PMIT-6102 Advanced Database Systems By- Jesmin Akhter Assistant Professor, IIT, Jahangirnagar University

Lecture 02 Overview of Relational DBMS

Outline Overview of Relational DBMS  Structure of Relational Databases  Relational Algebra

Why Relational DBMS Most of the distributed database technology has been developed using the relational model  Very simple model.  Often a good match for the way we think about our data. Example of a Relation: account (account-number, branch-name, balance)

Relational Design Simplest approach (not always best): convert each Entity Set to a relation and each relationship to a relation. Entity Set  Relation Entity Set attributes become relational attributes. Becomes : account (account-number, branch-name, balance) Slide 5 account account-number balance branch-name

Relational Model Table = relation. Column headers = attributes. Row = tuple Relation schema = name(attributes) + other structure info., e.g., keys, other constraints. Example: Account (account-number, branch-name, balance)  Order of attributes is arbitrary, but in practice we need to assume the order given in the relation schema. Relation instance is current set of rows for a relation schema. Database schema = collection of relation schemas. Slide 6 Account

Basic Structure Formally, given sets D 1, D 2, …. D n a relation r is a subset of D 1 x D 2 x … x D n Thus a relation is a set of n-tuples (a 1, a 2, …, a n ) where each a i  D i Example: if customer-name = {Jones, Smith, Curry, Lindsay} customer-street = {Main, North, Park} customer-city = {Harrison, Rye, Pittsfield} Then r = { (Jones, Main, Harrison), (Smith, North, Rye), (Curry, North, Rye), (Lindsay, Park, Pittsfield)} is a relation over customer-name, customer-street, customer-city Slide 7

A1 A2 A3... An a1 a2 a3 an b1 b2 a3 cn a1 c3 b3 bn. x1 v2 d3 wn Relational Data Model Set theoretic Domain — set of values like a data type n-tuples (V1,V2,...,Vn) s.t., V1  D1, V2  D2,...,Vn  Dn Tuples = members of a relation inst. Arity = number of domains Components = values in a tuple Domains — corresp. with attributes Cardinality = number of tuples Relation as table Rows = tuples Columns = components Names of columns = attributes Set of attribute names = schema REL (A1,A2,...,An) Arity CardinalityCardinality Attributes Component Tuple Slide 8

Name address tel # Cardinality of domain Domains N A T N1 A1 T1 N2 A2 T2 N3 A3 T3 N4 T4 N5 T5 T6 T7 Relation: Example Domain of Relation N A T N1 A1 T1 N1 A1 T2 N1 A1 T3. N1 A1 T7 N1 A2 T1 N1 A3 T1 N2 A1 T1 Arity3 Cardinality<=5x3x7 of relation Tuple Domain Component Attribute Slide 9

Attribute Types Each attribute of a relation has a name The set of allowed values for each attribute is called the domain of the attribute Attribute values are (normally) required to be atomic, that is, indivisible  E.g. multivalued attribute values are not atomic  E.g. composite attribute values are not atomic The special value null is a member of every domain Slide 10

Relation Schema A 1, A 2, …, A n are attributes R = (A 1, A 2, …, A n ) is a relation schema E.g. Customer-schema = (customer-name, customer-street, customer-city) r(R) is a relation on the relation schema R E.g.customer (Customer-schema) Slide 11

Relation Instance The current values (relation instance) of a relation are specified by a table An element t of r is a tuple, represented by a row in a table Jones Smith Curry Lindsay customer-name Main North Park customer-street Harrison Rye Pittsfield customer-city customer attributes (or columns) tuples (or rows) Slide 12

Relations are Unordered Order of tuples is irrelevant (tuples may be stored in an arbitrary order) E.g. account relation with unordered tuples 13

Database A database consists of multiple relations Information about an enterprise is broken up into parts, with each relation storing one part of the information E.g.: account : stores information about accounts depositor : stores information about which customer owns which account customer : stores information about customers Storing all information as a single relation such as bank(account-number, balance, customer-name,..) results in  repetition of information (e.g. customer own two account)  the need for null values (e.g. represent a customer without an account) Normalization theory deals with how to design relational schemas Slide 14

