Coloring Eulerian triangulations of the Klein
bottle
Daniel Kr′a l’?Bojan Mohar?Atsuhiro Nakamoto?
Ondˇr ej Pangr′a c§Yusuke Suzuki?
Abstract
We show that an Eulerian triangulation of the Klein bottle has chro-matic number equal to six if and only if it contains a complete graph of
order six,and it is5-colorable,otherwise.As a consequence of our proof,
we derive that every Eulerian triangulation of the Klein bottle with face-
width at least four is5-colorable.
1Introduction
A graph is said to be Eulerian if all of its vertices have even degree.In this paper we study colorings of Eulerian triangulations.These are triangulations of closed surfaces whose graph is Eulerian.
The celebrated Four Color Theorem[2,25]asserts that every planar graph, in particular,every triangulation of the plane,is4-colorable.However,only a handful of planar triangulations are3-colorable,and it is relatively easy to see that a planar triangulation is3-colorable if and only if it is Eulerian[26].
Eulerian triangulations are interesting because of their tight connection to 3-colorability of planar graphs.From the algorithmic point of view,it is trivial to decide whether a planar graph is4-colorable,and an algorithm based on the proof of the Four Color Theorem[24]?nds a4-coloring of a given planar graph in quadratic time.On the other hand,it is NP-complete to decide whether a planar graph is3-colorable[12].Knowing this,it is surprising that a planar graph is 3-colorable if and only if it is a subgraph of an Eulerian triangulation.This fact has appeared in[16]and[18].Other conditions for3-colorability of plane graphs can be found e.g.in[7].
In this paper we study Eulerian triangulations of more general surfaces,in particular those of the Klein bottle.We refer the reader for a detailed introduction into graph embeddings on surfaces to a recent monograph[21]and we provide here only the basic results that we need in our further exposition.
Going o?the plane,the situation concerning chromatic properties of Eulerian triangulations changes drastically.For example,the complete graph K7of order seven forms an Eulerian triangulation of the torus–and yet,it needs seven colors to be colored.Even more is true.It can be shown that every graph is a subgraph of some Eulerian triangulation,so the study of chromatic properties of Eulerian triangulations of arbitrary surfaces is as hard as the same problem for arbitrary graphs.However,there are two directions,where one can make some breakthrough.One is to study locally planar graphs,those with large edge-width (see[21]for more details),and the other possibility is to restrict to a?xed surface.
Eulerian triangulations of the projective plane are still well understood.A set C of vertices of a triangulation T is called a color class if every face of T has precisely one vertex in C.If C is a color class in the Eulerian triangulation T, then T?U is an even-faced map,i.e.an embedded graph whose faces all have even size.Fisk[11]proved that any Eulerian triangulation of the projective plane contains a color class.Mohar[19]extended this result by proving:
Theorem1(Mohar[19]).Let T be an Eulerian triangulation of the projective plane,and let C be a color class in T.Then the chromatic numberχ(T)of T is equal to3if and only if T?C is bipartite.If T?C is non-bipartite and contains a subgraph which is a quadrangulation of the projective plane,thenχ(T)=5; otherwiseχ(T)=4.Moreover,given T,one can?nd a color class in T and determineχ(T)in polynomial time.
The proof of Theorem1is based on a discovery of Youngs[31]that quad-rangulations of the projective plane are either bipartite or4-chromatic,but their chromatic number is never equal to3.It also uses an extension by Gimbel and Thomassen[13]who proved that a graph of girth at least4embedded in the projective plane is3-colorable if and only if it does not contain a non-bipartite quadrangulation of the projective plane.
As a corollary of Theorem1we have:
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Theorem2.Every Eulerian triangulation of the projective plane is5-colorable.
This result can be proved directly by?rst observing that every projective planar graph is6-colorable.Next,Dirac’s theorem[1,8]asserts that the only obstacle for5-colorability is the presence of a complete graph of order six:the chromatic number of a projective planar graph is six if and only if it contains a complete graph of order six as a subgraph.Finally,K6cannot be a subgraph of an Eulerian triangulation of the projective plane(this follows from Proposition4).
There are some extensions of these results to more general surfaces.The strongest results in this area have been obtained by DeVos et al.[9].Other results stem from the corresponding results about even-faced maps.Note that Eulerian triangulations can be obtained from even-faced maps by placing a single vertex into each face of it and joining it to all the other vertices incident with that face.
An important parameter that quantitatively measures local planarity of an embedded graph G is the length of a shortest non-contractible cycle in G,which is called the edge-width of G.As an example,let us mention the following result of Hutchinson[14](cf.also[9]for a strengthening of this result).For every genus g,there exists an integer r(g)such that every graph embedded on an orientable surface of genus g with all faces even and with edge-width at least r(g)is3-colorable.The claim trivially holds in the plane,since such a graph must be bipartite.For the torus,the original bound of25from[14]on r(1)has been improved to9by Archdeacon et al.[3].A recent improvement[17]of r(1)to6 is best possible since the Cayley graph C(Z13;1,5)has chromatic number four and can be embedded on the torus with edge-width?ve[3].In fact,this Cayley graph is the only obstacle for a triangle-free even-faced map on the torus to be3-colorable.The same phenomena appear for other classes of graphs:every Eulerian triangulation of an orientable surface that has su?ciently large edge-width is4-colorable[14]and every graph embedded on a?xed surface(orientable or non-orientable)with su?ciently large edge-width is5-colorable[29].
Note that the assumption in the results of Hutchinson[14]that a surface is orientable is essential since every non-orientable surface admits quadrangula-tions of arbitrarily large edge-width,whose chromatic number is four[3,9,20]. However,such non-3-colorable quadrangulations can be characterized[20].
Similarly,Eulerian triangulations on orientable surfaces having su?ciently large edge-width are4-colorable as proved by Hutchinson,Richter,and Sey-mour[15].On the other hand,every non-orientable surface has non-4-colorable Eulerian triangulations of arbitrarily large edge-width[3,9],and they can also be characterized[22].
As colorings of Eulerian triangulations of the projective plane are well-under-stood(cf.Theorem1),we focus on Eulerian triangulations of the Klein bottle,the next simplest non-orientable surface.Our goal is to characterize non-5-colorable Eulerian triangulations of the Klein bottle.In particular,we prove that an Eu-
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lerian triangulation of the Klein bottle is5-colorable if and only if it does not contain the complete graph of order six as a subgraph.Let us mention at this point that graphs considered in this paper can have parallel edges as long as they do not form bigons.As a corollary we prove that every Eulerian triangulation of the Klein bottle with edge-width at least four is5-colorable.It seems to be more di?cult to establish analogous results for4-colorable Eulerian triangulations of the Klein bottle since there are5-chromatic Eulerian triangulations with arbi-trarily large edge-width;in particular,the list of“forbidden”subgraphs in this case cannot be?nite.
Our main result is the following:
Theorem3.An Eulerian triangulation G of the Klein bottle is5-colorable if it does not contain a complete graph of order six.If G contains K6,then it can be obtained from one of the maps T A,T B,T C,T D and T E,which are depicted in Figures1,11,14and16,by adding vertices and edges and its chromatic number is equal to6.
Theorem3is a combination of Theorems28and31,which are proved in Sections8and10,respectively.
2Outline of the proof
Before we start proving our main result,we would like to explain the major steps of the proof.First,we realize that it is enough to prove our results for proper triangulations as de?ned in Section3.In Section3,we also de?ne a special type of contraction that can be applied to vertices of degree four.This operation has the property that if it is applied to an Eulerian triangulation,the resulting triangulation is also Eulerian and if the resulting triangulation is5-colorable,so is the original one.This leads us to the need to analyze Eulerian triangulations without contractible vertices.
Before we proceed with further analysis,we realize that unless the Eulerian triangulation of the Klein bottle contains a vertex of degree two,is6-regular or of a very special type(all these cases are easy to handle),it contains a vertex of degree four adjacent to a vertex of degree six(Lemma16).In Section5,we then analyze Eulerian triangulation with a non-contractible vertex of degree four adjacent to a vertex of degree six and identify one particular con?guration that requires a?ner analysis—we call such a con?guration the resistant con?guration and the vertex of degree four involved in it a resistant vertex.Eulerian triangu-lations with the resistant con?guration are then analyzed in Section6.All these results are combined in Section7where we prove that every minimal(under our operation of the contraction)non-5-colorable Eulerian triangulation of the Klein bottle is isomorphic to one of the?ve triangulations that are denoted by T A,T B, T C,T D and T E and are introduced later.
