**Conjecture**Every bridgeless cubic graph has two perfect matchings , so that does not contain an odd edge-cut.

Let be a bridgeless cubic graph. A *binary cycle* (henceforth called *cycle*) is a set so that every vertex of has even degree (equivalently, a cycle is any member of the binary cycle space). A *postman join* is a set so that is a cycle. Note that since is cubic, every perfect matching is a postman join. Next we state a well-known theorem of Jaeger in three equivalent forms.

**Theorem (Jaeger's 8-flow theorem)**

- \item has a nowhere-zero flow in the group . \item has three cycles so that . \item has three postman joins so that .

The last of these statements is interesting, since The Berge Fulkerson Conjecture (if true) implies the following:

**Conjecture**has three perfect matchings so that .

So, we know that has three postman joins with empty intersection, and it is conjectured that may be chosen so that each is a perfect matching, but now we see two statements in between the theorem and the conjecture. Namely, is it true that may be chosen so that one is a perfect matching? or two? The first of these was solved recently.

**Theorem (Macajova, Skoviera)**has two postman sets and one perfect matching so that

The second of these asks for two perfect matchings and one postman join so that . It is an easy exercise to show that a set contains a postman join if an only if has nonempty intersection with every odd edge-cut. Therefore, finding two perfect matchings and one postman join with empty common intersection is precisely equivalent to the conjecture at the start of this page - find two perfect matchings whose intersection contains no odd edge-cut.

## Bibliography

* Edita Macajova, Martin Skoviera, Fano colourings of cubic graphs and the Fulkerson conjecture. Theoret. Comput. Sci. 349 (2005), no. 1, 112--120. MathSciNet

* indicates original appearance(s) of problem.