# Relational Algebra by ert554898

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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!
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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?

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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
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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)
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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

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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.

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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)

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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.

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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
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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 ?

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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!
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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?)

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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 )

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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)

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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

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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

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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…

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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

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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: , , /

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Ch4-1: Relational Algebra   S. Pisitkasem

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