On the length of the longest subsequence avoiding an arbitrary pattern in a random permutat

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Free subalgebras of Lie algebras close to nilpotent

a r X i v :0805.0723v 2 [m a t h .R A ] 9 M a y 2008FREE SUBALGEBRAS OF LIE ALGEBRAS CLOSE TO NILPOTENTALEXEY BELOV AND ROMAN MIKHAILOVAbstract.We prove that for every automata algebra of exponential growth,the associatedLie algebra contains a free subalgebra.For n ≥1,let L n +2be a Lie algebra with generatorset x 1,...,x n +2and the following relations:for k ≤n ,any commutator of length k whichconsists of fewer than k diﬀerent symbols from {x 1,...,x n +2}is zero.As an application of thisresult about automata algebras,we prove that for every n ≥1,L n +2contains a free subalgebra.We also prove the similar result about groups deﬁned by commutator relations.Let G n +2bea group with n +2generators y 1,...,y n +2and the following relations:for k =3,...,n ,anyleft-normalized commutator of length k which consists of fewer than k diﬀerent symbols from{y 1,...,y n +2}is trivial.Then the group G n +2contains a 2-generated free subgroup.Main technical tool is combinatorics of periodical sequences and period switching.1.Introduction Let A be an associative algebra over a commutative ring with identity,generated by a set S .Denote by A ∼the Lie algebra with the same set of generators S and operation [u,v ]=uv −vu,u,v ∈A.In other words A ∼is the Lie subalgebra of A −generated by the given set S .The algebra A ∼clearly depends on the choice of the set of generators of A .For n ≥1,let L n +2be a Lie algebra with generator set x 1,...,x n +2and following relations:for k ≤n ,any commutator of length k which consists of fewer than k diﬀerent symbols from {x 1,...,x n +2}is zero.For example,the trivial commutators in L n ,which correspond to the case k =3,are:[[x i ,x j ],x i ],i =j.One of the main results of this paper is following:Theorem 1.For every n ≥1,L n +2contains a free Lie subalgebra.The proof of theorem 1is based on the theory of monomial algebras.An algebra with basis X is called monomial if all its deﬁning relations are of the form u =0,where u is a word written in X .Let A be a ﬁnitely generated algebra with generators x 1,...,x s .The Growth function V A (n )equal,the dimension of the space generated by words of length ≤n .If V A (n )grows exponentially,then A has exponential growth ;if polynomially,then A has polynomial growth .Intermediate growth is also possible.The polynomial or exponential growth property does not depend on the choice of the generator set.For n ≥1,let A n +2be the monomial algebra with generators x 1,...,x n +2and the followingrelations:u (x 1,...,x n +2)=0if |u |=k (k ≤n )and u consists of fewer than k symbols from {x 1,...,x n +2}.Clearly,the Lie algebra A ∼n +2is a quotient of L n +2.Hence,Theorem 1willfollow if we will be able to prove that A ∼n +2contains a free subalgebra.The algebra A n +2hasan alternative description,based on the following property:u (x 1,...,x n +2)=0in the algebra A n +2if the distance between two occurrences of the same letter in u (x 1,...,x n +2)is less than n +1.Consider a super-word w =(x 1···x n +1)∞.It is clear that w =0and any series of changes x n +1→x n +2will not yield zero,since the distance between two occurrences of the same letter is still ≤n .With the help of above changes it is possible to get 2M diﬀerent non–zero words.It follows that the number of diﬀerent non–zero words of length k in the monomial algebra A is not less1than2[kb)The positions of the occurrences of a word v of length≥|u|in u∞diﬀers by a period multiple.c)If|v|≥|u|and v2⊂u∞,then v is cyclically conjugate to a power of u.Therefore, nonnilpotent words in A u∞are exactly those words,which are cyclically conjugate to words of the form u k.Proposition3.If uW=W r,then uW is a subword of u∞and W=u n r,where r is an initial segment in u.Remark.The periodicity of an inﬁnite word means its invariance with respect to a shift.In the one-sided inﬁnite case a pre-period appears;in theﬁnite case there appear eﬀects related to the truncation.This,together with superword technique is the essence of a great many combinatorial arguments(see Proposition3)especially Bernside type problems.Proofs of the Shestakov hypothesis(nilpotency of subalgebra of n×n matrix algebra with all words of length ≤n are nilpotent),of the Shirshov height theorem(the normal basis of associative aﬃne P I-algebra A contains only piece-wise periodic words,number of periodic parts is less then h(A), and length of each period is≤n–maximal dimension of matrix algebra satisfying all identities of A)of the coincidence theorem of the nilradical and the Jacobson radical in a monomial algebra, are examples[4],[1],[3].Lemma1(on overlapping).If a subword of length m+n−1occurs simultaneously in two periodic words of periods m and n,then they are the same,up to a shift.Lemma1implies one technical statement,needed in sequelLemma2.Let r=u n v m,n>k,m>l.Then r has not common subwords with u∞of length ≥n|u|+|u|+|v|−1=(n+1)|u|+|v|−1and has not common subwords with v∞of length ≥(m+1)|v|+|u|−1.2.2.Periods switching.Proposition4.Let u,v are not powers of the same word,l|v|>2|u|and s|u|>2|v|.Then v l u s is not a subword of u∞or v∞.This proposition follows from the following assertion which is follows at once from Lemma 1(see[4]):if two periodical superwords of periods m and n have the common part of length >m+n−2,then these words are identical.In this case,u s v l is a subword v∞and u∞. Proposition5.Let u,v are not powers of the same word,l|v|>2|u|and s|u|>2|v|.Then v l u s is not a proper power.Proof.Without loss of generality we may suppose that both u and v are non-cyclic.Suppose that k>1and z k=v l u s for some non-cyclic word z.Without loss of generality we can assume that|v l|≥|u s|.If k≥4,then v l contains z2and from the other hand,v l is subword of z∞.It follows from overlapping lemma1that v is power of z and hence v=z(both z,v are non cyclic).Then u is also power of z and we are done.If k=2then u s is a subword of z and hence of v l.That contradicts overlapping lemma1. If k=3then because|v l|≥|u s|and s|u|>2|v|we have l≥3.In this case|v|<|z|/2and |v l|≥|v|+|z|.By overlapping lemma1we have that v∞=z∞and hence v=z because both are non cyclic.Then z3=v l u s=z l u s and u s=z3−l.Because u is noncyclic u=z.Hence u=v that contradicts conditions of the proposition5. Lemma3.Let r=u n v m,n>k,m>l.Then r is not a subword of W′=v∞/2u∞/2and hence of v p u q for all p,q.Proof.If r is a subword of W′then either u n(i.e.left part of r)is a subword of v∞or v m(i.e. right part of r)is a subword of u∞.Both cases are excluded by proposition4Proposition6.Consider superword W=u∞/2v∞/2,where u=v are diﬀerent noncyclic words. Let S=u k v l and suppose|u k−1|>2|v|,|v k−1|>2|u|,k,l≥2.Then S has just one occurrence in W,which is the obvious one(which we call the“standard occurrence”).Proof.Otherwise the extra occurrence of S is either to the left of the standard occurrence,or to the right.Without loss of generality it is enough to consider the left case.In this case,by Proposition2W is shifted respect to the standard occurrence by a distance divisible by|u|.Hence we have:u s W=W R,i.e.u s W starts with W.We can apply Proposition3and so we get that u∞/2starts with W.Then from combining Lemma1and Proposition2we get that v is cyclically conjugate to u,and|u|=|v|.But in that case W=u k v l is subword of u∞,implying that the relative shifts of u and v are divisible by|u|=|v|,and hence u=v.The proposition is proved. This proposition together with Lemmas2and3impliesCorollary1.Let R=r∞=(u n v m)∞,n>k,m>l.Then all the occurrences of S in R are separated by distances divisible by|r|=n|u|+m|v|.Proof.First of all,as in the proposition6one can deﬁne notion of standard occurrence of S in R.Consider an occurrence of S in R.Then only following cases are logically possible:(1)It naturally corresponds to occurrence of S in W(i.e.power of v in S starts in on thepower of v in W and similarly power of u in S ends in on the power of u in W).(2)It contains completely either u n or v m.(3)It lies on the position of period switching from v m to u m.Second possibility is excluded due to due to Lemma2,third–to due to Lemma3.First possibility due to proposition6corresponds only to standard occurrences and they are separated by distances divisible by|r|=|u n v m|=n|u|+m|v|. Note that r and t are cyclically conjugate,iﬀr∞=t∞.Using this observation and the previous corollary we get a proposition needed in the sequel:Proposition7.Let u,v be diﬀerent non-cyclic words,with|u n|>2|v|and|v n|>2|u|.For all k i,l i≥n,if(k1,l1)=(k2,l2),then u k1v l1and u k2v l2are not cyclically conjugate.Proof.Suppose that r=u k1v l1and t=u k2v l2are cyclically conjugate.Then r∞=t∞and because r,t are not cyclical,|r|=|t|=µ.Let us denote R=(u k2v l2)∞.Let S=u k v l and suppose|u k−1|>2|v|,|v k−1|>2|u|,k,l≥2and also k≤min k1,k2,l≤min(l1,l2).It is clear that such S exist and is a subword of booth r and t.Then due to corollary1all occurrences of S in W are shifted by distance divisible byµ–period of R.It means that any occurrence of S can be extended to occurrence of r as well as to occurrence of t.Hence there exists an occurrences of r=u n1Sv m1and t=u n2Sv m2in W with common part S.If r=t,then n1=n2because|s|=|t|.Without loss of generality we can suppose that n1<n2.In this case m1>m2.Word t is shifted to the left from the word r on the distance d=|u n1|−|u n2|=|v n2|−|v m1|.Consider a unionωof r and t.Thenω=er=ft,|e|=|f|=d.Note that r2is a subword of t∞=W,occurrence of v m1(which is end of r)precedes an occurrence of r.Because|v m2|>d, e=v m2−m1.Similarly f=u n1−n2.From other handωcan be also obtained by extending the subword S of W to the left on the dis-tance|v max(m1,m2)|and to the right on the distance|u max(n1,n2)|andω=u max(n1,n2)v max(m1,m2)= u n1−n2r=tv m2−m1.Hence u n1−n2=v m2−m1;n1=n2;m1=m2.It follows that u,v are powers of the same word s.Because u=v one of this powers is greater than1and booth u and v can not be non cyclic words.But this contradicts to their initial choice.3.Regular Words and Lie bracketsWe shall extend the relation≺by deﬁning the following£-relation(“Ufnarovsky order”): f£g,if,for any two right superwords W1,W2,such that W2(a,b)≻W1(a,b),when ever b≻a, the inequality W2(g,f)≻W1(g,f)holds.This condition is well deﬁned and equivalent to following:f£g iﬀf∞/2≻g∞/2(i.e.,f m≻g n,for some m and n).It is clear that if f≻g, then f£g.The relation£is a linear ordering on the following set of equivalence classes:f∼g,if for some s,f=s l,g=s k.Let us note that eachﬁnite word u uniquely corresponds to the right superword u∞.To equivalent words correspond the same superwords.The relation£corresponds to the relation ≻on the set of superwords.It is known([4],[6])that:A word u is called regular,if one of the following equivalent conditions holds:a)u word is greater all its cyclic conjugates:If u1u2=u,then u≻u2u1.b)If u1u2=u,then u£u2.c)If u1u2=u,then u1£u.A word u is called semi-regular in the following case:If u=u1u2,then,either u≻u2,or u2is a beginning of u.(An equivalent deﬁnition can be obtained if the relation¡is replaced by the relation¢in the deﬁnition of a regular word.)Every semi-regular word is a power of a regular one.It is well-known that every regular word u deﬁnes the unique bracket arrangement[u]such that after opening all Lie brackets u will be a highest term in this expression.Moreover,monomials of such type form a basis in the free Lie algebra(so called Hall–Shirshov basis)(see[2],[6]). We shall need some technical statements:Lemma4([4]).Suppose|u k|£|v2|and u k is a subword of v∞.Then there exists S′cyclically conjugate to S,such that u=(S′)m and v=(S2)n.If,moreover,the initial symbols of u and v are at a distance divisible by|S|in v∞,then S=S′.Corollary2.Let u£v be semi-regular words.Then,for suﬃciently large k and l,the words u k v l are regular andu k1v l1£u k2v l2for k1>k2.Proof.Letδbe a cyclic conjugate of u k v l.It is clear thatδ¤u k v l,we only need to prove inequalityδ=u k v l.In order to do this,we need only to show that u k v l is not cyclic word,but it follows from the proposition5.The next lemma follows from Lemma4and Corollary2.Lemma 5.Let k i>|d|,l i>|u|,for i=1,2.Then u k1d l1and u k2d l2are not cyclically conjugate,provided that u£d and u,d are not conjugate to proper powers of the same word.4.Words in automata algebrasByΦ x1,...,x s will be denoted the free associativeΦ-algebra with generators x1,...,x s.By A a1,...,a s will be denoted an arbitraryΦ-algebra with aﬁxed set of generators a1,...,a s.A word or a monomial from the set of generators M is an arbitrary product of elements in M. The set of all words constitutes a semigroup,which will be denoted by Wd M .The order a1≺···≺a s generates the lexicographic order on the set of words:The grater of two words is the one whoseﬁrst letter is greater;if theﬁrst symbols coincide,then the second letter are compared,then the third letters and so on.Two words are incomparable,only if one of them is initial in the other.By a word in an algebra we understand a nonzero word from its generators{a i}.We can-not speak about the value of a superword in an algebra,but can speak about its equality or nonequality to zero(and,in some cases,about linear dependence).A superword W is called zero superword,if it has aﬁnite zero subword,and it is called a nonzero superword,if it has noﬁnite zero subwords.An algebra A is called monomial,if it has a base of deﬁning relations of the type c=0,where c is a word from a1,...,a s.Obviously,a monomial algebra is a semigroup algebra(it coincides with the semigroup algebra over the semigroup of its words).4.1.Automata algebras.First we recall some well known deﬁnitions from[4].Suppose we are given an alphabet(i.e.,aﬁnite set)X.Byﬁnite automaton(FA)with the alphabet X of input symbols we shall understand an oriented graph G,whose edges are marked with the letters from X.One of the vertices of this graph is marked as initial,and some vertices are marked asﬁnal.A word w in the alphabet X is called accepted by aﬁnite automaton,if there exists a path in the graph,which begins at the initial vertex andﬁnishes in someﬁnal vertex,such that marks on the path edges in the order of passage constitute the word w.By a language in the alphabet X we understand some subset in the set of all words(chains)in X.A language L is called regular or automata,if there exists aﬁnite automaton which accepts all words from L and only them.An automaton is called deterministic,if all edges,which start from one vertex are marked by diﬀerent letters(and there are no edges,marked by the empty chain).If we reject such restriction and also allow edges,marked by the empty chain,then we shall come to the notion of a non-deterministicﬁnite automaton.Also we can allow an automaton to have several initial vertices.The following result from the theory ofﬁnite automata is well known:For each non-deterministic FA there exists a deterministic FA,which accepts the same set of words(i.e.the same language).It will be convenient for us to consider the class of FA,such that all vertices are initial and ﬁnal simultaneously.The reason of this is that the language of nonzero words in a monomial algebra has the following property:each subword of a word belonging to the language,also belongs to it.Suppose throughout that G is the graph of a deterministic FA,v is a vertex of G,and w is a word.If the corresponding path C starting from v exist in G,then one can deﬁne the vertex vw terminal vertex for C.Let A be a monomial algebra(not necessaryﬁnitely deﬁned).A is called an automata algebra, if the set of all of its nonzero words from A generators is a regular language.Obviously,a monomial algebra is an automata algebra,only if the set of its nonzero words is the set of all subwords of words of some regular language.It is known that every automata algebra can be given by a certain deterministic graph,and that everyﬁnitely deﬁned monomial algebra is automata([4],[6]).The Hilbert series for an automata algebra is rational(Proposition5.9[4]).An automata algebra has exponential growth if and only if G has two cycles C1and C2with common vertexv,such that the corresponding words w1,w2(we read them starting from v)are not powers of the same word.In this case the words w1and w2generate a free2-generated associative algebra. If there are no such cycles,A has a polynomial growth.No intermediate growth is possible. The following theorem is the aim of this section:Theorem3.Let A= a1,...,a n be an automata algebra of exponential growth.Then the Lie algebra A∼contains a free2-generator subalgebra.We continue to assume that G is the graph of a deterministic FA.Call a semi-regular word u well–based if it written on a certain cycle C with an initial vertex v;i.e.vu=v.Two semi–regular words u1and u2are pair–wisely well–based if u1and u2are written on cycles C1and C2 with a common initial vertex v and vC1=vC2=v.In this case,for any word W(a,b)the word W(u1,u2)=0in A;in particular u k11u k22=0.Main Lemma.The graph G contains two regular pairwise well–based words u=v. Deduction of Theorem3from the Main Lemma.We may always assume that u£v.Let a≻b and w a regular word.Then(see[4])w(u,v)also is a regular wold.For every regular word u we can choose a unique presentation u=u1u2with regular u1and regular u2of maximal length.In this case[u]=[[u1],[u2]](see[6]).Therefore,w(u,v)can be obtained by setting[u]→a,[v]→b to the word with brackets[w].Since u and v are well-based,for every word R(a,b),one has R(u,v)=0.Let[u],[v]be the results of the regular arrangement of the brackets for u and v respectively.Then[u]=0,[v]=0. Thus we have constructed a one-to-one correspondence between the Hall basis of a Lie algebra, generated by[u],[v]and the Hall basis of a free2-generated Lie algebra with generators a,b. The theorem follows.P4.2.Proof of the Main Lemma.Corollary2implies the following:Proposition8.Suppose the graph G contains two ordered(in the sense of the operation£) pairwise well-based words.Then G contains also two regular pairwise well-based words.It is suﬃcient toﬁnd two ordered semi-regular pairwise well-based words,i.e.with common ﬁnal and initial vertices.For that it is enough to prove the existence of a suﬃciently large number of well-based ordered semi-regular words.In this case,inﬁnitely many of them will have a common initial vertex,hence pairwise well–based,and the main lemma follows.Lemma6.Let u1be a well-based word and u2a cyclically conjugate word.Then u2is also well-based.Proof.Suppose u1=w1w2,u2=w2w1and v is a base vertex of u1.Then v′=vw1is a base vertex of u2.Indeed,v′u2=vw1(w2w1)=v(w1w2)w1=vw1=v′. Corollary3.If u is well-based,then semi-regular word conjugate to u is also well-based.Let u and d be ordered pairwise well-based words(necessarily not semi-regular).Then,for every w(a,b),the word w(u,d)is non-zero.In particular,u k d l are non-zero for all k,l.Now Lemma5,together with the fact that every non–cyclic word uniquely corresponds to a cyclically conjugated regular word,implies that there are inﬁnitely many well–based words. Inﬁnitely many of them will have the same initial vertex and so will be pairwise well based; hence the Main Lemma follows.5.Group theoretical applicationsId(S)denotes the ideal,generated by the set S.Lemma 7.Suppose a and b are homogenous elements of a graded associative algebra A ,such that the subalgebra generated by a,b is free associative algebra with free generators a,b .Let a ′(resp.b ′)be a linear combination of elements in A with degrees strictly greater than the degree of a (resp.b ).Let ˜a =a +a ′,˜b =b +b ′.Then the algebra generated by ˜a ,˜b is a free associative with free generators ˜a ,˜b .This lemma follows from the fact that for every polynomial h (u,v )with non-zero minimal component h ′(u,v ),the minimal component of h (˜a ,˜b )is h ′(a,b )=0.We call an algebra homogenous if all its deﬁning relations are homogenous respect to the set of generators.Let A be a homogenous algebra,and J be an idealof A ,generated by elements of degree ≥1.We call such algebra good .If A/J ≡k and nJ n =0.Every monomial algebra is good.For any x ∈A,the image of x in A/J n is not zero for some n ,so A can be embedded into the projective limit lim ←−A/J n .Lemma 8.Let B be a good homogenous algebra,such that 1+a and 1+b are invertible,a,b ∈J,and the elements a and b are free generators of a free associative subalgebra C of B .Then the group generated by 1+a and 1+b is free.Remark.Note that the pair of two diﬀerent pairwise well–based words in a monomial algebra generates a free associative subalgebra.Proof.Suppose W (1+a,1+b )=1for some non-trivial word W (x,y )=1in the free group.Consider free algebra k x,y and its localization by 1+x,1+y .Then W (1+x,1+y )=1,and,for some n 0=n 0(W ),W (1+¯x ,1+¯y )=1in π(k x,y )=k x,y /Id(x,y )n for all n ≥n 0.In each such image,the elements 1+x and 1+y are invertible,so there is no need for localization.On the other hand,because A is good homogenous,the image of J m ∩C under isomorphism φgenerated by a →x,b →y lies in Id(x,y )n 0,and1=π(φ(W (1+a,1+b )))=W (1+¯x ,1+¯y )=1.Contradiction. Let u and v be two pairwise well based words.They have canonical Lie bracket arrangement;let [u ]and [v ]be corresponding Lie elements (obtained via opening the Lie brackets).Notice that[u ]=u +lexicographically smaller terms ,[v ]=v +lexicographically smaller terms .Hence we have followingLemma 9.Let u and v be two pairwise well based words,and [u ]and [v ]be the corresponding Lie elements (obtained via opening Lie brackets).Then [u ],[v ]generate as a free generators 2-generated free associative algebra (and also free Lie algebra via commutator operation).For n ≥1,consider the monomial algebra A n +2with generators x 1,...,x n +2(see the In-troduction).Adjoint a unit A ′n +2=A n +2∪{1}.The elements ¯xi :=1+x i have inverses ¯x −1i :=1−x i .Consider the group A #n +2generated by the elements 1+x i .Consider the conﬁg-uration of brackets in the generators of the free subalgebra in the Lie algebra A ∼n +2and writethe correspondent Lie elements in the group A #n +2.Lemmas 7and 8,9imply that the subgroupA #n +2generated by these two elements will be free.It is clear from the construction that all left-normalized commutators of length k in A #n +2which consists of fewer that k diﬀerent symbols from {¯x 1,...,¯x n +2},are trivial.Hence,Theorem 2follows.References[1]S.A.Amitsur,L.W.Small:Aﬃne algebras with polynomial identities,Suppl.ai Rendiconti del Circolo Mat.di Palermo,Serie2,31(1992),9-43.[2]Bahturin Yu.,A.:Identities in Lie algebras.M.:Nauka,1985.,pages448(Russian).Engl.transl.(by Bakh-turin):Identical relations in Lie algebras.VNU Science Press,b.v.,Utrecht,1987.x+309pp.[3]Belov A.:About height m.in Algebra,1995,vol.23,N9,p.3551–3553.[4]A.Belov,V.Borisenko and tyshev:Monomial algebras.Algebra4,J,Math.Sci.(New York)87(1997),3463-3575.[5]M.Gromov:Entropy and Isoperimetry for Linear and non-Linear Group Actions,preprint.[6]V.Ufnarovskij:Combinatorial and asymptotic methods in algebra.Algebra,VI,1–196,Encyclopaedia Math.Sci.,57,Springer,Berlin,(1995)。

