# Berge-Fulkerson conjecture --- the planar case

A well-known conjecture of Berge and Fulkerson says that every bridgeless cubic graph has a collection of six perfect matchings that together cover every edge exactly twice. Is this still open for bridgeless cubic planar graphs?

Let $$G$$ be a bridgeless cubic planar graph. The dual graph $$G^*$$ is a triangulation. By the Four Colour Theorem, $$G^*$$ has a 4-colouring $$c$$. We will use $$\mathbb{Z}_2 \times \mathbb{Z}_2$$ as the set of colours for $$c$$. Now for each edge $$e \in E(G)$$, colour $$e$$ with colour $$c'(e):=c(f_1)+c(f_2)$$ where $$f_1$$ and $$f_2$$ are the two faces of $$G$$ incident to $$e$$. Since $$c$$ is a proper colouring of $$G^*$$, $$c'(e)$$ is a non-zero element of $$\mathbb{Z}_2 \times \mathbb{Z}_2$$ for all $$e \in E(G)$$. Moreover, if $$v \in V(G)$$ and $$e_1, e_2$$, and $$e_3$$ are the edges of $$G$$ incident to $$v$$, then the dual edges $$e_1^*, e_2^*$$, and $$e_3^*$$ are a triangular face $$\Delta$$ in $$G^*$$. Since the vertices of $$\Delta$$ receive different colours in $$c$$, $$c'(e_1), c'(e_2)$$, and $$c'(e_3)$$ are all distinct. That is, $$c'$$ is a proper 3-edge colouring of $$G$$. In other words, $$E(G)$$ can be partitioned into three perfect matchings. So you can just use each of these perfect matchings twice.