1. Relational Algebra

2. Query Languages
Language in which user requests information from the database.
Pure languages form underlying basis of query languages that people use.

3. Relational Algebra Basic operators
Select
Project
Union
Intersection
Set difference
Cartesian product
Rename
Assignment
Join
Division
Generalized Project
Aggregate functions
The operators take one or more relations as inputs and give a new relation as a result.

4. Select Operation Notation:  p(r) p is called the selection predicate
Example of selection:  branch-name=“Perryridge” (account)

5. Select Operation – Example
Account

6. Select Operation – Example
The result is the relation:
Account-number
Branch-name
balance
A-101
perryridge
400 6

7. Select Operation – Example
 Balance >“700” (account)
Account-number
Branch-name
balance
A-201
Brighton
900
A-217
750 7

8. Project Operation Notation: A1, A2, …, Ak (r)
where A1, A2 are attribute names and r is a relation name.
The result is defined as the relation of k columns obtained by erasing the columns that are not listed

9. Project Operation
E.g. To eliminate the branch-name attribute of account account-number, balance (account)
The result relation is:

10. Project Operation Account-number Balance A-101 500 A102 400 A-201 900
700
A-217
750
A-222
A-305
350 10

11. Union Operation Notation: r  s Defined as:
r  s = {t | t  r or t  s}
For r  s to be valid.
1. r, s must have the same number of attributes
2. The attribute domains must be compatible (e.g., 2nd column of r deals with the same type of values as does the 2nd column of s)

12. Union Operation
E.g. to find all customers with either an account or a loan customer-name (depositor)  customer-name (borrower)
Customer-name
Account-no.
Ali
A-101
Mahmood
A-201
Ahmid
A-217
Linda
A-222
Rana
A-305
Customer-name
Loan-no.
Ali
L-11
Kasim
Ahmid
L-25
Linda
Rana
L-34
depositor
borrower

13. Union Operation Customer-name Ali Mahmood Ahmid Linda Rana ….
customer-name (depositor) customer-name (borrower)
Customer-name
Ali
Mahmood
Ahmid
Linda
Rana ….
Customer-name
Ali
Kasim
Ahmid
Linda
Rana
…..

14. Union Operation Customer-name Ali Mahmood Ahmid Linda Rana Kasim
The result relation is:
Customer-name
Ali
Mahmood
Ahmid
Linda
Rana
Kasim

15. Intersection Operation
Notation: r  s
Defined as:
r  s ={ t | t  r and t  s }
Assume:
r, s have the same number of attributes
attributes of r and s are compatible

16. Intersection Operation
E.g. to find all customers with an account and a loan customer-name (depositor)  customer-name (borrower)
Customer-name
Account-no.
Ali
A-101
Mahmood
A-201
Ahmid
A-217
Linda
A-222
Rana
A-305
Customer-name
Loan-no.
Ali
L-11
Kasim
Ahmid
L-25
Linda
Rana
L-34
depositor
borrower

17. Intersection Operation
customer-name (depositor) customer-name (borrower)
Customer-name
Ali
Mahmood
Ahmid
Linda
Rana ….
Customer-name
Ali
Kasim
Ahmid
Linda
Rana
…..

18. Intersection Operation
The result relation is:
Customer-name
Ali
Ahmid
Linda
Rana

19. Set Difference Operation
Notation r – s
Defined as:
r – s = {t | t  r and t  s}
Set differences must be taken between compatible relations.
r and s must have the same number of attributes.
attribute domains of r and s must be compatible.

20. Set Difference Operation – Example
For example, find all customers who have an account and haven't a loan.
customer-name (depositor) - customer-name (borrower)
Customer-name
Account-no.
Ali
A-101
Mahmood
A-201
Ahmid
A-217
Linda
A-222
Rana
A-305
Customer-name
Loan-no.
Ali
L-11
Kasim
Ahmid
L-25
Linda
Rana
L-34
depositor
borrower 20

21. Set Difference Operation – Example
customer-name (depositor) customer-name (borrower)
Customer-name
Ali
Mahmood
Ahmid
Linda
Rana ….
Customer-name
Ali
Kasim
Ahmid
Linda
Rana
…..

