1. Algebraic and Logical Query Languages pp.54 is added
Chapter 5
Algebraic and Logical Query Languages
pp.54 is added

2. 5.1 Relational Operations on Bags
What is a bag?

3. Bags
What is a bag?
Bag is a relation that may( or may not ) have duplicate tuples.
Example: A B 1 2 3 4

4. Bags continue
Since the tuple (1,2) appear three times this is a bag

5. 5.1.1 Why Bags? Speed Example:
Suppose I want the projection of A and B from the following relation. A B C 1 2 5 3 4 6 7 8

6. I simply cut attribute C and get the result1 2 3 4
I created a table and copy A and B to it. Simple and fast!

7. Now suppose I wanted a set with no duplication
I will have to take the first tuple and put it in A B 1 2
I will then have to read the second tuple and compare it against the first. A B 1 2 3 4

8. Since they are different I will include this tuple in the result and getB 1 2 3 4
Now I will read the third tuple and compare it to the first. A B 1 2 3 4 1 2

9. Since they are the same I will not include this tuple.
The point is that I had to do a lot of work. Each new tuple has to be compared with all other tuples before I could add it to the set. Hence time consuming.

10. Another reason to use bags
Suppose I would like to calculate the average
Of attribute A. Suppose farther that A = revenue in million of dollars. A B C 1 2 5 3 4 6 7 8
Then the average of a set will be 2 and the actual average is 1.5. this is substantial difference.

11. 5.1 Relational Operations on Bags
5.1.2 Union intersection and Difference of bags

12. Union of bags
RUS that same as regular union only welcomes duplications.
Suppose we have two grocery bags. One has two boxes of Oreos the second has five boxes or Oreos. I consolidate the two bags into one bag with 2+5=7 Oreo boxes.

13. Intersection of Bags R ∩ S
If the tuple t appears n times in R and m times in S then the tuple t appear min(m, n) times in the intersection.
That is because the intersection it the common element in R and S and the relations has exactly min(m, n) in common.

14. The difference of R and S
Each occurrence of t in S will cancel one occurrence of t in R. Then output is the “left over” of t.

15. Examples of union, intersection and difference on bags

16. Let R be the relation (bag)1 2 3 4
* Let S be the relation bellow.(bag) A B 1 2 3 4 5 6

17. Then R U S in a bag is simply the two tables written together.1 2 3 4 1 2 3 4 5 6

18. Intersection of bags R ∩ S1 2 3 4

19. The difference of bags R and S, R-S1 2

20. 5.1 Relational Operations on Bags
Projection of Bags It has been explained previously (simply cut)

21. 5.1 Relational Operations on Bags
5.1.4 Selection on Bags

22. Selection on bags
Let R be the bag A B C 1 2 5 3 4 6 7 8
σC>=6(R)

23. Since it is a bag we allow duplication
σC>=6(R) A B C 3 4 6 1 2 7 8
Since it is a bag we allow duplication

24. 5.1 Relational Operations on Bags
5.1.5 Product of Bags

25. Product on bags R X S
Bag R A B 1 2
Bag S A B 2 3 4 5

26. Product on bags
As we learned earlier each row from R has to be paired with ALL rows in S. A
R.B
S.B C 1 2 3 4 5

27. Product on bags
As we learned earlier each row from R has to be paired with ALL rows in S. A
R.B
S.B C 1 2 3 4 5

28. Product on bags continue
Notice that in the above bag we again used the convention for the attribute name. B appear twice so we call it R.B and S.B.
Equivalent Relation

29. 5.1 Relational Operations on Bags
5.1.6 Joins of Bags

30. Joins of bags ∞
We compare each tuple of one relation with each tuple of the other, decide whether or not this pair of tuples joins successfully, and if so we put the resulting tuple in the answer. When constructing the answer we permit duplication.

31. Example of Joins in bags ∞
Relation R A B 1 2
Relation S A B 2 3 4 5

32. Result of R ∞ S A B C 1 2 3
Please notice that unlike the product we do not write B for each relation. We are “naturally” joining the relations. Think of it as the transitive rule.

