Relational Model and Relational Algebra. Slides by: Ms. Shree Jaswal

Transcription

1 Relational Model and Relational Algebra Slides by: Ms. Shree Jaswal

2 Topics: Introduction to Relational Model, Relational Model Constraints and Relational Database Schemas, Concept of Keys: Primary Kay, Secondary key, Foreign Key, Mapping the ER and EER Model to the Relational Model, Introduction to Relational Algebra, Relational Algebra expressions for Unary Relational Operations, Set Theory operations, Binary Relational operation Relational Algebra Queries

3 Which Chapter? Which Book? Chapter 3: The Relational Data Model and Relational Database Constraints, Elmasri and Navathe, Fundamentals of Database Systems, 6th Edition, PEARSON Education Chapter 8: Relational Database Design using ER to Relational Mapping, Elmasri and Navathe, Fundamentals of Database Systems, 6th Edition, PEARSON Education

4 Relational Model Concepts

5 Informal Definitions Informally, a relation looks like a table of values. A relation typically contains a set of rows. The data elements in each row represent certain facts that correspond to a real-world entity or relationship In the formal model, rows are called tuples Each column has a column header that gives an indication of the meaning of the data items in that column In the formal model, the column header is called an attribute name (or just attribute)

6 Example of a Relation

7 Informal Definitions Key of a Relation: Each row has a value of a data item (or set of items) that uniquely identifies that row in the table Called the key In the STUDENT table, SSN is the key Sometimes row-ids or sequential numbers are assigned as keys to identify the rows in a table Called artificial key or surrogate key

8 Formal Definitions - Schema The Schema (or description) of a Relation: Denoted by R(A1, A2,...An) R is the name of the relation The attributes of the relation are A1, A2,..., An Example: CUSTOMER (Cust-id, Cust-name, Address, Phone#) CUSTOMER is the relation name Defined over the four attributes: Cust-id, Cust-name, Address, Phone# Each attribute has a domain or a set of valid values. For example, the domain of Cust-id is 6 digit numbers.

9 Formal Definitions - Tuple A tuple is an ordered set of values (enclosed in angled brackets < > ) Each value is derived from an appropriate domain. A row in the CUSTOMER relation is a 4-tuple and would consist of four values, for example: <632895, "John Smith", "101 Main St. Atlanta, GA 30332", "(404) "> This is called a 4-tuple as it has 4 values A tuple (row) in the CUSTOMER relation. A relation is a set of such tuples (rows)

10 Formal Definitions - Domain A domain has a logical definition: Example: USA_phone_numbers are the set of 10 digit phone numbers valid in the U.S. A domain also has a data-type or a format defined for it. The USA_phone_numbers may have a format: (ddd)ddd-dddd where each d is a decimal digit. Dates have various formats such as year, month, date formatted as yyyy-mm-dd, or as dd mm,yyyy etc. The attribute name designates the role played by a domain in a relation: Used to interpret the meaning of the data elements corresponding to that attribute Example: The domain Date may be used to define two attributes named Invoice-date and Payment-date with different meanings

11 Formal Definitions - State The relation state is a subset of the Cartesian product of the domains of its attributes each domain contains the set of all possible values the attribute can take. Example: attribute Cust-name is defined over the domain of character strings of maximum length 25 dom(cust-name) is varchar(25) The role these strings play in the CUSTOMER relation is that of the name of a customer.

12 Definition Summary Informal Terms Table Column Header All possible Column Values Row Formal Terms Relation Attribute Domain Tuple Table Definition Populated Table Schema of a Relation State of the Relation

13 Characteristics of Relations Ordering of tuples in a relation Relation defined as a set of tuples Elements have no order among them Ordering of values within a tuple and an alternative definition of a relation Order of attributes and values is not that important As long as correspondence between attributes and values maintained

14 Example A relation STUDENT

15 Same state as previous Figure (but with different order of tuples)

16 Characteristics Of Relations Values in a tuple: All values are considered atomic (indivisible). Each value in a tuple must be from the domain of the attribute for that column If tuple t = <v1, v2,, vn> is a tuple (row) in the relation state r of R(A1, A2,, An) Then each vi must be a value from dom(ai) A special null value is used to represent values that are unknown or inapplicable to certain tuples.

17 Characteristics Of Relations Notation: We refer to component values of a tuple t by: t[ai] or t.ai This is the value vi of attribute Ai for tuple t Similarly, t[au, Av,..., Aw] refers to the subtuple of t containing the values of attributes Au, Av,..., Aw, respectively in t

18 Relational Model Notation Relation schema R of degree n Denoted by R(A 1, A 2,..., A n ) Uppercase letters Q, R, S Denote relation names Lowercase letters q, r, s Denote relation states Letters t, u, v Denote tuples

19 Relational Model Notation Name of a relation schema: STUDENT Indicates the current set of tuples in that relation Notation: STUDENT(Name, Ssn,...) Refers only to relation schema Attribute A can be qualified with the relation name R to which it belongs Using the dot notation R.A

20 Relational Integrity Constraints Constraints are conditions that must hold on all valid relation states. There are three main types of constraints in the relational model: Key constraints Entity integrity constraints Referential integrity constraints Another implicit constraint is the domain constraint Every value in a tuple must be from the domain of its attribute (or it could be null, if allowed for that attribute)

21 Key Constraints Superkey of R: Is a set of attributes SK of R with the following condition: No two tuples in any valid relation state r(r) will have the same value for SK That is, for any distinct tuples t1 and t2 in r(r), t1[sk] t2[sk] This condition must hold in any valid state r(r) Key of R: A "minimal" superkey That is, a key is a superkey K such that removal of any attribute from K results in a set of attributes that is not a superkey (does not possess the superkey uniqueness property)

