Unit03 dbms

INDEX UNIT-3 PPT SLIDES S.NO Module as per Lecture PPT Session planner No Slide NO ------------------------------------------------------------------------------------------ Introduction to relational model L1 L1- 1 to L1- 13 2. Enforcing integrity constraints L2 L2- 1 to L2- 3 3. Logical Database Design L3 L3- 1 to L3- 6 4. Logical Database Design L4 L4- 1 to L4 -6 5. Introduction to Views L5 L5- 1 to L5- 10 6. Relational Algebra L6 L6- 1 to L6- 17 7. Tuple Relational Calculus L7 L7- 1 to L7- 3 8. Domain Relational Calculus L8 L8- 1 to L8- 7

Relational Database: Definitions Relational database : a set of relations Relation: made up of 2 parts: Instance : a table , with rows and columns. #Rows = cardinality , #fields = degree / arity. Schema : specifies name of relation, plus name and type of each column. E.G. Students ( sid : string, name : string, login : string, age : integer, gpa : real). Can think of a relation as a set of rows or tuples (i.e., all rows are distinct).

Example Instance of Students Relation Cardinality = 3, degree = 5, all rows distinct Do all columns in a relation instance have to be distinct?

Relational Query Languages A major strength of the relational model: supports simple, powerful querying of data. Queries can be written intuitively, and the DBMS is responsible for efficient evaluation. The key: precise semantics for relational queries. Allows the optimizer to extensively re-order operations, and still ensure that the answer does not change.

The SQL Query Language SELECT * FROM Students S WHERE S.age=18 To find just names and logins, replace the first line: SELECT S.name, S.login

Querying Multiple Relations What does the following query compute? SELECT S.name, E.cid FROM Students S, Enrolled E WHERE S.sid=E.sid AND E.grade=“A” Given the following instances of Enrolled and Students: we get:

Creating Relations in SQL Creates the Students relation. Observe that the type of each field is specified, and enforced by the DBMS whenever tuples are added or modified. As another example, the Enrolled table holds information about courses that students take. CREATE TABLE Students (sid: CHAR(20) , name: CHAR(20) , login: CHAR(10), age: INTEGER , gpa: REAL ) CREATE TABLE Enrolled (sid: CHAR(20) , cid: CHAR(20) , grade: CHAR (2))

Destroying and Altering Relations Destroys the relation Students. The schema information and the tuples are deleted. DROP TABLE Students The schema of Students is altered by adding a new field; every tuple in the current instance is extended with a null value in the new field. ALTER TABLE Students ADD COLUMN firstYear: integer

Adding and Deleting Tuples Can insert a single tuple using: INSERT INTO Students (sid, name, login, age, gpa) VALUES (53688, ‘Smith’, ‘smith@ee’, 18, 3.2) Can delete all tuples satisfying some condition (e.g., name = Smith): DELETE FROM Students S WHERE S.name = ‘Smith’

Integrity Constraints (ICs) IC: condition that must be true for any instance of the database; e.g., domain constraints. ICs are specified when schema is defined. ICs are checked when relations are modified. A legal instance of a relation is one that satisfies all specified ICs. DBMS should not allow illegal instances. If the DBMS checks ICs, stored data is more faithful to real-world meaning. Avoids data entry errors, too!

Primary Key Constraints A set of fields is a key for a relation if : 1. No two distinct tuples can have same values in all key fields, and 2. This is not true for any subset of the key. Part 2 false? A superkey . If there’s >1 key for a relation, one of the keys is chosen (by DBA) to be the primary key . E.g., sid is a key for Students. (What about name ?) The set { sid, gpa } is a superkey .

Primary and Candidate Keys in SQL Possibly many candidate keys (specified using UNIQUE), one of which is chosen as the primary key . CREATE TABLE Enrolled (sid CHAR(20) cid CHAR(20), grade CHAR(2), PRIMARY KEY (sid,cid) ) “ For a given student and course, there is a single grade.” vs. “Students can take only one course, and receive a single grade for that course; further, no two students in a course receive the same grade.” Used carelessly, an IC can prevent the storage of database instances that arise in practice! CREATE TABLE Enrolled (sid CHAR(20) cid CHAR(20), grade CHAR(2), PRIMARY KEY (sid), UNIQUE (cid, grade) )

Foreign Keys, Referential Integrity Foreign key : Set of fields in one relation that is used to `refer’ to a tuple in another relation. (Must correspond to primary key of the second relation.) Like a `logical pointer’. E.g. sid is a foreign key referring to Students : Enrolled( sid : string, cid : string, grade : string) If all foreign key constraints are enforced, referential integrity is achieved, i.e., no dangling references. Can you name a data model w/o referential integrity? Links in HTML!

