Database Systems Chapter 3 PDF

  Database Systems Chapter 3 Relational Model  3.1   Relational Model Concepts (1)   Table is called relation  Row (record) is called tuple  Column header is called attribu...

  Database Systems Chapter 3 Relational Model  3.1   Relational Model Concepts (1)   Table is called relation  Row (record) is called tuple  Column header is called attribute Name of the relation Attributes Students SSN Name Age GPA head 123456789 John 20 3.2 23456789 Mary 18 2.9 tuples (rows) 345678912 Bill 19 2.7 columns  3.2   Relational Model Concepts (2)   Given a tuple t and an attribute A in a relation R,  t[A] represents the value of t under A in R  Example: If t is the second tuple in Students t[Name] = Mary t[Age] = 18 t[Name, Age] = (Mary, 18)  3.3   Domain of an Attribute   Definition:  The set of values that an attribute can take on is the domain of the attribute  dom(A) --- the domain of attribute A  A domain is usually represented by a type  Examples:  SSN char(9) --- character string of length 9  Name varchar(30) --- character string of variable length up to 30 characters  Age number --- a number  3.4   Relational Model Concepts (3)   Two aspects of a relation:  Schema --- the set of attributes of R.  State (or contents) --- the CURRENT set of tuples of R (denoted by r(R)).  Schema of a relation rarely changes. Some possible changes are:  Rename an attribute  Delete an attribute  Add an attribute  Delete the schema  3.5   Relational Model Concepts (4)   The state of a relation may change frequently. Some possible changes are:  Modify some attribute values  Delete an existing tuple  Insert a new tuple  A given schema may have different states at different times.  3.6   An Example Database  Students Departments SSN Name Major GPA Name Location Chairperson 1234 Jeff CS 3.2 CS N18 EB Aggarwal 2345 Mary Math 3.0 EE Q4 EB Sackman 3456 Bob CS 2.7 Math LN2200 Hanson 4567 Wang EE 2.9 Biology 210 S3 Smith Courses Sections Name Course# CreditHours Dept Course# Section# Semester Instructor Database CS432 4 CS CS432 01 Fall98 Meng Database CS532 4 CS CS532 01 Fall98 Meng Dis. Math Math314 4 Math Math314 02 Fall 97 Hanson Lin. Alg. Math304 4 Math Math304 01 Spring97 Brown  3.7   Relation Schema   A relation schema is used to describe a relation  Denoted by R(A1, A2, A3, …, An), where  R: Relation schema name  A1, …, An: attributes of R  The degree of a relation is the number of attributes in a relation schema.  3.8   Examples   STUDENT(Name, SSN, HomePhone, Address, OfficePhone, Age, GPA)  Degree(STUDENT) = 7  dom(Name) = all names which consists of at most 30 characters.  dom (SSN) is the set of valid 9-digit social security numbers.  dom(HomePhone): local phone numbers.  3.9   Relation Instance (1)   A relation (or relation instance) r of the relation schema R(A1, A2, …, An), denoted by r(R), is a set of n-tuples  r = {t1, t2, …, tm)  each n-tuple ti is an ordered list of n values,  ti =  Rejected, Primary Key is null (Entity Integrity Constraint)  3.35   The Insert Operation   Insert into EMPLOYEE >  Rejected, Primary Key is duplicate (Key Constraint)  Insert into EMPLOYEE >  Rejected, DEPARTMENT # 7 does not exist. (Referential Integrity Constraint)  Insert into EMPLOYEE >  Accepted  3.36   The Delete Operation   Deletes a tuple or tuples from r(R)  It can only violate referential integrity constraint.  “Delete from WORKS_ON where ESSN = ‘999887777’ and PNO = 10”  Accepted  “Delete from EMPLOYEE where SSN = ‘999887777’”  Rejected. Tuples in WORK_ON refer to this tuple, if the tuple is deleted, referential integrity violations will result  “Delete from EMPLOYEE where SSN = ‘333445555’”  Rejected. Tuples in EMPLOYEE, DEPARTMENT, WORK_ON, and DEPENDENT refer to this tuple, if the tuple is deleted, referential integrity violations will result  3.37   The Modify Operation   It is used to change the values of one or more attributes in a tuple or more of some relation r(R)  May violate all constraints  “Update EMPLOYEE, set Salary = 29000 where SSN = ‘999887777’”  Accepted  “Update EMPLOYEE, set DNO = 1 where SSN = ‘999887777’”  Accepted  3.38   The Modify Operation   “Update EMPLOYEE, set DNO = 7 where SSN = ‘999887777’”  Rejected, it violates referential integrity constraint  “Update EMPLOYEE, set SSN = ‘980000000’ where SSN = ‘999887777’”  Rejected, it violates referential integrity constraint  “Update EMPLOYEE, set SSN = ‘987654321’ where SSN = ‘999887777’”  Rejected, it violates unique row (key) and referential integrity constraints  3.39   Basic Relational Algebra Operations   Is a collection of operators that are used to manipulate entire relations. The result of each operation is a new relation which can be further manipulated.  3.40   SELECT Operation    --- Select (sigma)  Format: selection-condition(R)  Semantics:  returns all tuples of relation R that satisfy the selection-condition  Select operation is unary. It applies to a single relation  is used to select a subset of the tuples in a relation that satisfy a selection condition.  3.41   Formats of Selection Conditions   (a) A op v: A is an attribute, op is an operator (=, , , ), and v is a constant.  Age  20, Name = `Bill'  (b) A op B: A and B are two attributes in R.  Persons(SSN, Name, Birthplace, Residence)  Birthplace = Residence  (c) Combinations of (a) and (b) connected by and, or or not.  Age  20 and Birthplace = Residence  3.42   An Example of SELECT (1)   Example: Find all students who are 20 years old or younger, and whose birthplace is the same as his/her residence.  Age  20 and Birthplace = Residence(Students)  3.43    An Example of Select (2) If the current Students is: SSN Name Age GPA Birthplace Residence 123456789 John 20 3.2 Vestal Vestal 234567891 Mary 18 2.9 Binghamton Vestal 345678912 Bill 19 2.7 Endwell Endwell 456789123 Nancy 24 3.6 Binghamton NYC then the result is a new relation: SSN Name Age GPA Birthplace Residence 123456789 John 20 3.2 Vestal Vestal 345678912 Bill 19 2.7 Endwell Endwell  3.44   SELECT Operation   Commutativity of select:  condition-1(condition-2(R)) = condition-2(condition-1(R)) = condition-1 and condition-2(R)  city = “Irbid” AND GPA > 65 (STUDENT) or  city = “Irbid” (GPA > 65 (STUDENT)  3.45   PROJECT Operation    --- project (pi)  Format:  attribute-list(R),  where attribute-list is a subset of all attributes in R  Semantics:  Returns all tuples of relation R but for each tuple, only values under attribute-list are returned  Project removes duplicate tuples automatically  3.46   Project (2)   Selects certain columns from the table and discards other columns   ()  degree of resulting relation is equal to the number of attributes in the  The number of tuples of the result of project is less than or equal to the number of tuples in the original relation. (It removes the duplicates)  3.47    Project (3) Example: Find the name and GPA of all students. Name,GPA(Students) Students SSN Name Age GPA Name GPA 123456789 John 20 3.2 John 3.2 234567891 Mary 18 2.9 Mary 2.9 345678912 John 19 3.2 Input Relation Output Relation  3.48   Project (4)   If attribute-list-1  attribute-list-2, then  attribute-list-1(attribute-list-2(R)) = attribute-list-1(R)  The Project operation is not commutative  Retrieve all student numbers and names who live in Amman.  STNO, ST-Name(City = “Amman”(STUDENT))  3.49   Project (5)   In complex queries, it becomes necessary to store intermediate results, therefore we should know how to give names to relations and attributes  Amman-students = city = “Amman” (STUDENT)  Result =  STNO, ST-Name (Amman-Students)  or  Result(Number,Name) =  STNO, ST-Name (Amman-Students) Renaming of attributes  3.50   Select and Project   Example:  Find the name and GPA of all students who are 20 years old or younger and whose birthplace and residence are the same  Name, GPA(Age20 and Birthplace=Residence(Students))  3.51   RENAME Operation    --- rename (rho)  Format: S(R)  Semantics:  Make a copy of relation R and name the copy as S  S(R):  Rename R only  S(B1, B2, …, Bn)(R):  Rename R and its attributes  (B1, B2, …, Bn)(R):  Rename attributes only  3.52   Set Theoretic Operations   UNION, INTERSECTION, DIFFERENCE.  