1. The Relational Algebra
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2. Chapter Outline Fundamental Operator Deliverable & Additional Operator
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3. What is Relational Algebra?
Relational Algebra is a Procedural Paradigm You Need to Tell What/How to Construct the Result
Consists of a Set of Operators Which, When Applied to Relations, Yield Relations (Closed Algebra)
Basic Relational Operations:
Unary Operations
SELECT s
PROJECT  or P.
Binary Operations
Set operations:
UNION 
INTERSECTION 
DIFFERENCE –
CARTESIAN PRODUCT 
JOIN operations 
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4. Relational Algebra A1, A2, ..., An (R) projection F (R) selection
R S union
R S intersection
R \ S set difference
R S Cartesian product
A1, A2, ..., An (R) projection
F (R) selection
R S natural join
R S theta-join
RS division
[A1 B1,.., An Bn] rename
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5. Selection
Selects the Set of Tuples (Rows) From a Relation, Which Satisfy a Selection Condition
General Form s<selection condition> (R)
R is a Relation
Selection condition is a Boolean Expression on the Attributes of R
Resulting Relation Has the Same Schema as R
Select Finds and Retrieves All Relevant Rows (Tuples) of Table/Relation R which Includes ALL of its Columns (Attributes)
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6. Selection Example TITLE='Elect. Eng.'(EMP)
ENO
ENAME
TITLE E1
J. Doe
Elect. Eng. E2
M. Smith
Syst. Anal. E3
A. Lee
Mech. Eng. E4
J. Miller
Programmer E5
B. Casey E6
L. Chu E7
R. Davis E8
J. Jones
EMP
ENO
ENAME
TITLE E1
J. Doe
Elect. Eng E6
L. Chu
Elect. Eng.
TITLE='Elect. Eng.'(EMP)
TITLE='Elect. Eng.’ OR TITLE=‘Mech.Eng’(EMP)
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7. Another Selection Example
null W B
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8. Selection Condition A SELECT Condition is a Boolean Expression
Form F1 Y F2 Y ..., Y Fq (Q>=1), Where
Fi (I=1,…,q) are Atomic Boolean Expressions of the Form
aqc or aqb,
a, b are Attributes of R and c is a Constant.
The Operator q is one of the Arithmetic Comparison Operators: <, >, =, <>, >=, <=
The Operator Y is one of the Logical Operators: , , ¬
Nesting: ( )
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9. Projection
Select certain Columns (Attributes) Specified in an Attribute List X From a Relation R
General Form <attribute-list>(R)
R is a Relation
Attribute-list is a Subset of the Attributes of R Over Which the Projection is Performed
Project Retrieves Specified Columns of Table/Relation R which Includes ALL of its Rows (Tuples)
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10. Projection Example  PNO,BUDGET(PROJ) PROJ PNO BUDGET P2 135000 P3
250000 P4
310000 P5
500000
PNAME P1
150000
Instrumentation
Database Develop.
CAD/CAM
Maintenance
 PNO,BUDGET(PROJ)
PNO
BUDGET P1
150000 P2
135000 P3
250000 P4
310000 P5
500000
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11. Other Projection Examples
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12. Relational Algebra Expression
Several Operations can be Combined to form a Relational Algebra Expression (query)
Example: Retrieve all Customers over age 60?
Method 1:  CNAME, ADDRESS, AGE (s AGE>60(CUSTOMER) )
Method 2: Senior-CUST(C#, Addr, Age)
=  CNAME, ADDRESS, AGE (s AGE>60(CUSTOMER) )
Method 3:  CNAME, ADDRESS, AGE (C) where
C = s AGE>60(CUSTOMER)
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13. Characteristics of Projection
The PROJECT Operation Eliminates Duplicate Tuples in the Resulting Relation
Why?