33. Theta-join in bags Find the product of the two relations
select only these tuples that comply with the condition.
Allow duplications
It is the symbol ∞C with condition beneath it.

34. Example theta join
Relation R A B 1 2
Relation S A B 2 3 4 5

35. The theta-join of R and S with the condition R.B <S.B1 2 3 4 5
2< 4 hence selected

36. 5.1 Relational Operations on Bags
1. Exercise from previous section
(Team 3/7) P52: upload Fig into your oracle (submit the source codes: create and insert statements to grader
5.1.7 Exercises for Section 5.1
Ex (3/7)
Ex Assigned in the Class
List them into the algebraic law Table

37. 5.2 Extended Operators of Relational Algebra
5.2.1 Duplicate Elimination

38. The duplicate-elimination operator δ
Turns a bag into set.
Eliminate all but one copy of each tuple.
Relation R A B 1 2 3 4

39. Apply the duplication eliminator to R δ(R)B 1 2 3 4
( δ is the Greek letter Delta)

40. 5.2 Extended Operators of Relational Algebra
5.2.2 Aggregation Operators

41. Aggregation operators
Aggregation operators apply to attributes (columns ) of relations. Example of aggregation operators are sums and averages.

42. Example of aggregation operatorsB 1 2 3 4
Relation R
SUM(B)= =10
AVG(A)=( )/4
MIN(A)=1
MAX(B)=4
COUNT(A)=4  number of elements in A

43. 5.2 Extended Operators of Relational Algebra
5.2.3 Grouping

44. Grouping
Grouping of tuples according to their value in one or more attributes has the effect of partitioning the tuples of a relation into groups.

45. Example of grouping
Studio name
Length
Disney
123
MGM
345
Century fox
678
900 23
Suppose we use the aggregation, sum(length). This aggregation will give us the sum of the whole column.

46. Example of grouping continue
but suppose we want to know the total umber of minutes of movies produced by each studio.
Then we must have sub tables within the table. Each sub table represent a studio.
We will do that by grouping by studio name.
Now we can apply the aggregation operator sum( length) to each group.

47. Example of grouping continue
Studio name
Length
MGM 23
345
Century fox
678
900
Disney
123
Now the table is grouped by studio name and we can apply the aggregation operator sum(length)

48. 5.2 Extended Operators of Relational Algebra
5.2.4 The Grouping Operator

49. The grouping operator ϒ
Given the schema
StarsIn(title, year, StarName)
We would like to find the starName of each star who appeared in at least three movies and earliest year in which they appear.
How can we approach this problem?

50. Grouping operator continue
We must first group by StarName. It is very intuitive. We want to partition the table into stars and then we can do all the tests for each star
In relational algebra we write ϒ StarName
Bellow is the table grouped by starName
Group by

51. MOVIETITLE MOVIEYEAR STARNAME
Blood Diamond
2006
Leonardo Dicaprio
The Quick and the Dead
1995
Titanic
1997
The Departed
Body of lies
2008
Inception
2010
Somersault
2004
Samuel Henry
Macbeth
Love my Way
The Great Raid
2005
Terminator Salvation
2009
Avatar
Perseus
Autumn in
2000
Vera A Farmiga
Dust
2001
Mind the Gap

52. Notice that in the above table there are three groups one for each Star.
Now for each group we are interested in the first year in which the Star appeared, and we would like to know if he played in 3 or more movies.
We will use the aggregations min(year) and count(title)>=3

53. The grouping operator continue
How these aggregations works?
In each group separately we look for the min year
In each group we look for the number of titles in this group.
If the number of titles in a group is grate then 3, then this will be sent to the output otherwise this group is eliminated.

