1. Database Management Systems
Faculty of Computer Science
Technion – Israel Institute of Technology
Database Management Systems
Course
Lecture 3:
Relational Algebra

2. Outline

3. The Relational Model
A conceptual model for representing data, integrity constraints, and queries
All based on the notion of a schema
DBMS is responsible for translating specifications into the physical environment at hand
Storage in files, caches, indexes
Queries translated to query plans (high-level imperative programs)
Query plans translated to low-level execution over stored data
Student
sid
name
year
861
Alma 2
753
Amir 1
955
Ahuva
Course
cid
topic 23 PL 45 DB 76 OS
Studies
sid
cid
861 23 45
753
955 76

4. Querying: Which Courses Avia Took?ID
name
addr
1234
Avia
Haifa
2345
Boris
Nesher
number
name
lecturer
363 DB
Anna
319 PL
Barak
sID
cNum
grade
1234
363 95
319 82
2345 73
Assembly
...
mov $1, %rax
mov $1, %rdi        
mov $message, %rsi      
mov $13, %rdx              
syscall
mov $60, %rax
xor %rdi, %rdi         QL
SQL
SELECT C.name
FROM S,C,T
WHERE S.name = ‘Avia’ AND S.ID = T.sID
AND T.cNum = C.number
Logic (RC)
{⟨x⟩|∃y,n,z,l,g [S(y,n,'Avia')∧C(z,x,l)∧T(y,z,g)]}
Python
for s in S:
for c in C:
for t in T:
if s.sName==‘Avia’ and
s.ID==t.sID and
t.cNum == c.number:
print c.name
Logic Programming (Datalog)
Q(x)  S(y,n,‘Avia’),C(z,x,l),T(y,z,g)
Algebra (RA)
πC.name(σS.name=‘Avia’, number=cNum, ID=sID(S⨉C⨉T)))

5. The Relational Algebra (RA)
Mathematical query language
Introduced by Edgar Codd
[E. F. Codd: A Relational Model of Data for Large Shared Data Banks. Commun. ACM 13(6): (1970)]
[E. F. Codd: Relational Completeness of Data Base Sublanguages. In: R. Rustin (ed.): Database Systems: 65-98, Prentice Hall and IBM Research Report RJ 987, San Jose, California (1972)]
Since invention, developed and studied by Codd and many others
Edgar F. Codd ( )

6. Names of students who study DB:
RA Example
Names of students who study DB:
name(sid=sid1(sid/sid1Student × sid,cid(cid=cid1(topic=‘DB’Course × cid/cid1Studies))))
Student
sid
name
year
861
Alma 2
753
Amir 1
955
Ahuva
Course
cid
topic 23 PL 45 DB 76 OS
Studies
sid
cid
861 23 45
753
955 76
name
Alma
Amir

7. Why RA?
Understanding the relational algebra is a key understanding central concepts in databases: SQL, query evaluation, query optimization
Tool for building theoretical foundations of various query languages (e.g., SQL)
Tool for developing novel data/query models
See, e.g., [Fagin, Kimelfeld, Reiss, Vansummeren: Document Spanners: A Formal Approach to Information Extraction. J. ACM 62(2): 12 (2015)]

8. RA vs Other QLs
Some subtle (yet important) differences between RA and other languages
Can tables have duplicate records?
(RA vs. SQL)
Are missing (NULL) values allowed?
Is there any order among records?
Is the answer dependent on the domain from which values are taken (not just the DB)?
(RA vs. RC)

9. Relation Schema
A relation schema is a finite sequence of distinct attribute names att with a mapping of each to a domain dom of legal values
Notation: (att1:dom1,…,attk:domk)
Example: (sid:int, name:string, year:int)
sid
name
year

10. Tuples Let s be a relation schema (att1:dom1,…,attk:domk)
A tuple (over s) is a sequence (v1,…,vk) of values vi, where each vi is in domi
That is, a tuple is an element of dom1×...×domk
sid
name
year
861
Alma 2

11. Relations A relation R is a pair (s,r) s is a relation schema
Called the header of R
r is a finite set r of tuples over s
sid
name
year
861
Alma 2
753
Amir 1
955
Ahuva

