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Relations & Digraphs
Product Sets
Definition: An ordered pair π‘Žπ‘Ž, 𝑏𝑏 is a listing of the
objects/items π‘Žπ‘Ž and 𝑏𝑏 in a prescribed order: π‘Žπ‘Ž is the first
and 𝑏𝑏 is the second. (a sequence of length 2)
Definition: The ordered pairs π‘Žπ‘Ž1, 𝑏𝑏1 and π‘Žπ‘Ž2, 𝑏𝑏2 are
equal iff π‘Žπ‘Ž1 = π‘Žπ‘Ž2 and 𝑏𝑏1 = 𝑏𝑏2.
Definition: If 𝐴𝐴 and 𝐡𝐡 are two nonempty sets, we define
the product set or Cartesian product 𝐴𝐴 Γ— 𝐡𝐡 as the set of
all ordered pairs π‘Žπ‘Ž, 𝑏𝑏 with π‘Žπ‘Ž ∈ 𝐴𝐴 and 𝑏𝑏 ∈ 𝐡𝐡:
𝐴𝐴 Γ— 𝐡𝐡 = π‘Žπ‘Ž, 𝑏𝑏 π‘Žπ‘Ž ∈ 𝐴𝐴 and 𝑏𝑏 ∈ 𝐡𝐡}
Β© S. Turaev, CSC 1700 Discrete Mathematics 2
Product Sets
Example: Let 𝐴𝐴 = 1,2,3 and 𝐡𝐡 = π‘Ÿπ‘Ÿ, 𝑠𝑠 , then
𝐴𝐴 Γ— 𝐡𝐡 =
𝐡𝐡 Γ— 𝐴𝐴 =
Β© S. Turaev, CSC 1700 Discrete Mathematics 3
Product Sets
Theorem: For any two finite sets 𝐴𝐴 and 𝐡𝐡,
𝐴𝐴 Γ— 𝐡𝐡 = 𝐴𝐴 β‹… 𝐡𝐡 .
Proof: Use multiplication principle!
Β© S. Turaev, CSC 1700 Discrete Mathematics 4
Definitions:
 Let 𝐴𝐴 and 𝐡𝐡 be nonempty sets. A relation 𝑅𝑅 from 𝐴𝐴
to 𝐡𝐡 is a subset of 𝐴𝐴 Γ— 𝐡𝐡.
 If 𝑅𝑅 βŠ† 𝐴𝐴 Γ— 𝐡𝐡 and π‘Žπ‘Ž, 𝑏𝑏 ∈ 𝑅𝑅, we say that π‘Žπ‘Ž is related
to 𝑏𝑏 by 𝑅𝑅, and we write π‘Žπ‘Ž 𝑅𝑅 𝑏𝑏.
 If π‘Žπ‘Ž is not related to 𝑏𝑏 by 𝑅𝑅, we write π‘Žπ‘Ž 𝑅𝑅 𝑏𝑏.
 If 𝑅𝑅 βŠ† 𝐴𝐴 Γ— 𝐴𝐴, we say 𝑅𝑅 is a relation on 𝐴𝐴.
Relations & Digraphs
Β© S. Turaev, CSC 1700 Discrete Mathematics 5
Example 1: Let 𝐴𝐴 = 1,2,3 and 𝐡𝐡 = π‘Ÿπ‘Ÿ, 𝑠𝑠 . Then
𝑅𝑅 = 1, π‘Ÿπ‘Ÿ , 2, 𝑠𝑠 , 3, π‘Ÿπ‘Ÿ βŠ† 𝐴𝐴 Γ— 𝐡𝐡
is a relation from 𝐴𝐴 to 𝐡𝐡.
Example 2: Let 𝐴𝐴 and 𝐡𝐡 are sets of positive integer
numbers. We define the relation 𝑅𝑅 βŠ† 𝐴𝐴 Γ— 𝐡𝐡 by
π‘Žπ‘Ž 𝑅𝑅 𝑏𝑏 ⇔ π‘Žπ‘Ž = 𝑏𝑏
Relations & Digraphs
Β© S. Turaev, CSC 1700 Discrete Mathematics 6
Example 3: Let 𝐴𝐴 = 1,2,3,4,5 . The relation 𝑅𝑅 βŠ† 𝐴𝐴 Γ— 𝐴𝐴 is
defined by
π‘Žπ‘Ž 𝑅𝑅 𝑏𝑏 ⇔ π‘Žπ‘Ž < 𝑏𝑏
Then 𝑅𝑅 =
Example 4: Let 𝐴𝐴 = 1,2,3,4,5,6,7,8,9,10 . The relation
𝑅𝑅 βŠ† 𝐴𝐴 Γ— 𝐴𝐴 is defined by
π‘Žπ‘Ž 𝑅𝑅 𝑏𝑏 ⇔ π‘Žπ‘Ž|𝑏𝑏
Then 𝑅𝑅 =
Relations & Digraphs
Β© S. Turaev, CSC 1700 Discrete Mathematics 7
Definition: Let 𝑅𝑅 βŠ† 𝐴𝐴 Γ— 𝐡𝐡 be a relation from 𝐴𝐴 to 𝐡𝐡.
 The domain of 𝑅𝑅, denoted by Dom 𝑅𝑅 , is the set of
elements in 𝐴𝐴 that are related to some element in
𝐡𝐡.
 The range of 𝑅𝑅, denoted by Ran 𝑅𝑅 , is the set of
elements in 𝐡𝐡 that are second elements of pairs in
𝑅𝑅.
Relations & Digraphs
Β© S. Turaev, CSC 1700 Discrete Mathematics 8
Relations & Digraphs
Example 5: Let 𝐴𝐴 = 1,2,3 and 𝐡𝐡 = π‘Ÿπ‘Ÿ, 𝑠𝑠 .