binary (applied to two relations at a time)  To apply any of these operators to relations, relations should be union-compatible.  Two relations R(A1, A2, …, An) and S(B1, B2, …, Bm) are said to be union-compatible if:  they have the same degree (n = m) and  dom(Ai) = dom(Bi) for 1  i  n.  Both R and S have the same number of attributes and the corresponding attributes have the same domain  3.53   Union (1)    --- union  Format:  R1  R2  Semantics:  Returns all tuples that belong to either R1 or R2.  Formally:  R1  R2 = { t | t  R1 or t  R2 }  Condition of union:  R1 and R2 must be union compatible.  The union operator removes duplicate tuples automatically.  3.54   Union (2)  Example: R1 R2 R1  R2 A B C A B C A B C a1 b1 c1 a0 b0 c0 a1 b1 c1 a2 b2 c2 a1 b1 c1 a2 b2 c2 a3 b3 c3 a2 b2 c2 a3 b3 c3 a4 b4 c4 a0 b0 c0 a4 b4 c4  3.55   Set Difference (1)   - set difference  Format:  R1 - R2  Semantics:  Returns all tuples that belong to R1 but not R2.  Formally:  R1 - R2 = { t | t  R1 and t  R2 }  Set difference also requires union compatibility between R1 and R2  3.56    Set Difference (2) Example: R1 R2 R1 - R2 A B C A B C A B C a1 b1 c1 a0 b0 c0 a3 b3 c3 a2 b2 c2 a1 b1 c1 a3 b3 c3 a2 b2 c2 a4 b4 c4  3.57   INTERSECTION    --- set intersection  Format:  R1  R2  Semantics:  Returns all tuples that belong to both R1 and R2.  Formally:  R1  R2 = { t | t  R1 and t  R2 }  Derivation from existing operators:  R1  R2 = R1 - (R1 - R2) = R2 - (R2 - R1)  3.58   INTERSECTION   Union and Intersection are commutative operations  RS =SR and  RS=SR  Union and Intersection can be applied to any number of relations and both are associative:  R  (S  Q) = (S  R)  Q  R  (S  Q) = (S  R)  Q  Difference operator is not commutative:  R - S  S - R in general.  3.59   Cartesian Product (1)    --- Cartesian product  Format:  R1  R2  Semantics:  Returns every tuple that can be formed by concatenating a tuple in R1 with a tuple in R2  Binary operation, but the relations on which it is applied do not have to be union compatible  3.60    Cartesian Product (2) Example: R1 A B C R2 B D E a1 b1 c1 b1 d1 e1 a2 b2 c2 b2 d2 e2 a3 b3 c3 R1  R2 A R1.B C R2.B D E a1 b1 c1 b1 d1 e1 a1 b1 c1 b2 d2 e2 a2 b2 c2 b1 d1 e1 a2 b2 c2 b2 d2 e2 a3 b3 c3 b1 d1 e1 a3 b3 c3 b2 d2 e2  3.61   Cartesian Product (3)   If R1 and R2 have common attributes, then the full names of these attributes must be used  Example:  Use R.A instead of A  To prevent identical attribute names from occurring in the same relation schema, R  R is not allowed. However, R  S(R) is allowed  Commutativity: R1  R2 = R2  R1  3.62   Cartesian Product (4)   Given R(A1, A2, …, An) and S(B1, B2, …, Bm)  RS = Q(A1, A2, …, An, B1, B2, …, Bm)  degree of Q = n + m  If R1 has N tuples and R2 has M tuples, then  R1  R2 has N*M tuples  Cartesian product is extremely expensive  If R1 and R2 are both large, then each relation may need to be scanned many times to perform the Cartesian product.  Writing out the result can be very expensive due to the large size of the result  3.63   Example  Retrieve for each female employee a list of names of her dependents FEMALE-EMPS =  Sex = “F” (EMPLOYEE) EMP-NAMES = Fname,Lname,SSN(FEMALE-EMPS) EMP-DEPENDENTS = EMP-NAMES  DEPENDENT ACTUAL-DEP = SSN = ESSN (EMP-DEPENDENTS) RESULT = Fname,Lname,Dependent-name(ACTUAL-DEP)  3.64    Relational Algebra Example (1) Example: Find the names of each employee and his/her manager Employees: SSN Name Age Dept-Name 123456789 John 34 Sales 234567891 Mary 42 Service 345678912 Bill 39 null Departments: Name Location Manager Sales XYZ Bill Inventory