Foreign Keys in SQL Only students listed in the Students relation should be allowed to enroll for courses. CREATE TABLE Enrolled (sid CHAR(20), cid CHAR(20), grade CHAR(2), PRIMARY KEY (sid,cid), FOREIGN KEY (sid) REFERENCES Students ) Enrolled Students

Enforcing Referential Integrity Consider Students and Enrolled; sid in Enrolled is a foreign key that references Students. What should be done if an Enrolled tuple with a non-existent student id is inserted? ( Reject it! ) What should be done if a Students tuple is deleted? Also delete all Enrolled tuples that refer to it. Disallow deletion of a Students tuple that is referred to. Set sid in Enrolled tuples that refer to it to a default sid . (In SQL, also: Set sid in Enrolled tuples that refer to it to a special value null, denoting `unknown’ or `inapplicable’ .) Similar if primary key of Students tuple is updated.

Referential Integrity in SQL SQL/92 and SQL:1999 support all 4 options on deletes and updates. Default is NO ACTION ( delete/update is rejected ) CASCADE (also delete all tuples that refer to deleted tuple) SET NULL / SET DEFAULT (sets foreign key value of referencing tuple) CREATE TABLE Enrolled (sid CHAR (20), cid CHAR(20) , grade CHAR (2), PRIMARY KEY (sid,cid), FOREIGN KEY (sid) REFERENCES Students ON DELETE CASCADE ON UPDATE SET DEFAULT )

Where do ICs Come From? ICs are based upon the semantics of the real-world enterprise that is being described in the database relations. We can check a database instance to see if an IC is violated, but we can NEVER infer that an IC is true by looking at an instance. An IC is a statement about all possible instances! From example, we know name is not a key, but the assertion that sid is a key is given to us. Key and foreign key ICs are the most common; more general ICs supported too.

Logical DB Design: ER to Relational Entity sets to tables: CREATE TABLE Employees (ssn CHAR (11), name CHAR (20), lot INTEGER , PRIMARY KEY (ssn)) Employees ssn name lot

Relationship Sets to Tables In translating a relationship set to a relation, attributes of the relation must include: Keys for each participating entity set (as foreign keys). This set of attributes forms a superkey for the relation. All descriptive attributes. CREATE TABLE Works_In( ssn CHAR (11), did INTEGER , since DATE , PRIMARY KEY (ssn, did), FOREIGN KEY (ssn) REFERENCES Employees, FOREIGN KEY (did) REFERENCES Departments)

Review: Key Constraints Each dept has at most one manager, according to the key constraint on Manages. Translation to relational model? Many-to-Many 1-to-1 1-to Many Many-to-1 budget did Departments dname since lot name ssn Manages Employees

Translating ER Diagrams with Key Constraints Map relationship to a table: Note that did is the key now! Separate tables for Employees and Departments. Since each department has a unique manager, we could instead combine Manages and Departments. CREATE TABLE Manages( ssn CHAR(11) , did INTEGER , since DATE , PRIMARY KEY (did), FOREIGN KEY (ssn) REFERENCES Employees, FOREIGN KEY (did) REFERENCES Departments) CREATE TABLE Dept_Mgr( did INTEGER, dname CHAR(20), budget REAL, ssn CHAR(11) , since DATE , PRIMARY KEY (did), FOREIGN KEY (ssn) REFERENCES Employees)

Review: Participation Constraints Does every department have a manager? If so, this is a participation constraint : the participation of Departments in Manages is said to be total (vs. partial ). Every did value in Departments table must appear in a row of the Manages table (with a non-null ssn value!) lot name dname budget did since name dname budget did since Manages since Departments Employees ssn Works_In

