1. Relations and Their Properties
Sunday, April 23, 2017
Relations and Their Properties
Section 8.1
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2. Review: Cartesian Products
Sunday, April 23, 2017
Review: Cartesian Products
Definition: An ordered n-tuple (a1, a2, …, an) is the ordered collection that has a1 as its first element, a2 as its second element…, and an as its nth element.
2-tuples are called ordered pairs.
(a,b) = (c,d) if and only if a=c and b=d
(a,b) ≠ (b,a) unless a=b
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3. Sunday, April 23, 2017
Cartesian Product …
Definition: The Cartesian product of the sets A1, A2, …, An, denoted A1×A2×… ×An, is the set of ordered n-tuples (a1, a2, …, an), where ai belongs to set Ai for i =1, 2, …, n.
A1×A2× …×An = {(a1, a2, …, an) | aiAi for i=1, 2, …, n}
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4. Sunday, April 23, 2017
Binary Relations
Definition: Let A and B be sets. A binary relation R from A to B is a subset of the Cartesian product AB.
Notation: xRy means (x,y)R, and
x is said to be related to y under R.
xRy denotes (x,y)R.
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5. Sunday, April 23, 2017
Example
Let A be the set of all cities, and let B be the set of all states in the U.S.
Define the relation R from A to B by:
(a,b)R if city a is in state b.
Some ordered pairs belonging to relation R:
R= {(East Lansing, Michigan), (Boulder, Colorado),
(Chicago, Illinois), (Columbus, Ohio), …}
(East Lansing)R(Michigan), (Boulder)R(Colorado), …
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6. Another Example A= {a, b, c, d } B = {1, 2, 3}
Sunday, April 23, 2017
Another Example c b d 2 3 1 a
A= {a, b, c, d } B = {1, 2, 3}
R = {(a,2), (b,1), (b, 2), (c,3), (d,2)}. R 1 2 3 a x b c d
Is this relation a function?
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7. Functions As Relations
Sunday, April 23, 2017
Functions As Relations
Recall that: Let f: A → B be a function. The graph of f is the set of ordered pairs
Gf = {(x, f (x)) | x  A}.
The graph of a function from A to B is a relation from A to B.
Conversely, if R is a relation from A to B such that every element in A is the first element of exactly one ordered pair of R, then a function can be defined with R as its graph.
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8. Sunday, April 23, 2017
Relations on a Set
Definition: A relation on a set A is a relation from A to A, that is, a subset of AA.
Example: Let A = {1, 2, 3, 4}, and
R = {(a,b) | a evenly divides b}.
What are the elements of R?
R = {(1,1), (1,2), (1,3), (1,4),
(2,2), (2,4), (3,3), (4,4)}
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9. Sunday, April 23, 2017
Example
R can be displayed graphically and in tabular form/zero-one matrix form: R
1 x x x x
2 x x x x 3 2 4 1
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10. Sunday, April 23, 2017
Relations on a Set…
Let A be a set with n elements. How many distinct relations are there on A?
Answer: The number of relations is the same as the number of subsets of AA. There are n2 elements in AA. Thus, P(AA), the power set of AA, has 2n2 elements. Therefore, there are 2n2 relations on a set with n elements.
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11. Sunday, April 23, 2017
Types of Relations
Definition: A relation R on a set A is called reflexive if (a,a)R for every element a  A, that is,
Definition: A relation R on a set A is called irreflexive if (a,a)  R for every element a  A, that is,
Note that a relation can be both not reflexive and not irreflexive.
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12. Sunday, April 23, 2017
