转发张彧典的文章：四色猜想中“染色困局构形”的4染色（三）

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4- dyeing of "coloring dilemma configuration" in four-color conjecture
四色猜想中“染色困局构形”的4染色（三）
Yudian Zhang  Lichong Zhang（1633409368@qq.com）
张彧典      张利翀
Party School of Yuxian County, Shanxi Province, Chin
中国山西省盂县党校

当第k个五边形(粗虚线)的着色为ACDCD(或BACDA)时，第(k+1)个五边形的顶点着色必须为ABCDB。像图11(2)中的情况一样，特征环C-D(或A-B)仍然存在，Z着色程序仍然可行。   
At this point, the proof is completed.   
至此，证明完成。         
As for other enlarged forms of E configuration:
至于E构形的其他放大形式:

Figure 11(3)：Figure 11, Extended Classification of E1 and E4
图11（3），E1和E4的扩展分类
When some color edges are arbitrarily elongated in the minimum E- configuration, or the known four-color components are embedded in some triangles, quadrilaterals and pentagons, the Z-coloring scheme is still feasible as long as the geometric structure and large coloring graph with ten symmetry are not destroyed and the characteristic ring A-B or C-D exists.
当一些颜色边在最小E-构形中被任意拉长，或者已知的四色分量嵌入到一些三角形、四边形和五边形中时，只要不破坏几何结构和十折对称的大着色图，且特征环A-B或C-D存在，Z-着色程序仍然是可行的。
When the above-mentioned four-color components are embedded in the enlarged interface outside the minimum E- configuration and still present ten-fold symmetry, it can be seen from Figure 11(2)、11(3) in the inductive hypothesis proof that the added color edge is only the elongation of part of the color edge in the minimum E- configuration, and no further study is needed.
当上述四色分量嵌入最小E构型外的放大界面，仍呈现十折对称性时，从归纳假设证明中的图11(2)、11(3)可以看出，增加的色边只是最小E构型中部分色边的拉长，无需进一步研究。
If in doubt, please refer to the literature [8]
如有疑问，请参考文献[8]
6.Conclusions
6.结论
Combining 1 to 5, through four new theorems, the coloring dilemma configurations of geometric structures of arbitrary vertices and edges are divided into two categories according to whether they have 10-fold symmetric geometric structures, and then they are colored in four colors by using H coloring program or Z coloring program, thus making up the loophole of A. B Kempe's proof, completing the elementary and concise proof of the four-color conjecture and realizing Appel's scientific foresight.
结合1到5，通过4个新的定理，将任意顶点和边的几何结构的染色困境构形按照是否具有10折对称几何结构分为两类，然后利用H着色程序或Z着色程序对其进行四色着色，弥补了A. B Kempe证明的漏洞，完成了四色猜想的初等简洁证明，实现了Appel的科学预见。

【express one's thanks】 during the completion of the thesis, A.Lehoyd, a graph theorist at Lancaster university in England, sent document 6, and Mr. wan chunru of Singapore translated many proofs of four-color conjecture for me,Mr. gan feng of Beijing provided his research results "proofs of 4CC and 1+1", received sincere guidance from professor cuiqin Lin of Tsinghua University, professor wangming Wu of Shanghai normal university, professor Guizhen Liu of Shandong university, Here, we express our gratitude.
【致谢】论文完成期间，英国兰开斯特大学图理论家A.Lehoyd发来6号文件，新加坡万春如先生为我翻译了多份四色猜想的证明，北京的敢峰先生提供了他的研究成果“4CC与1+1的证明”，得到了清华大学林翠琴教授、上海师范大学吴名望教授、山东大学刘桂真教授的真诚指导，在此，我们表示感谢。

