1. Chapter 8 8.1 Relations and Their Properties 8.2 n-ary Relations and Their Applications 8.3 Representing Relations 8.4 Closures of Relations 8.5 Equivalence Relations 8.6 Partial Orderings 1

2. 8.1 Relations and Their Properties Definition: A binary relation R from a set A to a set B is a subset R  A  B.. Note: there are no constraints on relations as there are on functions. We have a common graphical representation of relations: Definition: A Directed graph or a Digraph D from A to B is a collection of vertices V  A  B and a collection of edges R  A  B. If there is an ordered pair e = in R then there is an arc or edge from x to y in D. The elements x and y are called the initial and terminal vertices of the edge e. 2

3. Relations and Their Properties Examples: Let A = { a, b, c} B = {1, 2, 3, 4} R is defined by the ordered pairs or edges {,, } can be represented by the digraph D: 3

4. Relations on a Set Definition: A binary relation R on a set A is a subset of A  A or a relation from A to A. Example: A = {a, b, c}, R = {,, }. Then a digraph representation of R is: Note: An arc of the form on a digraph is called a loop. Example : How many binary relations are there on a set A? Example 4: Let A be the set {1, 2, 3, 4}. Which ordered pairs are in the relation R={(a,b) | a divides b}? 4

5. Special Properties of Binary Relations Given: A Universe U A binary relation R on a subset A of U Definition: R is reflexive iff  x [ x  U  x, x  R ] Note: if U =  then the implication is true vacuously The void relation on a void Universe is reflexive! Note: If U is not void then all vertices in a reflexive relation must have loops! 5

6. Properties of Relations Example 7: Consider the following relations on {1, 2, 3, 4}: R 1 :{(1,1),(1, 2), (2, 1), (2, 2), (3, 4), (4, 1),(4, 4)}. R 2 :{(1,1),(1, 2), (2, 1)}. R 3 :{(1,1),(1, 2), (1, 4), (2, 1), (2, 2), (3, 3),(4, 1), (4, 4)}. R 4 :{(2, 1), (3, 1), (3, 2), (4, 1),(4, 2), (4, 3)}. R 5 :{(1,1),(1, 2), (1, 3), (1, 4), (2, 2), (2, 3),(2, 4), (3, 3), (3, 4), (4, 4)}. R 6 :{ (3, 4)}. Which of these relations are reflexive? 6

7. Properties of Relations Definition: R is symmetric iff  x  y [  x, y  R  y, x  R ] Note:If there is an arc there must be an arc. Definition: R is antisymmetric iff  x  y [  x, y  R  y, x  R  x  y ] Note: If there is an arc from x to y there cannot be one from y to x if x ≠ y. You should be able to show that logically: if is in R and x ≠ y then is not in R. 7

8. Properties of Relations Example 10: Which of the relations from Example 7 are symmetric and which are antisymmetric? R 1 :{(1,1),(1, 2), (2, 1), (2, 2), (3, 4), (4, 1),(4, 4)}. R 2 :{(1,1),(1, 2), (2, 1)}. R 3 :{(1,1),(1, 2), (1, 4), (2, 1), (2, 2), (3, 3),(4, 1), (4, 4)}. R 4 :{(2, 1), (3, 1), (3, 2), (4, 1),(4, 2), (4, 3)}. R 5 :{(1,1),(1, 2), (1, 3), (1, 4), (2, 2), (2, 3),(2, 4), (3, 3), (3, 4), (4, 4)}. R 6 :{ (3, 4)}. 8

9. Properties of Relations Definition: R is transitive iff  x  y  z [  x, y  R  y, z  R  x,z  R ] Note: if there is an arc from x to y and one from y to z then there must be one from x to z. This is the most difficult one to check. We will develop algorithms to check this later. Example 13: Which of the relations in Example 7 are transitive? R 1 :{(1,1),(1, 2), (2, 1), (2, 2), (3, 4), (4, 1),(4, 4)}. R 2 : {(1,1),(1, 2), (2, 1)}. R 3 :{(1,1),(1, 2), (1, 4), (2, 1), (2, 2), (3, 3),(4, 1), (4, 4)}. R 4 :{(2, 1), (3, 1), (3, 2), (4, 1),(4, 2), (4, 3)}. R 5 :{(1,1),(1, 2), (1, 3), (1, 4), (2, 2), (2, 3),(2, 4), (3, 3), (3, 4), (4,4)}. R 6 :{ (3, 4)}. 9

