# Decomposing eulerian graphs

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 Subject: Graph Theory » Basic Graph Theory » » Cycles
 Keywords: cover cycle Eulerian
 Posted by: mdevos on: March 7th, 2007

\begin{conjecture} If $G$ is a 6-\Def[edge-connected]{connectivity (graph theory)} \Def[Eulerian graph]{eulerian graph} and $P$ is a 2-transition system for $G$, then $(G,P)$ has a compaible decomposition. \end{conjecture}

Definition: Let $G$ be an Eulerian graph and for every vertex $v$, let $P(v)$ be a partition of the edges incident with $v$. We call $P$ a transition system. If every member of $P(v)$ has size at most $k$ (for every $v$), then we call $P$ a $k$-transition sytem. A compatible decomposition of $(G,P)$ is a list of edge-disjoint cycles $C_1,\ldots,C_n$ with union $G$ so that every $C_i$ contains at most one edge from every member of $P(v)$.

Let $G$ be a graph and let $G'$ be the graph obtained from $G$ by replacing each edge $e$ of G by two edges $e',e''$ in parallel. Let $P$ be the 2-transition system of $G$ with ${e',e''} \in P(v)$ whenever $e'$ and $e''$ are incident with $v$. Now, $G'$ is an Eulerian graph and every compatible decomposition of $(G',P)$ gives a cycle double cover of $G$. Since the cycle double cover conjecture can be reduced to graphs which are 3-edge-connected, the above conjecture would imply the cycle double cover conjecture.

We define a transition system $P$ to be admissable if every member of $P(v)$ contains no more than half of the edges in any edge-cut. It is easy to see that if there is a compatible decomposition of $(G,P)$, then $P$ must be admissable. The converse of this is not true; There is an admissable 2-transition system of the graph $K_5$ which does not admit a compatible decomposition. Recently, G. Fan and C.Q. Zhang [FZ] have proved that $(G,P)$ does have a compatible decomposition whenever $P$ is admissable and $G$ has no $K_5$ minor. This result imporoved upon an earlier theorem of Fleischner and Frank [FF]. Very recently, I have proved a weak version of the above conjecture, by showing that $(G,P)$ also has a compatible decomposition when P is a 2-transition system and G is 80-edge-connected. I'd quite like to see an improvement on this bound. Here is a related conjecture.

\begin{conjecture}[Sabidussi] Let $W$ be an Euler tour of the graph $G$. If $G$ has no vertex of degree two, then there is a cycle decomposition of $G$, say $F$, so that no two consecutive edges of $W$ are in a common circuit of $F$. \end{conjecture}

If $W$ is given by $v_1,e_1,v_2,e_2,...,e_{m-1},v_m$ then we may form a 2-transition system $P$ by putting $\{e_{i-1},e_i\}$ in $P(v_i)$ for every $i$ (working modulo $m$). Now a compatible decomposition of $(G,P)$ is precisely a cycle decomposition of $G$ satisfying the above conjecture. Thus, Sabidussi's conjecture is equivalent to the assertion that $(G,P)$ has a compatible decomposition whenever $G$ has no vertex of degree two and $P$ is a 2-transition system which comes from an Euler tour.

Let $G$ be a directed Eulerian graph and for every vertex $v$, let $P(v)$ be a partition of the edges incident with $v$ into pairs so that every in-edge is paired with an out-edge. We define a compatible decomposition to be a decomposition of $G$ into directed circuits so that every directed circuit contains at most one edge from every member of $P(v)$. Our current techniques don't seem to shed any light on the problem of finding compatible decompositions for Eulerian digraphs. Next I pose a very basic question which is still open.

\begin{problem}[DeVos] Does there exist a fixed integer $k$ such that $(G,P)$ has a compatible decomposition whenever $G$ is a $k$-edge-connected directed Eulerian graph and $P$ is a 2-transition system? \end{problem}