Relational Algebra
(continued)
Lecture 5
Relational Algebra
(Summary)
• Basic operations:
–
–
–
–
–
–
Selection (  ) Selects a subset of rows from relation.
Projection (  ) Deletes unwanted columns from relation.
Cross-product (  ) Allows us to combine two relations.
Set-difference ( - ) Tuples in reln. 1, but not in reln. 2.
Union (  ) Tuples either in reln. 1 or in reln. 2 or in both.
Rename (  ) Changes names of the attributes
• Since each operation returns a relation, operations can be
composed ! (Algebra is “closed”.)
• Use of temporary relations recommended.
Additional operators
We define additional operations that do not add
any power to the relational algebra, but that
simplify common queries.
– Natural join
– Conditional Join
– Equi-Join
– Division
All joins are really
special cases of
conditional join
Also, we’ve already seen “Set intersection”:
r  s = r - (r - s)
3
Quick note on notation
bad_customers
good_customers
customer-name
loan-number
customer-name
loan-number
Patty
1234
Seymour
3432
Apu
3421
Marge
3467
Selma
2342
Selma
7625
Ned
4531
Abraham
3597
If we have two or more relations which feature
the same attribute names, we could confuse them.
To prevent this we can use dot notation.
For example
good_customers.loan-number
4
Natural-Join Operation: Motivation
Very often, we have a query and the
answer is not contained in a single
relation. For example, I might wish to
know where Apu banks.
The classic relational algebra way to do
such queries is a cross product, followed
by a selection which tests for equality on
some pair of fields.
borrower
cust-name l-number
l-number branch
Patty
1234
1234
Dublin
Apu
3421
3421
Irvine
borrower.l-number = loan.l-number(borrower x loan)))
While this works…
• it is unintuitive
• it requires a lot of memory
• the notation is cumbersome
loan
cust-name
borrower.l-number
loan.l-number
branch
Patty
1234
1234
Dublin
Patty
1234
3421
Irvine
Apu
3421
1234
Dublin
Apu
3421
3421
Irvine
cust-name
borrower.l-number
loan.l-number
branch
Patty
1234
1234
Dublin
Apu
3421
3421
Irvine
Note that in this example the two relations are the
same size (2 by 2), this does not have to be the case.
So, we have a more intuitive way of achieving the same effect,
the natural join, denoted by the
symbol
5
Natural-Join Operation: Intuition
Natural join combines a cross product and a selection into one
operation. It performs a selection forcing equality on those attributes
that appear in both relation schemes. Duplicates are removed as in
all relation operations.
For the previous example (one attribute in common: “l-number”),
we have…
borrower
loan
= borrower.l-number = loan.l-number(borrower x loan)))
• If the two relations have no attributes in common, then their
natural join is simply their cross product.
• If the two relations have more than one attribute in common,
then the natural join selects only the rows where all pairs of
6
matching attributes match. (let’s see an example on the next slide).
A
l-name
f-name age
Bouvier
Selma
40
Bouvier
Patty
40
Smith
Maggie 2
Both the l-name and the
f-name match, so select.
Only the f-names match,
so don’t select.
Only the l-names match,
so don’t select.
We remove duplicate
attributes…
B
l-name
f-name ID
Bouvier
Selma
1232
Smith
Selma
4423
l-name
f-name
age
l-name
f-name
ID
Bouvier
Selma
40
Bouvier
Selma
1232
Bouvier
Selma
40
Smith
Selma
4423
Bouvier
Patty
2
Bouvier
Selma
1232
Bouvier
Patty
40
Smith
Selma
4423
Smith
Maggie 2
Bouvier
Selma
1232
Smith
Maggie 2
Smith
Selma
4423
l-name
f-name
age
l-name
f-name
ID
Bouvier
Selma
40
Bouvier
Selma
1232
The natural join of A and B
Note that this is just a way to visualize the natural
join, we don’t really have to do the cross product as
in this example
A
B=
l-name
f-name
age
ID
Bouvier
Selma
40
12327
Natural-Join Operation
• Notation: r s
• Let r and s be relation instances on schemas R and S
respectively.The result is a relation on schema R  S which is
obtained by considering each pair of tuples tr from r and ts from s.
• If tr and ts have the same value on each of the attributes in R  S, a
tuple t is added to the result, where
– t has the same value as tr on r
– t has the same value as ts on s
• Example:
R = (A, B, C, D)
S = (E, B, D)
• Result schema = (A, B, C, D, E)
• r s is defined as:
r.A, r.B, r.C, r.D, s.E (r.B = s.B
r.D = s.D (r x s))
8
Natural Join Operation – Example
• Relation instances r, s:
A
B
C
D
B
D
E





1
2
4
1
2





a
a
b
a
b
1
3
1
2
3
a
a
a
b
b





r
r
s
s
A
B
C
D
E
How did we get here?





