Tighter Bounds on the Degree-Truncated Choice Number of Planar Graphs

Assume $G$ is a graph and $k$ is a positive integer. Let $f:V(G)\to \mathbb{N}$ be defined as $f(v)=\min\{k,d_G(v)\}$. If $G$ is $f$-choosable, then we say $G$ is degree-truncated $k$-choosable. The degree-truncated choice number of $G$ is $\operatorname{ch}^{\text{\st{d}}}(G) = \min\{k: G \text{ is degree-truncated $k$-choosable}\}$. For a family $\mathcal{G}$ of graphs, $\operatorname{ch}^{\text{\st{d}}}(\mathcal{G}) = \max\{\operatorname{ch}^{\text{\st{d}}}(G):G \in \mathcal{G}\}$. Let $\mathcal{P}$ denote the family of 3-connected non-complete planar graphs. Richter asked in 2008 whether $ch^{\text{\st{d}}}(\mathcal{P}) \le 6$. In 2025, Zhou, Zhu and Zhu answered this question in negative and proved that $8 \le ch^{\text{\st{d}}}(\mathcal{P}) \le 16$. This result was improved by Jiang, Xu, Xu, and Zhu, who proved that $9 \le ch^{\text{\st{d}}}(\mathcal{P}) \le 12$. In this paper, we further improve the result and prove that $10 \le \operatorname{ch}^{\text{\st{d}}}(\mathcal{P}) \le 11$. We conjecture that $\operatorname{ch}^{\text{\st{d}}}(\mathcal{P}) =10$, and we confirm this conjecture for those planar graphs $G \in \mathcal{P}$ for which the subgraph induced by vertices of degree at least 11 is 4-choosable.