4.4 Properties of Relations

Relations Reflexive Irreflexive Symmetric Asymmetric Antisymmetric Transitive

Reflexive Reflexive relation has a cycle of 1 at every vertex A = {1,2,3} B=A R = {(1,1)(1,2),(2,2),(2,3),(3,3),(3,1)} Reflexive matrix has all 1’s on the main diagonal 1 3 2 a\b 1 2 3

Irreflexive Irreflexive relation has no cycles of length 1 at any vertex. A = {1,2,3} B=A R = {(1,2),(1,3),(2,3)} Irreflexive has all 0’s on its main diagonal 1 2 3 a\b 1 2 3

Symmetric If there is an edge from vertex i to vertex j, then there is an edge from vertex j to vertex i. A = {a,b,c} B = A R = {(a,a),(a,c),(b,c),(c,a),(c,b),(c,c)} a c b a\b a b c 1

R = {(a,b),(b,a),(a,c),(c,a),(b,c),(c,b)} This is a little messy R = {(a,b),(b,a),(a,c),(c,a),(b,c),(c,b)} Another way to draw this is: This is a connected digraph This digraph is not connected. R={(a,b),(b,a),(c,d),(d,c)} a b c a c b a c b d

Asymmetric aRb bRa A = {1,2,3,4} B = A R ={(1,2),(3,4),(2,3)} There is no (1,1) or (2,2) or (3,3) or (4,4) There is no (2,1), (4,3), or (3,2) IF mij = 1 then mji = 0 (i = row and j = column) There are no back arrows. All edges are 1 way streets 1 a\b 1 2 3 4 2 4 3

Antisymmetric Contains no back arrows. Can contain one node cycles 1 a\b 1 2 3 4 2 3 4

Transitive (tricky) Every time you have (a,b) and (b,c) you must also have (a,c). A = {1,2,3,4} This relation is not transitive: a,b b,c R = {(1,1),(2,2),(3,3),(4,4),(2,3),(3,4)} In order to make this relation transitive, you would a,c need (2,4). If you have R = {(1,1),(2,2),(3,3),(4,4),(2,3)} is it transitive? Yes, because it doesn’t break the rule. You have (a,b), you don’t have (b,c), therefore you do not need (a,c). Only when you have (a,b) AND (b,c), you must have (a,c).

A cycle is a path where you start and end with the same vertex. iff is shorthand for if and only if.