𝑅𝑅 = 1, π‘Ÿπ‘Ÿ , 2, 𝑠𝑠 , 3, π‘Ÿπ‘Ÿ
Dom R =
Ran R =
Example 6: Let 𝐴𝐴 = 1,2,3,4,5 . The relation 𝑅𝑅 βŠ† 𝐴𝐴 Γ— 𝐴𝐴 is
defined by π‘Žπ‘Ž 𝑅𝑅 𝑏𝑏 ⇔ π‘Žπ‘Ž < 𝑏𝑏
Dom R =
Ran R =
Β© S. Turaev, CSC 1700 Discrete Mathematics 9
The Matrix of a Relation
Definition: Let 𝐴𝐴 = π‘Žπ‘Ž1, π‘Žπ‘Ž2, … , π‘Žπ‘Ž π‘šπ‘š , 𝐡𝐡 = 𝑏𝑏1, 𝑏𝑏2, … , 𝑏𝑏𝑛𝑛
and 𝑅𝑅 βŠ† 𝐴𝐴 Γ— 𝐡𝐡 be a relation. We represent 𝑅𝑅 by the π‘šπ‘š Γ—
𝑛𝑛 matrix πŒπŒπ‘…π‘… = [π‘šπ‘šπ‘–π‘–π‘–π‘–], which is defined by
π‘šπ‘šπ‘–π‘–π‘–π‘– = οΏ½
1, π‘Žπ‘Žπ‘–π‘–, 𝑏𝑏𝑗𝑗 ∈ 𝑅𝑅
0, π‘Žπ‘Žπ‘–π‘–, 𝑏𝑏𝑗𝑗 βˆ‰ 𝑅𝑅
The matrix πŒπŒπ‘…π‘… is called the matrix of 𝑅𝑅.
Example: Let 𝐴𝐴 = 1,2,3 and 𝐡𝐡 = π‘Ÿπ‘Ÿ, 𝑠𝑠 .
𝑅𝑅 = 1, π‘Ÿπ‘Ÿ , 2, 𝑠𝑠 , 3, π‘Ÿπ‘Ÿ πŒπŒπ‘…π‘… =
Β© S. Turaev, CSC 1700 Discrete Mathematics 10
The Digraph of a Relation
Definition: If 𝐴𝐴 is finite and 𝑅𝑅 βŠ† 𝐴𝐴 Γ— 𝐴𝐴 is a relation. We
represent 𝑅𝑅 pictorially as follows:
 Draw a small circle, called a vertex/node, for each
element of 𝐴𝐴 and label the circle with the
corresponding element of 𝐴𝐴.
 Draw an arrow, called an edge, from vertex π‘Žπ‘Žπ‘–π‘– to
vertex π‘Žπ‘Žπ‘—π‘— iff π‘Žπ‘Žπ‘–π‘– 𝑅𝑅 π‘Žπ‘Žπ‘—π‘—.
The resulting pictorial representation of 𝑅𝑅 is called a
directed graph or digraph of 𝑅𝑅.
Β© S. Turaev, CSC 1700 Discrete Mathematics 11
The Digraph of a Relation
Example: Let 𝐴𝐴 = 1, 2, 3, 4 and
𝑅𝑅 = 1,1 , 1,2 , 2,1 , 2,2 , 2,3 , 2,4 , 3,4 , 4,1
The digraph of 𝑅𝑅:
Example: Let 𝐴𝐴 = 1, 2, 3, 4 and
Find the relation 𝑅𝑅:
Β© S. Turaev, CSC 1700 Discrete Mathematics
1
2
3
4
12
The Digraph of a Relation
Definition: If 𝑅𝑅 is a relation on a set 𝐴𝐴 and π‘Žπ‘Ž ∈ 𝐴𝐴, then
 the in-degree of π‘Žπ‘Ž is the number of 𝑏𝑏 ∈ 𝐴𝐴 such that
𝑏𝑏, π‘Žπ‘Ž ∈ 𝑅𝑅;
 the out-degree of π‘Žπ‘Ž is the number of 𝑏𝑏 ∈ 𝐴𝐴 such
that π‘Žπ‘Ž, 𝑏𝑏 ∈ 𝑅𝑅.
Example: Consider the digraph:
List in-degrees and out-degrees of all vertices.
Β© S. Turaev, CSC 1700 Discrete Mathematics
1
2
3
4
13
The Digraph of a Relation
Example: Let 𝐴𝐴 = π‘Žπ‘Ž, 𝑏𝑏, 𝑐𝑐, 𝑑𝑑 and let 𝑅𝑅 be the relation on
𝐴𝐴 that has the matrix
πŒπŒπ‘…π‘… =
1 0
0 1
0 0
0 0
1 1
0 1
1 0
0 1
Construct the digraph of 𝑅𝑅 and list in-degrees and out-
degrees of all vertices.
Β© S. Turaev, CSC 1700 Discrete Mathematics 14
The Digraph of a Relation
Example: Let 𝐴𝐴 = 1,4,5 and let 𝑅𝑅 be given the digraph
Find πŒπŒπ‘…π‘… and 𝑅𝑅.
Β© S. Turaev, CSC 1700 Discrete Mathematics
1 4
5
15
Paths in Relations & Digraphs
Definition: Suppose that 𝑅𝑅 is a relation on a set 𝐴𝐴.