22. Set Difference Operation – Example
The result relation is:
Customer-name
Mahmood

23. Set Difference Operation – Example
For example, find all customers who have a loan, but they haven't an account.
customer-name (borrower) - customer-name (depositor)
Customer-name
Account-no.
Ali
A-101
Mahmood
A-201
Ahmid
A-217
Linda
A-222
Rana
A-305
Customer-name
Loan-no.
Ali
L-11
Kasim
Ahmid
L-25
Linda
Rana
L-34
borrower
depositor 23

24. Set Difference Operation – Example
customer-name (borrower) customer-name (depositor)
Customer-name
Ali
Kasim
Ahmid
Linda
Rana
…..
Customer-name
Ali
Mahmood
Ahmid
Linda
Rana ….

25. Set Difference Operation – Example
The result relation is:
Customer-name
Kasim

26. Cartesian-Product Operation
Notation r x s
Defined as:
r x s = {t q | t  r and q  s}
Assume that attributes of r(R) and s(S) are disjoint. (That is, R  S = ).
If attributes of r(R) and s(S) are not disjoint, then renaming must be used.

27. Cartesian-Product Operation
Relations r and s : r s C D E c1 d1 1 c2 d2 2 c3 d3 3 c4 d4 A B a1 1 a2 2

28. Cartesian-Product Operation
The relation r x s is: A B C D E a1 1 c1 d1 c2 d2 2 c3 d3 3 c4 d4 a2 28

29. Rename Operation
Allows us to name, and therefore to refer to, the results of relational-algebra expressions.
Allows us to refer to a relation by more than one name.
Example:
 x (E)
returns the expression E under the name X
If a relational-algebra expression E has a number of attributes n, then
x (A1, A2, …, An) (E)
returns the result of expression E under the name X, and with the attributes renamed to A1, A2, …., An.

30. Assignment Operation
The assignment operation () provides a convenient way to express complex queries.
Write query as a sequential program consisting of
a series of assignments.
followed by an expression whose value is displayed as a result of the query.
Assignment must always be made to a temporary relation variable.

31. Assignment Operation Example: Write
customer-name (borrower)-customer-name (depositor)
temp1  customer-name (borrower) temp2  customer-name (depositor) result = temp1 – temp2
The result to the right of the  is assigned to the relation variable on the left of the .
May use variable in subsequent expressions.

32. Relational Schema for the Banking Enterprise
branch (branch-name, branch-city, assets)
account (account-number, branch-name, balance)
loan (loan-number, branch-name, amount)
depositor (customer-name, account-number)
borrower (customer-name, loan-number)

33. loan-number (amount > 1200 (loan))
Example Queries
Find all loans of over $1200.
amount > 1200 (loan)
Find the loan number for each loan of an amount greater than
$1200.
loan-number (amount > 1200 (loan))

34. Example Queries
Find the names of all customers who have a loan, an account, or both, from the bank.
customer-name (borrower)  customer-name (depositor)
Find the names of all customers who have a loan and an account at bank.
customer-name (borrower)  customer-name (depositor)

35. Example Queries
Find the names of all customers who have a loan at the Perryridge branch.
customer-name (branch-name=“Perryridge”
(borrower.loan-number = loan.loan-number(borrower x loan))) OR
customer-name(loan.loan-number = borrower.loan-number( (branch-name = “Perryridge”(loan)) x borrower)) OR
customer-name
(branch-name=“Perryridge” and borrower.loan-number = loan.loan-number
(borrower x loan))

36. Example Queries
Find the names of all customers who have a loan at the Perryridge branch but they are not depositors.
customer-name (branch-name = “Perryridge”
(borrower.loan-number = loan.loan-number(borrower x loan))) – customer-name(depositor)

37. Example Queries Find the largest account balance
Rename account relation as d
The query is:
balance(account) - account.balance
(account.balance < d.balance (account x rd (account)))

38. Natural-Join Operation
Notation: r s
Natural join () is an operator that is written as (R   S) where R and S are relations. The result of the natural join is the set of all combinations of tuples in R and S that are equal on their common attribute names.

