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Chapter 4
Relational Algebra
p
By relieving the brain of all unnecessary work, a
good notation sets it free to concentrate on more
advanced problems, and, in effect, increases the
mental power of the race.
-- Alfred North Whitehead (1861 - 1947)
Ch4-1: Relational Algebra S. Pisitkasem 1
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.
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Ch4-1: Relational Algebra S. Pisitkasem
Formal Relational Query Languages
Two mathematical Query Languages form the basis for
“real” languages (e.g. SQL), and for implementation:
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-procedural,
declarative.)
Understanding Algebra & Calculus is key to
understanding SQL, query processing!
3
Ch4-1: Relational Algebra S. Pisitkasem
Preliminaries
• A query is applied to relation instances, and the result
of a query is also a relation instance.
– Schemas of input relations for a query are fixed (but
query will run over any legal instance)
– The schema for the result of a given query is also
fixed. It is determined by the definitions of the query
language constructs.
• Positional vs. named-field notation:
– Positional notation easier for formal definitions,
named-field notation more readable.
– Both used in SQL
• Though positional notation is not encouraged
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Ch4-1: Relational Algebra S. Pisitkasem
Relational Algebra: 5 Basic Operations
• Selection ( ) Selects a subset of rows from relations.
(horizontal)
• Projection ( p ) Retains only wanted columns from
(vertical) relation.
• Cross-product ( ) Allows us to combine two relations.
• Set-difference ( — ) Tuples in r1, but not in r2.
• Union ( ) Tuples in r1 and/or in r2.
Since each operation returns a relation, operations can be
composed! (Algebra is “closed”.)
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Ch4-1: Relational Algebra S. Pisitkasem
Example Instances R1 sid bid day
22 101 10/10/96
58 103 11/12/96
S1 sid sname rating age
22 dustin 7 45.0
bid bname color
31 lubber 8 55.5
101 Interlake blue
102 Interlake red 58 rusty 10 35.0
103 Clipper green
104 Marine red
S2 sid sname rating age
Boats
28 yuppy 9 35.0
31 lubber 8 55.5
44 guppy 5 35.0
58 rusty 10 35.0
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Ch4-1: Relational Algebra S. Pisitkasem
Projection
•Examples:
p age(S2) p sname,rating(S2)
• Retains only attributes that are in the “projection list”.
• Schema of result:
– exactly the fields in the projection list, with the same
names that they had in the input relation.
• Projection operator has to eliminate duplicates (How
do they arise? Why remove them?)
– Note: real systems typically don’t do duplicate
elimination unless the user explicitly asks for it.
(Why not?)
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Ch4-1: Relational Algebra S. Pisitkasem
sname rating
Projection yuppy 9
lubber 8
guppy 5
rusty 10
sid sname rating age p sname,rating (S 2)
28 yuppy 9 35.0
31 lubber 8 55.5
44 guppy 5 35.0 age
58 rusty 10 35.0
35.0
S2
55.5
p age(S2) 8
Ch4-1: Relational Algebra S. Pisitkasem
Selection ()
• Selects rows that satisfy selection condition (=,<,>,NOT,AND, OR)
• Result is a relation.
Schema of result is same as that of the input relation.
• Do we need to do duplicate elimination?
sid sname rating age sname rating
28 yuppy 9 35.0
yuppy 9
31 lubber 8 55.5
44 guppy 5 35.0 rusty 10
58 rusty 10 35.0
rating 8(S2) p sname,rating( rating 8(S2))
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Ch4-1: Relational Algebra S. Pisitkasem
Union and Set-Difference
• All of these operations take two input relations,
which must be union-compatible:
– Same number of fields.
– `Corresponding’ fields have the same type.
• For which, if any, is duplicate elimination
required?