10. Properties of Relations Example: A: not reflexive symmetric antisymmetric transitive B: not reflexive not symmetric not antisymmetric not transitive C: not reflexive not symmetric antisymmetric not transitive D: not reflexive not symmetric antisymmetric transitive 10

11. Combining Relations A very large set of potential questions - Let R1 and R2 be binary relations on a set A: If R1 has property 1 and R2 has property 2, does R1 * R2 have property 3 where * represents an arbitrary binary set operation? 11

12. Combining Relations Example: If R1 is symmetric, and R2 is antisymmetric, does it follow that, R1 ∪ R2 is transitive? If so, prove it. Otherwise find a counterexample. Example : Let R1 and R2 be transitive on A. Does it follow that R1 ∪ R2 is transitive? 12

13. Composition Definition: Suppose R1 is a relation from A to B R2 is a relation from B to C. Then the composition of R2 with R1, denoted R2 。 R1 is the relation from A to C: If is a member of R1 and is a member of R2 then is a member of R2 。 R1. Note: For to be in the composite relation R2 。 R1 there must exist a y in B.... Note: We read them right to left as in functions. 13

14. Composition Example: Example 20: What is the composite of the relations R and S, where R is the relation form {1, 2, 3} to {1, 2, 3, 4} with R={(1, 1), (1, 4), (2, 3), (3, 1), (3, 4)} and S is the relation from {1, 2, 3, 4} to {0, 1, 2} with S={(1, 0), (2, 0), (3, 1), (3, 2), (4,1)}? 14

15. Composition Definition: Let R be a binary relation on A. Then Basis: R 1 = R Induction: R n + 1 = R n 。 R Note: an ordered pair is in R n iff there is a path of length n from x to y following the arcs (in the direction of the arrows) of R. 15

16. Composition Example : 16

17. Composition Theorem: R is transitive iff R n  R for all n > 0. 17

18. 8.2 n-ary Relations and Their Applications Definition 1: Let A 1, A 2,..., A n be sets. An n-ary relation on these sets is a subset of A 1  A 2 ...  A n The sets A 1, A 2,..., A n are called the domains of the relation, and n is called its degree. Example 4: Let R be the relation consisting of 5-ruples (A, N, S, D, T) representing airplane flights, where A is the airline, N is the starting point, D is the destination, and T is the departure time, for instance, if Nadir Express Airlines has flight 963 from Newark to Bangor at 15:00, then (Nadir, 963, Newark, Bangor, 15:00) belongs to R, the degree of this relation is 5, and its domains are the set of all airlines, the set of flight numbers, the set of cities, the set of cities(again), and the set of times. 18

19. Databases and Relations The time required to manipulate information in a database depends on how this information is stored. The operations of adding and deleting records, updating records, searching for records, and combining records from overlapping databases are performed millions of times each day in a large database. Because of the importance of these operations, various methods for representing databases have been developed. We will discuss one of these methods, called the relational data model, based on the concept of a relation. 19

20. Databases and Relations A database consists of records, which are n-tuples, made up of fields. For instance, a database of student records may be made up of fields containing the name, student number, major, and grade point average of the student. Thus, student records are represented as 4-tuples of the form (STUDENT NAME, ID NUMBER, MAJOR, GPA). A sample database of six such records is (Ackermann, 231455, Computer Science, 3.88) (Adams, 888323, physics, 3.45) (Chou, 102147, Computer Science, 3.49) (Goodfriend, 453876, Mathematics, 3.45) (Rao, 678543, Mathematics, 3.90) (Stevens, 786576, Psychology, 2.99) 20

21. Databases and Relations Relations used to represent databases are also called tables. A domain of an n-ary relation is called a primary key when the value of the n-tuple from this domain determines the n-tuple. The current collection of n-tuples in a relation is called the extension of the relation. The more permanent part of a database, including the name and attributes of the database, is called its intension. 21

