In the list coloring problem for two matroids, we are given matroids $M_1=(S,{\cal I}_1)$ and $M_2=(S,{\cal I}_2)$ on the same ground set $S$, and the goal is to determine the smallest number $k$ such that given arbitrary lists $L_s$ of $k$ colors for $s\in S$, it is possible to choose a color from each list so that every monochromatic set is independent in both $M_1$ and $M_2$. When both $M_1$ and $M_2$ are partition matroids, Galvin's list coloring theorem for bipartite graphs gives the answer. One of the main open questions is to decide if there exists a constant $c$ such that if the coloring number is $k$ (i.e., the ground set can be partitioned into $k$ common independent sets), then the list coloring number is at most $c\cdot k$. We consider matroid classes that appear naturally in combinatorial optimization problems, namely graphic matroids, paving matroids and gammoids. We show that if both matroids are from these fundamental classes, then the list coloring number is at most twice the coloring number. The proof is based on a new approach that reduces a matroid to a partition matroid without increasing its coloring number too much, and might be of independent combinatorial interest. In particular, we show that if $M=(S,{\cal I})$ is a matroid in which $S$ can be partitioned into $k$ independent sets, then there exists a partition matroid $N=(S,{\cal J})$ with ${\cal J}\subseteq{\cal I}$ in which $S$ can be partitioned into (A) $k$ independent sets if $M$ is a transversal matroid, (B) $2k-1$ independent sets if $M$ is a graphic matroid, (C) $\lceil kr/(r-1)\rceil$ independent sets if $M$ is a paving matroid of rank $r$, and (D) $2k-2$ independent sets if $M$ is a gammoid. We extend our results by showing that the existence of a matroid $N$ with $\chi(N)\leq 2\chi(M)$ implies the existence of a matroid $N'$ with $\chi(N')\leq 2\chi(M')$ for every truncation $M'$ of $M$.

中文翻译：

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通过归约划分拟阵列出两个拟阵的着色

在两个拟阵的列表着色问题中，我们在同一地面集 $S$ 上给出拟阵 $M_1=(S,{\cal I}_1)$ 和 $M_2=(S,{\cal I}_2)$ ，目标是确定最小的数字 $k$ 使得给定任意列表 $L_s$ 的 $k$ 颜色的 $s\in S$，可以从每个列表中选择一种颜色，以便每个单色集合是在 $M_1$ 和 $M_2$ 中独立。当 $M_1$ 和 $M_2$ 都是划分拟阵时，二部图的加尔文列表着色定理给出了答案。主要的开放性问题之一是确定是否存在一个常数 $c$ 使得如果着色数为 $k$（即，地面集可以划分为 $k$ 公共独立集），则列表着色数最多为 $c\cdot k$。我们考虑在组合优化问题中自然出现的拟阵类，即图形拟阵，铺路拟阵和gammoids。我们证明，如果两个拟阵都来自这些基本类，那么列表着色数最多是着色数的两倍。该证明基于一种新方法，该方法将拟阵简化为分区拟阵，而不会过多地增加其着色数，并且可能具有独立的组合兴趣。特别地，我们证明如果 $M=(S,{\cal I})$ 是一个 $S$ 可以被划分为 $k$ 个独立集的拟阵，那么存在一个分区拟阵 $N=(S, {\cal J})$ 和 ${\cal J}\subseteq{\cal I}$ 其中，如果 $M$ 是横向拟阵，则 $S$ 可以划分为 (A) $k$ 个独立集，（B ) 如果 $M$ 是图形拟阵，则 $2k-1$ 独立集， (C) $\lceil kr/(r-1)\rceil$ 独立集，如果 $M$ 是秩为 $r$ 的铺路拟阵，(D) 如果 $M$ 是 gammoid，则 $2k-2$ 独立集。我们通过显示具有 $\chi(N)\leq 2\chi(M)$ 的拟阵 $N$ 的存在来扩展我们的结果，这意味着存在具有 $\chi(N')\ 的拟阵 $N'$ leq 2\chi(M')$ 对于 $M$ 的每个截断 $M'$。