39. Natural-Join Operation
Notation: r s
Example:
R = (A, B, C, D)
S = (E, B, D)
Result schema = (A, B, C, D, E)
r s is defined as: r.A, r.B, r.C, r.D, s.E (r.B = s.B and r.D = s.D (r x s))
It is a special case of the “Cartesian Product”

40. Natural Join Operation – Example
Relations r, s: B D E A B C D 1 3 2 a b          1 2 4    a b r s
r s A B C D E   1 2    a b   

41. Division Operation r  s
Suited to queries that include the phrase “for all”.
The division is a binary operation that is written as R ÷ S. The result consists of the restrictions of tuples in the header of R but not in the header of S, and all their combinations with tuples in S are present in R.

42. Division Operation Example

43. Division Operation – ExampleB B
Relations r, s:      1 2 3 4 6 1 2 s
r  s: A r  

44. Division Operation Example
Relations r, s: A B C D E D E    a    a b 1 3 a b 1 s r A B C
r  s:   a 

45. Example Queries (1)
Find all customers who have an account from at least the “Downtown” and the Uptown” branches.
where CN denotes customer-name and BN denotes branch-name.
Query 1
CN(BN=“Downtown”(depositor account)) 
CN(BN=“Uptown”(depositor account))

46. Example Queries (2)
Find all customers who have an account from at least the “Downtown” and the Uptown” branches.
Query 2
customer-name, branch-name (depositor account)  temp(branch-name) ({(“Downtown”), (“Uptown”)})

47. Example Queries (3)
Find all customers who have an account at all branches located in Brooklyn city.
customer-name, branch-name (depositor account)  branch-name (branch-city = “Brooklyn” (branch))

48. Generalized Projection
Extends the projection operation by allowing arithmetic functions to be used in the projection list.  F1, F2, …, Fn(E)
E is any relational-algebra expression
Each of F1, F2, …, Fn are are arithmetic expressions involving constants and attributes in the schema of E.

49. Generalized Projection
Given relation credit-info(customer-name, limit, credit-balance).
find how much more each person can spend:
customer-name, limit – credit-balance (credit-info)

50. Generalized Projection
Credit-info
customer-name
Limit
credit-balance
Ali
1500
380
Ahmed
2000
1400
Rana
1000
500
Kasim
3500

51. Generalized Projection
The result relation is:
customer-name
Limit-credit-balance
Ali
1220
Ahmed
600
Rana
500
Kasim
2100

52. Aggregate Functions and Operations
Aggregation function takes a collection of values and returns a single value as a result.
avg: average value min: minimum value max: maximum value sum: sum of values count: number of values
Aggregate operation in relational algebra

53. Aggregate Functions and Operations
G1, G2, …, Gn g F1( A1), F2( A2),…, Fn( An) (E)
E is any relational-algebra expression
G1, G2 …, Gn is a list of attributes on which to group (can be empty)
Each Fi is an aggregate function
Each Ai is an attribute name

54. Aggregate Operation – ExampleB C
Relation r:     7 3 10
sum-C
g sum(c) (r) 27

55. Aggregate Operation – Example
account
branch-name
account-number
balance
Perryridge
Brighton
Redwood
A-102
A-201
A-217
A-215
A-222
400
900
750
700
branch-name g sum(balance) (account)
branch-name
balance
Perryridge
Brighton
Redwood
1300
1500
700

56. Aggregate Functions (Cont.)
Result of aggregation does not have a name
Can use rename operation to give it a name
For convenience, we permit renaming as part of aggregate operation
branch-name g sum(balance) as sum-balance (account)
Branch-name
Sum-balance
Perryridge
1300
Brighton
1500
Redwood
700

57. More Examples

58. Relational Instances for the Purchasing System
The Supplier relation:
S-number
S-name
S-city
S100
Ahmed
Amman
S200
Ali
Jarash
S300
Kasim
Irbid
S400
Jasim
Aqaba
S500
Rana

59. The Part relation: P-number P-name Color Price P-city P1 TV Silver 300
Amman P2
Camera
Black
100
Jarash P3
Video
200 P4 PC
400
Irbid P5
Printer
Red P6
Scanner
silver
150 59

