Chapter 8 8.1 Relations and Their Properties 8.2 n-ary Relations and Their Applications 8.3...

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Chapter 88.1 Relations and Their Properties8.2 n-ary Relations and Their Applications8.3 Representing Relations8.4 Closures of Relations8.5 Equivalence Relations8.6 Partial Orderings

8.1 Relations and Their Properties• Definition: A binary relation R from a set A to a set B is a subset R Í 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 Í A È B and a collection of edges R Í A B . If there is an ordered pair e = <x, y> 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.

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

{<a, 1>, <a, 2>, <c, 4>}• can be represented by the digraph D:

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 = {<a, a>, <a, b>, <a, c>}.

• Then a digraph representation of R is:

• Note: An arc of the form <x, x> 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}?

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 Î U ® < x, x >ÎR ]• Note: if U = Æ 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!

Properties of Relations• Example 7: Consider the following relations on {1, 2, 3,

4}: R1:{(1,1) ,(1, 2), (2, 1), (2, 2), (3, 4), (4, 1) ,(4, 4)}.

R2:{(1,1) ,(1, 2), (2, 1)}.

R3:{(1,1) ,(1, 2), (1, 4), (2, 1), (2, 2), (3, 3) ,(4, 1), (4, 4)}.

R4:{(2, 1), (3, 1), (3, 2), (4, 1) ,(4, 2), (4, 3)}.

R5:{(1,1) ,(1, 2), (1, 3), (1, 4), (2, 2),

(2, 3) ,(2, 4), (3, 3), (3, 4), (4, 4)}.R6:{ (3, 4)}.• Which of these relations are reflexive?

Properties of Relations• Definition: R is symmetric iff

"x"y [< x, y > Î R ® < y, x > Î R ]• Note:If there is an arc <x, y> there must be an arc <y ,

• Definition: R is antisymmetric iff"x"y [ < x, y > Î R < Ù y, x > Î 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 <x, y> is in R and x ≠ y then <y, x> is not in R.

Properties of Relations• Example 10: Which of the relations from Example 7 are symmetric

and which are antisymmetric?R1:{(1,1) ,(1, 2), (2, 1), (2, 2), (3, 4), (4, 1) ,(4, 4)}.

R2:{(1,1) ,(1, 2), (2, 1)}.

R3:{(1,1) ,(1, 2), (1, 4), (2, 1), (2, 2), (3, 3) ,(4, 1), (4, 4)}.

R4:{(2, 1), (3, 1), (3, 2), (4, 1) ,(4, 2), (4, 3)}.

R5:{(1,1) ,(1, 2), (1, 3), (1, 4), (2, 2),

(2, 3) ,(2, 4), (3, 3), (3, 4), (4, 4)}.R6:{ (3, 4)}.

Properties of Relations• Definition: R is transitive iff

"x"y"z [ < x, y > Î R < Ù y, z > Î R ® < x,z > Î 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?R1:{(1,1) ,(1, 2), (2, 1), (2, 2), (3, 4), (4, 1) ,(4, 4)}. R2: {(1,1) ,(1, 2), (2, 1)}.

R3:{(1,1) ,(1, 2), (1, 4), (2, 1), (2, 2), (3, 3) ,(4, 1), (4, 4)}.

R4:{(2, 1), (3, 1), (3, 2), (4, 1) ,(4, 2), (4, 3)}.

R5:{(1,1) ,(1, 2), (1, 3), (1, 4), (2, 2), (2, 3) ,(2, 4), (3, 3), (3, 4), (4,4)}.

R6:{ (3, 4)}.

Properties of Relations• Example:

A: not reflexive symmetric antisymmetrictransitive

B:not reflexivenot symmetricnot antisymmetricnot transitive

C: not reflexivenot symmetricantisymmetric not transitive

D: not reflexivenot symmetricantisymmetrictransitive

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 3where * represents an arbitrary binary set operation?

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?

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 <x. y> is a member of R1 and <y, z> is a member of R2 then <x, z> is a member of R2。 R1.• Note: For <x, z> 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.

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)}?

