转发张彧典的文章：四色猜想中“染色困局构形”的4染色（一）

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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（一）
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
  
Abstract: This paper attempts to successfully remedy Kemp's proof loophole, that is, 4-coloring of "coloring dilemma configuration". Our methods are as follows: (1) The existence theorem of four-color quadrilateral and its corollary have been discovered and proved, from which the non-10-fold symmetry transformation rule of the geometric structure of Erera configuration  has been produced, and the "coloring dilemma configuration" is divided into two categories according to whether it is 10-fold symmetry or not by this rule; (2) Using this rule to synthesize the different research results of several mathematicians on Errera diagram, a new theorem 3 is established by using four different classifications of true and false mathematical propositions; (3) Using Theorem 3, the theoretical proof that the non-10-fold symmetry "coloring dilemma configuration" can be 4- dyed is obtained, and using mathematical induction, the 4- dyeing proof of the 10-fold symmetry "coloring dilemma configuration" is given. Complete the complete and concise artificial proof of the four-color conjecture.
摘要:本文试图成功弥补肯普的证明漏洞，即“染色困局构形”的4-着色。我们的方法如下1)发现并证明了四色四边形的存在定理及其推论，由此产生了Erera构形几何结构的非10折对称变换规则，并以此规则将“染色困局构形”按是否为10折对称分为两类；(2)利用这一规则综合几位数学家对Errera图的不同研究成果，利用数学命题的真假四种不同分类，建立了新的定理3；(3)利用定理3，得到了非10折对称“染色困局构形”可以4染色的理论证明，并利用数学归纳法，给出了10折对称“染色困局构形”的4染色证明。完成了四色猜想的完整简洁的人工证明。
Key words: four-color conjecture, Kemp’s proof, coloring dilemma configuration, four-color vertex quadrilateral, Non-10 fold symmetric transformation
关键词:四色猜想，肯普证明，染色困局构形，四色顶点四边形，非10折对称变换
1.Introduction  引言
The four-color conjecture was formally put forward in 1852 and lasted for 169 years. In 1879, Kempe gave a short proof【1】, but there was a defect, that is, the correct 4- coloring of "coloring dilemma configuration "
四色猜想于1852年正式提出，历时169年。1879年，肯普给出了一个简短的证明【1】，但有一个缺陷，那就是“染色困局构形”的正确4-着色。
In 1976, American mathematicians Apel and Harken gave a machine proof. 【2】Robertson further simplified the machine proof in 1997, but it was not universally accepted in mathematics. Apel foresaw that one day someone would find a brief proof of the four-color conjecture.
1976年，美国数学家阿佩尔和哈肯给出了一个机器证明。【2】罗伯逊在1997年进一步简化了机器证明，但在数学界并没有被普遍接受。阿佩尔预见，有一天会有人找到四色猜想的简单证明。
In 2018, I found an important theorem (four-color vertex quadrilateral and its property theorem) in the 4-color maximum plane graph; At the same time, it is found that the "Erera diagram" given by Erera 1921 has three other homomorphic configurations (all 10-fold symmetric dyeing dilemma configurations), which constitute the E family 4 configuration. These two great discoveries produced two new theories and methods, which skillfully solved the classification of dyeing dilemma configuration and correct 4- dyeing, made up for Kemp's loophole, completed the brief proof of four-color conjecture, and realized Appel's foresight.
2018年，我在4色最大平面图中发现了一个重要定理(四色顶点四边形及其性质定理)；同时发现，Erera 1921给出的“Erera图”还有另外三种同态构型(均为10折对称染色困局构形)，构成了E族4构形。这两大发现产生了两种新的理论和方法，巧妙地解决了染色困局构形的分类和正确的4染色，弥补了肯普的漏洞，完成了四色猜想的简要证明，实现了阿佩尔的预见。
Next, follow the discussion method of combining practice with theory; The definitions and theorems are derived from 4-color maximal planar graph 【3】 (i.e., "triangulation"【4】). All configurations are maximal plan views, and the outer plane V adjacent to the Pentagon is omitted.
其次，遵循实践与理论相结合的讨论方法；定义和定理来源于4色最大平面图【3】(即“三角剖分”【4】)。所有构形都是最大平面图，与五边形相邻的外平面V被省略。
2. coloring dilemma configuration model and related definitions and theorems
2.着色困局构形模型及相关定义和定理
In 1935, “the Journal of the American Mathematical Society ”published “a set of operations on partially dyed maps” 【5】, in which " coloring dilemma  configuration" was defined and a basic model was given, as shown in the left of Figure 1. We simplified it as an 8-point minimum model with dual graph, as shown in the right of Figure 1; The author also gives the "Erera diagram" (abbreviated as "E configuration", that is, E1 in Figure 5).
1935年，《美国数学学会杂志》发表了《部分染色地图上的一组运算》【5】，其中定义了“染色困局构形”，并给出了基本模型，如图1左图所示。我们将其简化为具有对偶图的8点最小模型，如图1右侧所示；作者还给出了“Erera图”(缩写为“E构型”，即图5中的E1)。

