DOCSLIB.ORG

CS202 Fall 2011 Lecture (Lost Count) – 12/1 Relations

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
CS202 Fall 2011 Lecture (Lost Count) – 12/1 Relations 11/30/11 Relations CS202 Fall 2011 Lecture (lost count) – 12/1 Recall the definition of the Cartesian (Cross) Product: The Cartesian Product of sets A and B, A x B, is the set Relations A x B = {(a,b) : a∈A and b∈B}. Prof. Tanya Berger-Wolf aRb or (a,b) ∈ R means “a is related to b” A relation is just any subset of the Cartesian Product: R ⊆ AxB Ex1: A = {0,1,2}, B = {2,3} => AxB = {(0,2), (0,3), (1,2), (1,3), (2,2), (2,3)} R = {(a,b) | a < b}. So R = {(0,2), (0,3), (1,2), (1,3), (2,3)} = AxB-{(2,2)} Ex2: A = students at UIC; B = courses at UIC. R = {(a,b) | student a is enrolled in class b} Ex3: A = {3 letter strings}, B = {all English words} R = {(a,b) | a is a prefix of b} 1 Relations and Functions Properties of Relations Recall the definition of a function: f = {(a,b) : b = f(a) , a∈A and b∈B} Is every function a relation? Yes, a function is Reflexivity: a special kind of A relation R on AxA is reflexive if for all a∈A, (a,a) ∈R. relation. Symmetry: Draw Venn diagram of cross products, relations, Cross product A relation R on AxA is symmetric if functions (a,b) ∈ R implies (b,a) ∈ R . 1 11/30/11 Properties of Relations Properties of Relations - techniques… How can we check for transitivity? Draw a picture of the relation (called a “graph”) [Epp p. 580]. Transitivity: " ! Vertex for every element of A A relation on AxA is transitive if " ! Edge for every element of R (a,b) ∈ R and (b,c) ∈ R imply (a,c) ∈ R. R={(1,1),(1,2),(1,3),(1,4),(2,2),(2,3),(2,4),(3,3),(3,4),(4,4)} Anti-symmetry [Epp p.632]: A relation on AxA is anti-symmetric if (a,b) ∈ R implies (b,a) ∉ R. 1 A “short cut” 2 must be present for EVERY path of length 2. 3 4 Properties of Relations - techniques… Properties of Relations - techniques… How can we check for the reflexive property? How can we check for the symmetric property? Draw a picture of the relation (called a “graph”). Draw a picture of the relation (called a “graph”). " ! Vertex for every element of A " ! Vertex for every element of A " ! Edge for every element of R " ! Edge for every element of R R={(1,1),(1,2),(1,3),(1,4),(2,2),(2,3),(2,4),(3,3),(3,4),(4,4)} R={(1,1),(1,2),(1,3),(1,4),(2,2),(2,3),(2,4),(3,3),(3,4),(4,4)} 1 Loops must exist 1 EVERY edge 2 on EVERY vertex. 2 must have a return edge. 3 3 4 4 2 11/30/11 Properties of Relations - techniques… Properties of Relations - techniques… How can we check for the anti-symmetric property? Let R be a relation on People, Draw a picture of the relation (called a “graph”). R={(x,y): x and y have lived in the same country} " ! Vertex for every element of A " ! Edge for every element of R ? 1 2 R={(1,1),(1,2),(1,3),(1,4),(2,2),(2,3),(2,4),(3,3),(3,4),(4,4)} ? ? 3 Is R transitive? Is it symmetric? 1 NO edge can No Yes 2 have a return edge. Is it reflexive? Yes Is it anti-symmetric? No 3 4 Properties of Relations - techniques… Properties of Relations - techniques… Let R be a relation on positive integers, Let R be a relation on positive integers, R={(x,y): 3|(x-y)} R={(x,y): 3|(x-y)} Suppose (x,y) and (y,z) are in R. Is (x,x) in R, for all x? Definition of Definition of Then we can write 3j = (x-y) and 3k = (y-z) Does 3k = (x-x) for some k? “divides” “divides” Can we say 3m = (x-z)? Is (x,z) in R? Yes, for k=0. Add prev eqn to get: 3j + 3k = (x-y) + (y-z) 3(j + k) = (x-z) Is R transitive?! Yes Is R transitive? Yes Is it reflexive? Yes 3 11/30/11 Properties of Relations - techniques… Properties of Relations - techniques… Let R be a relation on positive integers, Let R be a relation on positive