Relational Model DEFINITION AND PROPERTIES CONSTRAINTS AND SCHEMAS UPDATE ER | EER RELATIONAL D. Christozov / G.Tuparov INF 280 Database Systems: Relational

Relational Model

DEFINITION AND PROPERTIES

CONSTRAINTS AND SCHEMAS UPDATE

ER | EER RELATIONAL

D. Christozov / G.Tuparov INF 280 Database Systems: Relational Model 1

Relational Model: Definition and Properties

Relational Model represents database as a collection of relations.• A relation may be considered as a table of values,

where each row represents a collection of related values, forming an instance, a fact, corresponding with an entity in an entity-set.

• Table-name and columns-names are used to interpret the meaning of values in each row and define relational schema.

• Each value in a column is drawn by from the same value-set (data-type).

D. Christozov / G.TuparovINF 280 Database Systems:

Relational Model 2

Relational Model: Definition and Properties

In original definition of Relational Model:Tables are called relations;Rows – tuples;Column-names – attributes;Data-type – domain.


Relational Model 3

Relational Model: Definition and Properties• Domain is a set of allowable values for one or more

attributes• Degree is a number of attributes in a relation.• Cardinality is a number of tuples in a relation.

Alternative terminologyFormal terms Alternative 1 Alternative 2Relation Table FileTuple Row RecordAttribute Column Field


Relational Model 4

Properties of Relations• Relation name is distinct from all other relations.• Each cell of relation contains exactly one atomic

(single) value.• Each attribute has a distinct name within relation.• Values of an attribute (column) are all from the

same domain.• Order of attributes has no significance.• Each tuple is distinct; there are no duplicate

tuples.• Order of tuples has no significance, theoretically.

To ensure efficiency, tuples in a table are ordered according to primary key.


Relational Model 5

Relational Constraints (1)

Key constraints (uniqueness constraint):• Superkey (Key) – – Formally - a set of attributes of a relation R is

called Superkey (SK), if no any two distinct tuples t1 and t2 in a state r of R have the same value for attributes belonging to SK:

t1[SK] <> t2[SK] – Informally - an attribute or a set of attributes that

uniquely identifies a tuple within a relation.


Relational Model 6

Relational Constraints (2)• Key (candidate key) is a minimal Superkey — a

Superkey from which we cannot remove an attribute and still, the remaining set of attributes, is Superkey (the uniqueness constraint holds).

• Primary Key - candidate key selected to identify tuples uniquely within relation.

• Alternate Keys - candidate keys that are not selected to be the primary key.

• Foreign Key - An attribute or set of attributes within one relation that matches candidate key of some (possibly the same) relation.


Relational Model 7

Selection of PK affects performance!

Defining foreign keys ensures database integrity!

Relational integrity (1)

• Null–Represents a value for an attribute that is

currently unknown or is not applicable for this tuple.–Deals with incomplete or exceptional data.–Null represents the absence of a value and

is not the same as zero or spaces, which are values.


Relational Model 8

Relational integrity (2)

2 principal rules for the relational model:• Entity Integrity– In a base relation, no attribute of a primary key

can be null.• Referential Integrity– If foreign key exists in a relation, either the foreign

key value must match a candidate key value of some tuple in its home relation or foreign key value must be wholly null.


Relational Model 9

Relational integrity (3)• Enterprise Constraints– Additional rules specified by users or database

administrators.– Eg., the maximum number of staff in a branch.


Relational Model 10

Company Database: Relational Schema


Relational Model 11

EMPLOYEE

Fname Minit Lname SSN Bdate Address Sex Salary SupSSN DNO

DEPARTMENTDname Dnumber MGRSSN MGRSDate

PROJECT

Pname Pnumber Plocation DNum

DEPENDENT

WORKS_ON

ESSN PNO Hours

ESSN D_name Sex Bdate Relationship

LOCATION

D_num D_location

Company Database: Relational State


Relational Model 12

EMPLOYEE

Fname Minit Lname SSN Bdate Address Sex Salary SupSSN DNO

DEPARTMENTDname Dnumber MGRSSN MGRSDate

Research 5 333445555 1988-05-22

Administration 4 987654321 1995-01-01

Headquarter 1 888665555 1981-06-19

John B Smith 123456789 1965-01-09 …TX… M 30000 3334455555 5

Franklin T Wong 333445555 1956-12-08 …TX… M 40000 888665555 5

Alicia J Zelaya 999887777 1968-01-19 …TX… F 25000 987654321 4

Jennifer S Wallace 987654321 1941-06-02 …TX… F 43000 888665555 4

Ramesh K Narayan 666884444 1962-09-15 …TX… M 38000 333445555 5

Joyce A English 453453453 1972-07-31 …TX… F 25000 333445555 5

Ahmad V Jabbar 987987987 1969-03-29 …TX… M 25000 987654321 4

James E Borg 888665555 1937-11-10 …TX… M 55000 Null 1

Transition Constraints: InsertThe Insert Operation can violate any of the four types of constraints discussed:• Domain constraints can be violated if an attribute is

given value that does not appear in the corresponding domain.