Example
Consider the following relations on {1,2,3,4}. Which are reflexive, irreflexive, neither?
R1 = {(1,1), (1,2), (2,1), (2,2), (3,3), (4,4)}
Reflexive
R2 = {(2,4), (4,2)}
Irreflexive
R3 = {(1,2), (2,3), (3,4)}
R4 = {(1,1), (2,2), (3,3), (4,4)}
R5 = {(1,3), (1,4), (2,3), (2,4), (3,1), (3,4)}
R6 = {(2,2), (2,3), (2,4), (3,2), (3,3), (3,4)}
Neither
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13. Sunday, April 23, 2017
Symmetric Relations
Definition: A relation R on a set A is called symmetric if (b,a)R whenever (a,b)R for a, b  A, that is,
Definition A relation R on a set A is called antisymmetric if whenever (a,b)R and (b,a)R, then a = b, for a, b in A, that is,
Note that these definitions are not complementary.
Give a relation that is both symmetric and antisymmetric.
Give a relation that is neither symmetrick not antisymmetric.
A relation can be both Symmetric and antisymmetric. A relation can be neither symmetric nor antisymmetric.
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14. Sunday, April 23, 2017
Example
Consider the following relations on {1,2,3,4}. Which are symmetric, antisymmetric, both, or neither?
R1 = {(1,1), (1,2), (2,1), (2,2), (3,3), (4,4)}
Symmetric
R2 = {(2,4), (4,2)}
R3 = {(1,2), (2,3), (3,4)}
Antisymmetric
R4 = {(1,1), (2,2), (3,3), (4,4)}
Both
R5 = {(1,3), (1,4), (2,3), (2,4), (3,1), (3,4)}
Neither
The “divides” relation on the set of positive integers
R6 = {(2,2), (2,3), (2,4), (3,2), (3,3), (3,4)}
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15. Sunday, April 23, 2017
Transitive Relations
Definition: A relation R on a set A is called transitive if whenever (a,b)R and (b,c)R, then (a,c)R, for a, b, c in A, that is,
Is the “football team relationship” transitive?
A defeats B and B defeats C implies A defeats C ?
Is the “subclass” relation on classes in an OO program transitive?
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16. Sunday, April 23, 2017
Example
Consider the following relations on {1,2,3,4}. Which are transitive?
R1 = {(2,2), (2,3), (2,4), (3,2), (3,3), (3,4)}
Transitive
R2 = {(1,1), (1,2), (2,1), (2,2), (3,3), (4,4)}
R3 = {(2,4), (4,2)}
Not Transitive
R4 = {(1,2), (2,3), (3,4)}
R5 = {(1,1), (2,2), (3,3), (4,4)}
R6 = {(1,3), (1,4), (2,3), (2,4), (3,1), (3,4)}
The “divides” relation on the set of positive integers
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17. Sunday, April 23, 2017
Example Summary
Consider the following relations on {1,2,3,4}. Name its properties?
R1 = {(2,2), (2,3), (2,4), (3,2), (3,3), (3,4)}
T/A
R2 = {(1,1), (1,2), (2,1), (2,2), (3,3), (4,4)}
R/S/T
R3 = {(2,4), (4,2)} S
R4 = {(1,2), (2,3), (3,4)} A
R5 = {(1,1), (2,2), (3,3), (4,4)}
R/S/A/T
R6 = {(1,3), (1,4), (2,3), (2,4), (3,1), (3,4)}
None
The “divides” relation on the set of positive integers
R/A/T
Keys: R = Reflexive, S = Symmetric, A = Antisymmetric, T=Transitive
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18. Pictorial Representation of Binary Relations on a Finite Set
Sunday, April 23, 2017
Pictorial Representation of Binary Relations on a Finite Set
Directed Graphs
Vertices, one for each element in the set
Arcs, one for each ordered-pair in the relation
Example: Set S = {1,2,3,4} and relation R on S
R = {(2,2), (2,3), (2,4), (3,2), (3,3), (3,4)} 1 4 3 2
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19. Boolean matrix representation of binary relations on a finite set
Recall: If R is a binary relation over set S
Order the elements of S = { x1, x2, …, xn }
Represent by matrix with a 1 in position (i, j) iff xiRxj
Example: Set S = {1,2,3,4} and relation R on S
R = {(2,2), (2,3), (2,4), (3,2), (3,3), (3,4)} 1