60. The shipment relation:
S-number
P-number
Quantity
S100 P1
100 P2
150 P3
200 P4
160
S200
S300
400
S400 80
S500 60

61. Q1- Find The cities for all suppliers .
 S-city (Supplier)

62. Q1- Find The cities for all suppliers .
 S-city (Supplier)
S-city
Amman
Jarash
Irbid
Aqaba
The result relation

63. Q2- Find city for Ahmed.

64. Q2- Find city for Ahmed.
Temp1   S-name = ‘Ahmed’ (Supplier)

65. Q2- Find city for Ahmed.
Temp1   S-name = ‘Ahmed’ (Supplier)
S-number
S-name
S-city
S100
Ahmed
Amman
Temp1

66. Q2- Find city for Ahmed.
Temp1   S-name = ‘Ahmed’ (Supplier)
S-number
S-name
S-city
S100
Ahmed
Amman
Temp1
Result   S-city (Temp1)
S-city
Amman
Result

67. Q3- Find the number and name for suppliers in Amman.

68. Q3- Find the number and name for suppliers in Amman.
Temp1   S-city= ‘Amman’ (Supplier)

69. Q3- Find the number and name for suppliers in Amman.
Temp1   S-city= ‘Amman’ (Supplier)
S-number
S-name
S-city
S100
Ahmed
Amman
S500
Rana
Temp1

70. Q3- Find the number and name for suppliers in Amman.
Temp1   S-city= ‘Amman’ (Supplier)
S-number
S-name
S-city
S100
Ahmed
Amman
S500
Rana
Temp1
Result   S-number, S-name (Temp1)

71. Q3- Find the number and name for suppliers in Amman.
Temp1   S-city= ‘Amman’ (Supplier)
S-number
S-name
S-city
S100
Ahmed
Amman
S500
Rana
Temp1
Result   S-number, S-name (Temp1)
S-number
S-name
S100
Ahmed
S500
Rana
Result

72. Q4- For each part shipped, find the part names and the names of all cities storing these parts.

73. Q4- For each part shipped, find the part names and the names of all cities storing these parts.
Temp1   P-number ( Shipment )

74. Q4- For each part shipped, find the part names and the names of all cities storing these parts.
Temp1   P-number ( Shipment )
P-number P1 P2 P3 P4
Temp1

75. Q4- For each part shipped, find the part names and the names of all cities storing these parts.
Temp1   P-number ( Shipment )
Temp2  Part
Temp1

76. Q4- For each part shipped, find the part names and the names of all cities storing these parts.
Temp1   P-number ( Shipment )
Temp2  Part
Temp1
P-number
P-name
Color
Price
P-city P1 TV
Silver
300
Amman P2
Camera
Black
100
Jarash P3
Video
200 P4 PC
400
Irbid
Temp2

77. Q4- For each part shipped, find the part names and the names of all cities storing these parts.
Temp2  Part
Shipment
Result   P-name, p-city(Temp2)
P-name
P-city TV
Amman
Camera
Jarash
Video PC
Irbid
Result

78. Q5- For each part shipped, find the part numbers and names of all cities supplying the parts.

79. Q5- For each part shipped, find the part numbers and names of all cities supplying the parts.
Temp1  Shipment
Supplier

80. Q5- For each part shipped, find the part numbers and names of all cities supplying the parts.
Temp1  Shipment
Supplier
S-number
P-number
Quantity
S-name
S-city
S100 P1
100
Ahmed
Amman P2
150 P3
200 P4
160
S200
Ali
Jarash
S300
400
Kasim
Irbid
S400
Jasim
Aqaba 80
S500
Rana
Temp1

81. Result   P-number, S-city ( Temp1 )
Q5- For each part shipped, find the part numbers and names of all cities supplying the parts.
Temp1  Shipment
Supplier
P-number
S-city P1
Amman P2 P3 P4
Jarash
Irbid
Aqaba
Result   P-number, S-city ( Temp1 )
Result