1
1
1
1
2





a
a
a
a
b





Lets do a trace over the
next few slides…
9
A
B
C
D
B
D
E





1
2
4
1
2





a
a
b
a
b
1
3
1
2
3
a
a
a
b
b





r
s
First we note which attributes the two relations have in common…
10
A
B
C
D
B
D
E





1
2
4
1
2





a
a
b
a
b
1
3
1
2
3
a
a
a
b
b





r
s
A
B
C
D
E


1
1


a
a


There are two rows in s that match our first row in r, (in the relevant
11
attributes) so both are joined to our first row…
A
B
C
D
B
D
E





1
2
4
1
2





a
a
b
a
b
1
3
1
2
3
a
a
a
b
b





r
s
A
B
C
D
E


1
1


a
a


…there are no rows in s that match our second row in r, so do
nothing…
12
A
B
C
D
B
D
E





1
2
4
1
2





a
a
b
a
b
1
3
1
2
3
a
a
a
b
b





r
s
A
B
C
D
E


1
1


a
a


…there are no rows in s that match our third row in r, so do
nothing…
13
A
B
C
D
B
D
E





1
2
4
1
2





a
a
b
a
b
1
3
1
2
3
a
a
a
b
b





r
s
A
B
C
D
E




1
1
1
1




a
a
a
a




There are two rows in s that match our fourth row in r, so both are
14
joined to our fourth row…
A
B
C
D
B
D
E





1
2
4
1
2





a
a
b
a
b
1
3
1
2
3
a
a
a
b
b





r
s
A
B
C
D
E





1
1
1
1
2





a
a
a
a
b





There is one row that matches our fifth row in r,.. so it is joined to
15
our fifth row and we are done!
Natural Join on Sailors Example
sid
22
S1
31
58
sname rating age
dustin
7
45.0
lubber
8
55.5
rusty
10 35.0
sid bid
day
R1 22 101 10/10/96
58 103 11/12/96
S1  R1
sid
22
58
sname
dustin
rusty
rating age
7
45.0
10
35.0
bid
101
103
day
10/10/96
11/12/96
16
Last week we saw…
Query: Find the name of the sailor who reserved boat 101.
Temp  (sidsid1, S1)  (sidsid2, R1)
Result 
(
(Temp))
Sname sid1sid2  bid101

* Note my use of “temporary” relation Temp.
17
Query revisited using natural join
Query: Find the name of the sailor who reserved boat 101.
Result 
(
(S1
Sname bid101
R1))
Or
Result 
(S1
Sname