A path of length 𝑛𝑛 in 𝑅𝑅 from π‘Žπ‘Ž to 𝑏𝑏 is a finite sequence
πœ‹πœ‹ ∢ π‘Žπ‘Ž, π‘₯π‘₯1, π‘₯π‘₯2, … , π‘₯π‘₯π‘›π‘›βˆ’1, 𝑏𝑏
beginning with π‘Žπ‘Ž and ending with 𝑏𝑏, such that
π‘Žπ‘Ž 𝑅𝑅 π‘₯π‘₯1, π‘₯π‘₯1 𝑅𝑅 π‘₯π‘₯2, … , π‘₯π‘₯π‘›π‘›βˆ’1 𝑅𝑅 𝑏𝑏.
Definition: A path that begins and ends at the same
vertex is called a cycle:
πœ‹πœ‹ ∢ π‘Žπ‘Ž, π‘₯π‘₯1, π‘₯π‘₯2, … , π‘₯π‘₯π‘›π‘›βˆ’1, π‘Žπ‘Ž
Β© S. Turaev, CSC 1700 Discrete Mathematics 16
Paths in Relations & Digraphs
Example: Give the examples for paths of length 1,2,3,4
and 5.
Β© S. Turaev, CSC 1700 Discrete Mathematics
1 2
43
5
17
Paths in Relations & Digraphs
Definition: If 𝑛𝑛 is a fixed number, we define a relation 𝑅𝑅 𝑛𝑛
as follows: π‘₯π‘₯ 𝑅𝑅𝑛𝑛
𝑦𝑦 means that there is a path of length 𝑛𝑛
from π‘₯π‘₯ to 𝑦𝑦.
Definition: We define a relation π‘…π‘…βˆž
(connectivity relation
for 𝑅𝑅) on 𝐴𝐴 by letting π‘₯π‘₯ π‘…π‘…βˆž
𝑦𝑦 mean that there is some
path from π‘₯π‘₯ to 𝑦𝑦.
Example: Let 𝐴𝐴 = π‘Žπ‘Ž, 𝑏𝑏, 𝑐𝑐, 𝑑𝑑, 𝑒𝑒 and
𝑅𝑅 = π‘Žπ‘Ž, π‘Žπ‘Ž , π‘Žπ‘Ž, 𝑏𝑏 , 𝑏𝑏, 𝑐𝑐 , 𝑐𝑐, 𝑒𝑒 , 𝑐𝑐, 𝑑𝑑 , 𝑑𝑑, 𝑒𝑒 .
Compute (a) 𝑅𝑅2
; (b) 𝑅𝑅3
; (c) π‘…π‘…βˆž
.
Β© S. Turaev, CSC 1700 Discrete Mathematics 18
Paths in Relations & Digraphs
Let 𝑅𝑅 be a relation on a finite set 𝐴𝐴 = π‘Žπ‘Ž1, π‘Žπ‘Ž2, … , π‘Žπ‘Žπ‘›π‘› , and
let πŒπŒπ‘…π‘… be the 𝑛𝑛 Γ— 𝑛𝑛 matrix representing 𝑅𝑅.
Theorem 1: If 𝑅𝑅 is a relation on 𝐴𝐴 = π‘Žπ‘Ž1, π‘Žπ‘Ž2, … , π‘Žπ‘Žπ‘›π‘› , then
πŒπŒπ‘…π‘…2 = πŒπŒπ‘…π‘… βŠ™ πŒπŒπ‘…π‘….
Example: Let 𝐴𝐴 = π‘Žπ‘Ž, 𝑏𝑏, 𝑐𝑐, 𝑑𝑑, 𝑒𝑒 and
𝑅𝑅 = π‘Žπ‘Ž, π‘Žπ‘Ž , π‘Žπ‘Ž, 𝑏𝑏 , 𝑏𝑏, 𝑐𝑐 , 𝑐𝑐, 𝑒𝑒 , 𝑐𝑐, 𝑑𝑑 , 𝑑𝑑, 𝑒𝑒 .
Β© S. Turaev, CSC 1700 Discrete Mathematics 19
Paths in Relations & Digraphs
Example: Let 𝐴𝐴 = π‘Žπ‘Ž, 𝑏𝑏, 𝑐𝑐, 𝑑𝑑, 𝑒𝑒 and
𝑅𝑅 = π‘Žπ‘Ž, π‘Žπ‘Ž , π‘Žπ‘Ž, 𝑏𝑏 , 𝑏𝑏, 𝑐𝑐 , 𝑐𝑐, 𝑒𝑒 , 𝑐𝑐, 𝑑𝑑 , 𝑑𝑑, 𝑒𝑒 .
πŒπŒπ‘…π‘… =
1 1
0 0
0 0
1 0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
1
1
0
Compute πŒπŒπ‘…π‘…2.
Β© S. Turaev, CSC 1700 Discrete Mathematics 20
Reflexive & Irreflexive Relations
Definition:
 A relation 𝑅𝑅 on a set 𝐴𝐴 is reflexive if π‘Žπ‘Ž, π‘Žπ‘Ž ∈ 𝑅𝑅 for
all π‘Žπ‘Ž ∈ 𝐴𝐴, i.e., if π‘Žπ‘Ž 𝑅𝑅 π‘Žπ‘Ž for all π‘Žπ‘Ž ∈ 𝐴𝐴.
 A relation 𝑅𝑅 on a set 𝐴𝐴 is irreflexive if π‘Žπ‘Ž 𝑅𝑅 π‘Žπ‘Ž for all
π‘Žπ‘Ž ∈ 𝐴𝐴.
Example:
 Ξ” = π‘Žπ‘Ž, π‘Žπ‘Ž | π‘Žπ‘Ž ∈ 𝐴𝐴 , the relation of equality on the
set 𝐴𝐴.
 𝑅𝑅 = π‘Žπ‘Ž, 𝑏𝑏 ∈ 𝐴𝐴 Γ— 𝐴𝐴| π‘Žπ‘Ž β‰  𝑏𝑏 , the relation of
inequality on the set 𝐴𝐴.