22. Databases and Relations 22

23. Databases and Relations Example 5: which domains are primary keys for the n-ary relation displayed in Table1, assuming that no n-tuples will be added in the future? Combinations of domains can also uniquely identify n-tuples in an n-ary relation. When the values of a set of domains determine an n-tuple in a relation, the Cartesian product of these domains is called a composite key. Example 6: Is the Cartesian product of the domain of major fields of study and the domain of GPAs a composite key for the n-ary relation from Table1, assuming that no n-tuples are ever added? 23

24. Operations on n-ary Relations Definition 2: Let R be an n-ary relation and C a condition that elements in R may satisfy. Then the selection operator s C maps the n-ary relation R to the n-ary relation of all n-tuples from R that satisfy the condition C. Example 7: To find the records of computer science majors in the n-ary relation R shown in Table 1, we use the operator s C 1, C 1 is the condition Major =“ Computer Science.” To find the records of students who have a grade point average above 3.5 in this database, we use the operator s C 2, where C 2 is the condition GPA > 3.5. To find the records of computer science majors who have a GPA above 3.5, we use the operator s C 3, C 3 is the condition (Major=“Computer Science” Λ GPA > 3.5 ). 24

25. Operations on n-ary Relations Definition 3: The projection, P i 1, P i 2,..., P i m where i 1 < i 2 <...< i m, maps the n-tuple (a 1, a 2,..., a n ) to the m-tuple (a i 1, a i 2,..., a i m ), m  n. Example 8: what results then the projection P 1,3 is applied to the 4-tuples (2, 3, 0, 4), (Jane Doe, 234111001, Geography, 3.14), and (a 1, a 2, a 3, a 4 )? Example 9: What relation results when the projection P 1,4 is applied to the relation in Table 1? 25

26. 8.3 Representing Relations Connection Matrices Let R be a relation from A = {a 1, a 2,..., a m } to B = {b 1, b 2,..., b n }. Definition: A n m  n connection matrix M for R is defined by M ij = 1 if is in R, = 0 otherwise. Example: We assume the rows are labeled with the elements of A and the columns are labeled with the elements of B. Let A = {a, b, c}, B = {e, f, g, h}; R = {, } Then the connection matrix M for R is Note: the order of the elements of A and B matters 26

27. Representing Relations Theorem: Let R be a binary relation on a set A and let M be its connection matrix. Then R is reflexive iff M ii = 1 for all i. R is symmetric iff M is a symmetric matrix: M = M T R is antisymetric if M ij = 0 or M ji = 0 for all i ≠ j. FIGURE 1 The Zero-One Matrix for a Reflexive Relation. FIGURE 2 The Zero-One Matrices for Symmetric and Antisymmetric Relations. 27

28. Combining Connection Matrices Example 3: Suppose that the relation R on a set is represented by the matrix Is R reflexive, symmetric and/or antisymmetric? Definition: the join of two matrices M 1, M 2, denoted M 1  M 2, is the component wise boolean ‘or’ of the two matrices. Fact: If M 1 is the connection matrix for R 1 and M 2 is the connection matrix for R 2 then the join of M 1 and M 2, M 1  M 2 is the connection matrix for R 1 ∪ R 2. 28

29. Combining Connection Matrices Definition: the meet of two matrices M 1, M 2, denoted M 1  M 2 is the componentwise boolean ‘and’ of the two matrices. Fact: If M 1 is the connection matrix for R 1 and M 2 is the connection matrix for R 2 then the meet of M 1 and M 2, M 1  M 2 is the connection matrix for R 1 ∩R 2. Example 4: Suppose that the relations R 1 and R 2 on a set A are represented by the matrices. and, What are the matrices representing R 1 ∪ R 2 and R 1 ∩R 2 ? 29

30. The Composition Definition: Let M 1 be the connection matrix for R 1 M 2 be the connection matrix for R 2. The boolean product of two connection matrices M 1 and M 2, denoted M 1  M 2, is the connection matrix for the composition of R 2 with R 1, R 2 。 R 1. (M 1  M 2 ) ij =  k=1 n [(M 1 ) ik  (M 2 ) kj ] Why? 30

31. The Composition In order for there to be an arc in the composition then there must be and arc in R 1 and an arc in R 2 for some y ! The Boolean product checkes all possible y’s. If at least one such path exists, that is sufficient. Note: the matrices M 1 and M 2 must be conformable: the number of columns of M 1 must equal the number of rows of M 2. If M 1 is m  n and M 2 is n  p then M 1  M 2 is m  p. 31

