Dense Eulerian graphs are $(1, 3)$-choosable

A graph $G$ is total weight $(k,k')$-choosable if for any total list assignment $L$ which assigns to each vertex $v$ a set $L(v)$ of $k$ real numbers, and each edge $e$ a set $L(e)$ of $k'$ real numbers, there is a proper total $L$-weighting, i.e., a mapping $f: V(G) \cup E(G) \to \mathbb{R}$ such that for each $z \in V(G) \cup E(G)$, $f(z) \in L(z)$, and for each edge $uv$ of $G$, $\sum_{e \in E(u)}f(e)+f(u) \ne \sum_{e \in E(v)}f(e) + f(v)$. This paper proves that if $G$ decomposes into complete graphs of odd order, then $G$ is total weight $(1,3)$-choosable. As a consequence, every Eulerian graph $G$ of large order and with minimum degree at least $0.91|V(G)|$ is total weight $(1,3)$-choosable. We also prove that any graph $G$ with minimum degree at least $0.999|V(G)|$ is total weight $(1,4)$-choosable.

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