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RESOLUTION OF IRREDUCIBLE INTEGRAL FLOWS ON A SIGNED

RESOLUTION OF IRREDUCIBLE INTEGRAL FLOWS ON A SIGNED
RESOLUTION OF IRREDUCIBLE INTEGRAL FLOWS ON A SIGNED

RESOLUTION OF IRREDUCIBLE INTEGRAL FLOWS ON A SIGNED

GRAPH

BEIFANG CHEN,JUE WANG,AND THOMAS ZASLAVSKY

Abstract.We completely describe the structure of irreducible integral?ows on a signed

graph by lifting them to the signed double covering graph.

A(real-valued)?ow(sometimes also called a circulation)on a graph or a signed graph(a graph with signed edges)is a real-valued function on oriented edges,f: E→R,such that the net in?ow to any vertex is zero.An integral?ow is a?ow whose values are integers.There are many reasons to be interested in?ows on graphs;an important one is their relationship to graph structure through the analysis of irreducible?ows,that is,integral?ows that cannot be decomposed as the sum of other?ows of lesser value.It is well known,and an important observation in the thoery of integral network?ows,that the irreducible?ows are identical to the circuit?ows,which have value1on the edges of a graph circuit(that is,a cycle)and0 on all other edges.Extending the theory of irreducible integral?ows to signed graphs,which was one of the topics of the doctoral dissertation of Wang[4],led to the remarkable discovery that there are,besides the anticipated circuit?ows(which in signed graphs are already more complicated than in unsigned graphs),also many‘strange’irreducible?ows with elaborate structure not describable by circuits.In this article we characterize that structure by lifting it to a simple cycle in the signed covering graph.(Indeed,this was how we discovered the correct characterization,though we were also guided by the partial result in Wang’s thesis.) We like to think of lifting as a combinatorial analog of resolution of singularities in contin-uous mathematics.The strange irreducible?ows are singular phenomena,which we resolve by lifting them to ordinary cycle?ows in a covering graph.This is not a precise statement but a philosophy that we believe will be fruitful.

1.Graphs and signed graphs

Graphs.A graph is(V,E),with vertex set V and edge set E.There may be loops and multiple edges.An edge e with endpoints v and w has two ends,which we symbolize by (v,e)and(w,e).A tricky technical point is that this notation does not distinguish the two ends of a loop;we take an easy way out by treating(v,e)and(w,e)as di?erent ends even when v=w.(There are more technically correct means of distinguishing the ends but they make the notation very complicated.)

A walk is a sequence W=v0e1v1e2···e l v l of vertices and edges such that the endpoints of e i are v i?1and v i.A walk is closed if l>0and v0=v l and open otherwise.A segment of W is a consecutive subwalk,i.e.,v i e i+1···e j v j.When W is closed we allow j>l,interpreting indices modulo l;thus,a segment may pass through v0.A circle is the edge set of a simple closed walk,i.e.,there is no repeated edge or vertex other than that v0=v l.A graph circuit Date:June23,2007.

2000Mathematics Subject Classi?cation.Primary05C22;Secondary.

The?rst author’s research was supported by???

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is a circle;the name comes from the fact that the circles of a graph form the circuits of the well known graphic matroid whose elements are the edges of the graph.

A isthmus is an edge whose deletion increases the number of connected components.

A cutpoint is a vertex whose deletion,with all incident edges,increases the number of components,or that supports a loop and is incident with at least one other edge.A block

is a maximal subgraph without cutpoints.Thus,a loop or isthmus or isolated vertex is a (trivial)block.We call blocks adjacent if they have a common vertex(which is necessarily

a cutpoint).An end block is a block adjacent to exactly one other block.

Signed graphs.A signed graphΣ=(V,E,σ)consists of a graph(V,E)and a signature σ:E→{+1,?1}.(Signs multiply;they do not add.)A walk inΣhas a signσ(W):=

σ(e1)σ(e2)···σ(e l).In particular,a circle has a sign(which is the sign of any walk that goes once around the circle),so it is either positive or negative.A subgraph or edge set is balanced if every circle in it is positive.

A signed circuit in a signed graph is a subgraph(or its edge set)of one of the following three types:

(I)a positive circle(that is,a balanced circle);

(II)a pair of negative circles whose intersection is one vertex(sometimes called a contra-balanced tight handcu?);and

(III)a pair of vertex-disjoint negative circles together with a connecting simple path(called the circuit path)that is internally disjoint from the circles(this type is sometimes

called a contrabalanced loose handcu?).

