Database Systems
Session 5 – Main Theme
Relational Algebra, Relational Calculus, and SQL
Dr. Jean-Claude Franchitti
New York University Computer Science Department Courant Institute of Mathematical Sciences
Presentation material partially based on textbook slides Fundamentals of Database Systems (7th Edition) by Ramez Elmasri and Shamkant Navathe Slides copyright © 2016 and on slides produced by Zvi Kedem copyight © 2014
1 Agenda
1 Session Overview
2 Relational Algebra and Relational Calculus
3 Relational Algebra Using SQL Syntax
5 Summary and Conclusion
2 Session Agenda
. Session Overview . Relational Algebra and Relational Calculus . Relational Algebra Using SQL Syntax . Summary & Conclusion
3 What is the class about?
. Course description and syllabus: » http://www.nyu.edu/classes/jcf/CSCI-GA.2433-001 » http://cs.nyu.edu/courses/spring16/CSCI-GA.2433-001/ . Textbooks: » Fundamentals of Database Systems (7th Edition) Ramez Elmasri and Shamkant Navathe Addition Wesley ISBN-10: 0133970779, ISBN-13: 978-0133970777 6th Edition (06/08/15)
4 Icons / Metaphors
Information
Common Realization
Knowledge/Competency Pattern
Governance
Alignment
Solution Approach
55 Agenda
1 Session Overview
2 Relational Algebra and Relational Calculus
3 Relational Algebra Using SQL Syntax
5 Summary and Conclusion
6 Agenda
. Relational Algebra » Unary Relational Operations » Relational Algebra Operations From Set Theory » Binary Relational Operations » Additional Relational Operations » Examples of Queries in Relational Algebra . Relational Calculus » Tuple Relational Calculus » Domain Relational Calculus . Example Database Application (COMPANY) . Overview of the QBE language (appendix D)
7 The Relational Algebra and Relational Calculus
. Relational algebra . Basic set of operations for the relational model . Relational algebra expression . Sequence of relational algebra operations . Relational calculus . Higher-level declarative language for specifying relational queries
8 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)
9 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)
10 Brief History of Origins of Algebra
. Muhammad ibn Musa al-Khwarizmi (800-847 CE) – from Morocco 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 (http://www.pbs.org/empires/islam/innoalgebra.html)
11 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)
12 Database State for Company
. All examples discussed below refer to the COMPANY database shown here.
13 Unary Relational Operations: SELECT (1/3)
. 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)
14 Unary Relational Operations: SELECT (2/3)
» In general, the select operation is denoted by
15 Unary Relational Operations: SELECT (3/3)
. SELECT Operation Properties
» The SELECT operation
16 Unary Relational Operations: SELECT and PROJECT (2/3)
. Example:
.
17 Unary Relational Operations: SELECT and PROJECT (3/3)
. Selectivity . Fraction of tuples selected by a selection condition . SELECT operation commutative . Cascade SELECT operations into a single operation with AND condition
18 The Following Query Results Refer to this Database State
19 Unary Relational Operations: PROJECT (1/3)
. 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)
20 Unary Relational Operations: PROJECT (2/3)
. The general form of the project operation is:
21 Unary Relational Operations: PROJECT (3/3)
. PROJECT Operation Properties » The number of tuples in the result of projection
(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
22 Examples of Applying SELECT and PROJECT Operations
23 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.
24 Single Expression vs. 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)
25 Unary Relational Operations: RENAME (1/3)
. 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)
26 Unary Relational Operations: RENAME (2/3)
. 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
27 Unary Relational Operations: RENAME (3/3)
. 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)
RESULT (F.M.L.S.B,A,SX,SAL,SU, DNO)(DEP5_EMPS) • The 10 attributes of DEP5_EMPS are renamed to F, M, L, S, B, A, SX, SAL, SU, DNO, respectively Note: the symbol is an assignment operator
28 Example of Applying Multiple Operations and RENAME
29 Relational Algebra Operations from Set Theory: UNION (1/2)
. 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 • Each pair of corresponding attributes must be type compatible (have same or compatible domains)
30 Relational Algebra Operations from Set Theory: UNION (2/2)
. 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
31 Result of the UNION Operation RESULT ← RESULT1 ∪ RESULT2
32 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 R1R2 (also for R1R2, or R1– R2, see next slides) has the same attribute names as the first operand relation R1 (by convention)
33 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”
34 Relational Algebra Operations from Set Theory: SET DIFFERENCE
. 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”
35 Example to Illustrate the Result of UNION/INTERSECTION/DIFFERENCE
36 Some Properties of UNION/INTERSECT/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
37 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 nR tuples (denoted as |R| = nR ), and S has nS tuples, then R x S will have nR * nS tuples. » The two operands do NOT have to be "type compatible”
38 Relational Algebra Operations from Set Theory: CARTESIAN PRODUCT
. 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
39 Relational Algebra Operations from Set Theory: CARTESIAN PRODUCT
. 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
40 The CARTESIAN PRODUCT (CROSS PRODUCT) Operation (1/3)
41 The CARTESIAN PRODUCT (CROSS PRODUCT) Operation (2/3)
42 The CARTESIAN PRODUCT (CROSS PRODUCT) Operation (3/3)
