisionA iAssignment



Division Operationp

r ÷ s

Suited to queries that include the phrase “for all”Suited to queries that include the phrase for all .Let r and s be relations on schemas R and S respectively where

R = (A1, …, Am, B1, …, Bn)S = (B1, …, Bn)

Th l f i l i hThe result of r ÷ s is a relation on schemaR – S = (A1, …, Am)

r ÷ s = { t | t ∈ ∏ R-S(r) ∧ ∀ u ∈ s ( tu ∈ r ) }



Division Operation – Examplep p

Relations r s: r ÷ s:

ABA B

Relations r, s: r ÷ s:

AB

α1

2

αα

12

β2αβγ

311

sγδδδ

1134δ

∈∈β

4612β 2

r



Division Operation – Examplep p

Relations r s: r ÷ s:

A B C D E D E A B C

Relations r, s: r ÷ s:

A B

α a

C D

α a

E

11

D

ab

E

11

A B

α a

C

γααβ

aaa

γγγ

aba

111

b 1 γ a γs

βγγ

aaa

γγγ

bab

311γ

γ aγβ b 1

r



Division Operationp

Property Let q = r ÷ sThen q is the largest relation satisfying q × s ⊆ r

Definition in terms of the basic algebra operationLet r(R) and s(S) be relations, and let S ⊆ R

r ÷ s = ∏R-S(r) –∏R-S ( (∏R-S (r) × s) – ∏R-S,S(r))

To see why∏R-S,S(r) simply reorders attributes of r

∏ (∏ (r) × s) ∏ (r)) gives those tuples t in ∏ (r)∏R-S(∏R-S (r) × s) – ∏R-S,S(r)) gives those tuples t in ∏R-S (r)such that for some tuple u∈s, tu ∉ r.



Example Queries

Find all customers who have an account from at least the

p Q

“Downtown” and the “Uptown” branches.Query 1

∏customer-name(σbranch-name=“Downtown”(depositor account)) ∩

∏ (σ (depositor account))∏customer-name(σbranch-name=“Uptown”(depositor account))

Query 2

∏customer-name, branch-name (depositor account)÷ ρtemp(branch-name) ({(“Downtown”), (“Uptown”)})



Example Queriesp Q

Find all customers who have an account at all branches located inFind all customers who have an account at all branches located in Brooklyn city.

∏customer-name, branch-name (depositor account)customer name, branch name ( p )÷ ∏branch-name (σbranch-city = “Brooklyn” (branch))



Extended Relational-Algebra Operationsg p

adds power and convenience to the relational algebrap g

Generalized ProjectionGeneralized ProjectionAggregate FunctionsOuter Join



Generalized Projectionj

Extends the projection operation by allowing arithmetic p j p y gfunctions to be used in the projection list.

∏F1, F2, …, Fn(E), , ,

E is any relational-algebra expressionEach of F1, F2, …, Fn are are arithmetic expressions involving constants and attributes in the schema of E.

Find how much more each person can spendcredit-info(customer-name, limit, credit-balance)f ( )

∏customer-name, limit – credit-balance (credit-info)



Aggregate Functions and Operations

Aggregation function

gg g p

gg gtakes a collection of values and returns a single value as a result.avg, min, max, sum, count

Aggregate operation in relational algebra

g (E)G1, G2, …, Gn g F1(A1), F2(A2), …, Fm(Am) (E)

E is any relational-algebra expressionG1, G2…, Gn is a list of attributes on which to group (can be empty)F i f i dFi is an aggregate function, and Ai is an attribute name, for i=1, …, m



Aggregate Operation – Examplegg g p p

Relation r:A B

α α

C

7ααβ

αββ

7

7

3ββ

ββ3

10

sum Cg sum(c) (r)sum-C

27



Aggregate Operation – Example

Relation account

gg g p p

branch-name account-number balancen h n o n n n

PerryridgePerryridge

A-102A-201

400900

BrightonBrightonRedwood

A-217A-215A-222

750750700

g (account) branch-name balancebranch-name g sum(balance) (account)

PerryridgeBrightonRedwood

13001500700

Result of aggregation does not have a nameC i i i ( i SQL)

Redwood 700

Can use rename operation to give it a name (as in SQL)

branch-name g sum(balance) as sum-balance (account)



branch-name g sum(balance) as sum-balance ( )

Outer JoinJ

An extension of the join operation that avoids loss of j pinformation

Compute the join add tuples that do not have matching tuples in the other relation

fill in undefined attribute values with null

Three typesleft outer joinright outer joinfull outer join

Conventional join is called inner join



Outer Join – ExampleJ p

loan-number amountbranch-nameRelation loan o n n o n

L-170L-230L 260

300040001700

DowntownRedwoodP idL-260 1700Perryridge

Relation borrowercustomer-name loan-number

JonesS i h

L-170L 230Smith

HayesL-230L-155

Inner join: loan BorrowerInner join: loan Borrower

loan-number amount customer-namebranch-name

L-170L-230

30004000

JonesSmith

DowntownRedwood



Outer Join – Example

Left outer join:

J p

loan-number amount customer-namebranch-nameloan borrower

o n n o n

L-170L-230L 260

300040001700

o n

JonesSmithll

n h n

DowntownRedwoodP idL-260 1700 nullPerryridge

Right outer join :loan borrower

loan-number amount

L-170L 230

30004000

customer-name

JonesSmith

branch-name

DowntownRedwoodL-230

L-1554000null

SmithHayes

Redwoodnull

Full outer join :loan borrower

loan-number amount

L-170 3000

customer-name

Jones

branch-name

Downtownloan borrower L 170L-230L-260L 155

300040001700null

JonesSmithnullHayes

DowntownRedwoodPerryridgenull



L-155 null Hayesnull

Null Values

null: an unknown or non-existing valuegThe result of any arithmetic expression involving null is nullAggregate functionsAggregate functions

simply ignore null values (SQL semantics)Is an arbitrary decision Could have returned null as result insteadIs an arbitrary decision. Could have returned null as result instead.

For duplicate elimination and groupingll i d lik h l d ll d b hnull is treated like any other value, and two nulls are assumed to be the

same (SQL semantics)Alternative: assume each null is different from each otherAlternative: assume each null is different from each otherBoth are arbitrary decisions



Null Values

Comparisons with null values return the special truth value unknownWhy do we need unknown?If false was used instead of unknown, thennot (A < 5) would not be equivalent to A >= 5not (A < 5) would not be equivalent to A > 5

Three-valued logic using the truth value unknown:OR: (unknown or true) = true,

(unknown or false) = unknown(unknown or unknown) = unknown

AND: (true and unknown) = unknown, ( )(false and unknown) = false,(unknown and unknown) = unknown

NOT: (not unknown) = unknown( )In SQL, “P is unknown” evaluates to true if predicate P evaluates to unknown

Result of select predicate is treated as false if it evaluates to unknown



Modification of the Database

The content of the database may be modified using the y gfollowing operations:

DeletionInsertionUpdating

All these operations are expressed using the assignment operatorAll these operations are expressed using the assignment operator.



Deletion

A delete request is expressed similarly to a query, except instead q p y q y pof displaying tuples to the user, the selected tuples are removed from the database.Deletion operation in relational algebra :

r ← r Er ← r – Ewhere r is a relation and E is a relational algebra query.

You can only delete whole tuples;cannot delete values on only particular attributes=> should use update operation



Deletion Examples

Delete all account records in the Perryridge branch.

p

account ← account – σ branch-name = “Perryridge” (account)Delete all loan records with amount in the range of 0 to 50

loan ← loan – σ amount ≥ 0 and amount ≤ 50 (loan)Delete all accounts at branches located in Needham.

r1 ← σ branch-city = “Needham” (account branch)

r ← ∏b h b b l (r )r2 ← ∏branch-name, account-number, balance (r1)

r3 ← ∏ customer-name, account-number (r2 depositor)

t ← taccount ← account – r2

depositor ← depositor – r3



Insertion

To insert data into a relation, we either:specify a tuple to be insertedwrite a query whose result is a set of tuples to be inserted

Insertion operation in relational algebra:r ← r ∪ Er ← r ∪ E

where r is a relation and E is a relational algebra expression.

The insertion of a single tuple is expressed by letting E be a constant relation containing one tuple.



Insertion Examplesp

Insert information in the database specifying that Smith has p y g$1200 in account A-973 at the Perryridge branch.

account ← account ∪ {(“Perryridge”, A-973, 1200)}account ← account ∪ {( Perryridge , A 973, 1200)}depositor ← depositor ∪ {(“Smith”, A-973)}

Provide as a gift for all loan customers in the Perryridge branch, a $200 savings account. Let the loan number serve as the account number for the new savings account.

r1 ← (σbranch-name = “Perryridge” (borrower loan))account ← account ∪ ∏branch-name account-number 200 (r1)branch-name, account-number,200 ( 1)depositor ← depositor ∪ ∏customer-name, loan-number,(r1)



Updatingp g

A mechanism to change a value in a tuple without changing allg p g gvalues in the tupleUse the generalized projection operator to do this taskUse the generalized projection operator to do this task

r ← ∏ F1, F2, …, FI, (r)

Each Fi, is eitherthe ith attribute of r: the attribute is not updated, or, an expression, involving only constants and the attributes of r, which gives h l f h ib h ib i d dthe new value for the attribute: the attribute is updated



Update Examplesp p

Make interest payments by increasing all balances by 5 percent.p y y g y paccount ← ∏ account#, branch-name, balance*1.05 (account)

Pay all accounts with balances over $10,000 6 percent interest d ll h 5and pay all others 5 percent

account ←∏ account#, branch-name, balance*1.06 (σ balance > 10000 (account) ) ∪, ,

∏account#, branch-name, balance*1.05 (σbalance ≤ 10000 (account) )



END OF CHAPTER 2END OF CHAPTER 2

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CHAPTER 2CHAPTER 2 RELATIONAL MODELRELATIONAL MODEL