Steinberg-type graphs, those planar graphs containing no 4-cycles or 5-cycles, became well known with the 1976 Steinberg Conjecture which stated that such graphs are properly 3-colorable. Recently, Steinberg’s Conjecture was demonstrated to be false (Cohen-Addad et al. in J Combin Theory Ser B 122: 452–456, 2016). However, Steinberg-type graphs are (3, 0, 0)-defective colorable (Hill et al. in Discrete Math 313:2312–2317, 2013), i. e. of the three colors, two are used properly and any vertex colored with the first color is allowed to be adjacent to up to three other veritces with the same color. In this paper, we introduce a stronger form of defective graph coloring that places limits on the permitted defects in a coloring. Using the strength of this new type of coloring, we prove the current closest result to Steinberg’s original conjecture and show that the counterexample given in Cohen-Addad et al. (J Combin Theory Ser B 122:452–456,2016) is colorable with this stronger form of defective 3-coloring.

中文翻译：

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对缺陷着色的新限制及其在steinberg型图上的应用

1976年的Steinberg猜想使Steinberg型图（那些不包含4个循环或5个循环的平面图）众所周知，该图表明此类图是3色的。最近，斯坦伯格的猜想被证明是错误的（Cohen-Addad等人在《 J Combin Theory Ser B 122：452–456，2016）》。但是，Steinberg型图是（3，0，0）缺陷的可着色（Hill等人在Discrete Math 313：2312-2317，2013），即。e。在这三种颜色中，有两种是正确使用的，并且使用第一种颜色着色的任何顶点都可以与最多三个具有相同颜色的顶点相邻。在本文中，我们介绍了缺陷图形着色的一种更强形式，它限制了着色中允许的缺陷。利用这种新型着色的力量，我们证明了目前最接近斯坦伯格最初猜想的结果，并证明了Cohen-Addad等人给出的反例。（J Combin Theory系列B 122：452–456，2016）可通过这种有力的有缺陷的3色着色而着色。