43 Binary Relational Operations: JOIN (1/2)
. 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
44 Binary Relational Operations: JOIN (2/2)
. 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
45 Result of the JOIN Operation |X| DEPT_MGR ← DEPARTMENT Mgr_ssn=SsnEMPLOYEE
46 Some Properties of JOIN (1/2)
. Consider the following JOIN operation: » R(A1, A2, . . ., An) S(B1, B2, . . ., Bm) R.Ai=S.Bj » 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 nR tuples, and S has nS tuples, then the join result will generally have less than nR * nS tuples. » Only related tuples (based on the join condition) will appear in the result
47 Binary Relational Operations: JOIN (2/2)
. The general case of JOIN operation is called a Theta-join: R S theta . The join condition is called theta . Theta can be any general boolean expression on the attributes of R and S; for example: » R.Ai . 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. 49 Binary Relational Operations: NATURAL JOIN (1/2) . 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. 50 Binary Relational Operations: NATURAL JOIN (2/2) . 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) 51 Example of NATURAL JOIN Operation 52 A Complete Set of Relational Algebra 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 53 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 tR appear in R with tR [Y] = t, and with • tR [X] = ts for every tuple ts 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. 54 Example of DIVISION 55 Operations of Relational Algebra (1/2) 56 Operations of Relational Algebra (2/2) 57 Query Tree Notation . Query Tree » An internal data structure to represent a query » Standard technique for estimating the work involved in executing the query, the generation of intermediate results, and the optimization of execution » Nodes stand for operations like selection, projection, join, renaming, division, …. » Leaf nodes represent base relations » A tree gives a good visual feel of the complexity of the query and the operations involved » Algebraic Query Optimization consists of rewriting the query or modifying the query tree into an equivalent tree » (See Textbook Chapter 15) 58 Example of Query Tree 59 Additional Relational Operations: Aggregate Functions & 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. 60 Aggregate Function Operation . Use of the Aggregate Functional operation ℱ » ℱMAX Salary (EMPLOYEE) retrieves the maximum salary value from the EMPLOYEE relation » ℱMIN Salary (EMPLOYEE) retrieves the minimum Salary value from the EMPLOYEE relation » ℱSUM Salary (EMPLOYEE) retrieves the sum of the Salary from the EMPLOYEE relation » ℱCOUNT 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 61 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 ℱCOUNT SSN, AVERAGE Salary (EMPLOYEE) . Above operation groups employees by DNO (department number) and computes the count of employees and average salary per department 62 The Aggregate Function Operation a.ρR(Dno, No_of_employees, Average_sal)(Dno ℑ COUNT Ssn, AVERAGE Salary (EMPLOYEE)). b. Dno ℑ alary(EMPLOYEE). c. ℑ COUNT Ssn, AVERAGE Salary(EMPLOYEE). 63 Results of GROUP BY and HAVING (in SQL) – Figure 7.1a (Q24) 64 Additional Relational Operations: Recursive Closure (1/2) . 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. 65 Additional Relational Operations: Recursive Closure (2/2) . 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. 66 A Two-Level Recursive Query (Figure 8.11) 67 Additional Relational Operations: OUTER JOIN (1/2) . 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. 68 Additional Relational Operations: OUTER JOIN (2/2) . 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. 69 The Result of a LEFT OUTER JOIN Operation 70 Additional Relational Operations: OUTER UNION (1/2) . 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). 71 Additional Relational Operations: OUTER UNION (2/2) . 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) 72 Examples of Queries in Relational Algebra – Procedural Form Q1: Retrieve the name and address of all employees who work for the ‘Research’ department. RESEARCH_DEPT DNAME=’Research’ (DEPARTMENT) RESEARCH_EMPS (RESEARCH_DEPT DNUMBER= DNOEMPLOYEEEMPLOYEE) RESULT FNAME, LNAME, ADDRESS (RESEARCH_EMPS) Q6: 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) 73 Examples of Queries in Relational Algebra – Single Expressions As a single expression, these queries become: Q1: Retrieve the name and address of all employees who work for the ‘Research’ department. Fname, Lname, Address (σ Dname= ‘Research’ (DEPARTMENT Dnumber=Dno(EMPLOYEE)) Q6: Retrieve the names of employees who have no dependents. Lname, Fname(( Ssn (EMPLOYEE) − ρ Ssn ( Essn (DEPENDENT))) ∗ EMPLOYEE) 74 Additional Examples of Queries in Relational Algebra 75 Relational Calculus (1/2) . A relational calculus expression creates a new relation, which is specified in terms of variables that range over rows of the stored database relations (in tuple calculus) or over columns of the stored relations (in domain calculus). . In a calculus expression, there is no order of operations to specify how to retrieve the query result—a calculus expression specifies only what information the result should contain. » This is the main distinguishing feature between relational algebra and relational calculus. 76 Relational Calculus (2/2) . Relational calculus is considered to be a nonprocedural or declarative language. . This differs from relational algebra, where we must write a sequence of operations to specify a retrieval request; hence relational algebra can be considered as a procedural way of stating a query. 