The signed circuits are the circuits of a matroid on the edge set of the signed graph[5].

An ordinary,unsigned graph can be treated as an all-positive signed graph.When all edges are positive,every circle is positive and the only signed circuits are those of Type I, i.e.,the graph circuits.

Orientation.A bidirection of a graph(a concept introduced by Edmonds[3])is a function from the edge ends to the sign group.On thinks of an end with sign+1as having an arrow directed away from the vertex and an end with sign?1as having an arrow directed toward the vertex(or vice versa;see[6]);thus a bidirected graph has two arrows,one at each end. Formally,we write a bidirection as a functionε:V×E→{+1,?1}such thatε(v,e)=0if and only if v is not an endpoint of e.(For loops this formalism is technically incorrect,but we trust the reader will be willing to understandε(v,e)andε(w,e)as independent values even when e is a loop so v=w.Otherwise we are forced into technical complications.)

An orientation of a signed graphΣis a bidirection of its edges such thatσ(e)=?ν(v,e)ν(w,e) [6].Thus,a positive edge has two arrows that are consistent and give e a direction,just as

in an ordinary directed graph.A negative edge has arrows that both point towards,or both away from,the endpoints.

A source in an oriented signed graph is a vertex v at which all edges are directed outwards; that is,ε(v,e)=+1for all edges at e.Conversely,if all edges point into v,v is a sink.

A walk W=v0e1v1e2···e l v l of an oriented signed graph is called coherent at v i ifε(v i,e i)=

?ε(e i+1);that is,if the walk has a consistent direction at v i.We apply this de?nition to v0,

if W is closed,by taking subscripts modulo l.A directed walk is coherent at every vertex except v0and v l.A directed closed walk is a closed walk that is coherent at every vertex with the possible exception of v0.Although each vertex has a direction,W itself need not have an overall direction,since at every negative edge the arrows reverse.A positive(or negative)

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directed closed walk must be coherent(or incoherent,respectively)at v0,by the following lemma.

Lemma1.The sign of a closed walk equals(?1)k,where k is the number of incoherent vertices in the walk,including the?nal vertex if it is incoherent.

Proof.We perform a short calculation that applies to open as well as closed walks.Note that if W is open,the?nal vertex cannot be incoherent.

(1)σ(W)=

l

i=1σ(e i)=l i=1[?ε(v i?1,e i)ε(v i,e i)]

=

l?1

j=1[?ε(v j,e j)ε(v j,e j+1)]·[?ε(v0,e1)ε(v l,e l)]

= (?1)k if W is closed,

?(?1)kε(v0,e1)ε(v l,e l)if W is open.

Reorienting S?E means reversing the orientations of the edges in S but not those outside S.Thus,εchanges toεS de?ned by

εS(v,e)= ?ε(v,e)if e∈S,

ε(v,e)if e/∈S.

The signed covering graph.Let?V:=V×{+1,?1}and let?E:=E×Z2,the union of two disjoint copies of E.For brevity we write(v,α)as vα.If an edge e ofΣhas endpoints v and w,then one copy of e in?Σhas endpoints v+1and wσ(e)while the other copy has endpoints v?1and w?σ(e).This de?nes?Σ,the signed covering graph ofΣ,which is a graph with unsigned edges and signed vertices.?Σhas a canonical involutory automorphism?de?ned by(vα)?:=v?αand(?e,k)?:=(?e,k+1)and projects toΣby the mapping p(vα)=v and p(e,k)=k,which is a2-to-1graph homomorphism.We write?e for an edge in the covering graph that projects to e and,following the notation of the canonical involution, p?1(e)={?e,?e?}.When e is a negative loop at v,?e and?e?are two parallel edges in?Σwith the endpoints v+1and v?1.

We can think of?V as having a positive level,V×{+1},and a negative level,V×{?1}. However,it is not possible to assign all edges to levels.A positive edge lifts to an edge that stays within a level,but a negative edge crosses between levels.

Suppose W=v0e1v1e2···e l v l is a walk inΣ.We can lift W to a walk?W in?Σ.First,we choose a vertex vα00that projects to v0.Then we follow edges that cover the edges of W,

getting a walk?W=vα00?e1vα11?e2···vαl?1

l?1?e l vαl

l

such thatαi=αi?1σ(e i).We call such a walk a

lift of W.There are at least two such lifts since the choice ofα0is arbitrary,and there are only two lifts if W contains no loops.