22 Key Constraints (continued) Example: Consider the CAR relation schema: CAR(State, Reg#, SerialNo, Make, Model, Year) CAR has two keys: Key1 = {State, Reg#} Key2 = {SerialNo} Both are also superkeys of CAR {SerialNo, Make} is a superkey but not a key. In general: Any key is a superkey (but not vice versa) Any set of attributes that includes a key is a superkey A minimal superkey is also a key

23 Key Constraints (continued) If a relation has several candidate keys, one is chosen arbitrarily to be the primary key. The primary key attributes are underlined. Example: Consider the CAR relation schema: CAR(State, Reg#, SerialNo, Make, Model, Year) We chose SerialNo as the primary key The primary key value is used to uniquely identify each tuple in a relation Provides the tuple identity Also used to reference the tuple from another tuple General rule: Choose as primary key the smallest of the candidate keys (in terms of size) Not always applicable choice is sometimes subjective

24 CAR table with two candidate keys LicenseNumber chosen as Primary Key

25 Relational Database Schema Relational Database Schema: A set S of relation schemas that belong to the same database. S is the name of the whole database schema S = {R1, R2,..., Rn} R1, R2,, Rn are the names of the individual relation schemas within the database S Following slide shows a COMPANY database schema with 6 relation schemas

26 COMPANY Database Schema

27 Entity Integrity Entity Integrity: The primary key attributes PK of each relation schema R in S cannot have null values in any tuple of r(r). This is because primary key values are used to identify the individual tuples. t[pk] null for any tuple t in r(r) If PK has several attributes, null is not allowed in any of these attributes Note: Other attributes of R may be constrained to disallow null values, even though they are not members of the primary key.

28 Referential Integrity A constraint involving two relations The previous constraints involve a single relation. Used to specify a relationship among tuples in two relations: The referencing relation and the referenced relation.

29 Referential Integrity Tuples in the referencing relation R1 have attributes FK (called foreign key attributes) that reference the primary key attributes PK of the referenced relation R2. A tuple t1 in R1 is said to reference a tuple t2 in R2 if t1[fk] = t2[pk]. A referential integrity constraint can be displayed in a relational database schema as a directed arc from R1.FK to R2.

30 Referential Integrity (or foreign key) Constraint Statement of the constraint The value in the foreign key column (or columns) FK of the the referencing relation R1 can be either: (1) a value of an existing primary key value of a corresponding primary key PK in the referenced relation R2, or (2) a null. In case (2), the FK in R1 should not be a part of its own primary key.

31 Displaying a relational database schema and its constraints Each relation schema can be displayed as a row of attribute names The name of the relation is written above the attribute names The primary key attribute (or attributes) will be underlined A foreign key (referential integrity) constraints is displayed as a directed arc (arrow) from the foreign key attributes to the referenced table Can also point the the primary key of the referenced relation for clarity Next slide shows the COMPANY relational schema diagram

32 Referential Integrity Constraints for COMPANY database

33 Domain Constraints Typically include: Numeric data types for integers and real numbers Characters Booleans Fixed-length strings Variable-length strings Date, time, timestamp Money Other special data types

34 Other Types of Constraints Semantic Integrity Constraints: based on application semantics and cannot be expressed by the model per se Example: the max. no. of hours per employee for all projects he or she works on is 56 hrs per week A constraint specification language may have to be used to express these SQL-99 allows triggers and ASSERTIONS to express for some of these

35 Populated database state Each relation will have many tuples in its current relation state The relational database state is a union of all the individual relation states Whenever the database is changed, a new state arises Basic operations for changing the database: INSERT a new tuple in a relation DELETE an existing tuple from a relation MODIFY an attribute of an existing tuple Next slide shows an example state for the COMPANY database

36 Populated database state for COMPANY

37 Update Operations on Relations INSERT a tuple. DELETE a tuple. MODIFY a tuple. Integrity constraints should not be violated by the update operations. Several update operations may have to be grouped together. Updates may propagate to cause other updates automatically. This may be necessary to maintain integrity constraints.

38 Update Operations on Relations In case of integrity violation, several actions can be taken: Cancel the operation that causes the violation (RESTRICT or REJECT option) Perform the operation but inform the user of the violation Trigger additional updates so the violation is corrected (CASCADE option, SET NULL option) Execute a user-specified error-correction routine

39 Possible violations for each operation INSERT may violate any of the constraints: Domain constraint: if one of the attribute values provided for the new tuple is not of the specified attribute domain Key constraint: if the value of a key attribute in the new tuple already exists in another tuple in the relation Referential integrity: if a foreign key value in the new tuple references a primary key value that does not exist in the referenced relation Entity integrity: if the primary key value is null in the new tuple

40 Possible violations for each operation DELETE may violate only referential integrity: If the primary key value of the tuple being deleted is referenced from other tuples in the database Can be remedied by several actions: RESTRICT, CASCADE, SET NULL RESTRICT option: reject the deletion CASCADE option: propagate the new primary key value into the foreign keys of the referencing tuples SET NULL option: set the foreign keys of the referencing tuples to NULL One of the above options must be specified during database design for each foreign key constraint

41 Possible violations for each operation UPDATE may violate domain constraint and NOT NULL constraint on an attribute being modified Any of the other constraints may also be violated, depending on the attribute being updated: Updating the primary key (PK): Similar to a DELETE followed by an INSERT Need to specify similar options to DELETE Updating a foreign key (FK): May violate referential integrity Updating an ordinary attribute (neither PK nor FK): Can only violate domain constraints

42 Mapping the ER and EER Model to the Relational Model

43 Outline ER-to-Relational Mapping Algorithm Step 1: Mapping of Regular Entity Types Step 2: Mapping of Weak Entity Types Step 3: Mapping of Binary 1:1 Relation 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. Mapping EER Model Constructs to Relations Step 8: Options for Mapping Specialization or Generalization. Step 9: Mapping of Union Types (Categories).