77 Tuple Relational Calculus (1/2) . The tuple relational calculus is based on specifying a number of tuple variables. . Each tuple variable usually ranges over a particular database relation, meaning that the variable may take as its value any individual tuple from that relation. . A simple tuple relational calculus query is of the form {t | COND(t)} » where t is a tuple variable and COND (t) is a conditional expression involving t. » The result of such a query is the set of all tuples t that satisfy COND (t). 78 Tuple Relational Calculus (2/2) . Example: To find the first and last names of all employees whose salary is above $50,000, we can write the following tuple calculus expression: {t.FNAME, t.LNAME | EMPLOYEE(t) AND t.SALARY>50000} . The condition EMPLOYEE(t) specifies that the range relation of tuple variable t is EMPLOYEE. . The first and last name (PROJECTION FNAME, LNAME) of each EMPLOYEE tuple t that satisfies the condition t.SALARY>50000 (SELECTION SALARY >50000) will be retrieved. 79 Tuple Variables and Range Relations . Tuple variables . Ranges over a particular database relation . Satisfy COND(t): . Specify: . Range relation R of t . Select particular combinations of tuples . Set of attributes to be retrieved (requested attributes) 80 Expressions and Formulas in Tuple Relational Calculus . General expression of tuple relational calculus is of the form: . Truth value of an atom . Evaluates to either TRUE or FALSE for a specific combination of tuples . Formula (Boolean condition) . Made up of one or more atoms connected via logical operators AND, OR, and NOT 81 The Existential and Universal Quantifiers (1/2) . Two special symbols called quantifiers can appear in formulas; these are the universal quantifier () and the existential quantifier (). . Informally, a tuple variable t is bound if it is quantified, meaning that it appears in an ( t) or ( t) clause; otherwise, it is free. . If F is a formula, then so are ( t)(F) and ( t)(F), where t is a tuple variable. » The formula ( t)(F) is true if the formula F evaluates to true for some (at least one) tuple assigned to free occurrences of t in F; otherwise ( t)(F) is false. » The formula ( t)(F) is true if the formula F evaluates to true for every tuple (in the universe) assigned to free occurrences of t in F; otherwise ( t)(F) is false. 82 The Existential and Universal Quantifiers (2/2) . is called the universal or “for all” quantifier because every tuple in “the universe of” tuples must make F true to make the quantified formula true. . is called the existential or “there exists” quantifier because any tuple that exists in “the universe of” tuples may make F true to make the quantified formula true. 83 Example Query Using Existential Quantifier . Retrieve the name and address of all employees who work for the ‘Research’ department. The query can be expressed as : {t.FNAME, t.LNAME, t.ADDRESS | EMPLOYEE(t) and ( d) (DEPARTMENT(d) and d.DNAME=‘Research’ and d.DNUMBER=t.DNO) } . The only free tuple variables in a relational calculus expression should be those that appear to the left of the bar ( | ). » In above query, t is the only free variable; it is then bound successively to each tuple. . If a tuple satisfies the conditions specified in the query, the attributes FNAME, LNAME, and ADDRESS are retrieved for each such tuple. » The conditions EMPLOYEE (t) and DEPARTMENT(d) specify the range relations for t and d. » The condition d.DNAME = ‘Research’ is a selection condition and corresponds to a SELECT operation in the relational algebra, whereas the condition d.DNUMBER = t.DNO is a JOIN condition. 84 Example Query Using Universal Quantifier . Find the names of employees who work on all the projects controlled by department number 5. The query can be: {e.LNAME, e.FNAME | EMPLOYEE(e) and ( ( x)(not(PROJECT(x)) or not(x.DNUM=5) OR ( ( w)(WORKS_ON(w) and w.ESSN=e.SSN and x.PNUMBER=w.PNO))))} . Exclude from the universal quantification all tuples that we are not interested in by making the condition true for all such tuples. » The first tuples to exclude (by making them evaluate automatically to true) are those that are not in the relation R of interest. . In query above, using the expression not(PROJECT(x)) inside the universally quantified formula evaluates to true all tuples x that are not in the PROJECT relation. » Then we exclude the tuples we are not interested in from R itself. The expression not(x.DNUM=5) evaluates to true all tuples x that are in the project relation but are not controlled by department 5. . Finally, we specify a condition that must hold on all the remaining tuples in R. ( ( w)(WORKS_ON(w) and w.ESSN=e.SSN and x.PNUMBER=w.PNO) 85 Sample Queries in Tuple Relational Calculus 86 Notation for Query Graphs 87 Transforming the Universal and Existential Quantifiers . Transform one type of quantifier into other with negation (preceded by NOT) . AND and OR replace one another . Negated formula becomes un-negated . Un-negated formula becomes negated 88 Using the Universal Quantifier in Queries 89 Safe Expressions . Guaranteed to yield a finite number of tuples as its result . Otherwise expression is called unsafe . Expression is safe . If all values in its result are from the domain of the expression 90 Languages Based on Tuple Relational Calculus (1/2) . The language SQL is based on tuple calculus. It uses the basic block structure to express the queries in tuple calculus: » SELECT 91 Languages Based on Tuple Relational Calculus (2/2) . Another language which is based on tuple calculus is QUEL which actually uses the range variables as in tuple calculus. Its syntax includes: » RANGE OF 92 The Domain Relational Calculus (1/2) . Another variation of relational calculus called the domain relational calculus, or simply, domain calculus is equivalent to tuple calculus and to relational algebra. . The language called QBE (Query-By-Example) that is related to domain calculus was developed almost concurrently to SQL at IBM Research, Yorktown Heights, New York. » Domain calculus was thought of as a way to explain what QBE does. . Domain calculus differs from tuple calculus in the type of variables used in formulas: » Rather than having variables range over tuples, the variables range over single values from domains of attributes. . To form a relation of degree n for a query result, we must have n of these domain variables— one for each attribute. 