Unit 03 dbms


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  • 3 The slides for this text are organized into several modules. Each lecture contains about enough material for a 1.25 hour class period. (The time estimate is very approximate--it will vary with the instructor, and lectures also differ in length; so use this as a rough guideline.) This covers Lectures 1 and 2 (of 6) in Module (5). Module (1): Introduction (DBMS, Relational Model) Module (2): Storage and File Organizations (Disks, Buffering, Indexes) Module (3): Database Concepts (Relational Queries, DDL/ICs, Views and Security) Module (4): Relational Implementation (Query Evaluation, Optimization) Module (5): Database Design (ER Model, Normalization, Physical Design, Tuning) Module (6): Transaction Processing (Concurrency Control, Recovery) Module (7): Advanced Topics
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  • Unit 03 dbms

    1. 1. DATABASE MANAGEMENT SYSTEMS MALLA REDDY ENGG. COLLEGE II B. Tech CSE II Semester UNIT-III PPT SLIDESText Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan and Korth
    2. 2. INDEX UNIT-3 PPT SLIDESS.NO Module as per Lecture PPT Session planner No Slide NO------------------------------------------------------------------------------------------1. Introduction to relational model L1 L1- 1 to L1- 132. Enforcing integrity constraints L2 L2- 1 to L2- 33. Logical Database Design L3 L3- 1 to L3- 64. Logical Database Design L4 L4- 1 to L4 -65. Introduction to Views L5 L5- 1 to L5- 106. Relational Algebra L6 L6- 1 to L6- 177. Tuple Relational Calculus L7 L7- 1 to L7- 38. Domain Relational Calculus L8 L8- 1 to L8- 7
    3. 3. 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). Slide No:L1-1
    4. 4. Example Instance of Students Relation sid name login age gpa 53666 Jones jones@cs 18 3.4 53688 Smith smith@eecs 18 3.2 53650 Smith smith@math 19 3.8 Cardinality = 3, degree = 5, all rows distinct Do all columns in a relation instance have to be distinct? Slide No:L1-2
    5. 5. 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. Slide No:L1-3
    6. 6. The SQL Query Language sid name login age gpaSELECT * 53666 Jones jones@cs 18 3.4FROM Students S 53688 Smith smith@ee 18 3.2WHERE S.age=18•To find just names and logins, replace the first line: SELECT S.name, S.login Slide No:L1-4
    7. 7. Querying Multiple Relations• What does the following SELECT S.name, E.cid query compute? FROM Students S, Enrolled E WHERE S.sid=E.sid AND E.grade=“A” Given the following instances sid cid grade of Enrolled and Students: 53831 Carnatic101 C sid name login age gpa 53831 Reggae203 B 53650 Topology112 A 53666 Jones jones@cs 18 3.4 53666 History105 B 53688 Smith smith@eecs 18 3.2 53650 Smith smith@math 19 3.8 S.name E.cid we get: Smith Topology112 Slide No:L1-5
    8. 8. Creating Relations in SQL• Creates the Students CREATE TABLE Students relation. Observe that the (sid: CHAR(20), type of each field is name: CHAR(20), specified, and enforced by login: CHAR(10), the DBMS whenever age: INTEGER, tuples are added or gpa: REAL) modified.• As another example, the CREATE TABLE Enrolled Enrolled table holds (sid: CHAR(20), information about courses cid: CHAR(20), that students take. grade: CHAR(2)) Slide No:L1-6
    9. 9. Destroying and Altering RelationsDROP TABLE Students• Destroys the relation Students. The schema information and the tuples are deleted. ALTER TABLE Students ADD COLUMN firstYear: integer 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. Slide No:L1-7
    10. 10. 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’ Slide No:L1-8
    11. 11. 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! Slide No:L1-9
    12. 12. 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. Slide No:L1-10
    13. 13. Primary and Candidate Keys in SQL• Possibly many candidate keys (specified using UNIQUE), one of which is chosen as the primary key. “For a given student and course, CREATE TABLE Enrolled there is a single grade.” vs. (sid CHAR(20) “Students can take only one cid CHAR(20), course, and receive a single grade grade CHAR(2), for that course; further, no two PRIMARY KEY (sid,cid) ) students in a course receive the same grade.” CREATE TABLE Enrolled Used carelessly, an IC can (sid CHAR(20) prevent the storage of database cid CHAR(20), instances that arise in practice! grade CHAR(2), PRIMARY KEY (sid), UNIQUE (cid, grade) ) Slide No:L1-11
    14. 14. 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! Slide No:L1-12
