1. Relations, Functions, and Matrices Mathematical Structures for Computer Science Chapter 4 Copyright © 2006 W.H. Freeman & Co.MSCS Slides Relations, Functions and Matrices

2. Binary Relations A relation establishes some kind of connection between an ordered pair of elements from some set or sets. Relations are defined in several ways One way is just to list a set of ordered pairs Another way is to define the relation in terms of x and y We write x  y to mean that the ordered pair (x, y) satisfies the relationship  Example 1: Let S = {1, 2, 3}; S  S = { (1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3) } List of ordered pairs: (1,1), (2,2), (3,3) Definition: x = y Example 2: Same set of ordered pairs List of pairs: (1, 2), (1, 3), (2, 3) Definition: x < y Section 4.1Relations1

3. Binary Operations v Binary Relations Both involve ordered pairs from some set. Operations represent an action using the elements of the ordered pair that produces a result; e.g., x ◦ y, where the operation is addition. Relations indicate that some relation exists between the elements; x  y is true if (x, y) is in the relation and false otherwise. In a way, relations are a special kind of operation where the result is Boolean. Compare to operators in a programming language. Arithmetic operators ( addition, modulus, etc.) produce an answer Relational operators ( =, >=, etc.) produce a true/false result. Section 4.1Relations2

4. Section 4.1Relations3 Binary Relations DEFINITION: BINARY RELATION on a set S: Given a set S, a binary relation on S is a subset of S  S (a set of ordered pairs of elements of S). A binary relation is always a subset of ordered pairs from some universal set with the property that: x  y  (x, y)   Example: What is the set where  on S is defined by x  y  x + y is odd? S = {1, 2}; S  S = { (1, 1), (1, 2), (2, 1), (2,2)} The set for  is {(1,2), (2,1)}. Example: S = {1, 2, 4} and the Cartesian product of S and S is: S  S = {(1,1), (1,2), (1,4), (2,1), (2,2), (2,4), (4,1), (4,2), (4,4)} Then the subset of S  S satisfying the relation x  y  x = 1/2y, is: {(1, 2), (2, 4)}

5. Section 4.1Relations4 Relations on Multiple Sets DEFINITION: RELATIONS ON MULTIPLE SETS Given two sets S and T, a binary relation from S to T is a subset of S  T. Given n sets S 1, S 2, …., S n for n > 2, an n-ary relation on S 1  S 2  …  S n is a subset of S 1  S 2  …  S n. S = {1, 2, 3} and T = {2, 4, 7}. Then x  y  x = y/2 is the set {(1,2), (2,4)}. S = {2, 4, 6, 8} and T = {2, 3, 4, 6, 7}. What is the set that satisfies the relation x  y  x = (y + 2)/2. The set is {(2,2), (4,6)}.

6. Section 4.1Relations5 Types of Relationships One-to-one: If each first component and each second component only appear once in the relation. (In examples below, S and T may be the same set) One-to-many: If a first component is paired with more than one second component. Many-to-one: If a second component is paired with more than one first component. Many-to-many: If at least one first component is paired with more than one second component and at least one second component is paired with more than one first component.

7. Section 4.1Relations6 Relationships: Examples If S = {2, 5, 7, 9}, then identify the types of the following relationships: {(5,2), (7,5), (9,2)}many-to-one {(2,5), (5,7), (7,2)}one-to-one {(7,9), (2,5), (9,9), (2,7)}many-to-many

8. Section 4.1Relations7 Properties of Relationships DEFINITION: REFLEXIVE, SYMMETRIC, AND TRANSITIVE RELATIONS Let  be a binary relation on a set S. Then:  is reflexive means (  x) (x  S  (x,x)  )  is symmetric means: (  x)(  y) (x  S  y  S  (x,y)    (y,x)   )  is transitive means: (  x)(  y)(  z) (x  S  y  S  z  S  (x,y)  (y,z)   (x,z)  )  is antisymmetric means: (  x)(  y) (x  S  y  S  (x,y)    (y,x)   x = y) Example: Consider the relation  on the set of natural numbers N. Is it reflexive? Yes, since for every nonnegative integer x, x  x. Is it symmetric? No, since x  y doesn’t imply y  x. If this was the case, then x = y. This property is called antisymmetric. Is it transitive? Yes, since if x  y and y  z, then x  z.

9. Symmetric and Antisymmetric A relation is symmetric if, for every (x, y) in the relation (y, x) must also be in the relation. If  is the relation “greater than” and S is the set of integers, is  symmetric? What about the relation “is equal to”? A relation is antisymmetric if, whenever (x, y) is in the relation and (y, x) is in the relation, then x = y. If  is the relation “greater than” and S is the set of integers, is  antisymmetric? Is it symmetric? What about the relation “is equal to”? What can you conclude about the equality relation? “Not symmetric”: informally, a relation is “not symmetric” if some (x, y) belongs to the relation but (y, x) doesn’t. Section 4.1Relations8

10. Section 4.1Relations9 Closures of Relations DEFINITION: CLOSURE OF A RELATION A binary relation  * on set S is the closure of a relation  on S with respect to property P if: 1.  * has the property P 2.    * 3.  * is a subset of any other relation on S that includes  and has the property P Example: Let S = {1, 2, 3} and  = {(1,1), (1,2), (1,3), (3,1), (2,3)}. This is not reflexive, symmetric, or transitive. The closure of  with respect to reflexivity is {(1,1),(1,2),(1,3), (3,1), (2,3), (2,2), (3,3)} and it contains . The closure of  with respect to symmetry is {(1,1), (1,2), (1,3), (3,1), (2,3), (2,1), (3,2)}. The closure of  with respect to transitivity is {(1,1), (1,2), (1,3), (3,1), (2,3), (3,2), (3,3), (2,1), (2,2)}.