93 The Domain Relational Calculus (2/2) . An expression of the domain calculus is of the form { x1, x2, . . ., xn | COND(x1, x2, . . ., xn, xn+1, xn+2, . . ., xn+m)} » where x1, x2, . . ., xn, xn+1, xn+2, . . ., xn+m are domain variables that range over domains (of attributes) » and COND is a condition or formula of the domain relational calculus. 94 Example Query Using Domain Calculus Retrieve the birthdate and address of the employee whose name is ‘John B. Smith’. . Query : {uv | ( q) ( r) ( s) ( t) ( w) ( x) ( y) ( z) (EMPLOYEE(qrstuvwxyz) and q=’John’ and r=’B’ and s=’Smith’)} . Abbreviated notation EMPLOYEE(qrstuvwxyz) uses the variables without the separating commas: EMPLOYEE(q,r,s,t,u,v,w,x,y,z) . Ten variables for the employee relation are needed, one to range over the domain of each attribute in order. » Of the ten variables q, r, s, . . ., z, only u and v are free. . Specify the requested attributes, BDATE and ADDRESS, by the free domain variables u for BDATE and v for ADDRESS. . Specify the condition for selecting a tuple following the bar ( | )— » namely, that the sequence of values assigned to the variables qrstuvwxyz be a tuple of the employee relation and that the values for q (FNAME), r (MINIT), and s (LNAME) be ‘John’, ‘B’, and ‘Smith’, respectively. 95 QBE: A Query Language Based on Domain Calculus (1/2) . This language is based on the idea of giving an example of a query using “example elements” which are nothing but domain variables. . Notation: An example element stands for a domain variable and is specified as an example value preceded by the underscore character. . P. (called P dot) operator (for “print”) is placed in those columns which are requested for the result of the query. . A user may initially start giving actual values as examples, but later can get used to providing a minimum number of variables as example elements. . See Textbook Appendix C 96 QBE: A Query Language Based on Domain Calculus (2/2) . The language is very user-friendly, because it uses minimal syntax. . QBE was fully developed further with facilities for grouping, aggregation, updating etc. and is shown to be equivalent to SQL. . The language is available under QMF (Query Management Facility) of DB2 of IBM and has been used in various ways by other products like ACCESS of Microsoft, and PARADOX. . For details, see Appendix C in the textbook. 97 QBE Examples . QBE initially presents a relational schema as a “blank schema” in which the user fills in the query as an example: 98 Example Schema as a QBE Query Interface 99 QBE Examples . The following domain calculus query can be successively minimized by the user as shown: . Query : {uv | ( q) ( r) ( s) ( t) ( w) ( x) ( y) ( z) (EMPLOYEE(qrstuvwxyz) and q=‘John’ and r=‘B’ and s=‘Smith’)} 100 Four Successive Ways to Specify a QBE Query 101 QBE Examples . Specifying complex conditions in QBE: . A technique called the “condition box” is used in QBE to state more involved Boolean expressions as conditions. . The C.4(a) gives employees who work on either project 1 or 2, whereas the query in C.4(b) gives those who work on both the projects. 102 Complex Conditions with and without a Condition Box in QBE Query 103 Handling AND conditions in a QBE Query 104 JOIN in QBE: Examples . The join is simply accomplished by using the same example element (variable with underscore) in the columns being joined from different (or same as in C.5 (b)) relation. . Note that the Result is set us as an independent table to show variables from multiple relations placed in the result. 105 Performing Join with Common Ex. Elements & Use of RESULT Relation 106 Aggregation in QBE . Aggregation is accomplished by using .CNT for count,.MAX, .MIN, .AVG for the corresponding aggregation functions . Grouping is accomplished by .G operator. . Condition Box may use conditions on groups (similar to HAVING clause in SQL – see Textbook Section 8.5.8) 107 Aggregation in QBE: Examples 108 Negation in QBE: Examples 109 UPDATING in QBE: Examples 110 Summary . Relational Algebra » Unary Relational Operations » Relational Algebra Operations From Set Theory » Binary Relational Operations » Additional Relational Operations » Examples of Queries in Relational Algebra . Relational Calculus » Tuple Relational Calculus » Domain Relational Calculus . Overview of the QBE language (see Textbook Appendix C) 111 Agenda 1 Session Overview 2 Relational Algebra and Relational Calculus 3 Relational Algebra Using SQL Syntax 4 Summary and Conclusion 112 Agenda . Relational Algebra and SQL . Basic Syntax Comparison . Sets and Operations on Relations . Relations in Relational Algebra . Empty Relations . Relational Algebra vs. Full SQL . Operations on Relations » Projection » Selection » Cartesian Product » Union » Difference » Intersection . From Relational Algebra to Queries (with Examples) . Microsoft Access Case Study . Pure Relational Algebra 113 Relational Algebra And SQL . SQL is based on relational algebra with many extensions » Some necessary » Some unnecessary . “Pure” relational algebra, use mathematical notation with Greek letters . It is covered here using SQL syntax; that is this unit covers relational algebra, but it looks like SQL And will be really valid SQL . Pure relational algebra is used in research, scientific papers, and some textbooks . So it is good to know it, and material is provided at the end of this unit material from which one can learn it . But in anything practical, including commercial systems, you will be using SQL 114 Key Differences Between SQL And “Pure” Relational Algebra . SQL data model is a multiset not a set; still rows in tables (we sometimes continue calling relations) » Still no order among rows: no such thing as 1st row » We can (if we want to) count how many times a particular row appears in the table » We can remove/not remove duplicates as we specify (most of the time) » There are some operators that specifically pay attention to duplicates » We must know whether duplicates are removed (and how) for each SQL operation; luckily, easy . Many redundant operators (relational algebra had only one: intersection) . SQL provides statistical operators, such as AVG (average) » Can be performed on subsets of rows; e.g. average salary per company branch 115 Key Differences Between Relational Algebra And SQL . Every domain is “enhanced” with a special element: NULL » Very strange semantics for handling these elements . “Pretty printing” of output: sorting, and similar . Operations for » Inserting » Deleting » Changing/updating (sometimes not easily reducible to deleting and inserting) 116 Basic Syntax Comparison (1/2) Relational Algebra SQL a, b SELECT a, b (d > e) (f =g ) WHERE d > e AND f = g p q FROM p, q a, b (d > e) (f =g ) (p q) SELECT a, b FROM p, q WHERE d > e AND f = g; {must always have SELECT even if all attributes are kept, can be written as: SELECT *} renaming AS {or blank space} p := result INSERT INTO p result {assuming p was empty} a, b (p) (assume a, b are the only SELECT * attributes) FROM p; 117 Basic Syntax Comparison (2/2) Relational Algebra SQL p q SELECT * FROM p UNION SELECT * FROM q p − q SELECT * FROM p EXCEPT SELECT * FROM q Sometimes, instead, we have DELETE FROM p q SELECT * FROM p INTERSECT SELECT * FROM q 118 Sets And Operations On Them . If A, B, and C are sets, then we have the operations . Union, A B = { x x A x B } . Intersection, A B = { x x A x B } . - Difference, A - B = { x x A x B } . Cartesian product, A B = { (x,y) x A y B }, A B C = { (x,y,z) x A y B z B }, etc. . The above operations form an algebra, that is you can perform operations on results of operations, such as (A B) (C A) So you can write expressions and not just programs! 119 Relations in Relational Algebra . Relations are sets of tuples, which we will also call rows, drawn from some domains . Relational algebra deals with relations (which look like tables with fixed number of columns and varying number of rows) . We assume that each domain is linearly ordered, so for each x and y from the domain, one of the following holds » x < y » x = y » x < y . Frequently, such comparisons will be meaningful even if x and y are drawn from different columns » For example, one column deals with income and another with expenditure: we may want to compare them 120 Reminder: Relations in Relational Algebra . The order of rows and whether a row appears once or many times does not matter . The order of columns matters, but as our columns will always be labeled, we will be able to reconstruct the order even if the columns are permuted. . The following two relations are equal: R A B R B A 1 10 20 2 2 20 10 1 20 2 20 2 121 Many Empty Relations . In set theory, there is only one empty set . For us, it is more convenient to think that for each relation schema, that for specific choice of column names and domains, there is a different empty relation . And of, course, two empty relations with different number of columns must be different . So for instance the two relations below are different . The above needs to be stated more precisely to be “completely correct,” but as this will be intuitively clear, we do not need to worry about this too much 122 Relational Algebra Versus Full SQL . Relational algebra is restricted to querying the database . Does not have support for » Primary keys » Foreign keys » Inserting data » Deleting data » Updating data » Indexing » Recovery » Concurrency » Security » … . Does not care about efficiency, only about specifications of what is needed 123 Operations on relations . There are several fundamental operations on relations . We will describe them in turn: » Projection » Selection » Cartesian product » Union » Difference » Intersection (technically not fundamental) . The very important property: Any operation on relations produces a relation . This is why we call this an algebra 124 Projection: Choice Of Columns R A B C D 1 10 100 1000 1 20 100 1000 1 20 200 1000 . SQL statement B A D SELECT B, A, D 10 1 1000 FROM R 20 1 1000 20 1 1000 . We could have removed the duplicate row, but did not have to 125 Selection: Choice Of Rows R A B C D 5 5 7 4 5 6 5 7 4 5 4 4 5 5 5 5 4 6 5 3 4 4 3 4 . SQL statement: SELECT * (this means all columns) 4 4 4 5 FROM R 4 6 4 6 WHERE A <= C AND D = 4; (this is a predicate, i.e., condition) A B C D 5 5 7 4 4 5 4 4 126 Selection . In general, the condition (predicate) can be specified by a Boolean formula with NOT, AND, OR on atomic conditions, where a condition is: » a comparison between two column names, » a comparison between a column name and a constant » Technically, a constant should be put in quotes » Even a number, such as 4, perhaps should be put in quotes, as ‘4’, so that it is distinguished from a column name, but as we will never use numbers for column names, this not necessary 127 Cartesian Product R A B S C B D 1 10 40 10 10 2 10 50 20 10 2 20 . SQL statement SELECT A, R.B, C, S.B, D FROM R, S; (comma stands for Cartesian product) A R.B C S.B D 1 10 40 