82. Q6- Get supplier numbers for suppliers who supply part p2

83. Q6- Get supplier numbers for suppliers who supply part p2
Temp1   p-number = ‘p2’ (Shipment)

84. Q6- Get supplier numbers for suppliers who supply part p2
Temp1   p-number = ‘p2’ (Shipment)
S-number
P-number
Quantity
S100 P2
150
S200
S300
400
S400
Temp1

85. Q6- Get supplier numbers for suppliers who supply part p2
Temp1   P-number = ‘p2’ (Shipment)
Result   S-number ( Temp1 )
S-number
S100
S200
S300
S400
Result

86. Example Queries
Q7- Get names of suppliers who supply at least one silver part.

87. Example Queries
Q7- Get names of suppliers who supply at least one silver part.
Temp1   P-color = ‘silver’ (Part)

88. Example Queries
Q7- Get names of suppliers who supply at least one silver part.
Temp1   P-color = ‘silver’ (Part)
P-number
P-name
Color
Price
P-city P1 TV
Silver
300
Amman P4 PC
400
Irbid P6
Scanner
silver
150
Jarash
Temp1

89. Example Queries
Q7- Get names of suppliers who supply at least one silver part.
Temp1   P-color = ‘silver’ (Part)
Temp2  Shipment
Temp1

90. Example Queries
Q7- Get names of suppliers who supply at least one silver part.
Temp1   P-color = ‘silver’ (Part)
Temp2  Shipment
Temp1
S-number
P-number
Quantity
P-name
Color
Price
P-city
S100 P1
100 TV
Silver
300
Amman P4
160 PC
400
Irbid
S200
200
S400 80
S500
Temp2

91. Example Queries
Q7- Get names of suppliers who supply at least one silver part.
Temp1   P-color = ‘silver’ (Part)
Temp2  Shipment
Temp1
Temp3   S-number (Temp2)
S-number
S100
S200
S400
S500
Temp3

92. Example Queries
Q7- Get names of suppliers who supply at least one silver part.
Temp1   P-color = ‘silver’ (Part)
Temp2  Shipment
Temp1
Temp3   S-number (Temp2)
Temp4  Supplier
Temp3
S-number
S-name
S-city
S100
Ahmed
Amman
S200
Ali
Jarash
S400
Jasim
Aqaba
S500
Rana
Temp4

93. Example Queries
Q7- Get names of suppliers who supply at least one silver part.
Temp1   P-color = ‘silver’ (Part)
Temp2  Shipment
Temp1
Temp3   S-number (Temp2)
Temp4  Supplier
Temp3
Result   S-name (Temp4)
S-name
Ahmed
Ali
Jasim
Rana
Result

94. Example Queries
Q8- Get the names and cities of suppliers who supply parts stored in Irbid.

95. Example Queries
Q8- Get the names and cities of suppliers who supply parts stored in Irbid.
Temp1   P-city = ‘Irbid’ (Part)

96. Example Queries
Q8- Get the names and cities of suppliers who supply parts stored in Irbid.
Temp1   P-city = ‘Irbid’ (Part)
P-number
P-name
Color
Price
P-city P4 PC
Silver
400
Irbid P5
Printer
Red
100
Temp1

97. Example Queries
Q8- Get the names and cities of suppliers who supply parts stored in Irbid.
Temp1   P-city = ‘Irbid’ (Part)
Temp2  Shipment
Temp1
Temp2
S-number
P-number
Quantity
P-name
Color
Price
P-city
S100 P4
160 PC
Silver
400
Irbid P5 50
Printer
Red
100
S400 80
S500

98. Example Queries
Q8- Get the names and cities of suppliers who supply parts stored in Irbid.
Temp1   P-city = ‘Irbid’ (Part)
Temp2  Shipment
Temp1
Temp3   S-number (Temp2)
S-number
S100
S400
S500
Temp3

99. Example Queries
Q8- Get the names and cities of suppliers who supply parts stored in Irbid.
Temp1   P-city = ‘Irbid’ (Part)
Temp2  Shipment
Temp1
Temp3   S-number (Temp2)
Temp4  Supplier
Temp3