32. The Composition Example : 32 (M 1  M 2 ) 12  [(M 1 ) 11  (M 2 ) 12 ]  [(M 1 ) 12  (M 2 ) 22 ]  [(M 1 ) 13  (M 2 ) 32 ]  [(M 1 ) 14  (M 2 ) 42 ] = [0  0]  [1  1]  [0  0]  [0  1]  1

33. The Composition Note: there is an arc in R 1 from node 1 in A to node 2 in B there is an arc in R 2 from node 2 in B to node 2 in C. Hence there is an arc in R 2 。 R 1 from node 1 in A to node 2 in C. 33

34. Representing Relations Using Digraphs Definition 1: A directed graph, or digraph, consists of a set V of vertices (or nodes ) together with a set E of ordered pairs of elements of V called edges (or arcs). The vertex a is called the initial vertex of the edge (a, b), and the vertex b is called the terminal vertex of this edge. An edge of the form (a, a) is represented using an arc from the vertex a back to itself. Such an edge is called a loop. 34

35. Representing Relations Using Digraphs Example 7: The directed graph with vertices a, b, c, and d, and edges (a, b), (a, d), (b, b), (b, d), (c, a), (c, b), and (d, b) is displayed in Figure 3. FIGURE 3 The Directed Graph. 35

36. Representing Relations Using Digraphs Example 8: The directed graph of the relation R={(1, 1), (1, 3), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (4, 1)} on the set {1, 2, 3, 4} is shown in figure 4. FIGURE 4 The Directed Graph of the Relations R. 36

37. Representing Relations Using Digraphs Example 9: What are the ordered pairs in the relation R represented by the directed graph shown in figure 5? FIGURE 5 The Directed Graph of the Relations R. 37

38. Representing Relations Using Digraphs Example 10: Determine whether the relations for the directed graphs shown in figure 6 are reflexive, symmetric, antisymmetric, and/or transitive. FIGURE 6 The Directed Graph of the Relations R and S. 38

39. 8.4 Closures of Relations Definition: The closure of a relation R with respect to property P is the relation obtained by adding the minimum number of ordered pairs to R to obtain property P. In terms of the digraph representation of R --To find the reflexive closure - add loops. --To find the symmetric closure - add arcs in the opposite direction. --To find the transitive closure - if there is a path from a to b, and a path from b to c, add an arc from a to c. Note: Reflexive and symmetric closures are easy. Transitive closures can be very complicated. 39

40. Closures of Relations Definition: Let A be a set and let △ = { | x in A}. △ is called the diagonal relation on A (sometimes called the equality relation E). Note that D is the smallest (has the fewest number of ordered pairs) relation which is reflexive on A. Theorem: Let R be a relation on A. The reflexive closure of R, denoted r(R), is R . Add loops to all vertices on the digraph representation of R. Put 1’s on the diagonal of the connection matrix of R. 40

41. Closures of Relations Example 1: What is the reflexive closure of the relation R={(a, b) | a< b} on the set of integers? Example 2: What is the symmetric closure of the relation R={(a, b) | a> b} on the set of positive integers? 41

42. Symmetric Closure Definition: Let R be a relation on A. Then R -1 or the inverse of R is the relation R -1 = { |  R} Note: to get R -1 reverse all the arcs in the digraph representation of R take the transpose M T of the connection matrix M of R. Note: This relation is sometimes denoted as R T or R c and called the converse of R The composition of the relation with its inverse does not necessarily produce the diagonal relation (recall that the composition of a bijective function with its inverse is the identity). 42

43. Symmetric Closure Theorem: Let R be a relation on A. The symmetric closure of R, denoted s(R ), is the relation R  R -1. Example 2: What is the symmetric closure of the relation R={(a, b) | a> b} on the set of positive integers? Example : 43

44. Symmetric Closure Examples: If A = Z, then r( ≠ ) = Z  Z If A = Z +, then s( < ) = ≠. What is the (infinite) connection matrix of s(<)? If A = Z, then s( ≦ ) = ? 44