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We observe that each of the?ve minimal triangulations contains the complete graph K6of order six as a subgraph.Moreover,there are only four non-isomorphic embeddings of K6in such triangulations.We call these embeddings bad.In Section8,we show that they are the only embeddings of K6that can be contained in an Eulerian triangulation of the Klein bottle.In Section9,we invert the operation of contraction and introduce expansions of vertices.We show that if an expansion of a vertex results in a triangulation with no complete graph of order six(and the original graph contained the complete graph of order six as a subgraph),then the obtained graph is5-colorable.Hence,every Eulerian triangulation of the Klein bottle that is not5-colorable must contain one of the four bad embeddings of K6.We formally combine the just explained results in Section10in which we state and prove our main result(Theorem31).
3Preliminaries
In this section,we introduce notation used throughout this paper.Graphs which we consider have no loops but they can contain multiple edges.If G is a graph and W a subset of its vertices,then G[W]is the subgraph of G induced by W.
A k-vertex is a vertex of degree k.Similarly,a k-cycle is a cycle of length k.If G is embedded on a surface and2-connected,then a k-face is a face bounded by a k-cycle.
Most of our proofs deal with graphs on surfaces.We refer to the mono-graph[21]for de?nitions related to graphs embedded on surfaces not mentioned here.If G is such a graph and C=v1...v?is a non-contractible?-cycle in G, we can cut the surface along C.This decreases the genus of the surface.In the obtained graph G′,the cycle C is either replaced by two copies C′and C′′of the original cycle C or by a2?-cycle C′.In the former case,the vertices of C′will always be denoted by v′1...v′?and the vertices of C′′by v′′1...v′′?.In the latter case,the vertices of C′are denoted by v′1...v′?v′′1...v′′?.The reader is referred for an example to Figure5where the Klein bottle was cut along a one-sided cycle vwu0.Note that the star in the?gure denotes a cross-cap.Let us now introduce some additional notation used in our?gures.The regions of a graph that are faces are always?lled with the white color and those that can contain additional vertices and edges of G with the gray color.
A triangulation of a surface is an embedding of a loopless graph such that each face is a3-face and an Eulerian triangulation of the surface is a triangulation in which all vertices have even degrees.A triangulation is proper if it has no contractible cycles of length two,every contractible3-cycle is facial,and every 4-cycle bounds a region that contains at most one vertex of G.
Besides Eulerian triangulations,we are also interested in near-triangulations and Eulerian near-triangulations.An embedding of a graph G in a surface is a near-triangulation if all its faces are3-faces with a possible exception of a single
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distinguished face.An Eulerian near-triangulation is a near-triangulation such that all the vertices not incident with the distinguished face have even degrees.
An embedding of a graph G is a defective Eulerian triangulation if it is a triangulation and all the vertices of G have even degrees except for a single pair of adjacent vertices.Similarly,a2-defective Eulerian triangulation is a triangulation with all the vertices of even degrees except for either two pairs of adjacent vertices or two non-adjacent vertices at distance two.
Proposition4.There is no defective Eulerian triangulation of the plane. Proof.Suppose that G is a defective Eulerian triangulation of the plane and let v and w be two adjacent vertices of odd degree.Remove the edge vw from G and consider the dual graph G?of G.Since the degrees of all the vertices of G\vw are even,G?is bipartite.On the other hand,all its vertices have degree three except for a single vertex of degree four(which corresponds to the face from which the edge vw was removed).Clearly,such a graph cannot exist.
On the other hand,defective Eulerian triangulations of the projective plane exist.In the next lemma,we show that defective and2-defective Eulerian trian-gulations of the projective plane are5-colorable.
Lemma5.Every defective or2-defective Eulerian triangulation of the projective plane is5-colorable.
Proof.Consider a defective or a2-defective triangulation G of the projective plane that is not5-colorable.By Dirac’s Theorem[1,8],G contains a complete graph of order six as a subgraph.Let H be such a subgraph of G.Note that the complete graph of order six has a unique embedding in the projective plane and this embedding is a triangulation.
If G is a defective triangulation,color the edge joining the vertices of odd degree red.If G is2-defective,color the two edges joining vertices of odd degree red or color a two-edge path between two such vertices red.Note that G contains at most two red edges and a vertex of G has odd degree if and only if it is incident with exactly one red edge.
We now observe that it can be assumed without loss of generality that G has no separating contractible3-cycle with all inner vertices of even degree.Assume that T is a separating3-cycle of G and let G T be the subgraph of G formed by the vertices and edges lying inside the closed disc bounded by T.By Proposition4, the degrees of all the vertices of G T are even.Hence,removing the interior of T yields a defective or a2-defective triangulation G′of the projective plane.Clearly, G is5-colorable if and only if G′is.
Since each facial cycle of H bounds a2-cell,each face of H either bounds a face in G or contains a vertex of odd degree(and thus a red edge)in its interior. On the other hand,every vertex v of H is incident either with a red edge(that can be an edge of G contained in H)or with a face containing a red edge in
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its interior.Otherwise,v would have odd degree and would be incident with no red edges.Since H contains six vertices and G contains at most two red edges,there are exactly two non-empty faces of H each containing a single red edge.Moreover,these faces of H are vertex-disjoint.However,there are no two disjoint facial triangles in the unique embedding of K6in the projective plane. This contradiction completes the proof.
Next,we establish a lemma on the existence of special5-colorings in plane graphs.This lemma straightforwardly follows from the Four Color Theorem,but we decided to provide a proof independent of it.
Lemma6.Let G be a plane graph.For every two non-adjacent vertices w1and w2of G,G has a5-coloring that assigns the vertices w1and w2the same color. Proof.Consider a counterexample G of the smallest size.Clearly,such a graph G is connected and simple.By Euler’s formula,G has at least three vertices of degree at most?ve.In particular,G has a vertex w of degree at most?ve that is neither w1and w2.By the choice of G,G?w has a5-coloring that assigns the vertices w1and w2the same color.If the neighbors of w in G are colored with at most four distinct colors,the coloring can be extended to G which is impossible. Hence,the degree of w is?ve and its?ve neighbors v1,...,v5are colored with mutually distinct colors.By symmetry,we can assume that v5is colored with the color of w1and w2.Let G ij be the subgraph of G induced by the vertices colored with the color of v i or v j,i,j∈{1,...,4}.Since G is plane,the vertices v1and v3are not contained in the same component of G13or the vertices v2and v4are not contained in the same component of G24.By symmetry,we can assume that the former is the case,and switch the two colors used in the component of G13 that contains v3.In this way,we obtain a5-coloring of G that assigns the vertices w1and w2the same color and such that the neighbors of w are colored with at most four distinct colors.Such a coloring can be extended to w.We conclude that G is not a counterexample to the statement of the lemma and thus there is no counterexample at all.
We next focus our attention to Eulerian near-triangulations and show that if the size of the distinguished face is small and G is embedded in the plane,then the parities of degrees of the vertices incident with the distinguished face are very restricted.
Proposition7.If G is an Eulerian near-triangulation of the plane with the distinguished face bounded by a two-cycle v1v2,then the degrees of both v1and v2 are odd.
Proof.By the hand-shaking lemma,the parities of the degrees of v1and v2are the same.If they are both even,we obtain by removing one of the two parallel edges v1v2a defective plane Eulerian triangulation which is impossible by Proposition4.
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Proposition8.If G is an Eulerian near-triangulation of the plane with the distinguished face bounded by a three-cycle v1v2v3,then the degrees of v1,v2and v3are even.
Proof.By the hand-shaking lemma,either the degrees of all the vertices v1,v2and v3are all even or exactly two of them have odd degree.In the latter case,G is a defective Eulerian triangulation of the plane which is impossible by Proposition4.
Arguments similar to those used in the proofs of Propositions7and8yield the proofs of the next two propositions.We have decided not to include their (short)proofs.
Proposition9.If G is an Eulerian near-triangulation of the plane with the distinguished face bounded by a four-cycle v1v2v3v4,then one of the following holds:
?the degrees of all the vertices v1,v2,v3and v4are odd,
?the degrees of v1and v3are even,and those of v2and v4are odd,or
?the degrees of v1and v3are odd,and those of v2and v4are even.