Β© S. Turaev, CSC 1700 Discrete Mathematics 21
Reflexive & Irreflexive Relations
Exercise: Let 𝐴𝐴 = 1, 2, 3 , and let 𝑅𝑅 = 1,1 , 1,2 .
Is 𝑅𝑅 reflexive or irreflexive?
Exercise: How is a reflexive or irreflexive relation
identified by its matrix?
Exercise: How is a reflexive or irreflexive relation
characterized by the digraph?
Β© S. Turaev, CSC 1700 Discrete Mathematics 22
(A-, Anti-) Symmetric Relations
Definition:
 A relation 𝑅𝑅 on a set 𝐴𝐴 is symmetric if whenever
π‘Žπ‘Ž 𝑅𝑅 𝑏𝑏, then 𝑏𝑏 𝑅𝑅 π‘Žπ‘Ž.
 A relation 𝑅𝑅 on a set 𝐴𝐴 is asymmetric if whenever
π‘Žπ‘Ž 𝑅𝑅 𝑏𝑏, then 𝑏𝑏 𝑅𝑅 π‘Žπ‘Ž.
 A relation 𝑅𝑅 on a set 𝐴𝐴 is antisymmetric if whenever
π‘Žπ‘Ž 𝑅𝑅 𝑏𝑏 and 𝑏𝑏 𝑅𝑅 π‘Žπ‘Ž, then π‘Žπ‘Ž = 𝑏𝑏.
Β© S. Turaev, CSC 1700 Discrete Mathematics 23
(A-, Anti-) Symmetric Relations
Example: Let 𝐴𝐴 = 1, 2, 3, 4, 5, 6 and let
𝑅𝑅 = π‘Žπ‘Ž, 𝑏𝑏 ∈ 𝐴𝐴 Γ— 𝐴𝐴 | π‘Žπ‘Ž < 𝑏𝑏
Is 𝑅𝑅 symmetric, asymmetric or antisymmetric?
 Symmetry:
 Asymmetry:
 Antisymmetry:
Β© S. Turaev, CSC 1700 Discrete Mathematics 24
(A-, Anti-) Symmetric Relations
Example: Let 𝐴𝐴 = 1, 2, 3, 4 and let
𝑅𝑅 = 1,2 , 2,2 , 3,4 , 4,1
Is 𝑅𝑅 symmetric, asymmetric or antisymmetric?
Example: Let 𝐴𝐴 = β„€+
and let
𝑅𝑅 = π‘Žπ‘Ž, 𝑏𝑏 ∈ 𝐴𝐴 Γ— 𝐴𝐴 | π‘Žπ‘Ž divides 𝑏𝑏
Is 𝑅𝑅 symmetric, asymmetric or antisymmetric?
Β© S. Turaev, CSC 1700 Discrete Mathematics 25
(A-, Anti-) Symmetric Relations
Exercise: How is a symmetric, asymmetric or
antisymmetric relation identified by its matrix?
Exercise: How is a symmetric, asymmetric or
antisymmetric relation characterized by the digraph?
Β© S. Turaev, CSC 1700 Discrete Mathematics 26
Transitive Relations
Definition: A relation 𝑅𝑅 on a set 𝐴𝐴 is transitive if
whenever π‘Žπ‘Ž 𝑅𝑅 𝑏𝑏 and 𝑏𝑏 𝑅𝑅 𝑐𝑐 then π‘Žπ‘Ž 𝑅𝑅 𝑐𝑐.
Example: Let 𝐴𝐴 = 1, 2, 3, 4 and let
𝑅𝑅 = 1,2 , 1,3 , 4,2
Is 𝑅𝑅 transitive?
Example: Let 𝐴𝐴 = β„€+
and let
𝑅𝑅 = π‘Žπ‘Ž, 𝑏𝑏 ∈ 𝐴𝐴 Γ— 𝐴𝐴 | π‘Žπ‘Ž divides 𝑏𝑏
Is 𝑅𝑅 transitive?
Β© S. Turaev, CSC 1700 Discrete Mathematics 27
Transitive Relations
Exercise: Let 𝐴𝐴 = 1,2,3 and 𝑅𝑅 be the relation on 𝐴𝐴
whose matrix is
πŒπŒπ‘…π‘… =
1 1 1
0 0 1
0 0 1
Show that 𝑅𝑅 is transitive. (Hint: Check if πŒπŒπ‘…π‘… βŠ™
2
= πŒπŒπ‘…π‘…)
Exercise: How is a transitive relation identified by its
matrix?
Exercise: How is a transitive relation characterized by the
digraph?
Β© S. Turaev, CSC 1700 Discrete Mathematics 28
Equivalence Relations
Definition: A relation 𝑅𝑅 on a set 𝐴𝐴 is called an equi-
valence relation if it is reflexive, symmetric and transitive.
Example: Let 𝐴𝐴 = 1, 2, 3, 4 and let
𝑅𝑅 = 1,1 , 1,2 , 2,1 , 2,2 , 3,4 , 4,3 , 3,3 , 4,4 .
Then 𝑅𝑅 is an equivalence relation.
Example: Let 𝐴𝐴 = β„€ and let
𝑅𝑅 = π‘Žπ‘Ž, 𝑏𝑏 ∈ 𝐴𝐴 Γ— 𝐴𝐴 ∢ π‘Žπ‘Ž ≑ 𝑏𝑏 mod 2 .
Show that 𝑅𝑅 is an equivalence relation.