20. Equivalence Relations
Sunday, April 23, 2017
Equivalence Relations
A relation R on a set A is called an equivalence relation if it is reflexive, symmetric and transitive.
In such a relation, for each element a  A, the set of all elements related to a under R is called the equivalence class of a, and is denoted by [a].
[a] = { (a,b) | a  A  b  A  aRb }
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21. Sunday, April 23, 2017
Example
Determine the properties of each of the following relations defined on the set of all real numbers R:
R = {(x,y) | x + y = 0}
R = {(x,y) | x = y  x = – y}
R = {(x,y) | x – y is a rational number}
R = {(x,y) | x = 2y}
R = {(x,y) | x y ≥ 0}
R = {(x,y) | x y = 0}
R = {(x,y) | x = 1}
R = {(x,y) | x = 1 or y = 1}
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22. R = {(x,y) | x + y = 0} Solution:
Sunday, April 23, 2017
R = {(x,y) | x + y = 0}
Solution:
It is not reflexive since, for example, (1,1)  R.
It is not irreflexive since, for example, (0,0)  R.
Since x + y = y + x, it follows that if x + y = 0 then y + x = 0, so the relation is symmetric.
It is not antisymmetric since, for example, (–1,1) and (1, –1) are both in R, but 1 ≠ –1
The relation is not transitive since, for example, (1,–1) R and (–1,1) R, but (1,1)  R.
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23. R = {(x,y) | x = y  x = – y} Solution:
Sunday, April 23, 2017
R = {(x,y) | x = y  x = – y}
Solution:
Since for each x, xRx then it is reflexive.
Since it is reflexive, it is not irreflexive.
Since x = ±y if and only if y = ±x, then it is symmetric.
It is not antisymmetric since, for example, (1,–1) and (–1,1) are both in R but 1 ≠ –1
It is also transitive because essentially the product of ±1 and ±1 is ±1.
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24. R = {(x,y) | x – y is a rational number}
Sunday, April 23, 2017
R = {(x,y) | x – y is a rational number}
Solution:
It is reflexive since, for all x, x – x = 0 is a rational number.
Since it is reflexive, it is not irreflexive.
It is symmetric because, if x – y is rational, then – (x – y) = y – x is also.
It is not antisymmetric because, for example, (1, –1) and (– 1,1) are both in R but 1 ≠ –1.
It is transitive because if x – y is a rational and y – z is a rational, so is x – y + y – z = x – z.
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25. R = {(x,y) | x = 2y} Solution:
Sunday, April 23, 2017
R = {(x,y) | x = 2y}
Solution:
It is not reflexive since, for example, (1,1)  R.
It is not irreflexive since, for example, (0,0)  R.
It is not symmetric since, for example, (2,1)  R but (1,2)  R.
It is antisymmetric because x = y =0 is the only time that (x,y) and (y,x) are both in R.
It is not transitive since, for example, (4,2)  R and (2,1)  R but (4,1)  R.
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26. R = {(x,y) | x y ≥ 0} Solution: It is reflexive since x·x ≥ 0.
Sunday, April 23, 2017
R = {(x,y) | x y ≥ 0}
Solution:
It is reflexive since x·x ≥ 0.
Since it is reflexive, it is not irreflexive.
It is symmetric as the role of x and y are interchangeable.
It is not antisymmetric since, for example, (2,3) and (3,2) are both in R, but 2 ≠ 3.
It is not transitive because, for example, (1,0) and (0, – 2) are both in R but (1, – 2) is not.
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27. R = {(x,y) | x y = 0} Solution: It is not reflexive since (1,1)  R.
Sunday, April 23, 2017
R = {(x,y) | x y = 0}
Solution:
It is not reflexive since (1,1)  R.
It is not irreflexive since (0,0)  R.
It is symmetric as the role of x and y are interchangeable.
It is not antisymmetric since, for example, (2,0) and (0,2) are both in R but 2 ≠ 0.
It is not transitive because, for example, (1,0) and (0, – 2) are both in R but (1, – 2) is not.
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28. R = {(x,y) | x = 1} Solution: It is not reflexive since (2,2)  R.
Sunday, April 23, 2017
R = {(x,y) | x = 1}
Solution:
It is not reflexive since (2,2)  R.
It is not irreflexive since (1,1)  R.
It is not symmetric since, for example, (1,2)  R but (2,1)  R.
It is antisymmetric because if (x,y)  R and (y,x)  R it means x = y = 1.