12. Ignoring Domains
In this lecture we ignore the attribute domains, since they play no special role
(Well, almost; they make a difference for query equivalence, but we do not get there…)
For example, we will write (sid, name, year) instead of (sid:int, name:string, year:int)

13. Notation Notation 1: Notation 2:
Let R be a relation with the header (att1,…,attk)
Let t=(v1,…,vk) be a tuple of R
We refer to vi by t.atti
Notation 2:
Let a1,…,am be attributes among att1,…,attk
We denote by t[a1,…,am] the tuple (t.a1,…,t.am)
t.sid = 861
t.name = Alma
t[sid,name] = (861,Alma)
t[name,sid,year] = (Alma,861,2)
sid
name
year
861
Alma 2
753
Amir 1
955
Ahuva t

14. Databases
A database schema is finite set of relation names, each mapped into a relation schema
Example: Student(sid,name,year) , Course(cid,topic) , Studies(sid,cid)
A database (or instance) over a schema consists of a relation for each relation schema
Student
sid
name
year
861
Alma 2
753
Amir 1
955
Ahuva
Course
cid
topic 23 PL 45 DB 76 OS
Studies
sid
cid
861 23 45
753
955 76

15. What is “Algebra”? An abstract algebra consists of: Each operator:
A class of elements
A collection of operators
Each operator:
Has an arity d
Has a domain of sequences (e1,…,ed) of elements
Maps every sequence in its domain to an element e
The definition of an operator allows for composition:
o1(o2(x),o1(y,o4(x,z)))
Examples:
Ring of integers: (𝕫,{+,·})
Boolean algebra: ({true,false},{∧,∨,¬})
Relational algebra

16. The Relational Algebra
In the relational algebra (RA) the elements are relations
Recall: a relation is a pair (s,r)
RA has 6 primitive operators:
Unary: projection, selection, renaming
Binary: union, difference, Cartesian product
Each of the six is essential (independent)—we cannot define it using the others
We will see what exactly this means and how this can be proved
We commonly allow many more useful operators that can be defined by the primitive ones
For example, intersection via union and difference

17. Outline

18. 6 Primitive (Basic) Operators
Projection ()
Selection ()
Renaming ()
Union (∪)
Difference (\)
Cartesian Product (×)

19. Task (end of this section)
Student
sid
name
year
861
Alma 2
753
Amir 1
955
Ahuva …
Course
cid
topic 23 PL 45 DB 76 OS …
Studies
sid
cid
861 23 45
753 …
Phrase a query that finds the names of students who get private lessons
(i.e., the student takes a course that no one else takes)

20. R = sid,name(R) = year(R) = Projection by Example sid name year 861
Alma 2
753
Amir 1
955
Ahuva
R =
sid
name
861
Alma
753
Amir
955
Ahuva
sid,name(R) =
year 2 1
year(R) =
Fewer tuples (why?)

21. Definition of Projection
Projection is a unary operator of the form A1,…,Ak where each Ai is an attribute name
A projection is parameterized by attributes, so we actually have infinitely many different projection operators
Legal input: relation R in with attributes A1,…,Ak (and possibly others)
A1,…,Ak(R) is the relation S with:
Header (A1,…,Ak)
Tuple set {t[A1,…,Ak] | t ∊ R}
Q: If R has 1000 tuples, how many tuples can A1,…,Ak(R) have?

22. year=1 ⋀ grade>84(R) =
Selection by Example
student
year
course
grade
Alma 1 DB 80 PL 94
Ahuva 2 72
R =
student
year
course
grade
Alma 1 DB 80
Ahuva 2 72
course=‘DB’(R) =
year=1 ⋀ grade>84(R) =
student
year
course
grade
Alma 1 PL 94

23. Definition of Selection
Selection is a unary operator of the form c where c is a logical condition (selection predicate) on attributes
c consists of comparisons and logical connectors (∧,∨,¬)
item1 = item2
price ≥ 500 ∧ price ≤ budget
Legal input: A relation with all the attributes mentioned in the selection predicate
The condition is applied to each tuple in the input, and each violating tuple is filtered out
Formally, c(R) is the relation S with the header of R and the tuple set {t | t ∊ R and t ⊨ c}
Q: If R has 1000 tuples, how many tuples can c(R) have?

