1. Chapter 5 Algebraic and Logical Query Languages pp.54 is added Pp 61 updated

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: AB 12 34 12 12

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. ABC 125 346 127 128

6. I simply cut attribute C and get the result AB 12 34 12 12 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 AB 12 I will then have to read the second tuple and compare it against the first. AB 12 34

8. Since they are different I will include this tuple in the result and get AB 12 34 Now I will read the third tuple and compare it to the first. AB 12 34 12

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. ABC 125 346 127 128 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. 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. Union of bags

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) AB 12 34 12 12 * Let S be the relation bellow.(bag) AB 12 34 34 56

17. Then R U S in a bag is simply the two tables written together. AB 12 34 12 12 12 34 34 56

18. Intersection of bags R ∩ S AB 12 34

19. The difference of bags R and S, R-S AB 12 12

20. 5.1 Relational Operations on Bags 5.1.3. 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 ABC 125 346 127 128 σ C>=6 (R)

23. ABC 346 127 128 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 AB 12 12 Bag R Bag S BC 23 45 45

26. Product on bags As we learned earlier each row from R has to be paired with ALL rows in S. AR.BS.BC 1223 1223 1245 1245 1245 1245

27. Product on bags As we learned earlier each row from R has to be paired with ALL rows in S. AR.BS.BC 1223 1223 1245 1245 1245 1245

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 ∞ AB 12 12 Relation R Relation S AB 23 45 45

32. Result of R ∞ S ABC 123 123 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 1.Find the product of the two relations 2. select only these tuples that comply with the condition. 3.Allow duplications It is the symbol ∞ C with condition beneath it.

34. Example theta join AB 12 12 Relation R Relation S AB 23 45 45

35. The theta-join of R and S with the condition R.B <S.B AR.BS.BC 1223 1223 1245 1245 1245 1245 2< 4 hence selected

36. 5.1 Relational Operations on Bags 1. Exercise from previous section (Team 3/7) P52: upload Fig 2.20-21 into your oracle (submit the source codes: create and insert statements to grader 5.1.7 Exercises for Section 5.1 Ex 5.1.1 (3/7) Ex 5.1.4 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 AB 12 34 12 12

39. Apply the duplication eliminator to R δ(R) AB 12 34 ( 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 operators Relation R AB 12 34 12 12 1.SUM(B)= 2+4+2+2=10 2.AVG(A)=(1+3+1+1)/4 3.MIN(A)=1 4.MAX(B)=4 5.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 nameLength Disney123 MGM345 Century fox678 Century fox900 MGM23 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 nameLength MGM23 MGM345 Century fox678 Century fox900 Disney123 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. MOVIETITLEMOVIEYEARSTARNAME Blood Diamond2006Leonardo Dicaprio The Quick and the Dead1995Leonardo Dicaprio Titanic1997Leonardo Dicaprio The Departed2006 Leonardo Dicaprio Body of lies2008Leonardo Dicaprio Inception2010Leonardo Dicaprio Somersault2004Samuel Henry Macbeth2006Samuel Henry Love my Way2006Samuel Henry The Great Raid2005Samuel Henry Terminator Salvation2009Samuel Henry Avatar2009Samuel Henry Perseus2010Samuel Henry Autumn in2000Vera A Farmiga Dust2001Vera A Farmiga Mind the Gap2004 Vera A Farmiga

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 Then Find the minimum of each group Count the number of title in each group

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 Projection Of attribute A AndAdd the value in B and C Rename it to X

59. Extending the projection operator continue Relation R ABC 012 012 345 Relation S AX 03 03 39 B+C=X

60. 5.2 Extended Operators of Relational Algebra 5.2.6 The Sorting Operator τ The sorting operator τ turns a relation into a list of tuples, sorted according to one or more attributes.

61. External Merge Sort How do you sorting 4000 students using only one class room (can hold only 40 students) 1.Fill in the class room with 40 students, let them line up alphabetically 2.So we have 100 sorted group 3.Line up two groups in front of class room 4.One of the two “head” students will walk into class room and sit at the first seat.

62. External Merge Sort 1.Once the 40 seats are full, let them go out. 2.Student continue to walk int until all sets are occupied. 3.Move then out, now we have a group of sorted students. 4.Continue...

63. 5.2 Extended Operators of Relational Algebra 5.2.7 Outerjoins

64. Outer join Youtube link http://www.youtube.com/watch?v=L5sKDSgPt 7M http://www.youtube.com/watch?v=L5sKDSgPt 7M

65. Outerjoins ABC BCD

66. 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:

67. ABC 123 456 789 BCD 2310 2311 6712 If I was doing natural join I would be done here. These are the only matching tuples

68. ABC 123 456 789 BCD 2310 2311 6712 But what about the dangling tuples. In the outer join we have to account for them too

69. Outer joins ABCD 12310 12311 456Null 789 6712

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

71. ABC 123 456 789 BCD 2310 2311 6712

72. Example of left outer join Step one do normal join. That is write all the tuples that pair correctly. ABCD 12310 12311

73. Left outer join Next look at the left relation and see that second and third tuples were not used. We must use them ABCD 12310 12311 456NULL 789 I have use the left relation fully

74. ABC 123 456 789 BCD 2310 2311 6712

75. 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. ABCD 12310 12311

76. 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. ABCD 12310 12311 NULL678

77. 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 5.2.1 (a), (b)

78. 5.3 A Logic for Relations 5.3.1 Predicates and Atoms

79. Relational Algebra Thanks

80. 5.3 A Logic for Relations 5.3.2 Arithmetic Atoms

81. 5.3 A Logic for Relations 5.3.3 Datalog Rules and Queries

82. 5.3 A Logic for Relations 5.3.4 Meaning of Datalog Rules

83. 5.3.5 Extensional and Intensional Predicates

84. 5.3 A Logic for Relations 5.3.6 Datalog Rules Applied to Bags

85. 5.3 A Logic for Relations 5.3.7 Exercises for Section 5.3

86. 5.4 Relational Algebra and Datalog 5.4.1 Boolean Operations

87. 5.4 Relational Algebra and Datalog 5.4.2 Projection

88. 5.4 Relational Algebra and Datalog 5.4.3 Selection

89. 5.4 Relational Algebra and Datalog 5.4.4 Product

90. 5.4 Relational Algebra and Datalog 5.4.5 Joins

91. 5.4 Relational Algebra and Datalog 5.4.6 Simulating Multiple Operations with Datalog

92. 5.4 Relational Algebra and Datalog 5.4.7 Comparison Between Datalog and Relational Algebra

93. 5.4 Relational Algebra and Datalog 5.4.8 Exercises for Section 5.4

94. 5.5 Summary of Chapter

95. 5 5.6 References for Chapter 5