It is transitive since if (1,y)  R and (y,z)  R, so is (1,z)  R .
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29. R = {(x,y) | x = 1 or y = 1} Solution:
Sunday, April 23, 2017
R = {(x,y) | x = 1 or y = 1}
Solution:
It is not reflexive since (2,2)  R.
It is not irreflexive since (1,1)  R.
It is symmetric as the roles of x and y are interchangeable.
It is not antisymmetric since, for example, (2,1) and (1,2) are both in R but 2 ≠ 1.
It is not transitive since, for example, (3,1) and (1,7) are both in R but (3,7)  R .
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30. Summary Relation R IR S A T E ✓ R = {(x,y) | x + y = 0}
Sunday, April 23, 2017
Summary
Relation R IR S A T E
R = {(x,y) | x + y = 0} ✓
R = {(x,y) | x = ±y}
{(x,y) | x – y is a rational number}
R = {(x,y) | x = 2y}
R = {(x,y) | x y ≥ 0}
R = {(x,y) | x y = 0}
R = {(x,y) | x = 1}
R = {(x,y) | x = 1 or y = 1}
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31. Sunday, April 23, 2017
Combining Relations
Since relations are special kind of sets, all of the operators we used for combining sets could be used to combine relations.
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32. Combining Relations – Examples
Sunday, April 23, 2017
Combining Relations – Examples
Let A={1, 2, 3} and B={1, 2, 3, 4}
Let R1 and R2 be relations from A to B
R1 = {(1,1), (2,2), (3,3)}
R2 = {(1,1), (1,2), (1,3), (1,4)}
R1  R2 = {(1,1), (2,2), (3,3), (1,2), (1,3), (1,4)}
R1  R2 = {(1,1)}
R1  R2 = {(2,2), (3,3)}
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33. Composition of Relations
Sunday, April 23, 2017
Composition of Relations
Definition: Let R be a relation from a set A to a set B, and S be a relation from B to a set C. The composition of R and S, denoted by S o R, is the relation consisting of ordered pairs (a,c), where:
aA, cC, and
bB | (a,b)R and (b,c)S. A B c R S
S o R
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34. Example Let A={1, 2, 3}, B={x, y, w, z}, C={0, 1, 2},
Sunday, April 23, 2017
Example
Let A={1, 2, 3}, B={x, y, w, z}, C={0, 1, 2},
R a relation from A to B:
R = {(1, x), (1, z), (2, w), (3, x), (3, z)}, and
S a relation from B to C:
S = {(x,0), (y,0), (w,1), (w,2), (z,1)}
What is S o R?
S o R = {(1,0), (1,1), (2,1), (2,2), (3,0), (3,1)}
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35. Sunday, April 23, 2017
Powers of Relations
Definition: Let R be a relation on the set A. The powers Rn, n = 1, 2, 3,… are defined recursively by:
R1 = R,
Rn+1=Rn o R, for n  1.
So:
R2 = RoR,
R3 = R2oR = (RoR) o R …
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36. Sunday, April 23, 2017
Rn+1
Rn+1=Rn o R R Rn A A A
In the digraph representation, Rn consists of all the order pairs (x,y) where y is reachable from x using a directed path of length n.
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37. Example Consider the following relations on {1,2,3,4}.
Sunday, April 23, 2017
Example
Consider the following relations on {1,2,3,4}.
R = {(1,2), (1,3), (3,4)}
Is R transitive? No
Find R2=R o R
R2 = {(1,4)}
Consider the following relation:
R = {(2,2), (2,3), (2,4), (3,2), (3,3), (3,4)}
Is R transitive?
Yes
R2 = {(2,2), (2,3), (2,4), (3,2), (3,3), (3,4)}
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38. Characterizing Transitive Relations
Sunday, April 23, 2017
Characterizing Transitive Relations
Theorem: The relation R on a set A is transitive if and only if Rn  R for n = 1, 2, 3, ...
If Part: If Rn  R for n = 1, 2, 3, ... then R is transitive
See exercise 6.1
Only If Part: If R is transitive then Rn  R for n = 1,2,3, ...
Proof By induction
Basis: If R is transitive then R1  R (trivially true)
Hypothesis: Assume true for n = 1, 2, 3, ... , k
Inductive: Given that if R is transitive then Rk  R, we want to show that if R is transitive then Rk+1  R
Proof: Assume that R is transitive, and let (a,b) be in Rk+1. So  x  A | (a,x)  R and (x,b)  Rk, by definition of Rk+1. Since R is transitive then Rk  R by inductive hypothesis. So (x,b)  R. Since R is transitive, then (a,b)  R. So Rk+1  R
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39. Applications in Relational Databases
Section 8.2