Projection Must Maintain a Mathematical Set (No Duplicate Elements)
ENO
ENAME
TITLE E1
J. Doe
Elect. Eng. E2
M. Smith
Syst. Anal. E3
A. Lee
Mech. Eng. E4
J. Miller
Programmer E5
B. Casey E6
L. Chu E7
R. Davis E8
J. Jones
EMP
 TITLE(PROJ)
Elect.Eng
Syst.Anal
Mech.Eng
Programmer
TITLE
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14. Selection with Projection Example
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15. Renaming The RENAME operator gives a new schema to a relation.
R1 := RENAMER1(A1,…,An)(R2) makes R1 be a relation with attributes A1,…,An and the same tuples as R2.
Simplified notation: R1(A1,…,An) := R2.
Other use () notation
R2  R1(A1,…,An)
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16. Sequences of Assignments
Create temporary relation names.
Renaming can be implied by giving relations a list of attributes.
Example: R3 := R1 JOINC R2 can be written:
R4 := R1 * R2
R3 := SELECTC (R4) OR
R4  R1 * R2
R3  SELECTC (R4)
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17. Union General Form R  S where R, S are Relations Example:
Result contains Tuples from both R and S
Duplications are Removed
The two Operands R, S should be union-compatible
Example:
“find students registered for course C1 or C3”
s#(CNO=‘C1’ (S-C)) s#(CNO=‘C3’ (S-C))
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18. Union Compatibility
Two Relations R1(A1, A2, ..., An) and R2(B1, B2, ..., Bn) are said Union-compatible If and Only If They Have
The Same Number of Attributes
The Domains of Corresponding Attributes are Compatible, i.e., Dom(Ai)=dom(Bi) for i=1, 2, ..., N
Names Do Not Have to be Same!
For Relational Union and Difference Operations, the Operand Relations Must Be Union Compatible
The Resulting Relation for Relational Set Operations
Has the Same Attribute Names as the First Operand Relation R1 (by Convention)
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19. Set Difference General Form R – S where R and S are Relations Example
Result Contains all tuples that are in R, but not in S.
R – S <> S – R
Again, there Must be Compatibility
Example
“Find the students who registered course C1 but not C3” s#(CNO=‘C1’ (S-C)) –s#(CNO=‘C3’ (S-C))
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20. Set Intersection s#(CNO=‘C1’ (S-C)) s#(CNO=‘C3’ (S-C))
General Form
R S
where R and S are Relations
Result Contains all Tuples that are in R and S.
R S = R – (R – S)
Again, there Must be Compatibility
Example
“find the students who registered for both C1 and C3”
s#(CNO=‘C1’ (S-C)) s#(CNO=‘C3’ (S-C))
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21. Union, Difference, Intersection Examples
What are these Other
Three Result Tables?
STUDENT  INSTRUCTOR
STUDENT - INSTRUCTOR
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INSTRUCTOR - STUDENT

22. Cartesian Product Given Relations Cartesian Product R  S
R of Degree k1 and Cardinality card1
S of Degree k2 and Cardinality card2
Cartesian Product
R  S
is a Relation of Degree (k1+ k2) and Consists of Tuples of Degree (k1+ k2) where each Tuple is a Concatenation of one Tuple of R with one Tuple of S
Cardinality of the Result of the Cartesian Product R  S is card1 * card2
What is One Problem with Cartesian Product w.r.t. the Result Set?
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23. Cartesian Product: ExampleB C E F a1 a2 a3 b1
b1b4 c3 c5 c7 e1 e2 f1 f5 A B C E F
R S  a1 a2 a3 b1 b4 c3 c5 c7 e1 e2 f1 f5
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24. Cartesian Product
Given R(A1, …,An) and S(B1,…,Bm), the result of a Cartesian product R  S is a relation of schema
R’(A1, …, An, B1, …, Bm).