10 10 1 10 50 20 10 2 10 40 10 10 2 10 50 20 10 2 20 40 10 10 2 20 50 20 10 128 A Typical Use Of Cartesian Product R Size Room# S ID# Room# YOB 140 1010 40 1010 1982 150 1020 50 1020 1985 140 1030 . SQL statement: SELECT ID#, R.Room#, Size FROM R, S WHERE R.Room# = S.Room#; ID# R.Room# Size 40 1010 140 50 1020 150 129 A Typical Use Of Cartesian Product . After the Cartesian product, we got Size R.Room# ID# S.Room YOB # 140 1010 40 1010 1982 140 1010 50 1020 1985 150 1020 40 1010 1982 150 1020 50 1020 1985 140 1030 40 1010 1982 140 1030 50 1020 1985 . This allowed us to correlate the information from the two original tables by examining each tuple in turn 130 A Typical Use Of Cartesian Product . This example showed how to correlate information from two tables » The first table had information about rooms and their sizes » The second table had information about employees including the rooms they sit in » The resulting table allows us to find out what are the sizes of the rooms the employees sit in . We had to specify R.Room# or S.Room#, even though they happen to be equal due to the specific equality condition . We could, as we will see later, rename a column, to get Room# ID# Room# Size 40 1010 140 50 1020 150 131 Union R A B S A B 1 10 1 10 2 20 3 20 . SQL statement (SELECT * A B FROM R) 1 10 UNION 2 20 (SELECT * FROM S); 3 20 . Note: We happened to choose to remove duplicate rows . Note: we could not just write R UNION S (syntax quirk) 132 Union Compatibility . We require same -arity (number of columns), otherwise the result is not a relation . Also, the operation “probably” should make sense, that is the values in corresponding columns should be drawn from the same domains . Actually, best to assume that the column names are the same and that is what we will do from now on . We refer to these as union compatibility of relations . Sometimes, just the term compatibility is used 133 Difference R A B S A B 1 10 1 10 2 20 3 20 . SQL statement (SELECT * FROM R) MINUS A B (SELECT * 2 20 FROM S); . Union compatibility required . EXCEPT is a synonym for MINUS 134 Intersection R A B S A B 1 10 1 10 2 20 3 20 . SQL statement (SELECT * FROM R) A B INTERSECT 1 10 (SELECT * FROM S); . Union compatibility required . Can be computed using differences only: R – (R – S) 135 From Relational Algebra to Queries . These operations allow us to define a large number of interesting queries for relational databases. . In order to be able to formulate our examples, we will assume standard programming language type of operations: » Assignment of an expression to a new variable; In our case assignment of a relational expression to a relational variable. » Renaming of a relations, to use another name to denote it » Renaming of a column, to use another name to denote it 136 A Small Example . The example consists of 3 relations: . Person(Name,Sex,Age) . This relation, whose primary key is Name, gives information about the human’s sex and age . Birth(Parent,Child) . This relation, whose primary key is the pair Parent,Child, with both being foreign keys referring to Person gives information about who is a parent of whom. (Both mother and father would be generally listed) . Marriage(Husband,Wife, Age) or . Marriage(Husband,Wife, Age) . This relation listing current marriages only, requires choosing which spouse will serve as primary key. For our exercise, it does not matter what the choice is. Both Husband and Wife are foreign keys referring to Person. Age specifies how long the marriage has lasted. . For each attribute above, we will frequently use its first letter to refer to it, to save space in the slides, unless it creates an ambiguity . Some ages do not make sense, but this is fine for our example 137 Relational Implementation . Two options for selecting the primary key of Marriage . The design is not necessarily good, but nice and simple for learning relational algebra Marriage Person Birth PK,FK2 W PK N PK,FK1 P PK,FK2 C FK1 H S A Marriage Person Birth PK,FK1 H PK N PK,FK1 P PK,FK2 C FK2 W S A . Because we want to focus on relational algebra, which does not understand keys, we will not specify keys in this unit 138 Microsoft Access Database . Microsoft Access Database with this example has been posted » The example suggests that you download and install Microsoft Access 2007 » The examples are in the Access 2000 format so that if you have an older version, you can work with it . Access is a very good tool for quickly learning basic constructs of SQL DML, although it is not suitable for anything other than personal databases 139 Microsoft Access Database . The database and our queries (other than the one with MINUS at the end) are the appropriate “extras” directory on the class web in “slides” » MINUS is frequently specified in commercial databases in a roundabout way » We will cover how it is done when we discuss commercial databases . Our sample Access database: People.mdb . The queries in Microsoft Access are copied and pasted in these notes, after reformatting them . Included copied and pasted screen shots of the results of the queries so that you can correlate the queries with the names of the resulting tables 140 Our Database With Sample Queries - Open In Microsoft Access 141 Our Database Person N S A Birth P C Albert M 20 Dennis Albert Dennis M 40 John Mary Evelyn F 20 Mary Albert John M 60 Robert Evelyn Mary F 40 Susan Evelyn Robert M 60 Susan Richard Susan F 40 Marriage H W A Dennis Mary 20 Robert Susan 30 142 Our Instance In Microsoft Access 143 A Query . Produce the relation Answer(A) consisting of all ages of people . Note that all the information required can be obtained from looking at a single relation, Person . Answer:= A SELECT A 20 40 FROM Person; 20 60 40 60 40 144 . Recall