References:
【1】 Zhongfu Zhang, the trap of mathematics-various "proofs" of four-color conjecture, Nature Journal, Volume 14, No.5, 1991,pp379-382
【2】 Shouchun Xu , Illustration of Four Colors, Peking University Press, January 2008,No.25.
【3】 Jiahe Jiang, Solving the Problem of Four Colors Map, World Science Translation Journal, 1979.04, pp49. Or scientific American, vol. 237, No.10, 1977.
【4】  Owen kittel, operation on partially dyed maps, Bulletin of  the American  Mathematical  Society, Volume  41, Number  6 (1935), pp407-413.
【5】A Tentative Four Coloring of Planar Graphs (khamis Carr and William Kokai), J. Comb. Math. Comb. Comp. 46 (2003), 97-112.
【6】Holryd,F.C and Miller,R.G, The example that Hewood shold have given Quart J Math.(1992),43(2),67-71
【7】 Kemp re-study, American Mathematical Monthly 105, No.2, February 1998, pp170-174
【8】 Enlargement of "E Family Configuration",see attachment.

Attachment:References material [8]
附件:参考资料[8]


Enlargement of"E- Family Configuration"
Yudian Zhang
“染色困局构形”的扩展 张彧典

In "A Tentative Four Coloring of Planar Graphs" [1], hamish Karl and William Cocay discussed the embedding amplification of CK graphs. They pointed out that:
在《平面图的一个试探性四着色》[1]中，hamish Karl和William Cocay讨论了CK图的嵌入放大。他们指出:
"All edges of the Kempe chain in G that caused the algorithm 2.1 loop still exist in G'. Therefore, for the input graph G', the cycle of algorithm 2.1 is a multiple of 20. Figure 6 shows an example.
显示“导致算法2.1循环的G中的Kempe链的所有边仍然存在于G’中。因此，对于输入图形G’，算法2.1的循环是20的倍数。图6显示了一个例子。
These structures can be used to generate a large number of plane graph to make algorithm 2.1 cycle. They are all based on the Kempe chain in the CK diagram. We have not found any other   plane graph that can loop algorithm 2.1.
这些结构可以用来生成大量的平面图，使算法2.1循环。它们都基于CK图中的肯普链。我们没有发现任何其他平面图可以循环算法
2.1。
Question: Is there any plan graph that is not based on CK graph that will make algorithm 2.1 cycle? "
问题:有没有不基于CK图的平面图会让算法2.1循环？
In the above discussion，
" 在上面的讨论中，
1. "Algorithm 2.1" is the H coloring program that Miller and his colleagues gave to cause the cycle of configuration.
1.“算法2.1”是米勒和他的同事为了引起配置循环而给出的H着色程序。
2. CK graph is the E1 configuration in the 4-coloring proof of the four-color conjecture "dye dilemma configuration";
2.CK图是四色猜想“染色困境构形”的4色证明中的E1构型；
3. Two problems need to be solved:
3.需要解决两个问题:
First, can we find "Kempe chain based on CK graph" and "other  plane graph that can make algorithm 2.1 cycle"?
一、能不能找到“基于CK图的Kempe链”和“其他能使算法2.1循环的平面图”？
For this problem, we have found 3 other configurations identical to E1 configuration, which are also "the plan   graph that makes algorithm 2.1 cycle".
对于这个问题，我们发现了3个与E1配置相同的其他配置，它们也是“使算法2.1循环的平面图”。
The second is "is there any plan graph that is not based on CK graph that will make algorithm 2.1 cycle?"Now, let's discuss the second question as follows.
第二个问题是“是否有任何不基于CK图的计划图会使算法2.1循环？” 现在我们来讨论第二个问题，如下。
We think the answer to this question can be divided into three types of analysis.
我们认为这个问题的答案可以分为三种分析。
In the first category, the four-color component is embedded inside any triangle in the CK diagram, as shown in fig. 1:
在第一类中，四色分量嵌入在CK图中的任何三角形内，如图1所示:
In this configuration, a four-color component (blue) is embedded in any triangulation (red) of CK graph. Because it does not destroy the ten-fold symmetric main component and its color diagram of the original configuration, the known A-B ring still exists, so it can be correctly 4-colored by Z coloring program.