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Each table must be in a separate file, and the files should be named by Table # (i.e., Table 1, Table 2, etc.).Electronic artwork. In general, each file must contain a single figure and be named by Figure # (i.e., Figure 1, Figure 2, etc.). All figures Submission should be original. It implies that the image or parts thereof have not been published elsewhere. Figures must be in TIFF or JPG. TIFF (or JPEG): Bitmapped (pure black & white pixels) line drawings and monochrome keep to a minimum of 1000 dpi. TIFF (or JPEG): Combinations bitmapped line/half-tone (color or grayscale) keep to a minimum of 300 dpi.Figure captions. All figures have a caption (with source information, when required). The captions have been in-cluded at the end of the text files. Please provide captions and illustrations separately. Please contains detailed expla-nation of all symbols and abbreviations.All manuscripts submitted to JNO must be submittedsolely to this journal, may not have been published in anoth-er publication of any type, professional or lay, and become the property of the publisher. No published material may be reproduced or published elsewhere without the written per-mission of the publisher and the author. The authors need to declare the above in the cover letter for the submission.。

新GRE阅读长难句解析汇总

新GRE阅读长难句解析汇总如何备考新GRE阅读呢？我今日整理了新GRE阅读长难句解析，快来一起学习吧，下面我就和大家共享，来观赏一下吧。

新GRE阅读长难句解析1. The historian Frederick J. Turner wrote in the 1890s that the agrarian discontent that had been developing steadily in the United States since about 1870 had been precipitated by the closing of the internal frontier——that is,the depletion of available new land neededfor further expansion of the American farming system. (4)史学家弗雷德里克.杰.特纳(Frederick J.Turner)于十九世纪九十年月著述道，US约自18世纪70年月以来始终在持续不断进展的农夫不满，由于国内边远地区(internal frontier)的封闭而更趋加剧——亦即是说，US农业系统进一步扩展所必需的可资利用的新土地几近耗竭。

难句类型：简单修饰解释：典型的句子套句子。

主句是F.J.T. wrote that，宾语从句中主干是the agrarian discontent had been precipitated by the closing of the internal frontier,在从句的主语the agrarian discontent后面又来了一个定语从句。

意群训练：The historian Frederick J. Turner wrote in the 1890s that the agrarian discontent that had been developing steadily in the United States since about 1870 had been precipitated by the closing of the internal frontier——that is,the depletion of available new land neededfor further expansion of the American farming system.2. In the early 1950s，historians who studied preindustrial Europe (which we may define here as Europe in the period from roughly 1300 to1800) began, for the first time in large numbers,to investigate more of the preindustrial European population than the 2 or 3 percent who comprised the political and social elite:thekings,generals,judges,nobles,bishops,and local magnates who had hitherto usually filled history books. (4)难句类型：简单修饰、插入语译文：二十世纪五十年月早期，讨论前工业化时代欧洲(此处我们可将其界定为约自1300年至1800年这一时期的欧洲)的史学家，首次以众多的人数(杨鹏的书中：第一次以大量的数据)，开头调查前工业化时代欧洲人口中的大多数，而非那些构成了政治与社会精英阶层的百分之二或三的人口，即国王、将军、法官、贵族、主教、以及地方上的达官显贵，而正是这部分人始终到那时为止普遍充斥于史学著作。

研究生公共英语教材阅读B第3、4、10、11、14课文原文及翻译

Unite 3 Doctor’s Dilemma: Treat or Let Die?Abigail Trafford1. Medical advances in wonder drugs, daring surgical procedures, radiation therapies, and intensive-care units have brought new life to thousands of people. Yet to many of them, modern medicine has become a double-edged sword.2. Doctor’s power to treat with an array of space-age techniques has outstripped the body’s capacity to heal. More medical problems can be treated, but for many patients, there is little hope of recovery. Even the fundamental distinction between life and death has been blurred.3. Many Americans are caught in medical limbo, as was the South Korean boxer Duk Koo Kim, who was kept alive by artificial means after he had been knocked unconscious in a fight and his brain ceased to function. With the permission of his family, doctors in Las Vegas disconnected the life-support machines and death quickly followed.4. In the wake of technology’s advances in medicine, a heated debate is taking place in hospitals and nursing homes across the country --- over whether survival or quality of life is the paramount goal of medicine.5. “It gets down to what medicine is all about, ” says Daniel Callahan, director of the Institute of Society, Ethics, and the Life Sciences in Hastings-on-Hudson, New York. “Is it really to save a life? Or is the larger goal the welfare of the patient?”6. Doctors, patients, relatives, and often the courts are being forced to make hard choices in medicine. Most often it is at the two extremes of life that these difficultyethical questions arise --- at the beginning for the very sick newborn and at the end for the dying patient.7. The dilemma posed by modern medical technology has created the growing new discipline or bioethics. Many of the country’s 127 medical s chools now offer courses in medical ethics, a field virtually ignored only a decade ago. Many hospitals have chaplains, philosophers, psychiatrists, and social workers on the staff to help patients make crucial decisions, and one in twenty institutions has a special ethics committee to resolve difficult cases.Death and Dying8. Of all the patients in intensive-care units who are at risk of dying, some 20 percent present difficult ethical choices --- whether to keep trying to save the life or to pull back and let the patient die. In many units, decisions regarding life-sustaining care are made about three times a week.9. Even the definition of death has been changed. Now that the heart-lung machine can take over the functions of breathing and pumping blood, death no longer always comes with the patient’s “last gasp” or when the heart stops beating. Thirty-one states and the District of Columbia have passed brain-death statutes that identify death as when the whole brain ceases to function.10. More than a do zen states recognize “living wills” in which the patients leave instructions to doctors not to prolong life by feeding them intravenously or by other methods if their illness becomes hopeless. A survey of California doctors showed that 20 to 30 percent were following instructions of such wills. Meanwhile, the hospicemovement, which its emphasis on providing comfort --- not cure --- to the dying patient, has gained momentum in many areas.11. Despite progress in society’s understanding of death and dying, t heory issues remain. Example: A woman, 87, afflicted by the nervous-system disorder of Parkinson’s disease, has a massive stroke and is found unconscious by her family. Their choices are to put her in a nursing home until she dies or to send her to a medical center for diagnosis and possible treatment. The family opts for a teaching hospital in New York city. Tests show the woman’s stroke resulted from a blood clot that is curable with surgery. After the operation, she says to her family: “Why did you bring me back to this agony?” Her health continues to worsen, and two years later she dies.12. On the other hand, doctors say prognosis is often uncertain and that patients, just because they are old and disabled, should not be denied life-saving therapy. Ethicists also fear that under the guise of medical decision not to treat certain patients, death may become too easy, pushing the country toward the acceptance of euthanasia.13. For some people, the agony of watching high-technology dying is too great. Earlier this year, Woodrow Wilson Collums, a retired dairyman from Poteet, Texas, was put on probation for the mercy killing of his older brother Jim, who lay hopeless in his bed at a nursing home, a victim of severe senility resul ting from Alzheimer’s disease. After the killing, the victim’s widow said: “I think God, Jim’s out of his misery. I hate to think it had to be done the way it was done, but I understand it. ”Crisis in Newborn Care14. At the other end of the life span, technology has so revolutionized newborn carethat it is no longer clear when human life is viable outside the womb. Newborn care has got huge progress, so it is absolutely clear that human being can survive independently outside the womb. Twenty-five years ago, infants weighting less than three and one-half pounds rarely survived. The current survival rate is 70 percent, and doctors are “salvaging” some babies that weigh only one and one-half pounds. Tremendous progress has been made in treating birth deformities such as spina bifida. Just ten years ago, only 5 percent of infants with transposition of the great arteries --- the congenital heart defect most commonly found in newborns --- survived. Today, 50 percent live.15. Yet, for many infants who owe their lives to new medical advances, survival has come at a price. A significant number emerge with permanent physical and mental handicaps.16. “The question of treatment and nontreatment of seriously ill newborns is not a single one,”says Thomas Murray of the Hastings Center. “But I feel strongly that retardation or the fact that someone is going to be less than perfect is not good grounds for allowing an infant to die.”17. For many parents, however, the experience of having a sick newborn becomes a lingering nightmare. Two years ago, an Atlanta mother gave birth to a baby suffering from Down’s Syndrome, a form of mental retardation; the child also had blocked intestines. The doctors rejected the parents’ plea not to operate, and today the child, severely retarded, still suffers intestinal problems.18. “Every time Melanie has a bowel movement, she cries,” explains her mother.“She’s not able to take care of herself, and we won’t live forever. I wanted to save her from sorrow, pain, and suffering. I don’t understand the emphasis on life at all costs, and I’m very angry at the doctors and the hospital. Who will take care of Melanie after we’re gone? Where will you doctors be then?”Changing Standards19. The choices posed by modern technology have profoundly changed the practice of medicine. Until now, most doctors have been activists, trained to use all the tools in their medical arsenals to treat disease. The current trend is toward nontreatment as doctors grapple with questions not just of who should get care but when to take therapy away.20. Always in the background is the threat of legal action. In August, two California doctors were charged with murdering a comatose patient by allegedly disconnecting the respirator and cutting off food and water. In 1981, a Massachusetts nurse was charged with murdering a cancer patient with massive doses of morphine but was subsequently acquitted.21. Between lawsuits, government regulations, and patients’ rights, many doctors feel they are under siege. Modern technology actually has limited their ability to make choices. More recently, these actions are resolved by committees.Public Policy22. In recent years, the debate on medical ethics has moved to the level of national policy. “It’s just beginning to hit us that we don’t have unlimited resources,” says Washington Hospital Center’s Dr. Lynch. “You can’t talk about ethics without talkingethics without talking about money.”23. Since 1972. Americans have enjoyed unlimited access to a taxpayer-supported, kidney dialysis program that offers life-prolonging therapy to all patients with kidney failure. To a number of police analysts, the program has grown out of control --- to a $1.4billion operation supporting 61,000 patients. The majority are over 50, and about a quarter have other illness, such as cancer or heart disease, conditions that could exclude them from dialysis in other countries.24. Some hospitals are pulling back from certain lifesaving treatment. Massachusetts General Hospital, for example, has decided not perform heart transplants on the ground that the high costs of providing such surgery help too few patients. Burn units --- through extremely effective --- also provide very expensive therapy for very few patients.25. As medical scientists push back the frontiers of therapy, the moral dilemma will continue to grow for doctors and patients alike, making the choice of to treat the basic question in modern medicine.1. 在特效药、风险性手术进程、放疗法以及特护病房方面的医学进展已为数千人带来新生。

TheLongTail(长尾理论)完整中译版

一、“长尾”的由来及含义：长尾理论这个概念是Wired 杂志主编Chris Anderson在2004 年提出。

最简单的例子: 在一个XY 的坐标系里面, y 对应销售收入, x 对应同一产业中不同品牌的产品或服务. 一般会出现名列前茅的几个品牌占据大部分的部分, 其他无数的小品牌占据小部分.举例来说，我们常用的汉字实际上不多，但因出现频次高，所以这些为数不多的汉字占据了上图广大的红区；绝大部分的汉字难得一用，它们就属于那长长的蓝尾。

二、成功的“长尾”案例：1、Google 是一个最典型的“长尾”公司，其成长历程就是把广告商和出版商的“长尾”商业化的过程。

数以百万计的小企业和个人，此前他们从未打过广告，或从没大规模地打过广告。

他们小得让广告商不屑，甚至连他们自己都不曾想过可以打广告。

但Google 的AdSense 把广告这一门槛降下来了：广告不再高不可攀，它是自助的，价廉的，谁都可以做的；另一方面，对成千上万的Blog 站点和小规模的商业网站来说，在自己的站点放上广告已成举手之劳。

Google 目前有一半的生意来自这些小网站而不是搜索结果中放置的广告。

数以百万计的中小企业代表了一个巨大的长尾广告市场。

这条长尾能有多长，恐怕谁也无法预知。

2、亚马逊：一个前亚马逊公司员工精辟地概述了公司的“长尾”本质：现在我们所卖的那些过去根本卖不动的书比我们现在所卖的那些过去可以卖得动的书多得多。

此外还有很多，诸如维基百科、Netflix 等等。

《长尾》一直在深刻地影响着全球各地互联网业的发展。

他所提出的推动型模式与拉动型模式的结合，广泛性与个性化的统一，已经成为网络产品设计开发的一个重要策略。

非常感谢”雷声大雨点大“和”拙尘“的辛勤工作，使广大中文读者可以完整领略长尾的妙处。

如果您觉得有价值，请把本文推荐给您的朋友、同事。

希望不久的将来，在Baidu 长尾时，不要在前十页中都找不到长尾的中文译文。

:)有兴趣的话可以加入我们，一起把有价值的外文资料带给中文读者。

GRE阅读长难句重磅解析

GRE阅读长难句重磅解析GRE阅读制胜法则,长句、难句重磅解析，盼望可以关心到大家，下面我就和大家共享，来观赏一下吧。

GRE阅读制胜法则：长句、难句重磅解析1. Although these molecules allow radiation at visible wavelengths, where most of the energy of sunlight is concentrated,to pass through,they absorb some of the longer-wavelength,infrared emissions radiated from the Earths surface,radiation that would otherwise be transmitted back into space. (4)虽然这些分子允许可见波长(visible wavelength)的辐射——阳光的绝大部分能量就汇合于此——不受阻挡地穿透，但它们却会汲取某些较长波长(longer- wavelength)，亦即从地球表面辐射出的红外放射(infrared emission)，这种辐射若不是二氧化碳的原因就会被重新输送回太空。

难句类型：简单修饰、抽象词解释：前面的分句中有一个不算很长的插入语，但是由于它插入的位置正好在固定搭配allow something to do somthing中间，将allow 和to分得很远，所以读起来让人感觉很不舒适。

后面的分句中的最终一行radiation that would otherwise be transmitted back into space是其前面的infrared emissions的同位语。

其中的otherwise是副词作状语，表示假如后面的分句所说的they absorb some of the longer-waverlength, infrared emissions不发生时的后果。

Mertens' Proof of Mertens' Theorem

3 Mertens’ Proof 3.1 A Sketch of the Proof . . . . . . . . . 3.2 Euler-Maclurin and Stirling . . . . . 3.3 The First Step of Mertens’ Proof . 3.4 Mertens’ Use of Partial Summation 3.5 Proof the the Grossehilfsatz 1 . . . . 3.6 The Grossehilfsatz 2 . . . . . . . . . 3.6.1 Merten’s proof . . . . . . . . 3.6.2 Modern Proof . . . . . . . . 3.7 The Formula for the Constant H . . 3.8 Completion of the Proof . . . . . . .
This is the wonderful identity which, today, we write [6], [8], [9]:
∞ 2
1 1− 1 p1+ρ
=1+
1 21+ρ
+
1 31+ρ
+
1 41+ρ
+··· ,
(1.1.1)
Here ρ > 0 and the product on the left is taken over all primes p 2, while the right hand side is the famous Riemann zeta function, ζ (1 + ρ). The modern statement is nice, but does not have the sense of wonder that Euler’s statement carries. Yes, it is not rigorous, but it is beautiful. Euler’s memoir is replete with extraordinary identities relating inﬁnite products and series of primes, but our interest is in his Theorema 19: “Summa seriei reciprocae numerorum primorum 1 1 1 1 1 1 + + + + + + etc. 2 3 5 7 11 13 est inﬁnite magna, inﬁnities tamen minor quam summa seriei harmonicae 1 1 1 1 + + + + etc. 2 3 4 5 Atque illius summa est huius summae quasi logarithmus.” 1+ We translate this as our ﬁrst formal theorem. Theorem 1. The sum of the reciprocals of the prime numbers 1 1 1 1 1 1 + + + + + + etc. 2 3 5 7 11 13 is inﬁnitely great but is inﬁnitely times less than the sum of the harmonic series 1 1 1 1 + + + + etc. 2 3 4 5 And the sum of the former is as the logarithm of the sum of the latter. 1+ 2