Lemma2.Let W be a closed walk inΣand?W a lift of it in?Σ.The sign of W isσ(W)=+1 if?W is closed and?1if?W is open.

Proof.Let W=v0e1v1e2···v l?1e l v l where v l=v0,lifted to?W=vα00?e1···?e l vαl

l

.Then

σ(W)=

l

i=1σ(e i)=l i=1αi?1αi=α0αl.

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This equals+1if and only if?W is closed. An orientationεofΣlifts to?Σby the rule

?ε(vα,?e)=αε(v,e),

where?e denotes the lift of e that is incident with vα.By this de?nition,the other lift edge,?e?,is incident with v?αand has the opposite orientation,?ε(v?α,?e?)=?αε(v,e).It is easy to see that?εorients?Σas an all-positive signed graph;that is,(?Σ,?ε)is an ordinary directed graph.For a positive edge,one lift lies in the positive level and has the same direction as e while the other lies in the negative level and has the direction opposite to that of e.For a negative edge,both lifts are directed from the positive to the negative level or the reverse.

2.Flows

Flows on a signed graph.An integral?ow on an oriented signed graph(Σ,ε)(or one could say,following[1],on a bidirected graph)is a function f:E→Z which is conservative at every vertex v,meaning that

?f(v):= e∈Eε(v,e)f(e)=0.

(We assume that a loop appears twice in this sum,once for each end.)The set of all integral ?ows on(Σ,ε)forms a Z-module,called the?ow lattice by Chen and Wang,who developed its basic theory in[2].One can de?ne?ows with values in any abelian group,such as the additive reals;many of the following remarks are applicable in general(so we omit the word ‘integral’).

The theory of?ows depends essentially on the graph and signature but it is not really oriented,since when one reorients an edge set S one gets a natural isomorphism of?ow lattices by taking a?ow f on(Σ,ε)to the?ow f′on(Σ,εS)that is de?ned by f′(e)=?f(e) for e∈S while f′(e)=f(e)for e/∈S.

The support of a function f:E→Z is the set supp f:=E\f(?1)(0).The?ow that is zero on all edges is the trivial?ow.A circuit?ow has support that is signed circuit;on a directed circuit it takes value1on edges in a circle of the circuit and,in Type III,2on edges in the circuit path,and one gets a circuit?ow on an arbitrary oriented circuit by reorienting it to be a directed circuit,applying the de?nition for directed circuits,and reorienting back to the original orientation while applying the natural?ow-lattice isomorphism;i.e.,negating the?ow values on the reoriented edges.

The usual theory of?ows on graphs is simply the all-positive case.There a circuit?ow is ±1on the edges of a circle and zero elsewhere;we omit further details.

We say an integral?ow f′conforms to the sign pattern of f if supp f′?supp f and, whenever f′(e)is nonzero,it has the same sign as f(e).

An integral?ow f on(Σ,ε)lifts to a?ow on the oriented signed covering graph,possibly in more than one way.The best way to see this is through the correspondence between?ows and walks,which exists if(and only if)the support is connected.

A positive,directed closed walk W on(Σ,ε)implies a unique corresponding?ow,which is the integral?ow de?ned by taking f W(e)(with e in an arbitrary?xed orientation)to be the number of times W traverses e in the?xed orientation less the number of times e appears in the opposite orientation.To prove f W is a?ow,consider the contribution to f of a pair of consecutive edges,e i v i e i+1,at the intervening vertex.Because W is coherent at v i,the

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contribution of these edges to?f(v i)is0.This same argument applies to the initial vertex if we take subscripts modulo the length of W.We can apply the same construction of a function f W:E→Z to any walk,but the result may not be a?ow.

Lemma3.For a directed walk W=v0e1v1e2···v l?1e l v l,the function f W satis?es?f(u)=0 if u=v0,v l,and

?f(v0)= 0if W is closed and positive,

±2if W is closed and negative;

and if W is open,then?f(v0)=?σ(W)?f(v l)=±1.

Proof.If W is closed and negative,it is incoherent at v0and therefore e1and e l contribute the same value±1to?f(v0).

If W is open,then e1contributesε(v0,e1)=±1to?f(v0)andε(v l,e l)=±1to?f(v l). These values are related byε(v l,e l)=?σ(W)ε(v0,e1),by Equation(1),because W is co-herent. In the other direction,the construction of a corresponding walk(which is not usually unique)from a?ow f is just like the usual one of an Eulerian tour of a connected digraph with equal in-and out-degrees,but more complicated because of negative edges. Proposition4.Let f be a nonnegative,nontrivial integral?ow on(Σ,ε).Then there is a positive,directed closed walk W such that f W=f.