Proposition10.If G is an Eulerian near-triangulation of the plane with the distinguished face bounded by a5-cycle v1v2v3v4v5,then there exists an index i such that the degrees of the vertices v i and v i+1are odd(the indices are modulo ?ve)and the degrees of the vertices v j,j=i,i+1,are even.
Since we are interested in colorings of Eulerian triangulations,we will also need some tools that allow us to extend precolorings of some of the vertices of a graph to the entire graph.
Proposition11.Let G be a plane Eulerian near-triangulation with the distin-guished face bounded by a k-cycle v1...v k,k≤5,and let W={v1,...,v k}. Every precoloring c of the vertices of W that is a proper coloring of G[W]can be extended to a proper5-coloring of G.
Proof.If k≤3,then all the vertices v1...v k have mutually distinct colors. Clearly,G is5-colorable and the colors of such a5-coloring of G can be permuted to match the colors of v1...v k.Suppose now that k=4.By Proposition9, G can be completed to an Eulerian triangulation either by adding an edge or a vertex of degree four.Hence,G is3-colorable.
Fix a3-coloring of G.If the vertices v1,...,v4should be colored with four mutually distinct colors,recolor v3and v4with the fourth and the?fth color.If two of the vertices,say v1and v3should have the same color,recolor v1and v3 with the fourth color and v4with the?fth color.If,in addition,the colors of v2
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and v4should be the same,also recolor v2with the?fth color.In this way we obtain a coloring that matches the precoloring of v1,...,v4(up to a permutation of the colors).
It remains to consider the case k=5.By Proposition10,we can assume that the degrees of v3and v4are odd and the degrees of the remaining vertices are even.Consider the plane graph G′obtained by adding edges v1v3and v1v4.G′is an Eulerian triangulation of the plane and it is thus3-colorable.Since all the vertices v2,...,v5must receive colors di?erent from the color of v1,the colors of v2and v4and the colors of v3and v5are the same.We now recolor some of the vertices v1,...,v5to match the colors in the precoloring.
Let A i,i=1,...,5,be the set of the vertices v1,...,v5precolored with the i-th color.Suppose?rst that the vertices v1,...,v5are precolored with three distinct colors.By symmetry,we can assume that|A1|=1and|A2|=|A3|=2. We proceed as follows:the vertex of A1keeps its color and the vertices of A2and A3are recolored with the fourth and the?fth color,respectively.
If the vertices v1,...,v5are precolored with four distinct colors,we may as-sume that|A1|=|A2|=|A3|=1,|A4|=2and the vertices contained in A1and A2are adjacent.In this case,the vertices of A1∪A2keep their colors,the vertex of A3is recolored with the fourth color and the vertices of A4with the?fth color.
Finally,if the vertices v1,...,v5are precolored with?ve distinct colors,the vertices v1,...,v3keep their colors and the vertices v4and v5are recolored with the fourth and the?fth colors.
Similarly,some special precolorings of plane Eulerian near-triangulations with the distinguished face of length six or seven can also be extended as we show in the next three propositions.Note that the vertex v6is not assumed to be precolored in the next proposition and its color is determined by the extended coloring. Proposition12.Let G be a plane Eulerian near-triangulation with the dis-tinguished face bounded by a6-cycle v1...v6and let W={v1,...,v5}.Every precoloring c of the vertices of W with?ve colors that is a proper coloring of G[W]such that c(v1)=c(v4)can be extended to a5-coloring of the entire graph G.
Proof.Let G′be an isomorphic copy of G,v′1...v′6the vertices bounding the dis-tinguished face of G′,and H a plane graph obtained from G and G′by identifying the vertices v i and v′i for i=1,...,6.Clearly,H is an Eulerian triangulation of the plane and thus it is3-colorable.Consider the3-coloring of G that is the restriction of a3-coloring of H.Recolor the vertices v1and v4with an unused (fourth)color and the vertex v5with another unused(?fth)color.If c(v i)=c(v5) for i∈{2,3},also recolor such a vertex v i with the?fth color.Note that the colors c(v2)and c(v3)are distinct from c(v1)=c(v4).The coloring of v1,...,v5 now matches the precoloring of G[W](up to a permutation of the colors).
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Proposition13.Let G be a plane Eulerian near-triangulation with the distin-guished face bounded by a6-cycle v1...v6and let W={v1,...,v5}.Every pre-coloring c of the vertices of W with?ve colors that is a proper coloring of G[W] such that c(v1)=c(v3)can be extended to a5-coloring of the entire graph G. Proof.Let G′and H be the graphs as in the proof of Proposition12and consider the3-coloring of G obtained by restricting a3-coloring of H.Recolor the vertices v1and v3with an unused(fourth)color and the vertex v2with another unused (?fth)color.If c(v i)=c(v1)for i∈{4,5},also recolor such a vertex v i with the fourth color,or if c(v i)=c(v2),recolor v i with the?fth color.The coloring of v1,...,v5now matches the precoloring.
Proposition14.Let G be a plane Eulerian near-triangulation with the distin-guished face bounded by a7-cycle v1...v7and let W={v1,...,v6}.Every pre-coloring c of the vertices of W with?ve colors that is a proper coloring of G[W] such that c(v1)=c(v5)and c(v2)=c(v6)can be extended to a5-coloring of the entire graph G.
Proof.As in the previous two proofs,we consider the3-coloring of G obtained by restricting a3-coloring of two copies of G pasted along the boundary of the distinguished face.Recolor now the vertices v1and v5with an unused(fourth) color and the vertices v2and v6with another unused(?fth)color.If c(v3)=c(v1), also recolor v3with the fourth color,and if c(v4)=c(v2),recolor v4with the?fth color.The coloring of the vertices v1,...,v6now matches the precoloring of G[W].
We now introduce a special type of contractions which will be used later in the paper.If G is a triangulation with a vertex v of degree four that is adjacent to vertices v1,v2,v3and v4(in this cyclic order around v),a triangulation obtained by a contraction of v1vv3is the triangulation G.v1vv3constructed from G in the following way:the edges vv1and vv3are contracted to a new vertex w,the edges v2v1,v2v and v2v3are identi?ed to a single edge v2w and the edges v4v1,v4v and v4v3to a single edge v4w.If v1=v3and the vertices v1and v3are not adjacent,G.v1vv3is again a triangulation.Moreover,if the original triangulation is Eulerian,then the obtained triangulation is also Eulerian:the degrees of the vertices v2and v4decrease by two and the degree of the vertex w is equal to the sum of the degrees of the vertices v1and v3decreased by four.Similarly,if G is a defective Eulerian triangulation,then G.v1vv3is also a defective Eulerian triangulation,and if G is2-defective,then G.v1vv3is either Eulerian,defective or2-defective Eulerian triangulation.
A vertex v of degree four such that G.v1vv3or G.v2vv4is a triangulation is said to be contractible;otherwise,v is called non-contractible.Note that if v is a non-contractible vertex,then v1=v3or the vertices v1and v3are adjacent,and similarly,v2=v4or the vertices v2and v4are adjacent.In the case of the Klein
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bottle,we call a non-contractible vertex v resistant if the degrees of v1,v2,v3 and v4are at least six,the cycle vv1v3is a two-sided non-separating3-cycle and vv2v4is a one-sided non-separating3-cycle.
At the end of this section,let us recall Theorem3.1from[6].The theorem is stated in[6]for simple plane graphs G but the proof readily translates to graphs with multiple edges as long as there are no2-faces.
Theorem15([6]).Let G be a plane near-triangulation with the distinguished face bounded by a k-cycle.If all the vertices that are not incident with the dis-tinguished face have degree at least six,then G contains at most k2/12+k/2+1 vertices.
In particular,if k=6,then G contains at most one vertex that is not incident with the distinguished face and if k=8,then G contains at most two such vertices. 4Degree structure of Eulerian triangulations of the Klein bottle
In this section,we?rst describe Eulerian triangulations of the Klein bottle with minimum degree at least four that do not contain a4-vertex adjacent to a6-vertex.Such triangulations are either6-regular or of a very special type: Lemma16.Let G be an Eulerian triangulation of the Klein bottle with minimum degree at least four.Then,either G is6-regular,or G contains a vertex of degree four adjacent to a vertex of degree at most six,or G contains only vertices of degree four and eight,no two vertices of degree four are adjacent and each vertex of degree eight is adjacent to four vertices of degree four.