24. Variants of Selection
Various variants of RA may allow different languages for specifying selection predicates
e.g., c2>a2+b2 ; name starts with ‘A’, etc.
Common to all predicate formalisms: a predicate applies to a single tuple
Cannot state cross-tuple conditions, e.g.,
“there is another tuple with the same name”
“contains at least 100 tuples”

25. R = year/level(R) = Renaming by Example student year course grade
Alma 1 DB 80 PL 94
Ahuva 2 72
R =
student
level
course
grade
Alma 1 DB 80 PL 94
Ahuva 2 72
year/level(R) =

26. Definition of Renaming
Renaming is a unary operator of the form A/B where A and B are attribute names
Legal input: A relation with a header that contains A and does not contain B
Renaming changes only the header—attribute A becomes B
Formally, A/B(R) is the relation S with
The header of R with A replaced by B
The tuple set of R
Q: If R has 1000 tuples, how many tuples can A/B(R) have?

27. Union and Difference by Example
student
year
Alma 1
Anna
Ahuva 2
student
year
Alma 1
Amir 3
R =
S =
student
year
Alma 1
Anna
Ahuva 2
Amir 3
student
year
Anna 1
Ahuva 2
R∪S =
R \ S =

28. Definition of Union and Difference
Binary operators, interpreted as operations over the tuple sets
Legal input: a pair of relations R and S with the exact same header
We then say that R and S are union compatible
Formally:
R∪S is the relation with the header of R (and S) and the union of the tuple sets
R\S is the relation with the header of R (and S) and the difference between the tuple sets
Q: If each of R and S have 1000 tuples, how many tuples can be in R∪S? R\S?

29. Cartesian Product by Example
sid
name
year
861
Alma 2
753
Amir 1
955
Ahuva
cid
topic 23 PL 45 DB 76 OS
R =
S =
sid
name
year
cid
topic
861
Alma 2 23 PL
753
Amir 1
955
Ahuva 45 DB 76 OS
R×S =

30. Definition of Cartesian Product
Binary operator, similar to set product, but each output pair is combined into a single tuple
Legal input: A pair of relations with disjoint sets of attributes
So how to cross-product Mom(ssn) with Dad(ssn)?
Formally, let R and S have the headers (A1,…,Ak) and (B1,…,Bm), respectively; then R×S is the relation T with:
Header (A1,…,Ak,B1,…,Bm)
Tuple set {r∘s | r ∊ R and s ∊ S}
∘ denotes concatenation
Q: If each of R and S have 1000 tuples, how many tuples can be in R×S?

31. R = S = R×S = Shorthand Notation
For Cartesian product of named relations (e.g., R, S), we actually allow common attributes, and implicitly assume their renaming to name.attribute
sid
name
year
861
Alma 2
753
Amir 1
955
Ahuva
sid
cid
861 23
753 45
R =
S =
R.sid
name
year
S.sid
cid
861
Alma 2 23
753
Amir 1
955
Ahuva 45
R×S =

32. Parentheses Convention
We have defined 3 unary operators and 3 binary operators
It is acceptable to omit the parentheses from o(R) when o is unary
Then, unary operators take precedence over binary ones
Example:
(course=‘DB’(Course)) ×(cid/cid1(Studies) ) becomes course=‘DB’Course × cid/cid1Studies

33. Names of students who study DB:
Composition Example
Names of students who study DB:
name(sid=sid1(sid/sid1Student × sid,cid(cid=cid1(topic=‘DB’Course × cid/cid1Studies))))
Student
sid
name
year
861
Alma 2
753
Amir 1
955
Ahuva
Course
cid
topic 23 PL 45 DB 76 OS
Studies
sid
cid
861 23 45
753
955 76