40. Records as n-tuples (with fields)
Record is an n-tuple of fields
e.g., student record might be 4-tuple of the form: (name, ID#, Major, GPA)
Relational Data Base: collection of records
e.g. student database { (Anderson, , CSE, 3.88), (Adams, , PHY, 3.45), (Chou, , CSE, 3.49), (Gould, , MTH, 3.45), (Rao, , MTH, 3.90), (Stevens, , PSY, 2.99) }
Field associates the attribute (name, ID#, etc.) with a value from a domain (set of student names, set of ID#s, etc.)

41. Table 1 Student Data: (Student_Name, PID, Major, GPA)
Students (ID# is the primary key)
Name
ID#
Major
GPA
Anderson
231455
CSE
3,88
Adams
888323
PHY
3.45
Chou
102147
3.49
Gould
453876
MTH
Rao
678543
3.90
Stevens
786576
PSY
2.99

42. Operations on relations (tables)
Projection : projects a table on one or more attributes (columns)
Selection : selects records (rows) that satisfy a condition
Join : combines two tables into one based on shared fields
Of course, you can also use the “usual” operations on relations (sets)
, , –, …

43. Projection of Students: P1,4 (Students)
Columns 1 and 4 are taken from Students
Columns 2 and 3 are ignored
Name is paired with GPA (perhaps so a department chair can send congratulations or condemnations to students with special GPAs)
Beware: the key ID# is lost so there may be some ambiguity
Perhaps there were 2 students “Smith”

44. Projection of Students: P1,4 (Students)
Name
ID#
Major
GPA
Anderson
231455
CSE
3.88
Adams
888323
PHY
3.45
Chou
102147
3.49
Gould
453876
MTH
Rao
678543
3.90
Stevens
786576
PSY
2.99
GPAs
Name
GPA
Anderson
3.88
Adams
3.45
Chou
3.49
Gould
Rao
3.90
Stevens
2.99
P1,4

45. P1,2 (Enrollments) Majors P1,2 Enrollments Name Major Course Glauser
BIO
MS 475
PY 410
MS 511
Marcus
MTH
MS 603
Miller
CSE
MS 575
CS 455
Majors
Name
Major
Glauser
BIO
Marcus
MTH
Miller
CSE
P1,2

46. Selection: SC
Select all records (rows) that satisfy the condition C

47. SMajor = “CSE” (Enrollments)
Name
Major
Course
Glauser
BIO
MS 475
PY 410
MS 511
Marcus
MTH
MS 603
Miller
CSE
MS 575
CS 455
CSE Enrollments
Name
Major
Course
Miller
CSE
MS 575
CS 455
SMajor = “CSE”