Proof.Assume that neither of the?rst two cases apply.Let Q be the set of vertices of degree four of G and O the set of vertices of degree at least eight and set q=|Q|and o=|O|.Note that all neighbors of each vertex of Q are in the set O by our assumption.Color now the edges incident with the vertices of degree four by blue.Next,the edges that are contained in a3-face incident with a vertex of degree four and that are not blue are colored red.Let b=4q be the number of blue edges and r the number of red edges.
Each3-face incident with a vertex of degree contains one red and two blue edges.Since each vertex of degree four is contained in four faces,there are b 3-faces incident with vertices of degree four.On the other hand,the number of such3-faces is at most2r(each red edge is contained in at most such faces).We conclude b≤2r.By Euler’s formula,the average degree of G is six.Therefore, o≤q and the equality holds if only if all the vertices in O have degree exactly eight.
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Figure1:The Eulerian triangulation T A of the Klein bottle.
The number of incidences of the vertices in O with red edges and blue edges is b+2r≥2b=8q.Hence,the average degree of G is at least
4q+8q+6s
=6(1)
2q+s
where s is the number of vertices of degree six of G.Since the average degree of G is six,all the inequalities b+2r≥2b and(1)are equalities.It follows that o=q and b+2r=2b=8q.Therefore,all the vertices of O have degree exactly eight,all the edges incident with vertices of O are red or blue and b=2r.Since two consecutive edges incident with a vertex of O cannot be blue,each vertex of O is incident with four blue and four red edges.In particular,it is not adjacent to a vertex of degree six.Since G is connected,we conclude that G contains no vertices of degree six.The statement of the lemma follows.
At the end of this section,we brie?y deal with one of the cases described in Lemma16.The structure of6-regular triangulations of the Klein bottle is well-understood[23,28]and in particular,the chromatic number of every6-regular triangulation of the Klein bottle has been determined by Sasanuma[27].The only6-chromatic6-regular triangulation of the Klein bottle is the one obtained from the complete graph of order six by adding a perfect matching(it is depicted in Figure1).We henceforth refer to this triangulation as the triangulation T A. Lemma17.A6-regular triangulation G of the Klein bottle is5-colorable unless G is the triangulation T A depicted in Figure1.
5Eulerian triangulations with non-contractible 4-vertices
In this section,we analyze Eulerian triangulations of the Klein bottle that contain a4-vertex that is neither contractible nor resistant.Those with resistant vertices are analyzed in the next section.We start with the simplest case that a4-vertex is non-contractible since it is contained in a short separating cycle.
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Figure2:The projective plane graph obtained by cutting along a one-sided2-cycle vw in the proof of Lemma19.
Lemma18.If G is an Eulerian triangulation of the Klein bottle that contains a surface-separating2-cycle or3-cycle,then G is5-colorable.
Proof.Cut the surface along the non-contractible2-or3-cycle.In case that the length of the cycle is two,remove one of the two parallel edges bounding a new2-face.In this way,we obtain two Eulerian triangulations of the projective plane or two defective Eulerian triangulations of the projective plane which are 5-colorable by Theorem2or by Lemma5.Their colorings can be easily combined to obtain a coloring of G with?ve colors.
We now deal with Eulerian triangulations with a4-vertex contained in a cycle of length two.
Lemma19.If G is a proper Eulerian triangulation of the Klein bottle that contains a surface-non-separating2-cycle vw such that the degree of v is four, then G is5-colorable.
Proof.Let u1and u2be the two neighbors of v di?erent from w.The2-cycle vw forms either a one-sided or a double-sided non-separating curve in the Klein bottle.We cut the surface along it.
If the2-cycle is one-sided,we obtain the projective plane graph G′depicted in Figure2.Remove the vertices v′and v′′and the edges w′u1and w′u2and identify the vertices w′and w′′.In this way,we obtain an Eulerian triangulation of the projective plane which is5-colorable by Theorem2.Assigning the same colors to the vertices of G?v yields a proper5-coloring of G that can be extended to the vertex v(since its degree is four).
If the2-cycle is double-sided,we obtain the plane graph G′as depicted in Figure3.By Lemma6,G′?{v′,v′′}has a5-coloring that assigns the vertices w′and w′′the same color.Assign now the vertices of G the colors of their counterparts in G′and extend the coloring to v.In this way,we obtain a5-coloring of G.
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Figure3:The plane graph obtained by cutting along a two-sided2-cycle vw in
19.
the proof of Lemma
Next,we analyze the case that one neighbor of a non-contractible4-vertex is also a4-vertex.
Lemma20.Let G be a proper Eulerian triangulation of the Klein bottle with no contractible vertices.If G contains two adjacent vertices of degree four,then G is5-colorable.
Proof.By Lemma18,we may assume that G does not contain separating2-or3-cycles.Let v and w be two adjacent vertices of degree four.If v or w is contained in a non-contractible2-cycle,then G is5-colorable by Lemma19.Otherwise, each of v and w is contained in two non-contractible3-cycles.Let u0,u1and u2be their neighbors as depicted in Figure4.Note that the homotopies of the cycles vu1u2and wu1u2are the same.
If both the cycles vu1u2and vwu0are one-sided,then their homotopies are the same.Cutting the surface along the cycle vwu0yields a projective plane graph G′.Since the cycles vu1u2and vwu0are homotopic,one of the areas bounded by u′0u1u2and u′′0u1u2is a2-cell while the other one contains a cross-cap(hence,we have obtained the left or the middle con?guration depicted in Figure5).Color the vertices u′0,u1and u2with three mutually di?erent colors and color the vertex u′′0with the same color as u′0.The coloring can be extended to both the graphs
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cycle vwu0in the proof of Lemma20.The left or the middle graph is obtained if the cycle vu1
Figure6:The plane graph G′obtained by cutting along the two-sided cycle vwu0 in the proof of Lemma20.
contained in the areas bounded by u′0u1u2and u′′0u1u2:these graphs are either Eulerian triangulations or defective Eulerian triangulations of the plane or the projective plane and thus they are5-colorable by Five Color Theorem,Theorem2 or Lemma5.The constructed5-coloring of G′readily yields a5-coloring of G.
If the cycles vu1u2and wu1u2are double-sided and the cycle vwu0is one-sided,we also cut the surface along the cycle vwu0.This yields the projective plane graph G′depicted in the right part in Figure5.Remove the vertices v′,v′′, w′and w′′and the edges u′0u1and u′0u2and identify the vertices u′0and u′′0to a vertex u′′′0.The obtained projective plane triangulation G′′contains two vertices of odd degree:u1and u2.Note that u1and u2are adjacent and thus G′′is a defective Eulerian triangulation which is5-colorable by Lemma5.The5-coloring of G′′can be extended to the vertices v and w to a5-coloring of G.
It remains to consider the case that the cycles vu1u2and wu1u2are one-sided and the cycle vwu0is double-sided.Let G′be the plane graph obtained by cutting along the cycle vwu0(see Figure6).By Lemma6,G′has a5-coloring that assigns the vertices u′0and u′′0the same color.This5-coloring can be easily extended to v and w yielding a5-coloring of G.
It remains to analyze one more case when a4-vertex is neither contractible nor resistant.We do so in the next lemma.
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Figure7:The two possible con?gurations contained in the graph G in the proof of Lemma21.
Lemma21.Let G be a proper Eulerian triangulation of the Klein bottle with no contractible vertices.If G contains a vertex v of degree four that is not resistant and that is adjacent to a vertex of degree six,then G is5-colorable.
Proof.By Lemmas18and19,we can assume that v is not contained in a2-cycle. In particular,the vertices u1,u2,u3and u4are mutually distinct.By symmetry, we assume that u1is the neighbor of v of degree six.By the assumption of the lemma,both the cycles vu1u3and vu2u4are one-sided.In particular,one of the neighbors of u1is the vertex u3,and G contains one of the two con?gurations depicted in Figure7.Hence,cutting along the cycle vu1u3yields one of the four projective plane graphs depicted in Figure8.Note that one of the two shaded areas in the?gure is a2-cell and one contains a cross-cap since the homotopies of the cycles vu1u3and vu2u4are the same.