• Key constraints can be violated if a key value in the new tuple t already exists in another tuple in the relation r(R).

• Entity integrity can be violated if the primary key of the new tuple t is null.

• Referential integrity can be violated if the value of any foreign key in t refers to a tuple that does not exist in the referenced relation.


Relational Model 13

Transition Constraints: Insert – Examples (1)

<‘Cecilia’, ‘F’, ‘Kolonsky’, null, ‘1960-04-05’, ‘6357 Windy Lane, Katy, TX’, F, 28000, null, 4> This insertion violates the entity integrity constraint (null for the primary key SSN) Rejected!

<‘Alicia’, ‘J’, ‘Zelaya’, ‘999887777’, ‘1960-04-05’, ‘6357 Windy Lane, Katy, TX’, F, 28000, ‘987654321’, 4>This insertion violates the key constraint because another tuple with the same SSN value already exists in the EMPLOYEE relation Rejected!


Relational Model 14

Transition Constraints: Insert – Examples (2)

<‘Cecilia’, ‘F’, ‘Kolonsky’, ‘677678989’, ‘1960-04-05’, ‘6357 Windy Lane, Katy, TX’, F, 28000, ‘987654321’, 7> This insertion violates the referential integrity constraint specified on DNO because no DEPARTMENT tuple exists with DNUMBER = 7 Rejected or cascade request for insertion a new tuple into Department!

<‘Cecilia’, ‘F’, ‘Kolonsky’, ‘677678989’, ‘1960-04-05’, ‘6357 Windy Lane, Katy, TX’, F, 28000, null, 4> This insertion satisfies all constraints Accepted!


Relational Model 15

Transition Constraints: Delete• The Delete Operation can violate only referential

integrity: If the tuple being deleted is referenced via a foreign key from other tuples in the database.

• To specify which tuples to delete, a conditions defined on relation’s attributes have to be specified

Example:Delete the tuple from WORKS_ON where ESSN = ‘999887777’ and PNO = 10

Accepted!


Relational Model 16

Transition Constraints: Delete - ExamplesDelete the EMPLOYEE tuple where SSN = ‘999887777’ Rejected!This deletion is not acceptable, because FK in WORKS_ON refers to this tuple. Executing deletion will result in violation of referential integrity.

Delete the EMPLOYEE tuple where SSN = ‘333445555’ Rejected!This deletion will result in referential integrity violations. The tuple involved is referenced by FK from the EMPLOYEE, DEPARTMENT, WORKS_ON, and DEPENDENT relations. D. Christozov / G.Tuparov

INF 280 Database Systems: Relational Model 17

Transition Constraints: UpdateThe Update Operation• The Update operation is used to change the values of

one or more attributes in a tuple (or tuples) of some relation R. It is necessary to specify a condition on the attributes of the relation to select the tuple (or tuples) to be modified.

• ExamplesUpdate the SALARY of the EMPLOYEE tuple where SSN = ‘999887777’ to 28000Acceptable! (Why?)


Relational Model 18

Transition Constraints: Update - ExamplesUpdate the DNO of the EMPLOYEE tuple with SSN = ‘999887777’ to 1 Acceptable! (Why?)Update the DNO of the EMPLOYEE tuple with SSN = ‘999887777’ to 7 Unacceptable, because it violates referential integrity. Update the SSN of the EMPLOYEE tuple with SSN = ‘999887777’ to ‘987654321’ Unacceptable, because it violates primary key and referential integrity constraints.