Composition • Definition: Let R be a binary relation on A. ThenBasis: R1 = RInduction: Rn + 1= Rn。 R• Note: an ordered pair <x, y> is in Rn iff there is a path

of length n from x to y following the arcs (in the direction of the arrows) of R.

Composition

Example :

Composition • Theorem: R is transitive iff Rn Í R for all n > 0 .

8.2 n-ary Relations and Their Applications• Definition 1: Let A1, A2, . . ., An be sets. An n-ary relation

on these sets is a subset of A1 A2 . . . An

The sets A1, A2, . . ., An 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.

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

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)

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.

Databases and Relations

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?

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 sC 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 sC1, C1 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 sC2, where C2 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 sC3 , C3 is the condition (Major=“Computer Science” Λ GPA > 3.5 ).

Operations on n-ary Relations• Definition 3: The projection, Pi1

, Pi2,. . . , Pim

i1 < i2 <. . .< im , maps the n-tuple (a1, a2,. . ., an) to the m-tuple (ai1

, ai2 , . . ., aim

) , m n.

• Example 8: what results then the projection P1,3 is applied to the 4-tuples (2, 3, 0, 4), (Jane Doe, 234111001, Geography, 3.14), and (a1, a2, a3, a4)?

• Example 9: What relation results when the projection P1,4 is applied to the relation in Table 1?

8.3 Representing Relations Connection Matrices

• Let R be a relation fromA = {a1, a2, . . . , am} to B = {b1, b2, . . . , bn}.

• Definition: An m n connection matrix M for R is

defined by Mij = 1 if <ai, bj> 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 = {<a, e>, <c, g>}• Then the connection matrix M for R is

• Note: the order of the elements of A and B matters

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 Mii = 1 for all i.• R is symmetric iff M is a symmetric matrix: M = MT

• R is antisymetric if Mij = 0 or Mji = 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

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 M1, M2, denoted

M1 Ú M2 , is the component wise boolean ‘or’ of the two

matrices.• Fact: If M1 is the connection matrix for R1 and M2 is the

connection matrix for R2 then the join of M1 and M2 ,

M1 Ú M2 is the connection matrix for R1 R∪ 2 .

Combining Connection Matrices• Definition: the meet of two matrices M1, M2, denoted

M1 Ù M2 is the componentwise boolean ‘and’ of the two matrices.

• Fact: If M1 is the connection matrix for R1 and M2 is the connection matrix for R2 then the meet of M1 and M2, M1 Ù M2 is the connection matrix for R1∩R2 .

• Example 4: Suppose that the relations R1 and R2 on a set A are represented by the matrices.

What are the matrices representing R1 R∪ 2 and R1∩R2 ?

The Composition• Definition: Let

M1 be the connection matrix for R1

M2 be the connection matrix for R2.

• The boolean product of two connection matrices M1 and M2, denoted M1 Ä M2 , is the connection matrix for the composition of R2 with R1 , R2。 R1.

(M1 Ä M2 )ij = Úk=1

n [(M1 )ik Ù (M2 )kj ]

• Why?

The Composition• In order for there to be an arc <x, z> in the composition then there must be and arc <x, y> in R1 and an arc <y, z>

in R2 for some y !• The Boolean product checkes all possible y’s. If at least one such path exists, that is sufficient.• Note: the matrices M1 and M2 must be conformable: the

number of columns of M1 must equal the number of rows

of M2.

• If M1 is m n and M2 is n p then M1 Ä M2 is m p.

The CompositionExample :

(M1 Ä M2 )12 = [(M1 )11 Ù (M2 )12 ]Ú[(M1 )12

Ù(M2 )22 ]

Ú[(M1 )13 Ù (M2 )32 ]Ú[(M1)14 Ù (M2 )42 ]

= [0 Ù 0] Ú [1 Ù 1] Ú [0 Ù 0] Ú [0 Ù 1] = 1

The Composition• Note:• there is an arc in R1 from node 1 in A to node 2 in B

• there is an arc in R2 from node 2 in B to node 2 in C.

• Hence there is an arc in R2。 R1 from node 1 in A to node 2 in C.