Figure 1: Dyeing dilemma configuration model
图1:染色困局构形模型
Definition 1. Configuration
On the basis of Kemp's definition, we add a detailed definition of the "configuration" , that is, the correct four-color dyed configuration consists of two parts: one is the geometric structure composed of points and edges, and the other is the "color distribution map" （abbreviated as"color diagrams,"）formed by a certain four-color dyeing scheme.
定义1。构形
我们在肯普定义的基础上，增加了对“构形”的详细定义，即正确的四色染色构形由两部分组成:一是由点和边组成的几何结构，二是由某种四色染色方案形成的“颜色分布图”(简称“色图”)。
Definition 2. 4-color vertex quadrilateral and 3-color vertex quadrilateral.        
定义2。四色顶点四边形和三色顶点四边形。
In the configuration correctly dyed with four colors, if four different dyed vertices and the edges connecting them form a minimum quadrilateral, this quadrilateral is called a four-color vertex quadrilateral; If three different colored vertices form a minimum quadrilateral, this quadrilateral is called a tricolor vertex quadrilateral.
在用四种颜色正确染色的配置中，如果四个不同的染色顶点和连接它们的边形成一个最小四边形，这个四边形称为四色顶点四边形；如果三个不同颜色的顶点形成一个最小四边形，这个四边形称为三色顶点四边形。
Theorem 1: There is inevitably at least one minimum four-color vertex quadrilateral in the configuration correctly colored with four colors.                           
Proofproof by contradiction)
In a configuration correctly dyed with four colors, if there is no at least one minimum four-color vertex quadrilateral, that is, all the minimum quadrilaterals are tricolor vertex quadrilaterals, then there must be two diagonal corners same color in this tricolor vertex quadrilateral, so the coloring of the two triangles in this tricolor vertex quadrilateral must be the same, then all triangle vertices in the whole configuration must also be dyed with such three colors, that is, the vertices in the whole configuration are also dyed with such three colors, which contradicts the conditions of dyeing with four colors. Use figure 2 for verification:
定理1:在用四种颜色正确着色的构形中，不可避免地至少有一个最小的四色顶点四边形。
证明反证法) 在用四种颜色正确染色的构形中，如果没有至少一个最小四色顶点四边形，也就是所有的最小四边形都是三色顶点四边形，那么这个三色顶点四边形中一定有两个对角颜色相同，那么这个三色顶点四边形中的两个三角形的着色一定是相同的，那么整个配置中的所有三角形顶点也用这样的三种颜色染色，也就是整个配置中的顶点也用这样的三种颜色染色，这与用四种颜色染色的条件相矛盾。使用图2进行验证：