Proof.We apply induction on f := e f(e).Choose a vertex of supp f,call it v0,and an edge e1∈supp f that is incident with v0;call its other endpoint v1.This gives a walk W1=v0e1v1of length1.

Now,suppose we have selected a partial walk,W k=v0e1v1···e k v k.

If W k is open or negative,then the lemma applied to W k,together with f≥f W

k ≥0,

shows that there must be an edge e k+1∈supp f?f W

k such thatε(v k,e k+1)=?ε(v k,e k).

Extend W k to W k+1:=W k e k+1v k+1,where v k+1is the other endpoint of e k+1.

If W k is closed and positive,we have a positive,directed closed walk,by Lemma1,such

that f W

k ≥0.If f?f W

k

=0,consider the restrictions of f?f W

k

to the components of its

support.Each restriction f′is a nonnegative?ow that has a corresponding positive,directed closed walk and has f′ < f .Each of these walks has a vertex in common with W k,so we can assemble them into a single positive,directed closed walk W such that f W=f. Lifted?ows.Consider a function f:?E→Z de?ned on the edge set of the signed covering graph?Σ.The projection is the function p(?f):E→Z de?ned by p(?f)(e)=?f(?e)+?f(?e?).A lift of an integral?ow on(Σ,ε)to the signed covering graph is a?ow?f on(?Σ,?ε)such that p(?f)=f.

Lemma5.A nonnegative integral?ow on(Σ,ε)lifts to a nonnegative integral?ow on(?Σ,?ε).

Proof.Construct a corresponding walk W to the?ow.Lift W to?W.By construction,?W is a directed closed walk on the signed covering graph.Thus,f?

W

is nonnegative.

If we have an integral?ow that is nonnegative,we apply the lemma after reorienting(Σ,ε) so f is nonnegative.

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Irreducible?ows.An integral?ow f onΣis called reducible if it is nontrivial and it can be represented as the sum of two other integral?ows,f=f1+f2,each of which is nontrivial and conforms to the sign pattern of f.It is easy to see that irreducibility is equivalent to minimality:the only nontrivial?ow that conforms to the sign pattern of f and satis?es 0≤f′≤f is f itself.

It is well known that the irreducible?ows on an unsigned graph are the circle?ows. Given a signed graphΣ,construct an integral?ow by the following process.

1.Choose a connected subgraphΣ′such that

(a)each block is a circle or an edge,

(b)each end block is a circle,

(c)each cutpoint is incident with exactly two blocks,and

(d)the sign of a circle block equals(?1)p,where p is the number of cutpoints on

the circle.

2.OrientΣ′so that

(a)within each circle block,each non-cutpoint is coherent and each cutpoint is

incoherent,and

(b)there are no sources or sinks.

3.Assign?ow values1to each circle edge and2to each isthmus.

Let us call the results of Step1essential signed graphs,the results of Step2essential orientations,the?ows of Step3essential?ows,and the corresponding closed walks essential walks.There are other ways to describe the essential orientations.

Proposition 6.Letεbe an orientation of an essential signed graphΣ.The following properties ofεare equivalent:

(i)It is essential.

(ii)(Σ,ε)has a coherently oriented essential walk.

(iii)Every essential walk in(Σ,ε)is coherently oriented.

Proof.Clearly,(iii)implies(ii)and(ii)implies(i).We show that(i)implies(iii).

Let W be an essential walk;we show it is coherent at every vertex v.This is obvious if v is not a cutpoint,since it has degree2and is neither a sink nor a source.If it is a cutpoint,v divides W into segments W1and W2.Each segment,by(i),has its ends at v directed similarly:both towards v,or both away from v.Also by(i),W1and W2are directed di?erently at v:one is towards v and the other away from it.The expression W=vW1vW2v shows that W is coherent at v. Theorem7.A?ow is irreducible if and only if it is essential.

Proof.We have two things to prove:that an essential?ow is irreducible,and that every irreducible?ow is essential.We begin with the latter.

