Partial Order Relations Lecture 50 Section 10 5

Partial Order Relations Lecture 50 Section 10. 5 Wed, Apr 20, 2005

Antisymmetry A relation R on a set A is antisymmetric if for all a, b A, (a, b) R and (b, a) R a = b. ¢ This is equivalent to (a, b) R and a b (b, a) R. ¢

Examples: Antisymmetry ¢ The following relations are antisymmetric. a b, on Z+. l A B, on (U). l x y, on R. l A B = A, on (U). l f(x)g(x) = f(x) on the set of all functions from R to R. l

Partial Order Relations ¢ A relation R on a set A is a partial order relation if R is reflexive. l R is antisymmetric. l R is transitive. l ¢ We use as the generic symbol for a partial order relation.

Examples: Partial Order Relations ¢ The following relations are partial order relations. a b, on Z+. l A B, on (U). l x y, on R. l ¢ Is the relation f(x)g(x) = f(x) on functions a partial order relation?

Lexicographic Order ¢ Given a finite set of symbols, let them be listed in a specific order. l For example, a < b < c. Let S be the set of all finite strings over the set of symbols. ¢ Then we may define the lexicographic order relation on S. ¢

Lexicographic Order Let s, t S and let s = a 1 a 2…am and t = b 1 b 2…bn. ¢ Then s t if ¢ s = , or l |s| |t| and ai = bi, for all 1 i m, or l There exists k 1 with k m and k n and ai = bi for all 1 i < k, but ak < bk. l

Lexicographic Order The first condition says that s for all strings s S. ¢ The secondition says that s is no longer than t and all of s matches the first part of t. ¢ The third condition says that s and t differ at some common position. ¢

Lexicographic Order Theorem: The lexicographic order relation is a partial order. ¢ Proof: ¢ Reflexive: Obvious (? ) l Antisymmetric: l • Suppose s t and t s for some s, t S. • There are potentially 9 combinations of reasons for this.

Lexicographic Order • Maybe both are true because of the first condition. Then s = and t = , so s = t. • Maybe both are true because of the secondition. Then |s| |t| and |t| |s|, implying that s and t have the same length. But also s matches “the first part” of t and t matches “the first part” of s. Therefore, s = t. • The second and third conditions are incompatible. • Etc.

Lexicographic Order l Transitivity: • Suppose that s t and t u, for some s, t, u S. • There are potentially 27 combinations of reasons for this. (!!!) • Maybe s = and t = . Then s u. • Etc.

Hasse Diagrams A Hasse diagram is a drawing that represents a partial order relation. ¢ Draw a diagram in which ¢ a b is represented by a b. l a is drawn below b. l If there exists c such that a c and c b, then we represent only a c and c b; a b is implied by transitivity. l

Example: Hasse Diagram ¢ Let the relation be on ({a, b, c}). {a , b , c} {a , b } {a , c} {b , c} {a } {b } {c} {}

Example: Hasse Diagram ¢ Let the relation be on {1, 2, 3, 4, 6, 12} 12 6 4 2 3 1

Example: Partial Order Relation Let F be the set of all functions f : R+ R+. ¢ Let E be the set of equivalence classes [f] of F, under the equivalence relation f ~ g if f(x) is (g(x)). ¢ Define on E by [f] [g] if f(x) is O(g(x)). ¢

[f] [g] is Well Defined First, we must show that is well defined on E. ¢ Let f 1, f 2 [f] and g 1, g 2 [g]. ¢ Then f 1~ f 2 and g 1 ~ g 2. ¢ To show that is well defined, we must show that if [f 1] [g 1], then [f 2] [g 2], meaning that it does not matter which function we select to represent the class. ¢

[f] [g] is Well Defined So suppose that [f 1] [g 1]. ¢ Then ¢ l f 2(x) is O(f 1(x)), and l f 1(x) is O(g 1(x)) and l g 1(x) is O(g 2(x)). So, f 2(x) is O(g 2(x)). ¢ Therefore, [f 2] [g 2]. ¢

Example: Partial Order Relation Theorem: is a partial order relation on E. ¢ Proof: ¢ Reflexivity ¢ l It is clear that [f]. (Use M = 1, x 0 = 0. )

Example: Partial Order Relation ¢ Antisymmetry Suppose that [f] [g] and [g] [f]. l Then f(x) is O(g(x)) and g(x) is O(f(x)). l Then f(x) is (g(x)). l Therefore, f ~ g. l Therefore, [f] = [g]. l

Example: Partial Order Relation ¢ Transitivity Suppose that [f] [g] and [g] [h]. l Then f(x) is O(g(x)) and g(x) is O(h(x)). l We have already shown that this implies that f(x) is O(h(x)). l Therefore, [f] [h]. l ¢ Thus, is a partial order relation on E.

Comparable Elements and Total Orders Given a partial order on a set A, two elements a, b A are comparable if a b or b a. ¢ A partial order relation is a total order relation if any two elements are comparable under that relation. ¢

Total Order Relations ¢ Which of the following partial orders are total orders? x y, on R. l A B, on (U). l a b, on Z+. l

Total Order Relations Define [f] [g] to mean f(x) is O(g(x)). Is this a total order relation on the set of equivalence classes of functions? ¢ Which are total orders on R R. ¢ (a, b) (c, d) if a c and b d. l (a, b) (c, d) if a c or b d. l (a, b) (c, d) if a + b c + d. l (a, b) (c, d) if a < c or (a = c and b d). l

Example: Total Ordering and Sorting ¢ In order to sort the list {(2, 3), (3, 2), (2, 2)}. it is necessary that be a total ordering of the Point class. If is not a total order relation, then the results of any sorting algorithm are unpredictable. ¢ Why? ¢

Example: Total Ordering and Sorting ¢ Define a Point object to consist of two doubles. class Point { double x; double y; };

Example: Total Ordering and Sorting ¢ Define operator<() on the Point class as follows. bool operator<(const Point& p, const Point& q) { if (p. x != q. x) return p. x < q. x; else return p. y < q. y; } ¢ Then define operator<=() to mean that a < b or a == b.