2022-2023学年辽宁省抚顺市重点高中六校协作体高一下学期期中考试英语试题

2022-2023学年辽宁省抚顺市重点高中六校协作体高一下学期期中考试英语试题1. Four Modern Poets Who Are Making a DifferenceShonto BegayShonto Begay was born in 1954. He is an artist, writer, poet and filmmaker. His experiences of growing up in Native American culture and spending 10 years in the National Park Service as a ranger (护林员) stir up all of his art and writing. Begay’s works try to keep the traditional culture alive while also expressing the realities and struggles of modern native Americans.Sarah KaySarah Kay was born in 1988 and lives in New York City. She is well known for her spoken word poetry. Kay has published four poetry collections. Her works have inspired people of all ages to be the best they can be. Young and extremely passionate, she is out there making the world a better place by doing charity work.Cathy Park HongCathy Park Hong is a Korean-American poet and professor. She has published three volumes (卷) of poetry: Translating Mo’um, Dance Dance Revolution, and Engine Empire. She is best known for her style of “code switching”—the use of mixed languages. She does this to create interactive possibilities and social experimentation.Amber TamblynAmber Tamblyn was born in 1983 and is an American actress and poet. Her work in both television and movies has influenced her poetry. Apart from two self-published books of poetry, she holds three poetry collections: Free Stallion, Bang Ditto, and Dark Sparkler, the last of which explores the lives and deaths of child star actresses. She also co-founded the Write Now Poetry Society which is devoted to creating poetry programs.1. What influenced Shonto Begay’s works?A．Native American culture.B．Modern young Americans.C．His work as a filmmaker.D．His childhood in the National Park service.2. Who is devoted to the public service?A．Shonto Begay. B．Sarah Kay.C．Cathy Park Hong. D．Amber Tamblyn.3. Which of the following books is about child stars?C．Dark Sparkler . D．Engine Empire .2. About 230 whales beached themselves on the western part of the Australian island of Tasmania. It is the second such incident in recent days. At least half are alive and animal officials are working to rescue them. They are trying to understand why the whales got stuck.The whales are stuck on Ocean Beach near Macquarie Harbor in the western part of Tasmania. About two years ago, 470 whales became stuck on sand in the same area. That time, about 100 whales were saved but the rest died.The whales are pilot whales, which weigh between 1,000 and 2,000 kilograms and measure over 6 meters. Pilot whales are the species that most often get stuck in large numbers. Some scientists believe it is because they travel in large groups and follow a leader.A team from the Marine Conservation Program was heading to the area to help the whales. The harbor does not have deep water and its entrance is known by locals as Hell’s Gate.Linton Kringle is a salmon farmer who once helped with the rescue in 2020. He told the Australian Broadcasting Corporation this year’s work would be more difficult. That time, he said that the whales were in the harbor, so people could reach them easily on boats. This time, they are outside of the safe area, on a beach, whic h means it will be harder to get boats near them. “It’s too shallow (浅的) and too rough. My thoughts would be to try to get them onto a vehicle if we can’t get them swim out,” Kringle added.Vanessa Pirotta is a wildlife scientist who studies animals like whales and dolphins. She said the fact that the whales have gotten stuck around the same place during the same part of the year, suggesting there might be something environmental here.1. What puzzles animal officials most?A．The number of beached whales. B．The cause of whales’ beaching.C．The times of whales’ beaching.D．The way to rescue beached whales.2. What can make pilot whales suffer beaching in large numbers?A．Their group travel habits. B．Their heavy weights.C．Their poor eyesight. D．Their long bodies.3. Why does Linton Kringle consider this year’s rescue work to be more difficult?A．There aren’t so many rescue workers.B．The entrance is known as Hell’s Gate.C．Making them swim back to the sea is almost impossible.D．The whales in danger are too far away to reach by boat.A．To introduce pilot whales.B．To call on people to protect whales.C．To recommend a suitable rescue method.D．To tell an incident about whales’ beaching.3. It is very important to stay healthy. A lot of people aren’t aware of many benefits of daily exercise and only begin to do so when health complications (并发症) happen. Why then is it important to stay healthy?Regular exercises will help you bum extra body fat and keep your body vigorous. There are many health complications related to having too much body fat. Heart diseases that are often very deadly are mostly a result of extra body fat.Exercising keeps you sharp at all times. Keeping your body active also improves the functioning of the brain. Any kind of exercise is terrific. Most sports are excellent for keeping the body in good health, such as, basketball, swimming, bicycling, running and so on. Sports are not only good for your body, but they are enjoyable and interesting, too. There is a reason why most athletes are usually happy. Maybe the reason why you stay angry easily all the time is due to lack of exercise. Fitness exercises are ideal for strength building. Regular daily exercise will build body muscles and improve your physical strength. They are also important in strengthening your back for good postures. People generally become less active as they grow older, especially when they have small kids that expect to play with them. Daily fitness will help you get the much needed strength and activity to keep up with the youthful energy of your children. Exercising also slows down the aging process by tightening your skin and increasing your body’s metabolic (新陈代谢的) rate.Exercising is very important and some of the results of not joining in daily fitness are dangerous. Having fitness goals that actually mean something to you will help you develop a controlled daily fitness and achieve your desired result. These are just some of the benefits of having a strong fitness program. Many people ignore them and just begin to suffer from lifestyle diseases in their later lives. Exercise is very significant.1. What can we learn from paragraph 1?A．Lots of people are too busy to exercise.B．Lots of people are used to staying quiet.C．Lots of people keep exercising when they are sick.D．Lots of people don’t realize the advantages of daily exercise.2. What does underlined word “vigorous” menu in paragraph 2?A．Fit. B．Thin. C．Young. D．Graceful.A．We can’t get ill.B．We will eat more food,C．We can become energetic. D．We will become beautiful.4. Which is the most suitable title for the text?A．What do people think about keeping fit? B．Why is exercise significant to keep fit?C．What kind of exercise are good for health? D．Why do people go to gymnasium toexercise?4. A seven-word sentence was discovered on a 3,700-year-old comb (梳子),which is likely the oldest known sentence written in letters,according to a new study. The sentence is in Canaanite (迦南语),which is the source (来源) of later letter systems in European languages.The comb was first discovered in 2016 in Israeli. However,it wasn’t until 2021 that a researcher from the Hebrew University of Jerusalem first noticed the words when checking the photo of the comb. The researchers were not able to directly date the comb. They believed it dated from around 1700 BC based on comparison of the letters with those on pottery (陶器) with a known age. Garfinkel,a professor at the Hebre w University of Jerusalem,said,“The Canaanites invented the letters. Nowadays every person in the world can read and write using the letter system. This is really one of the most important achievements of humankind. When you are writing in English,you’rer eally using Canaanite.”Small groups of Canaanite letters discovered on broken pieces of pottery before did not leave much room for further research on the lives of the Canaanites. But this find of a sentence written in the first letter-based language opens up the debate about the history during the ancient time period. Garfinkel said,“The sentence was found on an ivory comb in the ancient city’s palace and temple area,which could suggest that only wealthy men were able to read and write.” The earliest writ ing system started around 5,000 years ago,which relied on hundreds of pictographs (象形文字) to represent words,ideas and sounds. Canaanites were the first to use letters in their writing system. “It shows that even in the most ancient times there were full sentences,which further suggests that Canaanites stood out among the early civilizations in their use of the written word.”1. What do we know about Canaanite?A．It has a history of 3,700 years. B．It is the oldest known language.C．Its sentences are in many languages. D．Its letters are still in use at present.2. How was the complete sentence in Canaanite found?A．It was discovered on some pottery. B．It was identified on a comb in 2016.C．It was recognized when dating the comb. D．It was recognized on a photo of the comb.3. What does paragraph 3 mainly focus on?B．The rules of using Canaanite letters.C．The problem in identifying the Canaanite sentence.D．The explanation of the discovered Canaanite sentence.4. What does Garfinkel think of the new discovery in the last paragraph?A．Useless. B．Meaningful. C．Confusing. D．Unacceptable. 5. If you’re worried about being a self-centered person, that concern shows you’ve already taken an important step towards change. 1Focus on listening instead of talking. Give others your full attention and really hear them out. Self-centered people often make conversations center around themselves and they tend to get bored when the focus isn’t on them. 2 Give others a chance to speak their minds and do your best to show that you’re really listening.3 Imagining yourself in their situation helps you understand them better. If your friend is telling you about something that happened to him or her and you just don’t feel engaged (吸引住), it can help to imagine how you would feel in his or her situation. Try asking yourself how you would feel and what you would need if you were him or her. Then, keep those things in mind when you respond to your friend.Use fewer “I” and “me” statements. Fight the strong wish to talk about yourself in every conversation. It’s an easy habit to get into, but you can’t focus on anyone else if you’re always talking about yourself. Try to actively reduce the number of “I” and “me” statements you make in daily conversation. Studies show that talking about yourself less often can make you happier and healthier. 4Learn how to compromise (妥协). 5 Compromising means choosing to believe another person’s needs and desires are just as important as your own. Instead of demanding your way during a disagreement comes, try meeting the other person halfway so that each person gets some of their needs met.6. A 23-year-old woman just set a record for the fastest solo row across the Atlantic. Departing from the Canary Islands, Payne rowed 59 days, before arriving in Antigua. The _________ achievement was made for a charity. More than $14,200 was raised to help people _________.The challenge was described as one of the hardest ones on Earth. No _________ was allowed—all necessities must be brought along. But she was _________ enough to keep going without extra help.The last 8 miles was really _________ for Pa yne. It was that point where she knew she couldn’t give up and that she had to _________ it. Although 8 miles was _________ in the great plan of 3,000 miles, it felt like some of the longest for her. However, hunger, lack of sleep and the test of body and mind were _________ by sighting fantastic sea life and _________ the beauty of the sunrise and sunset. So the __________ was worthwhile.Payne rowed 15 hours per day alone in the Atlantic Ocean with strong winds. “In the last week, I thought I wasn’t going to break a __________ at all because the wind dropped and I was going nowhere,” she said. “There was one day when I rowed __________ for 18 hours, but I only rowed less than 10 miles. That was pretty__________. I could just feel the race record slipping a way.”But finally Payne __________ it with a strong will. She received the cheers of the crowd when approaching Antigua. She could hear the __________ from her family and friends waiting for her.1.A．remarkable B．artistic C．common D．industrial2.A．for fun B．for sale C．in charge D．in need3.A．error B．destination C．support D．profit4.A．strong B．tense C．ordinary D．generous5.A．free B．tough C．smooth D．brief6.A．measure B．finish C．celebrate D．extend7.A．something B．anything C．everything D．nothing8.A．interrupted B．matched C．balanced D．worsenedA．describing B．affecting C．imagining D．appreciating 10.A．struggle B．failure C．foundation D．tracking11.A．container B．commitment C．reform D．record12.A．globally B．carelessly C．determinedly D．instantly13.A．disappointing B．rewarding C．harmless D．formal14.A．lost B．deserted C．made D．refused15.A．regret B．screams C．permission D．complaints7. 阅读下面短文，在空白处填入1个适当的单词或括号内单词的正确形式。

kmp算法

the running time of the routine make_f is O(m).
Lemma. Let f be a non-negative integer valued function such that
f(1) = 0, and so that f is "continuous upwards", i.e., f(t+1) <= f(t) + 1.
starting array indices from 0 makes the code harder to follow than it
should be. To make it easier, we never use f[0]. */
int make_f(const char *target){
while loop can iterate more than once on given pass. For example,
if the target is abababb, then f(6)=4, f(f(6))=2, and f(f(f(6)))= 0.
The assignment is done 3 times when s = 6.) It is then easy to see that
Read stdin and find first occurence of target string in it. \n\
Print information about where it was found.\n\n"
int *f; /* Pointer to array containing values of failure function */

Great-Expectation远大前程

Great Expectations
CharlesDickens
(1812-1870) BY BOYS GROUP
Content:
Charles Dickens Literary Status Life
Bildungsroman (教育小说) Definition/Characteristics/Representatives
Great Expectations
Introduction
The protagonist Pip was once a poor orphan, but he lived a happy life with his brother-in-law Joe. Later, as soon as he met and fell in love with Estella, he always compared himself with her and felt ashamed of himself
Ⅱ Surroundings shape characters, both internally and externally.
THANKS
Married the daughter of a magazine owner/ Fallen love with a young actress/ Divorced his wife
Rose to his reputation/ public reading tour for money Died of a stroke in 1870 and buried at Westminster Abbey
Ⅲ When he realized money and social status is not the most

莎士比亚十四行诗第18首汉译

莎士比亚十四行诗第18首汉译1，朱湘译文我来比你作夏天，好不好？不，你比它更可爱，更温和：暮春的娇花有暴风侵扰，夏住在人间的时日不多：有时候天之目亮得太凌人，他的金容常被云霾掩蔽，有时因了意外，四季周行，今天的美明天已不再美丽：你的永存之夏却不黄萎，你的美丽亦将长寿万年，你不会死，死神无从夸嘴，因为你的名字入了诗篇：一天还有人活着，有眼睛，你的名字便将于此常新。

2，李霁野译文我来将你比作夏天吗?你比夏天更为可爱,更为温和:暴风摇落五月的柔嫩花芽,夏季的租赁期限要短得多:有的时候太阳照得太热,他的金色面孔常变阴暗;每种美有时都会凋零衰谢,由于机缘,或者由于自然变幻;但是你的永久夏季不会衰败,你的美也永远不会丧失;死亡不至夸口:你在他的阴影里徘徊, 当你在不朽的诗行中度日:——只要人还能呼吸，眼睛还能看望，这些诗行就会永存，使你万寿无疆。

3，梁宗岱译文我怎么能够把你来比作夏天？你不独比它可爱也比它温婉：狂风把五月宠爱的嫩蕊作践，夏天出赁的期限又未免太短：天上的眼睛有时照得太酷烈，它那炳耀的金颜又常遭掩蔽：被机缘或无常的天道所摧折，没有芳艳不终于雕残或销毁。

但是你的长夏永远不会雕落，也不会损失你这皎洁的红芳，或死神夸口你在他影里漂泊，当你在不朽的诗里与时同长。

只要一天有人类，或人有眼睛，这诗将长存，并且赐给你生命。

4，梁实秋译文我可能把你和夏天相比拟？你比夏天更可爱更温和：狂风会把五月的花苞吹落地，夏天也嫌太短促，匆匆而过：有时太阳照得太热，常常又遮暗他的金色的脸；美的事物总不免要凋落，偶然的，或是随自然变化而流转。

但是你的永恒之夏不会褪色；你不会失去你的俊美的仪容；死神不能夸说你在他的阴影里面走着，如果你在这不朽的诗句里获得了永生；只要人们能呼吸，眼睛能看东西，此诗就会不朽，使你永久生存下去。

5，屠岸译文能不能让我把你比拟作夏日？你可是更加温和，更加可爱：狂风会吹落五月的好花儿，夏季的生命又未免结束得太快：有时候苍天的巨眼照得太灼热，他那金彩的脸色也会被遮暗；每一样美呀，总会离开美而凋落，被时机或者自然的代谢所摧残；但是你永久的夏天决不会凋枯，你永远不会失去你美的仪态；死神夸不着你在他影子里踯躅，你将在不朽的诗中与时间同在；只要人类在呼吸，眼睛看得见，我这诗就活着，使你的生命绵延。