We now construct a5-coloring of G for each of the four cases depicted in Figure8.In the?rst case,the projective plane graph bounded by u2u′3u4is either an Eulerian triangulation or a defective Eulerian triangulation.Hence,it is5-colorable by Theorem2or Lemma5.Next,color the vertex u′′3by the same color as u′3and the vertices u′1and u′′1with the same color di?erent from the colors of u2,u′3and u4,and extend the coloring to the graph inside the4-cycle u′′1u2u4u′′3by Proposition11.In this way,we eventually obtain a5-coloring of G by assigning a color to the vertex v.
In the second case,let G′be the projective plane graph bounded by the cycle u2w1w2u′′3u4.If the vertices w1and u′′3are not adjacent,remove the edge w2u′′3and identify the vertices w1and u′′3.In this way,we obtain an Eulerian triangulation or a defective Eulerian triangulation of the projective plane which is5-colorable by Theorem2or Lemma5.Next,color the vertices u′1and u′′1with a color di?erent from the colors of u2,u′3=u′′3and u4and extend the obtained coloring to v.In this way,we obtain a5-coloring of G.We proceed analogously if the vertices w2and u2are not adjacent.
Assume now that both the vertices w1and u′′3and the vertices w2and u2are adjacent(see Figure9).If the vertices u4and w1are not adjacent,precolor them
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along the cycle vu 1u 3in the proof of Lemma 21.
u ′′1u 2u ′′3u 4w 1
w 2
u ′′1
u 2u ′′3
u 4w 1w 2Figure 9:The con?gurations that appear in G ′in the proof of Lemma 21in the analysis of the second con?guration depicted in Figure 9.
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Figure10:The resistant con?guration.Note that v1=v5,v2=v6,v3=v8,and v4=v7.
with the same color and color the vertices u2,w2and u′′3with mutually distinct colors di?erent from the color of u4and v1.Extend this coloring to the graphs inside the5-cycle u2u4u′′3w2w1and the4-cycle w1u′′3w2u2by Proposition11.Next, color the vertices u′1and u′′1with the same color and obtain a5-coloring of G by assigning a color to the vertex v.Again,we proceed similarly if the vertices w2 and u4are not adjacent.If G′contains both the edges w1u4and w2u4,then the degree of u4is seven—note that all the faces incident with u4in the graph depicted in Figure9are3-faces and G is a proper triangulation.However,this case is excluded by our assumption that G is an Eulerian triangulation.
The third and the fourth con?gurations depicted in Figure8are symmetric. Hence,we just analyze the third one.Let G′be the projective plane graph bounded by the4-cycle u2u′3wu4.Add an edge between the vertices u′3and u4.In this way,we obtain an Eulerian triangulation,a defective Eulerian triangulation or a2-defective Eulerian triangulation of the projective plane.In all the cases, G′with the added edge is5-colorable by Theorem2or Lemma5.Next,color the vertices u′1and u′′1with a color di?erent from the colors of u′3,u2and u4and extend the coloring inside the cycle u2u′′1u′′3u4by Proposition11.A5-coloring of G is then obtained by assigning a color to the vertex v.
6Eulerian triangulations with resistant vertices In this section,we analyze Eulerian triangulations of the Klein bottle that contain a resistant vertex.If G is such a triangulation,then it is of the form depicted in Figure10in which the pairs of vertices v1and v5,v2and v6,v3and v8,and v4and v7are the same.The resistant vertex is the vertex v.Throughout this section, we refer to the con?guration depicted in Figure10as to a resistant con?guration and all the arithmetic with indices of the vertices v i,i=1,...,8,throughout this section,is modulo eight.When referring to adjacencies between a vertex v i
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v v
v v
v 1=v 5
v 2=v 6
v 3=v 8v 4=v 7v 5v 6v 7v 8Figure 11:The Eulerian triangulation T B of the Klein bottle.
and a vertex w di?erent from all v i ,the vertices v i are treated as eight distinct vertices,i.e.,if the con?guration contains an edge wv 4but not wv 7,we say that w is adjacent to v 4and not adjacent to v 7.
Lemma 22.Let G be a proper Eulerian triangulation of the Klein bottle with a resistant con?guration.If G does not contain a vertex of degree four di?erent from v ,then G is either 5-colorable or it is the triangulation T B depicted in Figure 11.Proof.By Theorem 15,G contains at most two vertices di?erent from v and di?erent from all v i ,i =1,...,8.If G contains at most one such vertex,then a 5-coloring of G can be obtained as follows:?rst,color the neighbors of v by four mutually distinct colors and then extend the coloring to the remaining (non-adjacent)vertices of G .The same argument works if G contains two such non-adjacent vertices.Hence,we can assume that G contains adjacent vertices w 1and w 2that are di?erent from v and v i ,i =1, (8)
Since the degree of both w 1and w 2is at least six,each of them must be adjacent to ?ve consecutive vertices v i ,...,v i +4.There are four such triangula-tions G (see Figure 12):two of them are isomorphic to triangulation T B and the remaining ones are not Eulerian.
Next,we observe that each vertex of degree four di?erent from v is adjacent to at least two of the vertices v 1,...,v 8.
Lemma 23.Let G be a proper Eulerian triangulation of the Klein bottle with the resistant con?guration.If G contains a non-contractible vertex w =v ,then w is adjacent to at least two of the vertices v 1,...,v 8and such two vertices are consecutive in the cyclic order around w .
Proof.By Lemma 20,v and w are not adjacent.Let u 1,...,u 4be the neighbors of w in the cyclic order around w .Since w is resistant,all the four neighbors of w are mutually distinct and G contains non-contractible 3-cycles wu 1u 3and wu 2u 4.
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v v v
v
v 1=v 5v 2=v 6v 3=v 8v 4=v 7v 5v 6v 7v 8v v v v v 1=v 5
v 2=v 6
v 3=v 8v 4=v 7v 5v 6v 7v 8v v v v
v 1=v 5v 2=v 6v 3=v 8v 4=v 7v 5v 6v 7v 8v v
v v v 1=v 5
v 2=v 6
v 3=v 8v 4=v 7v 5v 6v 7v 8Figure 12:The four triangulations of the Klein bottle obtained in the proof of Lemma 22.