44 ER-to-Relational Mapping Algorithm Step 1: Mapping of Regular Entity Types. For each regular (strong) entity type E in the ER schema, create a relation R that includes all the simple attributes of E. Choose one of the key attributes of E as the 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. Example: We create the relations EMPLOYEE, DEPARTMENT, and PROJECT in the relational schema corresponding to the regular entities in the ER diagram. SSN, DNUMBER, and PNUMBER are the primary keys for the relations EMPLOYEE, DEPARTMENT, and PROJECT as shown.

45 FIGURE 7.1 The ER conceptual schema diagram for the COMPANY database.

46 FIGURE 7.2 Result of mapping the COMPANY ER schema into a relational schema.

47 ER-to-Relational Mapping Algorithm (contd.) Step 2: Mapping of Weak Entity Types For each weak entity type W in the ER schema with owner entity type E, create a relation R & include all simple attributes (or simple components of composite attributes) of W as attributes of R. Also, include as foreign key attributes of R the primary key attribute(s) of the relation(s) that correspond to the owner entity type(s). 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. Example: Create the relation DEPENDENT in this step to correspond to the weak entity type DEPENDENT. Include the primary key SSN of the EMPLOYEE relation as a foreign key attribute of DEPENDENT (renamed to ESSN). The primary key of the DEPENDENT relation is the combination {ESSN, DEPENDENT_NAME} because DEPENDENT_NAME is the partial key of DEPENDENT.

48 ER-to-Relational Mapping Algorithm (contd.) Step 3: Mapping of Binary 1:1 Relation Types For 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. There are three possible approaches: 1. Foreign Key approach: Choose one of the relations-say S-and include a 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. Example: 1:1 relation MANAGES is mapped by choosing the participating entity type DEPARTMENT to serve in the role of S, because its participation in the MANAGES relationship type is total. 2. Merged relation option: An alternate mapping of a 1:1 relationship type is possible by merging the two entity types and the relationship into a single relation. This may be appropriate when both participations are total. 3. Cross-reference or relationship relation option: The third alternative is to set up a third relation R for the purpose of cross-referencing the primary keys of the two relations S and T representing the entity types.

49 ER-to-Relational Mapping Algorithm (contd.) Step 4: Mapping of Binary 1:N Relationship Types. For each regular binary 1:N relationship type R, identify the relation S that represent 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. Include any simple attributes of the 1:N relation type as attributes of S. Example: 1:N relationship types WORKS_FOR, CONTROLS, and SUPERVISION in the figure. For WORKS_FOR we include the primary key DNUMBER of the DEPARTMENT relation as foreign key in the EMPLOYEE relation and call it DNO.

50 ER-to-Relational Mapping Algorithm (contd.) Step 5: Mapping of Binary M:N Relationship Types. For each regular 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 (or simple components of composite attributes) as attributes of S. Example: The M:N relationship type WORKS_ON from the ER diagram is mapped by creating a relation WORKS_ON in the relational database schema. The primary keys of the PROJECT and EMPLOYEE relations are included as foreign keys in WORKS_ON and renamed PNO and ESSN, respectively. Attribute HOURS in WORKS_ON represents the HOURS attribute of the relation type. The primary key of the WORKS_ON relation is Module the 3 combination Slides by: Ms. Shree of J. the foreign key attributes {ESSN, PNO}.

51 ER-to-Relational Mapping Algorithm (contd.) Step 6: Mapping of Multivalued attributes. For each multivalued attribute A, create a new relation R. This relation R will include an attribute corresponding to A, plus the primary key attribute K-as a foreign key in R-of the relation that represents the entity type of relationship type that has A as an attribute. The primary key of R is the combination of A and K. If the multivalued attribute is composite, we include its simple components. Example: The relation DEPT_LOCATIONS is created. The attribute DLOCATION represents the multivalued attribute LOCATIONS of DEPARTMENT, while DNUMBER-as foreign keyrepresents the primary key of the DEPARTMENT relation. The primary key of R is the combination of {DNUMBER, DLOCATION}.

52 ER-to-Relational Mapping Algorithm (contd.) Step 7: Mapping of N-ary Relationship Types. For each n-ary relationship type R, where n>2, create a new relationship 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 (or simple components of composite attributes) as attributes of S. Example: The relationship type SUPPY in the ER on the next slide. This can be mapped to the relation SUPPLY shown in the relational schema, whose primary key is the combination of the three foreign keys {SNAME, PARTNO, PROJNAME}

53 FIGURE 4.11 Ternary relationship types. (a) The SUPPLY relationship.

54 FIGURE 7.3 Mapping the n-ary relationship type SUPPLY from Figure 4.11a.

55 Summary of Mapping constructs and constraints Table 7.1 Correspondence between ER and Relational Models ER Model Relational Model Entity type Entity relation 1:1 or 1:N relationship type Foreign key (or relationship relation) M:N relationship type Relationship relation and two foreign keys n-ary relationship type Relationship relation and n foreign keys Simple attribute Attribute Composite attribute Set of simple component attributes Multivalued attribute Relation and foreign key Value set Domain Key attribute Primary (or secondary) key