On the anomalous red giant branch of the globular cluster Omega Cen

a rXiv:as tr o-ph/52585v128Fe b25ON THE ANOMALOUS RED GIANT BRANCH OF THE GLOBULAR CLUSTER ωCENTAURI 1L.M.Freyhammer 2,3,4,M.Monelli 5,6,G.Bono 5,P.Cunti 7,I.Ferraro 5,A.Calamida 5,6,S.Degl’Innocenti 7,8,P.G.Prada Moroni 7,8,9,M.Del Principe 9,A.Piersimoni 9,G.Iannicola 5,P.B.Stetson 10,M.I.Andersen 11,R.Buonanno 6,C.E.Corsi 5,M.Dall’Ora 5,6,J.O.Petersen 12,L.Pulone 5,C.Sterken 3,13,and J.Storm 11drafted February 2,2008/Received /Accepted ABSTRACT We present three diﬀerent optical and near-infrared (NIR)data sets for evolved stars in the Galactic Globular Cluster ωCentauri .The comparisonbetween observations and homogeneous sets of stellar isochrones and Zero-AgeHorizontal Branches provides two reasonableﬁts.Both of them suggest thatthe so-called anomalous branch has a metal-intermediate chemical composition(−1.1≤[Fe/H]≤−0.8)and is located∼500pc beyond the bulk ofωCen stars.Theseﬁndings are mainly supported by the shape of the subgiant branch in fourdiﬀerent color-magnitude diagrams(CMDs).The most plausibleﬁt requires ahigher reddening,E(B−V)=0.155vs.0.12,and suggests that the anomalousbranch is coeval,within empirical and theoretical uncertainties,to the bulk ofωCen stellar populations.This result is supported by the identiﬁcation of asample of faint horizontal branch stars that might be connected with the anoma-lous branch.Circumstantial empirical evidence seems to suggest that the starsin this branch form a clump of stars located beyond the cluster.Subject headings:globular clusters:general—globular clusters:individual(ωCentauri)1.IntroductionThe peculiar Galactic Globular Cluster(GGC)ωCentauri(NGC5139)is currently subject to substantial observational eﬀorts covering the whole wavelength spectrum.This gigantic star cluster,the most massive known in our Galaxy,has(at least)three separate stellar populations with a large undisputed spread in age,metallicity(Fe,Ca)and kinematics (e.g.Hilker&Richtler2000;Ferraro et al.2002;Smith2004).According to recent abundance measurements based on400medium resolution spectra collected by Hilker et al. (2004),it seems that the spread in metallicity among theωCen stars is of the order of one dex(−2 [Fe/H] −1).The metallicity distribution shows three well-deﬁned peaks around[Fe/H]=−1.7,−1.5and−1.2together with a few metal-rich stars at∼−0.8.On the basis of high-resolution spectra,it has been suggested by Pancino(2004)that stars in the anomalous branch(Lee et al.1999)might be more metal-rich,with a mean metallicity [Fe/H]∼−0.5.A similar metal-rich tail was also detected by Norris et al.(1996)and by Suntzeﬀ&Kraft(1996,and references therein).In the absence of conformity in the literature for theωCen RGB names,we here refer to the metal-poor component asω1,to the metal-intermediate asω2,and to the anomalous branch asω3.Results from new observations seem to pose as many new questions as they answer.It has been suggested(e.g.Lee et al.1999;Hughes et al.2004)that theω3population might be signiﬁcantly younger thanω2.However,in a recent investigation based on Very Large Telescope(VLT)and Hubble Space Telescope(HST)data,Ferraro et al.(2004)found thattheω3population is at least as old as theω2population and probably a few Gyrs older. The observational scenario was further complicated by the results brought forward by Bedin et al.(2004)on the basis of multi-band HST data.They not only conﬁrmed a bifurcation along the Main Sequence,but also found a series of diﬀerent Turn-Oﬀs(TOs)and sub-giant branches(SGBs).The proposed explanations in the literature for these peculiar stellar populations inωCen are many and include,e.g.,increased He content,or a separate group of stars located at a larger distance(Ferraro et al.2004;Bedin et al.2004;Norris2004). The latter hypothesis could be further supported by the occurrence of a tidal tail inωCen but we still lack aﬁrm empirical detection(see Leon et al.2000;Law et al.2003).The formation history and composition ofωCen thus form a complex puzzle that is being slowly pieced together by investigations based on the latest generation of telescopes.A well-known promising technique is to study evolved stars simultaneously in optical and NIR bands to limit subtle errors due to the absolute calibration,to crowding in the innermost regions,and to reddening corrections.The main aim of this Letter is to investigate whether diﬀerent assumptions concerning the spread in age and in chemical composition,or diﬀerences in distance and in reddening,account for observed optical(BRI)and NIR(JK)CMDs.2.Observations and data reductionNear-infrared J,K s images ofωCen were collected in2003with SOFI at the New Technology Telescope of ESO,La Silla.The seeing conditions were good and range from0.′′6 to1.′′1.Together with additional data from2001,available in the ESO archive,we end up with a total NIR sample of92J-and135K s-band images that cover a14×14arcmin2area centred on the cluster.These data were reduced with DAOPHOTII/ALLFRAME following the same technique adopted by Dall’Ora et al.(2004).The NIR catalogue includes∼1.4×105stars. With FORS1(standard-resolution mode)on the ANTU/VLT telescope,we also collected optical UV I images in1999.These data are from a2×2pointing mosaic centred on the cluster that covers an area slightly larger than13×13arcmin2.Exposure times range from 10(I)to30s(U)and seeing conditions were better than1.′′0.These data have also been reduced using DAOPHOTII/ALLFRAME and the photometric catalogue includes∼5×105stars. Optical F435W and F625W(hereinafter B and R bands)data were retrieved from the HST archive.These images were collected with the ACS instrument in9telescope pointings,each of which provides a BR-image pair with exposure times of12(B)and8s(R).Theﬁeld covered by these data is∼9×9arcmin2,centred on the cluster.The ACS data were reduced with ROMAFOTwo and provide BR photometry for∼4×105stars.The photometry was kept in the Vega system(see e.g.,/hst/acs/documents).A detailed description of observations and data reduction will be given in a future paper.Here we onlywish to mention that raw frames were prereduced using standard IRAF procedures and,in addition,the FORS1images were corrected for ampliﬁer cross-talk by following Freyhammer et al.(2001).To improve the photometric accuracy,carefully chosen selection criteria were applied to pinpoint a large number(>100)of point-spread function stars across the individual frames,and several diﬀerent reduction strategies were used to perform the photometry over the entire data set.The observedﬁelds were combined to the same geometrical system using iraf.immatch and DAOMATCH/DAOMASTER.The absolute photometric calibration of ground-based instrumental magnitudes was performed using standard stars observed during the same nights.The typical accuracy is0.02–0.03mag for both optical and NIR data.3.Results and DiscussionFigure1shows CMDs based on theﬁve diﬀerent photometric bands.The plotted stars were selected from individual catalogues by using the‘separation index’sep introduced by Stetson et al.(2003),since crowding errors dominate the photometric errors.The adopted sep ranges from6.5(NIR data)to8(ACS data),which corresponds to stars having less than0.3%of their measured light contaminated by neighbour stars;the higher the sep for a star,the less the severity of the crowding by neighbours.Error bars plotted at the left side of each panel account for photometric and calibration errors in magnitude and color. The input physics adopted in our evolutionary code has been discussed in detail in a series of papers(Cariulo et al.2004;Cassisi et al.1998).Here,we point out that the stellar models(partly available at http://astro.df.unipi.it/SAA/PEL/Z0.html)account for atomic diﬀusion,including the eﬀects of gravitational settling,and thermal diﬀusion with diﬀusion coeﬃcients given by Thoul et al.(1994).The amount of original He is based on a primordial He abundance Y P=0.23and a He-to-metals enrichment ratio of∆Y/∆Z∼2.5(Pagel& Portinari1998;Castellani&Degl’Innocenti1999).We adopt the solar mixture provided by Noels&Grevesse(1993).For details on the calibration of the mixing length parameter and on model validation,we refer to Cariulo et al.(2004)and Castellani et al.(2003).To avoid deceptive uncertainties in the comparison between theory and observations,the predictions were transformed into both the BR Vega system and the IJK Johnson-Cousins bands by adopting the atmosphere models provided by Castelli et al.(1997)and Castelli(1999).Data plotted in Fig.1display the comparison between observations and a set of four isochrones(solid lines)with the same age(12Gyr)and diﬀerent chemical compositions(see labels).A mean reddening of E(B−V)=0.12(E(B−V)=0.11±0.02,Lub(2002))was adopted and a true distance modulus ofµ=13.7((m−M)V=14.05±0.11;Thompson et al.(2001),(m−M)K=13.68±0.07,Del Principe et al.2004,private communication).Extinction parameters for both optical and NIR bands have been estimated using the extinc-tion model of Cardelli et al.(1989).The metal-poor and the metal-intermediate isochrones supply,within current empirical and theoretical uncertainties,a goodﬁt to the bulk of RGB and HB stars.Close to the RGB tip,the isochrones are slightly brighter than the observed stars,which is caused by the adopted mixing-length parameter(α=2).The SGB and the lower RGB are only marginally aﬀected by this parameter,since the empirical isochrone calibration is based on these evolutionary phases(Cariulo et al.2004).However,the most metal-rich isochrone appears to be systematically redder than the stars at the base of the ω3branch1(green dots)and fainter than the SGB stars.Clearly,the shape of theω3SGB does not support metal-abundances≥Z=0.004,since TO stars for more metal-rich popu-lations become brighter than SGB stars.To further constrain this evidence,Fig.2shows the comparison between observed HB stars(the entire sample includes more than2,300objects) and predicted Zero Age Horizontal Branches(ZAHBs)with a progenitor age of12Gyr and diﬀerent chemical compositions.For the sake of the comparison,the objects located between hot HB stars and RGB stars were selected in the B−R,B plane and plotted as blue dots. Theoretical predictions plotted in thisﬁgure show that the more metal-rich ZAHB appears to be systematically brighter than the observed HB stars for B−J≈B−K≈0.5.More-over,the same ZAHB crosses the RGB region,thus suggesting that the occurrence of HB stars more metal-rich than Z=0.002should appear as an anomalous bump along the RGB. Data plotted in theseﬁgures disclose that the stellar populations inωCen cannot be ex-plained with a ranking in metal abundance.In line with Ferraro et al.(2004)and Bedin et al. (2004),weﬁnd that plausible changes in the cluster distance,reddening,age,and chemical composition do not supply a reasonable simultaneousﬁt of theω1,ω2,andω3branches.To further investigate the nature of theω3branch,we performed a series of tests by changing the metallicity,the cluster age,and the distance.A goodﬁt to the anomalous branch is possible by adopting the same reddening as in Fig.1,a true distance modulus of µ=13.9,and an isochrone of15.5Gyr with Z=0.0025and Y=0.248.Figure3shows that these assumptions supply a goodﬁt in both the optical and the NIR bands,and indeed the current isochrone properlyﬁts the width in color of the sub-giant branch and the shape of a good fraction of the RGB.Moreover,the ZAHB for the same chemical composition(dashed line)agrees quite well with the faint component of HB stars.Note that more metal-poor ZAHBs at the canonical distance do not supply a reasonableﬁt of the yellow spur stars with 15≤B≤15.5.The yellow spur is visible in all planes,and in the NIR(c,d)it even splits up in brightness due to a stronger sensitivity to eﬀective temperature.The identiﬁcationof the entire sample was checked on individual images and,once conﬁrmed by independent measurements,the data indicate a separate HB sequence for theω3population.Although theω3ﬁt may appear good,it implies an increase in distance of∆µ=0.m2 and an unreasonable∼4Gyr increase in age.This estimate is at variance with the absolute age estimates of GGCs(Gratton et al.2003)and with CMB measurements by W-MAP (Bennett et al.2003).The discrepancy becomes even larger if we account for the fact that this isochrone was constructed by adopting a He abundance slightly higher(Y=0.248,vs 0.238)than estimated from a He-to-metals enrichment ratio∆Y/∆Z=2.5.This increase implies a decrease in age of∼1Gyr.Moreover,we are performing a diﬀerential age estimate, and therefore if we account for uncertainties:in the input physics of the evolutionary models (e.g.equation of state,opacity);in the eﬃciency of macroscopic mechanisms(like diﬀusion); in model atmospheres applied in the transformation of the models into the observational plane;and in the extinction models,we end up with an uncertainty of∼1−2Gyr in cluster age(Castellani&Degl’Innocenti1999;Krauss&Chaboyer2003).Owing to the wide range of chemical compositions and stellar ages adopted in the literature for explaining the morphology of theω3branch,we decided to investigate whether diﬀerent combinations of assumed values of distance,chemical composition,and reddening may also simultaneously account for theω3branch and the HB stars.We found that two isochrones of13Gyr for Z=0.0015and Z=0.003bracket theω3stars(see Fig.4),within empirical and theoretical uncertainties.Theﬁt was obtained using the same true distance modulus adopted in Fig.3,together with a mild increase in reddening,E(B−V)=0.155. Once again theory agrees reasonably well with observations in all color planes.Moreover, data plotted in Fig.5show that the predicted ZAHB with Z=0.0015and Z=0.003,for the adopted distance modulus and reddening correction,account for the yellow spur stars.The sameﬁgure shows a sample of53RR Lyrae stars selected from the variable-star catalogue by Kaluzny et al.(2004)for which we have a good coverage of J-and K-band light curves (Del Principe et al.2004,private communication).The RR Lyrae stars only account for a tiny fraction of the yellow-spur stars.Thisﬁnding together with the detection a well-deﬁned sequence in panels b,c,and d indicate that this spur might be the HB associated with the ω3population.Note that a few of these‘ω3-HB’spur stars have also been detected by Rey et al.(2004,see their Fig.7,for V≈14.75)and by Sollima et al.(2004a,see their Figs.6 and12).4.Final remarksWe have presented a new set of multi-band photometric data for the GGCωCen and—in agreement with previousﬁndings in the literature—weﬁnd no acceptableﬁt to the diﬀerent stellar populations for a single distance,reddening,and age.We found two reasonableﬁts for theω3stars:(1)by adopting a0.2higher distance modulus(≈500pc), a metal-intermediate composition(Z=0.0025,Fe/H≈−0.9),and an unreasonable increase in age of∼4Gyr;or(2)for the same∆µ=0.2shift,an increase in the reddening,metal-intermediate chemical compositions(0.0015≤Z≤0.003,−1.1≤Fe/H≤−0.8),and an age that,within current uncertainties,is coeval with the bulk of theωCen stars.Current ﬁndings indicate thatω3stars are not signiﬁcantly more metal-rich than Z=0.003.This evidence is supported by the shape of theω3SGB,as already suggested by Ferraro et al. (2004)and by theﬁt of HB stars.We are in favour of the latter solution for the following reasons:•The diﬀerence in distance between theω3branch and the bulk ofωCen stars is of the order of10%.This estimate is3–4times smaller than the estimate by Bedin et al. (2004)and in very good agreement with the distance of the density maxima detected by Odenkirchen et al.(2003)along the tidal tails of the GGC Pal5.Moreover,recent N-body simulations(Capuzzo Dolcetta et al.2004)indicate that clumps along the tidal tails can approximately include10%of the cluster mass.•We found that by artiﬁcially shifting theω3-branch stars to account for the assumed diﬀerence in distance and reddening,they overlap with theω2population.It has been recently suggested by Piotto et al.(2004),on the basis of low-resolution spectra,that the bluer main sequence detected by Bedin et al.(2004)is more metal-rich than the red main sequence.Unfortunately,current ACS photometry is only based on shallow ACS exposures, and therefore we cannot properly identify in our data these stellar populations located in the lower main sequence.The same outcome applies to the suspected extremely-hot HB progeny of the bluer main sequence,since they have not been detected in the NIR bands.•Current preliminaryﬁndings support recent N-body simulations by Chiba&Mizutani (2004)and by Ideta&Makino(2004).In particular,the latter authors found,by assuming that the progenitor of Omega Centauri is a dwarf galaxy,that more than90%of its stellar content was lost during theﬁrst few pericenter passages(see their Fig.2).•ωCen reddening estimates in the literature cluster around E(B−V)=0.12±0.02. However,the map from(Schlegel et al.1998)indicates reddening variations of0.02across the body of the cluster while,more importantly,2MASS data(Law et al.2003)show a very clumpy reddening distribution at distances beyond1◦(100pc)from the cluster centre,with large variations∆E(B−V)=0.18across a4degrees2area.However,the inference of reddening between the main body ofωCen and the supposed background population is very surprising.ωCen lies at the comparatively low Galactic latitude of+15◦,and the reddening variations seen in the Schlegel and2MASS maps likely originate in the foreground interstellar material of the Galactic disk.Any interstellar material behindωCen must lie at least1.4kpc from the Galactic plane,and therefore would most likely be associated with ωCen itself.Smith et al.(1990)have reported a signiﬁcant detection of H I in the direction ofωCen,blueshifted by∼40km s−1with respect to the cluster velocity.However,their interpretation is that this gas is associated with the northern extension of the Magellanic Stream far beyond the cluster.It would be a remarkable coincidence if this interstellar material happened to lie between the cluster and a clump in the tidal tail at a10%greater distance than the main cluster body,while traveling at the quoted relative speed.However, it is worth noting that Smith et al.(1990)estimated for this cloud a column density of N H≈3×1018atmos/cm2,and therefore a reddening E(B−V)≈0.07(Predehl&Schmitt 1995)that is at least a factor of two larger than required by our bestﬁts.This point is crucial for the proposed explanation and needs to be further investigated.Finally,we note that the comparison between predicted ZAHBs and HB stars indicates that the occurrence of an old stellar population with Z>0.002would imply the occurrence of an anomalous clump along the RGB.In fact,more metal-rich,red HB stars cover the same color range of metal-poor RGB stars.The detection of such a feature along the RGB can supply robust constraints on the progeny of theω3stellar population.These results, when independently conﬁrmed,would suggest that theω3branch might be a clump of stars located500pc beyond the bulk of the cluster.Noﬁrm conclusion can be drawn on the basis of current data,although this evidence together with the increase in radial velocity among ω3stars measured by Sollima et al.(2004b)and numerical simulations recently provided by Capuzzo Dolcetta et al.(2004)indicates that it could be a tidal tail.5.AcknowledgmentsIt is a pleasure to thank M.Sirianni and N.Panagia for fundamental suggestions on the calibration of ACS data.We wish to tank two anonymous referees for their suggestions and pertinent comments that helped us to improve the content and the readability of the manuscript.We are also grateful to V.Castellani for his critical reading of an early version of this manuscript.This work was supported by the Belgian Fund for Scientiﬁc Research (FWO)in the framework the project“IAP P5/36“of the Belgian Federal Science Policy,by MIUR/PRIN2003in the framework of the project:“Continuity and Discontinuity in theGalaxy Formation“,and by Danish NSRC in the framework of the project:“Structure and evolution of stars:new insight from eclipsing binaries and pulsating stars“.REFERENCESBedin,L.R.,Piotto,G.,Anderson,J.,Cassisi,S.et al.2004,ApJ,605,125Bennett,C.L.,Halpern,M.,Hinshaw,G.,Jarosik,N.et al.2003,ApJS,148,1Capuzzo Dolcetta,R.,Di Matteo,P.,&Miocchi,P.2004,AJ,accepted,astro-ph/0406313 Cardelli,J.A.,Clayton,G.C.,Mathis,J.S.1989,ApJ,345,245Cariulo,P.,Degl’Innocenti S.,Castellani,V.,2004,A&A421,1121Cassisi S.,Castellani V.,Degl’Innocenti S.&Weiss A.1998,A&AS,129,267 Castellani,V.,&Degl’Innocenti,S.1999,A&A,344,97Castellani,V.,Degl’Innocenti,S.,Marconi,M.,Prada Moroni,P.G.&Sestito,P.2003, A&A,404,645Castelli,F.1999,A&A,346,564Castelli,F.,Gratton,R.G.,&Kurucz,R.L.1997,A&A,318,841Chiba,M.,&Mizutani,A.2004,PASA,21,237Dall’Ora,M.,Storm,J.,Bono,G.,Ripepi,V.et al.,2004,ApJ,610,269Ferraro,F.R.,Bellazzini,M.,Pancino,E.2002,ApJ,573,95Ferraro,F.R.,Sollima,A.,Pancino,E.,Bellazzini,M.,et al.2004,ApJ,603,81 Freyhammer,L.M.,Andersen,M.I.,Arentoft,T.,Sterken,C.,Nørregaard,P.2001,Experi-mental Astronomy,12,147Gratton,R.G.,Bragaglia,A.,Carretta,E.,Clementini,G.,Desidera,S.,Grundahl,F., Lucatello,S.2003,A&A,408,529Hilker,M.&Richtler,T.,2000,A&A,362,895Hilker,M.,Kayser,A.,Richtler,T.,Willemsen,P.2004,A&A,422,9Hughes,J.;Wallerstein,G.;van Leeuwen,F.;&Hilker,M.2004,AJ,127,980Ideta,M.,&Makino,J.2004,ApJ,616,L107Kaluzny,J.,Olech,A.,Thompson,I.B.,Pych,W.,Krzeminski,W.,Schwarzenberg-Czerny,A.2004,A&A,424,1101Krauss,L.M.&Chaboyer,B.2003,Sci,299,65Law,D.R.,Majewski,S.R.,Skrutskie,M.F.,Carpenter,J.M.et al.2003,AJ,126,1871 Lee,Y.-W.,Joo,J.-M.,Sohn,Y.-J.,Rey,S.-C.et al.1999,Nature,402,55Leon,S.,Meylan,G.,Combes,F.2000,A&A,359,907Lub,J.,2002,in ASP Conf.Ser.265,Omega Centauri:A Unique Window into Astrophysics, ed.F.van Leeuwen,J.Hughes and G.Piotto(S.Francisco:ASP),95Noels,A.&Grevesse,N.1993in Origin and Evolution of the Elements,ed.N.Prantzos,E.Vangioni-Flam,M.Casse`(Cambridge:Cambridge Univ.Press),15Norris,J.E.,2004,ApJ,612,25Norris,J.E.,Freeman,K.C.;Mighell,K.J1996,ApJ,462,241Odenkirchen,M.et al.2003,AJ,126,2385Pagel,B.E.J.&Portinari,L.1998,MNRAS,298,747Pancino,E.,2004,in Carnegie Obs.Astrophysics Ser.,Vol.4,Origin and Evolution of the Elements,ed.A.McWilliam and M.Rauch(Cambridge:Cambridge Univ.Press),45 Predehl,P.,&Schmitt,J.H.M.M.1995,A&A,293,889Rey,S.-C.,Lee,Y.-W.,Ree,C.H.,Joo,J.-M.,Sohn,Y.-J.,Walker,A.R.,2004,AJ,127,958 Schlegel,D.J.,Finkbeiner,D.P.,Davis,M.,1998,ApJ,500,525Smith,V.V.2004,in Carnegie Obs.Astrophysics Ser.,Vol.4,Origin and Evolution of the Elements,ed.A.McWilliam and M.Rauch(Cambridge:Cambridge Univ.Press),1 Sollima,A.,Ferraro,F.R.,Pancino,E.,&Bellazzini,M.2004,MNRAS,astro-ph/0411546 Sollima,A.,Pancino,E.,Ferraro,F.R.,&Bellazzini,M.2004,in Chemical Abundances and Mixing in Stars in the Milky Way and its Satellites,ed.L.Pasquini,S.Randich (Berlin:Springer-Verlag),in pressSmith,G.H.,Wood,P.R.,Faulkner,D.J.,Wright,A.E1990,ApJ,353,168 Stetson,P.B.,Bruntt,H.,&Grundahl,F.2003,PASP,115,413Suntzeﬀ,N.B.,Kraft,R.P.1996,AJ,111,1913Thompson,I.B.,Kaluzny,J.,Pych,W.,Burley,G.et al.2001,AJ,121,3089 Thoul,A.,Bahcall,J.,&Loeb,A.1994,ApJ421,828Fig.1.—Optical(panels a,b),and NIR(panels c,d)CMDs for selected subsamples of the detected stars,compared with a set of12Gyr isochrones(solid lines)at diﬀerent chemical compositions(see color coding).The adopted true distance modulus and cluster reddening areµ=(m−M)0=13.7and E(B−V)=0.12,respectively.The number of stars selected (NS)is also indicated.Fig.2.—Same as in Fig.1,but the comparison between theory and observations is focused on Horizontal Branch stars.Blue objects mark stars located between hot HB stars and RGB stars.They have been selected in B−R,B plane.Fig.3.—Same as in Fig.1,but compared to a15.5Gyr isochrone for theω3population, constructed by adopting Z=0.0025,and Y=0.248.Theﬁt was performed by adopting a true distanceµ=13.9,and a reddening correction E(B−V)=0.12.Note that the corresponding ZAHB matches the selected HB stars(blue objects).Fig.4.—Same as Fig.3,but compared to13-Gyr isochrones constructed by adopting dif-ferent chemical compositions(see labels)and a higher reddening.Fig.5.—Same as in Fig.4,but the comparison between theory and observations is focused on Horizontal Branch stars.Blue objects mark stars located between hot HB stars and RGB stars,while red dots(panels c,d)in the HB region display RR Lyrae stars for which we have accurate mean NIR magnitudes.。

2006年考研英语答案解析和参考译文(二)

2006年考研英语答案解析和参考译文(二)SectionⅠUse of English篇章导读本文是一篇论说文。

文章的主题是"英才通才教育"。

作者在文章开头就提出了一个具有选择性的问题："如果我们只是需要决定是把基本的科学传授给每个人，还是找一些有才华的人，引领他们变得更出色，那么我们的工作将会相当容易。

"随后作者从"the education in public school，the balance among the branches of knowledge and the balance between current and classical knowledge"三个方面来论述在教育中保持知识平衡的重要性。

解读文章时注意作者的客观态度。

思路解析1「答案」[C]「解析」"选择"。

根据文章一致性原则，"choice"与文章第一句中的"decide决定"形成呼应，根据原文"decide whether......or......"所以下文就应该是对其有所"选择choice"或没有"选择choice"。

而选项[A]"（与属性区别的）本质：the entity of justice 正义的本质"，[B]"拍卖；（某些纸牌戏中的）叫牌；叫牌阶段"，[D]"结合体，联合；（政党、个人、国家等）临时结成的联盟"是本题的干扰，均不形成呼应，不符合题意。