Example
“Get a list containing (S#, C#) for all students who live in Storrs but are not registered for the database course”
(S#(city=‘Storrs’(STUDENT)) 
 C# (CNAME=‘Database’(COURSE))) – S#, C#(S-C)
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25. Cartesian Product Example
ENO
ENAME
TITLE E1
J. Doe
Elect. Eng E2
M. Smith
Syst. Anal. E3
A. Lee
Mech. Eng. E4
J. Miller
Programmer E5
B. Casey E6
L. Chu
Elect. Eng. E7
R. Davis E8
J. Jones
EMP
ENO
ENAME
EMP.TITLE
SAL.TITLE
SAL E1
J. Doe
Elect. Eng.
40000
Syst. Anal.
34000
Mech. Eng.
27000
Programmer
24000 E2
M. Smith E3
A. Lee E8
J. Jones
EMP SAL
TITLE
SAL
Elect. Eng.
40000
Syst. Anal.
34000
Mech. Eng.
27000
Programmer
24000
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26. The Join Operation
Used to combine related tuples from two relation into single tuples with some condition
Cartesian product all combination of tuples are included in the result, meanwhile in the join operation, only combination of tuples satisfying the join condition appear in the result
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27. Theta Join (-Join) General Form R  S where
R, S are Relations,
 is a Boolean Expression, called a Join Condition.
A Derivative of Cartesian Product
R  S =  (R  S)
R(A1, A2, ..., Am, B1, B2, ..., Bn) is the Resulting Schema of a -Join over R1 and R2:
R1(A1, A2, ..., Am)  R2 (B1, B2, ..., Bn)
-Join is derived from a Cartesian Product followed by a SELECT.
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28. -Join Condition
A -Join Condition is a Boolean Expression of the form F1 y1 F2 y2 ..., yn-1 Fq (q>=1), where
Fi (i=1,…,q) are Atomic Boolean Expressions of the form
Ai  Bj,
Ai, Bj are Attributes of R1 and R2 Respectively
q is one of the Algorithmic Comparison Operators =, <>, >, <. >=, <=
The Operator yi (i=1,…,n-1) is Either a Logical AND operator  or a logical OR operator 
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29. -Join Example EMP E.TITLE=SAL.TITLE SAL EMP SAL 40000 34000 27000
ENO
ENAME
TITLE E1
J. Doe
Elect. Eng E2
M. Smith
Syst. Anal. E3
A. Lee
Mech. Eng. E4
J. Miller
Programmer E5
B. Casey E6
L. Chu
Elect. Eng. E7
R. Davis E8
J. Jones
EMP
EMP E.TITLE=SAL.TITLE SAL
SAL
40000
34000
27000
24000
ENO
ENAME
TITLE
Elect. Eng.
Analyst
Mech. Eng.
Programmer
Syst. Anal.
SAL.TITLE
Elect. Eng.
Analyst
Mech. Eng.
Programmer
Syst. Anal. E1
J. Doe E2
M. Smith E3
A. Lee
TITLE
SAL
Elect. Eng.
40000
Syst. Anal.
34000
Mech. Eng.
27000
Programmer
24000 E4
J. Miller E5
B. Casey E6
L. Chu E7
R. Davis E8
J. Jones
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30. Other Types of Join Equi-join (EQUIJOIN) Natural Join
The  Expression only Contains one or more Equality Comparisons Involving Attributes from R1 and R2
Natural Join
Denoted as R S
Special Equi-join of Two Relations R and S Over a Set of Attributes Common to both R and S
By Common, it means that each Join Attribute in A has not only Compatible Domains but also the Same Name in both Relations R and S
Equi-join (EQUIJOIN)
The formula F only contains one or more equality comparisons involving attributes from R1 and R2.
Natural join
Equi-join of two relations R and S over a set of attributes (say A) common to both R and S
denoted as R A S.
Difference
In an EQUIJOIN, the join attribute of S appear redundantly in the result relation.
In a NATURAL JOIN, the redundant join attributes of S are eliminated from the result relation. The equality condition is implied.