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如何写先进个人事迹 篇一：如何写先进事迹材料 如何写先进事迹材料 一般有两种情况：一是先进个人，如先进工作者、优秀党员、劳动模范等；一是先进集体或先进单位，如先进党支部、先进车间或科室，抗洪抢险先进集体等。无论是先进个人还是先进集体，他们的先进事迹，内容各不相同，因此要整理材料，不可能固定一个模式。一般来说，可大体从以下方面进行整理。 (1)要拟定恰当的标题。先进事迹材料的标题，有两部分内容必不可少，一是要写明先进个人姓名和先进集体的名称，使人一眼便看出是哪个人或哪个集体、哪个单位的先进事迹。二是要概括标明先进事迹的主要内容或材料的用途。例如《王鬃同志端正党风的先进事迹》、《关于评选张鬃同志为全国新长征突击手的材料》、《关于评选鬃处党支部为省直机关先进党支部的材料》等。 (2)正文。正文的开头，要写明先进个人的简要情况，包括：姓名、性别、年龄、工作单位、职务、是否党团员等。此外，还要写明有关单位准备授予他(她)什么荣誉称号，或给予哪种形式的奖励。对先进集体、先进单位，要根据其先进事迹的主要内容，寥寥数语即应写明，不须用更多的文字。 然后，要写先进人物或先进集体的主要事迹。这部分内容是全篇材料
的主体，要下功夫写好，关键是要写得既具体，又不繁琐；既概括，又不抽象；既生动形象，又很实在。总之，就是要写得很有说服力，让人一看便可得出够得上先进的结论。比如，写一位端正党风先进人物的事迹材料，就应当着重写这位同志在发扬党的优良传统和作风方面都有哪些突出的先进事迹，在同不正之风作斗争中有哪些突出的表现。又如，写一位搞改革的先进人物的事迹材料，就应当着力写这位同志是从哪些方面进行改革的，已经取得了哪些突出的成果，特别是改革前后的．经济效益或社会效益都有了哪些明显的变化。在写这些先进事迹时，无论是先进个人还是先进集体的，都应选取那些具有代表性的具体事实来说明。必要时还可运用一些数字，以增强先进事迹材料的说服力。 为了使先进事迹的内容眉目清晰、更加条理化，在文字表述上还可分成若干自然段来写，特别是对那些涉及较多方面的先进事迹材料，采取这种写法尤为必要。如果将各方面内容材料都混在一起，是不易写明的。在分段写时，最好在每段之前根据内容标出小标题，或以明确的观点加以概括，使标题或观点与内容浑然一体。 最后，是先进事迹材料的署名。一般说，整理先进个人和先进集体的材料，都是以本级组织或上级组织的名义；是代表组织意见的。因此，材料整理完后，应经有关领导同志审定，以相应一级组织正式署名上报。这类材料不宜以个人名义署名。 写作典型经验材料-般包括以下几部分： (1)标题。有多种写法，通常是把典型经验高度集中地概括出来，一
How to manage time Time treats everyone fairly that we all have 24 hours per day. Some of us are capable to make good use of time while some find it hard to do so. Knowing how to manage them is essential in our life. Take myself as an example. When I was still a senior high student, I was fully occupied with my studies. Therefore, I hardly had spare time to have fun or develop my hobbies. But things were changed after I entered university. I got more free time than ever before. But ironically, I found it difficult to adjust this kind of brand-new school life and there was no such thing called time management on my mind. It was not until the second year that I realized I had wasted my whole year doing nothing. I could have taken up a Spanish course. I could have read ten books about the stories of successful people. I could have applied for a part-time job to earn some working experiences. B ut I didn’t spend my time on any of them. I felt guilty whenever I looked back to the moments that I just sat around doing nothing. It’s said that better late than never. At least I had the consciousness that I should stop wasting my time. Making up my mind is the first step for me to learn to manage my time. Next, I wrote a timetable, setting some targets that I had to finish each day. For instance, on Monday, I must read two pieces of news and review all the lessons that I have learnt on that day. By the way, the daily plan that I made was flexible. If there’s something unexpected that I had to finish first, I would reduce the time for resting or delay my target to the next day. Also, I would try to achieve those targets ahead of time that I planed so that I could reserve some more time to relax or do something out of my plan. At the beginning, it’s kind of difficult to s tick to the plan. But as time went by, having a plan for time in advance became a part of my life. At the same time, I gradually became a well-organized person. Now I’ve grasped the time management skill and I’m able to use my time efficiently.
英语演讲稿：未来的工作 这篇《英语演讲稿范文：未来的工作》，是特地，希望对大家有所帮助！ 热门演讲推荐：竞聘演讲稿 | 国旗下演讲稿 | 英语演讲稿 | 师德师风演讲稿 | 年会主持词 | 领导致辞 everybody good afternoon:. first of all thank the teacher gave me a story in my own future ideal job. everyone has a dream job. my dream is to bee a boss, own a pany. in order to achieve my dreams, i need to find a good job, to accumulate some experience and wealth, it is the necessary things of course, in the school good achievement and rich knowledge is also very important. good achievement and rich experience can let me work to make the right choice, have more opportunities and achievements. at the same time, munication is very important, because it determines whether my pany has a good future development. so i need to exercise their municative ability. i need to use all of the free time to learn
小学生个人读书事迹简介怎么写800字 书,是人类进步的阶梯,苏联作家高尔基的一句话道出了书的重要。书可谓是众多名人的“宠儿”。历来,名人说出关于书的名言数不胜数。今天小编在这给大家整理了小学生个人读书事迹,接下来随着小编一起来看看吧! 小学生个人读书事迹1 “万般皆下品,惟有读书高”、“书中自有颜如玉,书中自有黄金屋”,古往今来,读书的好处为人们所重视,有人“学而优则仕”,有人“满腹经纶”走上“传道授业解惑也”的道路……但是,从长远的角度看,笔者认为读书的好处在于增加了我们做事的成功率,改善了生活的质量。 三国时期的大将吕蒙,行伍出身,不重视文化的学习,行文时,常常要他人捉刀。经过主君孙权的劝导,吕蒙懂得了读书的重要性,从此手不释卷,成为了一代儒将,连东吴的智囊鲁肃都对他“刮目相待”。后来的事实证明,荆州之战的胜利,擒获“武圣”关羽,离不开吕蒙的“运筹帷幄,决胜千里”,而他的韬略离不开平时的读书。由此可见,一个人行事的成功率高低,与他的对读书,对知识的重视程度是密切相关的。 的物理学家牛顿曾近说过,“如果我比别人看得更远,那是因为我站在巨人的肩上”,鲜花和掌声面前,一代伟人没有迷失方向,自始至终对读书保持着热枕。牛顿的话语告诉我们,渊博的知识能让我们站在更高、更理性的角度来看问题,从而少犯错误,少走弯路。
读书的好处是显而易见的,但是,在社会发展日新月异的今天,依然不乏对读书,对知识缺乏认知的人,《今日说法》中我们反复看到农民工没有和用人单位签订劳动合同,最终讨薪无果;屠户不知道往牛肉里掺“巴西疯牛肉”是犯法的;某父母坚持“棍棒底下出孝子”,结果伤害了孩子的身心,也将自己送进了班房……对书本,对知识的零解读让他们付出了惨痛的代价,当他们奔波在讨薪的路上,当他们面对高墙电网时,幸福,从何谈起?高质量的生活,从何谈起? 读书,让我们体会到“锄禾日当午,汗滴禾下土”的艰辛;读书,让我们感知到“四海无闲田,农夫犹饿死”的无奈;读书,让我们感悟到“为报倾城随太守,西北望射天狼”的豪情壮志。 读书的好处在于提高了生活的质量,它填补了我们人生中的空白,让我们不至于在大好的年华里无所事事,从书本中,我们学会提炼出有用的信息,汲取成长所需的营养。所以,我们要认真读书,充分认识到读书对改善生活的重要意义,只有这样,才是一种负责任的生活态度。 小学生个人读书事迹2 所谓读一本好书就是交一个良师益友,但我认为读一本好书就是一次大冒险,大探究。一次体会书的过程,真的很有意思,咯咯的笑声,总是从书香里散发;沉思的目光也总是从书本里透露。是书给了我启示,是书填补了我无聊的夜空,也是书带我遨游整个古今中外。所以人活着就不能没有书,只要爱书你就是一个爱生活的人,只要爱书你就是一个大写的人,只要爱书你就是一个懂得珍惜与否的人。可真所谓
关于坚持的英语演讲稿 Results are not important, but they can persist for many years as a commemoration of. Many years ago, as a result of habits and overeating formed one of obesity, as well as indicators of overall physical disorders, so that affects my work and life. In friends to encourage and supervise, the participated in the team Now considered to have been more than three years, neither the fine rain, regardless of winter heat, a day out with 5:00 time. The beginning, have been discouraged, suffering, and disappointment, but in the end of the urging of friends, to re-get up, stand on the playground. 成绩并不重要，但可以作为坚持多年晨跑的一个纪念。多年前，由于庸懒习惯和暴饮暴食，形成了一身的肥胖，以及体检指标的全盘失常，以致于影响到了我的工作和生活。在好友的鼓励和督促下，参加了晨跑队伍。现在算来，已经三年多了，无论天晴下雨，不管寒冬酷暑，每天五点准时起来出门晨跑。开始时，也曾气馁过、痛苦过、失望过，但最后都在好友们的催促下，重新爬起来，站到了操场上。 In fact, I did not build big, nor strong muscles, not a sport-born people. Over the past few years to adhere to it, because I have a team behind, the strength of a strongteam here, very grateful to our team, for a long time, we encourage each other, and with sweat, enjoying common health happy. For example, Friends of the several run in order to maintain order and unable to attend the 10，000 meters race, and they are always concerned about the brothers and promptly inform the place and time, gives us confidence and courage. At the same time, also came on their own inner desire and pursuit for a good health, who wrote many of their own log in order to refuel for their own, and inspiring. 其实我没有高大身材，也没健壮肌肉，天生不属于运动型的人。几年来能够坚持下来，因为我的背后有一个团队，有着强大团队的力量，在这里，非常感谢我们的晨跑队，长期以来，我们相互鼓励着，一起流汗，共同享受着健康带来的快
1.How to build a business that lasts100years 0:11Imagine that you are a product designer.And you've designed a product,a new type of product,called the human immune system.You're pitching this product to a skeptical,strictly no-nonsense manager.Let's call him Bob.I think we all know at least one Bob,right?How would that go? 0:34Bob,I've got this incredible idea for a completely new type of personal health product.It's called the human immune system.I can see from your face that you're having some problems with this.Don't worry.I know it's very complicated.I don't want to take you through the gory details,I just want to tell you about some of the amazing features of this product.First of all,it cleverly uses redundancy by having millions of copies of each component--leukocytes,white blood cells--before they're actually needed,to create a massive buffer against the unexpected.And it cleverly leverages diversity by having not just leukocytes but B cells,T cells,natural killer cells,antibodies.The components don't really matter.The point is that together,this diversity of different approaches can cope with more or less anything that evolution has been able to throw up.And the design is completely modular.You have the surface barrier of the human skin,you have the very rapidly reacting innate immune system and then you have the highly targeted adaptive immune system.The point is,that if one system fails,another can take over,creating a virtually foolproof system. 1:54I can see I'm losing you,Bob,but stay with me,because here is the really killer feature.The product is completely adaptive.It's able to actually develop targeted antibodies to threats that it's never even met before.It actually also does this with incredible prudence,detecting and reacting to every tiny threat,and furthermore, remembering every previous threat,in case they are ever encountered again.What I'm pitching you today is actually not a stand-alone product.The product is embedded in the larger system of the human body,and it works in complete harmony with that system,to create this unprecedented level of biological protection.So Bob,just tell me honestly,what do you think of my product? 2:47And Bob may say something like,I sincerely appreciate the effort and passion that have gone into your presentation,blah blah blah-- 2:56(Laughter) 2:58But honestly,it's total nonsense.You seem to be saying that the key selling points of your product are that it is inefficient and complex.Didn't they teach you 80-20?And furthermore,you're saying that this product is siloed.It overreacts, makes things up as it goes along and is actually designed for somebody else's benefit. I'm sorry to break it to you,but I don't think this one is a winner.