54. Final statement in the grouping operator
ϒ starName, min(year)->minYear, count(title)->ctTitle(StarsIn)
Group by
Find the minimum of each group
Count the number of title in each group
Then

55. Final statement in the grouping operator
 starName (ctTitle>3(ϒ starName, min(year)->minYear, count(title)->ctTitle(StarsIn)))
See Fig 5.5 for tree expresion

56. Final statement in the grouping operator
 starName (ctTitle>3(ϒ starName, min(year)->minYear, count(title)->ctTitle(StarsIn)))

57. 5.2 Extended Operators of Relational Algebra
5.2.5 Extending the Projection Operator

58. Extending the projection operator
We can include, renaming and arithmetic operators in projection.
Example:
π A, B+C-->X
Rename it to X
Projection
Of attribute A
And
Add the value in B and C

59. Extending the projection operator continue
Relation R A B C 1 2 3 4 5
B+C=X
Relation S A X 3 9

60. 5.2 Extended Operators of Relational Algebra
5.2.6 The Sorting Operator

61. Sorting operator τ
The sorting operator turns a relation into a list of tuples, sorted according to one or more attributes.

62. 5.2 Extended Operators of Relational Algebra
5.2.7 Outerjoins

63. Outer join
Youtube link

64. Outerjoins A B C B C D

65. Simple outer join
First find all tuple that agree and pair them. Notice that unlike product the tuples that matches do not repeat.
Next we deal with tuples that do not agree. We call these dangling tuples.
Add the dangling tuples but what ever is messing add null.
example:

66. A B C 1 2 3 4 5 6 7 8 9 B C D 2 3 10 11 6 7 12
If I was doing natural join I would be done here. These are the only matching tuples

67. A B C 1 2 3 4 5 6 7 8 9 B C D 2 3 10 11 6 7 12
But what about the dangling tuples. In the outer join we have to account for them too

68. Outer joins A B C D 1 2 3 10 11 4 5 6
Null 7 8 9 12

69. Left outer join
The easier way of thinking of it is that we must keep all the tuples from the left relation.

70. A B C 1 2 3 4 5 6 7 8 9 B C D 2 3 10 11 6 7 12

71. Example of left outer join
Step one do normal join. That is write all the tuples that pair correctly. A B C D 1 2 3 10 11

72. Left outer join
Next look at the left relation and see that second and third tuples were not used. We must use them A B C D 1 2 3 10 11 4 5 6
NULL 7 8 9
I have use the left relation fully

73. A B C 1 2 3 4 5 6 7 8 9 B C D 2 3 10 11 6 7 12

74. Example of right outer join
Step one do natural join. That is write all the tuples that pair correctly. Same exact step as the left outer join. I actually copied and paste it. A B C D 1 2 3 10 11

75. Example right outer join continue
Now look for the tuple that were not used in the right relation. That is the third tuple. Add this tuple to complete the right outer join. A B C D 1 2 3 10 11
NULL 6 7 8

76. 5.2 Extended Operators of Relational Algebra
5.2.8 Exercises for Section 5.2
1. Show the commutate law for Cartesian Product by example in pp.25
2. Exercise (a), (b)

77. 5.3 A Logic for Relations
5.3.1 Predicates and Atoms

78. Relational Algebra
Thanks

79. 5.3 A Logic for Relations
5.3.2 Arithmetic Atoms

80. 5.3 A Logic for Relations
5.3.3 Datalog Rules and Queries

81. 5.3 A Logic for Relations
5.3.4 Meaning of Datalog Rules

82. 5.3.5 Extensional and Intensional Predicates

83. 5.3 A Logic for Relations
5.3.6 Datalog Rules Applied to Bags

84. 5.3 A Logic for Relations
5.3.7 Exercises for Section 5.3

85. 5.4 Relational Algebra and Datalog
5.4.1 Boolean Operations

86. 5.4 Relational Algebra and Datalog
5.4.2 Projection

87. 5.4 Relational Algebra and Datalog
5.4.3 Selection

88. 5.4 Relational Algebra and Datalog
5.4.4 Product

89. 5.4 Relational Algebra and Datalog
5.4.5 Joins

90. 5.4 Relational Algebra and Datalog
5.4.6 Simulating Multiple Operations with Datalog

91. 5.4 Relational Algebra and Datalog
5.4.7 Comparison Between Datalog and Relational Algebra

92. 5.4 Relational Algebra and Datalog
5.4.8 Exercises for Section 5.4

93. 5.5 Summary of Chapter

94. 5 5.6 References for Chapter 5