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.

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

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.

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.

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.

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.

Closures of Relations• Definition: Let A be a set and let △ = {<x, x> | 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 ÈD. • Add loops to all vertices on the digraph representation of R. • Put 1’s on the diagonal of the connection matrix of R.

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?

Symmetric Closure• Definition: Let R be a relation on A. Then R-1 or the inverse of R is the relation R-1 = {< y, x >|< x, y >

ÎR}• Note: to get R-1

• reverse all the arcs in the digraph representation of R • take the transpose MT of the connection matrix M of R.• Note: This relation is sometimes denoted as RT or Rc 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).

Symmetric Closure• Theorem: Let R be a relation on A. The symmetric closure of R, denoted s(R ), is the relation R È R-1 .Example 2: What is the symmetric closure of the

relation R={(a, b) | a> b} on the set of positive integers?

• Example :

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( ) = ?≦

Symmetric Closure• Theorem: Let R1 and R2

be relations from A to B. Then

• ( R -1) -1 = R

• (R1 R∪ 2) -1 = R1-1 R∪ 2

• (R1∩ R2)-1= R1 -1∩R2 -1

• (A x B) -1 = B x A

• Æ-1 = Æ

• (R1 - R2) -1 = R1 -1 - R2 -1

• If A = B, then (R1R2) -1 = R2

-1R1-1

• If R1 Í R2 then R1-1 Í R2

Paths in Directed Graphs• Definition 1: A path from a to b in the directed graph

G is a sequence of edges (x0, x1), (x1, x2), (x2, x3), . . ., (xn-1, xn) in G, where n is a nonnegative integer, and x0= a and x1=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 x0, x1, x2, . . .,xn-1 ,xn 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.

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;

Paths in Directed Graphs• Theorem: Let R be a relation

on A. There is a path of length n from a to b iff <a, b> Î Rn .• Proof: (by induction) • Basis: An arc from a to b is

a path of length 1. which is in R1 = 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, <a, x> ÎR

and since <x , b> is a path of length n,

• <x, b> Î Rn. If <a, x> Î R• And <x, b> Î Rn

, then

<a, b> ÎRn。 R = Rn+1

• by the inductive definition of the powers of R.

Useful Results for Transitive Closure• Theorem: If A Ì B and C Ì B, then A È C Ì B.• Theorem: If R Ì S and T Ì U then R。 T Ì S。 U.• Corollary: If R Ì S then Rn Ì Sn

• Theorem: If R is transitive then so is Rn

• Theorem: If Rk = Rj for some j > k, then Rj+m = Rn for some n £ j.– We don’t get any new relations beyond Rj.– As soon as you get a power of R that is the same

as one you had before, STOP.

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 Rn 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 <x, y> is in R.• Example: • If A = Z and R = {< i, i+1>} then t(R) = < Suppose R: is the following:

Transitive Closure• Definition: The connectivity relation or the star

closure of the relation R, denoted R*, is the set of ordered pairs <a, b> such that there is a path (in R) from a to b:

R*=⋃n=1∞ Rn

• Examples: • Let A = Z and R = {<i, i+1>}. R* = < . • Let A = the set of people, R = {<x, y> | person x is a parent of person y}. R* = ?

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 <x, y> and <y, z> are in R*. Show <x, z> is in R*.

Transitive Closure• By definition of R*, <x, y> is in Rm for some m and <y, z> is in Rn for some n.• Then <x, z> is in Rn Rm = Rm+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 Ì S so R2 Ì S2 Ì S since S is transitive

• Therefore Rn Ì Sn Ì S for all n. (why?)• Hence S must contain R* since it must also contain the

union of all the powers of R. 53

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 Rk for k > n doesn’t contain any arcs that don’t already appear in the first n powers of R.

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.

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:

Transitive Closure• Theorem 3: Let MR 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

][]3[]2[*

nRRRRR

Transitive Closure• Algorithm 1 : A Procedure for Computing the Transitive Closure procedure transitive closure (MR :zero-one nxn matrix)

A := MR

B := A for i :=2 to n begin A := A M⊙ R

B := B A end {B is the zero-one matrix for R* }

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

Equivalence Relations• It is easy to recognize equivalence relations using digraphs.• The subset of all elements related to a particularelement forms a universal relation (contains all possiblearcs) 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 theequivalence relation.