integers, R={(x,y): 3|(x-y)} R={(x,y): 3|(x-y)} Suppose (x,y) is in R. Suppose (x,y) is in R. Definition of Definition of Then 3j = (x-y) for some j. “divides” Then 3j = (x-y) for some j. “divides” Does 3k = (y-x) for some k? Does 3k = (y-x) for some k? Yes, for k=-j. Yes, for k=-j. Is R transitive?! Yes Is it symmetric?! Yes Is R transitive?! Yes Is it symmetric?! Yes Is it reflexive?! Yes Is it reflexive?! Yes Is it anti-symmetric?! No More than one relation More than one relation Let R be a relation from A to B (R ⊆ AxB), and let S be a relation from B to C (S ⊆ BxC). The composition of R and S is the relation from A to C (S°R ⊆ AxC): Suppose we have 2 relations, R1 and R2, and recall that relations are just sets! So we can take unions, intersections, complements, S°R = {(a,c): ∃ b∈B, (a,b) ∈ R, (b,c) ∈ S} symmetric differences, etc. There are other things we can do as well… A B C R S 1 x s 2 y t 3 z u 4 v S°R = {(1,u),(1,v),(2,t),(3,t),(4,u)} 4 11/30/11 More than one relation More than one relation Let R be a relation on A. Inductively define Let R be a relation on A. Inductively define R1 = R R1 = R Rn = Rn-1 ° R Rn = Rn-1 ° R A A A A A A R R1 R R2 1 1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 3 4 4 4 4 4 4 … = R4 = R5 R3 = R2°R = {(1,1),(1,2),(1,3),(2,3),(3,3),(4,1),(4,2),(4,3)} R2 = R1°R = {(1,1),(1,2),(1,3),(2,3),(3,3),(4,1), (4,2)} = R6… Relations - A Theorem: Relations - A Theorem: If R is a transitive relation, then Rn ⊆ R, ∀n. Proof by induction on n. If R is a transitive relation, then Rn ⊆ R, ∀n. Base case (n=1): R1 ⊆ R because by definition, R1 = R. Inductive Step: IH: if R is transitive, then Rn-1 ⊆ R. Prove: if R is transitive, then Rn ⊆ R.! Aside: notice that this theorem allows us to conclude that Typical the previous relation was NOT transitive. We are trying to prove that Rn ⊆ R. To do this, we select an way of Recall: “if p then q” ≡ “if not q then not p.” element of Rn and show that it is also an element of R. proving subset. We saw that Rn was not a subset of R (it was growing on Let (a,b) be an element of Rn. Since Rn = Rn-1 ° R, we know every iteration). there is an x so that (a,x) ∈ R and (x,b) ∈ Rn-1. Therefore, R is not transitive. By IH, since Rn-1 ⊆ R, (x,b) ∈ R. But wait, if (a,x) ∈ R, and (x,b) ∈ R, and R is transitive, then (a,b) ∈ R. 5 11/30/11 Relations - more techniques… Relations - more techniques… Suppose we have our old relation R on AxB, where A={1,2,3,4}, and B={u,v,w}, R={(1,u),(1,v),(2,w),(3,w),(4,u)}. Some things to think about. Then we can represent R as: Let R be a relation on a set A, and let MR be the matrix representation of R. u v w The labels on Then R is reflexive if, ______________. the outside are 1 1 1 0 for clarity. It’s really the ! 2 0 0 1 matrix in the u v w A. All entries in MR are 1. middle that’s 3 0 0 1 B.! The \ diagonal of MR contains important. u 1 1 0 only 1s. 4 1 0 0 v 0 1 1 C.! The first column of MR contains no 0s. This is a |A| x |B| matrix whose entries indicate membership w 0 0 1 D.! None of the above. in R. Relations - more techniques… Relations - more techniques… Suppose we have R1 and R2 defined on A: R1 u v w R2 u v w Some things to think about. u 1 0 1 u 1 1 0 Let R be a relation on a set A, and let MR be the matrix representation of R. Then R is symmetric if, ______________. v 0 0 1 v 0 1 1 w 1 1 0 w 0 0 1 1 1 1 A.! All entries above the \ are 1. Then R ∪ R is the bitwise “or” of the entries:! 0 1 1 u v w 1 2 B.! The first and last columns of 1 1 1 MR1∪R2 = MR1 v MR2 u 1 0 1 MR contain an equal # of 0s. C.! MR is visually symmetric about Then R ∩ R is the bitwise “and” of the entries: 1 0 0 v 0 0 1 the \ diagonal. 1 2 M = M ∧ M 0 0 1 D.! None of the above.