「解析」"因为"。

"for"与文章第一段第三句中的"Because we depend......"构成搭配，均表示解释原因。

而选项[A][B][C]均不用于解释原因，不符合原文意思。

Discovering Similar Multidimensional Trajectories

Discovering Similar Multidimensional TrajectoriesMichail VlachosUC Riverside mvlachos@George KolliosBoston Universitygkollios@Dimitrios GunopulosUC Riversidedg@AbstractWe investigate techniques for analysis and retrieval of object trajectories in a two or three dimensional space. Such kind of data usually contain a great amount of noise, that makes all previously used metrics fail.Therefore,here we formalize non-metric similarity functions based on the Longest Common Subsequence(LCSS),which are very ro-bust to noise and furthermore provide an intuitive notion of similarity between trajectories by giving more weight to the similar portions of the sequences.Stretching of sequences in time is allowed,as well as global translating of the se-quences in space.Efﬁcient approximate algorithms that compute these similarity measures are also provided.We compare these new methods to the widely used Euclidean and Time Warping distance functions(for real and synthetic data)and show the superiority of our approach,especially under the strong presence of noise.We prove a weaker ver-sion of the triangle inequality and employ it in an indexing structure to answer nearest neighbor queries.Finally,we present experimental results that validate the accuracy and efﬁciency of our approach.1IntroductionIn this paper we investigate the problem of discovering similar trajectories of moving objects.The trajectory of a moving object is typically modeled as a sequence of con-secutive locations in a multidimensional(generally two or three dimensional)Euclidean space.Such data types arise in many applications where the location of a given object is measured repeatedly over time.Examples include features extracted from video clips,animal mobility experiments, sign language recognition,mobile phone usage,multiple at-tribute response curves in drug therapy,and so on.Moreover,the recent advances in mobile computing, sensor and GPS technology have made it possible to collect large amounts of spatiotemporal data and there is increas-ing interest to perform data analysis tasks over this data [4].For example,in mobile computing,users equipped with mobile devices move in space and register their location at different time instants via wireless links to spatiotemporal databases.In environmental information systems,tracking animals and weather conditions is very common and large datasets can be created by storing locations of observed ob-jects over time.Data analysis in such data include deter-mining andﬁnding objects that moved in a similar way or followed a certain motion pattern.An appropriate and ef-ﬁcient model for deﬁning the similarity for trajectory data will be very important for the quality of the data analysis tasks.1.1Robust distance metrics for trajectoriesIn general these trajectories will be obtained during a tracking procedure,with the aid of various sensors.Here also lies the main obstacle of such data;they may contain a signiﬁcant amount of outliers or in other words incorrect data measurements(unlike for example,stock data which contain no errors whatsoever).Figure1.Examples of2D trajectories.Two in-stances of video-tracked time-series data representingthe word’athens’.Start&ending contain many out-liers.Athens 1Athens 2Boston 1Boston 2DTWLCSSFigure 2.Hierarchical clustering of 2D series (displayed as 1D for clariry).Left :The presence of many outliers in the beginning and the end of the sequences leads to incorrect clustering.DTW is not robust under noisy conditions.Right :The focusing on the common parts achieves the correct clustering.Our objective is the automatic classiﬁcation of trajec-tories using Nearest Neighbor Classiﬁcation.It has been shown that the one nearest neighbor rule has asymptotic er-ror rate that is at most twice the Bayes error rate[12].So,the problem is:given a database of trajectories and a query (not already in the database),we want to ﬁnd the trajectory that is closest to .We need to deﬁne the following:1.A realistic distance function,2.An efﬁcient indexing scheme.Previous approaches to model the similarity between time-series include the use of the Euclidean and the Dy-namic Time Warping (DTW)distance,which however are relatively sensitive to noise.Distance functions that are ro-bust to extremely noisy data will typically violate the trian-gular inequality.These functions achieve this by not consid-ering the most dissimilar parts of the objects.However,they are useful,because they represent an accurate model of the human perception,since when comparing any kind of data (images,trajectories etc),we mostly focus on the portions that are similar and we are willing to pay less attention to regions of great dissimilarity.For this kind of data we need distance functions that can address the following issues:Different Sampling Rates or different speeds.The time-series that we obtain,are not guaranteed to be the outcome of sampling at ﬁxed time intervals.The sensors collecting the data may fail for some period of time,leading to inconsistent sampling rates.Moreover,two time series moving at exactly the similar way,but one moving at twice the speed of the other will result (most probably)to a very large Euclidean distance.Similar motions in different space regions .Objectscan move similarly,but differ in the space they move.This can easily be observed in sign language recogni-tion,if the camera is centered at different positions.If we work in Euclidean space,usually subtracting the average value of the time-series,will move the similar series closer.Outliers.Might be introduced due to anomaly in the sensor collecting the data or can be attributed to hu-man ’failure’(e.g.jerky movement during a track-ing process).In this case the Euclidean distance will completely fail and result to very large distance,even though this difference may be found in only a few points.Different lengths.Euclidean distance deals with time-series of equal length.In the case of different lengths we have to decide whether to truncate the longer series,or pad with zeros the shorter etc.In general its use gets complicated and the distance notion more vague.Efﬁciency.It has to be adequately expressive but suf-ﬁciently simple,so as to allow efﬁcient computation of the similarity.To cope with these challenges we use the Longest Com-mon Subsequence (LCSS)model.The LCSS is a varia-tion of the edit distance.The basic idea is to match two sequences by allowing them to stretch,without rearranging the sequence of the elements but allowing some elements to be unmatched .The advantages of the LCSS method are twofold:1)Some elements may be unmatched,where in Eu-clidean and DTW all elements must be matched,even the outliers.2)The LCSS model allows a more efﬁcient approximatecomputation,as will be shown later(whereas in DTW you need to compute some costly Norm).Inﬁgure2we can see the clustering produced by the distance.The sequences represent data collected through a video tracking process.Originally they represent 2d series,but only one dimension is depicted here for clar-ity.The fails to distinguish the two classes of words, due to the great amount of outliers,especially in the begin-ning and in the end of the ing the Euclidean distance we obtain even worse results.The produces the most intuitive clustering as shown in the sameﬁgure. Generally,the Euclidean distance is very sensitive to small variations in the time axis,while the major drawback of the is that it has to pair all elements of the sequences.Therefore,we use the model to deﬁne similarity measures for trajectories.Nevertheless,a simple extension of this model into2or more dimensions is not sufﬁcient, because(for example)this model cannot deal with paral-lel movements.Therefore,we extend it in order to address similar problems.So,in our similarity model we consider a set of translations in2or more dimensions and weﬁnd the translation that yields the optimal solution to the problem.The rest of the paper is organized as follows.In section2 we formalize the new similarity functions by extending the model.Section3demonstrates efﬁcient algorithms to compute these functions and section4elaborates on the indexing structure.Section5provides the experimental validation of the accuracy and efﬁciency of the proposed approach and section6presents the related work.Finally, section7concludes the paper.2Similarity MeasuresIn this section we deﬁne similarity models that match the user perception of similar trajectories.First we give some useful deﬁnitions and then we proceed by presenting the similarity functions based on the appropriate models.We assume that objects are points that move on the-plane and time is discrete.Let and be two trajectories of moving objects with size and respectively,whereand.For a trajectory,let be the sequence.Deﬁnition1Given an integer and a real number,we deﬁne the as follows:A Ba and andThe constant controls how far in time we can go in order to match a given point from one trajectory to a point in another trajectory.The constant is the matching threshold(see ﬁgure3).Theﬁrst similarity function is based on the and the idea is to allow time stretching.Then,objects that are close in space at different time instants can be matched if the time instants are also close.Deﬁnition2We deﬁne the similarity function between two trajectories and,given and,as follows:Deﬁnition3Given,and the family of translations,we deﬁne the similarity function between two trajectories and,as follows:So the similarity functions and range from to. Therefore we can deﬁne the distance function between two trajectories as follows:Deﬁnition4Given,and two trajectories and we deﬁne the following distance functions:andNote that and are symmetric.is equal to and the transformation that we use in is translation which preserves the symmetric prop-erty.By allowing translations,we can detect similarities be-tween movements that are parallel in space,but not iden-tical.In addition,the model allows stretching and displacement in time,so we can detect similarities in move-ments that happen with different speeds,or at different times.Inﬁgure4we show an example where a trajectory matches another trajectory after a translation is applied. Note that the value of parameters and are also important since they give the distance of the trajectories in space.This can be useful information when we analyze trajectory data.XFigure4.Translation of trajectory.The similarity function is a signiﬁcant improvement over the,because:i)now we can detect parallel move-ments,ii)the use of normalization does not guarantee that we will get the best match between two u-ally,because of the signiﬁcant amount of noise,the average value and/or the standard deviation of the time-series,that are being used in the normalization process,can be distorted leading to improper translations.3Efﬁcient Algorithms to Compute the Simi-larity3.1Computing the similarity functionTo compute the similarity functions we have to run a computation for the two sequences.Thecan be computed by a dynamic programming algorithm in time.However we only allow matchings when the difference in the indices is at most,and this allows the use of a faster algorithm.The following lemma has been shown in[5],[11].Lemma1Given two trajectories and,with and,we canﬁnd the in time.If is small,the dynamic programming algorithm is very efﬁcient.However,for some applications may need to be large.For that case,we can speed-up the above computa-tion using random sampling.Given two trajectories and ,we compute two subsets and by sampling each trajectory.Then we use the dynamic programming algo-rithm to compute the on and.We can show that,with high probability,the result of the algorithm over the samples,is a good approximation of the actual value. We describe this technique in detail in[35].3.2Computing the similarity functionWe now consider the more complex similarity function .Here,given two sequences,and constants, we have toﬁnd the translation that maximizes the length of the longest common subsequence of()over all possible translations.Let the length of trajectories and be and re-spectively.Let us also assume that the translationis the translation that,when applied to,gives a longest common subsequence,and it is also the translation that maximizes the length of the longest common subsequence.The key observation is that,although there is an inﬁnite number of translations that we can apply to,each transla-tion results to a longest common subsequence between and,and there is aﬁnite set of possible longest common subsequences.In this section we show that we can efﬁciently enumerate aﬁnite set of translations,such that this set provably includes a translation that maximizes the length of the longest common subsequence of and .To give a bound on the number of transformations that we have to consider,we look at the projections of the two trajectories on the two axes separately.We deﬁne theprojection of a trajectoryto be the sequence of the valueson the -coordinate:.A one di-mensional translation is a function that adds a con-stant to all the elements of a 1-dimensional sequence:.Take the projections of and ,and respec-tively.We can show the following lemma:Lemma 2Given trajectories ,if ,then the length of the longest common subsequence of the one dimensional sequences and,is at least :.Also,.Now,consider and .A translation by ,applied to can be thought of as a linear transformation of the form .Such a transformation will allowto be matched to all for which ,and.It is instructive to view this as a stabbing problem:Con-sider the vertical line segments,where (Figure 5).Bx,i By,i+2Bx,i+3Bx,i+4Bx,i+5Ax,iAx,i+1Ax,i+2fc1(x) = x + c1fc2(x) = x + c2Ax axisBx axisFigure 5.An example of two translations.These line segments are on a two dimensional plane,where on the axis we put elements ofand on the axis we put elements of .For every pair of elementsin and that are within positions from eachother (and therefore can be matched by the algo-rithm if their values are within ),we create a vertical line segment that is centered at the point and extends above and below this point.Since each element in can be matched with at most elements in ,the total number of such line segments is .A translation in one dimension is a function of the form .Therefore,in the plane we de-scribed above,is a line of slope 1.After translatingby ,an element of can be matched to an el-ement of if and only if the line intersects the line segment .Therefore each line of slope 1deﬁnes a set of possi-ble matchings between the elements of sequences and.The number of intersected line segments is actually an upper bound on the length of the longest common sub-sequence because the ordering of the elements is ignored.However,two different translations can result to different longest common subsequences only if the respective lines intersect a different set of line segments.For example,the translations and in ﬁgure 5intersect different sets of line segments and result to longest common subsequences of different length.The following lemma gives a bound on the number of possible different longest common subsequences by bound-ing the number of possible different sets of line segments that are intersected by lines of slope 1.Lemma 3Given two one dimensional sequences ,,there are lines of slope 1that intersect different sets of line segments.Proof:Let be a line of slope 1.If we move this line slightly to the left or to the right,it still in-tersects the same number of line segments,unless we cross an endpoint of a line segment.In this case,the set of inter-sected line segments increases or decreases by one.There are endpoints.A line of slope 1that sweeps all the endpoints will therefore intersect at most different sets of line segments during the sweep.In addition,we can enumerate the trans-lations that produce different sets of potential matchings byﬁnding the lines of slope 1that pass through the endpoints.Each such translation corresponds to a line .This set of translations gives all possible matchings for a longest common subsequence of .By applying the same process on we can also ﬁnd a set of translations that give all matchings of.To ﬁnd the longest common subsequence of the se-quences we have to consider only thetwo dimensional translations that are created by taking the Cartesian product of the translations on and the trans-lations on .Since running the LCSS algorithm takeswe have shown the following theorem:Theorem 1Given two trajectories and,withand ,we can compute theintime.3.3An Efﬁcient Approximate AlgorithmTheorem 1gives an exact algorithm for computing ,but this algorithm runs in cubic time.In this section we present a much more efﬁcient approximate algorithm.The key in our technique is that we can bound the difference be-tween the sets of line segments that different lines of slope 1intersect,based on how far apart the lines are.Consider again the one dimensional projections. Lets us consider the translations that result to different sets of intersected line segments.Each translation is a line of the form.Let us sort these trans-lations by.For a given translation,let be the set of line segments it intersects.The following lemma shows that neighbor translations in this order intersect similar sets of line segments.Lemma4Let be the different translations for sequences and,where.Then the symmetric difference.We can now prove our main theorem:Theorem2Given two trajectories and,with and,and a constant,we canﬁnd an ap-proximation of the similaritysuch that intime.Proof:Let.We consider the projections of and into the and axes.There exists a translation on only such that is a superset of the matches in the optimal of and.In addition,by the previous lemma,there are translations()that have at most different matchings from the optimal. Therefore,if we use the translations,fortime if we runpairs of translations in the plane.Since there is one that is away from the optimal in each dimension,there is one that is away from the optimal in2dimensions.Setting-th quantiles for each set,pairs of translations.4.Return the highest result.4Indexing Trajectories for Similarity Re-trievalIn this section we show how to use the hierarchical tree of a clustering algorithm in order to efﬁciently answer near-est neighbor queries in a dataset of trajectories.The distance function is not a metric because it does not obey the triangle inequality.Indeed,it is easy to con-struct examples where we have trajectories and, where.This makes the use of traditional indexing techniques difﬁ-cult.We can however prove a weaker version of the triangle inequality,which can help us avoid examining a large por-tion of the database objects.First we deﬁne:Clearly,or in terms of distance:In order to provide a lower bound we have to maximize the expression.Therefore,for every node of the tree along with the medoid we have to keep the trajectory that maximizes this expression.If the length of the query is smaller than the shortest length of the trajec-tories we are currently considering we use that,otherwise we use the minimum and maximum lengths to obtain an approximate result.4.2Searching the Index tree for Nearest Trajec-toriesWe assume that we search an index tree that contains tra-jectories with minimum length and maximum length .For simplicity we discuss the algorithm for the1-Nearest Neighbor query,where given a query trajectory we try toﬁnd the trajectory in the set that is the most sim-ilar to.The search procedure takes as input a nodein the tree,the query and the distance to the closest tra-jectory found so far.For each of the children,we check if the child is a trajectory or a cluster.In case that it is a trajectory,we just compare its distance to with the current nearest trajectory.If it is a cluster,we check the length of the query and we choose the appropriate value for .Then we compute a lower bound to the distance of the query with any trajectory in the cluster and we compare the result with the distance of the current near-est neighbor.We need to examine this cluster only if is smaller than.In our scheme we use an approximate algorithm to compute the.Consequently,the value offrom the bound we compute for.Note that we don’t need to worry about the other terms since they have a negative sign and the approximation algorithm always underestimates the .5Experimental EvaluationWe implemented the proposed approximation and index-ing techniques as they are described in the previous sec-tions and here we present experimental results evaluating our techniques.We describe the datasets and then we con-tinue by presenting the results.The purpose of our experi-ments is twofold:ﬁrst,to evaluate the efﬁciency and accu-racy of the approximation algorithm presented in section3 and second to evaluate the indexing technique that we dis-cussed in the previous section.Our experiments were run on a PC AMD Athlon at1GHz with1GB RAM and60 GB hard disk.5.1Time and Accuracy ExperimentsHere we present the results of some experiments using the approximation algorithm to compute the similarity func-tion.Our dataset here comes from marine mammals’satellite tracking data.It consists of sequences of geo-graphic locations of various marine animals(dolphins,sea lions,whales,etc)tracked over different periods of time, that range from one to three months(SEALS dataset).The length of the trajectories is close to.Examples have been shown inﬁgure1.In table1we show the computed similarity between a pair of sequences in the SEALS dataset.We run the exact and the approximate algorithm for different values of and and we report here some indicative results.is the num-ber of times the approximate algorithm invokes the procedure(that is,the number of translations that we try).As we can see,for and we get very good results.We got similar results for synthetic datasets.Also, in table1we report the running times to compute the simi-larity measure between two trajectories of the same dataset. The approximation algorithm uses again from to differ-ent runs.The running time of the approximation algorithm is much faster even for.As can be observed from the experimental results,the running times of the approximation algorithm is not pro-portional to the number of runs().This is achieved by reusing the results of previous translations and terminat-ing early the execution of the current translation,if it is not going to yield a better result.The main conclusion of the above experiments is that the approximation algorithm can provide a very tractable time vs accuracy trade-off for computing the similarity between two trajectories,when the similarity is deﬁned using the model.5.2Classiﬁcation using the Approximation Algo-rithmWe compare the clustering performance of our method to the widely used Euclidean and DTW distance functions. Speciﬁcally:cover.htmlSimilarityApproximate for K tries Exact9494250.250.3160.18460.2530.00140.0022 0.50.5710.4100.5100.00140.0022 0.250.3870.1960.3060.00180.00281 0.50.6120.4880.5630.00180.00280 0.250.4080.2500.3570.001910.0031 0.50.6530.4400.5840.001930.0031 Table1.Similarity values and running times between two sequences from our SEALS dataset.1.The Euclidean distance is only deﬁned for sequencesof the same length(and the length of our sequences vary considerably).We tried to offer the best possible comparison between every pair of sequences,by slid-ing the shorter of the two trajectories across the longer one and recording their minimum distance.2.For DTW we modiﬁed the original algorithm in orderto match both x and y coordinates.In both DTW and Euclidean we normalized the data before computing the distances.Our method does not need any normal-ization,since it computes the necessary translations.3.For LCSS we used a randomized version with andwithout sampling,and for various values of.The time and the correct clusterings represent the average values of15runs of the experiment.This is necessary due to the randomized nature of our approach.5.2.1Determining the values for&The values we used for and are clearly dependent on the application and the dataset.For most datasets we had at our disposal we discovered that setting to more than of the trajectories length did not yield signiﬁcant improvement.Furthermore,after some point the similarity stabilizes to a certain value.The determination of is appli-cation dependent.In our experiments we used a value equal to the smallest standard deviation between the two trajec-tories that were examined at any time,which yielded good and intuitive results.Nevertheless,when we use the index the value of has to be the same for all pairs of trajectories.5.2.2Experiment1-Video tracking data.The2D time series obtained represent the X and Y position of a human tracking feature(e.g.tip ofﬁnger).In conjuc-tion with a”spelling program”the user can”write”various words[19].We used3recordings of5different words.The data correspond to the following words:’athens’,’berlin’,’london’,’boston’,’paris’.The average length of the series is around1100points.The shortest one is834points and the longest one1719points.To determine the efﬁciency of each method we per-formed hierarchical clustering after computing thepairwise distances for all three distance functions.We eval-uate the total time required by each method,as well as the quality of the clustering,based on our knowledge of whichword each trajectory actually represents.We take all possi-ble pairs of words(in this case pairs)and use the clustering algorithm to partition them into two classes.While at the lower levels of the dendrogram the clustering is subjective,the top level should provide an accurate divi-sion into two classes.We clustered using single,complete and average linkage.Since the best results for every dis-tance function are produced using the complete linkage,we report only the results for this approach(table2).The same experiment is conducted with the rest of the datasets.Exper-iments have been conducted for different sample sizes and values of(as a percentage of the original series length).The results with the Euclidean distance have many clas-siﬁcation errors and the DTW has some errors,too.For the LCSS the only real variations in the clustering are for sam-ple sizes.Still the average incorrect clusterings for these cases were constantly less than one().For 15%sampling or more,there were no errors.5.2.3Experiment2-Australian Sign LanguageDataset(ASL).The dataset consists of various parameters(such as the X,Y, Z hand position,azimuth etc)tracked while different writ-ers sign one the95words of the ASL.These series are rel-atively short(50-100points).We used only the X and Y parameters and collected5recordings of the following10 words:’Norway’,’cold’,’crazy’,’eat’,’forget’,’happy’,’innocent’,’later’,’lose’,’spend’.This is the experiment conducted also in[25](but there only one dimension was used).Examples of this dataset can be seen inﬁgure6.Correct Clusterings(out of10)Complete Linkage Euclidean34.96DTW237.6412.7338.04116.17328.85145.06565.203113.583266.753728.277Distance Time(sec)CorrectClusterings(out of45)ASL with noiseEuclidean 2.271520Figure 7.ASL data :Time required to compute the pairwise distances of the 45combinations(same for ASL and ASL withnoise)Figure 8.Noisy ASL data :The correct clusterings of the LCSS method using complete linkage.Figure 9.Performance for increasing number of Near-est Neighbors.Figure 10.The pruning power increases along with the database size.jectories.We executed a set of -Nearest Neighbor (K-NN)queries for ,,,and and we plot the fraction of the dataset that has to be examined in order to guarantee that we have found the best match for the K-NN query.Note that in this fraction we included the medoids that we check during the search since they are also part of the dataset.In ﬁgure 9we show some results for -Nearest Neigh-bor queries.We used datasets with ,and clusters.As we can see the results indicate that the algorithm has good performance even for queries with large K.We also per-formed similar experiments where we varied the number of clusters in the datasets.As the number of clusters increased the performance of the algorithm improved considerably.This behavior is expected and it is similar to the behavior of recent proposed index structures for high dimensional data [9,6,21].On the other hand if the dataset has no clusters,the performance of the algorithm degrades,since the major-ity of the trajectories have almost the same distance to the query.This behavior follows again the same pattern of high dimensional indexing methods [6,36].The last experiment evaluates the index performance,over sets of trajectories with increasing cardinality.We in-dexed from to trajectories.The pruning power of the inequality is evident in ﬁgure 10.As the size of the database increases,we can avoid examining a larger frac-tion of the database.6Related WorkThe simplest approach to deﬁne the similarity between two sequences is to map each sequence into a vector and then use a p-norm distance to deﬁne the similarity measure.The p-norm distance between two n-dimensional vectors and is deﬁned as。