Participation Constraints in SQL We can capture participation constraints involving one entity set in a binary relationship, but little else (without resorting to CHECK constraints). CREATE TABLE Dept_Mgr( did INTEGER, dname CHAR(20) , budget REAL , ssn CHAR(11) NOT NULL , since DATE , PRIMARY KEY (did), FOREIGN KEY (ssn) REFERENCES Employees, ON DELETE NO ACTION )

Review: Weak Entities A weak entity can be identified uniquely only by considering the primary key of another ( owner ) entity. Owner entity set and weak entity set must participate in a one-to-many relationship set (1 owner, many weak entities). Weak entity set must have total participation in this identifying relationship set. lot name age pname Dependents Employees ssn Policy cost

Translating Weak Entity Sets Weak entity set and identifying relationship set are translated into a single table. When the owner entity is deleted, all owned weak entities must also be deleted. CREATE TABLE Dep_Policy ( pname CHAR(20) , age INTEGER , cost REAL , ssn CHAR(11) NOT NULL , PRIMARY KEY (pname, ssn), FOREIGN KEY (ssn) REFERENCES Employees, ON DELETE CASCADE )

Review: ISA Hierarchies Overlap constraints : Can Joe be an Hourly_Emps as well as a Contract_Emps entity? ( Allowed/disallowed ) Covering constraints : Does every Employees entity also have to be an Hourly_Emps or a Contract_Emps entity? ( Yes/no) Contract_Emps name ssn Employees lot hourly_wages ISA Hourly_Emps contractid hours_worked As in C++, or other PLs, attributes are inherited. If we declare A ISA B, every A entity is also considered to be a B entity.

Translating ISA Hierarchies to Relations General approach: 3 relations: Employees, Hourly_Emps and Contract_Emps. Hourly_Emps : Every employee is recorded in Employees. For hourly emps, extra info recorded in Hourly_Emps ( hourly_wages , hours_worked , ssn ) ; must delete Hourly_Emps tuple if referenced Employees tuple is deleted). Queries involving all employees easy, those involving just Hourly_Emps require a join to get some attributes. Alternative: Just Hourly_Emps and Contract_Emps. Hourly_Emps : ssn , name, lot, hourly_wages, hours_worked. Each employee must be in one of these two subclasses .

Review: Binary vs. Ternary Relationships What are the additional constraints in the 2nd diagram? age pname Dependents Covers age pname Dependents Purchaser Bad design Better design name Employees ssn lot Policies policyid cost Beneficiary policyid cost Policies name Employees ssn lot

Binary vs. Ternary Relationships (Contd.) The key constraints allow us to combine Purchaser with Policies and Beneficiary with Dependents. Participation constraints lead to NOT NULL constraints. What if Policies is a weak entity set? CREATE TABLE Policies ( policyid INTEGER , cost REAL , ssn CHAR(11) NOT NULL , PRIMARY KEY (policyid). FOREIGN KEY (ssn) REFERENCES Employees, ON DELETE CASCADE ) CREATE TABLE Dependents ( pname CHAR(20) , age INTEGER , policyid INTEGER , PRIMARY KEY (pname, policyid). FOREIGN KEY (policyid) REFERENCES Policies, ON DELETE CASCADE )

Views A view is just a relation, but we store a definition , rather than a set of tuples. CREATE VIEW YoungActiveStudents (name, grade) AS SELECT S.name, E.grade FROM Students S, Enrolled E WHERE S.sid = E.sid and S.age<21 Views can be dropped using the DROP VIEW command. How to handle DROP TABLE if there’s a view on the table? DROP TABLE command has options to let the user specify this.

Views and Security Views can be used to present necessary information (or a summary), while hiding details in underlying relation(s). Given YoungStudents, but not Students or Enrolled, we can find students s who have are enrolled, but not the cid’s of the courses they are enrolled in.

View Definition A relation that is not of the conceptual model but is made visible to a user as a “virtual relation” is called a view . A view is defined using the create view statement which has the form create view v as < query expression > where <query expression> is any legal SQL expression. The view name is represented by v. Once a view is defined, the view name can be used to refer to the virtual relation that the view generates.