Β© S. Turaev, CSC 1700 Discrete Mathematics 29
Exercises : Relations
Β© S. Turaev, CSC 1700 Discrete Mathematics 30
Exercises : Relations
Β© S. Turaev, CSC 1700 Discrete Mathematics 31
Exercises : Relations
Β© S. Turaev, CSC 1700 Discrete Mathematics 32

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Relations digraphs

  • 2. Product Sets Definition: An ordered pair π‘Žπ‘Ž, 𝑏𝑏 is a listing of the objects/items π‘Žπ‘Ž and 𝑏𝑏 in a prescribed order: π‘Žπ‘Ž is the first and 𝑏𝑏 is the second. (a sequence of length 2) Definition: The ordered pairs π‘Žπ‘Ž1, 𝑏𝑏1 and π‘Žπ‘Ž2, 𝑏𝑏2 are equal iff π‘Žπ‘Ž1 = π‘Žπ‘Ž2 and 𝑏𝑏1 = 𝑏𝑏2. Definition: If 𝐴𝐴 and 𝐡𝐡 are two nonempty sets, we define the product set or Cartesian product 𝐴𝐴 Γ— 𝐡𝐡 as the set of all ordered pairs π‘Žπ‘Ž, 𝑏𝑏 with π‘Žπ‘Ž ∈ 𝐴𝐴 and 𝑏𝑏 ∈ 𝐡𝐡: 𝐴𝐴 Γ— 𝐡𝐡 = π‘Žπ‘Ž, 𝑏𝑏 π‘Žπ‘Ž ∈ 𝐴𝐴 and 𝑏𝑏 ∈ 𝐡𝐡} Β© S. Turaev, CSC 1700 Discrete Mathematics 2
  • 3. Product Sets Example: Let 𝐴𝐴 = 1,2,3 and 𝐡𝐡 = π‘Ÿπ‘Ÿ, 𝑠𝑠 , then 𝐴𝐴 Γ— 𝐡𝐡 = 𝐡𝐡 Γ— 𝐴𝐴 = Β© S. Turaev, CSC 1700 Discrete Mathematics 3
  • 4. Product Sets Theorem: For any two finite sets 𝐴𝐴 and 𝐡𝐡, 𝐴𝐴 Γ— 𝐡𝐡 = 𝐴𝐴 β‹… 𝐡𝐡 . Proof: Use multiplication principle! Β© S. Turaev, CSC 1700 Discrete Mathematics 4
  • 5. Definitions:  Let 𝐴𝐴 and 𝐡𝐡 be nonempty sets. A relation 𝑅𝑅 from 𝐴𝐴 to 𝐡𝐡 is a subset of 𝐴𝐴 Γ— 𝐡𝐡.  If 𝑅𝑅 βŠ† 𝐴𝐴 Γ— 𝐡𝐡 and π‘Žπ‘Ž, 𝑏𝑏 ∈ 𝑅𝑅, we say that π‘Žπ‘Ž is related to 𝑏𝑏 by 𝑅𝑅, and we write π‘Žπ‘Ž 𝑅𝑅 𝑏𝑏.  If π‘Žπ‘Ž is not related to 𝑏𝑏 by 𝑅𝑅, we write π‘Žπ‘Ž 𝑅𝑅 𝑏𝑏.  If 𝑅𝑅 βŠ† 𝐴𝐴 Γ— 𝐴𝐴, we say 𝑅𝑅 is a relation on 𝐴𝐴. Relations & Digraphs Β© S. Turaev, CSC 1700 Discrete Mathematics 5
  • 6. Example 1: Let 𝐴𝐴 = 1,2,3 and 𝐡𝐡 = π‘Ÿπ‘Ÿ, 𝑠𝑠 . Then 𝑅𝑅 = 1, π‘Ÿπ‘Ÿ , 2, 𝑠𝑠 , 3, π‘Ÿπ‘Ÿ βŠ† 𝐴𝐴 Γ— 𝐡𝐡 is a relation from 𝐴𝐴 to 𝐡𝐡. Example 2: Let 𝐴𝐴 and 𝐡𝐡 are sets of positive integer numbers. We define the relation 𝑅𝑅 βŠ† 𝐴𝐴 Γ— 𝐡𝐡 by π‘Žπ‘Ž 𝑅𝑅 𝑏𝑏 ⇔ π‘Žπ‘Ž = 𝑏𝑏 Relations & Digraphs Β© S. Turaev, CSC 1700 Discrete Mathematics 6
  • 7. Example 3: Let 𝐴𝐴 = 1,2,3,4,5 . The relation 𝑅𝑅 βŠ† 𝐴𝐴 Γ— 𝐴𝐴 is defined by π‘Žπ‘Ž 𝑅𝑅 𝑏𝑏 ⇔ π‘Žπ‘Ž < 𝑏𝑏 Then 𝑅𝑅 = Example 4: Let 𝐴𝐴 = 1,2,3,4,5,6,7,8,9,10 . The relation 𝑅𝑅 βŠ† 𝐴𝐴 Γ— 𝐴𝐴 is defined by π‘Žπ‘Ž 𝑅𝑅 𝑏𝑏 ⇔ π‘Žπ‘Ž|𝑏𝑏 Then 𝑅𝑅 = Relations & Digraphs Β© S. Turaev, CSC 1700 Discrete Mathematics 7
  • 8. Definition: Let 𝑅𝑅 βŠ† 𝐴𝐴 Γ— 𝐡𝐡 be a relation from 𝐴𝐴 to 𝐡𝐡.  The domain of 𝑅𝑅, denoted by Dom 𝑅𝑅 , is the set of elements in 𝐴𝐴 that are related to some element in 𝐡𝐡.  The range of 𝑅𝑅, denoted by Ran 𝑅𝑅 , is the set of elements in 𝐡𝐡 that are second elements of pairs in 𝑅𝑅. Relations & Digraphs Β© S. Turaev, CSC 1700 Discrete Mathematics 8