34. name(sid=sid1(sid/sid1Student × sid,cid(cid=cid1(topic=‘DB’Course × cid/cid1Studies))))
year
861
Alma 2
753
Amir 1
955
Ahuva
Course
cid
topic 23 PL 45 DB 76 OS
Studies
sid
cid
861 23 45
753
955 76

35. name(sid=sid1(sid/sid1Student × sid,cid(cid=cid1(topic=‘DB’Course × cid/cid1Studies))))
861 23 45
753
955 76
Student
sid
name
year
861
Alma 2
753
Amir 1
955
Ahuva
Course
cid
topic 23 PL 45 DB 76 OS
Studies
sid
cid
861 23 45
753
955 76

36. name(sid=sid1(sid/sid1Student × sid,cid(cid=cid1(topic=‘DB’Course × cid/cid1Studies))))45 DB
sid
cid1
861 23 45
753
955 76
Student
sid
name
year
861
Alma 2
753
Amir 1
955
Ahuva
Course
cid
topic 23 PL 45 DB 76 OS
Studies
sid
cid
861 23 45
753
955 76

37. name(sid=sid1(sid/sid1Student × sid,cid(cid=cid1(topic=‘DB’Course × cid/cid1Studies))))45 DB
861 23
753
955 76
Student
sid
name
year
861
Alma 2
753
Amir 1
955
Ahuva
Course
cid
topic 23 PL 45 DB 76 OS
Studies
sid
cid
861 23 45
753
955 76

38. name(sid=sid1(sid/sid1Student × sid,cid(cid=cid1(topic=‘DB’Course × cid/cid1Studies))))45 DB
861
753
Student
sid
name
year
861
Alma 2
753
Amir 1
955
Ahuva
Course
cid
topic 23 PL 45 DB 76 OS
Studies
sid
cid
861 23 45
753
955 76

39. name(sid=sid1(sid/sid1Student × sid,cid(cid=cid1(topic=‘DB’Course × cid/cid1Studies))))45
861
753
Student
sid
name
year
861
Alma 2
753
Amir 1
955
Ahuva
Course
cid
topic 23 PL 45 DB 76 OS
Studies
sid
cid
861 23 45
753
955 76

40. name(sid=sid1(sid/sid1Student × sid,cid(cid=cid1(topic=‘DB’Course × cid/cid1Studies))))
year
861
Alma 2
753
Amir 1
955
Ahuva
cid
sid 45
861
753
Student
sid
name
year
861
Alma 2
753
Amir 1
955
Ahuva
Course
cid
topic 23 PL 45 DB 76 OS
Studies
sid
cid
861 23 45
753
955 76

41. name(sid=sid1(sid/sid1Student × sid,cid(cid=cid1(topic=‘DB’Course × cid/cid1Studies))))
year
cid
sid
861
Alma 2 45
753
Amir 1
955
Ahuva
Student
sid
name
year
861
Alma 2
753
Amir 1
955
Ahuva
Course
cid
topic 23 PL 45 DB 76 OS
Studies
sid
cid
861 23 45
753
955 76

42. name(sid=sid1(sid/sid1Student × sid,cid(cid=cid1(topic=‘DB’Course × cid/cid1Studies))))
year
cid
sid
861
Alma 2 45
753
Amir 1
Student
sid
name
year
861
Alma 2
753
Amir 1
955
Ahuva
Course
cid
topic 23 PL 45 DB 76 OS
Studies
sid
cid
861 23 45
753
955 76

43. name(sid=sid1(sid/sid1Student × sid,cid(cid=cid1(topic=‘DB’Course × cid/cid1Studies))))
Alma
Amir
Student
sid
name
year
861
Alma 2
753
Amir 1
955
Ahuva
Course
cid
topic 23 PL 45 DB 76 OS
Studies
sid
cid
861 23 45
753
955 76

44. (i.e., the student takes a course that no one else takes)
Task
Student
sid
name
year
861
Alma 2
753
Amir 1
955
Ahuva …
Course
cid
topic 23 PL 45 DB 76 OS …
Studies
sid
cid
861 23 45
753 …
Phrase a query that finds the names of students who get private lessons
(i.e., the student takes a course that no one else takes)