56 Mapping EER Model Constructs to Relations Step8: Options for 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 relational schemas using one of the four following options: Option 8A: Multiple relations-superclass and subclasses Option 8B: Multiple relations-subclass relations only Option 8C: Single relation with one type attribute Option 8D: Single relation with multiple type attributes

57 Mapping EER Model Constructs to Relations Option 8A: Multiple relations-superclass and subclasses Create a relation L for C with attributes Attrs(L) = {k,a1, an} and PK(L) = k. Create a relation Li for each subclass Si, 1 < i < m, with the attributesattrs(li) = {k} U {attributes of Si} and PK(Li)=k. This option works for any specialization (total or partial, disjoint of over-lapping). Option 8B: Multiple relations-subclass relations only Create a relation Li for each subclass Si, 1 < i < m, with the attributes Attr(Li) = {attributes of Si} U {k,a1,an} and PK(Li) = k. This option only works for a specialization whose subclasses are total (every entity in the superclass must belong to (at least) one of the subclasses).

58 FIGURE 4.4 EER diagram notation for an attribute-defined specialization on JobType.

59 FIGURE 7.4 Options for mapping specialization or generalization. (a) Mapping the EER schema in Figure 4.4 using option 8A.

60 FIGURE 4.3 Generalization. (b) Generalizing CAR and TRUCK into the superclass VEHICLE.

61 FIGURE 7.4 Options for mapping specialization or generalization. (b) Mapping the EER schema in Figure 4.3b using option 8B. Tonnage

62 Mapping EER Model Constructs to Relations (contd.) Option 8C: Single relation with one type attribute Create a single relation L with attributes Attrs(L) = {k,a 1, a n } U {attributes of S 1 } U U {attributes of S m } U {t} and PK(L) = k. The attribute t is called a type (or discriminating) attribute that indicates the subclass to which each tuple belongs Option 8D: Single relation with multiple type attributes Create a single relation schema L with attributes Attrs(L) = {k,a 1, a n } U {attributes of S 1 } U U {attributes of S m } U {t 1, t 2,,t m } and PK(L) = k. Each t i, 1 < I < m, is a Boolean type attribute indicating whether a tuple belongs to the subclass S i.

63 FIGURE 4.4 EER diagram notation for an attribute-defined specialization on JobType.

64 FIGURE 7.4 Options for mapping specialization or generalization. (c) Mapping the EER schema in Figure 4.4 using option 8C. Eng_type

65 FIGURE 4.5 EER diagram notation for an overlapping (non-disjoint) specialization.

66 FIGURE 7.4 Options for mapping specialization or generalization. (d) Mapping Figure 4.5 using option 8D with Boolean type fields Mflag and Pflag.

67 Mapping EER Model Constructs to Relations (contd.) Mapping of Shared Subclasses (Multiple Inheritance) A shared subclass, such as STUDENT_ASSISTANT, is a subclass of several classes, indicating multiple inheritance. These classes must all have the same key attribute; otherwise, the shared subclass would be modeled as a category. We can apply any of the options discussed in Step 8 to a shared subclass, subject to the restriction discussed in Step 8 of the mapping algorithm. Below both 8C and 8D are used for the shared class STUDENT_ASSISTANT.

68 FIGURE 4.7 A specialization lattice with multiple inheritance for a UNIVERSITY database.

69 FIGURE 7.5 Mapping the EER specialization lattice in Figure 4.6 using multiple options. Eng_type Major

70 Mapping EER Model Constructs to Relations (contd.) Step 9: Mapping of Union Types (Categories). For mapping a category whose defining superclass have different keys, it is customary to specify a new key attribute, called a surrogate key, when creating a relation to correspond to the category. In the example below we can create a relation OWNER to correspond to the OWNER category and include any attributes of the category in this relation. The primary key of the OWNER relation is the surrogate key, which we called OwnerId.

71 FIGURE 4.8 Two categories (union types): OWNER and REGISTERED_VEHICLE.

72 FIGURE 7.6 Mapping the EER categories (union types) in Figure 4.7 to relations. OwnerId Cyear

73 Relational Algebra

74 Relational Algebra: Outline Unary Relational Operations Relational Algebra Operations From Set Theory Binary Relational Operations Additional Relational Operations Examples of Queries in Relational Algebra

75 Relational Algebra Overview Relational algebra is the basic set of operations for the relational model These operations enable a user to specify basic retrieval requests (or queries) The result of an operation is a new relation, which may have been formed from one or more input relations This property makes the algebra closed (all objects in relational algebra are relations)

76 Relational Algebra Overview (continued) The algebra operations thus produce new relations These can be further manipulated using operations of the same algebra A sequence of relational algebra operations forms a relational algebra expression The result of a relational algebra expression is also a relation that represents the result of a database query (or retrieval request)

77 Brief History of Origins of Algebra Muhammad ibn Musa al-khwarizmi ( CE) wrote a book titled al-jabr about arithmetic of variables Book was translated into Latin. Its title (al-jabr) gave Algebra its name. Al-Khwarizmi called variables shay Shay is Arabic for thing. Spanish transliterated shay as xay ( x was sh in Spain). In time this word was abbreviated as x. Where does the word Algorithm come from? Algorithm originates from al-khwarizmi" Reference: PBS (

