G)|^k, where the categorical graph product is used. This limit, surprisingly, lies in the range
(0,1/2∪{1\}. We can show that this limit can take any such rational number, but is there any
G for which A(G) is irrational? A useful technique for bounding A(G) is to consider special
spanning subgraphs. These bounds allow us to efficiently compute A(G) for many G. We
give a condition which if true for G shows that A(G)>β(G)/|V(G)|. This brings up the question; …