45. Symmetric Closure Theorem: Let R 1 and R 2 be relations from A to B. Then ( R -1 ) -1 = R (R 1 ∪ R 2 ) -1 = R 1 -1 ∪ R 2 -1 (R 1 ∩ R 2 ) -1 = R 1 -1 ∩R 2 -1 (A x B) -1 = B x A  -1 =  (R 1 - R 2 ) -1 = R 1 -1 - R 2 -1 If A = B, then (R 1 R 2 ) -1 = R 2 -1 R 1 -1 If R 1  R 2 then R 1 -1  R 2 -1 45

46. Paths in Directed Graphs Definition 1: A path from a to b in the directed graph G is a sequence of edges (x 0, x 1 ), (x 1, x 2 ), (x 2, x 3 ),..., (x n-1, x n ) in G, where n is a nonnegative integer, and x 0 = a and x 1 =b. that is, a sequence of edges where the terminal vertex of an edge is the same as the initial vertex in the next edge in the path. This path is denoted by x 0, x 1, x 2,...,x n-1,x n and has length n. We view the empty set of edges as a path from a to a. a path of length n ≧ 1 that begins and ends at the same vertex is called a circuit or cycle. 46

47. Symmetric ClosurePaths in Directed Graphs Examples 3: Which of the following are paths in the directed graph show below? What are the lengths of those that are paths? Which of the paths in this list are circuits? – a,b,e,d; – a,e,c,d,b; – b,a,c,b,a,a,b; – d,c; – c,b,a; – e,b,a,d,a,b,e; 47

48. Paths in Directed Graphs Theorem: Let R be a relation on A. There is a path of length n from a to b iff  R n. Proof: (by induction) Basis: An arc from a to b is a path of length 1. which is in R 1 = R. Hence the assertion is true for n = 1. Induction Hypothesis : Assume the assertion is true for n. Show it must be true for n+1. There is a path of length n+1 from a to b iff there is an x in A such that there is a path of length 1 from a to x and a path of length n from x to b. From the Induction Hypothesis,  R and since is a path of length n,  R n. If  R And  R n, then  R n 。 R = R n+1 by the inductive definition of the powers of R. 48

49. Useful Results for Transitive Closure Theorem: If A  B and C  B, then A  C  B. Theorem: If R  S and T  U then R 。 T  S 。 U. Corollary: If R  S then R n  S n Theorem: If R is transitive then so is R n Theorem: If R k = R j for some j > k, then R j+m = R n for some n  j. – We don’t get any new relations beyond R j. – As soon as you get a power of R that is the same as one you had before, STOP. 49

50. Transitive Closure Recall that the transitive closure of a relation R, t(R), is the smallest transitive relation containing R. Also recall: R is transitive iff R n is contained in R for all n. Hence, if there is a path from x to y then there must be an arc from x to y, or is in R. Example: If A = Z and R = { } then t(R) = < Suppose R: is the following: 50

51. Transitive Closure Definition: The connectivity relation or the star closure of the relation R, denoted R*, is the set of ordered pairs such that there is a path (in R) from a to b: R * = ⋃ n=1 ∞ R n Examples: Let A = Z and R = { }. R* = <. Let A = the set of people, R = { | person x is a parent of person y}. R* = ? 51

52. Transitive Closure Theorem 2: t(R) = R*. Proof: Note: this is not the same proof as in the text. We must show that R* 1) is a transitive relation 2) contains R 3) is the smallest transitive relation which contains R. Proof: Part 2): Easy from the definition of R*. Part 1): Suppose and are in R*. Show is in R*. 52

53. Transitive Closure By definition of R*, is in R m for some m and is in R n for some n. Then is in R n R m = R m+n which is contained in R*. Hence, R* must be transitive. Part 3): Now suppose S is any transitive relation that contains R. We must show S contains R* to show R* is the smallest such relation. R  S so R 2  S 2  S since S is transitive Therefore R n  S n  S for all n. (why?) Hence S must contain R* since it must also contain the union of all the powers of R. 53

54. Transitive Closure Theorem: If |A| = n, then any path of length > n must contain a cycle. Proof: If we write down a list of more than n vertices representing a path in R, some vertex must appear at least twice in the list (by the Pigeon Hole Principle). Thus R k for k > n doesn’t contain any arcs that don’t already appear in the first n powers of R. 54