YZX Charles Service ZXY Maria  3.65   Relational Algebra Example (2)   A relational algebra expression is: Employees.Name, Departments.Manager (Employees.Dept_Name = Departments.Name (Employees  Departments))  A simplified version (don't use full name when you don't have to): Employees.Name, Manager (Dept_Name =Departments.Name (Employees Departments))  3.66   Relational Algebra Example (3)   Use assignment operator ( = ) to save the intermediate result into a temporary relation  Example: The following expression Employees.Name, Manager(Dept_Name = Departments.Name (Employees  Departments) )  is equivalent to the following series of expressions: TEMP1 = Employees  Departments TEMP2 = Dept_Name=Departments.Name(TEMP1) RESULT = Employees.Name, Manager (TEMP2)  3.67    Relational Algebra Example (4) Example: Find the names of all students who have the highest GPA STUDENTS SSN Name GPA 123456789 John 3.8 234567891 Maria 3.2 345678912 Mike 3.0  3.68    Relational Algebra Example (5)  Step 1: Find the GPAs that are not the highest  TEMP1 = Students.GPA(Students.GPA < S2.GPA (Students  S2(Students))) Students  S2(Students) SSN Name GPA S2.SSN S2.Name S2.GPA 123456789 John 3.8 123456789 John 3.8 123456789 John 3.8 234567891 Maria 3.2 123456789 John 3.8 345678912 Mike 3.0 234567891 Maria 3.2 123456789 John 3.8 234567891 Maria 3.2 234567891 Maria 3.2 234567891 Maria 3.2 345678912 Mike 3.0 345678912 Mike 3.0 123456789 John 3.8 345678912 Mike 3.0 234567891 Maria 3.2 345678912 Mike 3.0 345678912 Mike 3.0  3.69   Relational Algebra Example (6)   Step 2: Find the highest GPA  TEMP2 = GPA(Students) - TEMP1  Step 3: Find the names of students who have the highest GPA  RESULT = Name(Students.GPA = EMP2.GPA (StudentsTEMP2))  3.70   Join (1)    --- join  Format:  R1 join-condition R2  Semantics:  Returns all tuples in R1  R2 which satisfy the join condition  Derivation from existing operators:  R1 join-condition R2 = join-condition(R1  R2)  Format of join condition:  R1.A op R2.B  R1.A1 op R2.B1 and R1.A2 op R2.B2...  Tuples whose join attributes are NULL do not appear in the result  3.71    Join (2) Example: Find the names of all employees and their department locations Employees: SSN Name Age Dept-Name 123456789 John 34 Sales 234567891 Mary 42 Service 345678912 Bill 39 null Departments: Name Location Manager Sales Binghamton Bill Inventory Endicott Charles Service Vestal Maria  3.72   Join (3)  Employees.Name, Location (Employees Dept-Name = Departments.Name Departments) Result Name Location John Binghamton Mary Vestal  3.73    Join (4) Example: Find the names of all employees who earn more than his/her manager Employees: SSN Name Salary Manager-SSN 123456789 John 34k 234567891 234567891 Bill 40k null 345678912 Mary 38k null 456789123 Mike 41k 345678912 Employees.Name (Employees Employees.Manager-SSN = EMP.SSN and Employees.Salary > EMP.Salary EMP(Employees))  3.74   Equijoin   Definition: A join is called an equijoin if only equality operator is used in all join conditions. R1 R2 R1 R1.B = R2.B R2 A B B C A R1.B R2.B C a b b c a b b c d b c d d b b c b c a d b c c d  Most joins in practice are equijoins.  3.75   Natural Join (1)   Definition:  A join between R1 and R2 is a natural join if  There is an equality comparison between every pair of identically named attributes from the two relations  Among each pair of identically named attributes from the two relations, only one remains in the result  Natural join is denoted by  with no join conditions explicitly specified  3.76   Natural Join (2)   Example:  R1(A, B, C)  R2(A, C, D)  has attributes (A, B, C, D) in the result  Questions:  How to express natural join in terms of equijoin and other relational operator?  R1  R2 = R1.A, B, R2.C, D (R1 R1.A=R2.A and R1.C= R2.C R2)  3.77   A Complete Set of Relational Algebra Operations  The relational algebra is a set of expressions as defined below:  A relation is an expression.  