  • 9. Relations & Digraphs Example 5: Let 𝐴𝐴 = 1,2,3 and 𝐡𝐡 = π‘Ÿπ‘Ÿ, 𝑠𝑠 . 𝑅𝑅 = 1, π‘Ÿπ‘Ÿ , 2, 𝑠𝑠 , 3, π‘Ÿπ‘Ÿ Dom R = Ran R = Example 6: Let 𝐴𝐴 = 1,2,3,4,5 . The relation 𝑅𝑅 βŠ† 𝐴𝐴 Γ— 𝐴𝐴 is defined by π‘Žπ‘Ž 𝑅𝑅 𝑏𝑏 ⇔ π‘Žπ‘Ž < 𝑏𝑏 Dom R = Ran R = Β© S. Turaev, CSC 1700 Discrete Mathematics 9
  • 10. The Matrix of a Relation Definition: Let 𝐴𝐴 = π‘Žπ‘Ž1, π‘Žπ‘Ž2, … , π‘Žπ‘Ž π‘šπ‘š , 𝐡𝐡 = 𝑏𝑏1, 𝑏𝑏2, … , 𝑏𝑏𝑛𝑛 and 𝑅𝑅 βŠ† 𝐴𝐴 Γ— 𝐡𝐡 be a relation. We represent 𝑅𝑅 by the π‘šπ‘š Γ— 𝑛𝑛 matrix πŒπŒπ‘…π‘… = [π‘šπ‘šπ‘–π‘–π‘–π‘–], which is defined by π‘šπ‘šπ‘–π‘–π‘–π‘– = οΏ½ 1, π‘Žπ‘Žπ‘–π‘–, 𝑏𝑏𝑗𝑗 ∈ 𝑅𝑅 0, π‘Žπ‘Žπ‘–π‘–, 𝑏𝑏𝑗𝑗 βˆ‰ 𝑅𝑅 The matrix πŒπŒπ‘…π‘… is called the matrix of 𝑅𝑅. Example: Let 𝐴𝐴 = 1,2,3 and 𝐡𝐡 = π‘Ÿπ‘Ÿ, 𝑠𝑠 . 𝑅𝑅 = 1, π‘Ÿπ‘Ÿ , 2, 𝑠𝑠 , 3, π‘Ÿπ‘Ÿ πŒπŒπ‘…π‘… = Β© S. Turaev, CSC 1700 Discrete Mathematics 10
  • 11. The Digraph of a Relation Definition: If 𝐴𝐴 is finite and 𝑅𝑅 βŠ† 𝐴𝐴 Γ— 𝐴𝐴 is a relation. We represent 𝑅𝑅 pictorially as follows:  Draw a small circle, called a vertex/node, for each element of 𝐴𝐴 and label the circle with the corresponding element of 𝐴𝐴.  Draw an arrow, called an edge, from vertex π‘Žπ‘Žπ‘–π‘– to vertex π‘Žπ‘Žπ‘—π‘— iff π‘Žπ‘Žπ‘–π‘– 𝑅𝑅 π‘Žπ‘Žπ‘—π‘—. The resulting pictorial representation of 𝑅𝑅 is called a directed graph or digraph of 𝑅𝑅. Β© S. Turaev, CSC 1700 Discrete Mathematics 11
  • 12. The Digraph of a Relation Example: Let 𝐴𝐴 = 1, 2, 3, 4 and 𝑅𝑅 = 1,1 , 1,2 , 2,1 , 2,2 , 2,3 , 2,4 , 3,4 , 4,1 The digraph of 𝑅𝑅: Example: Let 𝐴𝐴 = 1, 2, 3, 4 and Find the relation 𝑅𝑅: Β© S. Turaev, CSC 1700 Discrete Mathematics 1 2 3 4 12
  • 13. The Digraph of a Relation Definition: If 𝑅𝑅 is a relation on a set 𝐴𝐴 and π‘Žπ‘Ž ∈ 𝐴𝐴, then  the in-degree of π‘Žπ‘Ž is the number of 𝑏𝑏 ∈ 𝐴𝐴 such that 𝑏𝑏, π‘Žπ‘Ž ∈ 𝑅𝑅;  the out-degree of π‘Žπ‘Ž is the number of 𝑏𝑏 ∈ 𝐴𝐴 such that π‘Žπ‘Ž, 𝑏𝑏 ∈ 𝑅𝑅. Example: Consider the digraph: List in-degrees and out-degrees of all vertices. Β© S. Turaev, CSC 1700 Discrete Mathematics 1 2 3 4 13
  • 14. The Digraph of a Relation Example: Let 𝐴𝐴 = π‘Žπ‘Ž, 𝑏𝑏, 𝑐𝑐, 𝑑𝑑 and let 𝑅𝑅 be the relation on 𝐴𝐴 that has the matrix πŒπŒπ‘…π‘… = 1 0 0 1 0 0 0 0 1 1 0 1 1 0 0 1 Construct the digraph of 𝑅𝑅 and list in-degrees and out- degrees of all vertices. Β© S. Turaev, CSC 1700 Discrete Mathematics 14
  • 15. The Digraph of a Relation Example: Let 𝐴𝐴 = 1,4,5 and let 𝑅𝑅 be given the digraph Find πŒπŒπ‘…π‘… and 𝑅𝑅. Β© S. Turaev, CSC 1700 Discrete Mathematics 1 4 5 15