figure 1
在这种配置中，四色分量(蓝色)嵌入在CK图的任何三角测量(红色)中。由于不破坏原构型的十重对称主成分及其色图，已知的A-B环依然存在，因此可以通过Z着色程序正确地进行4色着色。

figure 2(fig. 6 in document 5)
The second type: four-color components are embedded in any quadrilateral of CK graph; As shown in fig. 2 (fig. 6 in document 5):
第二类:四色分量嵌入CK图中的任意四边形，如图2（文献5中的图6）所示:
In Figure 2, the four-color components shown in red, blue and pink are embedded outside the enlarged configuration of E1 configuration shown in red Pentagon and black quadrilateral. At this time, the existence of blue A-C chain or red B-D-B chain will not affect the cycle of the whole configuration, because:
在图2中，以红色、蓝色和粉色显示的四色组件嵌入在以红色五边形和黑色四边形显示的E1配置的放大配置之外。此时，蓝色A-C链或红色B-D-B链的存在不会影响整个构型的循环，因为:
As shown in fig. 3, when executing the h coloring program (the first cycle is 4 times), four color chains, a-d, b-d, b-c, and a-c, must be involved, which are distinguished by orange-green-blue-red in turn, and their trajectories in a 10-fold symmetric structure form a pattern, leaving only 12 uncovered edges, as shown by dashed lines.

Fig. 3 dyeing trace pattern     Fig. 4  10-fold symmetrical template
图3染色痕迹图案                      图4 10倍对称模板
如图3所示，当执行H着色程序(第一个周期4次)时，必须涉及四个颜色链，a-d、b-d、b-c、a-c，它们依次由橙-绿-蓝-红区分，并且它们在10折对称结构中的轨迹形成图案，仅留下12个未覆盖的边，如虚线所示。
Then, we put the pattern on the 10-fold symmetrical geometric structure template ① shown in fig. 4 and show it in orange.         
然后，我们把图案放在图4所示的10折对称几何结构模板①上，用橙色显示。
After we carry out the second period H dyeing procedure, the pattern is rotated 144 degrees counterclockwise from orange ① in Figure 4 to red ②. At this time, the other 8 edges are covered, as shown by the red edge marked 2 in fig. 4, and then there are still 4 edges uncovered.
当我们施行第二周期H染色程序以后，就是将图案从图4中的橙色①逆时针旋转144度至红色②。此时，另外8个边被覆盖，如图4中标记为2的红色边所示，然后仍有4个边未被覆盖。   
After the third period of H dyeing, the pattern is rotated 144 degrees counterclockwise from red ② to green ③. At this time, three edges are covered, as shown by the green edge marked 3 in Figure 4. Finally, only one edge is left uncovered.
当我们施行第三周期H染色程序以后，就是将这个图案从红色②逆时针旋转144度到绿色③，此时，三个边缘被覆盖，如图4中标记为3的绿色边缘所示，最后，只剩下一个边没有被覆盖。
After we carry out the fourth cycle H dyeing procedure, we will rotate the green patterns ③ to ④ counterclockwise by 144 degrees. At this time, the last edge is also covered, as shown by the blue edge marked 4 in fig. 2.
当我们施行第四周期H染色程序以后，就是将绿色③到④的图案逆时针旋转144度。此时，最后一条边也被覆盖，如图2中标记为4的蓝色边所示。
When we apply the above-mentioned covering theory (four H-staining procedures are reversed for 16 times) to cover the enlarged red E1 image shown in Figure 5, it is not difficult to find that the five sides of the green Pentagon are not covered, which means that any four-color component embedded in the quadrilateral where the five sides are located will not affect the cycle of this enlarged configuration.
当我们应用上述覆盖理论（4个H染色程序一共16次颠倒染色）覆盖图5所示的红色E1放大图时，不难发现绿色五边形的五条边都没有被覆盖，说明任何嵌入五条边所在四边形的四色分量都不会影响这种放大构型的循环。
Why is this happening?  Look at figure 5: The pentagonal fence with enlarged configuration and the pentagonal fence with minimum E1 configuration and the four vertices of the quadrilateral where they are located are dyed exactly the same, and each edge is the same, which is the outward extension of these edges in E1 configuration, so they will not participate in four times of reverse dyeing coverage and become useless edges. This shows that it is not necessary to discuss the case that the four-color components are embedded in the enlarged configuration of the E-family configuration, only the E-family configuration with the smallest embedding needs to be discussed.