If E1 and E2 are expressions, so are P(E1), A(E1), S(E1), E1E2, E1-E2, E1E2  That is, any expression that can be formed from base relations and the six relational operators is a relational algebra expression  3.78   Division (1)  A motivating example: Emp# Proj# Proj# 1 10 10 1 20 20 1 30 30 2 20 3 10 3 20 Which employee participates in all projects?  3.79   Division (2)    --- division  Format:  R1  R2  Restriction:  Every attribute in R2 is in R1  Semantics:  Consider: – R1(A1,..., An, B1,..., Bm)  R2(B1,..., Bm) – Let T =  A1,..., An (R1)  Returns those tuples in T such that for every tuple t returned, the concatenation of t with every tuple in R2 is in R1  R1  R2 = {t | t  A1,..., An(R1)   u  R2, tu  R1}  3.80   Division (3)  E.g.: R1 R2 T R1  R2 A B C D C D A B A B a b c d c d a b a b a b e f e f b c e d b c e f e d e d c d e d e f a b d e Derivation from existing operators: R1  R2 = T - A1,..., An ((T  R2) - R1)  3.81   Division (4)   Example:  Find the names and GPAs of all students who take all courses taken by a student with SSN = 123456789  Students (SSN, Name, GPA)  Takes (SSN, Course#, GRADE)  Step 1: Find all courses that are taken by the student with SSN = 123456789.  TEMP1 = Course#(SSN = ’123456789’(Takes))  3.82   Division (5)   Step 2: Find the SSNs of those students who take all courses in TEMP1.  TEMP2 = Takes  TEMP1  Step 3: Obtain the final result.  RESULT = Name, GPA(Students  TEMP2)  3.83   Division (6)   Find the names of employees who participate in every (all) project  Employees(SSN, Name, Department)  Projects(Proj#, Name, Budget)  Participation(SSN, Proj#, Hours)  Name(Employees(ParticipationProj#(Projects)))  Can we replace Projects by Participation? Only when Project has total participation  3.84   Aggregate Functions & Grouping   Well know Aggregate Functions:  SUM, AVERAGE, MAXIMUM, MINIMUM, COUNT  Format:  (Script F)   (R)  is a list of attributes in R  is a list of ( ) pairs  is one of aggregate functions  is an attribute in R  3.85   Examples   Retrieve for each department number, the number of employees in the department, and their average salary:  DNO COUNT( SSN), AVERAGE(SALARY) (EMPLOYEE) DNO COUNT_SSN AVERAGE_SALARY 1 1 55000 4 3 31000 5 4 33250  3.86   Examples   Retrieve for each department number, the number of employees in the department, and their average salary: R(DNO, NO_OF_EMP, AVERAGE_SAL)(DNO  COUNT( SSN), AVERAGE(SALARY) (EMPLOYEE)) DNO NO_OF_EMP AVERAGE_SAL 1 1 55000 4 3 31000 5 4 33250  3.87   Examples   If no grouping attributes are specified, the functions are applied to attribute values of all the tuples in the relation  Retrieve the number of employees, and their average salary:  COUNT( SSN), AVERAGE(SALARY) (EMPLOYEE)) COUNT_SSN AVERAGE_SALARY 8 35125  3.88    c R1 R2 R1  R2 A B C C D E A B C D E a1 b1 c1 c1 d1 e1 a1 b1 c1 d1 e1 a4 b3 c2 c6 d3 e2  The second tuples of R1 and R2 are not present in the result (called dangling tuples)  Applications exist that require to retain dangling tuples.  3.89    Outerjoin (2) O --- outer join Format: R1 O R2 Semantics: like join except  it retains dangling tuples from both R1 and R2  it uses null to fill out missing entries R1 O R2 A B C D E a1 b1 c1 d1 e1 a4 b3 c2 null null null null c6 d3 e2  3.90   Left Outerjoin and Right Outerjoin   LO --- left outer join  Format: R1 LO R2  Semantics: like outerjoin but retains only dangling tuples of the relation on the left  RO --- right outer join  Format: R1 RO R2  Semantics: like outerjoin but retains only dangling tuples of the relation on the right  3.91    Left Outerjoin and Right Outerjoin (2) R1 R2 R1  R2 A B C C D E A B C D E a1 b1 c1 c1 d1 e1 a1 b1 c1 d1 e1 a4 b3 c2 c6 d3 e2  3.92    Left Outerjoin and Right Outerjoin (3) R1 LO R2 A B C D E a1 b1 c1 d1 e1 a4 b3 c2 null null R1 RO R2 A B C D E a1 b1 c1 d1 e1 null null c6 d3 e2  3.93   Recursive Closure Operations   This operation is applied to a