The idea of the proof is to turn an irreducible?ow into a positive,directed closed walk W,which we lift into the signed covering graph,where it becomes a circle?W.Then we have to?nd out which circles project to irreducible?ows.Self-intersections in the base graph correspond to vertex pairs+v,?v—that is,vertex?bers—in?W;hence there are no triple self-intersections in W,and each half of W separated by a self-intersection is a negative closed walk.The self-intersection is a cutpoint because,if the two halves intersected at any other vertex,?W could be adjusted to become non-simple while still projecting to W;then

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?W,hence W,would be reducible.It follows that every block is a circle or an isthmus,and the sign of a circle block follows from the negativity of each half of W when split by a cutpoint. The proof itself is a series of lemmas.The irreducible?ow is f and a corresponding positive,directed closed walk(which exists,by Proposition4,because irreducibility implies that supp f is connected)is W=v0e1v1e2···e l v l.W lifts to a walk?W=vα00?e1vα11?e2···?e l vαl

l in the signed covering graph?Σ.A vertex in W is a double point if W passes through it exactly twice.A double point divides W into two closed subwalks.We can think of W as a subgraph,namely,as the subgraph induced by the edges of W;as a subgraph it has cutpoints and blocks.

Lemma8.A?ow inΣthat lifts to a reducible?ow in?Σis reducible.

Proof.Let f be the?ow inΣ,assumed nonnegative by choosing the right orientation ofΣ. Suppose f lifts to?f,which is the sum of nontrivial?ows?f1and?f2.Then the projections f1=p?f1and f2=p?f2are nontrivial,nonnegative?ows inΣ(in the chosen orientation) whose sum is f. Lemma9.?W is a circle.

Proof.Lemma8implies that the lift of an irreducible?ow is irreducible.?Σis an unsigned graph.An irreducible?ow in an unsigned graph is a circle?ow. Lemma10.W is a positive walk.

Proof.Since the lift?W is closed,W must be positive. Lemma11.Any self-intersection vertex of W is a double point and a cutpoint of W.The two closed subwalks into which it divides W are negative walks.

Proof.Suppose v i=v j=v with i

Now suppose there is a vertex w,other than v,that appears in both W1and W2.Then?W1 has wβas a vertex,and?W2has a vertex w?β.If we replace?W2in?W by(?W?2)?1(reversing the direction so it goes from vαi to v?αi,just like?W2),we get a new walk?W′that still is a lift of W but has a self-intersection at wβ.Thus,by Lemma8,f is reducible,contrary to the assumption.This shows that v is a cutpoint of W. Lemma12.Each block of W is a circle or an isthmus.An end block is a circle.

Proof.No block can have a vertex of degree greater than2.An end block which is an isthmus makes an edge with?ow0;this edge would not have been in W in the?rst place. Lemma13.The sign of a circle block of W equals(?1)p where p is the number of cutpoints of W on that circle.

Proof.Let C be a circle block with cutpoints v1,...,v k.Each v i separates W into a half walk that contains the edges of C and another half walk,which we call W i.We know W i is negative,W is positive,andσ(W)=σ(C) iσ(W i).The lemma follows at once. Lemma14.The edges of W are oriented so that no vertex is a source or a sink in its underlying oriented signed graph.The orientation of a circle block of W is incoherent at each cutpoint and coherent at each other vertex.

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The last statement is equivalent to saying that at a cutpoint v,which is necessarily incident with two blocks,the edges of one block are directed into v and the edges of the other block are directed away from v.This is true for both circle and isthmus blocks.

Proof.Since W is coherently oriented,no vertex can be a source or a sink.A cutpoint v divides W into segments W1and W2;the edges of each segment at v belong to di?erent blocks,B1and B2respectively.Each segment is negative,by Lemma16,and coherent,so its ends at v are directed similarly:both towards v,or both away from v;this shows v is incoherent in each B i.By coherence of W,W1and W2are directed di?erently at v:one is towards v and the other away from it. Lemma15.The?ow values on edges of W are1for an edge in a circle block and2for an isthmus.

Proof.These are the values obtained by projecting the circle?ow from?Σ.A doubly covered edge e has to have?ow value1+1or1+(?1)=0,but the latter case is impossible since then e would not have been in W in the?rst place. This completes the proof that every irreducible?ow is essential.We still have to prove that any essential?ow is irreducible.We require a lemma.

Lemma16.In an essential walk,a segment separated by a cutpoint is negative.

Proof.A cutpoint separates the walk W into two segments.Let W′be the one in question. The sign of W′is the product of the signs of its circle blocks.Thus,the sign is the number of times a cutpoint appears on a circle block of W′.