  • 16. Paths in Relations & Digraphs Definition: Suppose that 𝑅𝑅 is a relation on a set 𝐴𝐴. A path of length 𝑛𝑛 in 𝑅𝑅 from π‘Žπ‘Ž to 𝑏𝑏 is a finite sequence πœ‹πœ‹ ∢ π‘Žπ‘Ž, π‘₯π‘₯1, π‘₯π‘₯2, … , π‘₯π‘₯π‘›π‘›βˆ’1, 𝑏𝑏 beginning with π‘Žπ‘Ž and ending with 𝑏𝑏, such that π‘Žπ‘Ž 𝑅𝑅 π‘₯π‘₯1, π‘₯π‘₯1 𝑅𝑅 π‘₯π‘₯2, … , π‘₯π‘₯π‘›π‘›βˆ’1 𝑅𝑅 𝑏𝑏. Definition: A path that begins and ends at the same vertex is called a cycle: πœ‹πœ‹ ∢ π‘Žπ‘Ž, π‘₯π‘₯1, π‘₯π‘₯2, … , π‘₯π‘₯π‘›π‘›βˆ’1, π‘Žπ‘Ž Β© S. Turaev, CSC 1700 Discrete Mathematics 16
  • 17. Paths in Relations & Digraphs Example: Give the examples for paths of length 1,2,3,4 and 5. Β© S. Turaev, CSC 1700 Discrete Mathematics 1 2 43 5 17
  • 18. Paths in Relations & Digraphs Definition: If 𝑛𝑛 is a fixed number, we define a relation 𝑅𝑅 𝑛𝑛 as follows: π‘₯π‘₯ 𝑅𝑅𝑛𝑛 𝑦𝑦 means that there is a path of length 𝑛𝑛 from π‘₯π‘₯ to 𝑦𝑦. Definition: We define a relation π‘…π‘…βˆž (connectivity relation for 𝑅𝑅) on 𝐴𝐴 by letting π‘₯π‘₯ π‘…π‘…βˆž 𝑦𝑦 mean that there is some path from π‘₯π‘₯ to 𝑦𝑦. Example: Let 𝐴𝐴 = π‘Žπ‘Ž, 𝑏𝑏, 𝑐𝑐, 𝑑𝑑, 𝑒𝑒 and 𝑅𝑅 = π‘Žπ‘Ž, π‘Žπ‘Ž , π‘Žπ‘Ž, 𝑏𝑏 , 𝑏𝑏, 𝑐𝑐 , 𝑐𝑐, 𝑒𝑒 , 𝑐𝑐, 𝑑𝑑 , 𝑑𝑑, 𝑒𝑒 . Compute (a) 𝑅𝑅2 ; (b) 𝑅𝑅3 ; (c) π‘…π‘…βˆž . Β© S. Turaev, CSC 1700 Discrete Mathematics 18
  • 19. Paths in Relations & Digraphs Let 𝑅𝑅 be a relation on a finite set 𝐴𝐴 = π‘Žπ‘Ž1, π‘Žπ‘Ž2, … , π‘Žπ‘Žπ‘›π‘› , and let πŒπŒπ‘…π‘… be the 𝑛𝑛 Γ— 𝑛𝑛 matrix representing 𝑅𝑅. Theorem 1: If 𝑅𝑅 is a relation on 𝐴𝐴 = π‘Žπ‘Ž1, π‘Žπ‘Ž2, … , π‘Žπ‘Žπ‘›π‘› , then πŒπŒπ‘…π‘…2 = πŒπŒπ‘…π‘… βŠ™ πŒπŒπ‘…π‘…. Example: Let 𝐴𝐴 = π‘Žπ‘Ž, 𝑏𝑏, 𝑐𝑐, 𝑑𝑑, 𝑒𝑒 and 𝑅𝑅 = π‘Žπ‘Ž, π‘Žπ‘Ž , π‘Žπ‘Ž, 𝑏𝑏 , 𝑏𝑏, 𝑐𝑐 , 𝑐𝑐, 𝑒𝑒 , 𝑐𝑐, 𝑑𝑑 , 𝑑𝑑, 𝑒𝑒 . Β© S. Turaev, CSC 1700 Discrete Mathematics 19
  • 20. Paths in Relations & Digraphs Example: Let 𝐴𝐴 = π‘Žπ‘Ž, 𝑏𝑏, 𝑐𝑐, 𝑑𝑑, 𝑒𝑒 and 𝑅𝑅 = π‘Žπ‘Ž, π‘Žπ‘Ž , π‘Žπ‘Ž, 𝑏𝑏 , 𝑏𝑏, 𝑐𝑐 , 𝑐𝑐, 𝑒𝑒 , 𝑐𝑐, 𝑑𝑑 , 𝑑𝑑, 𝑒𝑒 . πŒπŒπ‘…π‘… = 1 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 Compute πŒπŒπ‘…π‘…2. Β© S. Turaev, CSC 1700 Discrete Mathematics 20
  • 21. Reflexive & Irreflexive Relations Definition:  A relation 𝑅𝑅 on a set 𝐴𝐴 is reflexive if π‘Žπ‘Ž, π‘Žπ‘Ž ∈ 𝑅𝑅 for all π‘Žπ‘Ž ∈ 𝐴𝐴, i.e., if π‘Žπ‘Ž 𝑅𝑅 π‘Žπ‘Ž for all π‘Žπ‘Ž ∈ 𝐴𝐴.  A relation 𝑅𝑅 on a set 𝐴𝐴 is irreflexive if π‘Žπ‘Ž 𝑅𝑅 π‘Žπ‘Ž for all π‘Žπ‘Ž ∈ 𝐴𝐴. Example:  Ξ” = π‘Žπ‘Ž, π‘Žπ‘Ž | π‘Žπ‘Ž ∈ 𝐴𝐴 , the relation of equality on the set 𝐴𝐴.  𝑅𝑅 = π‘Žπ‘Ž, 𝑏𝑏 ∈ 𝐴𝐴 Γ— 𝐴𝐴| π‘Žπ‘Ž β‰  𝑏𝑏 , the relation of inequality on the set 𝐴𝐴. Β© S. Turaev, CSC 1700 Discrete Mathematics 21
  • 22. Reflexive & Irreflexive Relations Exercise: Let 𝐴𝐴 = 1, 2, 3 , and let 𝑅𝑅 = 1,1 , 1,2 . Is 𝑅𝑅 reflexive or irreflexive? Exercise: How is a reflexive or irreflexive relation identified by its matrix? Exercise: How is a reflexive or irreflexive relation characterized by the digraph? Β© S. Turaev, CSC 1700 Discrete Mathematics 22