Relational Model 19

Q & AAttention ! Next class

Quiz 1 Basic concepts, ER and EER


Relational Model 20

ER-to-Relational Mapping Algorithms• Step 1: Mapping of Regular Entity Types• Step 2: Mapping of Weak Entity Types• Step 3: Mapping of Binary 1:1 Relationship Types• Step 4: Mapping of Binary 1:N Relationship Types• Step 5: Mapping of Binary M:N Relationship Types• Step 6: Mapping of Multivalued Attributes• Step 7: Mapping of N-ary Relationship Types• Step 8: Options for Mapping Specialization or

Generalization• Step 9: Mapping of Union Types (Categories)


Relational Model 21

Step 1: Mapping of Regular Entity TypesFor each regular (strong) entity type E in the ER schema, create a relation R that includes all the simple attributes of E. Include only the atomic component attributes of a composite attribute. Choose one of the key attributes of E as primary key for R. If the chosen key of E is composite, the set of simple attributes that form it will together form the primary key of R.


Relational Model 22

Sal

Fn

Nam

Min Ln

AddrSex

BD

SSN EMPLOYEE

Fname Minit Lname SSN Bdate Address Sex Salary

EMPLOYEE

Step 2: Mapping of Weak Entity TypesFor each weak entity type W in the ER schema with owner entity type E, create a relation R, and include all simple attributes of W as attributes of R. In addition, include as foreign key attributes of R the primary key attribute(s) of the relation(s) E. The primary key of R is the combination of the primary key(s) of the owner(s) and the partial key of the weak entity type W, if any.


Relational Model 23

Sal

Fn

Nam

Min Ln

AddrSex

BD

SSN EMPLOYEE

Name Sex BDate Relation

DEPENDENT

ESSN D_name Sex Bdate Relationship

DEPENDENT

Step 3: Mapping of Binary 1:1 Relationship TypesFor each binary 1:1 relationship type R in the ER schema, identify the relations S and T that correspond to the entity types participating in R. Choose one of the relations—S, and include as foreign key in S the primary key of T. It is better to choose an entity type with total participation in R in the role of S. Include all the simple attributes (or simple components of composite attributes) of the 1:1 relationship type R as attributes of S.


Relational Model 24

Dname Dnumber MGRSSN MGRSDate

DEPARTMENT

Sal

Fn

Nam

Min Ln

AddrSex

BD

SSN EMPLOYEE

Number

LocationName

DEPARTMENT

NumEmp

Manage11

Start_Date

TS

Step 4: Mapping of Binary 1:N Relationship TypesFor each regular binary 1:N relationship type R, identify the relation S that represents the participating entity type at the N-side of the relationship type. Include as foreign key in S the primary key of the relation T that represents the other entity type participating in R; this is because each entity instance on the N-side is related to at most one entity instance on the 1-side of the relationship type. Include any attributes of the relationship type as attributes of S.


Relational Model 25

Fname Minit Lname SSN Bdate Address Sex Salary DNO

EMPLOYEE

Sal

Fn

Nam

Min Ln

AddrSex

BD

SSN EMPLOYEE

Number

LocationName

DEPARTMENT

NumEmp

Works_forN

1

ST

Step 5: Mapping of Binary M:N Relationship TypesFor each binary M:N relationship type R, create a new relation S to represent R. Include as foreign key attributes in S the primary keys of the relations that represent the participating entity types; their combination will form the primary key of S. Also include any simple attributes of the M:N relationship type as attributes of S.


Relational Model 26

Fname Minit Lname SSN Bdate Address Sex Salary DNO

Pname Pnumber Plocation DNumESSN PNO Hours

EMPLOYEE

Works_onN M

hours

Name

Location

Number

PROJECTSal

Fn

Nam

Min Ln

AddrSex

BD

SSN EMPLOYEE

PROJECTWORKS_ON

Step 6: Mapping of Multivalued AttributesFor each multivalued attribute A of entity E, create a new relation R. This relation R will include an attribute(s) corresponding to A, plus the primary key K of E as a foreign key in R referencing the relation created for E. The primary key of R is the combination of A and K. If the multivalued attribute is composite, we include its atomic components.


Relational Model 27

Number

LocationName

DEPARTMENT

NumEmp

Dname Dnumber MGRSSN MGRSDate

DEPARTMENT

D_num D_location

LOCATION

Step 7: Mapping of N-ary Relationship Types

For each n-ary relationship type R, where n > 2, create a new relation S to represent R. Include as foreign key attributes in S the primary keys of the relations that represent the participating entity types. Also include any simple attributes of the n-ary relationship type as attributes of S. The primary key of S is usually a combination of all the foreign keys that reference the relations representing the participating entity types.