Figure 2: Minimum four-color and three-color vertex quadrilateral
图2:最小四色和三色顶点四边形
In Figure 2, the geometric structures of (1) and (2) are the same, except that one vertex of the color map is dyed differently.
(1)A quadrilateral with four-color vertices is included in the maximum plane graph, which is the minimum four-color map;
(2) A quadrilateral with a tricolor vertex is included in the maximum plane graph, which is the minimum tricolor map.
Obviously, (1) and (2) are contradictory, which is the best interpretation of Theorem 1.   
在图2中，(1)和(2)的几何结构是相同的，除了颜色图的一个顶点被不同地染色。 (1)具有四色顶点的四边形包含在最大平面图中，这是最小四色图； (2)具有三色顶点的四边形包含在最大平面图中，这是最小三色图。 显然，(1)和(2)是矛盾的，这是对定理1的最好解释。                              
Definition 3. Color chain and opposite color chain:
In the correct 4-coloring configuration, if at least two vertices of different coloring are connected by edges, they are called color chains, such as A-B chain, B-C chain, etc. When a color chain presents a closed loop, it is called a ring chain. If two color chains are dyed differently, they are said to be opposite to each other, such as A-B chain and C-D chain,                                   
定义3。颜色链和对比色链: 在正确的4-着色配置中，如果至少两个不同着色的顶点由边连接，则它们被称为颜色链，如A-B链、B-C链等。当颜色链呈现闭环时，称为环链。如果两种颜色链染色不同，就说是相反的，比如A-B链和C-D链，
Theorem 2. Four-color vertex quadrilateral property theorem:
定理2。四色顶点四边形性质定理；
（1）Two diagonal chains of a four-color vertex quadrilateral cannot exist in the same four-color vertex quadrilateral at the same time.
（1）一个四色顶点四边形的两条对角链不能同时存在于同一个四色顶点四边形中。
Proof: Because the four vertices in a four-color vertex quadrilateral are of different colors, diagonal chains connected by two groups of non-adjacent vertices are opposite color chains, so they cannot exist at the same time in the same four-color vertex quadrilateral plane.
证明：由于一个四色顶点四边形中的四个顶点颜色不同，由两组不相邻顶点连接的对角链是相反的颜色链，所以它们不能同时存在于同一个四色顶点四边形平面中。
（2）In the four-color vertex quadrilateral, changing the known diagonal chain can only destroy the geometric structure of the original configuration, but not the combination of color points of the original configuration.
（2）在四色顶点四边形中，改变已知的对角链只能破坏原构型的几何结构，而不能破坏原构型的色点组合。
Proof:
Changing the known diagonal chain in the quadrilateral with four-color vertices only changes the combination of edges in the original configuration (i.e., geometric structure), but does not change all vertices in the original configuration and their coloring. 证明:
改变具有四色顶点的四边形中已知的对角链，只会改变原始构形几何结构中边的组合，而不会改变原始构形中的所有顶点及其颜色。
3. Two kinds of proofs of an important lemma       
3.一个重要引理的两种证明
In "A Tentative Four Coloring of Planar Graphs" [5] published in 2003, the author expressed E1 configuration as its dual graph, called CK graph, and proved lemma 3.1: "When the initial staining is CK0, the algorithm 2.1 loops and takes 20 as a cycle", which is proved in detail in Literature 5.
在2003年发表的《平面图的一个试探性四染色》[5]中，作者将E1构形表示为它的对偶图，称为CK图，并证明了引理3.1:“当初始染色为CK0时，算法2.1循环并以20为循环”，这在文献5中有详细证明。
In 1992, in the article "The example of Hewood should be known", the author gave a "Hewood graph", that is, E1 configuration, and proved that "when Hewood inverted staining (H staining procedure for short) is carried out four times, E1 configuration dyeing cycle", which can be called lemma 3.2 here. See ref.6 for detailed proof. In this paper, "four times of Herwood inverted staining" is called "H staining procedure".
1992年，在《赫伍德的例子应该知道》一文中，作者给出了一个“赫伍德图”，即E1构形，并证明了“当赫伍德颠倒染色(简称H染色程序)进行四次时，E1构形染色循环”，这里可以称之为引理3.2。详细证明见参考文献6。本文将“四次赫伍德颠倒染色”称为“H染色法”。
The algorithm 2.1 in document 5 is essentially the same as the H-staining procedure in document 6, except that the four colors in the two proofs and the arrangement directions on the fence are different (the former 1, 2, 3 and 4 are arranged clockwise, while the latter B, R, Y and G are arranged counterclockwise), so the reverse staining of algorithm 2.1 is clockwise, and the H-staining procedure is counterclockwise, It shows that the reverse dyeing in two directions (counterclockwise and clockwise) is the same for the periodic cycle of E1 configuration. In this paper, A, B, C and D are used to represent four different stains, and the staining procedure is unified as "H staining procedure".   
文献5中的算法2.1与文献6中的H染色程序基本相同，只是两个样张中的四种颜色和栅栏上的排列方向不同(前者1、2、3、4为顺时针排列，后者B、R、Y、G为逆时针排列)，所以算法2.1的颠倒染色为顺时针，H染色程序为逆时针，说明对于E1组态的周期循环，两个方向(逆时针和顺时针)的颠倒染色是相同的。本文用A、B、C、D表示四种不同的染色，染色程序统一为“H染色程序”。                                                      
The proof of Lemma 3.1 and Lemma 3.2 :  shown in Figure 3:
引理3.1和引理3.2的证明:如图3所示:

Figure 3: Cyclic process of E1 configuration
图3:E1构形的循环过程
In fig. 3, (1) is the color diagram formed by the intersection of A-C chain (indicated by dashed line) and A-D chain (indicated by thick black line) when E1 configuration is in the initial position. In (2)-(5), the dotted line represents the known or generated maximum chain. Thick black lines represent color chains dyed upside down.
在图3中，(1)是当E1构形处于初始位置时，A-C链(用虚线表示)和A-D链(用粗黑线表示)相交形成的颜色图。在(2)-(5)中，虚线代表已知或生成的最大链， 粗黑线代表颠倒染色的颜色链。
Sequentially carrying out reverse dyeing for B-D, D-A, A-C and C-B four times counterclockwise along the arrow,(6)is the configuration generated after 4 times of reverse dyeing .Comparing (1) with (6), it is not difficult to find that the geometric structures of the two configurations are the same, and both have ten-fold symmetry, and color diagram is the same, except that the position of the color diagram  is rotated clockwise by 144° from (1).
沿箭头逆时针方向依次对B-D、D-A、A-C、C-B进行四次颠倒染色，(6)是经过四次颠倒染色后生成的构形。对比(1)和(6)不难发现，两种构形的几何结构是一样的，都有十折对称性，而色图也是一样的，只是色图的位置从(1)顺时针旋转了144°。
When the figure (6) undergoes four cycles of H dyeing, that is, a total of 20 times (4 times X5) reverse dyeing, the geometric structure and dyeing of E1 configuration completely rotate back to the position of the figure (1).   
当图(6)经过4次H染色循环，即总共20次(4次X5)颠倒染色时，E1构形的几何结构和染色完全旋转回到图(1)的位置。                                                                                                                                         4.The generation of E- family configuration and the perfection of two lemmas
4.E族构形的生成和两个引理的完善

Figure 4: Four Continuous Change Graphs of E1 Configuration
图4:E1构形的四个连续变化图
In fig. 3, when the H-dyeing procedure is carried out, the graph (1) and the continuously evolved graphs (3), (4) and (5) form a cycle due to the reverse dyeing for four consecutive times. after careful investigation of their color diagrams, they still belong to the dyeing dilemma configuration, as shown in fig. 4:
在图3中，当进行H染色程序时，由于连续四次颠倒染色，图(1)和连续演变的图(3)、(4)和(5)形成一个循环。仔细考察它们的色图后，它们仍然属于染色困局构形，如图4所示:
In fig. 4, their geometric structures have not changed,they all have 10-fold symmetry, but the color diagram has changed,  the combinations of two intersecting color chains (marked with dashed lines and solid lines) starting from peak points A1, C1, B1 and D1 are different.
在图4中，它们的几何结构没有变化，都具有10折对称性，但颜色图发生了变化，从峰值点A1、C1、B1和D1开始的两条相交颜色链(用虚线和实线标记)的组合不同。
When all of them are changed into "BAB" distribution of peak point A, the four different E configurations shown in Figure 4 are arranged in the order of peak points A1[ corresponding to Figure (1)], B1[ corresponding to Figure (4)], D1[ corresponding to Figure (5)] and C1[ corresponding to Figure (3)], and the four configurations shown in Figure 5 can be obtained, which is abbreviated E1、E2、E3、E4。
当全部变为峰点A的“BAB”分布时，图4所示的四种不同的E构形按照峰点A1[对应图(1)]、B1[对应图(4)]、D1[对应图(5)]和C1[对应图(3)]的顺序排列，即可得到图5所示的四种构形，缩写为E1、E2、E3、E4。