    15. 15. 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 sid cid grade Students53666 Carnatic101 C sid name login age gpa53666 Reggae203 B 53666 Jones jones@cs 18 3.453650 Topology112 A 53688 Smith smith@eecs 18 3.253666 History105 B 53650 Smith smith@math 19 3.8 Slide No:L1-13
    16. 16. 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. Slide No:L2-1
    17. 17. 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) CREATE TABLE Enrolled – CASCADE (also delete all (sid CHAR(20), tuples that refer to deleted cid CHAR(20), tuple) – SET NULL / SET DEFAULT grade CHAR(2), (sets foreign key value of PRIMARY KEY (sid,cid), referencing tuple) FOREIGN KEY (sid) REFERENCES Students ON DELETE CASCADE ON UPDATE SET DEFAULT ) Slide No:L2-2
    18. 18. 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. Slide No:L2-3
    19. 19. Logical DB Design: ER to Relational• Entity sets to tables: CREATE TABLE Employees name (ssn CHAR(11), ssn lot name CHAR(20), lot INTEGER, Employees PRIMARY KEY (ssn)) Slide No:L3-1
    20. 20. Relationship Sets to Tables• In translating a relationship set to a relation, attributes of the relation must include: CREATE TABLE Works_In( – Keys for each participating ssn CHAR(11), entity set (as foreign keys). did INTEGER, • This set of attributes since DATE, forms a superkey for the PRIMARY KEY (ssn, did), relation. – All descriptive attributes. FOREIGN KEY (ssn) REFERENCES Employees, FOREIGN KEY (did) REFERENCES Departments) Slide No:L3-2
    21. 21. Review: Key Constraints• Each dept has at most one manager, according to the key constraint since on Manages. name dname ssn lot did budget Employees Manages Departments Translation to relational model? 1-to-1 1-to Many Many-to-1 Many-to-Many Slide No:L3-3
    22. 22. Translating ER Diagrams with Key Constraints• Map relationship to a table: CREATE TABLE Manages( – Note that did is the ssn CHAR(11), key now! did INTEGER, – Separate tables for since DATE, Employees and PRIMARY KEY (did), Departments. FOREIGN KEY (ssn) REFERENCES Employees,• Since each department FOREIGN KEY (did) REFERENCES Departments) has a unique manager, we could instead combine Manages and CREATE TABLE Dept_Mgr( Departments. did INTEGER, dname CHAR(20), budget REAL, ssn CHAR(11), since DATE, PRIMARY KEY (did), FOREIGN KEY (ssn) REFERENCES Employees) Slide No:L3-4
    23. 23. 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!) since name dname ssn lot did budget Employees Manages Departments Works_In since Slide No:L3-5
    24. 24. 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) Slide No:L3-6
    25. 25. 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. name cost pname age ssn lot Employees Policy Dependents Slide No:L4-1
    26. 26. 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) Slide No:L4-2
    27. 27. Review: ISA Hierarchies name As in C++, or other ssn lotPLs, attributes areinherited. Employees If we declare A ISA B, hourly_wages hours_workedevery A entity is also ISA contractidconsidered to be a Bentity. Contract_Emps Hourly_Emps• 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) Slide No:L4-3
    28. 28. 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. Slide No:L4-4
    29. 29. Review: Binary vs. Ternary Relationships name ssn lot pname age• What are the additional Employees Covers constraints in the Dependents 2nd diagram? Bad design Policies policyid cost name pname age ssn lot Dependents Employees Purchaser Beneficiary Better design Policies Slide No:L4-5 policyid cost
    30. 30. Binary vs. Ternary Relationships (Contd.) CREATE TABLE Policies (• The key policyid INTEGER, constraints allow us to combine cost REAL, Purchaser with ssn CHAR(11) NOT NULL, Policies and PRIMARY KEY (policyid). Beneficiary with FOREIGN KEY (ssn) REFERENCES Employees, Dependents. ON DELETE CASCADE)• Participation CREATE TABLE Dependents ( constraints lead pname CHAR(20), to NOT NULL age INTEGER, constraints. policyid INTEGER,• What if Policies PRIMARY KEY (pname, policyid). is a weak entity set? FOREIGN KEY (policyid) REFERENCES Policies, ON DELETE CASCADE) Slide No:L4-6
    31. 31. 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. Slide No:L5-1
    32. 32. 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. Slide No:L5-2
    33. 33. 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. Slide No:L5-3
    34. 34. Example Queries• A view consisting of branches and their customers 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 ) Find all customers of the Perryridge branch select customer_name from all_customer where branch_name = Perryridge Slide No:L5-4
    35. 35. 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 Slide No:L5-5
