Let $G$ be a simple connected plane graph and let $C_1$ and $C_2$ be cycles in $G$ bounding distinct faces $f_1$ and $f_2$. For a positive integer $\ell$, let $r\left(\ell \right)$ denote the number of integers $n$ such that $-\ell \le n\le \ell$, $n$ is divisible by 3, and $n$ has the same parity as $\ell$; in particular, $r\left(4\right)=1$. Let $r_{f_1,f_2}\left(G\right)={\prod }_{f}r\left(\left|f\right|\right)$, where the product is over the faces $f$ of $G$ distinct from $f_1$ and $f_2$, and let $q\left(G\right)=1+{\sum }_{f:\left|f\right|\ne 4}\left|f\right|$, where the sum is over all faces $f$ of $G$ (of length other than four). We give an algorithm with time complexity $O\left(r_{f_1,f_2}\left(G\right)q\left(G\right)\left|G\right|\right)$ which, given a 3-coloring $\psi$ of $C_1\cup C_2$, either finds an extension of $\psi$ to a 3-coloring of $G$, or correctly decides no such extension exists.

The algorithm is based on a min–max theorem for a variant of integer 2-commodity flows, and consequently in the negative case produces an obstruction to the existence of the extension. As a corollary, we show that every triangle-free graph drawn in the torus with edge-width at least 21 is 3-colorable.

让 $G$ 是一个简单的连接平面图，让 $C_{\mathrm{1个}}$ 和 $C_2$ 被循环 $G$ 界定不同的面孔 $F_{\mathrm{1个}}$ 和 $F_2$。对于正整数$\ell$，让 $\mathrm{[R}（\ell ）$ 表示整数数 $ñ$ 这样 $-\ell \le ñ\le \ell$， $ñ$ 被3整除 $ñ$ 具有与 $\ell$; 尤其是，$\mathrm{[R}（4）=\mathrm{1个}$。让${\mathrm{[R}}_{F_{\mathrm{1个}}，F_2}（G）={\prod }_F\mathrm{[R}（\left|F\right|）$，产品放在脸上 $F$ 的 $G$ 有别于 $F_{\mathrm{1个}}$ 和 $F_2$， 然后让 $q（G）=\mathrm{1个}+{\sum }_{F：\left|F\right|\ne 4}\left|F\right|$，总和在所有面上 $F$ 的 $G$（长度不为四个）。我们给出了一个具有时间复杂度的算法$Ø（{\mathrm{[R}}_{F_{\mathrm{1个}}，F_2}（G）q（G）\left|G\right|）$ 给定三色 $\psi$ 的 $C_{\mathrm{1个}}\cup C_2$，或者找到的扩展名 $\psi$ 到3色 $G$，或正确地确定不存在此类扩展名。

该算法基于整数最大商品定理的最小-最大定理，因此，在否定情况下，该扩展的存在受到阻碍。作为推论，我们表明在圆环中绘制的每个无边三角形的边宽至少为21的无色图形都是3色的。