(R1))
bid101

18
Conditional-Join Operation:
The conditional join is actually the most general type of join. We
introduced the natural join first only because it is more intuitive
and... natural!
Just like natural join, conditional join combines a cross product and
a selection into one operation. However instead of only selecting
rows that have equality on those attributes that appear in both
relation schemes, we allow selection based on any predicate.
r
c
s = c(r x s)
Where c is any predicate on
the attributes of r and/or s
Duplicate rows are removed as always, but duplicate columns are
not removed!
19
Conditional-Join Example:
We want to find all women that are younger than their husbands…
r
l-name
f-name marr-Lic
age
l-name
Simpson
Marge
777
35
Simpson Homer
Lovejoy
Helen
234
38
Flanders
Maude 555
24
Krabappel
Edna
40
r
978
s
Lovejoy
f-name
marr-Lic age
777
36
Timothy 234
36
Simpson Bart
null
r.age < s.age AND r.Marr-Lic = s.Marr-Lic
9
s
r.l-name
r.f-name
r.Marr-Lic
r.age
s.l-name s.f-name
s.marr-Lic
s.age
Simpson
Marge
777
35
Simpson
777
36
Homer
Note we have removed ambiguity of attribute names by using “dot” notation
20
Also note the redundant information in the marr-lic attributes
Equi-Join
• Equi-Join: Special case of conditional join
where the conditions consist only of
equalities.
• Natural Join: Special case of equi-join in
which equalities are specified on ALL fields
having the same names in both relations.
21
Equi-Join
r
l-name
f-name marr-Lic
age
l-name
Simpson
Marge
777
35
Simpson Homer
Lovejoy
Helen
234
38
Flanders
Maude 555
24
Krabappel
Edna
40
r
978
s
Lovejoy
f-name
777
36
Timothy 234
36
Simpson Bart
r.Marr-Lic = s.Marr-Lic
marr-Lic age
null
9
s
r.l-name
r.f-name
Marr-Lic
r.age
s.l-name s.f-name
s.age
Simpson
Marge
777
35
Simpson
Homer
36
Lovejoy
Helen
234
38
Lovejoy
Timothy
36
22
Review on Joins
• All joins combine a cross product and a selection
into one operation.
• Conditional Join
– the selection condition can be of any predicate (e.g.
rating1 > rating2)
• Equi-Join:
– Special case of conditional join where the conditions
consist only of equalities.
• Natural Join
– Special case of equi-join in which equalities are
specified on ALL fields having the same names in both
relations.
23
Banking Examples
branch (branch-name, branch-city, assets)
customer (customer-name, customer-street, customer-only)
account (account-number, branch-name, balance)
loan (loan-number, branch-name, amount)
depositor (customer-name, account-number)
borrower (customer-name, loan-number)
Note that I have not indicated primary keys here for simplicity.
24
Example Queries
• Find all loans of over $1200
amount > 1200 (loan)
loan
amount > 1200 (loan)
“select from the relation loan,
only the rows which have an
amount greater than 1200”
loan-number
1234
3421
2342
4531
branch-name amount
Riverside
1,923.03
Irvine
123.00
Dublin
56.25
Prague
120.03
1234
Riverside
1,923.03
25
Example Queries
• Find the loan number for each loan of an amount greater than $1200
loan-number (amount > 1200 (loan))
“select from the relation loan,
only the rows which have an
amount greater than 1200, then
project out just the
loan_number”
loan
amount > 1200 (loan)
loan-number (amount > 1200 (loan))
loan-number branch-name
amount
1234
Riverside
1,923.03
3421
Irvine
123.00
2342
Dublin
56.25
4531
Prague
120.03
1234
Riverside
1,923.03
1234
26
Example Queries
• Find all loans greater than $1200 or less than $75
amount > 1200  amount < 75(loan)
“select from the relation loan, only
the rows which have an amount
greater than 1200 or an amount less
than 75
loan
amount > 1200  amount < 75(loan)
loan-number branch-name
amount
1234
Riverside
1,923.03
3421
Irvine
123.00
2342
Dublin
56.25
4531
Prague
120.03
1234
Riverside
2342
Dublin
1,923.03
56.25
27
Example Queries
• Find the names of all customers who have a loan, an account, or both, from the
bank
customer-name (borrower)  customer-name (depositor)
borrower
depositor
customer-name
loan-number
customer-name
account-number
Patty
1234
Moe
3467
Apu
3421
Apu
2312
Selma
2342
Patty
9999
Ned
4531
Krusty
3423
customer-name (borrower)
Moe
Apu
customer-name (depositor)
Patty
Patty
Moe
Apu
Krusty
Apu
Selma
Selma
Patty
Ned
Ned
Krusty
28
Example Queries
Note this example is
split over two slides!
Find the names of all customers who have a loan at the Riverside branch.
customer-name (branch-name=“Riverside ” (borrower.loan-number = loan.loan-number(borrower x loan)))
