Rosen, Section 8.5 Equivalence Relations

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Rosen, Section 8.5Equivalence Relations

• Longin Jan Latecki
• Temple University, Philadelphia
• latecki_at_temple.edu
• Some slides from Aaron Bloomfield

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Introduction

• Certain combinations of relation properties are
  very useful
• In this set we will study equivalence
  relationsA relation that is reflexive,
  symmetric and transitive
• Next slide set we will study partial orderingA
  relation that is reflexive, antisymmetric, and
  transitive
• The difference is whether the relation is
  symmetric or antisymmetric

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Outline

• What is an equivalence relation
• Equivalence relation examples
• Related items
• Equivalence class
• Partitions

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We can group properties of relations together to
define new types of important relations. ________
_________ Definition A relation R on a set A is
an equivalence relation iff R is reflexive
symmetric transitive Two elements related by
an equivalence relation are called equivalent.
Consider relation R (a,b) len(a) len(b)
, where len(a) means the length of string a It
is reflexive len(a) len(a) It is symmetric if
len(a) len(b), then len(b) len(a) It is
transitive if len(a) len(b) and len(b)
len(c), then len(a) len(c) Thus, R is a
equivalence relation
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Equivalence relation example

• Consider the relation R (a,b) a b (mod m)
  
• Remember that this means that m a-b
• Called congruence modulo m
• Is it reflexive (a,a) ? R means that m a-a
• a-a 0, which is divisible by m
• Is it symmetric if (a,b) ? R then (b,a) ? R
• (a,b) means that m a-b
• Or that km a-b. Negating that, we get b-a
  -km
• Thus, m b-a, so (b,a) ? R
• Is it transitive if (a,b) ? R and (b,c) ? R then
  (a,c) ? R
• (a,b) means that m a-b, or that km a-b
• (b,c) means that m b-c, or that lm b-c
• (a,c) means that m a-c, or that nm a-c
• Adding these two, we get kmlm (a-b) (b-c)
• Or (kl)m a-c
• Thus, m divides a-c, where n kl
• Thus, congruence modulo m is an equivalence
  relation

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Rosen, section 8.5, question 1

• Which of these relations on 0, 1, 2, 3 are
  equivalence relations? Determine the properties
  of an equivalence relation that the others lack
• (0,0), (1,1), (2,2), (3,3)
• Has all the properties, thus, is an equivalence
  relation
• (0,0), (0,2), (2,0), (2,2), (2,3), (3,2), (3,3)
  
• Not reflexive (1,1) is missing
• Not transitive (0,2) and (2,3) are in the
  relation, but not (0,3)
• (0,0), (1,1), (1,2), (2,1), (2,2), (3,3)
• Has all the properties, thus, is an equivalence
  relation
• (0,0), (1,1), (1,3), (2,2), (2,3), (3,1), (3,2)
  (3,3)
• Not transitive (1,3) and (3,2) are in the
  relation, but not (1,2)
• (0,0), (0,1) (0,2), (1,0), (1,1), (1,2), (2,0),
  (2,2), (3,3)
• Not symmetric (1,2) is present, but not (2,1)
• Not transitive (2,0) and (0,1) are in the
  relation, but not (2,1)

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Rosen, Section 8.5, question 9

• Suppose that A is a non-empty set, and f is a
  function that has A as its domain. Let R be the
  relation on A consisting of all ordered pairs
  (x,y) where f(x) f(y)
• Meaning that x and y are related if and only if
  f(x) f(y)
• Show that R is an equivalence relation on A
• Reflexivity f(x) f(x)
• True, as given the same input, a function always
  produces the same output
• Symmetry if f(x) f(y) then f(y) f(x)
• True, by the definition of equality
• Transitivity if f(x) f(y) and f(y) f(z) then
  f(x) f(z)
• True, by the definition of equality

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Rosen, Section 8.5, question 11

• Show that the relation R, consisting of all pairs
  (x,y) where x and y are bit strings of length
  three or more that agree except perhaps in their
  first three bits, is an equivalence relation on
  the set of all bit strings
• Let f(x) the bit string formed by the last n-3
  bits of the bit string x (where n is the length
  of the string)
• Thus, we want to show let R be the relation on A
  consisting of all ordered pairs (x,y) where f(x)
  f(y)
• This has been shown in question 9 on the previous
  slide

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An equivalence class of an element x x y
ltx, ygt is in R x is the subset of all
elements related to x by R. The element in the
bracket is called a representative of the
equivalence class. We could have chosen any
one. Theorem Let R be an equivalence relation
on A. Then either a b or a nb
F The number of equivalence classes is called
the rank of the equivalence relation.
Let Aa,b,c and R be given by a digraph
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More on equivalence classes

• Consider the relation R (a,b) a mod 2 b
  mod 2 on the set of integers
• Thus, all the even numbers are related to each
  other
• As are the odd numbers
• The even numbers form an equivalence class
• As do the odd numbers
• The equivalence class for the even numbers is
  denoted by 2 (or 4, or 784, etc.)
• 2 , -4, -2, 0, 2, 4,
• 2 is a representative of its equivalence class
• There are only 2 equivalence classes formed by
  this equivalence relation

