# Clarifications regarding conformability in graph colorings

As an outgrowth of this question, I have another question, that is, why not the definition of conformability includes a $$\Delta$$ vertex coloring also, instead of only $$\Delta+1$$ coloring of vertices. This is because, the square of the $$8$$ cycle, $$C_8^2$$ is easily seen to have a $$\Delta$$ conformable coloring(each color class has only two vertices), whereas I do not see a $$\Delta+1$$ conformable coloring( the number of vertices in each color class cannot exceed two, hence, a five coloring should have one vertex in at least two color classes). But, the graph is Type 1( total colorable with $$\Delta+1$$ colors) Am I missing something here? Thanks beforehand.

There is nothing to be missed. In fact, a $$\Delta$$-conformable(or any $$k$$-conformable coloring, where $$k\le\Delta+1$$) induces a $$\Delta+1$$-conformable coloring in case of even order graphs, as one(or more) color class(es) can be taken to have $$0$$ vertices, which has same parity as the total number of vertices. However, this might not be true for odd order graphs. In this case, the square of $$8$$-cycle is a graph of even order.