45. Outline

46. Implied Operators We now discuss relational operators that are:
Not among the 6 basic operators
Can be expressed in RA (implied)
Very common in practice
Enhancing the available operator set with the implied operators is a good idea!
Easier to write queries
Easier to understand/maintain queries
Easier for DBMS to apply specialized optimizations

47. Outline

48. Joins
Cartesian product is rarely standalone without selection, and is commonly followed by projection
The combination π× is referred to generally as “join”
There are several common cases that apply specific selections and projections, which we introduce here

49. Conditional Join
Binary operator R ⨝c S where c is a condition over the header of R×S
Shorthand notation for:
c(R×S)
Example: R ⨝a=b ⋀ c<d S

50. Theta Join and Equijoin
Theta join is a special case of conditional join ⨝c where c has the form AθB or Aθv where A and B are attributes, v a constant value, and θ a comparison operator
Example: R ⨝c<d S
Equijoin is the special case where c has the form A=B where A and B belong to the left and right operands, respectively
Example: Course⨝name=courseStudies

51. S = T = S⨝sid=studT = Equijoin Example sid name year 861 Alma 2 753
Amir 1
955
Ahuva
stud
course
861 PL DB
762 OS
955
S =
T =
sid
name
year
stud
course
861
Alma 2 PL DB
955
Ahuva OS
S⨝sid=studT =

52. πB1,...,Bm,...,C1...,ClA1=A'1,…,Ak=A'k (R×A1/A'1,…,Ak/A’kS)
Natural Join ⨝
Cartesian product, equality on all common attributes, projection on unique attributes
Formally, R ⨝ S is equivalent to:
πB1,...,Bm,...,C1...,ClA1=A'1,…,Ak=A'k (R×A1/A'1,…,Ak/A’kS)
where:
(B1,...,Bm) is the header of R
A1,...,Ak are the attributes common to R and S
(C1,...,Cl) is the header of S with A1,...,Ak removed
Should we care about which new names are defined by renaming?

53. S = T = S⨝T = Natural Join Example sid name year 861 Alma 2 753 Amir 1
955
Ahuva
sid
course
861 PL DB
762 OS
955
S =
T =
sid
name
year
course
861
Alma 2 PL DB
955
Ahuva OS
S⨝T =

54. where (A1,...,Am) is the header of R
Semijoin
Semijoin of R and S is the restriction of R to the tuples that can naturally join with S
Formally: R⋉S is the operator equivalent to
πA1,...,Am(R⨝S)
where (A1,...,Am) is the header of R

55. S = T = Semijoin Example S⋉T = sid name year 861 Alma 2 753 Amir 1 955
Ahuva
sid
course
861 PL DB
762 OS
955
S =
T =
sid
name
year
861
Alma 2
955
Ahuva
S⋉T =

56. Intersection The usual binary set-theoretic operator ∩
Legal input: a pair of relations that are union compatible (i.e., same header)
Special case of natural join and semijoin
If R and S have the same header, then R ⨝ S and R ⋉ S are equal to R ∩ S

57. Outline

58. Who took all core courses?
Studies
CourseType
sid
student
course
861
Alma DB PL
753
Amir AI
955
Ahuva
course
type DB
core PL AI
elective DC
Who took all core courses?

59. Consider two relations R(X,Y) and S(Y)
Division
Consider two relations R(X,Y) and S(Y)
Here, X and Y are tuples of attributes
R  S is the relation T(X) that contains all the Xs that occur with every Y in S

60. {t[X]∊πXR | t[X,Y]∊R for all s[Y]∊S}
Formal Definition
Legal input: (R,S) such that R has all the attributes of S
RS is the relation T with:
The header of R, with all attributes of S removed
Tuple set
{t[X]∊πXR | t[X,Y]∊R for all s[Y]∊S}
This is an abuse of notation, since the attributes in X need not necessarily come before those of Y

61. ? ? ? ?  =  = = = Questions (RxS)S (RxS)R sid student course 861
Alma DB PL
753
Amir AI
955
Ahuva 
course AI =
course DB PL AI ?  =
Q: If R has 1000 tuples and S has 100 tuples, how many tuples can be in RS? ?
(RxS)S =
Q: If R has 1000 tuples and S has 1001 tuples, how many tuples can be in RS? ?
(RxS)R =