The customer Relation 15 The branch Relation The depositor Relation Account Relation

borrower Relation 16 The Loan Relation Loan-numberBranch-nameamount L-11Round Hill900 L-14Downtown1500 L-15Perryridge1500 L-16Perryridge1300 L-17Downtown1000 L-23Redwood2000 L-93Mianus500

E-R Diagram for the Banking Enterprise 17 Total Participation mean Every account must be related via account-branch to some branch Arrow from account-branch to branch mean Each account is for a single branch

Keys A super key of an entity set is a set of one or more attributes whose values uniquely determine each entity. A candidate key of an entity set is a minimal super key  Customer-id is candidate key of customer  account-number is candidate key of account Although several candidate keys may exist, one of the candidate keys is selected to be the primary key.

A superkey is defined in the relational model of database organization as a set of attributes of a relation variable for which it holds that in all relations assigned to that variable, there are no two distinct tuples (rows) that have the same values for the attributes in this set. [1] Equivalently a superkey can also be defined as a set of attributes of a relation schema upon which all attributes of the schema are functionally dependent.relational model databaseset tuples [1]relation schemafunctionally dependent Note that the set of all attributes is a trivial superkey, because in relational algebra duplicate rows are not permitted.relational algebra Also note that if attribute set K is a superkey of relation R, then at all times it is the case that the projection of R over K has the same cardinality as R itself.projectioncardinality Informally, a superkey is a set of attributes within a table whose values can be used to uniquely identify a tuple. A candidate key is a minimal set of attributes necessary to identify a tuple, this is also called a minimal superkey. For example, given an employee schema, consisting of the attributes employeeID, name, job, and departmentID, we could use the employeeID in combination with any or all other attributes of this table to uniquely identify a tuple in the table. Examples of superkeys in this schema would be {employeeID, Name}, {employeeID, Name, job}, and {employeeID, Name, job, departmentID}. The last example is known as trivial superkey, because it uses all attributes of this table to identify the tuple.candidate key In a real database we do not need values for all of those attributes to identify a tuple. We only need, per our example, the set {employeeID}. This is a minimal superkey – that is, a minimal set of attributes that can be used to identify a single tuple. So, employeeID is a candidate key. candidate key Slide 19

Determining Keys from E-R Sets Strong entity set. The primary key of the entity set becomes the primary key of the relation. Weak entity set. The primary key of the relation consists of the union of the primary key of the strong entity set and the discriminator of the weak entity set. Relationship set. The union of the primary keys of the related entity sets becomes a super key of the relation.  For binary many-to-one relationship sets, the primary key of the “many” entity set becomes the relation’s primary key.  For one-to-one relationship sets, the relation’s primary key can be that of either entity set.  For many-to-many relationship sets, the union of the primary keys becomes the relation’s primary key Slide 20

Schema Diagram for the Banking Enterprise 21

Query Languages Language in which user requests information from the database. Categories of languages  Procedural o User instructs the system to perform a sequence of operations on the database to compute the desired result.  non-procedural o User describes the desired information without giving a specific procedure for obtaining that information. “Pure” languages:  Relational Algebra  Tuple Relational Calculus  Domain Relational Calculus Pure languages form underlying basis of query languages that people use. Slide 22

Relational Algebra Procedural language Six basic operators  select  project  union  set difference  Cartesian product  rename The operators take two or more relations as inputs and give a new relation as a result. Slide 23

Select Operation – Example Relation r ABCD    A=B ^ D > 5 (r) ABCD  

Select Operation Notation:  p (r) p is called the selection predicate Defined as:  p (r) = {t | t  r and p(t)} Where p is a formula in propositional calculus consisting of terms connected by :  (and),  (or),  (not) Each term is one of: op or where op is one of: =, , >, . <.  Slide 25

Example of selection:  branch-name = “Perryridge” (loan) 26  branch-name=“Perryridge” (loan)