100. Example Queries
Q8- Get the names and cities of suppliers who supply parts stored in Irbid.
Temp1   P-city = ‘Irbid’ (Part)
Temp2  Shipment
Temp1
Temp3   S-number (Temp2)
Temp4  Supplier
Temp3
S-number
S-name
S-city
S100
Ahmed
Amman
S400
Jasim
Aqaba
S500
Rana
Temp4

101. Example Queries
Q8- Get the names and cities of suppliers who supply parts stored in Irbid.
Temp1   P-city = ‘Irbid’ (Part)
Temp2  Shipment
Temp1
Temp3   S-number (Temp2)
Temp4  Supplier
Temp3
Result   S-name , S-city (Temp4)

102. Example Queries
Q8- Get the names and cities of suppliers who supply parts stored in Irbid.
Temp1   P-city = ‘Irbid’ (Part)
Temp2  Shipment
Temp1
Temp3   S-number (Temp2)
Temp4  Supplier
Temp3
Result   S-name , S-city (Temp4)
S-name
S-city
Ahmed
Amman
Jasim
Aqaba
Rana
Result

103. Example Queries
Q9- Find supplier names for suppliers who supply all parts.

104. Example Queries
Q9- Find supplier names for suppliers who supply all parts.
Temp1   P-number (Part)

105. Example Queries P-number
Q9- Find supplier names for suppliers who supply all parts.
Temp1   P-number (Part)
P-number P1 P2 P3 P4 P5 P6
Temp1

106. Example Queries
Q9- Find supplier names for suppliers who supply all parts.
Temp1   P-number (Part)
Temp2   S-number , P-number (Shipment)

107. Example Queries
Q9- Find supplier names for suppliers who supply all parts.
Temp1   P-number (Part)
S-number
P-number
S100 P1 P2 P3 P4 P5 P6
S200
S300
S400
S500
Temp2   S-number , P-number (Shipment)
Temp2

108. Example Queries
Q9- Find supplier names for suppliers who supply all parts.
Temp1   P-number (Part)
Temp2   S-number , P-number (Shipment)
Temp3  Temp2  Temp1

109. Example Queries S-number S100
Q9- Find supplier names for suppliers who supply all parts.
Temp1   P-number (Part)
Temp2   S-number , P-number (Shipment)
Temp3  Temp2  Temp1
S-number
S100
Temp3

110. Example Queries
Q9- Find supplier names for suppliers who supply all parts.
Temp1   P-number (Part)
Temp2   S-number , P-number (Shipment)
Temp3  Temp2  Temp1
Temp4  Supplier
Temp3

111. Example Queries
Q9- Find supplier names for suppliers who supply all parts.
Temp1   P-number (Part)
Temp2   S-number , P-number (Shipment)
Temp3  Temp2  Temp1
Temp4  Supplier
Temp3
Temp4
S-number
S-name
S-city
S100
Ahmed
Amman

112. Example Queries
Q9- Find supplier names for suppliers who supply all parts.
Temp1   P-number (Part)
Temp2   S-number , P-number (Shipment)
Temp3  Temp2  Temp1
Temp4  Supplier
Temp3
Result   S-name (Temp4)

113. Example Queries
Q9- Find supplier names for suppliers who supply all parts.
Temp1   P-number (Part)
Temp2   S-number , P-number (Shipment)
Temp3  Temp2  Temp1
Temp4  Supplier
Temp3
Result   S-name (Temp4)
S-name
Ahmed
Result

114. Example Queries
Q10- How many parts they have?

115. Example Queries P-number 6 Q10- How many parts they have?
g count (P-number) (Part)
P-number 6

116. Example Queries
Q11- Find the total quantities supplied by each supplier.

117. Example Queries
Q11- Find the total quantities supplied by each supplier.
S-number g sum (Quantity) (Shipment)

118. Example Queries
Q11- Find the total quantities supplied by each supplier.
S-number g sum (Quantity) (Shipment)
S-number
Quantity
S100
730
S200
350
S300
400
S400
250
S500
100

119. Example Queries
Q11- Find the total quantities supplied by each supplier.
S-number g sum (Quantity) as ( sum-quantity) (Shipment)
S-number
Sum-Quantity
S100
730
S200
350
S300
400
S400
250
S500
100

120. END