    36. 36. 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 v1 is said to depend directly on a view relation v2 if v2 is used in the expression defining v1 – A view relation v1 is said to depend on view relation v2 if either v1 depends directly to v2 or there is a path of dependencies from v1 to v2 – A view relation v is said to be recursive if it depends on itself. Slide No:L5-6
    37. 37. View Expansion• A way to define the meaning of views defined in terms of other views.• Let view v1 be defined by an expression e1 that may itself contain uses of view relations.• View expansion of an expression repeats the following replacement step: repeat Find any view relation vi in e1 Replace the view relation vi by the expression defining vi until no more view relations are present in e1• As long as the view definitions are not recursive, this loop will terminate Slide No:L5-7
    38. 38. 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 Slide No:L5-8
    39. 39. 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 … Slide No:L5-9
    40. 40. 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 Slide No:L5-10
    41. 41. 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.) Slide No:L6-1
    42. 42. 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 Slide No:L6-2
    43. 43. Example Instances R1 sid bid day 22 101 10/10/96 58 103 11/12/96• “Sailors” and “Reserves” relations for our examples. sid sname rating age• We’ll use positional or S1 named field notation, 22 dustin 7 45.0 assume that names of fields in query results are 31 lubber 8 55.5 `inherited’ from names of 58 rusty 10 35.0 fields in query input relations. sid sname rating age S2 28 yuppy 9 35.0 31 lubber 8 55.5 44 guppy 5 35.0 58 rusty 10 35.0 Slide No:L6-3
    44. 44. Relational Algebra• Basic operations: – Selection (σ ) Selects a subset of rows from relation. – Projection ( π ) Deletes unwanted columns from relation. – Cross-product ( – Set-difference ( × ) Allows us to combine two relations. ) 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”.) Slide No:L6-4
    45. 45. Projection sname rating yuppy 9• Deletes attributes that are not in projection list. lubber 8• Schema of result contains guppy 5 exactly the fields in the rusty 10 projection list, with the same names that they had in the π sname,rating(S2) (only) input relation.• Projection operator has to eliminate duplicates! (Why??) age – Note: real systems typically 35.0 don’t do duplicate elimination unless the user explicitly asks 55.5 π age(S2) for it. (Why not?) Slide No:L6-5
    46. 46. Selection sid sname rating age• Selects rows that satisfy 28 yuppy 9 35.0 selection condition. 58 rusty 10 35.0 σrating >8(S2)• No duplicates in result! (Why?)• Schema of result identical to schema of (only) input relation. sname rating• Result relation can be the input for another relational yuppy 9 algebra operation! (Operator rusty 10 composition.) π sname,rating(σ rating >8(S2)) Slide No:L6-6
    47. 47. Union, Intersection, Set-Difference• All of these operations take two sid sname rating age input relations, which must be union-compatible: 22 dustin 7 45.0 – Same number of fields. 31 lubber 8 55.5 – `Corresponding’ fields have the same type. 58 rusty 10 35.0• What is the schema of result? 44 guppy 5 35.0 28 yuppy 9 35.0 S1∪ S2 sid sname rating age sid sname rating age 22 dustin 7 45.0 31 lubber 8 55.5 58 rusty 10 35.0 S1− S2 S1∩ S2 Slide No:L6-7
    48. 48. 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. (sid) sname rating age (sid) bid day 22 dustin 7 45.0 22 101 10/10/96 22 dustin 7 45.0 58 103 11/12/96 31 lubber 8 55.5 22 101 10/10/96 31 lubber 8 55.5 58 103 11/12/96 58 rusty 10 35.0 22 101 10/10/96 58 rusty 10 35.0 58 103 11/12/96 Renaming operator: ρ (C(1→ sid1, 5 → sid 2), S1× R1) Slide No:L6-8
    49. 49. Joins R  c S = σ c ( R × S) • Condition Join:(sid) sname rating age (sid) bid day22 dustin 7 45.0 58 103 11/12/9631 lubber 8 55.5 58 103 11/12/96 S1  R1 S1. sid < R1. sid• 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. Slide No:L6-9
    50. 50. Joins• Equi-Join: A special case of condition join where the condition c contains only equalities. sid sname rating age bid day 22 dustin 7 45.0 101 10/10/96 58 rusty 10 35.0 103 11/12/96 S1  R1 sid• Result schema similar to cross-product, but only one copy of fields for which equality is specified.• Natural Join: Equijoin on all common fields. Slide No:L6-10
    51. 51. 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, , y ∈ A ∀ have only field y: – A/B = x | ∃ x x and y; B y ∈ 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. Slide No:L6-11
