4- dyeing of "coloring dilemma configuration" in four-color conjecture

zhangyd2007@soh

                                   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（中英文）
                                                   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
                                                              Ming  Lei    ( lm85639720@163.com )
                                                                                               雷  明
                                                            No.4, Unit 6, Building 9, Jinduicheng Community,
                                              West Chang'an Street, Chang'an District, Xi'an, Shaanxi Province
                                                                   中国：金堆城钼业集团有限公司教育处退修职工  
       Abstract: This paper attempts to successfully remedy Kempe'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.
        
          Key words: four-color conjecture, Kempe’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 Appel and Haken gave a machine proof. 【2】Robertson further simplified the machine proof in 1997, but it was not universally accepted in mathematics. Appel foresaw that one day someone would find a brief proof of the four-color conjecture.
     
       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 "Errera diagram" given by Errera 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.
      
      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.
     

                                                 2. coloring dilemma configuration model and related definitions and theorems
                                                  2.染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 "Errera 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操作》【5】，其中定义了“染色困局构形”，并给出了基本模型。我们将其简化为具有对偶图的8点最小模型；作者还给出了“Errera图”(缩写为“E构形”，即图5中的E1)。

       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 a configuration that is correctly colored with four colors, if four different colored vertices and the edges connecting them form a minimal quadrilateral, this quadrilateral is called a four-color vertex quadrilateral; if the four vertices of the minimal quadrilateral only use three different Color dyeing, then, this quadrilateral is called a three-color vertex quadrilateral.
在用四种颜色正确染色的构形中，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；如果最小四边形的4个顶点只用三种不同颜色染色，那么，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，那么整个构形中的所有三角形顶点也用这样的三种颜色染色，也就是整个构形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。
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。颜色链和相反色链: 在正确的4-染色构形中，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.
      
       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.
      （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。
        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.
      
      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.
    （2）在四色顶点四边形中，改变已知的对角链只能破坏原构形的几何结构，而不能破坏原构形的色点组合。
      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 Heawood should be known", the author gave a "Heawood graph", that is, E1 configuration, and proved that "when Heawood 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 Heawood 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构型，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 two staining methods are called "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构形的周期循环，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，两种染色方法统一称为“H染色程序”。  
                                         
                                              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

        In two documents, it is proved that the E1 configuration obtains 4 configurations continuously transformed during the periodic cycle. We found that among these 4 configurations, the latter 3 configurations are homomorphic with E1. Check these 4 isomorphisms State configuration, it is not difficult to find that their geometric structure is 10-fold symmetrical, but their color diagrams are different, that is, the intersection of the A-C and A-D chains is different. When we apply the H coloring procedure to them separately, the configuration will cycle periodically regardless of the direction (clockwise or counterclockwise). What factors determine their cycle? We believe that the 10-fold symmetry of the geometric structure is the main factor, while the color map is only a 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:
     
       When E1 configuration  has ten fold symmetry and the initial coloring is CK0, the algorithm 2.1 loops and takes 20 as a cycle.
       当E1构形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 is carried out four times, E1 configuration staining with ten symmetry presents a cycle.
      当赫伍德颠倒染色进行四次时，具有十折对称性的E1构形染色呈现一个循环。 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 true and false sex , 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，根据四个命题真假性的四种不同组合，如图6所示，可以判断这四个命题都是真命题。因此，文献5中引理3.1的否定理也是成立的，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。

     Obviously, the following inference can be drawn from Theorem 3:
     As long as the geometric structure of the dyeing dilemma configuration of any combination of points and edges is not 10-fold symmetric, there will be no periodic cycle when the H dyeing procedure is carried out, that is, 4- dyeing can be carried out correctly through a limited number of reverse dyeing.
     
The inference of Theorem 3 realizes the research direction given by Professor Lin Cuiqin of Tsinghua University in 1996: If it can be proved that 4- coloring can be successfully achieved for any maximal plane graph after finite inverse coloring, it will be a great success and a shock at home and abroad
定理3的推论实现了清华大学林翠琴教授1996年给出的研究方向：如果能证明对于任意极大平面图，在有限次颠倒染色之后，都可以成功地实现4-染色，这将是一个巨大的成功，也是国内外的一个震惊。

                                                    5. Verification of Theorem 3 and its corollaries.
                                                                    定理 3 及其推论的验证。   
                                                  (Because this part has nothing to do with the proof, it is omitted here)
                                                 （因为这部分与证明没有多少关系，所以这里从略）


                                                    6.  4-Coloring proof of two kinds of dyeing dilemma configurations
                                                  
        Dyeing dilemma configuration can be divided into two categories according to whether its geometric structure has ten times symmetry. One is a 10-fold symmetric configuration (E-family four configurations and their extended configurations), and the other is a non-10-fold symmetric configuration. The 4-staining proofs of these two configurations are given below.
      
                                  6.1  The 4-coloring proof of the dilemma configuration of non-tenfold symmetric dyeing
                                
        Since these configurations already have Theorem 3, their 4-coloring problem has a pure theoretical proof.
       由于这些构形已经有了定理3，所以它们的4-染色问题有了一个纯粹的理论证明。

                                   6.2   4-staining proof of dyeing dilemma configuration with ten-fold symmetry
                                   6.2   具有十折对称性的染色困局构形的4-染色证明
       We use mathematical induction to prove it.
       我们用数学归纳法来证明。
       6.2.1 Inductive base proof:
      Because the E- group 4 configuration has a periodic cycle in the process of H dyeing, it cannot be proved that they can be 4- dyed by this method. In the four minimum E configurations, E1 and E2 contain characteristic chain A-B rings, and E3 and E4 contain characteristic chain C-D rings. Therefore, we use this dyeing property to give a special "Zhang dyeing program", called "Z dyeing program" for short (some called "tangent chain" method [7], some called "closed chain" method 4).
       6.2.1   归纳基础证明:
       由于E-群4构形在H染色过程中存在周期循环，不能证明它们可以用这种方法进行4染色。在四个最小E构形中，E1和E2包含特征链A-B环，E3和E4包含特征链C-D环。因此，我们利用这种染色性质给出一种特殊的“张染色程序”，简称“Z染色程序”(有的叫“切线链”法[7]，有的叫“闭链”法[4])。
        The z- coloring program of E-family 4 configuration can also be called Theorem 4.
        
         6.2.2   归纳假设证明;
                    分3种情形，都证明了定理4可行。


                                                           
                                                                    7. Conclusion
                                                                  
         
         
          Through the discussion of 1-6, establishes 4 new theorems, which solves the 4-coloring problem of the four-color conjecture coloring dilemma configuration, makes up for the loopholes in Kemp’s proof, and completes the short proof of the four-color conjecture .
        
      【注】如果想详细评读这篇论文，可以参考雷明先生转发在本栏目中的《四色猜想中染色困局构形的4-染色》（连续3篇）。
     

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