62. Who took all core courses?
Studies
CourseType
sid
student
course
861
Alma DB PL
753
Amir AI
955
Ahuva
course
type DB
core PL AI
elective DC
Who took all core courses?
Studies  πcoursetype=‘core’CourseType

63. RS in Primitive RA πXR \ πX( (πXR × S) \ R ) R(X,Y) S(Y) RS
Each X of R w/ each Y of S
(X,Y) s.t. X in R, Y in S, but (X,Y) not in R
Xs in R where for some Y in S, (X,Y) is not in R
RS

64. Examples of Inexpressible Queries
Some very useful queries cannot be expressed in RA!
follower
followed
Amir
Alma
Ahuva
Anna
Aggregates: How many followers does Ahuva have? How many persons does one follow on average?
Transitive closure: Is there a follower path from Anna to Amir? Is there a cycle?
(How can one prove inexpressiveness?)

65. Outline

66. RA Expressions (Queries)
Let S be a relation schema
Recall: S is a finite set of named relation schemas
An RA expression (RA query) over S is an expression in RA, applied to the relation names of S
For example:
πsid(sid=stud(Student × sid/studStudies))
Student
sid
name
year
Course
cid
topic
Studies
sid
cid

67. Query Result Let S be a database schema Let φ be an RA query over S
Let I be a database over S
The result of evaluating φ over I, denoted φ(I), is the relation obtained by applying φ to the relations of I
That is, every relation name is replaced with the corresponding relation in I

68. Equivalence of RA Expressions
Let S be a database schema, and let φ and ψ be two RA queries over S
We say that φ and ψ are equivalent, denoted φ ≡ ψ, if:
for every database I over S it holds that φ(I) = ψ(I)

69. Who Cares?
Query optimization: we wish to allow DBMS to replace a query with an equivalent one that is more efficient to evaluate
Expressiveness: do different sets of operators “give the same” class of expressible questions?
Examples on R(A,B), S(A,B), T(A,B)
A=‘a’ (R⨝S) ≡ (A=‘a’ R)⨝(A=‘a’ S) (selection push)
πA(R∪S) ≡ πA(R)∪πA(S)
(R⨝S)⨝T ≡ (T⨝S)⨝R
Is it true that R.A/AπR.A(R×S) ≡ πAR ?

70. Q: How does containment relate to equivalence?
Let S be a database schema, and let φ and ψ be two RA queries over S
We say that φ is contained in ψ, denoted φ⊆ψ, if for every instance I over S we have φ(I) ⊆ ψ(I)
Q: How does containment relate to equivalence?

71. ? ? πsid(Student ⨝ Studies)) ≡ πsidStudent ∩ πsidStudies
name
year
Course
cid
topic
Studies
sid
cid ?
πsid(Student ⨝ Studies)) ≡ πsidStudent ∩ πsidStudies ?
πsid(Student ⨝ Studies)) ⊆ πsidStudent ∩ πsidStudies
Q: How do we prove containment? equivalence?

72. ? ? ? πsid(Student ∩ TA)) ≡ πsidStudent ∩ πsidTA
cid
year TA
sid
cid
year ?
πsid(Student ∩ TA)) ≡ πsidStudent ∩ πsidTA ?
πsid(Student ∩ TA)) ⊆ πsidStudent ∩ πsidTA ?
πsid(Student ∩ TA)) ⊇ πsidStudent ∩ πsidTA
Q: How do we prove non-containment? non-equivalence?

73. 6 Primitive Operators Projection () Selection () Renaming ()
Union (∪)
Difference (\)
Cartesian Product (×)
Q: Is this a "good" set of primitives? Could we drop an operator “without losing anything”?