78 Relational Algebra Overview Relational Algebra consists of several groups of operations Unary Relational Operations SELECT (symbol: (sigma)) PROJECT (symbol: (pi)) RENAME (symbol: (rho)) Relational Algebra Operations From Set Theory UNION ( ), INTERSECTION ( ), DIFFERENCE (or MINUS, ) CARTESIAN PRODUCT ( x ) Binary Relational Operations JOIN (several variations of JOIN exist) DIVISION Additional Relational Operations OUTER JOINS, OUTER UNION AGGREGATE FUNCTIONS (These compute summary of information: for example, SUM, COUNT, AVG, MIN, MAX)

79 Database State for COMPANY All examples discussed below refer to the COMPANY database shown here.

80 Unary Relational Operations: SELECT The SELECT operation (denoted by (sigma)) is used to select a subset of the tuples from a relation based on a selection condition. The selection condition acts as a filter Keeps only those tuples that satisfy the qualifying condition Tuples satisfying the condition are selected whereas the other tuples are discarded (filtered out) Examples: Select the EMPLOYEE tuples whose department number is 4: DNO = 4 (EMPLOYEE) Select the employee tuples whose salary is greater than $30,000: SALARY > 30,000 (EMPLOYEE)

81 Unary Relational Operations: SELECT In general, the select operation is denoted by <selection condition>(r) where the symbol (sigma) is used to denote the select operator the selection condition is a Boolean (conditional) expression specified on the attributes of relation R tuples that make the condition true are selected appear in the result of the operation tuples that make the condition false are filtered out discarded from the result of the operation

82 Unary Relational Operations: SELECT (contd.) SELECT Operation Properties The SELECT operation <selection condition> (R) produces a relation S that has the same schema (same attributes) as R SELECT is commutative: <condition1> ( < condition2> (R)) = <condition2> ( < condition1> (R)) Because of commutativity property, a cascade (sequence) of SELECT operations may be applied in any order: <cond1> ( <cond2> ( <cond3> (R)) = <cond2> ( <cond3> ( <cond1> ( R))) A cascade of SELECT operations may be replaced by a single selection with a conjunction of all the conditions: <cond1> ( < cond2> ( <cond3> (R)) = <cond1> AND < cond2> AND < cond3>(r))) The number of tuples in the result of a SELECT is less than (or equal to) the number of tuples in the input relation R

83

84 Unary Relational Operations: PROJECT PROJECT Operation is denoted by (pi) This operation keeps certain columns (attributes) from a relation and discards the other columns. PROJECT creates a vertical partitioning The list of specified columns (attributes) is kept in each tuple The other attributes in each tuple are discarded Example: To list each employee s first and last name and salary, the following is used: LNAME, FNAME,SALARY (EMPLOYEE)

85 Unary Relational Operations: PROJECT (cont.) The general form of the project operation is: <attribute list> (R) (pi) is the symbol used to represent the project operation <attribute list> is the desired list of attributes from relation R. The project operation removes any duplicate tuples This is because the result of the project operation must be a set of tuples Mathematical sets do not allow duplicate elements.

86 Unary Relational Operations: PROJECT (contd.) PROJECT Operation Properties The number of tuples in the result of projection <list> (R) is always less or equal to the number of tuples in R If the list of attributes includes a key of R, then the number of tuples in the result of PROJECT is equal to the number of tuples in R PROJECT is not commutative <list1> ( <list2> (R) ) = <list1> (R) as long as <list2> contains the attributes in <list1>

87 Examples of applying SELECT and PROJECT operations

88

89 Relational Algebra Expressions We may want to apply several relational algebra operations one after the other Either we can write the operations as a single relational algebra expression by nesting the operations, or We can apply one operation at a time and create intermediate result relations. In the latter case, we must give names to the relations that hold the intermediate results.

90 Single expression versus sequence of relational operations (Example) To retrieve the first name, last name, and salary of all employees who work in department number 5, we must apply a select and a project operation We can write a single relational algebra expression as follows: FNAME, LNAME, SALARY ( DNO=5 (EMPLOYEE)) OR We can explicitly show the sequence of operations, giving a name to each intermediate relation: DEP5_EMPS DNO=5 (EMPLOYEE) RESULT FNAME, LNAME, SALARY (DEP5_EMPS)

91 Unary Relational Operations: RENAME The RENAME operator is denoted by (rho) In some cases, we may want to rename the attributes of a relation or the relation name or both Useful when a query requires multiple operations Necessary in some cases (see JOIN operation later)

92 Unary Relational Operations: RENAME (contd.) The general RENAME operation can be expressed by any of the following forms: S (B1, B2,, Bn ) (R) changes both: the relation name to S, and the column (attribute) names to B1, B1,..Bn S (R) changes: the relation name only to S (B1, B2,, Bn ) (R) changes: the column (attribute) names only to B1, B1,..Bn

93 Unary Relational Operations: RENAME (contd.) For convenience, we also use a shorthand for renaming attributes in an intermediate relation: If we write: RESULT FNAME, LNAME, SALARY (DEP5_EMPS) RESULT will have the same attribute names as DEP5_EMPS (same attributes as EMPLOYEE) If we write: RESULT (F, M, L, S, B, A, SX, SAL, SU, DNO) FNAME, LNAME, SALARY (DEP5_EMPS) The 10 attributes of DEP5_EMPS are renamed to F, M, L, S, B, A, SX, SAL, SU, DNO, respectively

94 Example of applying multiple operations and RENAME

95

96 Relational Algebra Operations from Set Theory: UNION UNION Operation Binary operation, denoted by The result of R S, is a relation that includes all tuples that are either in R or in S or in both R and S Duplicate tuples are eliminated The two operand relations R and S must be type compatible (or UNION compatible) R and S must have same number of attributes (same arity) Each pair of corresponding attributes must be type compatible (have same or compatible domains)