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31. Examples R S A B C B E a1 a2 a3 b1 b1b4 c3 c5 c7 b1 b5 e1 e2
R R.B=S.B S A
R.B a1 a2 b1 C E c3 c5 e1
S.B
EQUIJOIN
R S
Natural Join
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32. Natural Join
Natural Join Combines Relations on Attributes with the Same Names
STUDENT(S#, SN, CITY, )
SC(S#, C#, G)
Example Query 1:
“list of students with complete course grade info”
STUDENT SC
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33. Natural Join
All Natural Joins can be Expressed by a Combination of Primitive Operators
Example Query 2:
“print all students info (courses taken and grades)”
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34. Natural Join Example EMP EMP SAL Database System-dww ENO ENAME TITLE
J. Doe
Elect. Eng E2
M. Smith
Syst. Anal. E3
A. Lee
Mech. Eng. E4
J. Miller
Programmer E5
B. Casey E6
L. Chu
Elect. Eng. E7
R. Davis E8
J. Jones
EMP
EMP SAL
ENO
ENAME
E.TITLE
SAL E1
J. Doe
Elect. Eng.
70000 E2
M. Smith
Syst. Anal.
80000 E3
A. Lee
Mech. Eng.
56000 E4
J. Miller
Programmer
60000 E5
B.Casey
Syst.Anal
80000
TITLE
SAL
Elect. Eng.
70000
Syst. Anal.
80000
Mech. Eng.
56000
Programmer
60000 E6
L. Chu
Elect.Eng
70000 E7
R.Davis
Mech.Eng
56000 E8
J. Jones
Syst. Anal.
80000
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36. Another Join Example
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37. Quotient (Division) Given Relations The Division of R by S, R ÷ S
R(T,U) of degree r
S(U) of degree s
The Division of R by S, R ÷ S
Results is a Relation of Degree (rs)
Consists of all (rs)-tuples t such that for all s-tuples u in S, the tuple tu is in R.
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38. Division Example
ENO
PNO
PNAME E1 P1
Instrumentation
150000
BUDGET E2 P2
Database Develop.
135000 E3 P4
Maintenance E4 E5 E6 E7 P3
CAD/CAM E8
310000
250000 R
Find the employees who work for both project P1 and project P4? S
PNO
PNAME
BUDGET P1
Instrumentation
150000 P4
Maintenance
310000
R ÷ S
ENO E3
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40. Database System-dww

41. Division: Another Example
“list the S# of students that have taken all of the courses listed in SC”
S# (SCC# (SC))
“list the S# of students who have taken all of the courses taught by instructor Smith”
S# (SCC# (Instructor=‘Smith’(SC)))
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42. Relational Algebra Fundamental Operators
Selection
Projection
Union
Difference
Cartesian Product
Intersection
Join, Equi-join, Natural Join
Quotient (Division)
Fundamental Operators
Derivable from the fundamental operators
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43. All Relational Algebra Operations
A Set of Relational Algebra Operations Is Called a Complete Set, If and Only If
Any Relational Algebra Operator in the Set Cannot be Derived in Terms of a Sequence of Others in Set
Any Relational Algebra Operator Not in the Set Can Be Derived in Terms of a Sequence of Only the Operators in the Set
All the operations discussed so far (Join, Natural Join) can be described as a sequence of only the operations SELECT, PROJECT
, UNION, SET DIFFERENCE, and CARTESIAN PRODUCT.
Hence, the set {s ,P , U, - , X } is called a complete
set of relational algebra operations. Any query
language equivalent to these operations is called
relationally complete.
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44. All Relational Algebra Operations
Important Concepts:
The Set of Algebra Operations {, P , , –, } is a Complete Set of Relational Algebra Operations
Any Query Language Equivalent to These Five Operations is Called Relationally Complete
All the operations discussed so far (Join, Natural Join) can be described as a sequence of only the operations SELECT, PROJECT
, UNION, SET DIFFERENCE, and CARTESIAN PRODUCT.