Fig. 5: Four Homomorphic Configurations in E- Family
图5：E族中的四种同态构形
Examining the four homomorphic configurations in Figure 5, it is not difficult to find that their geometric structures are 10 times symmetrical, but their color diagrams are different. When we apply H staining programs to them respectively, the configuration will cycle periodically regardless of the direction (clockwise or counterclockwise). What factors determine their circulation? We think that the 10-fold symmetry of geometric structure is the main factor, while the color chart is only the secondary factor. Therefore, the propositional conditions of lemma 3.1 and lemma 3.2 are imperfect. Therefore,
Lemma 3.1 should be perfected as follows:
检查图5中的四个同态构形，不难发现，它们的几何结构是10折对称的，但是它们的色图是不同的。当我们分别对它们应用H染色程序时，无论方向如何(顺时针或逆时针)，构形都会周期性地循环。是什么因素决定了它们的循环？我们认为几何结构的10折对称性是主要因素，而色图只是次要因素。因此，引理3.1和引理3.2的命题条件是不完善的。因此， 引理3.1应完善如下:
When it has ten fold symmetry and the initial coloring is CK0, the algorithm 2.1 loops and takes 20 as a cycle.
当它具有十折对称性并且初始着色为CK0时，算法2.1循环并以20为周期。
Lemma 3.2 should be perfected as follows: 引理3.2应完善如下:
When Herwood inverted staining (H staining for short) is carried out four times, E1 configuration staining with ten symmetry presents a cycle.
当赫伍德倒置染色(简称H染色)进行四次时，具有十折对称性的E1构型染色呈现一个循环。 上述两个引理显然是可逆的。
The above two lemmas are obviously reversible. If lemma 3.1 is taken as the original theorem, then lemma 3.2 is the inverse theorem of lemma 3.1. At this time, according to the four different combinations of the four propositions, as shown in Figure 6, it can be judged that these four propositions are true propositions. Therefore, the negative principle of lemma 3.1 in literature 5 is also true, that is,
如果把引理3.1作为原定理，那么引理3.2就是引理3.1的逆定理。这时，根据四个命题的四种不同组合，如图6所示（图6实际上是一个表格），可以判断这四个命题是真命题。因此，文献5中引理3.1的否定原理也是成立的，即，
Theorem 3: When the configuration is not 10-fold symmetric and the initial color is CK0. Algorithm 2.1 does not loop.
定理3：当构形不是10折对称，初始颜色为CK0时，算法2.1不循环。
Generally speaking,there are four situations in which the four propes are true or false.
一般来说，有四种情况，这四个属性是真或假。

一般来说，四种合题的真假性，有且仅有以下四种情况

●两个命题互为逆否命题，他们有相同的真假性；
●两个命题互为逆命题或互为否命题，他们的真假性没有关系。
Figure 6: Classification of the truth and false of four propositions
图6：四个命题的真假分类
Obviously, the following inference can be drawn from Theorem 3: As long as the geometric structure of the dye dilemma configuration with any combination of points and edges is not 10-fold symmetrical, there will be no periodic cycle in the process of H dyeing, and this configuration can be correctly 4- dyed by a limited number of reverse dyeing.
显然，从定理3可以得出以下推论: 只要任意点和边的组合的染色困境构型的几何结构不是10折对称，H染色过程中就不会出现周期循环，这种构型可以通过有限次数的反向染色正确地进行4染色。

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