10
Ch4-1: Relational Algebra S. Pisitkasem
Union
sid sname rating age sid sname rating age
22 dustin 7 45.0 22 dustin 7 45.0
31 lubber 8 55.5
31 lubber 8 55.5 58 rusty 10 35.0
58 rusty 10 35.0 44 guppy 5 35.0
S1 28 yuppy 9 35.0
sid sname rating age S1 S2
28 yuppy 9 35.0
31 lubber 8 55.5
44 guppy 5 35.0
58 rusty 10 35.0
S2 11
Ch4-1: Relational Algebra S. Pisitkasem
Set Difference
sid sname rating age sid sname rating age
22 dustin 7 45.0 22 dustin 7 45.0
31 lubber 8 55.5 S1 S2
58 rusty 10 35.0
S1
sid sname rating age sid sname rating age
28 yuppy 9 35.0 28 yuppy 9 35.0
31 lubber 8 55.5 44 guppy 5 35.0
44 guppy 5 35.0
58 rusty 10 35.0 S2 – S1
S2
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Ch4-1: Relational Algebra S. Pisitkasem
Cross-Product
• S1 R1: Each row of S1 paired with each
row of R1.
• Q: How many rows in the result?
• Result schema has one field per field of S1
and R1, with field names `inherited’ if
possible.
– May have a naming conflict: Both S1 and
R1 have a field with the same name.
– In this case, can use the renaming
operator:
(C(1 sid1, 5 sid2), S1 R1)
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Ch4-1: Relational Algebra S. Pisitkasem
Cross Product Example
sid bid day sid sname rating age
22 101 10/10/96 22 dustin 7 45.0
58 103 11/12/96 31 lubber 8 55.5
58 rusty 10 35.0
R1 S1
(sid) sname rating age (sid) bid day
22 dustin 7 45.0 22 101 10/10/96
22 dustin 7 45.0 58 103 11/12/96
R1 X S1 = 31 lubber 8 55.5 22 101 10/10/96
31 lubber 8 55.5 58 103 11/12/96
58 rusty 10 35.0 22 101 10/10/96
58 rusty 10 35.0 58 103 11/12/96
14
Ch4-1: Relational Algebra S. Pisitkasem
Compound Operator: Intersection
• In addition to the 5 basic operators, there are several
additional “Compound Operators”
– These add no computational power to the language,
but are useful shorthands.
– Can be expressed solely with the basic ops.
• Intersection takes two input relations, which must be
union-compatible.
• Q: How to express it using basic operators?
R S = R (R S)
15
Ch4-1: Relational Algebra S. Pisitkasem
Intersection
sid sname rating age
22 dustin 7 45.0
31 lubber 8 55.5
58 rusty 10 35.0 sid sname rating age
31 lubber 8 55.5
S1
58 rusty 10 35.0
sid sname rating age
28
31
yuppy
lubber
9
8
35.0
55.5
S1 S2
44 guppy 5 35.0
58 rusty 10 35.0
S2 16
Ch4-1: Relational Algebra S. Pisitkasem
Compound Operator: Join
• Joins are compound operators involving cross
product, selection, and (sometimes) projection.
• Most common type of join is a “natural join” (often
just called “join”). R S conceptually is:
– Compute R S
– Select rows where attributes that appear in both
relations have equal values
– Project all unique atttributes and one copy of each
of the common ones.
• Note: Usually done much more efficiently than this.
• Useful for putting “normalized” relations back
together.
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Ch4-1: Relational Algebra S. Pisitkasem
Natural Join Example
sid bid day sid sname rating age
22 101 10/10/96 22 dustin 7 45.0
58 103 11/12/96 31 lubber 8 55.5
58 rusty 10 35.0
R1
S1
R1 S1 =
sid sname rating age bid day
22 dustin 7 45.0 101 10/10/96
58 rusty 10 35.0 103 11/12/96
18
Ch4-1: Relational Algebra S. Pisitkasem
Other Types of Joins
• Condition Join (or “theta-join”):
R c S = c ( R S)
(sid) sname rating age (sid) bid day
22 dustin 7 45.0 58 103 11/12/96
31 lubber 8 55.5 58 103 11/12/96
S1 < R1.sid R1
S1.sid
• Result schema same as that of cross-product.
