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A relation R on a set A is called a partial order if R is Reflexive, Antisymmetric and Transitive. Reflexive – has a cycle of 1 at every vertex. Example: (1,1), (2,2) Antisymmetric – contains no back arrows – all one way paths and can contain cycles of 1. Example: (1,1), (1,2), (2,3) Transitive – if you have (a,b) and (b,c), you must have (a,c). Example: a,b b,c a,c (1,2), (2,3), (1,3) The set A together with the partial order R is called a partially ordered set or poset. This is denoted (A,R) or (A, ≤). The relation of divisibility (aRb if and only if a divides b) is a partial order on Z+ . A = {1,2,3,4} x ≤ y iff x|y B = {1,7,11,14} x ≤ y in the regular way x y x y a,b a,b a,b a,b (1,7) ≤ (3,14) (2,7) ≤ (3,14) Because 1|3 and 7 ≤ 14. Because 2 does not divide 3. a and b of set A are said to be comparable if a ≤ b or b ≤ a. Every pair of elements does not need to be comparable aRb iff a|b is a partial order on Z+ . 2 and 7 are not comparable since 2|7 and 7|2. Partial means some elements may not be comparable. Lexicographic (lex sick o graphic) Means alphabetic order. park < part help < hind Jump < mump Hasse Diagrams Hasse diagrams, for simplicity, remove one cycles, remove transitive property aRc, remove arrows, and all edges point upwards. Jt does not mean one cycles, transitive property and arrows don’t exist….they still exist. A = {1,2,3,4,12} a and b are ∈ of A a ≤ b if and only if a|b Draw the Hasse diagram of the poset(A, ≤). (1,1),(1,2),(1,3),(1,4),(1,12), (2,2),(2,4),(2,12), (3,3),(3,12), (4,4),(4,12), (12,12) Cross off one cycles and (a,c) of transitive property. (1,1),(1,2),(1,3),(1,4),(1,12), (2,2),(2,4),(2,12), (3,3),(3,12), (4,4),(4,12), (12,12) 12 12 4 4 3 2 3 1 2 1 Finite linear ordered set 1 2 3 1 2 3 (1,1), (1,2), (1,3), (2,2), (2,3) (3,3) 3 3 2 2 1 1 The process of constructing a linear order is called topological sorting. With topological sorting, you need to be able to insert a relation in the linear order without breaking an existing relation. 3 1 is related to 2 2 1 is also related to 3 1 3 4 could be placed after 1 if 4 is related to 1 and this would not break the relations of 1 to 2 and 1 to 3, meaning 4 is related to 2 and 3 2 4 1 Isomorphism A = {1,2,3,6} ≅ f Under divisibility, x ≤ y iff x|y B = {{}, {x},{y}, {x,y}} Under subset, (a ≤ b) iff a ≤ b Hasse diagrams of A & B 1 2 3 6 ≅ {} {x} {y} {x,y} {x,y} 6 3 2 1 {y} {x} {} The function A B is one to one. The function f is an isomorphism. Incomparable a ≤ b and b ≤ a. There is no relation between a and b so we cannot compare them. b a c d A poset (A, R -1) is called a dual poset…..it is the inverse. f b a e d e d f a b c c