关于工作的优秀英语演讲稿 Different people have various ambitions. Some want to be engineers or doctors in the future. Some want to be scientists or businessmen. Still some wish to be teachers or lawers when they grow up in the days to come. Unlike other people, I prefer to be a farmer. However, it is not easy to be a farmer for Iwill be looked upon by others. Anyway,what I am trying to do is to make great contributions to agriculture. It is well known that farming is the basic of the country. Above all, farming is not only a challenge but also a good opportunity for the young. We can also make a big profit by growing vegetables and food in a scientific way. Besides we can apply what we have learned in school to farming. Thus our countryside will become more and more properous. I believe that any man with knowledge can do whatever they can so long as this job can meet his or her interest. All the working position can provide him with a good chance to become a talent. 1 ————来源网络整理，仅供供参考
个人先进事迹简介 01 在思想政治方面,xxxx同学积极向上,热爱祖国、热爱中国共产党,拥护中国共产党的领导.利用课余时间和党课机会认真学习政治理论,积极向党组织靠拢. 在学习上,xxxx同学认为只有把学习成绩确实提高才能为将来的实践打下扎实的基础,成为社会有用人才.学习努力、成绩优良. 在生活中,善于与人沟通,乐观向上,乐于助人.有健全的人格意识和良好的心理素质和从容、坦诚、乐观、快乐的生活态度,乐于帮助身边的同学,受到师生的好评. 02 xxx同学认真学习政治理论,积极上进,在校期间获得原院级三好生,和校级三好生,优秀团员称号,并获得三等奖学金. 在学习上遇到不理解的地方也常常向老师请教,还勇于向老师提出质疑.在完成自己学业的同时,能主动帮助其他同学解决学习上的难题,和其他同学共同探讨,共同进步. 在社会实践方面,xxxx同学参与了中国儿童文学精品“悦”读书系,插画绘制工作,xxxx同学在班中担任宣传委员,工作积极主动,认真负责,有较强的组织能力.能够在老师、班主任的指导下独立完成学院、班级布置的各项工作. 03 xxx同学在政治思想方面积极进取,严格要求自己.在学习方面刻苦努力,不断钻研,学习成绩优异,连续两年荣获国家励志奖学金；作
为一名学生干部,她总是充满激情的迎接并完成各项工作,荣获优秀团干部称号.在社会实践和志愿者活动中起到模范带头作用. 04 xxxx同学在思想方面,积极要求进步,为人诚实,尊敬师长.严格 要求自己.在大一期间就积极参加了党课初、高级班的学习,拥护中国共产党的领导,并积极向党组织靠拢. 在工作上,作为班中的学习委员,对待工作兢兢业业、尽职尽责 的完成班集体的各项工作任务.并在班级和系里能够起骨干带头作用.热心为同学服务,工作责任心强. 在学习上,学习目的明确、态度端正、刻苦努力,连续两学年在 班级的综合测评排名中获得第1.并荣获院级二等奖学金、三好生、优秀班干部、优秀团员等奖项. 在社会实践方面,积极参加学校和班级组织的各项政治活动,并 在志愿者活动中起到模范带头作用.积极锻炼身体.能够处理好学习与工作的关系,乐于助人,团结班中每一位同学,谦虚好学,受到师生的好评. 05 在思想方面,xxxx同学积极向上,热爱祖国、热爱中国共产党,拥护中国共产党的领导.作为一名共产党员时刻起到积极的带头作用,利用课余时间和党课机会认真学习政治理论. 在工作上,作为班中的团支部书记,xxxx同学积极策划组织各类 团活动,具有良好的组织能力. 在学习上,xxxx同学学习努力、成绩优良、并热心帮助在学习上有困难的同学,连续两年获得二等奖学金. 在生活中,善于与人沟通,乐观向上,乐于助人.有健全的人格意 识和良好的心理素质.