97 Relational Algebra Operations from Set Theory: UNION Example: To retrieve the social security numbers of all employees who either work in department 5 (RESULT1 below) or directly supervise an employee who works in department 5 (RESULT2 below) We can use the UNION operation as follows: DEP5_EMPS DNO=5 (EMPLOYEE) RESULT1 SSN (DEP5_EMPS) RESULT2(SSN) SUPERSSN (DEP5_EMPS) RESULT RESULT1 RESULT2 The union operation produces the tuples that are in either RESULT1 or RESULT2 or both

98 Example of the result of a UNION operation UNION Example

99

100 Relational Algebra Operations from Set Theory Type Compatibility of operands is required for the binary set operation UNION, (also for INTERSECTION, and SET DIFFERENCE, see next slides) R1(A1, A2,..., An) and R2(B1, B2,..., Bn) are type compatible if: they have the same number of attributes, and the domains of corresponding attributes are type compatible (i.e. dom(ai)=dom(bi) for i=1, 2,..., n). The resulting relation for R1 R2 (also for R1 R2, or R1 R2, see next slides) has the same attribute names as the first operand relation R1 (by convention)

101 Relational Algebra Operations from Set Theory: INTERSECTION INTERSECTION is denoted by The result of the operation R S, is a relation that includes all tuples that are in both R and S The attribute names in the result will be the same as the attribute names in R The two operand relations R and S must be type compatible

102 Relational Algebra Operations from Set Theory: SET DIFFERENCE (cont.) SET DIFFERENCE (also called MINUS or EXCEPT) is denoted by The result of R S, is a relation that includes all tuples that are in R but not in S The attribute names in the result will be the same as the attribute names in R The two operand relations R and S must be type compatible

103 Example to illustrate the result of UNION, INTERSECT, and DIFFERENCE

104 Some properties of UNION, INTERSECT, and DIFFERENCE Notice that both union and intersection are commutative operations; that is R S = S R, and R S = S R Both union and intersection can be treated as n-ary operations applicable to any number of relations as both are associative operations; that is R (S T) = (R S) T (R S) T = R (S T) The minus operation is not commutative; that is, in general R S S R

105 Relational Algebra Operations from Set Theory: CARTESIAN PRODUCT CARTESIAN (or CROSS) PRODUCT Operation This operation is used to combine tuples from two relations in a combinatorial fashion. Denoted by R(A1, A2,..., An) x S(B1, B2,..., Bm) Result is a relation Q with degree n + m attributes: Q(A1, A2,..., An, B1, B2,..., Bm), in that order. The resulting relation state has one tuple for each combination of tuples one from R and one from S. Hence, if R has n R tuples (denoted as R = n R ), and S has n S tuples, then R x S will have n R * n S tuples. The two operands do NOT have to be "type compatible

106 Relational Algebra Operations from Set Theory: CARTESIAN PRODUCT (cont.) Generally, CROSS PRODUCT is not a meaningful operation Can become meaningful when followed by other operations Example (not meaningful): FEMALE_EMPS SEX= F (EMPLOYEE) EMPNAMES FNAME, LNAME, SSN (FEMALE_EMPS) EMP_DEPENDENTS EMPNAMES x DEPENDENT EMP_DEPENDENTS will contain every combination of EMPNAMES and DEPENDENT whether or not they are actually related

107 Relational Algebra Operations from Set Theory: CARTESIAN PRODUCT (cont.) To keep only combinations where the DEPENDENT is related to the EMPLOYEE, we add a SELECT operation as follows Example (meaningful): FEMALE_EMPS SEX= F (EMPLOYEE) EMPNAMES FNAME, LNAME, SSN (FEMALE_EMPS) EMP_DEPENDENTS EMPNAMES x DEPENDENT ACTUAL_DEPS SSN=ESSN (EMP_DEPENDENTS) RESULT FNAME, LNAME, DEPENDENT_NAME (ACTUAL_DEPS) RESULT will now contain the name of female employees and their dependents

108 Example of applying CARTESIAN PRODUCT

109 Binary Relational Operations: JOIN JOIN Operation (denoted by ) The sequence of CARTESIAN PRODECT followed by SELECT is used quite commonly to identify and select related tuples from two relations A special operation, called JOIN combines this sequence into a single operation This operation is very important for any relational database with more than a single relation, because it allows us combine related tuples from various relations The general form of a join operation on two relations R(A1, A2,..., An) and S(B1, B2,..., Bm) is: R <join condition>s where R and S can be any relations that result from general relational algebra expressions.

110 Binary Relational Operations: JOIN (cont.) Example: Suppose that we want to retrieve the name of the manager of each department. To get the manager s name, we need to combine each DEPARTMENT tuple with the EMPLOYEE tuple whose SSN value matches the MGRSSN value in the department tuple. We do this by using the join operation. DEPT_MGR DEPARTMENT MGRSSN=SSN EMPLOYEE MGRSSN=SSN is the join condition Combines each department record with the employee who manages the department The join condition can also be specified as DEPARTMENT.MGRSSN= EMPLOYEE.SSN

111

112 Example of applying the JOIN operation

113 Some properties of JOIN Consider the following JOIN operation: R(A1, A2,..., An) R.Ai=S.Bj S(B1, B2,..., Bm) Result is a relation Q with degree n + m attributes: Q(A1, A2,..., An, B1, B2,..., Bm), in that order. The resulting relation state has one tuple for each combination of tuples r from R and s from S, but only if they satisfy the join condition r[ai]=s[bj] Hence, if R has n R tuples, and S has n S tuples, then the join result will generally have less than n R * n S tuples. Only related tuples (based on the join condition) will appear in the result