11. Section 4.1Relations10 Exercise: Closures of Relations Find the reflexive, symmetric and transitive closure of the relation {(a,a), (b,b), (c,c), (a,c), (a,d), (b,d), (c,a), (d,a)} on the set S = {a, b, c, d} Practice 7 – see solution in back of book

12. Section 4.1Relations11 Partial Ordering and Equivalence Relations DEFINITION: PARTIAL ORDERING A binary relation on a set S that is reflexive, antisymmetric, and transitive is called a partial ordering on S. If  is a partial ordering on S, then the ordered pair (S,  ) is called a partially ordered set (also known as a poset). Denote an arbitrary, partially ordered set by (S,  ); in any particular case,  has some definite meaning such as “less than or equal to,” “is a subset of,” “divides,” and so on. It’s a generic symbol representing some relation. Examples: On N, x  y  x  y. On {0,1}, x  y  x = y 2  

13. Terminology – Partial Ordering Let (S,  ) be a partially ordered set. For x, y in S, if x  y (x is related to y) but x  y we say that x is a predecessor of y, (x < y ) or y is a successor of x. If x < y and there is no z such that x < z < y, then x is an immediate predecessor of y. The partial ordering defined by this kind of relation connects some, but not necessarily all, of the elements Two elements may be unrelated to each other, other elements may not be related to any other element. If x  y for all x, then x is a least element of the partially ordered set. If there’s no element x in S with x < y then y is said to be minimal. Easier to understand by looking at a picture. Section 4.1Relations12

14. Section 4.1Relations13 Hasse Diagram Hasse Diagram: A diagram used to visually depict a partially ordered set if S is finite. Each of the elements of S is represented by a dot, called a node, or vertex, of the diagram. If x is an immediate predecessor of y, then the node for y is placed above the node for x and the two nodes are connected by a straight-line segment. Example: Given the partial ordering on a set S = {a, b, c, d, e, f } as {(a,a), (b,b), (c,c), (d,d), (e,e), (f, f), (a, b), (a,c), (a,d), (a,e), (d,e)}, the Hasse diagram is:

15. Section 4.1Relations14 Equivalence Relation DEFINITION: EQUIVALENCE RELATION A binary relation on a set S that is reflexive, symmetric, and transitive is called an equivalence relation on S. Examples: On N, x  y  x + y is even. On {1, 2, 3}, 

16. Section 4.1Relations15 Partitioning a Set DEFINITION: PARTITION OF A SET It is a collection of nonempty disjoint subsets of S whose union equals S. For  an equivalence relation on set S and x  S, then [x] is the set of all members of S to which x is related, called the equivalence class of x. Thus: [x] = {y | y  S  x  y} Hence, for  = {(a,a), (b,b), (c,c), (a,c), (c,a)} [a] = {a, c} = [c]

17. Section 4.1Relations16 Congruence Modulo n DEFINITION: CONGRUENCE MODULO n For integers x and y and positive integer n, x = y(mod n) if x  y is an integral multiple of n. This binary relation is always an equivalence relation Congruence modulo 4 is an equivalence relation on Z. Construct the equivalence classes [0], [1], [2], and [3]. Note that [0], for example, will contain all integers differing from 0 by a multiple of 4, such as 4, 8, 12, and so on. The distinct equivalence classes are: [0] = {..., -8, -4, 0, 4, 8,...} [1] = {..., -7, -3, 1, 5, 9,...} [2] = {..., -6, -2, 2, 6, 10,...} [3] = {..., -5, -1, 3, 7, 11,...}

18. Section 4.1Relations17 Partial Ordering and Equivalence Relations

19. Section 4.1Relations18 Exercises 1. Which of the following ordered pairs belongs to the binary relation  on N? x  y  x + y < 7; (1,3), (2,5), (3,3), (4,4) x  y  2x + 3y = 10; (5,0), (2,2), (3,1), (1,3) 2. Show the region on the Cartesian plane such that for a binary relation  on R: x  y  x 2 + y 2  25 x  y  x  y 3. Identify each relation on N as one-to-one, one-to-many, many- to-one or many-to-many:   

20. Section 4.1Relations19 Exercises 4. S = {0, 1, 2, 4, 6}. Which of the following relations are reflexive, symmetric, antisymmetric, and transitive. Find the closures for each category for all of them:       5. For the relation {(1,1), (2,2), (1,2), (2,1), (1,3), (3,1), (3,2), (2,3), (3,3), (4,4), (5,5), (4,5), (5,4)} What is [3] and [4]?