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Rosen, Section 8.5 Equivalence Relations Longin Jan Latecki Temple University, Philadelphia [email protected] Some slides from Aaron Bloomfield Introduction • Certain combinations of relation properties are very useful • In this set we will study equivalence relations: A relation that is reflexive, symmetric and transitive • Next slide set we will study partial ordering: A relation that is reflexive, antisymmetric, and transitive • The difference is whether the relation is symmetric or antisymmetric Outline • What is an equivalence relation • Equivalence relation examples • Related items – Equivalence class – Partitions 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 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 km+lm = (a-b) + (b-c) Or (k+l)m = a-c Thus, m divides a-c, where n = k+l • Thus, congruence modulo m is an equivalence relation Rosen, section 8.5, question 1 • a) 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 b) { (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) c) { (0,0), (1,1), (1,2), (2,1), (2,2), (3,3) } Has all the properties, thus, is an equivalence relation d) { (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) e) { (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) 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 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 An equivalence class of an element x: [x] = {y | <x, y> 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] ∩[b] = Φ The number of equivalence classes is called the rank of the equivalence relation. Let A={a,b,c} and R be given by a digraph: 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 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 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. 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: Rosen, section 8.5, question 44 • Which of the following are partitions of the set of integers? a) The set of even integers and the set of odd integers Yes, it’s a valid partition b) The set of positive integers and the set of negative integers No: 0 is in neither set c) 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, it’s a valid partition d) 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, it’s a valid partition e) 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 it’s not a valid partition 1. Determine whether the relations represented by these zero-one matrices are equivalence relations. If yes, with how many equivalence classes? 1 1 1 1 0 M 0 1 1 P 1 1 1 1 0 0 1 0 1 1 0 1 0 0 1 1 1 R 1 0 1 0 1 1 1 0 1 1 1 0 0 0 0 1 2. What are the equivalence classes (sets in the partition) of the integers arising 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? 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 n 1 p(n) C (n 1, j ) p(n j 1) j 0 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)=1+1=2, 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. n 1 p(n) C (n 1, j ) p(n j 1) j 0 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. ■ Theorem: If R1 and R2 are equivalence relations on A, then R1∩R2 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. 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: 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 <x, x> on the diagraph must be present when constructing r(R). • If there is an arc <x, y> then the symmetric closure of r(R) ensures there is an arc <y, x>. • Now argue that if we construct the transitive closure of sr(R) and we add an edge <x, z> 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 <z, x>. Hence the transitive closure of sr(R) is symmetric. What we wish computers could do Quick survey • a) b) c) d) I felt I understood the material in this slide set… Very well With some review, I’ll be good Not really Not at all Quick survey • a) b) c) d) The pace of the lecture for this slide set was… Fast About right A little slow Too slow Quick survey • a) b) c) d) How interesting was the material in this slide set? Be honest! Wow! That was SOOOOOO cool! Somewhat interesting Rather boring Zzzzzzzzzzz

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