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More on equivalence classes

• Consider the relation R (a,b) a b or a
  -b
• Thus, every number is related to additive inverse
• The equivalence class for an integer a
• 7 7, -7
• 0 0
• a a, -a
• There are an infinite number of equivalence
  classes formed by this equivalence relation

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Theorem Let R be an equivalence relation on a
set A. The equivalence classes of R partition the
set A into disjoint nonempty subsets whose union
is the entire set. This partition is denoted A/R
and called the quotient set, or the partition
of A induced by R, or, A modulo R. Definition
Let S1, S2, . . ., Sn be a collection of subsets
of a set A. Then the collection forms a partition
of A if the subsets are nonempty, disjoint and
exhaust A
Note that , 1,3, 2 is not a partition
(it contains the empty set). 1,2, 2, 3
is not a partition because . 1, 2 is
not a partition of 1, 2, 3 because none of its
blocks contains 3.
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It is easy to recognize equivalence relations
using digraphs The equivalence class of a
particular element forms a universal relation
(contains all possible edges) between the
elements in the equivalence class. The
(sub)digraph representing the subset is called a
complete (sub)digraph, since all arcs are
present. Example All possible equivalence
relations on a set A with 3 elements
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Rosen, section 8.5, question 44

• Which of the following are partitions of the set
  of integers?
• The set of even integers and the set of odd
  integers
• Yes, its a valid partition
• The set of positive integers and the set of
  negative integers
• No 0 is in neither set
• The set of integers divisible by 3, the set of
  integers leaving a remainder of 1 when divided by
  3, and the set of integers leaving a remaineder
  of 2 when divided by 3
• Yes, its a valid partition
• The set of integers less than -100, the set of
  integers with absolute value not exceeding 100,
  and the set of integers greater than 100
• Yes, its a valid partition
• The set of integers not divisible by 3, the set
  of even integers, and the set of integers that
  leave a remainder of 3 when divided by 6
• The first two sets are not disjoint (2 is in
  both), so its not a valid partition

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1. Determine whether the relations represented by
these zero-one matricesare equivalence
relations. If yes, with how many equivalence
classes?
2. What are the equivalence classes (sets in the
partition) of the integersarising from
congruence modulo 4? 3. Can you count the number
of equivalence relations on a set A with n
elements. Can you find a recurrence relation? The
answers are 1 for n 1 2 for n 2 5 for n
3 How many for n 4?
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Theorem (Bell number) Let p(n) denotes the number
of different equivalence relations on a set with
n elements (which is equivalent to the number of
partitions of the set with n elements). Then
p(n) is called Bell number, named in honor of
Eric Temple Bell Examples p(0)1, since there
is only one partition of the empty set into the
empty collection of subsets p(1)C(0,0)p(0)1,
since 1 is the only partition of
1 p(2)C(1,0)p(1)C(1,1)p(0)115, since
portions of 1,2 are 1,2 and
1,2 p(3)5, since, the set 1, 2, 3 has
these five partitions. 1, 2, 3 ,
sometimes denoted by 1/2/3. 1, 2, 3 ,
sometimes denoted by 12/3. 1, 3, 2 ,
sometimes denoted by 13/2. 1, 2, 3 ,
sometimes denoted by 1/23. 1, 2, 3 ,
sometimes denoted by 123.
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Proof (Bell number)
We want to portion 1, 2, , n. For a fixed j, A
is a subset of j elements from 1, 2, , n-1
union n. Note that j can have values from 0 to
n-1. We can select a subset of j elements from
1, 2, , n-1 in C(n-1,j) ways, and we have
p(n-1-j) partitions of the remaining n-1-j
elements.
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Theorem If R1 and R2 are equivalence relations
on A, then R1nR2 is an equivalence relation on
A. Proof It suffices to show that the
intersection of reflexive relations is
reflexive, symmetric relations is symmetric,
and transitive relations is transitive.
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Definition Let R be a relation on A. Then the
reflexive, symmetric, transitive closure of R,
tsr(R), is an equivalence relation on A, called
the equivalence relation induced by R. Example
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Theorem tsr(R) is an equivalence
relation. Proof We need to show that tsr(R) is
still symmetric and reflexive. Since we only
add arcs vs. deleting arcs when computing
closures it must be that tsr(R) is reflexive
since all loops ltx, xgt on the diagraph must be
present when constructing r(R). If there is an
arc ltx, ygt then the symmetric closure of r(R)
ensures there is an arc lty, xgt. Now argue that
if we construct the transitive closure of sr(R)
and we add an edge ltx, zgt because there is a path
from x to z, then there must also exist a path
from z to x (why?) and hence we also must add an
edge ltz, xgt. Hence the transitive closure of
sr(R) is symmetric.
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What we wish computers could do
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Quick survey

• I felt I understood the material in this slide
  set
• Very well
• With some review, Ill be good
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• Not at all

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Quick survey

• The pace of the lecture for this slide set was
• Fast
• About right
• A little slow
• Too slow

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Quick survey

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