Hence, the set {s ,P , U, - , X } is called a complete
set of relational algebra operations. Any query
language equivalent to these operations is called
relationally complete.
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45. 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.
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46. Additional Relational Operations (cont.)
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47. Additional Relational Operations (cont.)
Use of the Functional operator 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
DNO ℱCOUNT SSN, AVERAGE Salary (Employee) groups employees by DNO (department number) and computes the count of employees and average salary per department.[ Note: count just counts the number of rows, without removing duplicates]
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50. 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.
Example: 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 .
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.
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51. Additional Relational Operations (cont.)
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52. Additional Relational Operations (cont.)
The OUTER JOIN Operation
In NATURAL JOIN 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.
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.
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53. Additional Relational Operations (cont.)
TEMP  (EMPLOYEE SSN=MGRSSN DEPARTMENT
RESULT  P FNAME, MINIT, LNAME, DNAME (TEMP)
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54. 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 union 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 union compatible.
The attributes that are union compatible are represented only once in the result, and those attributes that are not union compatible from either relation are also kept in the result relation T(X, Y, Z).
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57. Additional Relational Operations (cont.)
OUTER UNION Operations
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)
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58. Database System-dww

59. 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= DNOEMPLOYEE)
RESULT   FNAME, LNAME, ADDRESS (RESEARCH_EMPS)
This query could be specified in other ways; for example, the order of the JOIN and SELECT could be reversed, or the JOIN could be replaced by a NATURAL JOIN after renaming one of the join attributes.
RESULT   FNAME, LNAME, ADDRESS ( DNAME=’Research’ (DEPARTMENT) DNUMBER= DNOEMPLOYEE))
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60. Q2: 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)
RESULT   LNAME, FNAME (( SSN(EMPLOYEE) -  ESSN(DEPENDENT)) * EMPLOYEE)
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61. Q3: List the name of all employees with two or more dependents
T1 (SSN, NO_OF_DEPT) 
ESSN ℱ COUNT DEPENDENT_NAME (DEPENDENT)
T2   NO_OF_DEPS >= 2 (T1)
RESULT   LNAME, FNAME (T2 * EMPLOYEE)
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62. Exercises
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63. From the above schema,
(a) Retrieve the names of employees in department 5 who work more than 10 hours per week on the 'ProductX' project.
(b) List the names of employees who have a dependent with the same first name as themselves.
(c) Find the names of employees that are directly supervised by 'Franklin Wong'. 
(d) For each project, list the project name and the total hours per week (by all employees) spent on that project.
(e) Retrieve the names of employees who work on every project.
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64. Exercises
(f) Retrieve the names of employees who do not work on any project.
(g) For each department, retrieve the department name, and the average salary of
employees working in that department.
(h) Retrieve the average salary of all female employees.
(i) Find the names and addresses of employees who work on at least one project located in Houston but whose department has no location in Houston.
List the last names of department managers who have no dependents.
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65. Retrieve the names of employees in department 5 who work more than 10 hours per week on the 'ProductX' project.
EMPLOYEE_WORKS_ON_ProductX 
(PNAME=‘ProductX’(PROJECT) PNUMBER=PNO(WORKS_ON))
EMPLOYEE_WORKS_MORE_10 
((EMPLOYEE) SSN=ESSN( HOURS>10(EMPLOYEE_WORKS_ON_ProductX)))
RESULT  FNAME,LNAME,ADDRESS( EMPLOYEE_WORKS_MORE_10)
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66. List the names of employees who have a dependent with the same first name as themselves
EMPLOYEE_FNAME_SAME_DEPENDENT_NAME 
((EMPLOYEE) (SSN,FNAME)=(ESSN,DEPENDENT_NAME)(DEPENDENT))
RESULT  FNAME,LNAME(EMPLOYEE_FNAME_SAME_DEPENDENT_NAME)
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