Project Operation – Example Relation r: ABC  AC  = AC   A,C (r) Slide 27 Duplicate rows removed

Project Operation Notation:  A1, A2, …, Ak (r) where A 1, A 2 are attribute names and r is a relation name. The result is defined as the relation of k columns obtained by erasing the columns that are not listed Duplicate rows removed from result, since relations are sets E.g. To eliminate the branch-name attribute of account  account-number, balance (account) Slide 28

Union Operation – Example Relations r, s: r  s: AB  AB  2323 r s AB  Slide 29

Union Operation Notation: r  s Defined as: r  s = {t | t  r or t  s} For r  s to be valid. 1. r, s must have the same arity (same number of attributes) 2. The attribute domains must be compatible (e.g., 2nd column of r deals with the same type of values as does the 2nd column of s) E.g. to find all customers with either an account or a loan  customer-name (depositor)   customer-name (borrower) Slide 30

Names of All Customers Who Have Either a Loan or an Account 31  customer-name (depositor)   customer-name (borrower) Union Operation

Set Difference Operation Notation r – s Defined as: r – s = {t | t  r and t  s} Set differences must be taken between compatible relations.  r and s must have the same arity  attribute domains of r and s must be compatible Slide 32

Set Difference Operation – Example Relations r, s: r – s : AB  AB  2323 r s AB  1111 Slide 33

Cartesian-Product Operation Notation r x s Defined as: r x s = {t q | t  r and q  s} Assume that attributes of r(R) and s(S) are disjoint. (That is, R  S =  ). If attributes of r(R) and s(S) are not disjoint, then renaming must be used. Slide 34

Cartesian-Product Operation-Example Relations r, s: r x s: AB  1212 AB  CD  E aabbaabbaabbaabb CD  E aabbaabb r s 35

Composition of Operations Can build expressions using multiple operations Example:  A=C (r x s) r x s  A=C (r x s) AB  CD  E aabbaabbaabbaabb ABCDE   20 aabaab Slide 36

Rename Operation Allows us to refer to a relation by more than one name. Example:  x (E) returns the expression E under the name X If a relational-algebra expression E has arity n, then  x (A1, A2, …, An) (E) returns the result of expression E under the name X, and with the attributes renamed to A 1, A2, …., An. Slide 37

Banking Example branch (branch-name, branch-city, assets) customer (customer-name, customer-street, customer-city) account (account-number, branch-name, balance) loan (loan-number, branch-name, amount) depositor (customer-name, account-number) borrower (customer-name, loan-number) Slide 38

Example Queries Find all loans of over $1200 Find the loan number for each loan of an amount greater than $1200  amount > 1200 (loan)  loan-number (  amount > 1200 (loan)) Slide 39 Loan-numberBranch-nameamount L-14Downtown1500 L-15Perryridge1500 L-16Perryridge1300 L-23Redwood2000 Loan-number L-14 L-15 L-16 L-23 loan

Example Queries Find the names of all customers who have a loan, an account, or both, from the bank Find the names of all customers who have a loan and an account at bank.  customer-name (borrower)   customer-name (depositor)  customer-name (borrower)   customer-name (depositor) Slide 40

Example Queries Find the names of all customers who have a loan at the Perryridge branch. Find 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.  customer-name (  branch-name = “Perryridge” (  borrower.loan-number = loan.loan-number ( borrower x loan ))) –  customer-name ( depositor )  customer-name (  branch-name=“Perryridge ” (  borrower.loan-number = loan.loan-number (borrower x loan))) Slide 41

Result of borrower  loan 42

Result of  branch-name = “Perryridge” (borrower  loan) 43  customer-name (  branch-name = “Perryridge” (  borrower.loan-number = loan.loan-number (borrower x loan)))  customer-name (  branch-name = “Perryridge” (  borrower.loan-number = loan.loan-number (borrower x loan))) –  customer-name (depositor) Customer-name Adams

Customers With An Account But No Loan 44  customer-name (depositor)-  customer-name (borrower)

Thank You