C Library Functions There are two standard library functions, bsearch() and qsort(), that perform a binary search and a quicksort on an array, respectively. ¢ For each function, one parameter is a function compare(a, b) that returns ¢ < 0 if a < b l = 0 if a == b l > 0 if a > b l

C Library Functions Those three requirements are not sufficient to guarantee that operator<=() is a total order relation. ¢ It is the programmer’s responsibility to be sure that it is. ¢ Otherwise, neither bsearch() nor qsort() is guaranteed to work properly. ¢

Chains of Elements Let a partial order be defined on a set A. ¢ A subset B A is called a chain if any two elements of B are comparable. That is, is a total order on B (although not on A). ¢

Chains of Elements ¢ Find a chain of subsets in ({a, b, c}) under the subset relation. l ¢ What is the longest possible chain? Find a chain of integers in Z+ under the divides relation. l What is the longest possible chain?

Maximal and Greatest Elements Let A be a partially ordered set under . ¢ An element a A is a maximal element if, for all b A, either b a or a and b are not comparable. ¢ An element a A is a greatest element if, for all b A, b a. ¢

Example Let A = {1, 2, 3, …, 20} and let the relation be a b. ¢ Are there any maximal elements? ¢ l ¢ If so, what are they? Are there any greatest elements? l If so, what are they?

Example Let B = {1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60} and let the relation be a b. ¢ Are there any maximal elements? ¢ l ¢ If so, what are they? Are there any greatest elements? l If so, what are they?

Maximal and Greatest Elements Must maximal elements be exist? ¢ Must greatest elements be exist? ¢ If maximal elements exist, must they be unique? ¢ If greatest elements exist, must they be unique? ¢

Minimal and Least Elements Minimal and least elements are similar to maximal and greatest elements, respectively. ¢ An element a A is a minimal element if, for all b A, either a b or a and b are not comparable. ¢ An element a A is a least element if, for all b A, a b. ¢

Examples ¢ What are the minimal and least elements (if any) of the sets A = {1, 2, 3, …, 20} and B = {1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60} under the relation a b?

Compatible Order Relations Let 1 and 2 be partial order relations defined on a set A. ¢ We call 2 compatible with 1 if, whenever a 1 b, it is also the case that a 2 b. ¢ In other words, a, b A, a 1 b a 2 b. ¢

Topological Sorting Given a partial order relation 1 defined on a set A, a topological sorting for 1 is a total order relation 2 that is compatible with 1. ¢ For example, the ordinary “less than or equal to” relation is a topological sorting for the “divides” relation because ¢ is a total order, and l If a b, then a b. l

Application: Computer Graphics In the “old days, ” a difficult problem in computer graphics was to draw the objects of a scene in such a way that when two objects were on the same line of sight, the closer object obscured the farther object. ¢ One solution was the Painter’s Algorithm: sort the shapes from front to back, and then draw them from back to front. ¢

Application: Computer Graphics Define the relation a 1 b on the shapes to mean that shape a hides part of shape b. ¢ Define the relation a 2 b on the shapes to mean that shape a is closer to the viewer than is shape b. ¢ Clearly, 2 is a topological sorting for 1. ¢

Application: Computer Graphics ¢ Therefore, the Painter’s Algorithm was to Determine which shapes hide which shapes (relation 1). l Find a topological sorting for 1 (relation 2). l Draw the shapes in reverse order according to the total ordering 2. l

Example ¢ Consider the following scene, with the viewpoint inside the house, looking out at the tree.

Example ¢ The scene might be rendered like this:

Topological Sorting Theorem: If A is a finite set with a partial order 1 defined on it, then there exists a topological sorting for 1. ¢ The proof is the following algorithm. ¢ A partially ordered finite set must have a minimal element. l Let x 0 be a minimal element of A. l Let A 1 = A – {x 0}. l

Topological Sorting A 1 is again a partially ordered finite set, so it must have a minimal element. l Let x 1 be a minimal element of A 1 and define A 2 = A 1 – {x 1}. l Continue in this manner, defining x 3, x 4, x 5, and so on, until no elements remain. l Then define 2 as follows: xi 2 xj if and only if i j. l

Example ¢ Let A = {1, 2, 3, …, 10} under the “divides” relation. 8 10 4 9 6 2 3 5 1 7

Example ¢ Let x 0 = 1 and remove 1 from the set. 8 10 4 9 6 2 3 5 7

Example ¢ Let x 1 = 7 and remove 7 from the set. 8 10 4 9 6 2 3 5

Example ¢ Let x 2 = 5 and remove 5 from the set. 8 10 4 9 6 2 3

Example ¢ Let x 3 = 3 and remove 3 from the set. 8 10 4 6 2 9

Example ¢ Let x 4 = 9 and remove 9 from the set. 8 10 4 6 2

Example ¢ Let x 5 = 2 and remove 2 from the set. 8 10 4 6

Example ¢ Let x 6 = 10 and remove 10 from the set. 8 4 6

Example ¢ Let x 7 = 6 and remove 6 from the set. 8 4

Example ¢ Let x 8 = 4 and remove 4 from the set. 8

Example ¢ Let x 9 = 8 and remove 8 from the set.

Example The total ordering is 1, 7, 5, 3, 9, 2, 10, 6, 4, 8. ¢ This ordering is compatible with the divide relation. ¢ A number of the choices were arbitrary, so there are many other topological sortings for the divide relation on this set. ¢