114 Some properties of JOIN The general case of JOIN operation is called a Thetajoin: R theta S The join condition is called theta Theta can be any general boolean expression on the attributes of R and S; for example: R.Ai<S.Bj AND (R.Ak=S.Bl OR R.Ap<S.Bq) Most join conditions involve one or more equality conditions AND ed together; for example: R.Ai=S.Bj AND R.Ak=S.Bl AND R.Ap=S.Bq

115 Binary Relational Operations: EQUIJOIN EQUIJOIN Operation The most common use of join involves join conditions with equality comparisons only Such a join, where the only comparison operator used is =, is called an EQUIJOIN. In the result of an EQUIJOIN we always have one or more pairs of attributes (whose names need not be identical) that have identical values in every tuple. The JOIN seen in the previous example was an EQUIJOIN.

116 Binary Relational Operations: NATURAL JOIN Operation NATURAL JOIN Operation Another variation of JOIN called NATURAL JOIN denoted by * was created to get rid of the second (superfluous) attribute in an EQUIJOIN condition. because one of each pair of attributes with identical values is superfluous The standard definition of natural join requires that the two join attributes, or each pair of corresponding join attributes, have the same name in both relations If this is not the case, a renaming operation is applied first.

117 Binary Relational Operations NATURAL JOIN (contd.) Example: To apply a natural join on the DNUMBER attributes of DEPARTMENT and DEPT_LOCATIONS, it is sufficient to write: DEPT_LOCS DEPARTMENT * DEPT_LOCATIONS Only attribute with the same name is DNUMBER An implicit join condition is created based on this attribute: DEPARTMENT.DNUMBER=DEPT_LOCATIONS.DNUMBER Another example: Q R(A,B,C,D) * S(C,D,E) The implicit join condition includes each pair of attributes with the same name, AND ed together: R.C=S.C AND R.D.S.D Result keeps only one attribute of each such pair: Q(A,B,C,D,E)

118 Example of NATURAL JOIN operation

119

120 Complete Set of Relational Operations The set of operations including SELECT, PROJECT, UNION, DIFFERENCE -, RENAME, and CARTESIAN PRODUCT X is called a complete set because any other relational algebra expression can be expressed by a combination of these five operations. For example: R S = (R S ) ((R - S) (S - R)) R <join condition>s = <join condition> (R X S)

121 Binary Relational Operations: DIVISION DIVISION Operation The division operation is applied to two relations R(Z) S(X), where X subset Z. Let Y = Z - X (and hence Z = X Y); that is, let Y be the set of attributes of R that are not attributes of S. The result of DIVISION is a relation T(Y) that includes a tuple t if tuples t R appear in R with t R [Y] = t, and with t R [X] = t s for every tuple t s in S. For a tuple t to appear in the result T of the DIVISION, the values in t must appear in R in combination with every tuple in S.

122 Example of DIVISION

123 Recap of Relational Algebra Operations

124 Additional Relational Operations: Aggregate Functions and Grouping A type of request that cannot be expressed in the basic relational algebra is to specify mathematical aggregate functions on collections of values from the database. Examples of such functions include retrieving the average or total salary of all employees or the total number of employee tuples. These functions are used in simple statistical queries that summarize information from the database tuples. Common functions applied to collections of numeric values include SUM, AVERAGE, MAXIMUM, and MINIMUM. The COUNT function is used for counting tuples or values.

125 Aggregate Function Operation Use of the Aggregate Functional operation F F MAX Salary (EMPLOYEE) retrieves the maximum salary value from the EMPLOYEE relation F MIN Salary (EMPLOYEE) retrieves the minimum Salary value from the EMPLOYEE relation F SUM Salary (EMPLOYEE) retrieves the sum of the Salary from the EMPLOYEE relation F COUNT SSN, AVERAGE Salary (EMPLOYEE) computes the count (number) of employees and their average salary Note: count just counts the number of rows, without removing duplicates

126 Using Grouping with Aggregation The previous examples all summarized one or more attributes for a set of tuples Maximum Salary or Count (number of) Ssn Grouping can be combined with Aggregate Functions Example: For each department, retrieve the DNO, COUNT SSN, and AVERAGE SALARY A variation of aggregate operation F allows this: Grouping attribute placed to left of symbol Aggregate functions to right of symbol DNO F COUNT SSN, AVERAGE Salary (EMPLOYEE) Above operation groups employees by DNO (department number) and computes the count of employees and average salary per department

127

128 Examples of applying aggregate functions and grouping

129 Additional Relational Operations (cont.) Recursive Closure Operations Another type of operation that, in general, cannot be specified in the basic original relational algebra is recursive closure. This operation is applied to a recursive relationship. An example of a recursive operation is to retrieve all SUPERVISEES of an EMPLOYEE e at all levels that is, all EMPLOYEE e directly supervised by e; all employees e directly supervised by each employee e ; all employees e directly supervised by each employee e ; and so on.