borrower
We retrieve
borrower and
loan…
…we calculate
their cross
product…
loan
customer-name
loan-number
loan-number branch-name
amount
Patty
1234
1234
Riverside
1,923.03
Apu
3421
3421
Irvine
customer-name
borrower.loannumber
loan.loannumber
branch-name
Patty
1234
1234
Riverside
Patty
1234
3421
Irvine
Apu
3421
1234
Riverside
1,923.03
Apu
3421
3421
Irvine
29123.00
123.00
amount
1,923.03
123.00
customer-name (branch-name=“Riverside ” (borrower.loan-number = loan.loan-number(borrower x loan)))
…we calculate
their cross
product…
…we select the
rows where
borrower.loannumber is equal to
loan.loan-number…
…we select the
rows where
branch-name is
equal to
customer-name
borrower.loan loan.loan-number
number
branch-name
Patty
1234
1234
Riverside
Patty
1234
3421
Irvine
Apu
3421
1234
Riverside
Apu
3421
3421
Irvine
customer-name
borrower.loan loan.loan-number
number
branch-name
Patty
1234
1234
Riverside
Apu
3421
3421
Irvine
customer-name
borrower.loan loan.loan-number
number
branch-name
Patty
1234
Riverside
“Riverside”
…we project out
the customer-name.
Patty
1234
amount
1,923.03
123.00
1,923.03
123.00
amount
1,923.03
123.00
amount
1,923.03
30
Now Using Natural Join
Find the names of all customers who have a loan at the Riverside branch.
customer-name (branch-name=“Riverside ” (borrower.loan-number = loan.loan-number(borrower x loan)))
=> customer-name (branch-name=“Riverside ” borrower
We retrieve
borrower and
loan…
1234 in borrower
is matched with
1234 in loan…
3421 in borrower
is matched with
3421 in loan…
The rest is the
same.
loan))
borrower
loan
customer-name
loan-number
loan-number branch-name
amount
Patty
1234
1234
Riverside
1,923.03
Apu
3421
3421
Irvine
customer-name
borrower.loan loan.loan-number
number
branch-name
Patty
1234
1234
Riverside
Apu
3421
3421
Irvine
customer-name
borrower.loan loan.loan-number
number
branch-name
Patty
1234
Riverside
1234
123.00
amount
1,923.03
123.00
amount
31
1,923.03
Example Queries
Note this example is
split over three slides!
Find the largest account balance
...we will need to rename account relation as d...
balance(account) - account.balance(account.balance < d.balance (account x (d,account)))
d
account
We do a rename to
get a “copy” of
account which we
call d…
… next we will do
a cross product…
account- balance
number
accountnumber
Apu
100.30
Apu
100.30
Patty
12.34
Patty
12.34
Lenny
45.34
Lenny
45.34
balance
32
balance(account) - account.balance(account.balance < d.balance (account x (d,account)))
account.accountnumber
… do a cross
product…
…select out all rows
where account.balance
is less than
d.balance…
.. next we project…
account. d.account
balance -number
d.balance
Apu
100.30 Apu
100.30
Apu
100.30 Patty
12.34
Apu
100.30 Lenny
45.34
Patty
12.34 Apu
100.30
Patty
12.34 Patty
12.34
Patty
12.34 Lenny
45.34
Lenny
45.34 Apu
100.30
Lenny
45.34 Patty
12.34
Lenny
45.34 Lenny
45.34
account.accountnumber
account. d.account
balance -number
Patty
12.34 Apu
Patty
12.34 Lenny
Lenny
45.34 Apu
d.balance
100.30
45.34
100.30
33
balance(account) - account.balance(account.balance < d.balance (account x (d,account)))
account.accountnumber
.. next we project out
account.balance…
…then we do a set
difference between it
and the original
account.balance from
the account relation…
… the set difference
leaves us with one
number, the largest
value!
account. d.account
balance -number
Patty
12.34 Apu
Patty
12.34 Lenny
Lenny
45.34 Apu
d.balance
100.30
45.34
100.30
account.
balance
12.34
account
12.34
account- balance
number
45.34
Apu
100.30
Patty
12.34
Lenny
45.34
100.30
34
Now Using Conditional Join
Find the largest account balance
...we will need to rename account relation as d...
balance(account) - account.balance(account.balance < d.balance (account x (d,account)))
 (d,account)
balance(account) - account.balance(account
account.balance < d.balance d)
35
More Examples on Sailors
Relations
Sailors(sid, sname, rating, age)
Boats(bid, bname, color)
Reserves(sid, bid, day)
36
Find names of sailors who’ve reserved boat
#103
• Solution 1:  sname(( bid 103(Reserves)) Sailors)
• Solution 2: Temp1 bid 103(Reserves)
Temp2Temp1 Sailors
Result  sname(Temp2)
• Solution 3:  sname ( bid 103 (Re serves  Sailors))
37
Find names of sailors who’ve reserved a
red boat
• Information about boat color only available
in Boats; so need an extra join:
 sname ((
Boats)  Re serves  Sailors)
color ' red '
• A more efficient solution:
 sname ( ((