55. Transitive Closure Lemma 1: Let A be a set with n elements, and let R be a relation on A. If there is a path of length at least one in R from a to b, then there is such a path with length not exceeding n. Moreover, when a ≠ b, if there is a path of length at least one in R from a to b, then there is such a path with length not exceeding n-1. 55

56. Transitive Closure Corollary: We can find the connection matrix of t(R) by computing the join of the first n powers of the connection matrix of R. Powerful Algorithm! Example: 56

57. Transitive Closure Theorem 3: Let M R be the zero-one matrix of the relation R on the relation R on a set with n elements. Then the zero-one matrix of the transitive closure R* is Example: Find the zero-one matrix of the transitive closure of the relation R where 57

58. Transitive Closure Algorithm 1 : A Procedure for Computing the Transitive Closure procedure transitive closure (M R :zero-one nxn matrix) A := M R B := A for i :=2 to n begin A := A ⊙ M R B := B  A end {B is the zero-one matrix for R* } 58

59. 8.5 Equivalence Relations Now we 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 59

60. Equivalence Relations It is easy to recognize equivalence relations using digraphs. The subset of all elements related to a particular element forms a universal relation (contains all possible arcs) on that subset. The (sub)digraph representing the subset is called a complete (sub)digraph. All arcs are present. The number of such subsets is called the rank of the equivalence relation. 60

61. Equivalence Relations Example: A has 3 elements: 61

62. Equivalence Relations Each of the subsets is called an equivalence class. A bracket around an element means the equivalence class in which the element lies. [x] = {y | is in R} The element in the bracket is called a representative of the equivalence class. We could have chosen any one. Example: [a] = {a, c}, [c] = {a, c}, [b] = {b}. rank = 2 62

63. Equivalence Relations Definition: Let S 1, S 2,..., S n be a collection of subsets of A. Then the collection forms a partition of A if the subsets are nonempty, disjoint and exhaust A: S i  S i  S j =  if i  j ∪ S i = A FIGURE 1 A Partition of a Set. 63

64. Equivalence Relations Theorem: The equivalence classes of an equivalence relation 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. Example : A = [a]  [b] = [a]  [c] = {a}  {b,c} rank = 2 64

65. Equivalence Relations Theorem: Let R be an equivalence relation on A. Then either [a] = [b] Or [a]  [b] =  Theorem: If R 1 and R 2 are equivalence relations on A then R 1  R 2 is an equivalence relation on A. 65

66. Equivalence Relations 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: tsr(R) rank = 2 A = [a]  [b] = {a}  {b, c, d} ; A/R = {{a}, {b, c, d}} 66

67. Equivalence Relations Theorem: tsr(R) is an equivalence relation Proof: We have to be careful and 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 on the diagraph must be present when constructing r(R). If there is an arc then the symmetric closure of r(R) ensures there is an arc. 67

68. Equivalence Relations Now argue that if we construct the transitive closure of sr(R) and we add an edge 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. Hence the transitive closure of sr(R) is symmetric. Q.E.D 68

69. 8.6 Partial Orderings Definition: Let R be a relation on A. Then R is a partial order iff R is reflexive antisymmetric transitive (A, R) is called a partially ordered set or a poset. 69

70. Partial Orderings Note: It is not required that two things be related under a partial order. That's the partial part of it. If two objects are always related in a poset, it is called a total order or linear order or simple order. In this case (A, R) is called a chain. 70

71. Partial Orderings Examples: (Z  ) is a poset. In this case either a  b or b  a so two things are always related. Hence,  is a total order and (Z,  ) is a chain. If S is a set then (P(S),  ) is a poset. It may not be the case that A  B or B  A. Hence,  is not a total order. (Z+, 'divides') is a poset which is not a chain. 71

72. Partial Orderings Definition 2: The elements a and b of a poset (S, ≼ ) are called comparable if either a ≼ b or b ≼ a. When a and b are elements of S such that neither a ≼ b nor b ≼ a, a and b are called incomparable. Example 5: In the poset (Z +, |), are the integers 3 and 9 comparable? Are 5 and 7 are incomparable, because 5 ∤ 7 and 7 ∤ 5. 72

73. Partial Orderings Definition: Let R be a total order on A and suppose S  A. An element s in S is a least element of S iff sRb for every b in S. Similarly for greatest element. Note: this implies that is not in R for any a unless a = s. (There is nothing smaller than s under the order R). 73