    52. 52. Examples of Division A/Bsno pno pno pno pnos1 p1 p2 p2 p1s1 p2 p4 p2s1 p3 B p4s1 p4 B2 1 B3s2 p1 snos2 p2 s1 snos3 p2 sno s2 s1 s1s4 p2 s3 s4s4 p4 s4 A A/B1 A/B2 A/B3 Slide No:L6-12
    53. 53. 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: π x ((π x ( A) × B) − A) A/B: π x ( A) − all disqualified tuples Slide No:L6-13
    54. 54. Find names of sailors who’ve reserved boat #103 • Solution 1: π sname((σ Reserves)  Sailors)  bid =103 Solution 2: ρ (Temp1, σ Re serves) bid = 103 ρ ( Temp2, Temp1  Sailors)  π sname (Temp2) Solution 3: π sname (σ (Re serves  Sailors))  bid =103 Slide No:L6-14
    55. 55. Find names of sailors who’ve reserved a red boat• Information about boat color only available in Boats; so need an extra join: π sname ((σ Boats)  Re serves  Sailors)  color = red  A more efficient solution: π sname (π ((π σ Boats)  Re s)  Sailors)   sid bid color = red A query optimizer can find this, given the first solution! Slide No:L6-15
    56. 56. 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:ρ (Tempboats, (σ Boats)) color = red ∨ color = green π sname(Tempboats  Re serves  Sailors)   Can also define Tempboats using union! (How?) What happens if ∨ is replaced by ∧ in this query? Slide No:L6-16
    57. 57. 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):ρ (Tempred, π ((σ Boats)  Re serves))  sid color = red ρ (Tempgreen, π ((σ Boats)  Re serves))  sid color = greenπ sname((Tempred ∩ Tempgreen)  Sailors)  Slide No:L6-17
    58. 58. 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. Slide No:L7-1
    59. 59. Domain Relational Calculus• Query has the form:      x1, x2,..., xn | p x1, x2,..., xn           Answer includes all tuples x1, x2,..., xn that make the formula p x1, x2,..., xn  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. Slide No:L7-2
    60. 60. DRC Formulas• Atomic formula: – x1, x2,..., xn ∈ Rname, or X op Y, or X op constant – op is one of <, >, =, ≤, ≥, ≠• Formula: – an atomic formula, or ¬ p, p ∧ q, p ∨ q – , where p and q are formulas, or –∃X ( p( X )) , where variable X is free in p(X), or ∀ X ( p( X )) – , where variable X is free in p(X)• ∃X The use of quantifiers and ∀X is said to bind X. – A variable that is not bound is free. Slide No:L7-3
    61. 61. Free and Bound Variables• The use of quantifiers and ∃X ∀X in a formula is said to bind X. – A variable that is not bound is free.• Let us revisit the definition of a query:      x1, x2,..., xn | p x1, x2,..., xn           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(...). Slide No:L8-1
    62. 62. Find all sailors with a rating above 7     I, N,T, A | I, N,T, A ∈ Sailors ∧ T > 7      • The condition I, N,T, A ∈ Sailors ensures that the domain variables I, N, T and A are bound to fields of the same Sailors tuple.• The term I, N,T, A to the left of `|’ (which should be read as such that) says that every tuple I, N,T, A 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’. Slide No:L8-2
    63. 63. Find sailors rated > 7 who have reserved boat #103 I, N,T, A | I, N, T, A ∈ Sailors ∧ T > 7 ∧ ∃ Ir, Br, D Ir, Br, D ∈ Re serves ∧ Ir = I ∧ Br = 103             • We have used ∃ Ir , Br , D ( . . .) as a shorthand for ∃ Ir ( ∃ Br ( ∃ D ( . . .) ) )• Note the use of ∃ to find a tuple in Reserves that `joins with’ the Sailors tuple under consideration. Slide No:L8-3
    64. 64. Find sailors rated > 7 who’ve reserved a red boat I, N, T, A | I, N, T, A ∈Sailors ∧T >7 ∧ ∃ Ir, Br, D Ir, Br, D ∈Re serves ∧ Ir = I ∧     ∃B, BN, C B, BN, C ∈Boats ∧ B =Br ∧C = red          • 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) Slide No:L8-4
    65. 65. Find sailors who’ve reserved all boats     I, N,T, A | I, N, T, A ∈Sailors ∧  ∀ B, BN,C ¬ B, BN,C ∈Boats ∨                   ∃ Ir, Br, D    Ir, Br, D ∈Re serves ∧ I = Ir ∧ Br = B            • Find all sailors I such that for each 3-tuple B, BN,C either it is not a tuple in Boats or there is a tuple in Reserves showing that sailor I has reserved it. Slide No:L8-5
    66. 66. Find sailors who’ve reserved all boats (again!)     I, N,T, A | I, N, T, A ∈ Sailors ∧  ∀ B, BN, C ∈ Boats   ∃ Ir, Br, D ∈Re serves I = Ir ∧ Br = B              • Simpler notation, same query. (Much clearer!)• To find sailors who’ve reserved all red boats: ....    C ≠ red ∨ ∃ Ir, Br, D ∈ Re serves I = Ir ∧ Br = B             .   Slide No:L8-6
    67. 67. 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.,  S | ¬  S ∈ Sailors               • 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. Slide No:L8-7