  • 23. (A-, Anti-) Symmetric Relations Definition:  A relation 𝑅𝑅 on a set 𝐴𝐴 is symmetric if whenever π‘Žπ‘Ž 𝑅𝑅 𝑏𝑏, then 𝑏𝑏 𝑅𝑅 π‘Žπ‘Ž.  A relation 𝑅𝑅 on a set 𝐴𝐴 is asymmetric if whenever π‘Žπ‘Ž 𝑅𝑅 𝑏𝑏, then 𝑏𝑏 𝑅𝑅 π‘Žπ‘Ž.  A relation 𝑅𝑅 on a set 𝐴𝐴 is antisymmetric if whenever π‘Žπ‘Ž 𝑅𝑅 𝑏𝑏 and 𝑏𝑏 𝑅𝑅 π‘Žπ‘Ž, then π‘Žπ‘Ž = 𝑏𝑏. Β© S. Turaev, CSC 1700 Discrete Mathematics 23
  • 24. (A-, Anti-) Symmetric Relations Example: Let 𝐴𝐴 = 1, 2, 3, 4, 5, 6 and let 𝑅𝑅 = π‘Žπ‘Ž, 𝑏𝑏 ∈ 𝐴𝐴 Γ— 𝐴𝐴 | π‘Žπ‘Ž < 𝑏𝑏 Is 𝑅𝑅 symmetric, asymmetric or antisymmetric?  Symmetry:  Asymmetry:  Antisymmetry: Β© S. Turaev, CSC 1700 Discrete Mathematics 24
  • 25. (A-, Anti-) Symmetric Relations Example: Let 𝐴𝐴 = 1, 2, 3, 4 and let 𝑅𝑅 = 1,2 , 2,2 , 3,4 , 4,1 Is 𝑅𝑅 symmetric, asymmetric or antisymmetric? Example: Let 𝐴𝐴 = β„€+ and let 𝑅𝑅 = π‘Žπ‘Ž, 𝑏𝑏 ∈ 𝐴𝐴 Γ— 𝐴𝐴 | π‘Žπ‘Ž divides 𝑏𝑏 Is 𝑅𝑅 symmetric, asymmetric or antisymmetric? Β© S. Turaev, CSC 1700 Discrete Mathematics 25
  • 26. (A-, Anti-) Symmetric Relations Exercise: How is a symmetric, asymmetric or antisymmetric relation identified by its matrix? Exercise: How is a symmetric, asymmetric or antisymmetric relation characterized by the digraph? Β© S. Turaev, CSC 1700 Discrete Mathematics 26
  • 27. Transitive Relations Definition: A relation 𝑅𝑅 on a set 𝐴𝐴 is transitive if whenever π‘Žπ‘Ž 𝑅𝑅 𝑏𝑏 and 𝑏𝑏 𝑅𝑅 𝑐𝑐 then π‘Žπ‘Ž 𝑅𝑅 𝑐𝑐. Example: Let 𝐴𝐴 = 1, 2, 3, 4 and let 𝑅𝑅 = 1,2 , 1,3 , 4,2 Is 𝑅𝑅 transitive? Example: Let 𝐴𝐴 = β„€+ and let 𝑅𝑅 = π‘Žπ‘Ž, 𝑏𝑏 ∈ 𝐴𝐴 Γ— 𝐴𝐴 | π‘Žπ‘Ž divides 𝑏𝑏 Is 𝑅𝑅 transitive? Β© S. Turaev, CSC 1700 Discrete Mathematics 27
  • 28. Transitive Relations Exercise: Let 𝐴𝐴 = 1,2,3 and 𝑅𝑅 be the relation on 𝐴𝐴 whose matrix is πŒπŒπ‘…π‘… = 1 1 1 0 0 1 0 0 1 Show that 𝑅𝑅 is transitive. (Hint: Check if πŒπŒπ‘…π‘… βŠ™ 2 = πŒπŒπ‘…π‘…) Exercise: How is a transitive relation identified by its matrix? Exercise: How is a transitive relation characterized by the digraph? Β© S. Turaev, CSC 1700 Discrete Mathematics 28
  • 29. Equivalence Relations Definition: A relation 𝑅𝑅 on a set 𝐴𝐴 is called an equi- valence relation if it is reflexive, symmetric and transitive. Example: Let 𝐴𝐴 = 1, 2, 3, 4 and let 𝑅𝑅 = 1,1 , 1,2 , 2,1 , 2,2 , 3,4 , 4,3 , 3,3 , 4,4 . Then 𝑅𝑅 is an equivalence relation. Example: Let 𝐴𝐴 = β„€ and let 𝑅𝑅 = π‘Žπ‘Ž, 𝑏𝑏 ∈ 𝐴𝐴 Γ— 𝐴𝐴 ∢ π‘Žπ‘Ž ≑ 𝑏𝑏 mod 2 . Show that 𝑅𝑅 is an equivalence relation. Β© S. Turaev, CSC 1700 Discrete Mathematics 29
  • 30. Exercises : Relations Β© S. Turaev, CSC 1700 Discrete Mathematics 30
  • 31. Exercises : Relations Β© S. Turaev, CSC 1700 Discrete Mathematics 31
  • 32. Exercises : Relations Β© S. Turaev, CSC 1700 Discrete Mathematics 32