1. Relational Algebra
Chapter 4 - part I

2. Relational Query Languages
Query languages: Allow manipulation and retrieval of data from a database.
Relational model supports simple powerful QLs:
Strong formal foundation based on logic.
Allows for much optimization.
Query Languages != Programming languages!
QLs not expected to be “Turing complete”.
QLs not intended to be used for complex calculations.
QLs support easy, efficient access to large data sets. 2

3. Formal Relational Query Languages
Relational Algebra:
More operational, very useful for representing execution plans.
Relational Calculus:
Lets users describe what they want, rather than how to compute it. (Non-operational, rather declarative.)

4. Preliminaries A query is applied to relation instances.
The result of a query is also a relation instance.
Schemas of input relations for a query are fixed ; (but query will run regardless of instance!)
The schema for result of given query is also fixed! Determined by definition of query language . 4

5. Preliminaries Positional field notation e.g., S.1
Named field notation e.g., S.sid 4

6. Example Instances
Sailors
Reserves R1 S1
Sailors S2 5

7. Relational Algebra Basic operations: Additional operations:
Selection ( ) Selects a subset of rows from relation.
Projection ( ) Deletes unwanted columns from relation.
Cartesian-product ( ) Allows us to combine two relations.
Set-difference ( ) Tuples in reln. 1, but not in reln. 2.
Union ( ) Tuples in reln. 1 and in reln. 2.
Additional operations:
Intersection, joins, division, renaming: Not essential, but (very!) useful.
Since each operation returns a relation, operations can be composed! (Algebra is “closed”.) 6

8. Selection
Sailors S2 5

9. Projection
Sailors S2 5

10. Union, Intersection, Set-Difference
All of these operations take two input relations, which must be union-compatible:
Same number of fields.
`Corresponding’ fields have same type.
What is the schema of result? 9

11. Example Instances : Union
Sailors S1
Sailors S2 5

12. Difference Operation
Sailors S1
Sailors S2 5

13. Intersection Operation
Sailors S1
Sailors S2 5

14. Cross-Product (Cartesian Product)
S1 R1 : Each row of S1 is paired with each row of R1.
Reserves R1
Sailors S1 10

15. Cross-Product (Cartesian Product)
S1 R1 : Result schema has one field per field of S1 and R1, with field names `inherited’ if possible.
Conflict: Both S1 and R1 have a field called sid.
Renaming operator: 10

16. Why we need a Join Operator ?
In many cases,
Join = Cross-Product + Select + Project
However :
Cross-product is too large to materialize
Apply Select and Project "On-the-fly" 11

17. Condition Join / Theta Join
Result schema same as that of cross-product.
Fewer tuples than cross-product, more efficient. 11

18. EquiJoin
Equi-Join: A special case of condition join where the condition c contains only equalities.
Result schema similar to cross-product, but only one copy of fields for which equality is specified.
An extra project: PROJECT ( THETA-JOIN) 12

19. Natural Join
Natural Join: Equijoin on all common fields.

20. Relational Algebra: Outer Join
R fojC S, Full Outer Join, where C is equality condition
Also Left Outer Join and Right Outer Join
R.A
R.B A 1 B 2 D 3 F 4 E 5
S.A
S.B A 1 C 2 D 3 E 4
R lojR.A=S.A S A 1 D 3 E 5 4 B 2 - F
R fojR.A=S.A S A 1 D 3 E 5 4 B 2 - F C
R rojR.A=S.A S A 1 D 3 E 5 4 - C 2

21. Relational Algebra: Semi Join
Def: Rows from R ( ONCE !)
if at least one matching row in S.
R ⋉R.A=S.A S
R ⋉R.A=S.A and 2*R.B=S.B S
Use : test equality on just one column
R.A
R.B A 1 B 2 D 3 F 4 E
S.A
S.B A 3 B 4 C 5 E 8
R ⋉ S A 1 3 B 2 4 E 8
R ⋉ S B 2 4 E 8

22. One Application of Semi Join
When R & S reside on different machines
minimizing data movement
Step 1: Obtain all values of S.A and move them to the machine with R
Step 2: Calculate R ⋉ S: All tuples of R that join with at least one S
Step 3: Move R ⋉ S to the machine with S and calculate R ⋈ S.
R.A
R.B A 1 B 2 D 3 F 4 E 5
S.A
S.B A 1 C 2 D 3 E 4
S.A A C D E
R ⋉ S A 1 D 3 E 5
R ⋈R.A=S.A S A 1 D 3 E 5 4

23. Relational Algebra: Inner Join
R.B A 1 B 2 D 3 F 4 E 5
S.A
S.B A 1 C 2 D 3 E 4 7
R ⋉ S A 1 D 3 E 5
Same as Semi Join, except it contains duplicates of R tuples for each match with S.