130 Additional Relational Operations (cont.) Although it is possible to retrieve employees at each level and then take their union, we cannot, in general, specify a query such as retrieve the supervisees of James Borg at all levels without utilizing a looping mechanism. The SQL3 standard includes syntax for recursive closure. BORG_SSN SSN ( Fname= James AND Lname= Borg (EMPLOYEE)) SUPERVISION(SSN1,SSN2) SSN,Super_ssn (EMPLOYEE) RESULT1(SSN) SSN1 (SUPERVISION ssn2=ssn BORG_SSN) RESULT2(SSN) SSN1 (SUPERVISION ssn2=ssn RESULT1) RESULT RESULT2 U RESULT1

131 Additional Relational Operations (cont.)

132 Additional Relational Operations (cont.) The OUTER JOIN Operation In NATURAL JOIN and EQUIJOIN, tuples without a matching (or related) tuple are eliminated from the join result Tuples with null in the join attributes are also eliminated This amounts to loss of information. A set of operations, called OUTER joins, can be used when we want to keep all the tuples in R, or all those in S, or all those in both relations in the result of the join, regardless of whether or not they have matching tuples in the other relation.

133 Additional Relational Operations (cont.) The left outer join operation keeps every tuple in the first or left relation R in R S; if no matching tuple is found in S, then the attributes of S in the join result are filled or padded with null values. A similar operation, right outer join, keeps every tuple in the second or right relation S in the result of R S. A third operation, full outer join, denoted by keeps all tuples in both the left and the right relations when no matching tuples are found, padding them with null values as needed.

134 Additional Relational Operations (cont.) Ssn=Mgr_ssnDEPARTMENT) TEMP (EMPLOYEE RESULT Fname, Minit, Lname, Dname(TEMP)

135 Additional Relational Operations (cont.) OUTER UNION Operations The outer union operation was developed to take the union of tuples from two relations if the relations are not type compatible. This operation will take the union of tuples in two relations R(X, Y) and S(X, Z) that are partially compatible, meaning that only some of their attributes, say X, are type compatible. The attributes that are type compatible are represented only once in the result, and those attributes that are not type compatible from either relation are also kept in the result relation T(X, Y, Z).

136 Additional Relational Operations (cont.) Example: An outer union can be applied to two relations whose schemas are STUDENT(Name, SSN, Department, Advisor) and INSTRUCTOR(Name, SSN, Department, Rank). Tuples from the two relations are matched based on having the same combination of values of the shared attributes Name, SSN, Department. If a student is also an instructor, both Advisor and Rank will have a value; otherwise, one of these two attributes will be null. The result relation STUDENT_OR_INSTRUCTOR will have the following attributes: STUDENT_OR_INSTRUCTOR (Name, SSN, Department, Advisor, Rank)

137 Examples of Queries in Relational Algebra Query 1: Retrieve the name and address of all employees who work for the Research department. RESEARCH_DEPT DNAME= Research (DEPARTMENT) RESEARCH_EMPS (RESEARCH_DEPT DNUMBER= DNOEMPLOYEE) RESULT FNAME, LNAME, ADDRESS (RESEARCH_EMPS)

138 Query 2. For every project located in Stafford, list the project number, the controlling department number, and the department manager s last name, address, and birth date. STAFFORD_PROJS σ Plocation= Stafford ( PROJECT) CONTR_DEPTS (STAFFORD_PROJS PROJ_DEPT_MGRS (CONTR_DEPTS Dnum=DnumberDEPARTMENT) Mgr_ssn=SsnEMPLOYEE) RESULT π Pnumber, Dnum, Lname, Address, Bdate (PROJ_DEPT_MGRS)

139 Query 3. Find the names of employees who work on all the projects controlled by department number 5. DEPT5_PROJS ρ (Pno) (π Pnumber (σ Dnum=5 (PROJECT))) EMP_PROJ ρ (Ssn, Pno) (π Essn, Pno (WORKS_ON)) RESULT_EMP_SSNS EMP_PROJ DEPT5_PROJS RESULT π Lname, Fname (RESULT_EMP_SSNS * EMPLOYEE)

140 Query 4. Make a list of project numbers for projects that involve an employee whose last name is Smith, either as a worker or as a manager of the department that controls the project. SMITHS(Essn) π Ssn (σ Lname= Smith (EMPLOYEE)) SMITH_WORKER_PROJS π Pno (WORKS_ON * SMITHS) MGRS π Lname, Dnumber (EMPLOYEE Ssn=Mgr_ssnDEPARTMENT) SMITH_MANAGED_DEPTS(Dnum) π Dnumber (σ Lname= Smith (MGRS)) SMITH_MGR_PROJS(Pno) π Pnumber (SMITH_MANAGED_DEPTS * PROJECT) RESULT (SMITH_WORKER_PROJS SMITH_MGR_PROJS)

141 Query 5. List the names of all employees with two or more dependents. Strictly speaking, this query cannot be done in the basic (original) relational algebra. We have to use the AGGREGATE FUNCTION operation with the COUNT aggregate function. We assume that dependents of the same employee have distinct Dependent_name values. T1(Ssn, No_of_dependents) Essn I COUNT Dependent_name(DEPENDENT) T2 σno_of_dependents>2(t1) RESULT πlname, Fname(T2 * EMPLOYEE)

142 Query 6: Retrieve the names of employees who have no dependents. ALL_EMPS SSN(EMPLOYEE) EMPS_WITH_DEPS(SSN) ESSN(DEPENDENT) EMPS_WITHOUT_DEPS (ALL_EMPS - EMPS_WITH_DEPS) RESULT LNAME, FNAME (EMPS_WITHOUT_DEPS * EMPLOYEE)

143 Query 7. List the names of managers who have at least one dependent. MGRS(Ssn) π Mgr_ssn (DEPARTMENT) EMPS_WITH_DEPS(Ssn) π Essn (DEPENDENT) MGRS_WITH_DEPS (MGRS EMPS_WITH_DEPS) RESULT π Lname, Fname (MGRS_WITH_DEPS * EMPLOYEE)