Example Queries A view consisting of branches and their customers Find all customers of the Perryridge branch create view all_customer as ( select branch_name, customer_name from depositor, account where depositor.account_number = account.account_number ) union ( select branch_name, customer_name from borrower, loan where borrower.loan_number = loan.loan_number ) select customer_name from all_customer where branch_name = 'Perryridge'

Uses of Views Hiding some information from some users Consider a user who needs to know a customer’s name, loan number and branch name, but has no need to see the loan amount. Define a view ( create view cust_loan_data as select customer_name, borrower.loan_number, branch_name from borrower, loan where borrower.loan_number = loan.loan_number ) Grant the user permission to read cust_loan_data, but not borrower or loan Predefined queries to make writing of other queries easier Common example: Aggregate queries used for statistical analysis of data

Processing of Views When a view is created the query expression is stored in the database along with the view name the expression is substituted into any query using the view Views definitions containing views One view may be used in the expression defining another view A view relation v 1 is said to depend directly on a view relation v 2 if v 2 is used in the expression defining v 1 A view relation v 1 is said to depend on view relation v 2 if either v 1 depends directly to v 2 or there is a path of dependencies from v 1 to v 2 A view relation v is said to be recursive if it depends on itself.

View Expansion A way to define the meaning of views defined in terms of other views. Let view v 1 be defined by an expression e 1 that may itself contain uses of view relations. View expansion of an expression repeats the following replacement step: repeat Find any view relation v i in e 1 Replace the view relation v i by the expression defining v i until no more view relations are present in e 1 As long as the view definitions are not recursive, this loop will terminate

With Clause The with clause provides a way of defining a temporary view whose definition is available only to the query in which the with clause occurs. Find all accounts with the maximum balance with max_balance ( value ) as select max ( balance ) from account select account_number from account, max_balance where account.balance = max_balance.value

Complex Queries using With Clause Find all branches where the total account deposit is greater than the average of the total account deposits at all branches. with branch_total ( branch _ name , value ) as select branch _ name , sum ( balance ) from account group by branch _ name with branch _ total _ avg ( value ) as select avg ( value ) from branch _ total select branch _ name from branch _ total , branch _ total_avg where branch_total.value >= branch_total_avg.value Note: the exact syntax supported by your database may vary slightly. E.g. Oracle syntax is of the form with branch_total as ( select .. ), branch_total_avg as ( select .. ) select …

Update of a View Create a view of all loan data in the loan relation, hiding the amount attribute create view loan_branch as select loan_number, branch_name from loan Add a new tuple to loan_branch insert into loan_branch values ('L-37‘, 'Perryridge‘) This insertion must be represented by the insertion of the tuple ('L-37', 'Perryridge', null ) into the loan relation

Formal Relational Query Languages Two mathematical Query Languages form the basis for “real” languages (e.g. SQL), and for implementation: Relational Algebra : More operational , very useful for representing execution plans. Relational Calculus : Lets users describe what they want, rather than how to compute it. (Non-operational, declarative .)

Preliminaries A query is applied to relation instances , and the result of a query is also a relation instance. Schemas of input relations for a query are fixed (but query will run regardless of instance!) The schema for the result of a given query is also fixed! Determined by definition of query language constructs. Positional vs. named-field notation: Positional notation easier for formal definitions, named-field notation more readable. Both used in SQL

Example Instances “ Sailors” and “Reserves” relations for our examples. We’ll use positional or named field notation, assume that names of fields in query results are `inherited’ from names of fields in query input relations. R1 S1 S2

Relational Algebra 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 in reln. 1 and in reln. 2. Additional operations: Intersection, join , division, renaming: Not essential, but (very!) useful. Since each operation returns a relation, operations can be composed ! (Algebra is “closed”.)

Projection Deletes attributes that are not in projection list . Schema of result contains exactly the fields in the projection list, with the same names that they had in the (only) input relation. Projection operator has to eliminate duplicates ! (Why??) Note: real systems typically don’t do duplicate elimination unless the user explicitly asks for it. (Why not?)

Selection Selects rows that satisfy selection condition . No duplicates in result! (Why?) Schema of result identical to schema of (only) input relation. Result relation can be the input for another relational algebra operation! ( Operator composition. )

Union, Intersection, Set-Difference All of these operations take two input relations, which must be union-compatible : Same number of fields. `Corresponding’ fields have the same type. What is the schema of result?