Note that isthmi appear in paths that connect two circle blocks,since each end block is a circle.Every cutpoint appears twice in a circle block,with the following exceptions:A cutpoint between two isthmi appears no times and consequently does not contribute to the sign of W′.A cutpoint that connects a circle block to an isthmus contributes?1to the sign but pairs with the cutpoint at the other end of the path to which that isthmus belongs; the two cutpoints contribute a total of+1to the sign.Finally,v appears only once,so it contributes?1.The signσ(W′)is the product of these signs,hence is?1. Suppose f is an essential?ow which is reducible,say f=f1+f2+···where f i corresponds to a walk W i.Then each supp f i forms a connected subgraph of supp f which is joined to the rest of supp f only at cutpoints of supp f.There must be a supp f i that is joined at only one cutpoint v.Then W i is a closed segment of W.By Lemma16,W i has negative sign, but the corresponding walk of a?ow is positive.Therefore,f cannot be reducible.

References

[1]A.Bouchet,Nowhere-zero integral?ows on a bidirected https://www.wendangku.net/doc/af5432052.html,bin.Theory Ser.B34(1983),

279–292.MR85d:05109.Zbl.518.05058.

[2]Beifang Chen and Jue Wang,The circuit and bond space and minimal?ows,tensions,and potentials

of signed graphs.Submitted.

[3]Jack Edmonds,Maximum matching and a polyhedron with0,1-vertices.J.Res.Nat.Bur.Standards

Sect.B69B(1965),125–130.MR32#1012.Zbl.141,218b(e:141.21802).

[4]Jue Wang,Algebraic Structures of Signed Graphs.Doctoral dissertation,Hong Kong University of

Science and Technology,2007.

[5]Thomas Zaslavsky,Signed graphs.Discrete Appl.Math.4(1982),47–74.Erratum.Discrete Appl.Math.

5(1983),248.MR84e:05095.Zbl.503.05060.

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[6]Thomas Zaslavsky,Orientation of signed graphs.European https://www.wendangku.net/doc/af5432052.html,bin.12(1991),361–375.MR

93a:05065.Zbl.761.05095.

Department of Mathematics,Hong Kong University of Science and Technology,Clear Water Bay,Kowloon,Hong Kong

Department of Mathematics,Hong Kong University of Science and Technology,Clear Water Bay,Kowloon,Hong Kong

Department of Mathematical Sciences,Binghamton University(SUNY),Binghamton,NY 13902-6000,U.S.A.

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李阳:我热爱丢脸_名人故事

李阳:我热爱丢脸 我热爱丢脸 李阳和他的公司10年来只做一件事,仍然是10年前提出的——让3亿中国人讲一口流利的英语! 其实疯狂英语最初的动机,只是针对当时传统英语教学中聋哑英语的很多问题,反其道提出的将英语脱口“说”出的办法。其次,李阳本人是一个性格内向的人,小时候,他认为自己能够去买一瓶酱油,就是很大的成功了,因为他不敢和陌生人说话。所以至今他的父母跟同学都会感到纳闷:那是李阳吗?是什么力量使他突破自己最早禁锢的心?李阳说,他是通过喊英语喊出自信来的。为了克服自己的弱点,他找了一个同学跟他一起喊,这个人绝对认死理,一旦制定计划就一定要做到,他们两个相互鼓励坚持喊英语。一段时间后,他们到外语

角讲英语,得到了别人的夸奖,尝到甜头后他们又继续努力。就这样一步一步地,小小的成就感,渐渐地坚定了信念,获得更大的成就。 当问他进展如何时,他说当时提出3亿人觉得还是容易做到的,现在才发现,这是一个庞大的工作:“第一,中华民族是一个要面子的民族,不但自己要面子,而且要求别人要面子,自己不但不能练英语,别人也不能练,你要练他还会给你泼冷水。比如你在单位讲英语,别人会把你当作异类。在中国学习英语口语的环境非常恶劣,这是个巨大的障碍。第二,我们有三四亿人在为了考试而学习英语,每年为考试而出版的书有几十亿册,要砍掉多少林木,这太可怕了,更重要的是没有去用。语言的目的就是交流,为了交流,为了完成一个国际交流的任务,而我们却把语言当作一个游戏,喜欢出选择题,三个错误答案,一个正确答案,中国人四分之三的时间在复习错误答案,这是一个非常严肃的问题。就这样,中国人一张口就是错误的英语,因为他每天都在看错误的答案,正确的东西反而记不清楚了。”

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