that whether duplicates are removed is not important (at least for the time being in our course) The Query In Microsoft Access . The actual query was copied and pasted from Microsoft Access and reformatted for readability . The result is below 145 A Query . Produce the relation Answer(N) consisting of all women who are less or equal than 32 years old. . Note that all the information required can be obtained from looking at a single relation, Person . Answer:= SELECT N FROM Person WHERE A <= 32 AND S =‘F’; N Evelyn 146 The Query In Microsoft Access . The actual query was copied and pasted from Microsoft Access and reformatted for readability . The result is below 147 A Query . Produce a relation Answer(P, Daughter) with the obvious meaning . Here, even though the answer comes only from the single relation Birth, we still have to check in the relation Person what the S of the C is . To do that, we create the Cartesian product of the two relations: Person and Birth. This gives us “long tuples,” consisting of a tuple in Person and a tuple in Birth . For our purpose, the two tuples matched if N in Person is C in Birth and the S of the N is F 148 A Query Answer:= SELECT P, C AS Daughter FROM Person, Birth WHERE C = N AND S = ‘F’; P Daughter John Mary Robert Evelyn Susan Evelyn 149 . Note that AS was the attribute renaming operator Cartesian Product With Condition: Matching Tuples Indicated Person N S A Albert M 20 Dennis M 40 Evelyn F 20 John M 60 Mary F 40 Robert M 60 Susan F 40 Birth P C Dennis Albert John Mary Mary Albert Robert Evelyn Susan Evelyn Susan Richard 150 The Query In Microsoft Access . The actual query was copied and pasted from Microsoft Access and reformatted for readability . The result is below 151 A Query . Produce a relation Answer(Father, Daughter) with the obvious meaning. . Here we have to simultaneously look at two copies of the relation Person, as we have to determine both the S of the Parent and the S of the C . We need to have two distinct copies of Person in our SQL query . But, they have to have different names so we can specify to which we refer . Again, we use AS as a renaming operator, these time for relations . Note: We could have used what we have already computer: Parent,Daughter 152 A Query . Answer := SELECT P AS Father, C AS Daughter FROM Person, Birth, Person AS Person1 WHERE P = Person.N AND C = Person1.N AND Person.S = ‘M’ AND Person1.S = ‘F’; Father Daughter John Mary Robert Evelyn 153 Cartesian Product With Condition: Matching Tuples Indicated Person N S A Person N S A Albert M 20 Albert M 20 Dennis M 40 Dennis M 40 Evelyn F 20 Evelyn F 20 John M 60 John M 60 Mary F 40 Mary F 40 Robert M 60 Robert M 60 Susan F 40 Susan F 40 Birth P C Dennis Albert John Mary Mary Albert Robert Evelyn Susan Evelyn Susan Richard 154 The Query In Microsoft Access . The actual query was copied and pasted from Microsoft Access and reformatted for readability . The result is below 155 A Query . Produce a relation: Answer(Father_in_law, Son_in_law). . A classroom exercise, but you can see the solution in the posted database. . Hint: you need to compute the Cartesian product of several relations if you start from scratch, or of two relations if you use the previously computed (Father, Daughter) relation F_I_L S_I_L John Dennis 156 The Query In Microsoft Access . The actual query was copied and pasted from Microsoft Access and reformatted for readability . The result is below 157 A Query . Produce a relation: Answer(Grandparent,Grandchild) . A classroom exercise, but you can see the solution in the posted database G_P G_C John Albert 158 Cartesian Product With Condition: Matching Tuples Indicated Birth P C Dennis Albert John Mary Mary Albert Robert Evelyn Susan Evelyn Susan Richard 159 The Query In Microsoft Access . The actual query was copied and pasted from Microsoft Access and reformatted for readability . The result is below 160 Further Distance . How to compute (Great-grandparent,Great-grandchild)? . Easy, just take the Cartesian product of the (Grandparent, Grandchild) table with (Parent,Child) table and specify equality on the “intermediate” person . How to compute (Great-great-grandparent,Great-great- grandchild)? . Easy, just take the Cartesian product of the (Grandparent, Grandchild) table with itself and specify equality on the “intermediate” person . Similarly, can compute (Greatx-grandparent,Greatx- grandchild), for any x . Ultimately, may want (Ancestor,Descendant) 161 Relational Algebra Is Not Universal:Cannot Compute (Ancestor,Descendant) . Standard programming languages are universal . This roughly means that they are as powerful as Turing machines, if unbounded amount of storage is permitted (you will never run out of memory) . This roughly means that they can compute anything that can be computed by any computational machine we can (at least currently) imagine . Relational algebra is weaker than a standard programming language . It is impossible in relational algebra (or standard SQL) to compute the relation Answer(Ancestor, Descendant) 162 Relational Algebra Is Not Universal: Cannot Compute (Ancestor,Descendant) . It is impossible in relational algebra (or standard SQL) to compute the relation Answer(Ancestor, Descendant) . Why? . The proof is a reasonably simple, but uses cumbersome induction. . The general idea is: » Any relational algebra query is limited in how many relations or copies of relations it can refer to » Computing arbitrary (ancestor, descendant) pairs cannot be done, if the query is limited in advance as to the number of relations and copies of relations (including intermediate results) it can specify . This is not a contrived example because it shows that we cannot compute the transitive closure of a directed graph: the set of all paths in the graph 163 A Sample Query . Produce a relation Answer(A) consisting of all ages of visitors that are not ages of marriages SELECT A FROM Person MINUS SELECT A FROM MARRIAGE; 164 The Query In Microsoft Access . We do not show this here, as it is done in a roundabout way and we will do it later 165 It Does Not Matter If We Remove Duplicates . Removing duplicates A A A 20 20 40 . 