Equivalence RelationsIntroductionEquivalence relationsEquivalence relation exampleSample questionsSlide 6Slide 7Equivalence classesMore on equivalence classesSlide 10PartitionsSlide 121Equivalence RelationsAaron BloomfieldCS 202Epp, section ???2Introduction•Certain combinations of relation properties are very useful–We won’t have a chance to see many applications in this course•In this set we will study equivalence relations–A relation that is reflexive, symmetric and transitive•Next slide set we will study partial orderings–A relation that is reflexive, antisymmetric, and transitive•The difference is whether the relation is symmetric or antisymmetric3Equivalence relations•A relation on a set A is called an equivalence relation if it is reflexive, symmetric, and transitive•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 relation4Equivalence relation example•Consider the relation R = { (a,b) | 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 relation5Sample questions•Which of these relations on {0, 1, 2, 3} are equivalence relations? Determine the properties of an equivalence relation that the others lacka) { (0,0), (1,1), (2,2), (3,3) }Has all the properties, thus, is an equivalence relationb) { (0,0), (0,2), (2,0), (2,2), (2,3), (3,2), (3,3) }Not reflexive: (1,1) is missingNot 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 relationd) { (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)6Sample questions•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 equality7Sample questions•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 5 on the previous slide8Equivalence classes•Let R be an equivalence relation on a set A. The set of all elements that are related to an element a of A is called the equivalence class of a.•The equivalence class of a with respect to R is denoted by [a]R•When only one relation is under consideration, the subscript is often deleted, and [a] is used to denote the equivalence class•Note that these classes are disjoint!–As the equivalence relation is symmetric9More on equivalence classes•Consider the relation R = { (a,b) | a mod 2 = b mod 2 }–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 it’s equivalence class•There are only 2 equivalence classes formed by this equivalence relation10More 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 relation11Partitions•Consider the relation R = { (a,b) | a mod 2 = b mod 2 }•This splits the integers into two equivalence classes: even numbers and odd numbers•Those two sets together form a partition of the integers•Formally, a partition of a set S is a collection of non-empty disjoint subsets of S whose union is S•In this example, the partition is { [0], [1] }–Or { {…, -3, -1, 1, 3, …}, {…, -4, -2, 0, 2, 4, …} }12Sample questions•Which of the following are partitions of the set of integers?a) The set of even integers and the set of odd integersYes, it’s a valid partitionb) The set of positive integers and the set of negative integersNo: 0 is in neither setc) 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 3Yes, it’s a valid partitiond) The set of integers less than -100, the set of integers with absolute value not exceeding 100, and the set of integers greater than 100Yes, it’s a valid partitione) 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 6The first two sets are not disjoint (2 is in both), so it’s not a valid