A simple logic proof of the Four Color Theorem
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A simple logic proof of the Four Color TheoremHuang XingbinCollege of Physics Science and Technology,Heilongjiang University,China,Harbin,Xuefu Road74,150086,AbstractThe Four Color Conjecture already was solved under the computer help.In this paper, we unify A.B.Kempe exchange color method and the diagonaltransformation method. Proposed and proved the using diagonaltransformation certainly can change every standard graph into all maximum planar graph with same vertex;if a random maximum planar graph is4colorable,then the new maximum planar graph which obtained by one time random diagonaltransformation was4colorable still.Further proved all maximum planar graphs are4colourable.Therefore has given The Four Color Conjecture simple and direct logical proof.1IntroductionThe story of the Four Color Problem comes from England.In1852,F.Guthrie graduate from University College London was coloring in a map showing the counties of England.He puts forward an interesting problem that the maximum number of colors required coloring any map seemed likely to be four.The coloring has to meet the obvious requirement that no two regions sharing a length of common boundary should be given the same color.In1872,England at that time wellknown mathematicianA.Cayley put forward this problem toward the London Mathematical Society formally. From then on Guthrie's question became known as the Four Color Problem but also caused the extensive attention of mathematics field in the world.In the world many topranking mathematicians attended to prove the Four Color Problem in succession,but nobody can give a right proof.And it grew to be the second most famous unsolvedproblem in mathematics after Fermat's last theorem.In1976,two American mathematicians,K.Appel and W.Haken,announced that they had solved the problem. But there was a twist.Much of their proof was carried out on a computer,and was far too long for humans to check.To this day it is not known if there can be a short proof of the Four Color Theorem that a human could follow.Most mathematicians think there is not.Here we give a simple logic proof by the method of the diagonaltransformation.2Map’s Dual Graph and All Maximum Planar GraphsWhat makes the four color problem so hard is that it refers to all mapsnot just all the maps in all the atlases around the world,but all conceivable maps,maps with millions (and more)of countries of all shapes and sizes.Knowing that you can color some particular map using four colors does not help you at all.You need to produce an argument that will work in all cases[1].Since the sizes and shapes of the map regions do not matter,only the way they join together,the Four Color Problem is a question in topology.We shall see momentarily,it is easily reformulated as a problem about maximum planar graph.Within each region of the given map,you place a single point,known as a vertex of the graph.(You can think of these vertices as the capital cities of the countries,if you wish.)You then join up the vertices to form a graph,in much the same way that you might link two cities by a rail,known as an edge of the graph.The rule is that two vertices are joined together if,and only if,their respective map regions share a common boundary,in which case the line joining them has to lie entirely within the two regions. (In terms of a rail link this would mean that the line couldn’t cross the territory of any third country.)Like this obtains graph named map’s dual graph.Indeed,the Four Color Theorem can be expressed in terms of the dual graph as follows:the dual graph’svertices can be colored with at most four colors such that no two vertices connected by an edge are colored the same color.A map is a planar graph with no“bridges”.Moreover,a planar graph’s dual is also a planar graph.Notice that no map’s dual will have a loop or more than one edge between any two vertices.From now on,we will ignore all graphs that have loops or more than one edge connecting any two vertices.Planar graphs are graphs that can be represented in the plane without any edges intersecting away from the vertices.A simple graph is called maximum planar graph if it is planar but adding any an edge would destroy that planar property.All faces(even the outer one)are then bounded by three edges,explaining the alternative term triangular for these graphs.The Four Color Theorem asserts that any map can have its faces colored with at most four different colors such that no two faces that share an edge are the same color. Thus,to prove the Four Color Theorem,it is equivalent to prove that no maximum planar graph requires more than four colors to color its vertices.For the sake of simplicity,from now on,we will refer to a graph as being4colorable if its vertices can be colored with4colors.Similarly,we called it is the4colored graph if its all vertices have colored by4colors(an admissible vertex4coloring).In this article,uses the maximum planar graph to prove the Four Color Theorem. If all maximum planar graphs can be colored using four colors,so can all planar graphs. The Four Color Theorem is true.All maximum planar graphs meaning be:The vertex quantity changes arbitrarily;if the vertex quantity is a constant so the edge quantity is also the constant but the shape of connecting the vertices with the edges changes arbitrarily in certain scope.In other word,there will be a lot of different configurations of graphs with the vertex of thesame quantity.Therefore the proof all maximum planar graphs are4colorable is extremely difficult.3Standard Graph and DiagonaltransformationOur proof method and other is completely different.But we can use the predecessor to solve the question experience.It mainly divides three steps.First,Let us choose a special configuration of maximum planar