Load more

Recommended publications
  • Section 4.1 Relations
    Binary Relations (Donny, Mary) (cousins, brother and sister, or whatever) Section 4.1 Relations - to distinguish certain ordered pairs of objects from other ordered pairs because the components of the distinguished pairs satisfy some relationship that the components of the other pairs do not. 1 2 The Cartesian product of a set S with itself, S x S or S2, is the set e.g. Let S = {1, 2, 4}. of all ordered pairs of elements of S. On the set S x S = {(1, 1), (1, 2), (1, 4), (2, 1), (2, 2), Let S = {1, 2, 3}; then (2, 4), (4, 1), (4, 2), (4, 4)} S x S = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2) , (3, 3)} For relationship of equality, then (1, 1), (2, 2), (3, 3) would be the A binary relation can be defined by: distinguished elements of S x S, that is, the only ordered pairs whose components are equal. x y x = y 1. Describing the relation x y if and only if x = y/2 x y x < y/2 For relationship of one number being less than another, we Thus (1, 2) and (2, 4) satisfy . would choose (1, 2), (1, 3), and (2, 3) as the distinguished ordered pairs of S x S. x y x < y 2. Specifying a subset of S x S {(1, 2), (2, 4)} is the set of ordered pairs satisfying The notation x y indicates that the ordered pair (x, y) satisfies a relation .
    [Show full text]
  • Binary Relations.Pdf
    Binary Relations Any set of ordered pairs defines a binary relation. We express a particular ordered pair, (x, y) R, where R is a binary relation, as xRy. Properties Properties of a binary relation R on a set X: a. reflexive: if for every x X, xRx holds, i.e. (x, x) R. b. symmetric: if for every x and y in X, whenever xRy holds, then yRx holds. c. transitive: if for every x, y, and z in X, whenever xRy holds and yRz holds, then xRz holds. Graphical Representation Suppose that X = {A, B, C, D} and that R = {(A, B), (B, D), (C, C), (D, B)}. We can represent the binary relation R as a set, as a graph or as a Boolean matrix. Graphically, we have Boolean Matrix Representation As a Boolean matrix we have A B C D A 0 1 0 0 B 0 0 0 1 C 0 0 1 0 D 0 1 0 0 What would a Boolean matrix look like if it represented a reflexive binary relation, a symmetric binary relation, a transitive binary relation, any combination of the above? Reflexive Relation A B C D A - 1 - - B - - - 1 C - - 1 - D - 1 - - Symmetric Relation A B C D A - 1 - - B - - - 1 C - - 1 - D - 1 - - Transitive Relation A B C D A - 1 - - B - - - 1 C - - 1 - D - 1 - - Click here for Answers Composition of Relations: Let R be a relation from X to Y, and S be a relation from Y to Z. Then a relation written as R o S is called a composite relation of R and S where R o S = { (x, z) | x X z Z ( y) (y Y (x, y) R (y, z) S) } We can also write the composition as R o S = { (x, z) | x X z Z ( y) (y Y xRy ySz) } Note: Relational composition can be realized as matrix multiplication.
    [Show full text]
  • Discrete Mathematics
    Discrete Mathematics Chapter 7 Relations 7.1 Relations and their properties. ※The most direct way to express a relationship between elements of two sets is to use ordered pairs. For this reason, sets of ordered pairs are called binary relations. Example 1. A : the set of students in your school. B : the set of courses. R = { (a, b) : a∈A, b∈B, a is enrolled in course b } 7.1.1 Def 1. Let A and B be sets. A binary relation from A to B is a subset R of A×B. ( Note A×B = { (a,b) : a∈A and b∈B } ) Def 1’. We use the notation aRb to denote that (a, b)∈R, and aRb to denote that (a,b)∉R. Moreover, a is said to be related to b by R if aRb. 7.1.2 Example 3. Let A={0, 1, 2} and B={a, b}, then {(0,a),(0,b),(1,a),(2,b)} is a relation R from A to B. This means, for instance, that 0Ra, but that 1Rb A B R ⊆ A×B = { (0,a) , (0,b) , (1,a) (1,b) , (2,a) , (2,b)} 0 a ∈R ∈R 1 b 2 R 7.1.3 Example (上例) : A : 男生, B : 女生, R : 夫妻關系 A : 城市, B : 州, 省 R : 屬於 (Example 2) Note. Relations vs. Functions A relation can be used to express a 1-to-many relationship between the elements of the sets A and B. ( function 不可一對多，只可多對一 ) Def 2. A relation on the set A is a subset of A × A ( i.e., a relation from A to A ).