Relational Model 28

Correspondence between ER and Relational Models


SQL - Single Table queries 29D. Christozov / G.TuparovINF 280 Database Systems:

Relational Model 29

ER MODEL RELATIONAL MODELEntity type Entity relation1:1 or 1:N relationship type Foreign key (or relationship relation)M:N relationship type Relationship relation and two foreign keysn-ary relationship type Relationship relation and n foreign keysSimple attribute AttributeComposite attribute Set of simple component attributesMultivalued attribute Relation and foreign keyValue set DomainKey attribute Primary (or secondary) key

Step 8: Mapping Specialization or Generalization

Convert each specialization with m subclasses {S1, S2, . . ., Sm} and (generalized) superclass C, where the attributes of C are {k, a1, . . ., an} and k is the (primary) key, into relation schemas using one of the four following options:



Relational Model 30

Step 8: Option A Multiple relations—superclass and subclasses



Relational Model 31

Create a relation L for C

Create a relation Li

for each subclass Si, with the attributes

Step 8: Option 8BMultiple relations—subclass relations only



Relational Model 32

Create a relation Li for each subclass Si, with the attributes

Step 8: Option 8CSingle relation with one type attribute



Relational Model 33

Create a single relation L, such that

This option is for a specialization whose subclasses are disjoint, and t is defining attribute, that indicates the subclass to which each tuple belongs.

This option has the potential for generating a

large number of null values.

Step 8: Option 8DSingle relation with multiple type attributes



Relational Model 34

Create a single relation schema L such that

This option is for a specialization whose subclasses are overlapping (but may also work for a disjoint) specialization, and each ti, 1 i m, is a Boolean attribute indicating whether a tuple belongs to subclass Si.

Step 9: Mapping of Union Types (Categories)



Relational Model 35

Mapping Shared Subclasses (Multiple Inheritance):Any of options 8A – 8D may apply.

STEP 9: For mapping of UNION Types (Categories), a new surrogate key attribute is specified, because the primary keys of the super-classes are different and may not use any of them to identify all entities in the category. The surrogate key is introduce in any of the super-classes relations as a foreign key to specify the correspondence.

u

owner

bank company person

OwnerID

OWNER

PERSON

SSN … OwnerID

BName … OwnerID CName … OwnerID

BANK COMPANY

Example from Lecture 2: Map ER R



Relational Model 36

Quack Consulting is a computer consulting firm. It keeps the list of its clients (id, name, phone). Each of them may have many different projects (id, description, start_date, due_to, team_leader). A project may consists of several tasks (project_id, task_id, consultant, description, start, due_to). Only one consultant is ever assigned to a given task. The time a consultant works for a project is recorded in time table. In order to choose the appropriate consultant for a client information about consultants and his/her specialty (including the billing_rate) is maintain as well as some comments about his/her qualities important for Quack Consulting.




Relational Model 37

Fancy Fruits is a specialty grocery store supplying a variety of boxed fruits to its customers. The entities are customers, orders, stock, and vendors. The stock keeps track to current inventory (quantities on-hand of the items and current price). The vendors lists the wholesalers who supply Fancy Fruits with items, it includes the cost of item for Fancy Fruit and geographical region, where the wholesaler is located (Hint: Consider that one vendor could supply a number of items and one item could be supplied by a number of vendors - different regions and costs). The customer also store information about the region of the customer as well as customer_name and phone. Orders defined the quantities of given item ordered by a customer




Relational Model 38

Retailers

M

Id

Name

Price

On_hand

Products

Codes

D_codeD_Price

PurchaseDate

Quantity

C_Id

Name

address Customers Dealersd

N

Quantity

D_code

orders Date

N

M

N

M




Relational Model 39

RecipesR_Id

Description

Activities

IngredientsI_Id

Name

On_hand

GoodsName

Price

Cook

N

1 M

Suppliers

Supplies

N

M

Price

Quantity

Name

Address




Relational Model 40

BANK BRANCH BANK OFFICE

LOAN

CUSTOMER

ACCOUNT DEBIT CARD

A - C

GRANTS

L - C

OPENS

BRANCHES

Code

Name Address Address Name

ISSUES

Loan No

Type

Amount

Rate

Period

Clinum

SSN Telephone

Name

Employer

Home Mobile Business

Acct No

Comfort InsAcct No

Type

Start Date

N1

1

1

1

N

N

N

M

N

M

N

Entity-Relationship

Database Diagram

Address

Q & AAttention ! Next class

Quiz 2: ER R


Relational Model 41

Documents

Relational Model DEFINITION AND PROPERTIES CONSTRAINTS AND SCHEMAS UPDATE ER | EER RELATIONAL D. Christozov / G.Tuparov INF 280 Database Systems: Relational