• May have fewer tuples than cross-product.
• Equi-Join: Special case: condition c contains only
conjunction of equalities.
19
Ch4-1: Relational Algebra S. Pisitkasem
Semijoin
Applications in distributed database:
• Product(pid,cid,pname,…) at site 1
• Company(cid,cname,…) at site 2
•Query: Find product which price > 1000 which
company producer ay site 2.
price>1000(Product) < Company
• Compute as follows:
T1 = price>1000(Product) site 1
T2 = pcid(T1) site 1
send T2 to site 2 (T2 smaller than T1)
T3 = T2 < Company site 2 (semijoin)
send T3 to site 1 (T3 smaller than Company)
Answer = T1 < T3 site 1 (semijoin)
20
Ch4-1: Relational Algebra S. Pisitkasem
Complex Queries
Product (pid, name, price, category, maker-cid)
Purchase(ssn, storeID ,pid)
Company (cid, name, stock_price,country)
Person(ssn, name, phone number, city)
Store(storeID, name, phone, city)
Note
• in purchase : ssn, storeID and pid are foreign key in Person,
Store and Product
• in Product make-cid is a foreign key in Company
Query
•Find phone numbers of people who bought TV
from ‘Bestbuy’
•Find IT products that somebody bought
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Ch4-1: Relational Algebra S. Pisitkasem
Exercises
Product (pid, name, price, category, maker-cid)
Purchase(ssn, storeID ,pid)
Company (cid, name, stock_price,country)
Person(ssn, name, phone number, city)
Store(storeID, name, phone, city)
Ex#1: Find people who bought IT product
Ex#2: Find names of people who bought Thai products
Ex#3: Find names of people who bought Thai products and did not buy
American products
Ex#4: Find name of people who bought American products from Depo
Ex#5: Which store in his living city which “smith” bought stuff from
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Ch4-1: Relational Algebra S. Pisitkasem
Compound Operator: Division
• Useful for expressing “for all” queries like:
Find sids of sailors who have reserved all boats.
• For A/B attributes of B are subset of attrs of A.
– May need to “project” to make this happen.
• E.g., let A have 2 fields, x and y; B have only field y:
AB = x y
B( x, y A)
A/B contains all tuples (x) such that for every y tuple in
B, there is an xy tuple in A.
23
Ch4-1: Relational Algebra S. Pisitkasem
Examples of Division A/B
sno pno pno pno pno
s1 p1 p2 p2 p1
s1 p2 p4
B1 p2
s1 p3
B2 p4
s1 p4
s2 p1 sno
B3
s2 p2 s1
s3 p2 s2 sno
s4 p2 s3 s1 sno
s4 p4 s4 s4 s1
A A/B1 A/B2 A/B3
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Ch4-1: Relational Algebra S. Pisitkasem
Expressing A/B Using Basic Operators
• Division is not essential op; just a useful shorthand.
– (Also true of joins, but joins are so common that systems
implement joins specially.)
• Idea: For A/B, compute all x values that are not `disqualified’
by some y value in B.
– x value is disqualified if by attaching y value from B, we obtain
an xy tuple that is not in A.
Disqualified x values: p x ((p x ( A) B) A)
A/B: p x ( A) Disqualified x values
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Ch4-1: Relational Algebra S. Pisitkasem
sid bid day
Reserves
Examples 22 101 10/10/96
58 103 11/12/96
sid sname rating age
Sailors 22 dustin 7 45.0
31 lubber 8 55.5
58 rusty 10 35.0
bid bname color
Boats
101 Interlake Blue
102 Interlake Red
103 Clipper Green
104 Marine Red
26
Ch4-1: Relational Algebra S. Pisitkasem
Find names of sailors who’ve reserved boat #103
Solution 1:
p sname ((
Re serves) Sailors)
bid =103
Solution 2:
p sname (
(Re serves Sailors))
bid =103
Which solution is more efficient ?