recursive relationship  Such as the relationship between an employee and a supervisor  The relationship is described by the foreign key SUPERSSN of the EMPLOYEE relation  Example:  Retrieve all supervisees of an employee e at all levels  All employees e’ directly supervised by employee e  All employees e’’ directly supervised by each employee e’  All employees e’’’ directly supervised by each employee e’’; and so on  3.94   Recursive Closure Operations   Retrieve the SSNs of all employees e’ directly supervised – at level one - by the employee e whose name is ‘James Borg’ Borg_SSN = SSN(FNAME=‘James AND LNAME=‘Borg’(EMPLOYEE)) SUPERVISSION(SSN1, SSN2) = SSN, SUPERSSN(EMPLOYEE) RESULT1(SSN) = SSN1(SUPERVISION SSN2=SSN BORG_SSN)  3.95   Recursive Closure Operations   To retrieve all employees supervised by ‘Borg’ at level 2 – that is, all employees e’’ supervised by some employee e’ whose is directly supervised by ‘Borg’ RESULT2(SSN) = SSN1(SUPERVISION SSN2=SSN RESULT1)  To get both sets of employees supervised at levels 1 and 2 by ‘James Borg’ RESULT = RESULT1  RESULT2  3.96    Examples of Queries in Relational Algebra (1)  Many relational algebra queries can be expressed using selection, projection and join operators by following steps:  (1) Determine necessary relations to answer the query.  If R1,..., Rn are all the relations needed,  P are all conditions and  T are all (target) attributes to be output,  then form the initial query:  T(P(R1 ...  Rn))  3.97   Relational Algebra Example (2)   (2) If P contains a condition, say Ci, that involves only attributes in Ri  replace Ri by Ci(Ri) and remove Ci from P  (3) If P contains a condition, say C, that involves attributes from both Ri and Rj  replace Ri  Rj by Ri C Rj (or a natural join) and remove C from P  3.98   Relational Algebra Example (3)   Consider the following database schema:  Students(SSN, Name, GPA, Age, Dept-Name)  Enrollment(SSN, Course#, Grade)  Courses(Course#, Title, Dept-Name)  Departments(Name, Location, Phone)  3.99   Relational Algebra Example (5)   Query: Find the SSNs and names of all students who are CS major and who take CIS328  (1) Relations Students and Enrollment are needed  T = {Students.SSN, Students.Name}  P = {Students.Dept-Name = ‘CS’, Enrollment.Course# = ‘CIS328’, Students.SSN = Enrollment.SSN}  The initial relational algebra query is:  Students.SSN, Name (Dept-Name=`CS’ and Course#=`CIS328’ and Students.SSN= Enrollment.SSN (Students  Enrollment))  3.100   Relational Algebra Example (6)   (2) Replace  Students by Dept-Name = `CS’(Students) and  Enrollment by Course#=`CIS328’(Enrollment)  Remove the two conditions from the initial expression  3.101   Relational Algebra Example (7)   (3) Replace  Students  Enrollment by Students  Enrollment and  remove Students.SSN = Enrollment.SSN from the initial expression.  The final expression:  Students.SSN,Name(Dept-Name = `CS’(Students)   Course# = `CIS328’ (Enrollment))  3.102   Relational Algebra Example (8)   Query: Find the SSN and name of each student who is CS major together with the titles of the courses taken by the student   Students.SSN, Name, Title (Students.Dept-Name = `CS’ and Students.SSN = Enrollment.SSN and Enrollment.Course# = Courses.Course# (Students  Enrollment  Courses))  3.103   Relational Algebra Example (9)   =  Students.SSN, Name, Title (((Students.Dept-Name = `CS’ (Students))  Enrollment)  Courses)  =  Students.SSN, Name, Title (( Students.Dept-Name = `CS’ (Students))  (Enrollment  Courses))  3.104   Relational Algebra Summary (1)  Relational algebra operators:  Fundamental operators: C(R), A(R), S(R), R1  R2, R1 - R2, R1  R2  Other traditional operators: R1  R2, R1 C R2, R1  R2  3.105   Relational Algebra Summary (2)  Some identities:  C1(C2(R)) = C2(C1(R)) = C1 and C2(R)  L1(L2(R)) = L1(R) , if L1  L2  R1  R2 = R2  R1  R1  (R2  R3) = (R1  R2)  R3  R1  R2 = R2  R1  R1  (R2  R3) = (R1  R2)  R3  3.106 