瓦尔登湖英文原文文笔水平

Walden is melting apace.There is a canal two rods wide along the northerly and westerly sides, and wider still at the east end. A great field of ice has cracked off from the main body.I hear a song sparrow singing from the bushes on the shore - olit, olit, olit - chip, chip, chip, che char - che wiss, wiss, wiss. He too is helping to crack it. How handsome the great sweeping curves in the edge of the ice, answering somewhat to those of the shore, but more regular!It is unusually hard, owing to the recent severe but transient cold, and all watered or waved like a palace floor. But the wind slides eastward over its opaque surface in vain, till it reaches the living surface beyond. It is glorious to behold this ribbon of water sparkling in the sun!the bare face of the pond full of glee and youth, as if it spoke the joy of the fishes within it, and of the sands on its shore - a silvery sheen as from the scales of a leuciscus, as it were all one active fish. Such is the contrast between winter and spring.Walden was dead and is alive again. But this spring it broke up more steadily, as I have said.。

托福阅读素材之埃及罗塞塔石碑整理

托福阅读素材之埃及罗塞塔石碑整理积累多一点素材可以关心我们更简单理解托福阅读题，为了便利大家备考，下面我给大家带来托福阅读素材之埃及罗塞塔石碑，望喜爱。

托福阅读素材之埃及罗塞塔石碑A Rosetta Stone for a lost language0:11Id like to begin with a thought experiment. Imagine that its 4,000 years into the future. Civilization as we know it has ceased to exist -- no books, no electronic devices, no Facebook or Twitter. All knowledge of the English language and the English alphabet has been lost. Now imagine archeologistsdigging through the rubble of one of our cities. What might they find? Well perhaps some rectangular pieces of plastic with strange symbols on them. Perhaps some circular pieces of metal. Maybe some cylindrical containers with some symbols on them. And perhaps one archeologist becomes an instant celebrity when she discovers -- buried in the hills somewhere in North America -- massive versions of these same symbols. Now lets ask ourselves, what could such artifacts say about us to people 4,000 years into the future?1:14This is no hypothetical question. In fact, this is exactly the kind of question were faced with when we try to understand the Indus V alley civilization, which existed 4,000 years ago. The Indus civilization was roughly contemporaneous with the much better known Egyptian and theMesopotamian civilizations,but it was actually much larger than either of these two civilizations. It occupied the area of approximately one million square kilometers, covering what is now Pakistan, Northwestern India and parts of Afghanistan and Iran. Given that it was such a vast civilization, you might expect to find really powerful rulers, kings, and huge monuments glorifying these powerful kings. In fact, what archeologists have found is none of that. Theyve found small objects such as these.1:59Heres an example of one of these objects. Well obviously this is a replica. But who is this person? A king? A god? A priest? Or perhaps an ordinary person like you or me? We dont know. But the Indus people also left behind artifacts with writing on them. Well no, not pieces of plastic, but stone seals, copper tablets, pottery and, surprisingly, one large sign board, which was found buried near the gate of a city. Now we dont know if it says Hollywood, or even Bollywood for that matter. In fact, we dont even know what any of these objects say, and thats because the Indus script is undeciphered. We dont know what any of these symbols mean.2:44The symbols are most commonly found on seals. So you see up there one such object. Its the square object with the unicorn-like animal on it. Now thats a magnificent piece of art. So how big do you think that is? Perhaps that big? Or maybe that big? Well let me show you. Heres a replica of one such seal.Its only about one inch by one inch in size --pretty tiny. So what were these used for? We know that these were used for stamping clay tags that were attached to bundles of goods that were sent from one place to the other. So you know those packing slips you geton your FedEx boxes? These were used to make those kinds of packing slips. You might wonder what these objects contain in terms of their text.Perhaps theyre the name of the sender or some information about the goods that are being sent from one place to the other -- we dont know. We need to decipher the script to answer that question.3:38Deciphering the script is not just an intellectual puzzle; its actually become a question thats become deeply intertwined with the politics and the cultural history of South Asia. In fact, the script has become a battleground of sorts between three different groups of people. First, theres a group of people who are very passionate in their belief that the Indus script does not represent a language at all. These people believe that the symbols are very similar to the kind of symbols you find on traffic signs or the emblems you find on shields. Theres a second group of people who believe that the Indus script represents an Indo-European language. If you look at a map of India today, youll see that most of the languages spoken in North India belong to the Indo-European language family. So some people believe that the Indus script represents an ancient Indo-European language such as Sanskrit.4:28Theres a last group of people who believe that the Indus people were the ancestors of people living in South India today. These people believe that the Indus script represents an ancient form of the Dravidian language family, which is the language family spoken in much of South India today. And the proponents of this theory point to that small pocket of Dravidian-speaking people in the North, actually near Afghanistan, and they say that perhaps, sometime in the past, Dravidian languages werespoken all over India and that this suggests that the Indus civilization is perhaps also Dravidian.5:03Which of these hypotheses can be true? We dont know, but perhaps if you deciphered the script, you would be able to answer this question. But deciphering the script is a very challenging task. First, theres no Rosetta Stone. I dont mean the software; I mean an ancient artifact that contains in the same text both a known text and an unknown text. We dont have such an artifact for the Indus script.And furthermore, we dont even know what language they spoke. And to make matters even worse,most of the text that we have are extremely short. So as I showed you, theyre usually found on these seals that are very, very tiny.5:37And so given these formidable obstacles, one might wonder and worry whether one will ever be able to decipher the Indus script. In the rest of my talk, Id like to tell you about how I learned to stop worryingand love the challenge posed by the Indus script. Ive always been fascinated by the Indus script ever since I read about it in a middle school textbook. And why was I fascinated? Well its the last major undeciphered script in the ancient world. My career path led me to become a computational neuroscientist, so in my day job, I create computer models of the brain to try to understand how the brain makes predictions, how the brain makes decisions, how the brain learns and so on.6:15But in 2023, my path crossed again with the Indus script. Thats when I was in India, and I had the wonderful opportunity to meet with some Indian scientists who were using computer models to try to analyze the script. And so it was then that I realized there was an opportunity for me to collaborate with these scientists, and so I jumped at that opportunity. And Id like to describe some of the results that we have found. Or better yet, lets all collectively decipher. Are you ready?6:41The first thing that you need to do when you have an undeciphered script is try to figure out the direction of writing. Here are two texts that contain some symbols on them. Can you tell me if the direction of writing is right to left or left to right? Ill give you a couple of seconds. Okay. Right to left, how many? Okay. Okay. Left to right? Oh, its almost 50/50. Okay. The answer is: if you look at the left-hand side of the two texts, youll notice that theres a cramping of signs, and it seems like 4,000 years ago, when the scribe was writing from right to left, they ran out of space. And so they had to cram the sign. One of the signs is also below the text on the top. This suggests the direction of writing was probably from right to left, and so thats one of the first things we know, that directionality is a very key aspect of linguistic scripts. And the Indus script now has this particular property.7:34What other properties of language does the script show? Languages contain patterns. If I give you the letter Q and ask you to predict the next letter, what do you think that would be? Most of you said U, which is right. Now if I asked you to predict one more letter, what do you think that would be? Now theres several thoughts. Theres E. It could be I. Itcould be A, but certainly not B, C or D, right? The Indus script also exhibits similar kinds of patterns. Theres a lot of text that start with this diamond-shaped symbol. And this in turn tends to be followed by this quotation marks-like symbol. And this is very similar to a Q and U example. This symbol can in turn be followed by these fish-like symbols and some other signs, but never by these other signs at the bottom. And furthermore, theres some signsthat really prefer the end of texts, such as this jar-shaped sign, and this sign, in fact, happens to be the most frequently occurring sign in the script.8:24Given such patterns, here was our idea. The idea was to use a computer to learn these patterns, and so we gave the computer the existing texts. And the computer learned a statistical model of which symbols tend to occur together and which symbols tend to follow each other. Given the computer model, we can test the model by essentially quizzing it. So we could deliberately erase some symbols,and we can ask it to predict the missing symbols. Here are some examples. You may regard this as perhaps the most ancient game of Wheel of Fortune.9:04What we found was that the computer was successful in 75 percent of the cases in predicting the correct symbol. In the rest of the cases, typically the second best guess or third best guess was the right answer. Theres also practical use for this particular procedure. Theres a lot of these texts that are damaged. Heres an example of one such text. And we can use the computer model now to try to complete this text and make a best guess prediction. Heres an example of a symbol that was predicted. And this could be really useful as we try to decipher the script bygenerating more data that we can analyze.9:36Now heres one other thing you can do with the computer model. So imagine a monkey sitting at a keyboard. I think you might get a random jumble of letters that looks like this. Such a random jumble of letters is said to have a very high entropy. This is a physics and information theory term. But just imagine its a really random jumble of letters. How many of you have ever spilled coffee on a keyboard?You might have encountered the stuck-key problem -- so basically the same symbol being repeated over and over again. This kind of a sequence is said to have a very low entropy because theres no variation at all. Language, on the other hand, has an intermediate level of entropy; its neither too rigid,nor is it too random. What about the Indus script? Heres a graph that plots the entropies of a whole bunch of sequences. At the very top you find the uniformly random sequence, which is a random jumble of letters -- and interestingly, we also find the DNA sequence from the human genome and instrumental music. And both of these are very, very flexible, which is why you find them in the very high range. At the lower end of the scale, you find a rigid sequence, a sequence of all As, and you also find a computer program, in this case in the language Fortran, which obeys really strict rules. Linguistic scripts occupy the middle range.10:49Now what about the Indus script? We found that the Indus script actually falls within the range of the linguistic scripts. When this result was first published, it was highly controversial. There were people who raised a hue and cry, and these people were the ones who believed that the Indus script does not represent language. I even started to get somehate mail. My students said that I should really seriously consider getting some protection. Whod have thought that deciphering could be a dangerous profession? What does this result really show? It shows that the Indus script shares an important property of language. So, as the old saying goes, if it looks like a linguistic script and it acts like a linguistic script, then perhaps we may have a linguistic script on our hands. What other evidence is there that the script could actually encode language?11:38Well linguistic scripts can actually encode multiple languages. So for example, heres the same sentence written in English and the same sentence written in Dutch using the same letters of the alphabet. If you dont know Dutch and you only know English and I give you some words in Dutch,youll tell me that these words contain some very unusual patterns. Some things are not right, and youll say these words are probably not English words. The same thing happens in the case of the Indus script. The computer found several texts -- two of them are shown here -- that have very unusual patterns. So for example the first text: theres a doubling of this jar-shaped sign. This sign is the most frequently-occurring sign in the Indus script, and its only in this text that it occurs as a doubling pair.12:23Why is that the case? We went back and looked at where these particular texts were found, and it turns out that they were found very, very far away from the Indus Valley. They were found in present day Iraq and Iran. And why were they found there? What I havent told you is that the Indus people were very, very enterprising. They used to trade with people pretty far away from where they lived, and so in this case, theywere traveling by sea all the way to Mesopotamia, present-day Iraq. And what seems to have happened here is that the Indus traders, the merchants, were using this script to write a foreign language. Its just like our English and Dutch example. And that would explain why we have these strange patterns that are very different from the kinds of patterns you see in the text that are found within the Indus Valley. This suggests that the same script, the Indus script, could be used to write different languages. The results we have so far seem to point to the conclusion that the Indus script probably does represent language.13:19If it does represent language, then how do we read the symbols? Thats our next big challenge. So youll notice that many of the symbols look like pictures of humans, of insects, of fishes, of birds. Most ancient scripts use the rebus principle, which is, using pictures to represent words. So as an example, heres a word. Can you write it using pictures? Ill give you a couple seconds. Got it? Okay. Great.Heres my solution. You could use the picture of a bee followed by a picture of a leaf -- and thats belief, right. There could be other solutions. In the case of the Indus script, the problem is the reverse.You have to figure out the sounds of each of these pictures such that the entire sequence makes sense. So this is just like a crossword puzzle, except that this is the mother of all crossword puzzlesbecause the stakes are so high if you solve it.14:17My colleagues, Iravatham Mahadevan and Asko Parpola, have been making some headway on this particular problem. And Id like to give you a quick example of Parpolas work. Heres a really short text.It contains seven vertical strokes followed by this fish-like sign. And I want tomention that these seals were used for stamping clay tags that were attached to bundles of goods, so its quite likely that these tags, at least some of them, contain names of merchants. And it turns out that in India theres a long tradition of names being based on horoscopes and star constellations present at the time of birth. In Dravidian languages, the word for fish is meen which happens to sound just like the word for star.And so seven stars would stand for elu meen, which is the Dravidian word for the Big Dipper star constellation. Similarly, theres another sequence of six stars, and that translates to aru meen, which is the old Dravidian name for the star constellation Pleiades. And finally, theres other combinations,such as this fish sign with something that looks like a roof on top of it. And that could be translated into mey meen, which is the old Dravidian name for the planet Saturn. So that was pretty exciting. It looks like were getting somewhere.15:29But does this prove that these seals contain Dravidian names based on planets and star constellations?Well not yet. So we have no way of validating these particular readings, but if more and more of these readings start making sense, and if longer and longer sequences appear to be correct, then we know that we are on the right track. Today, we can write a word such as TED in Egyptian hieroglyphics and in cuneiform script, because both of these were deciphered in the 19th century. The decipherment of these two scripts enabled these civilizations to speak to us again directly. The Mayans started speaking to us in the 20th century, but the Indus civilization remains silent.16:14Why should we care? The Indus civilization does not belong to justthe South Indians or the North Indians or the Pakistanis; it belongs to all of us. These are our ancestors -- yours and mine. They were silenced by an unfortunate accident of history. If we decipher the script, we would enable them to speak to us again. What would they tell us? What would we find out about them? About us? I cant wait to find out.16:45Thank you.16:47(Applause)托福阅读备考之代词的应用1、指代题例题1：Paragraph 3: At the upper timberline the trees begin to become twisted and deformed. This is particularly true for trees in the middle and upper latitudes, which tend to attain greater heights on ridges, whereas in the tropics the trees reach their greater heights in the valleys. This is because middle- and upper- latitude timberlines are strongly influenced by the duration and depth of the snow cover. As the snow is deeper and lasts longer in the valleys, trees tend to attain greater heights on the ridges, even though they are more exposed to high-velocity winds and poor, thin soils there. In the tropics, the valleys appear to be more favorable because they are less prone to dry out, they have less frost, and they have deeper soils.The word they in the passage refers to○valleys○trees○heights○ridges首先，在考察指代题时，代词“they”在题干中打上阴影，并且原文中代词“they”也打上了阴影，所以指代题不需要定位。

GRE真题考试阅读部分关键句整合

GRE真题考试阅读部分关键句整合我给大家带来了GRE真题考试阅读部分关键句，快来学习一下吧，下面我就和大家共享，来观赏一下吧。

GRE真题考试阅读部分关键句英文原句31. Given the nature of government and private employers, it seems most likely that discrimination by private employers would be greater.32. The release of the carbon in these compounds for recycling depends almost entirely on the action of both aerobic and anaerobic bacteria and certain types of fungi.33. A spirited discussion springs up between a young girl who says that women have out grown the jumping-on-a-chair-at-the-sight-of-a mouse era and a major who says that they haven’t.34. They are trying to find out whether there is something about the way we teach language to children which in fact prevents children from learning sooner.35. Mathematicians who have tried to use the computers to copy the way the brain works have found that even using the latest electronic equipment they would have to build a computer which weighed over 10,000 kilos.36. Since different people like to do so many different things in their spare time, we could make a long list of hobbies, taking in everything from collecting matchboxes and raising rare fish, to learning about the stars and making model ships.37. They know that a seal swimming under the ice will keep abreathing hole open by its warm breath, so they will wait beside the hole and kill it.38. We may be able to decide whether someone is white only by seeing if they have none of the features that would mark them clearly as a member of another race.39. Although signs of dishonesty in school , business and government seem much more numerous in years than in the past, could it be that we are getting better at revealing such dishonesty?40. It is not quite a matter of disagreeing with the theory of independence, but of rejecting its implications: that the romances may be taken in any or no particular order, that they have no cumulative effect, and that they are as separate as the works of a modern novelist.GRE真题考试阅读部分关键句译文31、依据政府和私人雇主的性质来看，私人雇主更有可能实行卑视。