尊敬的领导，老师，亲爱的同学们， 大家好！我是5班的梁浩东。今天早上我坐车来学校的路上，我仔细观察了路上形形色色的人，有开着小车衣着精致的叔叔阿姨，有市场带着倦容的卖各种早点的阿姨，还有偶尔穿梭于人群中衣衫褴褛的乞丐。于是我问自己，十几年后我会成为怎样的自己，想成为社会成功人士还是碌碌无为的人呢，答案肯定是前者。那么十几年后我怎样才能如愿以偿呢，成为一个受人尊重，有价值的人呢？正如我今天演讲的题目是：自主管理。 大家都知道爱玩是我们孩子的天性，学习也是我们的责任和义务。要怎样处理好这些矛盾，提高自主管理呢？ 首先，我们要有小主人翁思想，自己做自己的主人，要认识到我们学习，生活这一切都是我们自己走自己的人生路，并不是为了报答父母，更不是为了敷衍老师。 我认为自主管理又可以理解为自我管理，在学习和生活中无处不在，比如通过老师，小组长来管理约束行为和同学们对自身行为的管理都属于自我管理。比如我们到一个旅游景点，看到一块大石头，有的同学特别兴奋，会想在上面刻上：某某某到此一游话。这时你就需要自我管理，你需要提醒自己，这样做会破坏景点，而且是一种素质低下的表现。你设想一下，如果别人家小孩去你家墙上乱涂乱画，你是何种感受。同样我们把自主管理放到学习上，在我们想偷懒，想逃避，想放弃的时候，我们可以通过自主管理来避免这些，通过他人或者自己的力量来完成。例如我会制定作息时间计划表，里面包括学习，运动，玩耍等内容的完成时间。那些学校学习尖子，他们学习好是智商高于我们吗，其实不然，在我所了解的哪些优秀的学霸传授经验里，就提到要能够自我管理，规范好学习时间的分分秒秒，只有辛勤的付出，才能取得优异成绩。 在现实生活中，无数成功人士告诉我们自主管理的重要性。十几年后我想成为一位优秀的，为国家多做贡献的人。亲爱的同学们，你们们？让我们从现在开始重视和执行自主管理，十几年后成为那个你想成为的人。 谢谢大家！
关于工作的英语演讲稿 【篇一：关于工作的英语演讲稿】 关于工作的英语演讲稿 different people have various ambitions. some want to be engineers or doctors in the future. some want to be scientists or businessmen. still some wish to be teachers or lawers when they grow up in the days to come. unlike other people, i prefer to be a farmer. however, it is not easy to be a farmer for iwill be looked upon by others. anyway,what i am trying to do is to make great contributions to agriculture. it is well known that farming is the basic of the country. above all, farming is not only a challenge but also a good opportunity for the young. we can also make a big profit by growing vegetables and food in a scientific way. besides we can apply what we have learned in school to farming. thus our countryside will become more and more properous. i believe that any man with knowledge can do whatever they can so long as this job can meet his or her interest. all the working position can provide him with a good chance to become a talent. 【篇二：关于责任感的英语演讲稿】 im grateful that ive been given this opportunity to stand here as a spokesman. facing all of you on the stage, i have the exciting feeling of participating in this speech competition. the topic today is what we cannot afford to lose. if you ask me this question, i must tell you that i think the answer is a word---- responsibility. in my elementary years, there was a little girl in the class who worked very hard, however she could never do satisfactorily in her lessons. the teacher asked me to help her, and it was obvious that she expected a lot from me. but as a young boy, i was so restless and thoughtless, i always tried to get more time to play and enjoy myself. so she was always slighted over by me. one day before the final exam, she came up to me and said, could you please explain this to me? i can not understand it. i
Dear teacher and colleagues: my topic is on “spare time”. It is a huge blessing that we can work 996. Jack Ma said at an Ali's internal communication activity, That means we should work at 9am to 9pm, 6 days a week .I question the entire premise of this piece. but I'm always interested in hearing what successful and especially rich people come up with time .So I finally found out Jack Ma also had said :”i f you don’t put out more time and energy than others ,how can you achieve the success you want? If you do not do 996 when you are young ,when will you ?”I quite agree with the idea that young people should fight for success .But there are a lot of survival activities to do in a day ,I want to focus on how much time they take from us and what can we do with the rest of the time. As all we known ,There are 168 hours in a week .We sleep roughly seven-and-a-half and eight hours a day .so around 56 hours a week . maybe it is slightly different for someone . We do our personal things like eating and bathing and maybe looking after kids -about three hours a day .so around 21 hours a week .And if you are working a full time job ,so 40 hours a week , Oh! Maybe it is impossible for us at
关于人英语演讲稿(精选多篇) 关于人的优美句子 1、“黑皮小子”是我对在公交车上偶遇两次的一个男孩的称呼代号。一听这个外号，你也定会知道他极黑了。他的脸总是黑黑的；裸露在短袖外的胳膊也是黑黑的；就连两只有厚厚耳垂的耳朵也那么黑黑的，时不时像黑色的猎犬竖起来倾听着什么；黑黑的扁鼻子时不时地深呼吸着，像是在警觉地嗅着什么异样的味道。 2、我不知道，如何诠释我的母亲，因为母亲淡淡的生活中却常常跳动着不一样的间最无私、最伟大、最崇高的爱，莫过于母爱。无私，因为她的爱只有付出，无需回报；伟大，因为她的爱寓于
普通、平凡和简单之中；崇高，是因为她的爱是用生命化作乳汁，哺育着我，使我的生命得以延续，得以蓬勃，得以灿烂。 3、我的左撇子伙伴是用左手写字的，就像我们的右手一样挥洒自如。在日常生活中，曾见过用左手拿筷子的，也有像超级林丹用左手打羽毛球的，但很少碰见用左手写字的。中国汉字笔画笔顺是左起右收，适合用右手写字。但我的左撇子伙伴写字是右起左收的，像鸡爪一样迈出田字格，左看右看，上看下看，每一个字都很难看。平时考试时间终了，他总是做不完试卷。于是老师就跟家长商量，决定让他左改右写。经过老师引导，家长配合，他自己刻苦练字，考试能够提前完成了。现在他的字像他本人一样阳光、帅气。 4、老师，他们是辛勤的园丁，帮助着那些幼苗茁壮成长。他们不怕辛苦地给我们改厚厚一叠的作业，给我们上课。一步一步一点一点地给我们知识。虽然
他们有时也会批评一些人，但是他们的批评是对我们有帮助的，我们也要理解他们。那些学习差的同学，老师会逐一地耐心教导，使他们的学习突飞猛进。使他们的耐心教导培养出了一批批优秀的人才。他们不怕辛苦、不怕劳累地教育着我们这些幼苗，难道不是美吗？ 5、我有一个表妹，还不到十岁，她那圆圆的小脸蛋儿，粉白中透着粉红，她的头发很浓密，而且好像马鬓毛一样的粗硬，但还是保留着孩子一样的蓬乱的美，卷曲的环绕着她那小小的耳朵。说起她，她可是一个古灵精怪的小女孩。 6、黑皮小子是一个善良的人，他要跟所有见过的人成为最好的朋友！这样人人都是他的好朋友，那么人人都是好友一样坦诚、关爱相交，这样人与人自然会和谐起来，少了许多争执了。 7、有人说，老师是土壤，把知识化作养分，传授给祖国的花朵，让他们茁壮成长。亦有人说，老师是一座知识的桥梁，把我们带进奇妙的科学世界，让
优秀党务工作者事迹简介范文 优秀党务工作者事迹简介范文 ***，男，198*年**月出生，200*年加入党组织，现为***支部书记。从事党务工作以来，兢兢业业、恪尽职守、辛勤工作，出色地完成了各项任务，在思想上、政治上同党中央保持高度一致，在业务上不断进取，团结同事，在工作岗位上取得了一定成绩。 一、严于律己，勤于学习 作为一名党务工作者，平时十分注重知识的更新，不断加强党的理论知识的学习，坚持把学习摆在重要位置，学习领会和及时掌握党和国家的路线、方针、政策，特别是党的十九大精神，注重政治理论水平的提高，具有坚定的理论信念;坚持党的基本路线，坚决执行党的各项方针政策，自觉履行党员义务，正确行使党员权利。平时注重加强业务和管理知识的学习，并运用到工作中去，不断提升自身工作能力，具有开拓创新精神，在思想上、政治上和行动上时刻同党中央保持高度一致。 二、求真务实，开拓进取 在工作中任劳任怨，踏实肯干，坚持原则，认真做好学院的党务工作，按照党章的要求，严格发展党员的每一个步骤，认真细致的对待每一份材料。配合党总支书记做好学院的党建工作，完善党总支建设方面的文件、材料和工作制度、管理制度等。
三、生活朴素，乐于助人 平时重视与同事间的关系，主动与同事打成一片，善于发现他人的难处，及时妥善地给予帮助。在其它同志遇到困难时，积极主动伸出援助之手，尽自己最大努力帮助有需要的人。养成了批评与自我批评的优良作风，时常反省自己的工作，学习和生活。不但能够真诚的指出同事的缺点，也能够正确的对待他人的批评和意见。面对误解，总是一笑而过，不会因为误解和批评而耿耿于怀，而是诚恳的接受，从而不断的提高自己。在生活上勤俭节朴，不铺张浪费。 身为一名老党员，我感到责任重大，应该做出表率，挤出更多的时间来投入到**党总支的工作中，不找借口，不讲条件，不畏困难，将总支建设摆在更重要的位置，解开工作中的思想疙瘩，为攻坚克难铺平道路，以支部为纽带，像战友一样团结，像家庭一样维系，像亲人一样关怀，践行入党誓言。把握机遇，迎接挑战，不负初心。
关于时间的英语演讲稿范文_演讲稿 and organizing people to learn from advanced areas to broaden their horizons in order to understand the team of cadres working conditions in schools, the ministry of education has traveled a number of primary and secondary schools to conduct research, listen to the views of the school party and government leaders to make school leadership cadres receive attention and guidance of the ministry of education to carry out a variety of practical activities to actively lead the majority of young teachers work hard to become qualified personnel and for them to put up the cast talent stage, a single sail swaying, after numerous twists and turns arrived in port, if there is wind, a hand, and naturally smooth arrival and guide students to strive to e xcel, need to nazhen “wind” - teacher. teachers should be ideological and moral education, culture education and the needs of students organically combining various activities for the students or students to carry out their own. for example: school quiz competitions, essay contests, ke benju performances and other activities to enable students to give full play to their talents. teachers rush toil, in order to that will enable students to continue to draw nutrients, to help them grow up healthily and become pillars of the country before. for all students in general education, the government departments have also not forget those who cared about the