Equivalence Relations

Example:A has 3 elements:

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 | <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

Equivalence Relations• Definition: Let S1, S2, . . ., Sn be a collection of

subsets of A. Then the collection forms a partition of A if the subsets are nonempty, disjoint and exhaust A:

• Si ¹ Æ

• Si Ç Sj = Æ if i ¹ j

• ∪Si = A

FIGURE 1 A Partition of a Set.

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] È[b] = [a] È[c] = {a} È {b,c}• rank = 2

Equivalence Relations• Theorem: Let R be an equivalence relation on A.

Then either[a] = [b]

Or [a] Ç[b] = Æ

• Theorem: If R1 and R2 are equivalence relations on A

then R1 Ç R2 is an equivalence relation on A.

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] È[b] = {a} È{b, c, d} ; A/R = {{a}, {b, c, d}}

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 <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>.

Equivalence Relations• 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.• Q.E.D

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.

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.

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), Í) is a poset. It may not be the case that A Í B or B Í A. Hence, Í is not a total order.

• (Z+, 'divides') is a poset which is not a chain.

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 .

Partial Orderings• Definition: Let R be a total order on A and suppose S Í 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 <a, s> is not in R for any a

unless a = s. (There is nothing smaller than s under the order R).

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, ³) 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 Î S, if • Inductive Step: For every y Î S , if P(x) is true for all x Î S with x ≺ y , then P(y) is true.

Lexicographic Order• Given two posets (A1, R1) and (A2, R2) we construct an

induced partial order R on A1×A2:

< x1, y1> R <x2, y2> iff

x1 R1 x2 ,Or x1 = x2 and y1 R2 y2.

• Example: Let A1 = A2 = Z+ and R1 = R2 = 'divides‘ . Then

• <2, 4> R <2, 8> since x1 = x2 and y1 R2 y2.

• <2, 4> is not related under R to<2, 6> since x1 = x2

but 4 does not divide 6.• <2, 4> R <4, 5> since x1 R1 x2.

(Note that 4 is not related to 5). 75

Lexicographic Order• This definition extends naturally to multiple Cartesian

products of partially ordered sets:A1 × A2 × A3 × . . .× An.

• Example: Using the same definitions of Ai and Ri as

above,• < 2, 3, 4, 5> R < 2, 3, 8, 2> since x1 = x2, y1 = y2 and 4

divides 8.• <2, 3, 4, 5> is not related to <3, 6, 8, 10> since 2 does not divide 3.

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.

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

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.

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},≦).

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},｜ ).

Hasse Diagrams• Example 13: Draw the Hasse diagram representing

the partial ordering {(A, B)| A Í B} on the power set S={a ,b, c}.

FIGURE 4 The Hasse Diagram of (P({a,b,c}), ). 82

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, Æ is a minimal element and {a, b, c} is a maximal element.

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

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. Æ is the least element.

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

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. Æ is a lower bound for all other subsets.

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.

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.

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.

Lattices• Definition: A poset is a

lattice if every pair of elements has a lub and a glb.

• Examples: In the poset (P(S), Í ), lub(A, B) = AÈ 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

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

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:A1 × A2 × A3 × . . .× An. 93

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, Ti and Tj , are related, Ti R Tj, if Ti is

more distant from the viewer than Tj .

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

Topological Sorting• Algorithm 1 Topological Sorting procedure Topological Sorting ((S, ≼ ) : finite poset) k :=1 while S ≠ begin ak := a minimal element of S { such an element

exists by Lemma 1} S := S - {ak }

k := k + 1 end {a1, a2, . . ., an is a compatible total ordering of S}

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},｜ ).

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.

Topological Sorting

FIGURE 11 A Topological Sort of the Tasks.

Chapter 8 8.1 Relations and Their Properties 8.2 n-ary Relations and Their Applications 8.3...

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