数学证明的重要性

WHAT ARE MATHEMATICAL PROOFS AND WHYTHEY ARE IMPORTANT?introductionMany students seem to have trouble with the notion of a mathemat-ical proof.People that come to a course like Math216,who certainly know a great deal of mathematics-Calculus,Trigonometry,Geometry and Algebra,all of the sudden come to meet a new kind of mathemat-ics,an abstract mathematics that requires proofs.In this document we will try to explain the importance of proofs in mathematics,and to give a you an idea what are mathematical proofs.This will give you some reference to check if your proofs are correct.We begin by describing the role of proofs in mathematics,then we deﬁne the logical language which serves as the basis for proofs and logical deductions. Next we discuss brieﬂy the role of axioms in mathematics.Finally we give several examples of mathematical proofs using various techniques. There is also an excellent document on proofs written by Prof.Jim Hurley(liknked to the course homepage),and I recommend to all of you to read his document as well.You willﬁnd there more examples and additional explanations.It is my hope that reading these docu-ments will make yourﬁrst steps in writing proofs easier.1.The importance of Proofs in mathematicsIt is diﬃcult to overestimate the importance of proofs in mathemat-ics.If you have a conjecture,the only way that you can safely be sure that it is true,is by presenting a valid mathematical proof.For example,consider the following well known mathematical theorem:Theorem1(Euclid).There are inﬁnitely many prime numbers.For those of you who don’t remember,a prime number is a posi-tive integer p>1that cannot be written as a product of two strictly smaller positive integers a and b.For example,the number91is not prime since it can be written as91=13·7,but67is a prime.Although most people have a hunch that there are inﬁnitely many primes,it is not obvious at all.Even today with the powerful computers,all we can do is to verify that there are very large prime numbers,but we12WHAT ARE PROOFSstill can say nothing about the existence of prime numbers whom size is beyond the ability of the current computers.The only reason that we are one hundred percent sure that the theorem is true,is because a mathematical proof was presented by Euclid some2300years ago.We shall give his proof later.Another importance of a mathematical proof is the insight that it may oﬀer.Being able to write down a valid proof may indicate that you have a thorough understanding of the problem.But there is more than this to it.The eﬀorts to prove a conjecture,may sometimes require a deeper understanding of the theory in question.A mathematician that tries to prove something may gain a great deal of understanding and knowledge,even if his/her eﬀorts to prove that conjecture will end with failure.The following theorem,is due to the giant German Mathematician Karl Friedrich Gauss,and is called the Fundamental Theorem of Algebra.Theorem2(the Fundamental Theorem of Algebra).Let p(x)be a non constant polynomial whose coeﬃcients are complex numbers.Then the equation f(x)=0has a solution in complex numbers.In particular this means that equations like x6+x5−3x2+1=0have a complex solution,a non trivial fact.The proof that Gauss gave relies heavily on the fact that the complex numbers form a two dimensional plane.In fact,to understand the idea behind the theorem,we need to incorporate knowledge from a seemingly unrelated area of mathemat-ics–Topology.Topology,like usual geometry deals with shapes,but unlike geometry,the shapes are not rigid and may be deforemd(This is a very unproﬀesional deﬁnition).For example,a cube and a ball are topologically equivalent,but both are not equivalent to a donut. The topological fact used by the proof is that if you have two rubber bands lying on a plain,one of them is surrounding a nail stuck in the plane,and the other isn’t,then you cannot bring theﬁrst rubber band to the situation of the second one without lifting it oﬀthe plane or opening ing this fact in an ingenious way,where the plane is the plane of complex numbers,gave a proof to the above algebraic. What was demonstrated here is that trying to prove something may lead to a deeper knowledge and to relations to otherﬁelds of mathe-matics.Interestingly,there are additional proofs to the same theorem, each coming from a completely diﬀerent approach and mathematical knowledge,and it is a challenge to try to understand them all as parts of a bigger picture.The fundamental theorem itself can be used to giveWHAT ARE PROOFS3 a complete explanation why people failed to create a three or more di-mensional number system with the+,−,·,÷operations(what we will call a’Field’later in the course).Understanding the ideas behind the theorem lead to the understanding why the complex numbers are so fundamental in mathematics(this was not recognized by the contem-poraries of Gauss).Part of the diﬃculties of people in understanding the notion of proofs stem from the fact that people do not have the right picture of what mathematics is.From elementary school through theﬁrst years of col-lege we teach people that the goal is to solve an equation or toﬁnd a minimum of a function or toﬁnd how much wheat we should grow. This is of course something that mathematics can do,but it is not what mathematics is about.Mathematics is about understanding the laws behind numbers,algebra and geometry.It is aboutﬁnding new and non routine ways to look at these systems and to explain strange phenomena that we may encounter.To make it more interesting,we can even change the laws to create new systems and then study them. An example for such a new system is the notion of a’Group’,that we shall study in this course.Some of these new systems,like groups, shed light on the old ones.Groups have been found to be a necessary utility in understanding questions in number theory,algebra,,topology, geometry,and modern physics.There is a whole new world of ideas, understanding and discoveries that is invisible to people who only know how to diﬀerentiate a function.To enter to this world,it is necessary to use the ideas of abstraction and mathematical proof.2.What are Mathematical Proofs?2.1.The rules of the game.All of you are aware of the fact that in mathematics’we should follow the rules’.This is indeed the case of writing a mathematical proof.Before we see how proofs work,let us introduce the’rules of the game’.Mathematics is composed of statements.The Law of the excluded middle says that every statement must be either true of false,never both or none.If it is not true,then it is considered to be false.For example,consider the statement y=x2.In real life’s terminology,like with most statements,it is sometimes true and sometimes false.In4WHAT ARE PROOFSmathematics it is false since it is not completely true.Take for exam-ple y=1and x=2.We can make up new statements from old ones.If P and Q are statements,then we have.Statement Notation P and Q P∧Q(or P&Q) P or Q P∨Q If P Then Q(or P implies Q)P=⇒Q P if and only if Q(or P and Q are equivalent)P⇐⇒Q not P¬P The statement P∧Q is true if and only if both P and Q are true.The statement P∨Q is true if and only if at least one of the statements P, Q is true.This is what we called the inclusive or as opposed to the exclusive or that we use in everyday’s life.For example,the statement:’We will have class in the morning or in the afternoon’means in real life that only one of the alternatives will take places(exclusive or). In mathematics however,this includes the possibility that we will have class in the morning as well as in the afternoon(inclusive or).P=⇒Q is considered to be false only in the case that P is true and Q is false. Otherwise it is true.This is also in contrast to the plain English.For example,a statement like:’If it rains now then2is a prime number’is mathematically true,despite the fact that there is no relation between the two parts of the statement,and regardless of whether theﬁrst part is true or not.The statement P⇐⇒Q is true if and only if P and Q have the same truth values.We also say that P and Q are equivalent.Finally,¬P is true if and only if P is false.Using the above operations we can make more complex statements like ((¬(P=⇒Q)∨(T∧P))⇐⇒(R=⇒¬T).Two additional players in this game are the quantiﬁers,∀which means’for all’,and∃which means’there exists’.For example,the statement∀x>0∃y(y2=x),reads:For all x>0there exists y such that y2=x.Let us denote by Z+the set of all positive integers.The following statement aboutWHAT ARE PROOFS5 p∈Z+says that p is a prime:∀a∈Z+∀b∈Z+[p=ab=⇒(a=1)∨(b=1)].2.2.Tautologies and Contradictions.Logical deductions are based tautologies.If we try to give an informal deﬁnition,a tautology is a general statement that is true under all possible circumstances.Exam-ples are P=⇒P,P∨¬P,Modus Ponens:[P∧(P=⇒Q)]=⇒Q or another form of Modus Ponens:[(P=⇒Q)∧(Q=⇒R)]=⇒(P=⇒R).Notice that the truth of each of these statements is independent of whatever are P and Q and R and whether they are true or false.We can establish a tautology by our understanding of the statement or by constructing a truth table.A logical deduction is obtained by substituting in a tautology.For example,look at the following deduction:Allﬁsh are living creatures;All living creatures canmove;Therefore allﬁsh can move.This logical deduction is a substitution in the second form of Modus Ponens where P=’x is aﬁsh’;Q=’x is an living creature’;R=’x can move’.On the other extreme of logical statements are the contradictions. Contradictions are false under all circumstances.In fact every contra-diction is the negation of a tautology,and conversely,every tautology is the negation of a contradiction.A typical contradiction is P∧¬P. Try to think yourself of some other examples.2.3.Axioms.As it turns out,to prove something requires the knowl-edge of some previous truths.Logic just supplies the ways that we can deduce a statement from others,but we need some statements to begin with.These initial statements are called axioms.There are two layers of axioms:The axioms of set theory,and the axioms of the mathemat-ical theory in question.Modern mathematics is based on the foundation of set theory and logic.Most mathematical objects,like points,lines,numbers,func-tions,sequences,groups etc.are really sets.Therefore it is necessary to begin with axioms of set theory.Very fundamental to set theory is the set of positive integers Z+,which has the natural order relation on it.The Well ordering principle is an axiom which says that every nonempty subset of Z+has a smallest element.You willﬁnd it very6WHAT ARE PROOFSdiﬃcult to resist to this statement,however,it is considered an axiom without proof.This axiom turns out to be equivalent to the principle of mathematical induction (cf.your textbook p.4).The axioms of set theory form the basic layer of axioms.It needs to be said that some of the axioms of set theory,like the well ordering principle,are accepted only for being highly intuitive,and have been debated among mathematicians.On top of this,we have the axioms that deﬁne the theory in ques-tions.For example,in geometry we have the axiom which states that one straight line and only one straight line can pass through two given distinct points .In modern mathematics there is no point with arguing with such an axiom.If it does not hold,we simply cannot call the objects by the names ’straight line’and ’point’.The axioms serve here as deﬁnitions that characterize that mathematical system in question.A mathematical theory is like a chess game and the axioms correspond to the rules of the game.If you don’t accept a rule,this is not chess any more.The only kind of debates is to whether the theory is math-ematically fruitful or interesting.2.4.Proofs.A proof of a theorem is a ﬁnite sequence of claims,each claim being derived logically (i.e.by substituting in some tautology)from the previous claims,as well as theorems whose truth has been already established.The last claim in the sequence is the statement of the theorem,or a statement that clearly implies the theorem.We wish now to give some examples that will illustrate how this works in prac-tice,as well as some techniques of proofs.You can ﬁnd more examples in the document of Prof.HurleyDirect Proofs:Many theorems are of the form P =⇒Q ,that is,of the form If ...Then ....We begin by assuming P (as well as other established truths)and proceed by using a sequence of Modus Ponenses to derive new statements,the last statement being Q .Alternatively we use the second form of Modus Ponens and a sequence of implications to derive P =⇒Q .Here is an example:Theorem 3.For any two positive numbers x and y ,(∗)√xy ≤x +y 2.Proof.Suppose that x and y are positive numbers.Then (√x −√y )2≥0.By algebra this implies that x +y −2√x √y ≥0.Moving 2√x √yWHAT ARE PROOFS 7to the other side we get that x +y ≥2√xy .Dividing both sides by 2yields the inequality (∗)of the theorem as desired. Notice that we have used implicitly some general truths from algebra,like √x √y =√xy ,in conjunction with the statements of the proofs.The use of external knowledge becomes more and more necessary as we go further in developing a theory.More complex proofs require nested sequences of Modus Ponenses.Theorem 4.Let A and B be two sets.If A ∪B =A ∩B then A ⊆B .Proof.Assume that A ∪B =A ∩B .We shall prove that x ∈A =⇒x ∈B ,which by deﬁnition is equivalent to the consequence of the theorem.Assume that x ∈A .Since A ⊆A ∪B ,then x ∈A ∪B .We assumed that A ∪B =A ∩B ,so x ∈A ∩B .Finally,A ∩B ⊆B ,so consequently,x ∈B .This concludes the proof Notice that we did not end the proof with the consequence of the theorem,but rather with a statement that suﬃces to imply the theo-rem by our earlier understanding.Here also we used implicitly some deﬁnitions from set theory,like (x ∈D )∧(D ⊆C )=⇒(x ∈C ).Notice that we have made a substential use of the English language.To be absolutely sure that our proofs are valid,we must use the sym-bolic language and list down all the claims that we used,including the knowledge from algebra or set theory,and to check that the implica-tions really follow from tautologies.This however,will make the proofs cumbersome and more diﬃcult to understand.We choose to resolve this dilemma by using the English language,but in a very limited way,so that the deductions will include only the necessary ’logical’words like ’if’,’then’,’suppose’,’consequently’,’or’etc.This,together with some intelligence,makes us quite convinced that our proofs are with-out ﬂaws,and that if we wish,it would be possible to rewrite the proof in the symbolic structural language.Proofs by Negation:This is another strategy of proving a theorem P =⇒Q .We begin by assuming that P is true and Q is false i.e.we assume P ∧¬Q .The proof proceeds until we derive contradiction F .This completes the proof,since something must be wrong and the only questionable thing was our assumption P ∧¬Q .Thus if P is true,Q must be true too.Technically,we have based our proof scheme on the tautology:[(P ∧¬Q )=⇒F ]=⇒[P =⇒Q ],where F is a contradiction.Here is an example:8WHAT ARE PROOFSTheorem5.There are inﬁnitely many prime numbersNotice that we can view the theorem as an If-Then theorem by claim-ing that if S is the set of prime numbers then S is an inﬁnite set. Proof.Suppose,by negation,that the set S of prime numbers isﬁnite. Then we can write it as the set{p1,p2,...p n}where p1,p2,...,p n aredistinct numbers.The product N=p1p2···p n is clearly a multiple of each of the numbers p1,p2,...,p n.Therefore,the number N+1 is not a multiple of neither p1,p2,...,p n.On the other hand,by the fundamental theorem of arithmetic(see textbook p.6),N+1is a product of prime numbers so it is a multiple of at least one prime number.This is a contradiction because N+1is not a multiple of p1,p2,...,p n which are all the prime numbers.The proof is complete.Mathematical Induction Suppose that we have a sequence of claims P(1),P(2),...and we wish to prove them all at once.This amounts to the claim∀nP(n)(i.e.For all n,P(n)is true).Rather then proving this directly,we can use the principle of mathematical induction:P(1)∧∀n[P(n)=⇒P(n+1)]=⇒∀nP(n).In words,this means as follows:Suppose that we have proved the following two statements:(1)[The induction basis]P(1)is true,and(2)[The induction step]For all n,if P(n)is true,then P(n+1)is trueThen for all n,P(n)is true.The induction principle makes sense because upon establishing the statements P(1),P(1)=⇒P(2),P(2)=⇒P(3),etc.,we can use a sequence of Modus Ponenses to establish P(2),P(3)and so on.As we mentioned above,the principle of mathematical induction may be seen as a consequence of the well ordering principal.It turns out that the converse is also true,namely,that the well ordering principal can be proved from mathematical induction.In some mathematical literature the induction principle is taken to be an axiom rather than the equivalent well ordering principle.We shall now give an example.Suppose that we want to show that for every n∈Z+,the number17n−10n is divisible by7.We will prove this by induction.Let P(n)be the claim that17n−10n is divisible byWHAT ARE PROOFS9 7.First,if n=1,then171−101=7so P(1)is true.Secondly,assume that P(n)is true.We shall now prove that this implies that P(n+1) is true.We compute:17n+1−10n+1=17·17n−10·10n=(7+10)·17n−10·10n= =7·17n+10·17n−10·10n=7·17n+10·(17n−10n). Now,7·17n is clearly divisible by7.By the induction hypothesis, 17n−10n and so10·(17n−10n)is divisible by7.By the computation above we have shown that17n+1−10n+1is divisible by7.By the in-duction principle we are done.There is a slightly diﬀerent version of induction,where the induction step(2)is replaced with(2)’If P(k)is true for all k≤n,then P(n+1)is trueIn some situations like the following one,the latter version applies rather than the usual version.Suppose that we play the following game.You are given a pile of N matches.You break the pile into two smaller piles of m and n matches.Then you form the product2mn and remember it.Next,you take one of the piles and break it into two smaller piles(if possible),say of m and n matches.You form the product2m n and add it to the2mn that you had before,so now you have2mn+2m n .You proceed again by breaking one of the piles into two and adding the resulting product.The process isﬁnished when youﬁnally have N piles of one match in each.By convention,if N=1 then you don’t do anything and the result is0.Try to take a pile ofﬁve matches and play this game several times,each time breaking to piles in a diﬀerent way.What do you see?In fact the following theorem is true:Theorem6.If you start from a pile of N matches,no matter how you break it,the sum of the computed products will always be N2−N. Proof.We will use the second version of induction on N.First,if N=1 then the resulting sum is0=N2−N.If N=2then there is only one way to break it,namely to m=1and n=1so that we end up with2mn=2=N2−N.Assume that N≥2and the theorem is true for all k≤N.To prove the theorem for N+1,we begin with a pile of N+1matches and break it into two plies of m and n matches,so that N+1=m+n.By the induction hypothesis,breaking up the m pile will contribute to the sum m2−m,and similarly the second pile will contribute n2−n,no matter how we continue to break them. Therefore the total sum is2mn+(n2−n)+(m2−m).A little algebra10WHAT ARE PROOFSshows that this equals to(m+n)2−(m+n)=(N+1)2−(N+1)as desired. The Pigeon Hole Principle:is a frequently used principle inﬁ-nite mathematics and is often used in proofs of existence.To illustrate the principle(and explain its name),suppose that you have n pigeon holes,but more than n pigeons.If you are to put all your pigeons in the holes,then the pigeon hole principle says that in at least one pigeon hole you will have to put more than one pigeon.To state this in a more mathematical language,recall that if A is aﬁnite set,then the order of A,which is denoted by|A|,is the number of elements in A.The pigeon hole principle will say that if A and B areﬁnite sets with|A|>|B|and f:A→B is a function,then f is not one to one, i.e.there exist x,y∈A,x=y,such that f(x)=f(y).Here is an example.Suppose that you take the numbers1,2,...,10 and rearrange them as you like.As you know there are10!=3,628,800 diﬀerent rearrangements.After you have rearranged,you look for the longest partial sequence,which is rearranged by ascending or descend-ing order.For example,if you have rearranged the numbers to have 4,3,6,5,9,10,1,7,2,8you notice that the partial sequence4,3,5,7,8 is in ascending order.If you look hard enough you will notﬁnd a longer ascending/descending partial sequence.Your goal is to rear-range the numbers in the’most eﬀective way’so that the longest as-cending/descending partial sequence is the shortest possible.How well can you do?The following theorem gives the answer:Theorem7.In any rearrangement of the numbers1,2,...,10,there is always a partial sequence of four numbers which is ordered in an ascending or descending order.Proof.We will prove this existence theorem using the pigeon hole prin-ciple.Pick up any rearrangement and let the rearranged numbers be x1,...,x10.To each index1≤i≤10,let a(i)be the length of the longest ascending partial sequence that ends with x i,and let d(i)be the length of the longest descending partial sequence that begins with x i.We claim that if for two indices i and j,a(i)=a(j)and d(i)=d(j), then necessarily i=j.Indeed suppose by negation that i=j.Then of course x i=x j and there are two cases:(i)x i<x j,in which we could append x j to the longest ascending sequence ending in x i and thus get yet a longer ascending sequence ending in x j which shows thatWHAT ARE PROOFS11 a(j)>a(i).In case(ii),x i>x j and a similar argument shows that now d(i)>d(j).In either cases there is a contradiction to our assump-tion that a(i)=a(j)and d(i)=d(j).Suppose by negation that in our rearrangement the longest ascend-ing/descending partial sequence consists of less than4numbers.Then clearly for each i,a(i)and d(i)can only have the values of1or2or3. Then we have a functionf:{1,...,10}→{(1,1),(1,2),(1,3),...,(3,2),(3,3)},given by f(i)=(a(i),d(i)).Now f is a function from a set of10 elements to a set of9elements.By the pigeon hole principle,f is not one to one,and there exist indices i=j such the pairs(a(i),d(i))= f(i)=f(j)=(a(j),d(j)).As explained above this is impossible. Exercise.Try toﬁnd a rearrangement of the numbers1,...,10,such that the longest ascending/descending partial sequence is of length4.。

21_flexray

[FlexRay04]
Dynamic segment frames are similar
• Adds a DTS = dynamic trailing sequence field; helps line up minislots
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FlexRay Frame Format
[FlexRay04]
• This data is encoded into NRZ bytes per the encoding format
Preview
FlexRay – main competitor to TTP for X-by-Wire applications
• Created by industry consortium founded in 2000 • Core members: BMW, DaimlerChrysler, General Motors, Motorola, Philips, Volkswagen, and Robert Bosch. • Hyundai Kia Motors is a “premium associate” member
– Less critical/less expensive nodes might only connect to one channel
• Slots are lock-stepped in order on both channels
TDMA order is by ascending frame ID number
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FlexRay Frame Fields
Frame ID
• Frame’s slot number (1 .. 2047); unique within channel in communication cycle