27
Ch4-1: Relational Algebra S. Pisitkasem
Find names of sailors who’ve reserved a red boat
•Information about boat color only available in Boats;
so need an extra join;
p sname(( color = ‘red’Boats) Reserves Sailors)
p sname
Sailors
color = ‘red’ Reserves
Boats
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Ch4-1: Relational Algebra S. Pisitkasem
Find names of sailors who’ve reserved a red boat
•A more efficient solution
p sname(p sid((p bid color = ‘red’ Boats) Res) Sailors)
p sname
p sid
Sailors
p bid
Reserves
color = ‘red’
Boats
A query optimizer can find this given the first solution!
29
Ch4-1: Relational Algebra S. Pisitkasem
Find names of 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:
(Tempboats,( color = ‘red’ V color = ‘green’ Boats))
p sname (Tempboats Reserves Sailors)
Can also define Tempboats using union! (How?)
30
Ch4-1: Relational Algebra S. Pisitkasem
Find names of sailors who’ve reserved a red and a
green boat
• Previous approach won’t work! Must identify sailors who’ve
reserved red boats, sailors who’ve reserved green boats, then
find the intersection (note that sid is a key for Sailors):
(Tempred , p ( ( Boats) Re serves))
sid color =' red '
(Tempgreen, p (( Boats) Reserves))
sid color =' green'
p sname ((Tempred Tempgreen ) Sailors )
31
Ch4-1: Relational Algebra S. Pisitkasem
Find the names of sailors who’ve
reserved all boats
• Use division; schemas of the input relations to /
must be carefully chosen:
(Tempsids, (p sid, bid Reserves) / (p bid Boats))
p sname (Tempsids Sailors)
To find sailors who’ve reserved all ‘red’ boats:
…. / p bid ( color = ‘red’ Boats)
32
Ch4-1: Relational Algebra S. Pisitkasem
Relational Algebra
• Why bother ? Can write any RA expression directly
in C++/Java , seem easy.
• Two reasons:
- Each Operator admits sophisticated implementations
(think of , C)
- Expressions in relational algebra can be rewritten :
optimized
33
Ch4-1: Relational Algebra S. Pisitkasem
Efficient Implementations of Operators
• (age>=30 AND age<=35)(Employees)
- Method 1 : scan the file,test each employee
- Method 2 : use an index on age
- Which one is better ? Well, depends….
• Employees Relatives
- Iterate over Employees, then over Relatives
- Iterate over Relatives, then over Employees
- Sort Employees, Relatives, do “merge-join”
- “hash-join”
- etc
34
Ch4-1: Relational Algebra S. Pisitkasem
Optimizations
Product (pid, name, price, category, maker-cid)
Purchase(ssn, storeID ,pid)
Person(ssn, name, phone number, city)
• Which is better :
price>100(Product) (Purchase city=sea People)
(price>100(Product) Purchase) city=sea People
• Depends! This is the optimizer’s job…
35
Ch4-1: Relational Algebra S. Pisitkasem
Finally : RA has Limitations!
• Cannot compute “transitive closure”
Name1 Name2 Name3
Fred Mary Father
Mary Joe Cousin
Mary Bill Spouse
Nancy Lou Sister
• Find all direct and indirect relatives of Fred
• Cannot express in RA!!! Need to write C program
36
Ch4-1: Relational Algebra S. Pisitkasem
Summary
• Relational Algebra: a small set of operators
mapping relations to relations
– Operational, in the sense that you specify the
explicit order of operations
– A closed set of operators! Can mix and match.
• Basic ops include: s, p, , , —
• Important compound ops: , , /
37
Ch4-1: Relational Algebra S. Pisitkasem