74. Independence
Let o be an RA operator, and let A be a set of RA operators
We say that o is independent of A if o cannot be expressed in A; that is, no expression in A is equivalent to o

75. Independence among Primitives
π σ ρ ×∪-
Theorem: Each of the six primitives is independent of the other five
Proof:
Separate argument for each of the six
Arguments follow a common pattern (next slide)
We will do one operator here (union)

76. Recipe for Proving Independence
Proving that operator o is independent:
Fix a schema S and an instance I over S
Find a property P over relations
Prove that for every expression φ over S that does not use o, the relation φ(I) satisfies P
Such proofs are typically by induction on the size of the expression, since operators compose
Find an expression ψ such that ψ uses o and ψ(I) violates P

77. Independence of Union Fix a schema S and an instance I over S
S: R(A), S(A) I: {R(0), S(1)}
Find a property P over relations
#tuples < 2
Prove that for every expression φ that does not use o, the relation φ(I) satisfies P
Induction base: R and S have #tuples<2
Inductive: If φ1(I) and φ2(I) have #tuples<2, then so do σc(φ1(I)), A(φ1(I)), ρA/B(φ1(I)), φ1(I)×φ2(I), φ1(I)\φ2(I)
Find an expression ψ such that ψ uses o and ψ(I) violates P
ψ=R∪S

78. Outline

79. Task: find Israelis who like albums with dog pictures
Person(ssn,country)
Picture(pid,topic,album)
Likes(ssn,album)
Task: find Israelis who like albums with dog pictures
Which of the equivalent expressions is more efficient to apply?
πssntopic=‘dog’⋀country=‘Israel’ (Likes ⨝ (Person ⨝ Picture)) ×
πssntopic=‘dog’⋀country=‘Israel’ ((Likes ⨝ Person) ⨝ Picture))
πssn ((Likes ⨝ country=‘Israel’Person) ⨝ topic=‘dog’Picture))
πssn ((Likes ⨝ country=‘Israel’Person) ⨝ πalbumtopic=‘dog’Picture)) ∩
(πssncountry=‘Israel’Person) ⨝ (πssn (Likes ⨝ πalbumtopic=‘dog’Picture)))

80. Rules of Thumb for Optimization
Main computational challenges in RA:
Large intermediate results
Join is expensive
Make intermediate results as small as possible before joining (while preserving equivalence)
Apply selection and projection as early as possible (“push select/projection”)
Reorder joins to minimize intermediate relations
Some optimization decisions are “always beneficial” (e.g., push selection) while others require knowledge on the data (e.g., join order)

81. Pushing Projection πX (πZ1R1 ⨝ πZ2R2)
Projection reduces the length of each row, and can substantially reduce the number of rows
Example: Person(ssn,country)
Consider the query πX(R1 ⨝ R2); denote:
Y = R1∩R2 (i.e. the attributes in both R1 and R2)
X1 = X∩R1
X2 = X∩R2
(Note the abuse of notation – we mix attribute sequences with attributes sets)
We would like to push projections into the join, that is:
πX (πZ1R1 ⨝ πZ2R2)
Which Z1 and Z2 can work (equivalence preserved)?

82. Correct Projection Push
Y = R1∩R2
X1 = X∩R1
X2 = X∩R2 ?
πX (R1 ⨝ R2) ≡
πX1(R1) ⨝ πX2(R2) ? ≡
πX (R1 ⨝ R2)
πX1Y(R1) ⨝ πX2Y(R2) ? ≡
πX (R1 ⨝ R2)
πX (πX1Y(R1) ⨝ πX2Y(R2))
When we push projection, we need to retain all the attributes that are used for (1) joining, and (2) operations outside the join

83. Pushing Down the Expression Tree
Y = R1∩R2
X1 = X∩R1
X2 = X∩R2
πX (R1 ⨝ R2)
πX (πX1Y(R1) ⨝ πX2Y(R2)) πX πX ⨝ ⨝
πX1Y
πX2Y R1 R2 R1 R2

84. Selection Push Can we rewrite c(R1 ⨝ R2) as (cR1 ⨝ cR2)?
If all the attributes of C are in R1, then
c(R1 ⨝ R2) ≡ (cR1 ⨝ R2)
If all the attributes of C are in R2, then
c(R1 ⨝ R2) ≡ (R1 ⨝ cR2)
If all the attributes of C in both R1 and R2, then
c(R1 ⨝ R2) ≡ (cR1 ⨝ cR2)
Pushing selection is generally beneficial; we may need some rewriting to get opportunities...