Boats)  Re s)  Sailors)
sid bid color ' red '
 A query optimizer can find this given the first solution!
38
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:
Tempboats
color 'red 'color 'green'
(Boats)
Re sult  sname(Tempboats  Re serves  Sailors)

Can also define Tempboats using union! (How?)

What happens if  is replaced by  in this query?
Find 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 
Tempgreen 
sid
sid
((
(Boats)) Reserves)
color 'red '
((
(Boats)) Re serves)
color ' green'
Re sult  sname((Tempred Tempgreen) Sailors)
Consider yet another query
• Find the sailor(s) (sid) who
reserved all the red boats.
R1
sid
22
22
58
B
bid
day
101 10/10/96
103 10/11/96
102 11/12/96
bid
color
101 Red
102 Green
103 Red
Start an attempt
• who reserved what
sid bid
boat:
22 101
S1
(R1)  22 103
sid,bid
58

All the red boats:
S 2 
(
bid color  red
(B)) 
102
bid
101
103
Division Operation
r /s
• Suited to queries that include the phrase “for all”, e.g. Find sailors who have
reserved all red boats.
• Let S1 have 2 fields, x and y; S2 have only field y:
–
S1/S2 =
–
i.e., S1/S2 contains all x tuples (sailors) such that for every y tuple (redboat) in S2, there
is an xy tuple in S1 (i.e, x reserved y).





x | y in S2 ( x,y in S1)






• In general, x and y can be any lists of fields; y is the list of fields in S2, and xy
is the list of fields of S1.
• Let r and s be relations on schemas R and S respectively where
– R = (A1, …, Am, B1, …, Bn)
– S = (B1, …, Bn)
The result of r / s is a relation on schema
R – S = (A1, …, Am)
43
Division Operation – Example
Relations r, s:
A
B
B











1
2
3
1
1
1
3
4
6
1
2
1
2
s
A
r / s:


r
 occurs in the presence of both 1 and 2, so it is returned.
 occurs in the presence of both 1 and 2, so it is returned.
 does not occur in the presence of both 1 and 2, so is ignored.
...
44
Another Division Example
Relations r, s:
A
B
C
D
E
D
E








a
a
a
a
a
a
a
a








a
a
b
a
b
a
b
b
1
1
1
1
3
1
1
1
a
b
1
1
r
s
r /s:
A
B
C


a
a


<, a , > occurs in the presence of both <a,1> and <b,1>, so it is returned.
< , a , > occurs in the presence of both <a,1> and <b,1>, so it is returned.
<, a , > does not occur in the presence of both <a,1> and <b,1>, so it is ignored.
45
Examples of Division A/B
sno
s1
s1
s1
s1
s2
s2
s3
s4
s4
pno
p1
p2
p3
p4
p1
p2
p2
p2
p4
A
pno
p2
B1
pno
p2
p4
B2
pno
p1
p2
p4
B3
sno
s1
s2
s3
s4
sno
s1
s4
sno
s1
A/B1
A/B2
A/B3
Find the sailor(s) who reserved
ALL red boats
• who reserved what boat:
S1
sid,bid
sid
22
22
58
bid
101
103
102
(B)) 
bid
101
103
(R1) 
• All the red boats:
S 2 
(
bid color  red
=> S1/S2
47
Find the names of sailors who’ve reserved
all boats
• Uses division; schemas of the input
relations must be carefully chosen:
Tempsids (
(Reserves))/ (
sid,bid
bid
Result  sname (Tempsids  Sailors)