40 - = 30 60 60 . Not removing duplicates A 20 A A 40 40 20 20 60 - 30 = 60 40 40 60 60 40 40 166 It Does Not Matter If We Remove Duplicates . The resulting set contains precisely ages: 40, 60 . So we do not have to be concerned with whether the implementation removes duplicates from the result or not . In both cases we can answer correctly » Is 50 a number that is an age of a marriage but not of a person » Is 40 a number that is an age of a marriage but not of a person . Just like we do not have to be concerned with whether it sorts (orders) the result . This is the consequence of us not insisting that an element in a set appears only once, as we discussed earlier . Note, if we had said that an element in a set appears once, we would have to spend effort removing duplicates! 167 Now To “Pure” Relational Algebra . This was described in several slides . But it is really the same as before, just the notation is more mathematical . Looks like mathematical expressions, not snippets of programs . It is useful to know this because many articles use this instead of SQL . This notation came first, before SQL was invented, and when relational databases were just a theoretical construct 168 : Projection: Choice Of Columns R A B C D 1 10 100 1000 1 20 100 1000 1 20 200 1000 . SQL statement Relational Algebra B A D 10 1 1000 SELECT B, A, D B,A,D (R) FROM R 20 1 1000 20 1 1000 . We could have removed the duplicate row, but did not have to 169 : Selection: Choice Of Rows R A B C D 5 5 7 4 5 6 5 7 4 5 4 4 5 5 5 5 4 6 5 3 4 4 3 4 4 4 4 5 4 6 4 6 . SQL statement: Relational Algebra SELECT * A C D=4 (R) Note: no need for FROM R A B C D WHERE A <= C AND D = 4; 5 5 7 4 4 5 4 4 170 Selection . In general, the condition (predicate) can be specified by a Boolean formula with and on atomic conditions, where a condition is: » a comparison between two column names, » a comparison between a column name and a constant » Technically, a constant should be put in quotes » Even a number, such as 4, perhaps should be put in quotes, as ‘4’ so that it is distinguished from a column name, but as we will never use numbers for column names, this not necessary 171 : Cartesian Product R A B S C B D 1 10 40 10 10 2 10 50 20 10 2 20 . SQL statement Relational Algebra SELECT A, R.B, C, S.B, D R S FROM R, S A R.B C S.B D 1 10 40 10 10 1 10 50 20 10 2 10 40 10 10 2 10 50 20 10 2 20 40 10 10 2 20 50 20 10 172 A Typical Use Of Cartesian Product R Size Room# S ID# Room# YOB 140 1010 150 1020 40 1010 1982 50 1020 1985 140 1030 . SQL statement: Relational Algebra SELECT ID#, R.Room#, Size ID#, R.Room#, Size (R.B = S.B (R S)) FROM R, S WHERE R.Room# = S.Room# ID# R.Room# Size 40 1010 140 50 1020 150 173 : Union R A B S A B 1 10 1 10 2 20 3 20 . SQL statement Relational Algebra (SELECT * A B R S FROM R) 1 10 UNION (SELECT * 2 20 FROM S) 3 20 . Note: We happened to choose to remove duplicate rows . Union compatibility required 174 -: Difference R A B S A B 1 10 1 10 2 20 3 20 . SQL statement Relational Algebra (SELECT * A B R - S FROM R) 2 20 MINUS (SELECT * FROM S) . Union compatibility required 175 : Intersection R A B S A B 1 10 1 10 2 20 3 20 . SQL statement Relational Algebra A B (SELECT * R S FROM R) 1 10 INTERSECT (SELECT * FROM S) . Union compatibility required 176 Summary (1/2) . A relation is a set of rows in a table with labeled columns . Relational algebra as the basis for SQL . Basic operations: » Union (requires union compatibility) » Difference (requires union compatibility) » Intersection (requires union compatibility); technically not a basic operation » Selection of rows » Selection of columns » Cartesian product . These operations define an algebra: given an expression on relations, the result is a relation (this is a “closed” system) . Combining this operations allows production of sophisticated queries 177 Summary (2/2) . Relational algebra is not universal: We can write programs producing queries that cannot be specified using relational algebra . We focused on relational algebra specified using SQL syntax, as this is more important in practice . The other, “more mathematical” notation came first and is used in research and other venues, but not commercially 178 Agenda 1 Session Overview 2 Relational Algebra and Relational Calculus 3 Relational Algebra Using SQL Syntax 4 Summary and Conclusion 179 Summary . Relational algebra and relational calculus are formal languages for the relational model of data . A relation is a set of rows in a table with labeled columns . Relational algebra and associated operations are the basis for SQL . These relational operations define an algebra . Relational algebra is not universal as it is possible to write programs producing queries that cannot be specified using relational algebra . Relational algebra can be specified using SQL syntax . The other, “more mathematical” notation came first and is used in research and other venues, but not commercially 180 Assignments & Readings . Readings » Slides and Handouts posted on the course web site » Textbook: Chapters 6, 8 . Assignment #5 » Textbook exercises: Textbook exercises: 8.15, 8.18, 8.22, 8.30, 6.8, 6.12 . Project Framework Setup (ongoing) 181 Next Session: Standard Query Language (SQL) Features . SQL as implemented in commercial databases 182 Any Questions? 183 » FROM
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