Unit6KeytoExercises

Unit 6KEY TO EXERCISESText comprehension 1I. Decide which of the following best states the speaker's purpose of his speech.AII. Judge, according to the text, whether the following statements are true or false.1. T. Refer to Paragraph 1.2. F. Refer to Paragraph3. The question "Are fleets and armies necessary to a work of love and reconciliation?" is obviously a rhetorical question.3. T. Refer to Paragraph 3.4. T. Refer to Paragraphs 5 and 6.5. F. Refer to Paragraph 7. This is how the British colonists and a few gentlemen in the house think.6. T. Refer to the whole speech.III. Answer the following questions.1. Refer to Paragraph 1. They expect Patrick Henry to express in the strongest and most explicit terms his opinions on the nation's choice between freedom and slavery, which are very different from those expressed by some previous speakers.2. Refer to Paragraph3. Lamp metaphorically refers to one's experience that guides one's course of action. The past conduct of the British ministry has proved itself untrustworthy. The recent reception of our petition will prove a snare to mislead us. As a matter of fact, the British are preparingwar against us. We must call in our own forces to fight back. This is whatthe lamp guides us to do.3. Refer to Paragraph4. It has accumulated its navies and armies for war against the Americans in an attempt to subdue them.4. Refer to Paragraph 6. He thinks that their petitions have been treated rudely, their remonstrations have brought about more violence and insult, and that their solicitations have been disregarded. After they have done all that can be done, there is no longer any possibility for hope of peace and reconciliation.5. Refer to Paragraph6. They have done everything that could be done to prevent the coming war, but in vain. If they wish to be free, if they meanto preserve those inestimable rights of national independence and dignity,for which they have long been engaged in the noble struggle, they must fight.6. Refer to Paragraphs 7 and 8. There is no way for them to retreat, or submission and slavery will be in store for them! The clanging of the chains the British colonialists have long forged for them may be heard on the plains of Boston! War has actually begun and is approaching fast. If theyare unwilling to be subdued and enslaved, they must get ready for a war of resistance against the British colonists!7. Refer to Paragraph 8. Sweet as peace, dear as life might be, he willgive both up providing they should be won at the sacrifice of dignity and liberty.8. Refer to the whole speech. The speech is filled with patriotic passion and urgency, severe criticism, and appealing sensation.IV. Explain in your own words the following sentences taken from the text. 1. The same object may be observed and judged from different perspectives by different people.2. No time should be wasted on ceremonial procedures because the house, at present, is encountering an extremely crucial problem for the nation.3. We tend to close our eyes when facing a painful truth, and be intoxicated by the song of the sea nymph that will eventually turn us into animals.4. As for me, I'm willing to know the whole truth and be prepared for the worst that might happen, no matter how much pain I may endure.5. The cunning smile, with which the British recently received our petition, will be a trap for you to fall into.6. These are the tools for war and suppression, the last means kings will turn to when all arguments fall flat.7. We have been humble and submissive in front of the British King, and have begged his Majesty to intervene and stop the cruelty and injustice ofthe British colonial ministry and Parliament.8. The victory of the battle is determined not just by strength, but by vigilance, activeness, and courage.9. It is useless to underestimate the severity of the situation.Writing strategies .(1) Both the content and the tone of the beginning make it very appealing to the audience. The first two sentences are particularly eye-catching and capable of arousing the readers' interest, because they present a sharp contrast of different opinions. The first sentence makes it clear that the speaker greatly admires the worthy gentlemen for their patriotism and abilities, while the second sentence points out that he entertains entirely different opinions. Next, the speaker defines the question before the houseas one of freedom or slavery, which clearly states the critical importanceof the question. Then, he renders it crystal clear why he is duty-bound to speak forth his sentiments. The last sentence of the first paragraphindirectly highlights the great importance of his opinions. The tone of thefirst paragraph seems ironical and solemn as well.(2) The last paragraph expresses the speaker's sentiments quite clearly and effectively, and it impressively calls on people to fight for freedom. It creates an atmosphere of urgency, urging people to throw themselves into the battle to fight for freedom. It is to be noted that the second sentence andthe following three present a sharp contrast: the gentlemen keep crying, "Peace, peace", but as a matter of fact, no peace exists. The last sentenceis particularly stimulating, for it not only expresses the orator's ownattitude towards life, but also succeeds in calling on people to fight for liberty.(3) The beginning is closely related to the conclusion in the following two ways. First, both the beginning and the conclusion present the sharp contrast: the actual situation is quite different from what the gentlemenclaim. Secondly, the beginning is connected with the conclusion by the topicof freedom.(4) Examples of metonymy: The question before the house ?I should consider myself as guilty of treason toward my country, and of an act of disloyalty toward the Majesty of Heaven ?Examples of metaphor: 1) I have but one lamp by which my feet are guided, and that is the lamp of experience. 2) ?it will prove a snare to your feet.3) They are sent over to bind and rivet upon us those chains which theBritish ministry have been so long forging. 4) Sir, we have done everythingthat could be done to avert the storm which is now coming on. 5) The nextgale that sweeps from the north will bring to our ears the clash ofresounding arms!Examples of rhetorical questions: 1) Is this the part of wise men, engaged in a great and arduous struggle for liberty? Are we disposed to be of the number of those who having eyes see not, and having ears hear not, thethings which so nearly concern their temporal salvation? 2) Are fleets and armies necessary to a work of love and reconciliation? Have we shown ourselves so unwilling to be reconciled that force must be called in to winback our love? 3) Shall we resort to entreaty and humble supplication? What terms shall we find which have not already been exhausted? 4) Will it bewhen we are totally disarmed, and when a British guard shall be stationed in every house? Shall we gather strength by irresolution and inaction? Shall we acquire the means of effectual resistance by lying supinely on our backs,and hugging the delusive phantom of hope until our enemies shall have bound us hand and foot? 5) Why stand we here idle? 6) Is life so dear, or peace so sweet, as to be purchased at the price of chains and slavery?Examples of parallelism: The above rhetorical questions are also parallel sentences. More instances of parallelism: 1) They are meant for us: they canbe meant for no other. They are sent over to bind and rivet upon us those chains which the British ministry have been so long forging. 2) We have petitioned; we have remonstrated; we have supplicated; we have prostrated ourselves before the throne, and have implored its interposition to arrestthe tyrannical hands of the ministry and Parliament. Our petitions have been slighted; our remonstrances have produced additional violence and insult;our supplications have been disregarded; and we have been spurned with contempt from the foot of the throne! 3) If we wish to be free, if we meanto preserve inviolate those inestimable privileges for which we have been so long contending, if we mean not basely to abandon the noble struggle inwhich we have been so long engaged, and which we have pledged ourselves never to abandon until the glorious object of our contest shall be obtained-- we must fight! I repeat it, sir, we must fight! 4) There is no retreatbut in submission and slavery! Our chains are forged! Their clanging may be heard on the plains of Boston! The war is inevitable -- and let it come! Irepeat it, sir, let it come! 5) Why stand we here idle? What is it that gentlemen wish? What would they have? Is life so dear, or peace so sweet, asto be purchased at the price of chains and slavery?As is seen clearly, the rhetorical device of repetition is often usedtogether with parallelism. The two rhetorical devices usually accompany each other.Language work .I. Explain the underlined part in each sentence in your own words.1. the very same thing as2. appropriately in agreement with3. considered and tried every means in order to deal with the subject4. has given us at our disposal5. with liberty as our ultimate aim to fight for6. is not destined to be won byII. Fill in each blank with one of the two words from each pair in their appropriate forms and note the difference of meaning between them.solace consoleExplanation: Console suggests the attempt to make up for a loss by offering something in its place, as well as the effort of one person to mitigate the serious grief felt by another. Solace might sound more precious than comfort and it suggests a tender intensity of fellow-feeling.1. consoled2. solace3. console4. solacedpetition pleadExplanations: Both words refer to a humble, deferential, urgent, or formal request for help. Plead suggests a dignified humility, stressing an urgency. Petition suggests a formal address to authority, usually referring to the backing up of a request by the signed approval of others.1. pleaded2. pleaded3. petitioned4. petitioningbase (adj.) meanExplanation: Both words are alike in describing persons or actions regarded as being far below common worth or dignity. Mean suggests a contemptible smallness of mind, or a petty, ungenerous nature, while base is used to condemn what is openly evil, selfish, dishonourable or otherwise immoral.1. meanest2. mean3. base4. basedelusive misleadingExplanation: The two words refer to the giving or receiving of mistaken impressions. Misleading is restricted to something that is apt to give afalse impression, and it can also apply to great or small potential misapprehensions, whether fostered intentionally, unintentionally, orwithout any intent whatever. Delusive suggests mistaken impressions or aself-imposed belief that corresponds to one's own wishes or needs.1. delusive2. delusive3. misleading4. misleading III. Fill in the blank in each sentence with a word or phrase taken from thebox, using its appropriate form.1. Her manner is friendly and relaxed and much less formidable than she appears at her after-game press conference.2. Nothing has ever equaled the magnitude and speed with which the human species is altering the physical and chemical world and demolishing the environment.3. When heated, the mixture becomes soft and malleable and can be formed by various techniques into a vast array of shapes and sizes.4. Where I part company with him, however, is over the link he forges between science and liberalism.5. Percy was lying prostrate, his arms outstretched and his eyes closed.6. Given data which are free from bias, there are further snares to avoidin statistical work.7. In pragmatics, the study of speech, one is able to see how specific actsare related to a temporal and spatial context.8. His dad might have been able to say something solacing, had he not been fighting back his own flood of anguish.IV. Make a sentence of your own for each of the given words with meanings other than those used in the text. Y ou may change the part of speech ofthese words.1. No matter how we, my mum and I, protest, my dad chain-smokes as long as he is awake.2. The meeting ended on reasonably amicable terms.3. That doesn't interest me in the slightest.4. It is plain truth that we can't afford a deluxe car, so we have to makedo with this old gas guzzler.5. Children's programs on TV should aim to both educate and entertain atthe same time.6. The causes are a blend of local and national tensions.V. Rewrite the following sentences by transforming the finite clauses (in italics) into prepositional phrases.1. He painted so well, to the astonishment of every one of us.2. Every precaution was taken against the failure of the plan.3. But for Jack, they would have lost the football game.4. I must remind you of your responsibility towards your children.5. The committee has decided on postponing the meeting.6. No one is sorry about Peter's resignation.7. They will never get there without walking a long way.8. We were amused at your meeting the Harrisons there.9. For all / Despite his immense fortune, he died a most unhappy man.10. In spite of / Despite the clear evidence showing that smoking was harmful to one's health, people still refused to believe it.Note: A prepositional phrase consists of a preposition followed by a prepositional complement, which is characteristically a noun phrase or awh-clause or V-ing clause. A prepositional phrase may be used in place of anon-finite clause, simplifying the structure of the sentence.VI. Put a word in each blank that is appropriate for the context.1.version2. how3. Not4. instead5. did6. when7. top8. patience9. like 10. followTranslation .I. Translate each of the following sentences into English, using the wordsor expressions given in the brackets.1. If you look at this painting in a different light, you'd feel muchbetter about it.2. The guest speaker will address the students on the importance of harmonyin our society.3. The intensity of work leaves little room for personal grief andpleasure.4. The inviolate learner-centered teaching is to let students speak out inclass freely and without reserve.5. It is equally important to bind and rivet onto teachers a sense ofhumour which is not beseeched but instead comes out naturally.6. All attempts and overtures to woo him back have been spurned with contempt.7. He is generously disposed to admit the validity of opposing views rather than consider reconciliation.8. He seems to advise us not to hug the delusive phantom of hope for anative-like foreign language ability.II. Translate the following into English.Dear Students of the Graduating Class:As you are leaving your alma mater, I have nothing to offer you as a gift except a word of advice.My advice is, "Never give up the pursuit of learning." Y ou have perhaps finished your college courses mostly for obtaining the diploma, or, in other words, out of sheer necessity. However, from now on you are free to followyour own bent in the choice of studies. While you are in the prime of life,why not devote yourselves to a special field of study? Y outh will soon begone never to return. And it will be too late for you to go into scholarshipwhen in your declining years. Knowledge will do you a good turn even as a means of subsistence. If you give up studies while holding a job, you willin a couple of years have had yourselves replaced by younger people. It will then be too late to remedy the situation by picking up studies again.。

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ON THE LENGTH OF THE LONGEST SUBSEQUENCEA VOIDING AN ARBITRARY PATTERN IN ARANDOM PERMUTATIONM.H.ALBERTAbstract.We consider the distribution of the length of the longest subsequence avoiding an arbitrary pattern,π,in a random permu-tation of length n .The well-studied case of a longest increasingsubsequence corresponds to π=21.We show that there is someconstant c πsuch that as n →∞the mean value of this length is asymptotic to 2√c πn and that the distribution of the length istightly concentrated around its mean.We observe some apparentconnections between c πand the Stanley-Wilf limit of the class ofpermutations avoiding the pattern π.Consider an arbitrary pattern avoidance class A .Given any permuta-tion πdeﬁne the longest A subsequences of πor L A S(π)to be the set of those subsequences of πof maximum length,subject to the condition that their patterns belong to A .Also deﬁne L A (π)to be the length of any sequence in L A S(π).Let I =Av(21)be the class of increasing permutations.Apparently Ulam [10]was the ﬁrst to ask the question:What can be said about the distribution of values ofL I (Πn )when Πn is a random variable whose value is apermutation πchosen uniformly at random from amongthe elements of S n ?We intend to address the generalization of this problem to the random variable L A (Πn )deﬁned in a similar fashion.The history of the analysis of Ulam’s problem is well documented in [2].We repeat here a few details relevant to our investigations of the more general problem.For convenience let L n =L I (Πn ).Ulam conjectured that for some constant c lim n →∞E L n √n=c.This conjecture was proven by Hammersley [6]who showed also that n −1/2L n →c in probability,and who conjectured that c =2.This12M.H.ALBERTfurther conjecture was proven in part by Logan and Shepp[8]and simultaneously in whole by Kerov and Verˇs ik[11].Frieze[5]and Bollob´a s and Brightwell[4]used martingale methods to establish tight concentration of L n about its mean.Subsequently, Baik,Deift and Johansson[3]obtained complete asymptotic informa-tion about the distribution of L n.Our main theorems are analogs of the results of Hammersley and Frieze for the more general case of L A(Πn).For theﬁrst of these we need to impose a mild additional restriction on A.The proofs for the general case are then essentially identical to the originals.Theorem1.Let A be an inﬁnite and proper pattern avoidance class which is either sum-closed or diﬀerence-closed.There exists a constant 1≤c A<∞such that(1)limn→∞E L A(Πn)√n=2√c A.Theorem2.Let A be a proper pattern class.Forα>1/3andβ< min(α,3α−1)(2)Pr(|L A(Πn)−E L A(Πn)|≥nα)<exp(−nβ).The Marcus-Tardos theorem is a key ingredient in the proof of Theo-rems1and2.We will now present observations which provide some evidence for a connection between the constants c A ands A:=lim supn→∞|S n∩A|1/n.Conjecture1.For any proper pattern avoidance class A,the limits superior deﬁnining c A and s A are in fact limits,and c A=s A.The evidence for this conjecture is somewhat fragmentary at this point. It is supported by the following results.Proposition3.For each positive integer k,the classesAv(k(k−1)(k−2)···21)and Av(123···k)satisfy Conjecture1.Proposition4.Let A and B be two pattern avoidance classes which satisfy Conjecture1.Then their union,direct sum and juxtaposition also satisfy Conjecture1.If,additionally,A∩B is aﬁnite class then their merge also satisﬁes Conjecture1.LONGEST PATTERN AVOIDING SUBSEQUENCES 3The results of the preceding proposition apply in part to the weaker version of Conjecture 1which only asserts the equality of two limits superior.Let A be any pattern avoidance class and deﬁne Rot(A )to be the set of permutations obtained by taking all the cyclic rotations of elements of A .It is easily veriﬁed that Rot(A )is also a pattern avoidance class.Proposition 5.If A satisﬁes Conjecture 1then so does Rot(A ).Again,the “weak”form of this proposition is valid.We have no signiﬁcant evidence in favour of Conjecture 1based on an exact computation of c A for any classes other than those which can be produced from the classes Av(k (k −1)···321)or Av(123···(m −1)m )by the constructions described in the previous section.An obvious starting point for the investigation of Conjecture 1would be the col-lection of pattern avoidance classes whose basis consists entirely of permutations of length 3.These classes,as a group,were ﬁrst analysed by Simion and Schmidt [9].Starting from the “easy”end we note that almost all such classes having three or more basis elements sat-isfy Conjecture 1as a consequence of Propositions 4and 5or trivial modiﬁcations of them.The exceptional cases are the classes whose growth is governed by the Fibonacci numbers,and we consider these below.Also,most of the cases having two basis elements are covered by Proposition 4.Up to symmetry there are two exceptions which we consider below.•The layered permutations,L =Av(231,312),consisting of all permutations of the form D 1⊕D 2⊕···⊕D k where each of D 1through D k is a descending permutation.The number of permutations of length n in L is 2n −1and so s L =2.•The subclass L (2)=Av(231,312,321)of L formed by requiring that each D i contain at most two elements.The number of permutations of length n in L (2)is equal to the n th Fibonacci number,so s L (2)=(1+√5)/2.Kaiser and Klazar [7]proved that L (2)is the smallest pattern avoidance class whose Stanley-Wilf limit is strictly greater than 1.•The class C =Av(321,312)whose elements are those permuta-tions that can be written as direct sums C 1⊕C 2⊕···⊕C k where each C i is of the form 234···n 1for some n >1,or simply 1.Results in [1]give dynamic programming algorithms for solving the longest subsequence problem for both L and L (2)whose complexity is O (n 2log n )where n is the length of the input permutation.We4M.H.ALBERTLengthµσ∼c L(2)1×104239.3 4.5 1.4312×104340.7 5.2 1.4514×104484.7 6.1 1.4688×104688.4 6.4 1.48116×104978.17.1 1.49532×1041386.88.3 1.50364×1041965.39.3 1.510128×1042785.310.2 1.515Table1.Summary data for the mean,µ,and standarddeviationσof the length of the longest L(2)subsequencesin a sample of1000random permutations together withcorresponding estimates of c L(2).have been able to improve the latter algorithm,based on a tableau style method to result in a complexity of O(n log n).The algorithm provided in[1]for C has worst case complexity O(n3log n)though in practice with some minor optimisations it performs signiﬁcantly bet-ter than this on random permutations.All three algorithms were im-plemented and a long period random number generator was used to provide experimental data concerning the values c L,c L(2),and c C.For L(2)we present data based on permutations of length2k×104for 0≤k≤7.For each value of k,1000random permutations of that length were generated and the length of the longest L(2)subsequences was computed.Table1shows the mean,sample standard deviation, and resulting estimates of c L(2)based on these simulations.We would be forced to classify a person who believed in the truth of Conjecture1 based on this data for L(2)as an optimist.If the estimates are indeed converging to s L(2)then they are not yet within6%of theirﬁnal limit at n=128×104.By contrast,for this value of n the estimate for c I (whose actual value is1)is approximately0.985.Because of the slower running time and increased space requirements required by the algorithm forﬁnding longest layered subsequences data for L is based on permutations of length2k×102for0≤k≤7.As for L(2),1000random permutations of each length were analysed and the results are presented in Table2.The data for this class do not require as much optimism as the L(2)data to be viewed as support for Conjecture1.LONGEST PATTERN AVOIDING SUBSEQUENCES5Lengthµσ∼c L1×10223.8 1.8 1.4182×10234.8 2.2 1.5174×10250.6 2.5 1.6028×10273.4 3.0 1.68216×102105.2 3.3 1.73032×102150.7 4.0 1.77464×102215.9 4.4 1.821128×102307.5 4.9 1.847Table2.Summary data for the mean,µ,and standarddeviationσof the length of the longest L subsequencesin a sample of1000random permutations together withcorresponding estimates of c L.Lengthµσ∼c L1×10222.9 2.0 1.3062×10233.5 2.3 1.4064×10248.5 2.4 1.4708×10270.5 3.1 1.55516×102101.2 3.3 1.60132×102145.2 3.9 1.647Table3.Summary data for the mean,µ,and standarddeviationσof the length of the longest C subsequencesin a sample of1000random permutations together withcorresponding estimates of c C.Finally,the data for the class C presented in Table3is even more limited,but again it seems to provide qualiﬁed support for Conjecture 1.Notable by its omission from our discussion is the class Av(312).This class has Stanley-Wilf limit4.A polynomial time algorithm for the longest subsequence problem based on this class is given in[1]but its complexity on permutations of length n is O(n5)which makes it impractical for experiments of the size required to produce even vaguely convincing evidence.The goal of producing such evidence would seem to requireﬁnding,even on an ad hoc basis some collection of classes for which the longest subsequence problem can be solved algorithmically in reasonable time(basically,at worst quadratic)and/or developing better algorithms for classes such as Av(312).6M.H.ALBERTReferences[1]M.H.Albert,R.E.L.Aldred,M.D.Atkinson,H.P.van Ditmarsch,B.D.Handley,C.C.Handley,and J.Opatrny.Longest subsequences in bin.,28:225–238,2003.[2]David Aldous and Persi Diaconis.Longest increasing subsequences:from pa-tience sorting to the Baik-Deift-Johansson theorem.Bull.Amer.Math.Soc.(N.S.),36(4):413–432,1999.[3]Jinho Baik,Percy Deift,and Kurt Johansson.On the distribution of the lengthof the longest increasing subsequence of random permutations.J.Amer.Math.Soc.,12(4):1119–1178,1999.[4]B´e la Bollob´a s and Graham Brightwell.The height of a random partial order:concentration of measure.Ann.Appl.Probab.,2(4):1009–1018,1992.[5]Alan Frieze.On the length of the longest monotone subsequence in a randompermutation.Ann.Appl.Probab.,1(2):301–305,1991.[6]J.M.Hammersley.A few seedlings of research.In Proceedings of the SixthBerkeley Symposium on Mathematical Statistics and Probability(Univ.Cali-fornia,Berkeley,Calif.,1970/1971),Vol.I:Theory of statistics,pages345–394,Berkeley,Calif.,1972.Univ.California Press.[7]Tom´aˇs Kaiser and Martin Klazar.On growth rates of closed permutationbin.,9(2):Research paper10,20pp.(electronic), 2002/03.Permutation patterns(Otago,2003).[8]B. F.Logan and L. A.Shepp.A variational problem for random Youngtableaux.Advances in Math.,26(2):206–222,1977.[9]Rodica Simion and Frank W.Schmidt.Restricted permutations.European J.Combin.,6(4):383–406,1985.[10]Stanislaw M.Ulam.Monte Carlo calculations in problems of mathematicalphysics.In Modern mathematics for the engineer:Second series,pages261–281.McGraw-Hill,New York,1961.[11]A.M.Verˇs ik and S.V.Kerov.Asymptotic behavior of the Plancherel measureof the symmetric group and the limit form of Young tableaux.Dokl.Akad.Nauk SSSR,233(6):1024–1027,1977.Department of Computer Science,University of Otago,Dunedin,New ZealandE-mail address:malbert@。