24. who have reserved all boats. Example : “Reservations / Boats ”
Division
Not supported as a primitive operator, but useful for expressing queries like:
Find sailors
who have reserved all boats.
Example :
“Reservations / Boats ” 13

25. Division Let A have 2 fields x and y; B have only field y: A/B =
A/B contains all x tuples (sailors) such that for every y tuple (boat) in B, there is an xy tuple in A. 13

26. Division Example A = Supplies relation of supplied by suppliers,
B= Parts relation
A/B = suppliers who supply all parts listed in B 13

27. Examples of Division A/B14

28. Expressing A/B Using Basic Operators
Idea: For A/B, compute all x values that are not `disqualified’ by some y value in B.
x is disqualified if by attaching y value from B,
we obtain an xy tuple that is not in A.
Disqualified x values:
A/B: 15

29. A few example queries

30. Example Instances
Sailors
Reserves R1 S1
Sailors S2 5

31. Find names of sailors who’ve reserved boat #103
Solution 1
Solution 2
Solution 3 16

32. Find names of sailors who’ve reserved a red boat
Reserves R1 S1 17

33. Find names of sailors who’ve reserved a red boat
Information about boat color only available in Boats; so need an extra join:
A more efficient solution:
A query optimizer can find this given the first solution! 17

34. Find sailors who’ve reserved a red or a green boat
Can identify all red or green boats, then find sailors who’ve reserved one of these boats:
Can also define Tempboats using union! (How?)
What happens if is replaced by in this query? 18

35. Find sailors who’ve reserved a red and a green boat
Previous approach won’t work!
Must identify sailors who reserved red boats, sailors who’ve reserved green boats, then find their intersection 19

36. Find the names of sailors who’ve reserved all boats
Uses division; schemas of the input relations to / must be carefully chosen:
To find sailors who’ve reserved all ‘Interlake’ boats:
..... 20

37. Relational Algebra : Some More Operators
Beyond Chapter 4

38. Generalized Projection
 sname, (ratings * 2 as myrating) ( Sailors ) 7

39. Aggregation
Use G operator to indicate application of an aggregate function; only one result tuple is being returned for each set of tuples on which function is applied.
COUNT (*)
COUNT ( [DISTINCT] A)
SUM ( [DISTINCT] A)
AVG ( [DISTINCT] A)
MAX (A)
MIN (A) 7

40. Aggregate Operator on Whole Relation
Example:
G sum (rating) as myrating … ( Sailor )
G sum-DISTINCT (rating), avg (age) ( Sailor ) 7

41. Find name and age of the oldest sailor(s)
Rename ( Tmp1, G max (rating) as max-age ( Sailor ))
Project [sname,sage] ( Tmp1 JOIN[max-age=age] Sailor ) Or
Project [sname,sage] ( SELECT [age IN Tmp1] Sailor ) 7

42. Find name and age of the oldest sailor(s)
So far, aggregate operators to all (qualifying) tuples.
Question:
What if want to apply aggregate to each group of tuples ?
Example :
Find the age of the youngest sailor for each rating level.
Example Procedure :
Suppose rating values  {1,2,…,10}, 10 queries:
For i = 1, 2, ... , 10:
G[rating] [min(sage)] (Select[S.rating=i] (Sailors)) 7

43. Motivation for Grouping
SELECT MIN (S.age)
FROM Sailors S
WHERE S.rating = i
For i = 1, 2, ... , 10:
What are the problems with above ?
We may not know how many rating levels exist.
Nor what the rating values for these levels are.
Performance issue (why ?)

44. Aggregate Operator G on Groups
G operator first partitions of relation into groups, then apply aggregate function to each group.
Then one result tuple is being returned per group.
G [age] [ sum (rating) as sum-rating ] ( Sailor ) 7

45. Running Example
Instances of the Sailors and Reserves relations in our examples. R1 S1 S2

46. Find age of the youngest sailor with age 18, for each rating with at least 2 such sailors
PROJECT[min-age](
SELECT[s-count > 1]
( G[S.rating][count(*) as s-count; min(S.age) as min-age ]
(SELECT[S.age>=18](Sailors))))

47. Find age of the youngest sailor with age 18, for each rating with at least 2 such sailors
Sailors instance:

48. Find age of the youngest sailor with age 18, for each rating with at least 2 such sailors.

49. Summary
The relational model has rigorously defined query languages that are simple and powerful.
Relational algebra is operational; useful as internal representation for query evaluation plans.
Several ways of expressing a given query; a query optimizer should choose most efficient version.