    [Show full text]
  • Packet #5: Binary Relations Applied Discrete Mathematics
    CSC224/226 – Packet 5 – Binary Relations Packet #5: Binary Relations Applied Discrete Mathematics Table of Contents Binary Relations Summary Page 1 Binary Relations Examples Page 2 Properties of Relations Page 3 Examples Pages 4-5 Representations Page 6 Composites Page 8 Closures Pages 10-11 Equivalence Relations Pages 12-13 Posets & Hasse Diagrams Pages 14-17 CSC224 – Packet 5 – Binary Relations CSC224/226 – Packet 5 – Binary Relations Binary Relations Summary I. Definition: A binary relation is a set of points, or ordered pairs. The ordered pairs are of the form (a,b) where a is a member of a set A (a ∈ A) and b is a member of a set B (b ∈ B). Any subset R ⊆ A x B is a binary relation. (Remember that A x B is the set of ordered pairs {(a,b) | a∈A ∧ b ∈ B}) II. Properties: Consider the binary relation R ⊆ A x A. A. Reflexive:R is reflexive if ∀ a ∈ A, (a,a) ∈ R B. Irreflexive: R is irreflexive if ∀ a ∈ A, (a,a) ∉ R. Is ¬ (R is reflexive) ⇔ (R is irreflexive)? C. Non-reflexive: ¬ (R is reflexive) D. Symmetric: R is symmetric if ∀ a, b ∈ A, [ ((a,b) ∈ R) ⇔ ((b,a) ∈ R) ] E. Antisymmetric: R is antisymmetric if ∀ a, b ∈ A, [(a,b) ∈ R ∧ (b,a) ∈ R] ⇒ (b = a) F. Asymmetric: R is asymmetric if ∀ a, b ∈ A, [(a,b) ∈ R] ⇒ [(b,a) ∉ R] Asymmetric = irreflexive AND antisymmetric G. Transitive: R is transitive if ∀ a, b ∈ A, [(a,b) ∈ R ∧ (b,c) ∈ R] ⇒ (a,c) ∈ R Also: R2 ⊆ R if and only if R is transitive.
    [Show full text]
  • Chapter 5 Partial Orders, Lattices, Well Founded Orderings
    Chapter 5 Partial Orders, Lattices, Well Founded Orderings, Equivalence Relations, Distributive Lattices, Boolean Algebras, Heyting Algebras 5.1 Partial Orders There are two main kinds of relations that play a very important role in mathematics and computer science: 1. Partial orders 2. Equivalence relations. In this section and the next few ones, we define partial orders and investigate some of their properties. 459 460 CHAPTER 5. PARTIAL ORDERS, EQUIVALENCE RELATIONS, LATTICES As we will see, the ability to use induction is intimately related to a very special property of partial orders known as well-foundedness. Intuitively, the notion of order among elements of a set, X, captures the fact some elements are bigger than oth- ers, perhaps more important, or perhaps that they carry more information. For example, we are all familiar with the natural ordering, , of the integers ≤ , 3 2 1 0 1 2 3 , ··· − ≤− ≤− ≤ ≤ ≤ ≤ ≤··· the ordering of the rationals (where p1 p2 iff q1 q2 p2q1 p1q2 ≤ − 0, i.e., p2q1 p1q2 0ifq1q2 > 0 else q1q2 ≥ − ≥ p2q1 p1q2 0ifq1q2 < 0), and the ordering of the real numbers.− ≤ In all of the above orderings, note that for any two number a and b, either a b or b a. ≤ ≤ We say that such orderings are total orderings. 5.1. PARTIAL ORDERS 461 A natural example of an ordering which is not total is provided by the subset ordering. Given a set, X, we can order the subsets of X by the subset relation: A B, where A, B are any subsets of X. ⊆ For example, if X = a, b, c , we have a a, b .