graphs,such as figure1.We called it is the standard graph.It is obvious conclusion that every standard graph that vertex quantity can change arbitrarily is certainly4colorable.Figure1:The Standard graph is composed by n vertex.Its all vertices are colored into the red,the green,the black and the white such that no two vertices connected by an edge are colored the same color.In the graph n is an even number.If n is an odd number,we can remove vertex n,with all has connectedits edge.Next,consider essences relation of the standard graph and the other graph.We use the conclusion that Professor LiuYanpei proved:The step by step using diagonaltransformation[2]change certainly all maximum planar graphs into its standard graph. The diagonaltransformation is shown in Figure2.Because the diagonaltransformation is the reversible transformation,therefore use it to change certainly every standardgraph into its all maximum planar graphs.We will prove the conclusion in the later word.Finally,let us assume a maximum planar graph is4colorable,using random one time diagonaltransformation change it into a new random maximum planar graph(See figure2).Is the new graph4colorable?If the answer is affirmative,we could prove that all maximum planar graphs are4colorable,so are all planar graphs.Figure2:Using diagonaltransformation once to be able to change to a new maximum planar graph from its standard graph(figure1).It is also4colorable. The socalled diagonaltransformation is:after elimination connection vertex②with④edge after,lays a new edge to connect the vertex③with⑤.4Diagonaltransformation does not Increase the Coloring NumberWe use the conventional method to solve the above these problems successfully. Proving these problems are very easy because had many scientists to do a lot of contributions for proving the Four Color Conjecture.The key part we proved is using Kempe exchange colors the method to complete;this method successfully has proven the Five Color Theorem.Proposition1:All standard graphs are4colorable.Proof:Standard graph with n vertices is constitutes by two chains.Such as figure1.A chain is from vertex①to vertex②,another chain is from vertex③to vertex n.Firstchain only has two vertices colored with the colors red and green;second chain is composed which by n2vertices are(alternately)colored with the colors black,white. Two standard graph’s difference is the vertex quantity different,but the vertex quantity change only in second chain.Therefore the vertex quantity change does not affect a standard graph is4colorable.Hence all standard graphs are4colorable.Proposition2:The step by step using diagonaltransformation certainly changes every standard graph into its all maximum planar graphs with same vertices.Proof:We proceed by induction on the number of vertices n.Let G(n)is any of the all maximum planar graph with n vertices,G s(n)is their standard graph with n vertices.In fact we will prove using diagonaltransformation can change certainly Gs(n)into allG(n).If the n is smaller,then using diagonaltransformation can change certainly Gs(n) into all G(n).Because it is verify easily.Assume:Step by step using diagonaltransformation can certainly change a standard graph with n1vertices into all maximum planar graphs with n1vertices.According to the conclusion of graph’s theory,any planar graph contains a vertex of degree at most5.The degree of a vertex is defined as the number of edges with that vertex as an end[3].Therefore G(n)contains a vertex v of degree at most5.If the degree of v is3,let them respectively is v1、v2and v3.Remove v and all the edges connected to it.The graph left has n1vertices and the mark is G(n1).According to assume,step by step using diagonal transformation can certainly change the standard graph Gs(n1)into G(n1).See figure3.Place the vertex v now in the v1,v2with v3constitutes of triangle inside in theGs(n1)and connects them,then getting the Gs(n).Let Gs(n)to duplicate the aboveGs(n1)diagonaltransformation process,therefore change certainly Gs(n)into theG(n).Figure3:The standard graph with n1vertices,the mark is G S(n1).If the degree of v is4,let them respectively is v1,v2,v3and v4,such as figure4A. Then use one time diagonaltransformation changes G(n)into G1(n),such as figure4B. In the G1(n),the degree of v is3.In this case,we run same argument on with above. Hence step by step using the diagonaltransformation can change certainly Gs(n)into G1(n).Next step using the inverse diagonaltransformation change the G1(n)into theG(n).Figure4:The A shows the degree of v is4in G(n).After the diagonaltransformation the B shows the degree of v is3in G1(n).If the degree of v is5,let them respectively is v1,v2,v3,v4and v5.Then use twice diagonaltransformation change G(n)into G2(n).In the G2(n)the degree of v is3.In this case,we run same argument on with above.Hence one time or many times(step by step)using the diagonaltransformation can certainly change Gs(n)into G2(n).Next step using twice the inverse diagonaltransformation change G2(n)into G(n).Hence step by step using the diagonaltransformation certainly changes every standard graph into its all maximum planar graphs with same vertices.Proposition3:Using random diagonaltransformation certainly changes any4colored maximum planar graphs into another4colored maximum planar graph.In order to proving the Proposition3we need an arbitrary diagonaltransformation so that it can replace all situations.Hence choose a quadrangle made of arbitrary four vertices in random4colored maximum planar graph G(n),such as figure5.In an arbitrary4colored maximum planar graph G(n)figure5A shows an arbitrary diagonaltransformation that certainly can replace all situations.After the diagonaltransformation figure5B shows the quadrangle in another maximum planar graph G d(n).Using a diagonaltransformation once change G(n)into G d(n).If G(n)is4colorable,so long