Barnette identified two interesting classes of cubic polyhedral graphs for which he conjectured the existence of a Hamiltonian cycle. We examine these classes from the point of view of distance-two colourings. A distance-two $r$-colouring of a graph $G$ is an assignment of $r$ colours to the vertices of $G$ so that any two vertices at distance at most two have different colours. A cubic graph obviously needs at least four colours, and the distance-two four-colouring problem for cubic planar graphs is known to be NP-complete. We prove the problem remains NP-complete for tri-connected bipartite cubic planar graphs, which we call type-one Barnette graphs, since they are the first class identified by Barnette. By contrast, we show that the problem is polynomial for cubic plane graphs with face sizes $3,4,5,$ or 6, which we call type-two Barnette graphs, because of their relation to Barnette’s second conjecture. In fact, the colourable instances that are bipartite can be fully described, and those that are non-bipartite can be characterized by their face sizes. We have similar results for quartic plane graphs: the analogue of type-two Barnette graphs are graphs with face sizes 3 or 4. For this class, the corresponding distance-two five-colouring problem is also polynomial; in fact, we can again fully describe all colourable instances — there are exactly two such graphs. It has recently been proved that every planar subcubic graph can be distance-two coloured with at most seven colours, and conjectured that six colours suffice if the planar subcubic graph can be drawn without faces of size five. We consider a weaker version of the conjecture, stating that six colours suffice for a bipartite cubic planar graph, i.e., a cubic plane graph with all faces even. We prove this conjecture in the case when all faces have sizes divisible by four, and in another more general case, when the faces are coloured in three colours, of which two colours consist of faces with sizes divisible by four.

巴尼特（Barnette）确定了两类有趣的三次多面体图，他推测哈密顿环的存在。我们从距离两种颜色的角度检查这些类。距离二$\mathrm{[R}$图的着色 $G$ 是 $\mathrm{[R}$ 顶点的颜色 $G$因此，相距不远的两个顶点最多具有两个不同的颜色。立方图显然至少需要四种颜色，而立方平面图的距离为二的四色问题是已知的NP完全的。我们证明了对于三连通二分立方平面图（我们将其称为一类Barnette图）来说，问题仍然是NP完全的，因为它们是Barnette识别的第一类。相比之下，我们表明问题是具有面大小的立方平面图的多项式$3，4，5，$或6，我们将其称为第二类Barnette图，因为它们与Barnette的第二个猜想有关。实际上，可以完全描述两部分的可着色实例，而非两部分的可着色实例可以通过其面部大小来表征。对于四次平面图，我们得到相似的结果：第二类Barnette图的类似物是具有3或4号面的图。对于此类，相应的距离二五色问题也是多项式；实际上，我们可以再次完全描述所有可着色实例-恰好有两个这样的图。最近已经证明，每个平面次立方图都可以是距离为两个的两个颜色，最多可以有7种颜色，并且可以推测，如果可以绘制平面次立方图而没有大小为5的面，则六种颜色就足够了。我们考虑这个猜想的弱版本，指出六色足以满足二方立方平面图，即，所有面均相等的立方平面图。我们在所有面孔的大小可被四分之一的情况下证明这种猜想，而在另一种更普遍的情况下，当面孔被三种颜色上色时，其中两种颜色由大小被四分之一的面孔组成。