    [Show full text]
  • Relations III
    CS 441 Discrete Mathematics for CS Lecture 23 Relations III. Milos Hauskrecht [email protected] 5329 Sennott Square CS 441 Discrete mathematics for CS M. Hauskrecht Composite of relations Definition: Let R be a relation from a set A to a set B and S a relation from B to a set C. The composite of R and S is the relation consisting of the ordered pairs (a,c) where a A and c C, and for which there is a b B such that (a,b) R and (b,c) S. We denote the composite of R and S by S o R. Example: AB C c a b RS CS 441 Discrete mathematics for CS M. Hauskrecht 1 Composite of relations Definition: Let R be a relation from a set A to a set B and S a relation from B to a set C. The composite of R and S is the relation consisting of the ordered pairs (a,c) where a A and c C, and for which there is a b B such that (a,b) R and (b,c) S. We denote the composite of R and S by S o R. (a,c) S o R Example: AB C c a b RS CS 441 Discrete mathematics for CS M. Hauskrecht Composite of relations Definition: Let R be a relation from a set A to a set B and S a relation from B to a set C. The composite of R and S is the relation consisting of the ordered pairs (a,c) where a A and c C, and for which there is a b B such that (a,b) R and (b,c) S.
    [Show full text]
  • Today's Lecture Chapter 8: Section
    Announcements ICS 6B } Regrades for Quiz #3 and Homeworks #4 & Boolean Algebra & Logic 5 are due Thursday Lecture Notes for Summer Quarter, 2008 Michele Rousseau Set 7 – Ch. 8.4, 8.5, 8.6 Lecture Set 7 - Chpts 8.4, 8.5, 8.6 2 Grades Overall Grades & Quiz #3… Quiz #3 90-100 } Max: 100% } Min: 42% 80-89 } Median: 83% 70-79 Overall 60-69 less } Max: 99% than 50 } Min 45% 01234567 } Median: 84% Overall Quiz #3 Lecture Set 7 - Chpts 8.4, 8.5, 8.6 3 Lecture Set 7 - Chpts 8.4, 8.5, 8.6 4 Today’s Lecture } Chapter 8 8.4, 8.5, 8.6 ● Closures of Relations 8.4 Chapter 8: Section 8.4 ● Equivalence Relations 8.5 ● Partial Orderings 8.6 Closures of Relations Lecture Set 7 - Chpts 8.4, 8.5, 8.6 5 1 Closure of Relations Reflexive Closure Let R be a relation on set A. Example Let P be a property (reflexive, symmetric, etc.) A1,2,3 The closure of R with respect to the property P R1,1,1,2,1,3 is the smallest relation containing R which has this property. P“being reflexive” } In other words, add the minimum number of pairs to obtain R is not reflexive, b/c its missing 2,2, 3,3 property P. The smallest reflexive relation containing R is } Note: This may not be possible. S1,1,1,2,1,3,2,2,3,3 Example: This is the reflexive closure of R & it’s the intersection A1,2,3,4, R1,1,1,3,14 of all of the reflexive relations that contain R P is being “irreflexive” Any relation on A which is reflexive and contains R } If the closure S of R w.r.t.
    [Show full text]
  • Relations and Functions
    2 Relations and Functions 2.1 INTRODUCTION The word relation implies an association of two objects, two persons, etc., according to some property possessed by them, e.g. (i) A is the son of B. The relation R here is “is the son of ”. (ii) Delhi is the capital of INDIA. The relation R here is “is the capital of ”. In Mathematics we can study relationship between a real number and a number which is smaller than it, between a program and a variable it uses. Relationship between elements of sets is represented using a structure called relation. 2.2 RELATION OR BINARY RELATION Let A and B be two non-empty sets. A binary relation or simply a relation from A to B is a sub- set of A × B. Given x ∈A and y ∈ B, we write x R y if (x, y) ∈ R and x R y if (x, y) ∉ R. If R is relation from A to A, then R is said to be relation on A. A binary relation on a set A is a subset of A × A. This definition needs some explanation, since the connection between it and the informal idea of a relation might not be obvious. Note: The word ‘binary’ refers to the fact that the relation is between two elements of A. As this is the only kind of relation we will be studying, we will omit the word ‘binary’ in what follows. The easiest way to do this is by means of an example. Let A = {1, 3, 5} be a set.
    [Show full text]