Fig. 5(same as basic frame of fig. 1)
图5(与图1的基本框架相同)
为什么会这样？看图5:放大配置的五边形栅栏和最小E1配置的五边形栅栏以及它们所在四边形的四个顶点染色完全一样，每条边都是一样的，是这些边在E1配置中的向外延伸，所以不会参与四次反向染色覆盖，成为无用边。这说明没有必要讨论E族配置的放大配置中嵌入四色分量的情况，只需要讨论嵌入量最小的E族配置。
Now, when the E- family configurations expand inward, are their periodic cycles affected by any four-color components embedded in pentagons?
现在，当E族配置向内扩展时，它们的周期循环是否受到嵌入五边形的任何四色分量的影响？
After proving the lemma 3.1 correct, document 1 points out that the center vertex of CK has the same color in CK0, CK4(because it is not affected by the color exchange of any Kempe component).
在证明引理3.1正确后，文献1指出CK的中心顶点在CK0、CK4中具有相同的颜色(因为它不受任何Kempe分量的颜色交换的影响)。
The reason for this is that it is not involved in the dyeing of the B-D (or B-C) chain when the A-C (or A-D) chain side is reversed, and When the dyeing of A-C (or A-D) chain is reversed on one side of the generated B-C (or B-D) chain, it will not be involved.
其原因是，当A-C(或A-D)链侧反转时，它不参与B-D(或B-C)链的染色，而当A-C(或A-D)链的染色在生成的B-C(或B-D)链的一侧反转时，它将不参与。
This conclusion shows that the existence of the central vertex of a 10-fold symmetric geometric structure is indispensable for E- family configurations in periodic cycles.
这个结论表明，10折对称几何结构的中心顶点的存在对于周期环中的E族构型是不可缺少的。
From this, we can discuss the different situations of inward expansion shown in figs. 4, 5 and 7 (only 4 and 7 are given here) given in document 1. In fact, G and H in the three configurations have the same meaning(any four-color component is embedded in the central pentagon).

Fig. 6 (fig. 4 in reference 1)
图6(参考文献1中的图4)
由此，我们可以讨论文献1中给出的图4、5和7(这里仅给出4和7)中所示的向内膨胀的不同情况。实际上，三种配置中的G和H有着相同的含义(任何四色分量都嵌入在中央五边形中)。
See if the embedded four-color component G or H breaks the 10-fold symmetry and color map of the original image.
看嵌入的四色分量G或H是否破坏了原图像的10折对称性和色图。
If the 10-fold symmetry and color diagram of the original image are not destroyed, then the new configuration still has a periodic cycle, which can be reduced by z-dyeing program.
如果原图的10折对称性和色图没有被破坏，那么新的构形还是有周期循环的，可以通过Z染色程序减少。
The third category: any dichromatic edge extension in CK graph
shown in fig. 6:
第三类:CK图中的任意双色边延拓 如图6所示:   
This kind of configuration is that two colors are embedded in the same color chain, any color chain except C2-D2 chain in an arbitrarily elongated CK graph (because the elongation of this color chain will simplify the original graph into the simple case where A1-C1 and A1-D1 are separated from each other), which is actually an embedded quadrilateral case. Because the basic framework of the original CK graph is not broken, , z staining procedure can also be correctly 4-colored.
这种构形是两种颜色嵌入在同一个颜色链中，任意拉长的CK图中除了C2-D2链之外的任何一个颜色链(因为这个颜色链的拉长会把原来的图简化成A1-C1和A1-D1相互分离的简单情况)，这其实是一种嵌入四边形的情况。因为原始CK图的基本框架没有被破坏，Z染色程序也可以正确地进行4色染色。

Fig. 7 (also fig. 7 in reference 1)
图7(参考文献1中的图7)

Figure 8

References: 参考文献:
Chunru Wan , "A Tentative Four Coloring of Plan", Mathematics Forum of China Doctor Network, August 26, 2017.
万春如，“一个试探性的四着色方案”，中国博士网数学论坛，2017年8月26日。

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