Cross-Product Each row of S1 is paired with each row of R1. Result schema has one field per field of S1 and R1, with field names `inherited’ if possible. Conflict : Both S1 and R1 have a field called sid . Renaming operator :

Joins Condition Join : Result schema same as that of cross-product. Fewer tuples than cross-product, might be able to compute more efficiently Sometimes called a theta-join .

Joins Equi-Join : A special case of condition join where the condition c contains only equalities . Result schema similar to cross-product, but only one copy of fields for which equality is specified. Natural Join : Equijoin on all common fields.

Division Not supported as a primitive operator, but useful for expressing queries like: Find sailors who have reserved all boats . Let A have 2 fields, x and y ; B have only field y : A/B = i.e., A/B contains all x tuples (sailors) such that for every y tuple (boat) in B , there is an xy tuple in A . Or : If the set of y values (boats) associated with an x value (sailor) in A contains all y values in B , the x value is in A/B . In general, x and y can be any lists of fields; y is the list of fields in B , and x y is the list of fields of A .

Examples of Division A/B A B1 B2 B3 A/B1 A/B2 A/B3

Expressing A/B Using Basic Operators Division is not essential op; just a useful shorthand. (Also true of joins, but joins are so common that systems implement joins specially.) Idea : For A/B , compute all x values that are not `disqualified’ by some y value in B . x value is disqualified if by attaching y value from B , we obtain an xy tuple that is not in A . Disqualified x values : A/B: all disqualified tuples

Find names of sailors who’ve reserved boat #103 Solution 1: Solution 2 : Solution 3 :

Find names of sailors who’ve reserved a red boat Information about boat color only available in Boats; so need an extra join: A query optimizer can find this, given the first solution! A more efficient 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: 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):

Relational Calculus Comes in two flavors: Tuple relational calculus (TRC) and Domain relational calculus (DRC). Calculus has variables, constants, comparison ops , logical connectives and quantifiers . TRC : Variables range over (i.e., get bound to) tuples . DRC : Variables range over domain elements (= field values). Both TRC and DRC are simple subsets of first-order logic. Expressions in the calculus are called formulas . An answer tuple is essentially an assignment of constants to variables that make the formula evaluate to true .

Domain Relational Calculus Query has the form: Answer includes all tuples that make the formula be true . Formula is recursively defined, starting with simple atomic formulas (getting tuples from relations or making comparisons of values), and building bigger and better formulas using the logical connectives .

DRC Formulas Atomic formula : , or X op Y, or X op constant op is one of Formula : an atomic formula, or , where p and q are formulas, or , where variable X is free in p(X), or , where variable X is free in p(X) The use of quantifiers and is said to bind X. A variable that is not bound is free .

Free and Bound Variables The use of quantifiers and in a formula is said to bind X. A variable that is not bound is free . Let us revisit the definition of a query : There is an important restriction: the variables x1, ..., xn that appear to the left of `|’ must be the only free variables in the formula p(...).

Find all sailors with a rating above 7 The condition ensures that the domain variables I, N, T and A are bound to fields of the same Sailors tuple. The term to the left of `|’ (which should be read as such that ) says that every tuple that satisfies T> 7 is in the answer. Modify this query to answer: Find sailors who are older than 18 or have a rating under 9, and are called ‘Joe’.

Find sailors rated > 7 who have reserved boat #103 We have used as a shorthand for Note the use of to find a tuple in Reserves that `joins with’ the Sailors tuple under consideration.

Find sailors rated > 7 who’ve reserved a red boat Observe how the parentheses control the scope of each quantifier’s binding. This may look cumbersome, but with a good user interface, it is very intuitive. (MS Access, QBE)

Find sailors who’ve reserved all boats Find all sailors I such that for each 3-tuple either it is not a tuple in Boats or there is a tuple in Reserves showing that sailor I has reserved it.

Find sailors who’ve reserved all boats (again!) Simpler notation, same query. (Much clearer!) To find sailors who’ve reserved all red boats: .....

Unsafe Queries, Expressive Power It is possible to write syntactically correct calculus queries that have an infinite number of answers! Such queries are called unsafe . e.g., It is known that every query that can be expressed in relational algebra can be expressed as a safe query in DRC / TRC; the converse is also true. Relational Completeness : Query language (e.g., SQL) can express every query that is expressible in relational algebra/calculus.

Unit03 dbms

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