1. Lecture 4: Relational algebra

2. Today’s lecture What’s the (core) relational algebra?
How can we write queries using the relational algebra?
How powerful is the relational algebra?

3. Relational query languages
Query languages allow the manipulation and retrieval of data from a database
The relational model supports simple, powerful query languages
Strong formal foundation
Allows for much (provably correct) optimisation
NOTE: Query languages are not (necessarily) programming languages

4. Formal relational query languages
Two formal query languages
Relational Algebra
Simple ‘operational’ model, useful for expressing execution plans
Relational Calculus
Logical model (‘declarative’), useful for theoretical results
Both languages were introduced by Codd in a series of papers
They have equivalent expressive power
They are the key to understanding SQL query processing!

5. Preliminaries
A query is applied to relation instances, and the result of a query is also a relation instance
Schema of relations are fixed (cf. types)
The query will then execute over any valid instance
The schema of the result can also be determined

6. Example relation instances
A database of boats, sailors, and reservations S2
sid
sname
rating
age 10
Myleene 6 23 22
Tim 8 26 99
Julia
100 20 88
Gavin 21 R1
sid
bid
day 22
101
101001 99
103
111201 S1
sid
sname
rating
age 11
Sue 7 26 22
Tim 8 33
Bob 9 28 55
Kim 10 B1
bid
colour
101
red
102
blue
103
green

7. Core relational algebra
Five basic operator classes:
Selection
Selects a subset of rows
Projection
Picking certain columns
Renaming
Renaming attributes
Set theoretic operations
The familiar operations: union, intersection, difference, …
Products and joins
Combining relations in useful ways

8. Selection Selects rows that satisfy a condition, written R1 = c(R2)
where c is a condition involving the attributes of R2, e.g.
rating>8(S2)
returns the relation instance
sid
sname
rating
age 99
Julia
100 20 88
Gavin 21

9. sname=“Julia”(rating>8(S2))
Selection cont.
Note:
The schema of the result is exactly the same as the schema of the input relation instance
There are no duplicates in the resulting relation instance (why?)
The resulting relation instance can be used as the input for another relational algebra operator, e.g.
sname=“Julia”(rating>8(S2))

10. Projection Deletes fields that are not in the projection list
R1=A(R2)
where A is a list of attributes from the
schema of R2, e.g.
sname,rating(S2)
returns the relation instance
sname
rating
Myleene 6
Tim 8
Julia
100
Gavin

11. Projection cont.
Note:
Projection operator has to eliminate duplicates (why?)
Aside: Real systems don’t normally perform duplicate elimination unless the user explicitly asks for it (why not?)

12. Renaming
R1= A:=B(R2)
Returns a relation instance identical to R2 except that field A is renamed B
For example, sname:=nom(S1)
sid
nom
rating
age 11
Sue 7 26 22
Tim 8 33
Bob 9 28 55
Kim 10

13. Familiar set operations
We have the familiar set-theoretic operators, e.g. , , -
There is a restriction on their input relation instances: they must be union compatible
Same number of fields
Same field names and domains
E.g. S1S2 is valid, but S1R1 is not

14. Cartesian products AB Concatenate every row of A with every row of B
What do we do if A and B have some field names in common?
Several choices, but we’ll simply assume that the resulting duplicate field names will have the suffix 1 and 2

15. Example S1R1 Note! sid.1 sname rating age sid.2 bid day 11 Sue 7 2622
101
101001 99
103
111201
Tim 8 33
Bob 9 28 55
Kim 10

16. Theta join Theoretically, it is a derived operator
R1 Vc c(R1R2)
E.g., S1 Vsid.1<=sid.2R1
sid.1
sname
rating
age
sid.2
bid
day 11
Sue 7 26 22
101
101001 99
103
111201
Tim 8 33
Bob 9 28 55
Kim 10

17. Theta join cont. The result schema is the same as for a cross-product
Sometimes this operator is called a conditional join
Most commonly the condition is an equality on field names, e.g. S1 Vsid.1=sid.2R1

18. Equi- and natural join
Equi-join is a special case of theta join where the condition is equality of field names, e.g. S1 Vsid R1
Natural join is an equi-join on all common fields where the duplicate fields are removed. It is written simply A V B
sid.1
sname
rating
age
sid.2
bid
day 22
Tim 8 26
101
101001

19. Natural join cont.
Note that the common fields appear only once in the resulting relation instance
This operator appears very frequently in real-life queries
It is always implemented directly by the query engine (why?)