85. Examples of Rewriting Operations
Splitting conjunctions:
c⋀d(R) ≡ c(d(R)) ≡ d(c(R))
Applies to disjunction as well?
Pushing through selection:
c(d(R)) ≡ d(c(R))
Pushing through projection:
c(πA(R)) ≡ πA(c(R))
Assuming that c uses only attributes from A!

86. Pushing Down the Expression Tree
c⋀dπX (R1 ⨝ R2)
πXc⋀d (R1 ⨝ R2)
πXcd (R1 ⨝ R2)
πXc (R1 ⨝ dR2)
c⋀d πX πX πX πX
c⋀d c c ⨝ ⨝ d ⨝ R1 R2 R1 R2 ⨝ R1 d R1 R2 R2

87. Rewriting Joins
Up to order of attributes, the natural join is commutative and associative
Commutative: R ⨝ S ≡ S ⨝ R
Associative: (R ⨝ S) ⨝ T = R ⨝ (S ⨝ T)
Proof: straightforward
So, given an RA query that involves only natural joins, apply the joins in whatever order you want (similarly to addition)
We may need to reorder attributes... nonissue

88. Example Person(ssn,country) Picture(pid,topic,album) Likes(ssn,album)
πssntopic=‘dog’⋀country=‘Israel’ (Likes ⨝ (Person ⨝ Picture))
Reorder joins
πssntopic=‘dog’⋀country=‘Israel’ (Person ⨝ (Likes ⨝ Picture))
Split selection
πssntopic=‘dog’country=‘Israel’ (Person ⨝ (Likes ⨝ Picture))
Push selection
πssntopic=‘dog’ ((country=‘Israel’Person) ⨝ (Likes ⨝ Picture))
Push selection (x2)
πssn ((country=‘Israel’Person) ⨝ (Likes ⨝ (topic=‘dog’Picture)))

89. Example (cont’d)
Person(ssn,country)
Picture(pid,topic,album)
Likes(ssn,album)
πssn ((country=‘Israel’Person) ⨝ (Likes ⨝ (topic=‘dog’Picture)))
Push projection
πssn ((country=‘Israel’Person) ⨝ πssn(Likes ⨝ (topic=‘dog’Picture)))
Push projection
πssn ((country=‘Israel’Person) ⨝ πssn(Likes ⨝(πalbumtopic=‘dog’Picture)))
Push projection
πssn ((πssncountry=‘Israel’Person) ⨝ πssn(Likes ⨝ (πalbumtopic=‘dog’Picture)))
Remove redundant projection
(πssncountry=‘Israel’Person) ⨝ πssn(Likes ⨝ (πalbumtopic=‘dog’Picture))

90. Perspective on Query-Plan Optimization
Algorithms for RA query-plan optimization have been the subject of much research
One of the first and common algorithms is the “Sellinger algorithm” from IBM Almaden
[Patricia G. Selinger, Morton M. Astrahan, Donald D. Chamberlin, Raymond A. Lorie, Thomas G. Price: Access Path Selection in a Relational Database Management System. SIGMOD Conference 1979: 23-34]
Idea: dynamic programming; compute cost & size estimation for every possible subquery, using the costs of smaller subqueries
General toolkit and concepts apply to many data/query models: algebra, equivalence, cost, plan optimization

91. Note on Alternative Approaches
In a recent line of research, several alternative algorithms for RA computation are developed
These algorithm do not construct intermediate results from sub-queries
Rather, compute answers by simultaneously scanning all input relations
More reading:
LogicBlox’s Leapfrog Trie Join
[Todd L. Veldhuizen: Triejoin: A Simple, Worst-Case Optimal Join Algorithm. ICDT 2014: ]
Stanford’s Minesweeper
[Hung Q. Ngo, Dung T. Nguyen, Christopher Re, Atri Rudra: Beyond worst-case analysis for joins with minesweeper. PODS 2014: ]
Not discussed in this course