as we can prove G d(n)also is4colorable.Hence we had also proven the proposition3.Figure5:The A shows a quadrangle made of arbitrary four vertices in random 4colored maximum planar graph G(n).The B shows its diagonaltransformationProof:The proposition3will be proved by Kempe method.Let G(n)is a4colored graph with red,green,black and white(an admissible vertex4coloring).The vertex①,②with③is the white in turn;red and the black vertex(see figure5A).Our goal removes connection vertex first①with③edge,and then lays a new edge connection vertex②With④.Regardless of vertex④color is green or red,we must guarantee all G d(n)is4colorable.First we can eliminate an easy case.If the④is green,removes connection the vertex①with③edge,then we achieve the goal immediately.Next if vertex④is red,we must transform the color of vertex④to with the vertex②color is different.Let us take a small detour,and introduce Kempe method.We denote the subgraph of G(n)by G rg.G rg subgraph by all red,the green vertex as well as connects their all edge is composed.A Kempe chain is a chain whose vertices are colored with only two colors.In particular,we speak of an rgchain if it is a Kempe chain whose vertices are(alternately)colored with the red and green.A Kempe net is a connected component of a subgraph.In particular,we speak of an rgnet if it is a Kempe net whose vertices are colored with the red and green.Each pair of vertices of a Kempe net can be joined by a Kempe chain.In fact;Kempe nets can be characterized using Kempe chain.We easy to understand on a Kempe net exchange colors can obtain another admissible vertex4coloring.Now we return to the proof:If vertex④is red but the vertex②with④must be in different rgnets.If so,and then we exchange the colors on the rgnet of containing vertex④.That assigns green to the④,and the②is red yet.Hence we can connect the ②with the④.Finally if the vertex④is red;the vertex②with④must be in the same rgnet, therefore there certainly is at least one rgchain of containing the vertex②with④. According to figure5there certainly is at the same time at least one rbchain of containing the vertex②with④.The rw–chain is also such.We consider the most complicated case now such as figure6show.Random exists in a lot of the rxchains of containing the vertex②with④(x is random one of in the g,b and w).If so,we can exchange the colors on the inside of any rxchains.Our goal is to change all rkchains into rpchains(k and p are random one of in the g,b and w but the p is not equal to the k)so that only retains two kinds of rxchains.Figure6:Random exist in a lot of the rxchains of containing the vertex②with④(x is random one of in the g,b and w).When the rxchains are divided into two parts,some encircle vertex①,else encircle vertex③.In order to let the rxchains changed and disappeared in the two parts is consistent colors.First considers left and right outmost rxchain color.If the left and right have same colors(e.g.they all are the rw–chain),we can choose any one of retains two kinds of rxchains(e.g.the rg–chain or rb–chain).If the left and right have different colors(e.g.the rw–chain and the rb–chain),we have no chosen,only can change a surplus rxchain(e.g.the rg–chain).According to abovementioned principle we handle the left rx–chain in figure6. Our goal is to change all the rbchains into the rgchains or the rw–chains.Let us consider the chain of②C2④,they is constituted alternately by the red and green vertices.Attention:Have already removed the edge connected the①with③now.So the②C2④②chain is equal to an rgcycle.Therefore we can in②C2④②inside exchange the black and white vertex the color.Now the C3change into the rw–chain from the rbchain,else chain one by one in order is:C1(rw–chain),C2(rg–chain),C3(rw),C4(rb),C5(rw),C6(rg),··①(rb).Next we can in②C3④②inside exchanges black and the green vertex color. Has:C1(rw),C2(rg),C3(rw),C4(rg),C5(rw),C6(rb),··①(rg).Repeat abovementioned process thus,the certainly finally has:C1(rw),C2(rg),C3(rw),C4(rg),C5(rw),C6(rg),··①(rg or rw).According to abovementioned principle we handle the right rx–chain in figure6. Our goal is same with the left side.We can run same argument on with above,the certainly finally has:C/1(rg),C/2(rw),C/3(rg),C/4(rw),C/5(rg),C/6(rw),··③(rg or rw).Now there is no the rbchain connected the vertex②with④.So the vertex②with④must be in different rbnet of G rb.If so,and then we flip the colors on the component containing the vertex④of G rb.That assigns black to the vertex④,and the vertex②is red yet.Hence we can connect the vertex②with④.Hence,G d(n)also is 4colored graph.Thus using random diagonaltransformation certainly changes any4colored maximum planar graphs into another random4colored maximum planar graph.5All Maps are4colorableConclusion:Every standard graph is four colorable(see figure1).Using step by step the diagonaltransformation must change every standard graph into its all maximum planar graphs with same vertices.Therefore,from a4colored standard graph transforms one time obtains a new4colored maximum planar graph.Again transformsone time also obtains another4colored maximum planar graph.So,continuously transforms it,finally we certainly can obtain all4colored maximum planar graph.In other word,all maximum planar graphs are4colorable,so are all maps.Now we can certainly say that the four colors were sufficient.The Four Color Conjecture is true.References[1]Keith st doubts removed about the proof of the Four Color Theorem From:/devlin/devlin_01_05.html[2]Liu Yanpei.Planar graph theory and the Four Color Problem(Ⅰ).Journal of mathematical research and exposition.1983,Vol.3(3):pp.123136.[3]K.Appel and W.Haken.The solution of the fourcolormap problem.Scientific American.1977,Vol.27(4):pp.108121AcknowledgementsWe thank Mr.Wang Xunye and Miss.Huang Yumeng for translating the paper. Correspondence should be addressed to H.X.(email:huangxingbin@)。