20. Find sailors who have reserved all the boats
Division
Not a primitive operator, but useful to express queries such as
Find sailors who have reserved all the boats
Consider the simple case, where relation A has fields x and y, and relation B has field y
A/B is the set of xs (sailors) such that for every y (boat) in B, there is a row (x,y) in A

21. Division cont.
Can you code this up in the relational algebra?

22. Division cont. Can you code this up in the relational algebra?
x’s that are disqualified:
x((x(A)  B) – A)
Thus:
x(A)-x((x(A)  B) – A)

23. Find names of sailors who’ve reserved boat 103
Example 1
Find names of sailors who’ve reserved boat 103
Solution 1: sname(bid=103(Reserves) V Sailors)
Solution 2: sname(bid=103(Reserves V Sailors))
Which is more efficient?
Query
optimisation

24. Find names of sailors who’ve reserved a red boat
Example 2
Find names of sailors who’ve reserved a red boat
DO THIS ON SLIDE MYSELF

25. Find names of sailors who’ve reserved a red boat
Example 2
Find names of sailors who’ve reserved a red boat
sname(colour=“red”(Boats) V Reserves V Sailors)
Better:
sname(sid(bid(colour=“red”(Boats)) V Reserves) V Sailors)
DO THIS ON SLIDE MYSELF

26. Find sailors who’ve reserved a red or a green boat
Example 3
Find sailors who’ve reserved a red or a green boat
MARS Bar challenge!!!
Need to know that SQL has logical AND and OR operators

27. Find sailors who’ve reserved a red or a green boat
Example 3
Find sailors who’ve reserved a red or a green boat
let T = colour=“red”colour=“green”(Boats) in
sname(T V Reserves V Sailors)
MARS Bar challenge!!!
Need to know that SQL has logical AND and OR operators

28. Find sailors who’ve reserved a red and a green boat
Example 4
Find sailors who’ve reserved a red and a green boat
Mars Bar challenge

29. Find sailors who’ve reserved a red and a green boat
Example 4
Find sailors who’ve reserved a red and a green boat
NOTE: Can’t just trivially modify last solution!
let T1 = sid (colour=“red”(Boats) V Reserves)
T2 = sid (colour=“green”(Boats) V Reserves) in
sname((T1  T2) V Sailors)
Mars Bar challenge

30. Find the names of sailors who’ve reserved at least two boats
Example 5
Find the names of sailors who’ve reserved at least two boats
let T = sid.1:=sid (sid.1,sname,bid (Sailors V Reserves)) in
sname.1 (sid.1=sid.2bid.1bid.2(T  T))
Mars bar challenge

31. Find the names of sailors who’ve reserved all boats
Example 6
Find the names of sailors who’ve reserved all boats
let T = sid,bid (Reserves) / bid (Boats) in
sname(T V Sailors)
But this is pretty much given on the division slides

32. Computational limitations
Suppose we have a relation SequelOf of movies and their immediate sequels
We want to compute the relation ‘isFollowedBy’ …
movie
sequel
Naked Gun
Naked Gun 2½
Naked Gun 33 1/3
Rocky
Rocky II
Rocky III
Rocky IV
Rocky V

33. Computational limitations
We could compute
fst,thd(movie:=fst,sequel:=snd(SequelOf)
V movie:=snd,sequel:=thd(SequelOf))
This provides us with sequels-of-sequels
We could write three joins to get sequels-of- sequels-of-sequels and union the results
What about Friday the 13th (9 sequels)? 
In general we need to be able to write an arbitrarily large union…
The relational algebra needs to be extended to handle these sorts of queries

34. Summary You should now understand: The core relational algebra
Operations and semantics
Union compatibility
Computational limitations of the relational algebra
Next lecture: Relational calculus