Coloring graph with 3 colors

Question

The nodes of the graph $[V, E]$ are the lists $\left[ {{s_1},\,{s_2},\,{s_3}} \right]$ where ${s_j} \in \left\{ {1,2} \right\},\,\,j = 1,2,3$. There is an undirected edge between the nodes $\left[ {{s_1},\,{s_2},\,{s_3}} \right]$ and $\left[ {{t_1},\,{t_2},\,{t_3}} \right]$ if and only if $|{s_1} - {t_1}| + |{s_2} - {t_2}| + |{s_3} - {t_3}|\,\, = \,\,1$. Draw this graph and put the nodes in an order so that the greedy coloring algorithm uses $3$ colors.

My attempt:

I got the following graph:

Now i don't know how to put the nodes in an order so that the greedy coloring algorithm will use $3$ colors instead of obvious $2$ colors.

Answer 1

Lets order the vertices $(211), (111), (122),(112),(212), (222), (221), (121)$

Using the colours $1,2,3$, the greedy algorithm assigns the colours $1$ to $(221)$, $2$ to $(111)$, $1$ to $(122)$, $3$ to $(112)$, $2$ to $(212)$, $3$ to $(222)$, $2$ to $(221)$, and $3$ to $(121)$, giving you a $3$-colouring.

Answer 2

As I understand the greedy coloring algorithm, and as the question implies, there is no fixed order of the vertices. However, you are required to fix the order of the colors before you begin to apply the algorithm - considering the vertices of the graph in sequence and assigning each vertex its first available color (as per the color order you have chosen).

In this context it seems you might have chosen each next vertex from those connected to the current. In such a case you will get two alternating colors for this graph. What happens if you pick a first vertex and color, and then pick a "distant" vertex for the second vertex?