(Boats))
To find sailors who’ve reserved all ‘Interlake’ boats:
.....
/
bid
(
(Boats))
bname'Interlake'
Some Properties
• Selection
 c1 ...  cn ( R)   c1 ( . . .  cn ( R))
 c1 ( c 2 (R))   c 2 ( c1 (R))
• Projection
 ( R)   l
l
1
1
( . . . (
ln
(R)))
(Cascade)
(Commute)
(Cascade)
where li is a subset of attributes in R, li  li+1
• A projection commutes with a selection that only
uses attributes retained by the projection:
a1,..,an(c(a1,..,an) (R))= c(a1,..,an) (a1,..,an( R ))
where c(a1,…,an) is a condition on attributes a1,…,an
Properties of Joins
Joins
• R
c1
• (R
cS)
• c(R
(S
c2T)
 (S
c
 (R
c1S)
(Associative)
(Commute)
cR)
S)  c(R)
c2T
c
S
where c(a1,…,an) is a condition on attributes only in R
Additional Operator: Outer Join
• Joins: match tuples satisfying a join
condition.
– Tuples without a matching are discarded.
– Tuples with Null value in at least one join
attribute are discarded.
• Outer Joins: useful when we want to keep
tuples in one or both of the involved relations
in a join operation that do not satisfy the join
condition.
– Left outer join
– Right outer join
– Full outer join
• Schema of any Outer Join: Schema of
Natural Join
Additional Operator: Outer Join
• An extension of the join operation that avoids loss of
information.
• Computes the join and then adds tuples from one
relation that does not match tuples in the other
relation to the result of the join.
• Uses null values:
– null signifies that the value is unknown or does not exist
– All comparisons involving null are (roughly speaking) false
by definition.
• Will study precise meaning of comparisons with nulls later
52
Outer Join – Example
• Relation loan
 Relation borrower
loan-number branch-name
L-170
L-230
L-260
Springfield
Shelbyville
Dublin
amount
3000
4000
1700
customer-name loan-number
Simpson
Wiggum
Flanders
L-170
L-230
L-155
53
Outer Join – Example
loan-number branch-name
• Inner Join
L-170
L-230
Springfield
Shelbyville
loan-number
loan
Borrower
• Left Outer Join
loan
L-170
L-230
L-260
amount customer-name
3000
4000
Simpson
Wiggum
branch-name
amount
3000
4000
1700
Springfield
Shelbyville
Dublin
customer-name
loan-number
Simpson
Wiggum
Flanders
L-170
L-230
L-155
borrower
loan-number branch-name
L-170
L-230
L-260
Springfield
Shelbyville
Dublin
amount customer-name
3000
4000
1700
Simpson
Wiggum
null
54
Outer Join – Example
loan-number branch-name
Right Outer Join
loan
borrower
Full Outer Join
loan
L-170
L-230
L-155
borrower
Springfield
Shelbyville
null
3000
4000
null
loan-number
L-170
L-230
L-260
Simpson
Wiggum
Flanders
amount
branch-name
3000
4000
1700
Springfield
Shelbyville
Dublin
customer-name
loan-number
Simpson
Wiggum
Flanders
L-170
L-230
L-155
loan-number branch-name
L-170
L-230
L-260
L-155
amount customer-name
Springfield
Shelbyville
Dublin
null
amount
3000
4000
1700
null
customer-name
Simpson
Wiggum
null
Flanders
55
Full Outer Join
• Full Outer Join S
R : keeps every tuple in
both R and S, if not matching tuples are found in
R (S), then the attributes in R (S) are set to Null
(S
R)  (S
R)
Cost of Relational Algebra Operations
• Given two relations R, S, such as :
Cardinality (R) = n, Cardinality(S) = m
• Cost O(n)
c ( R )
a1,…,an ( R )
• Cost O(n x m)
RxS
 When working with Cross-product (Joins),
keep input relations as small as possible!
Cost of Relational Algebra Operations
Let q= max(m,n),
Set Operations: R  S, R  S, R – S:Cost O(q log q)
Join Operations: R
S, R
Outer Joins Operations:
R
S, R
cS
S, R
S
Best Case : attributes in the join condition include the
primary key in S: Cost O(q log q)
Otherwise : worst case : Cost O(m x n)
When working with these operations, keep input
relations as small as possible
Summary
• The relational model has rigorously defined query
languages that are simple and powerful.
• Relational algebra is more operational; useful as
internal representation for query evaluation plans.
• Several ways of expressing a given query; a query
optimizer should choose the most efficient
version.
• Operations covered: 5 basic operations (selection,
projection, union, set difference, cross product),
rename, joins (natural join, equi-join, conditional
join, outer joins), division
59