74. Partial Orderings A Chain (A,R) is well-ordered iff every subset of A has a least element. Examples: (Z,  ) is a chain but not well-ordered. Z does not have least element. (N,  ) is well-ordered. (N,  ) is not well-ordered. Theorem 1: The Principle of Well-Ordered Induction Suppose that S is a well-ordered set. Then P(x) is true for all x  S, if Inductive Step: For every y  S, if P(x) is true for all x  S with x ≺ y, then P(y) is true. 74

75. Lexicographic Order Given two posets (A 1, R 1 ) and (A 2, R 2 ) we construct an induced partial order R on A 1 ×A 2 : R iff x 1 R 1 x 2,Or x 1 = x 2 and y 1 R 2 y 2. Example: Let A 1 = A 2 = Z + and R 1 = R 2 = 'divides‘. Then R since x 1 = x 2 and y 1 R 2 y 2. is not related under R to since x 1 = x 2 but 4 does not divide 6. R since x 1 R 1 x 2. (Note that 4 is not related to 5). 75

76. Lexicographic Order This definition extends naturally to multiple Cartesian products of partially ordered sets: A 1 × A 2 × A 3 ×...× A n. Example: Using the same definitions of A i and R i as above, R since x 1 = x 2, y 1 = y 2 and 4 divides 8. is not related to since 2 does not divide 3. 76

77. Lexicographic Order In Figure 1 the ordered pairs in Z + × Z + that are less than (3, 4) are highlighted. FIGURE 1 The Ordered Pairs Less Than (3,4) in Lexicographic Order. 77

78. Strings We apply this ordering to strings of symbols where there is an underlying 'alphabetical' or partial order (which is a total order in this case). Example: Let A = { a, b, c} and suppose R is the natural alphabetical order on A: a R b and b R c. Then Any shorter string is related to any longer string(comes before it in the ordering). If two strings have the same length then use the induced partial order from the alphabetical order: aabc R abac 78

79. Hasse or Poset Diagrams To construct a Hasse diagram: 1) Construct a digraph representation of the poset (A, R) so that all arcs point up (except the loops). 2) Eliminate all loops 3) Eliminate all arcs that are redundant because of transitivity 4) eliminate the arrows at the ends of arcs since everything points up. 79

80. Hasse Diagrams For instance, consider the directed graph for the partial ordering {(a, b)| a  b} on the set {1, 2, 3, 4}. Figure 2(a). we do not have to show these loops because they must be present. Figure 2(a). We do not have to show those edges that must be present because of transitivity. Figure 2(c). FIGURE 2 Constructing the Hasse Diagram for ({1,2,3,4}, ≦ ). 80

81. Hasse Diagrams Example 12: Draw the Hasse diagram representing the partial ordering {(a, b)| a divides b} on {1, 2, 3, 4, 6, 8, 12}. FIGURE 3 Constructing the Hasse Diagram of ({1,2,3,4,6,8,12}, ｜ ). 81

82. Hasse Diagrams Example 13: Draw the Hasse diagram representing the partial ordering {(A, B)| A  } on the power set S={a,b, c}. FIGURE 4 The Hasse Diagram of (P({a,b,c}), ). 82

83. Maximal and Minimal Elements Definition: Let (A, R) be a poset. Then a in A is a minimal element if there does not exist an element b in A such that bRa. Similarly for a maximal element. Example: In the above Hasse diagram,  is a minimal element and {a, b, c} is a maximal element. 83

84. Maximal and Minimal Elements Example 14: Which elements of the poset ({2, 4, 5, 10, 12, 20, 25}, |) are maximal, and which are minimal? FIGURE 5 The Hasse Diagram of a Poset. 84

85. Least and Greatest Elements Definition: Let (A, R) be a poset. Then a in A is the least element if for every element b in A, aRb and b is the greatest element if for every element a in A, aRb. Theorem: Least and greatest elements are unique. Proof: Assume they are not...’ Example: In the poset above {a, b, c} is the greatest element.  is the least element. 85

86. Maximal and Minimal Elements Example 15: Determine whether the posets represented by each of the Hasse diagrams in Figure 6 have a greatest element and a least element. FIGURE 6 Hasse Diagrams of Four Posets. 86