48. SMajor = “CSE”  GPA > 3.5 (Students)
Name
ID#
Major
GPA
Anderson
231455
CSE
3.88
Adams
888323
PHY
3.45
Chou
102147
3.49
Gould
453876
MTH
Rao
678543
3.90
Stevens
786576
PSY
2.99
SMajor = “CSE”  GPA > 3.5
CSE Students on Dean’s list
Name
ID#
Major
GPA
Anderson
231455
CSE
3.88

49. Join: Jp (R, S)
Joins records (rows) from R and S on the last p fields (columns) of R and the first p fields (columns) of S
To join two records corresponding fields must be identical
The join of (a1, …, am ) and (b1, …, bn ) is (a1, …, am, bp+1 , …, bn ), if am – p + 1 = b1 and am – p + 2 = b2 and and am = bp

50. Example: Joining two tables (on 2 cols) …
Teaching Assignments
Professor
Dept.
Course
No.
Cruz
ZOO
335
412
Farber
PSY
501
617
Grammer
PHY
544
551
Rosen
CSE
518
MTH
575
Class Schedule
Dept.
Course
No.
Room
Time
CSE
518
N521
2:00 pm
MTH
575
N502
3:00 pm
611
4:00 pm
PHY
544
B505
PSY
501
A100
617
A110
11:00 am
ZOO
335
9:00 am
412
8:00 am

51. Tables 5 and 6 joined over Dept&Course
Tuples of Table 5: cols 2,3 are matched to tuples of Table 6: cols 1, 2.
The joined 5-tuples create a new relation
The new relation adds (joins)
the teacher information to the Class-schedule information
the room and time information to the Teaching_assignment information

52. …yields a new table with 5 columns:
Professor
Dept.
Course
No.
Room
Time
Cruz
ZOO
335
A100
9:00 am
412
8:00 am
Farber
PSY
501
3:00 pm
617
A110
11:00 am
Grammer
PHY
544
B505
4:00 pm
Rosen
CSE
518
N521
2:00 pm
MTH
575
N502
Pure join: Keep only rows that can be joined (inner join)

53. One useful variation on “pure join”:
Professor
Dept.
Course
No.
Room
Time
Cruz
ZOO
335
A100
9:00 am
412
8:00 am
Farber
PSY
501
3:00 pm
617
A110
11:00 am
Grammer
PHY
544
B505
4:00 pm
551 ?
Rosen
CSE
518
N521
2:00 pm
MTH
575
N502
611
Outer join: keep tuples that don’t join; fill missing entries with special “undefined” value

54. Table 7 Teaching_schedule
Professor
Department
Course Number
Room
Time
Cruz
Zoology
335
A100
9:00 am
412
8:00 am
Farber
Psychology
501
3:00 pm
617
A110
11:00 am
Grammer
Physics
544
B505
4:00 pm
Rosen
Computer science
518
N521
2:00 pm
Mathematics
575
N502

55. SQL: Structured Query Language
Query language references Tables and their Attributes and applies conditions to them
SELECT Departure_time
FROM Flights
WHERE Destination = “Detroit”
name of Table or relation
project P5 (Flights)
adds a select condition

56. Table 8 Flights Flights Airline Flight# Gate Destination Departure
Nadir
122 34
Detroit
08:10
Acme
221 22
Denver
08:17 33
Anchorage
08:22
323
Honolulu
08:30
199 13
08:47
222
09:10
322
09:44
SELECT Departure_time FROM Flights WHERE Destination = “Detroit”
Result should be: 08:10
08:47
09:44

57. SQL example SELECT Professor, Time
FROM Teaching_assignments, Class_schedule
WHERE Department = ‘Mathematics’
project these columns
from the join of these tables
provided this condition holds
The “answer table” is a single 2-tuple: (Rosen, 3:00 pm)