87. Upper and Lower Bounds Definition: Let S be a subset of A in the poset (A, R). If there exists an element a in A such that sRa for all s in S, then a is called an upper bound. Similarly for lower bounds. Note: to be an upper bound you must be related to every element in the set. Similarly for lower bounds. Example: In the poset in Example 13, {a, b, c}, is an upper bound for all other subsets.  is a lower bound for all other subsets. 87

88. Upper and Lower Bounds Example 18: Find the lower and upper bounds of the subsets {a, b, c}, {j, h}, and {a, c, d, f} in the poset with the Hasse diagram shown in Figure 7. FIGURE 7 The Hasse Diagram of a Poset. 88

89. Least Upper and Greatest Lower Bounds Definition: If a is an upper bound for S which is related to all other upper bounds then it is the least upper bound, denoted lub(S). Similarly for the greatest lower bound, glb(S). Example: Consider the element {a} in Example 13. Since {a, b, c}, {a, b} {a, c} and {a} are upper bounds and {a} is related to all of them, {a} must be the lub. It is also the glb. 89

90. Least Upper and Greatest Lower Bounds Example 19: Find the greatest lower bound and the least upper bound of {b, d, g}, if they exist, in the poset shown in Figure 7. FIGURE 7 The Hasse Diagram of a Poset. 90

91. Lattices Definition: A poset is a lattice if every pair of elements has a lub and a glb. Examples: In the poset (P(S),  ), lub(A, B) = A  B. What is the glb(A, B)? Consider the elements 1 and 3. Upper bounds of 1 are 1, 2, 4 and 5. Upper bounds of 3 are 3, 2, 4 and 5. 2, 4 and 5 are upper bounds for the pair 1 and 3. There is no lub since -- 2 is not related to 4 -- 4 is not related to 2 -- 2 and 4 are both related to 5. There is no glb either. The poset is not a lattice. 91

92. Lattices Example 30: Determine whether the posets represented by each of the Hasse diagrams in Figure 8 are lattices. FIGURE 8 Hasse Diagrams of Three Posets. 92

93. Topological Sorting We impose a total ordering R on a poset compatible with the partial order. Useful in PERT charts to determine an ordering of tasks. Useful in rendering in graphics to render objects from back to front to obscure hidden surfaces. A painter uses a topological sort when applying paint to a canvas - he/she paints parts of the scene furthest from the view first. This definition extends naturally to multiple Cartesian products of partially ordered sets: A 1 × A 2 × A 3 ×...× A n. 93

94. Topological Sorting Example: Consider the rectangles T and the relation R = “is more distant than.” Then R is a partial order on the set of rectangles. Two rectangles, T i and T j, are related, T i R T j, if T i is more distant from the viewer than T j. 94

95. Topological Sorting Then 1R2, 1R4, 1R3, 4R9, 4R5, 3R2, 3R9, 3R6, 8R7. The Hasse diagram for R is Draw 1 (or 8) and delete 1 from the diagram to get Now draw 4 (or 3 or 8) and delete from the diagram. Always choose a minimal element. Any one will do....and so forth. 95

96. Topological Sorting Algorithm 1 Topological Sorting procedure Topological Sorting ((S, ≼ ) : finite poset) k :=1 while S ≠  begin a k := a minimal element of S { such an element exists by Lemma 1} S := S - {a k } k := k + 1 end {a 1, a 2,..., a n is a compatible total ordering of S} 96

97. Topological Sorting Example 26: Find a compatible total ordering for the poset ({1, 2, 4, 5, 12, 20}, | ). FIGURE 9 A Topological Sort of ({1,2,4,5,12,20}, ｜ ). 97

98. Topological Sorting Example 27 : A development project at a computer company requires the completion of seven tasks. Some of these tasks can be started only after other tasks are finished. A partial ordering on tasks is set up by considering task X ≺ task Y if task Y cannot be started until task X has been completed. The Hasse diagram for the seven tasks, with respect to this partial ordering, is shown in Figure 10. Find an order in which these tasks can be carried out to complete the project. FIGURE 10 The Hasse Diagram for Seven Tasks. 98

99. Topological Sorting FIGURE 11 A Topological Sort of the Tasks. 99